American House-Carpenter, 1883

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11 oe iji il oe I а == шыр р? X <= =e Е "ui EDI =т= МЕН; LL | | ji | ! ШІ | | || E = NS Aet - T. a a ini ^h ІШІШІП А T ІШІ "n EROR Ny, 7. TEMPLE OF NEPTUNE, AT PAESTUM.

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THE American House CARPENTER. A. TREATISE ON THE ART OF BUILDING. COMPRISING STYLES OF ARCHITECTURE, STRENGTH OF MATERIALS, ҚЫ рон OF FLOORS, FRAMED GIRDERS, ROOF T OE ROLLEDIRON BE М5, TUBULAR-IRON oy IRDERS,| STAIR SE DOORS, OULDINGS, AND D» CES; ЖЕТІСУ; ? А COMPEND ОҒ MATHEMATICS. A MANUAL FOR THE ing USE OF ARCHITECTS, CARPENTERS, STAIR-BUILDERS, AND OTHERS. EIGHTH EDITION, REWRITTEN AND ENLARGED. BY R. 6. HATFIELD, ARCHITECT, LATE FELLOW OF THE AMERICAN INSTITUTE OF ARCHITECTS, MEMBER OF THE AMERICAN SOCIETY OF CIVIL ENGINEERS, ETC. AUTHOR OF “TRANSVERSE STRAINS? | EprrED BY О. P. HATFIELD, Ғ.А.І.А., ARCHITECT. NINTH EDITION. NEW YORK: JOHN WILEY & SONS, 15 ASTOR PLACE 1883;

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CorvniGHT 1880, Bv THE ESTATE OF R. G. HATFIELD. 2062 & 2_—

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PREFACE. SINCE the publication of the first edition of this work, six subsequent editions have been issued ; but; although from time to time many additions to its pages and revisions of its subject-matter have been made, still its sev- eral issues have always been printed substantially from the original stereotype plates. In this edition, however, the book has been extensively remodelled and expanded, the greater portion of it rewritten, and the whole put in a new dress by being newly set up in type uniform in style with that of the late author's recent work, Transverse Strains. To this revision—a labor of love to him—he devoted all the time he could spare from his other pressing engage- ments for a year or more, and by close and arduous application brought the book- to a successful termination, notwithstanding the engrossing nature of his customary business avocations. Although essentially an elementary work, and intended originally for a class of minds not generally favored with oppor- tunities for securing a very extended form of éducation, either in the store of information acquired or in the diteipiise ой mind which culture confers, still it has been his aim to embody in its pages so complete and exhaustive a treat- ment of the various subjects discussed, and so practical and useful a collection of data and the rules governing their application, as to make it also not un- worthy the attention of those who have been more highly favored in that respect. In all the various trades connected with building it is the intelligent workman that commands the greatest respect, and who receives in all cases the highest remuneration. As apprentice, journeyman, and master-builder, his course is upward and onward, and success crowns his efforts in all that he undertakes. "There is a kind of freemasonry in the very air that surrounds the skilful, intelligent man, that gives him a pass at once into the appreciation and recognition of all those whose regard is valuable. We admire and respect the plodding toil of the honest, patient laborer, whose humble task may tax his muscles though not his mind, but we yield a deeper homage to the skilful hand and tutored eye that accomplish wonders in art and science through per- severance in aspiring studies. It was to excite in the minds of workmen like - these an ambition to excel in their calling, and to point out to them the surest path to 4һаї consummation, that the preparation of this volume was under- taken ; that all its tendencies аге in that direction, and that it cannot well fail

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li PREFACE. of its purpose when judiciously used, must be the conviction of all who will take the trouble to examine its pages. In the first part of the book matters more particularly relating to building are treated of. Тһе first section is in the nature of an introduction, serving by its historical references to excite an interest in the general subject, while in the second are presented the methods of erecting edifices in accordance with the acknowledged principles of sound construction. In the remaining sections of Part I. the several well-defined branches of house-building, as stairs, doors and windows, etc., are illustrated and explained. In the second part the more useful rules and simple problems of mathematics are reduced to an easily acquired form, and adapted to the necessities of the ordinary workman. Ву studying the latter, the young mechanic may not only improve and strengthen his mind, but grow more self-reliant daily, demonstrating in his own experience that scientific knowledge gives power. By carefully studying this part of the book he will see how easy it is to acquire the knowl- edge of solving problems by signs and symbols, commonly called Algebra (although looked upon by the uninitiated as almost incomprehensible), and thus find it easy to understand all the illustrations of the various subjects wherein those condensed forms of expression are used. Useful problems in geometry, described in simple language, and hints upon the subject of draw- ing and shading, are also to be found in Part II. A glossary of architectural terms and many useful tables are provided in the Appendix, and finally, an Index is added to aid in referring to special subjects. Тһе full-plate illustra- tions are inserted to make it attractive to the general reader, and at the same time to serve as explanatory of the historical portion of the volume. It will not be denied that the class of information herein furnished is one of the most instructive and useful that can. be presented to the practical mind of a workingman, or to any mind engaged in mechanical pursuits. The im- press stamped upon it by the author's peculiar line of study is not to be effaced, but this has given it characteristics of originality and strength not to be found in a mere compilation. THE EDITOR. New York, 31 Pine Street, January 6, 1880.

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CONTENTS. foe For a Table of Contents more in detail, see p. 613, and for Index, see p. 657. PART 7. PAGE ОО L = ArCHMICCCHITE.... l.u ooon onon cereni noae Жу... Mee 5 е II.—Construction ..... NC. EET oe v ooo 57 ТШ 5g .............................- е... 240 “ IV Doors апа ТҮ/іпботв............................... 315 i V.—MoudiNes. апа Согпсев............................ 323 PART II. TON VL-Gmometm. ...................... "——— 347 С ЭЛИТУ MOOR RO EOn.... .......................... 366 МОЕТ СОННА te 378 “ (cd! Oko o oa aa a e = 392 is 2%- РО”... uet Т 439 «ELS Cirele,............. "MT M 468 277 lui ПЕ _....................... uem. m. 481 “ОО. ЭЕ Батамба,................... КЕЕН 492 Ss EEEE 2 510 OR —— Uo 75 ..........................-:............ 536 “ XVI.—Practical Сеотеѓгу..................: Е 544 ео 2... ke eee ene CT TEES... 596

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iv CONTENTS. APPENDIX. (^0 259957 E ERETEQI — — TE 11 ETE ЕЧ 627 Table of Squares; Cubes and ”Кооб5................. T ST 638 Rules for the Reduction of Әесітздіс............................ 647 pg SO” .22.................2.27042............ 649 Table showing the Capacity/of Wells, Cisterns, etc.............. 653 Mable of the Weights of Materials........ ................ ES. m 654 РГ ................ ............ М СТЯ 657

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| AE Ot "HE ФА UNIVERSITY CA rp oA

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г o | А i 1! | 11 LU util | ІШІ, ІШ | THE PARTHENON, ATHENS.

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UNIVERSITY) КРОВ PART SECTION I.—ARCHITECTURE. ART. l.— Building Defined.—Building and Architecture are technical terms by some thought to be synonymous; but there is a distinction. Architecture has. been defined to ре—“ е art of building ;" but more correctly it is—“ the art of designing and constructing buildings, in accordance with such principles as constitute stability; utility; and beauty." The literal signification of the Greek word archi- tecton, from which the word architect is derived, is chief- carpenter; and the architect who designs and builds well may truly be considered the chief duzlder. Of the three classes into which architecture has been divided—viz., Civil, Military, and Naval—the first is that which refers to the construction of edifices known as dwellings, churches, and other public buildings, bridges, etc., for the accommodation of civilized man—and is the subject of the remarks which follow. e 2.—Antique Buildings; Tower of Babel.— Building is one of the most.ancient of the arts: the Scriptures inform us of its existence at a very early period. Cain, the son of Adam, “builded a city, and called the name of the city after the name of his son, Enoch ;" but of the peculiar style or manner of building we are not informed. It is presumed thatit was not remarkable for beauty, but that utility and perhaps stability were its characteristics. Soon after the deluge — that memorable event, which removed from ex- istence all traces of the works of man—the Tower of Babel

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6 ARCHITECTURE. was commenced. This was a work of such magnitude that the gathering of the materials, according to some writers, occupied three years; the period from its commencement until the work was abandoned was twenty-two years; and the bricks were like blocks of stone, being twenty feet long, fifteen broad, and seven thick. Learned men have given it as their opinion that the tower in the temple of Belus at Babylon was the same as that which in the Scriptures is called the Tower of Babel. Тһе tower of the temple of Belus was square at its base, each side measuring one furlong, and consequently half a mile in circumference. Its form was that of a pyramid, and its height was 660 feet. It had a winding passage on the outside from the base to the summit, which was wide enough for two carriages. 3. of ancient cities, such as Babylon, Palmyra, and Nineveh of the Assyrians; Sidon, Tyre, Aradus, and Serepta of the Phoenicians; and Jerusalem, with its splendid temple, of the Israelites —show that architecture among them had made great advances. Ancient monuments of the art are found also among other nations; the subterraneous temples of the Hindoos upon the islands Elephanta and Salsetta ; the ruins of Persepolis in Persia; pyramids, obelisks, tem- ples, palaces, and sepulchres in Egypt—all prove that the architects of those early times were possessed of skill and judgment highly cultivated. The principal characteristics of their works are gigantic dimensions, immovable solidity, and, in some instances, harmonious splendor. ‘The extra- ordinary size of some is illustrated in the pyramids of Egypt. The largest of these stands not far from the city of Cairo: its base, which is square, covers about 113 acres, and its . height is nearly 500 feet. The stones of which it is built are immense—the smallest being full thirty feet long. 4. tecture was cultivated as a fine art. Dignity and grace were added to stability and magnificence. In the Doric order, their first style of building, this is fully exemplified. Phidias, Ictinus, and Calicrates are spoken of as masters in

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| W u | m B | ГІ! | | | | p up P = ІШІ ы ІШУ i fii ПІН ІШ ҮШҮҮ "қ қ ue n THE PANTHEON ROME.

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GRECIAN AND ROMAN BUILDINGS. y the art at this period: the encouragement and support of Pericles stimulated them to a noble emulation. The beauti- ful temple of Minerva, called the Parthenon, erected upon the acropolis of Athens, the Propvleum, the Odeum, and others, were lasting monuments of their success. The Ionic and Corinthian orders were added to the Doric, and many magnificent edifices arose. These exemplified, in their chaste proportions, the elegant refinement of Grecian taste. Improvement in Grecian architecture continued to advance until perfection seems to have been attained. The speci- mens which have been partially preserved exhibit a com- bination of elegant proportion, dignified simplicity, and majestic grandeur. Architecture among the Greeks was at the height of its glory at the period immediately preceding the Peloponnesian war; after which the art declined. An excess of enrichment succeeded its former simple grandeur ; yet a strict regularity was maintained amid the profusion of ornament. After the death of Alexander, 323 B.C., a love of gaudy splendor increased: the consequent decline of the art was visible, and the Greeks afterwards paid but little attention to the science. 5.—Architecture in Rome.— While the Greeks illustrated their knowledge of architecture in the erection of their temples and other public buildings, the Romans gave their attention to the science in the construction of the many aqueducts and sewers with which Rome abounded ; build- ing no such splendid edifices as adorned Athens, Corinth, and Ephesus, until about 200 years B.C., when their inter- course with the Greeks became more extended. Grecian architecture was introduced into Rome by Sylla; by whom, as also by Marius and Cesar, many large edifices were erected in various cities of Italy. But under Cæsar Augus- tus, at about the beginning of the Christian era, the art arose to the greatest perfection it ever attained in Italy. Under his patronage Grecian artists were encouraged, and many emigrated to Rome. It was at about this time that Solomon's temple at Jerusalem was rebuilt by Herod —a Roman. This was 46, years іп the erection, and was most probably of'the Grecian style of building— perhaps of the:

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THE GOTHS AND VANDALS. 9 provement; but very soon after his reign the art began rapidly to decline, as particularly evidenced in the mean and trifling character of the ornaments. 7.—Architecture Debased.—The Goths and Vandals overran Italy, Greece, Asia, and Africa, destroying most of their works of ancient architecture. - Cultivating no art but that of war, these savage hordes could not be expected to take any interest in the beautiful forms:and proportions of their habitations. From this time architecture assumed an entirely different aspect. The celebrated styles of Greece were. unappreciated and forgotten; and modern architec- ture made its first appearance on the stage of existence. The Goths, in their conquering invasions, gradually ex- tended it over Italy, France, Spain, Portugal, and Ger- many, into England. From the reign of Galienus.may be reckoned thé total extinction of the arts among the Romans. From this time until the sixth or seventh century, architec- ture was almost entirely neglected. Тһе buildings which were erected during this suspension of the arts were very rude. Being constructed :оЁ the fragments of the edifices which had been demolished by tke Visigoths in their unre- strained fury, and the builders being destitute of a proper knowledge of architecture, many sad blunders and exten- sive patch-work might have been seen in their construction —entablatures inverted, columns standing on their wrong ends, and other ridiculous arrangements characterized their clumsy work. Тһе vast number of columns which the ruins around them afforded they used as piers in the construction of arcades—which by some is thought, after having passed through various changes, to have been the origin of the plan of the Gothic cathedral. Buildings generally, which are not of the classical styles, and which were erected after the fall of the Roman empire, have by some been indiscrim- inately included under the term Gothic. But the changes which architecture underwent during the Mediæval age show that there were then several distinct modes of building. 8.—The Ostrogoths.—Theodoric, a friend of the arts, who reigned in Italy from A.D. 493 to 525, endeavored to

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IO ARCHITECTURE. restore and preserve some of the ancient buildings ; and erected others, the ruins of which are still seen at Verona and Ravenna. Simplicity and strength are the character- istics of the structures erected by him ; they are, however, devoid of grandeur and elegance. or fine proportions. These are properly of the Goruic style; by some called the o/d Gothic, to distinguish it from the pointed Gothic. 9.—The Lombards, who ruled in Italy from A.D. 568, had no taste for arcbitecture nor respect for antiquities. Accordingly, they pulled down the splendid monuments of classic architecture which they found standing, and erected in their stead huge buildings of stone which were greatly destitute of proportion, elegance, or utility—their charac- teristics being scarcely anything more than stability and immensity combined with ornaments of a puerile character. Their churches were decorated with rows of small columns along the cornice of the pediment, small doors and win- dows with circular heads, roofs supported by arches having arched buttresses to resist their thrust, and a lavish display of incongruous ornaments. This kind of architecture 15 called the LoMBARD style, and was employed in the seventh century in Pavia, the chief city of the Lombards ; at which ` city, as also at many other places, a great many edifices were erected in accordance with its peculiar forms. 10.—The Byzantine Architects, of Byzantium, Constan- tinople, erected many spacious edifices; among which are included the cathedrals of Bamberg, Worms, and Mentz, and the most ancient part of the minster at Strassburg ; in all of these they combined the classic styles with the crude Lombardian. This style is called the LOMBARD-BYZANTINE. To the last style there were afterwards added cupolas sim- ilar to those used in the East, together with numerous slen- der pillars with elaborate capitals, and the many minarets’ which are the characteristics of the proper MR PN Or Oriental style. 11.— Тһе Moors.— When the Arabs and Moors destroyed the kingdom of the Goths, the arts and sciences were mostly

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KASE LIB UNIVERSITY! TH P ТИЛЕШ H Dx nm; MOSQUE AT CAIRO.

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THE MEDI/EVAL STYLES. II in possession of the Musselmen-conquerors ; at which time there were three kinds of architecture practised; viz.: the Arabian, the Moorish, and the Lombardian. The ARABIAN style was formed from Greek models, having circular arches added, and towers which terminated with globes and mina- rets. The MooRisH is very similar to the Arabian, being distinguished from it by arches in the form of a horseshoe. It originated in Spain in the erection of buildings with the ruins of Roman architecture, and is seen in all its splendor in the ancient palace of the Mohammedan monarchs at Grenada, called the А/Латбта, or red-house. The style which was originated by the Visigoths in Spain by a combination of the Arabian and Moorish styles, was introduced by Charle-: magne into Germany. Оп account of the changes and im- provements it there underwent, it was, at about the 13th or 14th century, termed the German or romantic style. It 15 ex- hibited in great perfection in the towers of the minster of Strassburg, the cathedral of Cologne and other edifices. The most remarkable features of this lofty and aspiring style are the lancet or pointed arch, clustered pillars, lofty towers, and flying buttresses. It was principally employed in eccle- siastical architecture, and in this capacity introduced into France, Italy, Spain, and England. (2.—The Architecture of England: is divided into the Norman, the Early-English, the Decorated, and the Perpendic- ular styles. The Norman is principally distinguished by the character of its ornaments—the chevron, or zigzag, being the most common. Buildings in this style were erected in the 12th century. The Early-English is celebrated for the beauty of its edifices, the chaste simplicity and purity of design which they display, and the peculiarly graceful char- acter of its foliage. This style is of the 13th century. The Decorated style, as its name implies, is characterized by а. _great profusion of enrichment, which consists principally of the crocket, or feathered-ornament, and ball-flower. It was mostly in use in the 14th century. The Perpendicular style, which dates from the 15th century, is distinguished by its high towers, and parapets surmounted with spires similar in number and grouping to oriental minarets.

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12 ARCHITECTURE. - 13.—Arechitecture Progressive.— [he styles erroneously termed Gothic were distinguished by peculiar characteris- tics as well as by different names. The first symptoms of a desire to return to a pure:style in architecture, after the ruin caused by the Goths, was manifested in the character of the art as displayed in the church of St. Sophia at Con- stantinople, which was erected by Justinian in the 6th century. The church of St. Mark at Venice, which arose in the roth or 11th century, is a most remarkable building; a compound of. many of the forms of ancient architecture. The cathedral at Pisa, a. wonderful structure for the age, was erected by a Grecian architect in 1016. The marble with which the. walls of this building were faced, and of which the four rows of columns that support the roof are composed, is said to be of an excellent character. The Campanile, or leaning-tower as it is usually called, was erected near the cathedral in the 12th century. Its inclina- tion is generally supposed to have arisen from a poor foun- dation ; although by some it is said to have been. thus con- structed originally, in order to inspire in the minds of the beholder sensations of sublimity and awe. Іп the 13th cen- tury, the science in Italy was slowly progressing ; many fine churches were erected, the style of which displayed a de- cided advance in the progress towards pure classical archi- tecture. In other parts of Europe, the Gothic, or pointed style was prevalent. The cathedral at Strassburg, designed by Irwin Steinbeck, was erected in the 13th and 14th cen- turies. In France and England during the 14th century, many very superior edifices were erected in this style. 14.—Architecture in Italy.—In the 14th and 15th cen- turies, architecture in Italy was greatly revived. The mas- ters began to study the remains of ancient Roman edifices ; and many splendid buildings were erected, which displayed a purer taste in the science. Among others, St. Peter's of Rome, which was built about this time, is a lasting monu- ment of the architectural skill of the age. Giocondo, Mi- chael Angelo, Palladio, Vignola, and other celebrated archi- tects, each in their turn, did much to restore the art to its

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INTERIOR OF ST. SOPHIA, CONSTANTINOPLE.

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ORIGIN OF STYLES. I3 former excellence. In the edifices which were erected under their direction, however, it is plainly to be seen tbat they studied not from the pure models of Greece, but from the remains of the deteriorated architecture of Rome. The high pedestal, the coupled columns, the rounded pediment, the many curved-and-twisted enrichments, and the convex frieze, were unknown to pure Grecian architecture. Yet their efforts were serviceable in correcting, to a good de- gree, the very impure taste that had prevailed since the over- throw of the Roman empire. [5.—The Renaissance.—The Italian masters and numer- ous artists who had visited Italy for the purpose, spread the Roman style over various countries of Europe; which was gradually received into favor in place of the pointed Gothic. _ This fell into. disuse; although it has of late years been again cultivated. It requires a building of great magnitude and complexity for à perfect display of its beauties. Іп America, the pure Grecian style was at first more or less studied; and perhaps the simplicity of its principles would be better adapted to a republican country than the more intricate medizval styles; yet these, during the last quarter of a century, have been extensively studied, and now wholly supersede the Grecian styles. 16.—Styles of Architecture.—It is generally acknowl- edged that the várious styles in architecture were the results of necessity, and originated in accordance with the different pursuits of the early inhabitants of the earth; and were brought by their descendants to their present state of per- fection, through the propensity for imitation and desire of emulation which are found more or less among all nations. Those that followed agricultural pursuits, from being em- ployed constantly upon the same piece of land, needed a permanent residence, and the wooden йи? was the offspring of their wants ; while the shepherd, who followed his flocks and was compelled to traverse large tracts of country for pasture, found the 227 to be the most portable habitation ; again, the man devoted to hunting and fishing—an idle and vagabond way of living—is naturally supposed to have been

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14 ARCHITECTURE. content with the caverz as a place of shelter. .Тһе latter is said to have been the origin of'the Egyptian style; while the curved roof of Chinese structures gives a strong indica- tion of their having had the tent for their model; and the simplicity of the original style of the Greeks (the Doric) shows quite conclusively, as is generally conceded, that its original was of wood. The pointed, or ecclesiastical style, is said to have originated in an attempt to imitate the bower, or grove of trees, in which the ancients performed their idol- worship. But it is more probably the result of repeated scientific attempts to secure real strength with apparent lightness; thus giving a graceful, aspiring effect. (7.—Orders: or styles, in architecture are numerous; and a knowledge of the peculiarities of each is important to the student in the art. An ORDER, in architecture, is com- posed of three principal parts, viz.: the Stylobate, the Col- uinn, and the Entablature. This appertains chiefly to the classic styles. | (8..—The Stylobate: is the substructure, or basement, upon which the columns of an order are arranged. Іп Roman architecture—especially in the interior of an edi- fice—it frequently occurs that each column has a separate substructure; this is called а pedestal. If possible, the ped- estal should be avoided in all cases; because it gives to the column the appearance of having been originally designed for a small building, and afterwards pieced out to make it' long enough for a larger one. |9.—The Column: is composed of the base, shaft, and capital. 20.—The Entablature: above and supported by the columns, is horizontal ; and is composed of the architrave, frieze, and cornice. These principal parts are again divided into various members and mouldings. 21.—The Base: of a column is so called from basis, а foundation or footing.

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INTERIOR OF ST. STEPHENS, PARIS.

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PARTS OF AN ORDER. 15 . 22.—The Shaft: the upright part of a column standing upon the base and crowned with the capital, is from sha/fto, to dig—in the manner of a well, whose inside is not unlike the form of a column. 23.—The Capital: from kephale or caput, the head, is the uppermost and crowning part of the column. 24.—The Arehitrave: from archi, chief or principal, and ¢rabs, a beam, is that part of the entablature which lies in immediate connection with the column. | 25.—' The Frieze: from fibron, a fringe or border, is that part of the entablature which is immediately above the architrave and beneath the cornice. It was called by some of the ancients zophorus, because it was usually enriched with sculptured animals. 26.—The Cornice: from corona, a crown, is the upper and projecting part of the entablature—being also the upper- most and crowning part of the whole order. 27.—The Pediment: above the entablature, is the tri- angular portion which is formed by the inclined edges of the roof at the end of the building. In Gothic architecture, the pediment is called a gae. ee 28.—The Tympanum: is the perpendicular triangular surface which is enclosed by the cornice of the pediment. 29.—The Attic: is a small order, consisting of pilasters and entablature, raised above a larger order, instead of a pediment. An attic story is the upper story, its windows being usually square. 30.—Proportions іп an Order.—An order has its several members proportioned to one another by a scale of 60 equal parts, which are called minutes. If the height of buildings were always the same, the scale of equal parts would be a fixed quantity—an exact number of feet and inches. But as buildings are erected of different heights, the column and

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16 ARCHITECTURE. its accompaniments are required to be of different dimen- sions. To ascertain the scale of equal parts, it is necessary to know the height to which the whole order is to be erected. This must be divided by the number of diameters which is directed for the order under consideration. Then the quotient obtained by such division is the length of the scale of equal parts—and is, also, the diameter of the column next above the base. For instance, in the. Grecian Doric order the whole height, including column and entablature, is 8 diameters. Suppose now it were desirable to construct an example of this order, forty feet high. Then до feet divided. by 8 gives 5 feet for the length of the scale; and this being divided by 60, the scale is completed. The up- right columns of figures, marked H and P, by the side of the drawings illustrating the orders, designate the height and the projection of the members. The projection of each member is reckoned from a line passing through the axis of the column, and extending above it tó the top of the entab- lature. The figures represent minutes, or 6oths, of the major diameter of the shaft of the column. 31.—Grecian Styles.—The original method of building among the Greeks was in what is called the Doric order: to this were afterwards added the /олтес and the Corinthian. These three were the only styles known among them. Each is distinguished from the other two by not only a peculiar- ity of some one or more of its principal parts, but also by a particular destination. The character of the Doric is robust, manly, and Herculean-like ; that of the Ionic is more deli- cate, feminine, таќгопу ; while that of the Corinthian is extremely delicate, youthful, апа virgin-like. However they may differ in their general character, they are alike famous for grace and digu:ty, elegance and grandeur, to a high degree of perfection. 32.—The Dorie Order: (Fig. 2,) is so ancient that its origin is unknown—although some have pretended to have discovered it. But the most general opinion is, that it is an improvement upon the original wooden buildings of the

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FANCIFUL ‘ORIGIN OF THE DORIC; I Grecians. These no doubt were very rude, and perhaps not unlike the following figure. ~ d EN US SSS, [m -U F уст, hl pi J-N OL 2, ЭРТА Ad » Б ү ^, v. Ай; dM IE iine ^ RIN. M PP gp P VX, r4 3 >. 2 ЕРУ СУЯ Airra “2 25. Re- 5 7 FIG. 1.—SUPPOSED ORIGIN oF Doric TEMPLE. The trunks of trees, set perpendicularly to support the roof, may be taken for columns; the tree laid upon the tops of the perpendicular ones, the architrave; the ends of the cross-beams which rest upon the architrave, the triglyphs ; the tree laid on the cross-beams as a support for the ends of the rafters, the bed-moulding of the cornice ; the ends of the rafters which project beyond the bed-moulding, the mutules; and perhaps the projection of the roof in front, to screen the entrance from the weather, gave origin to the portico. The peculiarities of the Doric order are the triglyphs— those parts of the frieze which have perpendicular channels cut in their surface ; the absence of a base to the column— as also of fillets between the flutings of the column ; and the plainness of the capital. The triglyphs should be so dis- posed that the width of the metopes—the space between the trigly phs—shall be equal to their height. 33. — The Intercolumniation: or space between the col- umn, is regulated by placing the centres of the columns under the centres of the triglyphs—except at the angle of the building ; where, as шау be seen in Fig. 2, one edge of

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18: ARCHITECTURE. ы `| col aa y S Y IMP WU ЧЫ EN MED EIS I2 1} 20) orare pg Epa WE FIG. 2.—GnECIAN DORIC.

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PECULIARITIES OF THE DORIC. I9 the triglyph must be over the centre of the column.* Where the columns are so disposed that one of them stands beneath every other triglyph, the arrangement is called mono-triglyph and is most common. When a column is placed beneath every third triglyph, the arrangement is called dzastyle ; and when beneath every fourth, aræostyle. This last style is the worst, and is seldom adopted. 34.—The Doric Order: is suitable for buildings that are destined for national purposes, for banking-houses, etc. Its appearance, though massive and grand, is nevertheless rich and graceful. The Patent Office at Washington, and ‘the Treasury at New York, are good specimens of this order. 35.—The Ionic Order. (/7g. 3.) — The Doric was for some time the only order in use among the Greeks. They gave their attention to the cultivation of it, until perfection seems to have been attained. Their temples were the prin- . Ф 2 * GRECIAN Doric ORDER. When the width to be occupied by the whole front ts limited, to determine the diameter of the column. The relation between the parts may be expressed thus : 60 a ab + c) + (60 — с) Where а equals the width in feet occupied by the columns, and their inter- columniations taken collectively, measured at the base ; 4 equals the width of the metope, in minutes ; c equals the width of the triglyphs in minutes; d. equals the number of metopes, and x equals the diameter in feet. Example.—A front of six columns—hexastyle—61 feet wide; the frieze having one triglyph over each intercolumniation, or mono-triglyph. In this case, there being five intercolumniations and two metopes over each, therefore there are 5 x 2— ro metopes. Let the metope equal 42 minutes and the triglyph equal 28. Then a = 61; 6 = 42; с = 28; and d = 10; and the formula above becomes (96 mmm ШЕ Тл ТЕТІ ЛҒ мәй ЖАРТЫ" БА” Дені: feet — the diameter required. Example.—An octastyle front, 8 columns, 184 feet wide, three metopes over each, intercolumniation, 21 in all, and the metope and triglyph 42 and 28 as before. Then 60 x 184 __ 11040 T жа Жайы ; feet — the diameter required. 21 (42 + 28) + (60 — 28) 1502 7.351502 fee e dia q

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20 ; ARCHITECTURE. cipal objects upon which their skill in the art was displayed ; and as the Doric order seems to have been well fitted, by its massive proportions, to represent the character of their male deities rather than the female, there seems to have been а necessity for another style which should be emble- matical of feminine graces, and with which they might decorate such temples as were dedicated to the goddesses. Hence the origin of the Ionic order. This was invented, according to historians, by Hermogenes of Alabanda ; and he being a native of Caria, then in the possession of the Ionians, the order was called the Ionic. The distinguishing features of this order arethe volutes or spirals of the capital; and the denti/s among the bed- mouldings of the cornice: although in some instances dentils are wanting. The volutes are said to have been designed as a representation of curls of hair on the head of a matron, of whom the whole column is taken as a sem- blance. The Ionic order is appropriate for churches, colleges, seminaries, libraries, all edifices dedicated to literature and the arts, and all places of peace and tranquillity. Тһе front of the Custom-House, New York City,is a good specimen of this order. | 36.—The Intercolumniation: of this and the other orders—both Roman and Grecian, with the exception of the Doric—are distinguished as follows. When the interval is one and a half diameters, it is called pycnostyle, or columns thick-set; when two diameters, syszy/e; when two and a quarter diameters, eustyle ; when three diameters, dastyle ; and when more than three diameters, aræostyle, or columns thin-set. In all the orders, when there are four columns in one row, the arrangement is called Ze£raszy/e; when there are six in a row, hexastyle ; and when eight, octastyle. 37.--То Describe the Ionic Volute,—Draw а perpen- dicular from a to s (Fig. 4), and make a s equal to 20 min. or to 4 of the whole height, ас; draw s o at right angles to sa, and equal to 1} min.; upon o, with 24 min. for radius;

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PROPORTIONS OF GRECIAN IONIC. { ; Pa asus D (v NIV ERSITY| Ne O) УМА * dia. 33m. Fic. 3.—GRECIAN IONIC.

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%2 ARCHITECTURE. describe the eye of the volute; about o, the centre of the eye, draw the square, ғ 7 1 2, with sides equal to half the diameter of the eye, viz. 24 min., and divide it into 144 equal parts, as shown at А70. 5. The several centres in rotation аге at the angles formed by the heavy lines, as figured, 1, 2, 3, 4, 5, 6, etc. The position of these angles is determined by commencing at the point, 1, and making each heavy line one part less in length than the preceding one. No.1 is the See с Fic. 4.—Ionic VOLUTE.

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THE IONIC VOLUTE. 23 centre for the arc a 2 (Fig. 4;) 2 is the centre for the arc бс; and so on to the last. The inside spiral line is to be described from the centres, ғ, x, x, etc. (Fig. 5), being the centre of the first small square towards the middle of the eye from the centre for the outside arc. The breadth of the fillet at æ 7 isto be made equal to 245; min. This 15 for a spiral of three revolutions; but one of any number of revolutions, as 4 ог 6, may be drawn, by dividing o f (Fig. 5) into a cor- responding number of equal parts. Then divide the part nearest the centre, o, into two parts, as at 2; join o and 1, also o and 2; draw 2 3 parallel to o 1, and 2 4 parallel to o Fic. 5.--Еүе or VOLUTE. 2; then the lines o 1, 0 2, й 3, £ 4 will determine the length of the heavy lines, and the place of the centres. (See 477. 288.) 38.—The Corinthian Order: (Fig. 7,) is in general like | the Ionic, though the proportions are lighter. The Corin- thian displays a more airy elegance, a richer appearance ; but its distinguishing feature is its beautiful capital. This is generally supposed to have had its origin in the capitals

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24 * ARCHITECTURE. of the columns of Egyptian temples, which, though not ap- proaching it in elegance, have yet a similarity of form with the Corinthian. The oft-repeated story of its origin which is told by Vitruvius—an architect who flourished in Rome in the days of Augustus Caesar—though pretty generally considered to be fabu- lous, is nevertheless worthy of be- ing again recited. It is this: A young lady of Corinth was sick, and . finally died. Нег nurse gathered =, .into a deep basket such trinkets and keepsakes as the lady had been fond of when alive, and placed them upon her grave, cover- ing the basket with a flat stone or tile, that its contents might not be disturbed. The basket was placed accident- ally upon the stém of an acanthus plant, which, shooting forth, enclosed,the basket with its foliage, some of which, reaching the tile, turned gracefully over in the form of a volute. / A celebrated sculptor, Calimachus; saw the basket thus ‘decorated, and from the hint which it suggested conceived and constructed a capital for a column. This was called Corinthian, from the fact that it was invented and first made use of at Corinth. The Corinthian being the gayest, the richest, the most lovely of all the orders, it is appropriate for edifices which are dedicated to amusement, banqueting, and festivity—for all places where delicacy, gayety, and. splendor are desir- able. 39.—Persians and Caryatides.—In addition to the three regular orders of architecture, it was customary among the Greeks and other nations to employ representations of the human form, instead of columns, to support entablatures ; these were called Persians апа Caryatides. : are statues of men, and are so called in : commemoration of a victory gained over thc Persians by Pausanias. The Persian prisoners were brought to Athens

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PROPORTIONS OF GRECIAN CORINTHIAN. 7 (ХА SKA mu "ы ee жос Ls ee i с сс А mee _ ©, 393] >. 39b x 34 a NUR EAS | © \ | Ed ІП ДШ ТТТ | | | w 28. =. ) m 325. ax FARA «9: Fa т 5 434 ка 55-ы. ЕСІКТЕ = Gus аа ЕГ (^. 2 3 қыса» ғ 75 x Fic. 7.—GRECIAN CORINTHIAN. 25

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26 ARCHITECTURE. and condemned to abject slavery ; and in order to represent. them in the lowest state of servitude and degradation, the statues were loaded with the heaviest entablature, the Doric. 41.—Caryatides: are statues of women dressed in long robes after the Asiatic manner. Their origin is as follows: In a war between the Greeks and the Caryans, the latter were totally vanquished, their male population extinguished, and their females carried to Athens. То perpetuate the memory of this event, statues of females, having the form and dress of the Caryans, were erected, and crowned with the Ionic or Corinthian entablature. Тһе caryatides were generally formed of about the human size, but the persians much larger, in order to produce the greater awe and astonishment in the beholder. The entablatures were pro- portioned to a statue in like manner as to a column of the same height. These semblances of slavery have been in frequent use among moderns as well as ancients; and, as a relief from the stateliness and formality of the regular orders, are capa- ble of forming a thousand varieties; yet in a land of liberty such marks of human degradation ought not to be perpetu- ated. 42.—Roman Styles, —Strictly speaking, Rome had no architecture of her own; all she possessed was borrowed {тот other nations. Before the Romans exchanged inter- course with the Greeks, they possessed some edifices of considerable extent and merit, which were erected by archi- tects from Etruria; but Rome was principally indebted to Greece for what she acquired of the art. Although there is no such thing as an architecture of Roman invention, yet no nation, perhaps, ever was so devoted to the cultivation of the art as the Roman. Whether we consider the number and extent of their structures, or the lavish richness and splendor with which thcy were adorned, we are compelled to yield to them our admiration and praise. At one time, under the consuls and emperors, Rome employed 400 ar- chitects. Тһе public works— such as theatres, circuses, baths, aqueducts, ctc. — were, in extent and grandeur, be-

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PORTICO OF THE ERECTHEUM, ATHENS.

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CHANGE OF. STYLES. BY THE ROMANS. 27 yond anything attempted in modern times. Aqueducts were built to convey water from a distance of 60 miles or more. In the prosecution of this work rocks and mountains were tunnelled, and valleys bridged. Some of the latter descended 200 feet below the level of the water; апа іп passing them the canals were supported by an arcade, or succession of arches. Public baths are spoken of as large as cities, being fitted up with numerous conveniences for ex- ercise and amusement. Their decorations were most splen- did; indeed, the exuberance of the ornaments alone was offensive to good taste. So overloaded with enrichments were the baths of Diocletian that on one occasion of public festivity great quantities of sculpture fell from the ceilings and entablatures, killing many of the people. 43.— Grecian Orders modified by the Romans.—The orders of Greece were introduced into Rome in all their perfection. But the luxurious Romans, not satisfied with the simple elegance of their refined proportions, sought to . improve upon them by lavish displays of ornament. They transformed in many instances the true elegance of the Grecian art into a gaudy splendor, better suited to their less refined taste. The Romans remodelled each of the orders: the Doric (276. 8) was modified by increasing the height of the column to 8 diameters; by changing the echinus of the capital for an ovolo, or quarter-round, and adding an astragal and neck below it; by placing the centre, instead of one edge, of the first triglyph over the centre of the column; and introducing horizontal instead of inclined mutules in the cornice, and in some instances dispensing with them altogether. Тһе Ionic was modified by diminish- ing the size of the volutes, and, in some specimens, intro- ducing a new capital in which the volutes were diagonally arranged (Fig. 9). This new capital has been termed dern Ionic. The favorite order at Rome and her colonies was the Corinthian (Fig. 10). But this order the Roman artists, in their search for novelty, subjected to many alterations— especially in the foliage of its capital. Into the upper part of this they introduced the modified Ionic capital; thus

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- 28 ARCHITECTURE. . combining the two in one. This change was dignified with the importance of an order, and received the appellation Ызаа аз-аз m4 AS RA ER BAR ERAN ONES ERES NERO AGRO A OE ORE A Алаш ШЕ э MAMM, RO CR MEALS ES EP » Sy) 45) oed ELSE RE X - Б c cpu TNI UAI Мз ммс Xa br i = e а 208 ASS Fic. 8.—Roman Doric, of COMPOSITE, or Roman: the best specimen of which is found in the Arch of Titus (Fig. 11). This style was not

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PROPORTIONS OF THE ROMAN IONIC. ke Eb SIS EF NEC nS LY ARMENIA SAIL LIIS GBS ZEB САЛИ И Ж ДИЛ Ja NU 25] s | | х КҮТТІ ee cake: PS RNG SN LOUIE: су P ae I 25 > сы ъс» Ии” DS SN 2 157. pig %% DES DU Fic, 9.—ROMAN IONIC.

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30 ARCHITECTURE. much used among the Romans themselves, and is but slightly appreciated now. 44.—The Tuscan Order: is said to have been intro- duced to the Romans by the Etruscan architects, and to have been the only style used in Italy before the introduc- tion of the Grecian orders. However this may be, its simi- larity to the Doric order gives strong indications of its having been a rude imitation of that style: this is very prob- able, since history informs us that the Etruscans held inter- course with the Greeks at a remote period. The rudeness of this order prevented its extensive use in Italy. All that is known concerning it is from Vitruvius, no remains of buildings in this style being found among ancient ruins. For mills, factories, markets, barns, stables, etc., where utility and strength are of more importance than ST the improved modification of this order, called the modern Tuscan (Fig. 12), will be useful; and its simplicity recom- mends it where economy is маре. 45.— Egyptian Style.—The architecture of the ancient Egyptians—to which that of the ancient Hindoos bears some resemblance—is characterized by boldness of outline, solidity, and grandeur. < The amazing labyrinths and exten- sive artificial lakes, the splendid palaces and gloomy ceme- teries, the gigantic pyramids and towering obelisks, of the : Egyptians were works of immensity and durability ; and their extensive remains are enduring proofs of the enlight- ened skill of this once-powerful but long since extinct na- tion. The principal features of the Egyptian style of archi- tecture are—uniformity of plan, never deviating from right lines and angles; thick walls, having the outer surface slightly deviating inwardly from the perpendicular; the whole building low ; roof flat, composed of stones reaching in one piece from pier to pier, these being supported by enormous columns, very stout in proportion to their height ; the shaft sometimes polygonal, having no base but with a great variety of handsome capitals, the foliage of these being of the palm, lotus, and other leaves; eatablatures having simply an architrave, crowned with a huge cavetto orna- |

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PROPORTIONS OF THE ROMAN CORINTHIAN. 31 „Өлөт ORO WT S U TENTS TETS nE T TENTE E y mS SPA SA c. UNNA Aa Hee VACA AONAN ISISININI ANININ PEAR ECA NCAA Eu ge RO At АССИ xe 314 Lom Уа NT NEN ERE SN ROSE ТАК CAM RR АТО уе ги Ге LEO Ј Lnd UNIVERSITY AESTATE ASINO NET NAE АУАҒА AYE NEY M EZ S EAS T MAD И (ү V ыз : 253 тен - = eS кыз —— EEUU. A2 DAA SRR NE Fic. 10.—ROMAN CORINTHIAN.

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38 ARCHITECTURE. OS ҚАЗДЫ Sa SS Bok а Sous Быз сыш a Pi E | E 8 ЕЕ S EET REORUM SUA BUE КИ КБК НУРУ PARS NOMEN em NONSE IY 2 ЗЕҢ ЕКА ЕГЕТЕ 28 © IRT badii: GEND i) ВР + aD i: OEDD: aD ЛР 5: D 5, GD 5. CD >: GERD +: ОЬ 5 GED с OD + р, D ic ae Ги E É etii а) > 2, Ad ip $s ‚е > % е SS PU U^ Ths 1 Ж j Mi. э її T ЧЕ) = he 29) ( EE Ad ; kx a т ( " SS 26% MT ыз! ХЗ. WA. Ес. 11.—CowMrostiTE ORDER—ARCI or TITUS.

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MASSIVENESS OF EGYPTIAN STRUCTURES. 33 mented with sculpture; and the Meera i em very nar- row, usually 1$ diameters and seldom exceeding 24. Іп the remains оға temple the walls were found to be 24 feet thick ; and at the gates of Thebes, the walls at the foundation were бо feet thick and perfectly solid. The immense stones of which these, as well as Egyptian walls generally, were built, had both their inside and outside surfaces faced, and the oints throughout the body of the wall as perfectly close as upon the outer surface. For this reason, as well as that the buildings generally partake of the pyramidal form, arise their great solidity and durability. Тһе dimensions and ex- tent of the buildings may be judged from the temple of Jupiter at Thebes, which was 1400 feet long and 300 feet wide—exclusive of the porticos, of which there was a great number. It is estimated by Mr. Gliddon, U. S. Consul in Egypt, that not less than 25,000,000 tons of hewn stone were em- ployed in the erection of the Pyramids of Memphis alone— or enough to construct 3000 Bunker Hill monuments. Some of the blocks are 40 feet long, and polished with emery to a surprising degree. It is conjectured that the stone for these pyramids was brought, by rafts and canals, from a distance of six or seven hundred miles. The general appearance of the Egyptian style of archi- tecture is that of solemn grandeur—amounting sometimes to sepulchral gloom. For this reason it is appropriate for cem- eteries, prisons, etc.; and being adopted for these gijpeses it is gradually gaining favor. A great dissimilarity exists in the proportion, form, and ‘ general features of Egyptian columns. Іп some instances, there is no uniformity even in those of the same building, each differing from the others either in its shaft or capital. For practical use in this country, Fig. 13 may be taken as а standard of this style. The Halls of Justice in Centre Street, New York City, is a building in general accordance with the principles of Egyptian architecture. 46. in General.—In selecting a style for an edifice, its peculiar requirements must be allowed to govern.

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ARCHITECTURE. 34 "eK 9 Ес. 12.—MopirFIiED TUSCAN ORDER.

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FITNESS OF STYLES. 35 That style of architecture is to be preferred in which utility, stability, and regularity are gracefully blended with gran- deur and elegance. Ви as ап arrangement designed for a warm country would be inappropriate for a colder climate, it would seem that the style of building ought to be modified to suit the wants of the people for whom it is, designed. High roofs to resist the pressure of heavy snows, and ar- rangements for artificial heat, are indispensable in northern climes; while they would be regarded as entirely out of place in buildings at the equator. | Among the Greeks, architecture was employed chiefly upon their temples and other large buildings; and the pro- portions of the orders, as determined by them, when execu- ted to such large dimensions, have the happiest effect. But when used for small buildings, porticos, porches, etc., espe- cially in country places, they are rather heavy and clumsy ; in such cases, more slender proportions will be found to pro- duce a better effect. The English cottage-style is rather more appropriate, and is becoming extensively practised for small buildings in the country. .47.—Expression.—Every building should manifest its destination. If it be intended for national purposes, it should be magnificent—grand ; for a private residence, neat and modest ; for a banqueting-house, gay and splendid; for a monument or cemetery, gloomy—melancholy ; or, if for a church, majestic and graceful—by some it has been said, * somewhat dark and gloomy, as being favorable to a devo- tional state of feeling ;" but such impressions can only re- sult from a misapprehension of the nature of true devotion. “ Нег ways are ways of /easantness, and all her paths are peace." The church should rather be a type of that brighter world to which it leads. Simply for purposes of contemplation, however, the glare of the noonday light should be excluded, that the worshipper may, with Milton— “ Love the high, embowéd roof, With antique pillars massy р 9, And storied windows richly t, Casting a dim, religious light.”

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36 ARCHITECTURE. Fig. Т ін ARCHITECTURE. ^

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PREVALENCE OF WOODEN DWELLINGS. 37 However happily the several parts of an edifice may be disposed, and however pleasing it may appear as a whole, yet much depends upon its séée, as also upon the character and style of the structures in its immediate vicinity, and the degree of cultivation of the adjacent country. А splendid country-seat should have the out-houses and fences in the same style with itself, the trees and shrubbery neatly trimmed, and the grounds well cultivated. 48.—Durability.—Europeans express surprise that we build so much with wood. Апа yet, in а new country, where wood is plenty, that this should be so is no cause for wonder. Still the practice should not be encouraged. Build- ings erected with brick or stone are far preferable to those of wood: they are more durable; not so liable to injury by fire, nor to need repairs; and will be found in the end quite ав economical. А wooden house is suitable for a temporary residence опу ; and those who would bequeath a dwelling to their children will endeavor to build with a more dura- ble material. Wooden cornices and gutters, attached to brick houses, are objectionable—not only on account of their frail nature, but also because they render the building liable to destruction by fire. 49.—Dwelling-Houses: are built of various dimensions and styles, according to their destination; and to give de- signs and directions for their erection, it is necessary to know their situation and object. A dwelling intended for a gar- dener would require very different dimensions and arrange- ments from one intended for a retired gentleman—with his servants, horses, etc.; nor would a house designed for the city be appropriate for the country. For city houses, ar- rangements that would be convenient for one family might be very inconvenient for two or more. Figs. 14, 15, 16, 17, 18, and 19 represent.the ichnographical projection, or ground- plan, of the floors of an ordinary city house, designed to be occupied by one family only. Fig. 21 is an elevation, or front view, of the same house. All these plans are drawn at the same scale—which is that at the bottom of 7zg.gg1.

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38 ARCHITECTURE. Fig. 14 is a Plan of the Under-Cellar. a, is the coal-vault, 6 by то feet. à, is the furnace for heating the house. c, d, are front and rear areas. Fig. 15 is a Plan of the Basement. a, is the library, or ordinary dining-room, 15 by 20 feet. à, is the kitchen, 15 by 22 feet. c, is the store-room, 6 by 9 feet. d, is the pantry, 4 by 7 feet. е, 15 the china closet, 4 by 7 feet. У, is the servants’ water-closet. g, is a closet. h, is a closet with a dumb-waiter to the first story above. 7, 15 an ash closet under the front stoop. J, is the kitchen-range. 2, is the sink for washing and ien water. l, are wash-trays. Fig. 16 is a Plan of the First Story. а, is the parlor, 15 by 34 feet. 0, is the dining-room, 16 by 23 feet. с, is the vestibule. é, is the closet containing the ти from the base- ment. f, is the closet containing butler's sink. 2, £; are closets. h, is a closet for hats and cloaks. 1, у, are front and rear balconies. Fig. 17 is the Second Story. a, a, are chambers, 15 by 13 feet. б, is a bed-room, 7% by 13 feet. с, is the bath-room, 73 by 13 feet. а,жа, ate dressing-rooms, 6 by 74 feet. е, е, are closets. f, f, are wardrobes. 2, 2, «е cupboards.

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PLANS OF A CITY HOUSE. Fic. 15. FiG. 14. UNDER-CELLAR. BASEMENT. City DWELLING.

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ARCHITECTURE. FA INARA Fic. 16. FIG. 17. FIRST STORY. SECOND STORY. City DWELLING.

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a UPPER STORIES OF А CITY HOUSE. 41 Fig. 18 is the Third Story. “а, a, are chambers, 15 by 19 feet. b, 6, are bed-rooms, 74 by 13 feet. c, c, are closets. d, is a linen-closet, 5 by 7 feet. PIG. 18. Ғіс. 19. FOURTH STORY. THIRD STORY. Сїтү DWELLING. е, е, are dressing-closets. f, f, are wardrobes. £, g, are cupboards. ~ Fig. 19 is the Fourth Story. a, a, are chambers, 14 by 17 feet.

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42 ARCHITECTURE. б, b, are bed-rooms, 84 by 17 feet. с, с, с, are Closets. d, is the step-ladder to the roof. Fig. 20 is the Section of the House showing the heights of the several stories. Fig.21 is the Front Elevation. The size of the house is 25 feet front by 55 feet deep; this is about the average depth, although some are extended to бо and 65 feet in depth. , These are introduced to give some general ideas of the principles to be followed in designing city houses. In plac- ing the chimneys in the parlors, set the chimney-breasts equidistant from the ends of the room. The basement chimney-breasts may be placed nearly in the middle of the side of the room, as there is but one flue to pass through the chimney-breast above ; but in the second story, as there are two flues, one from the basement and one from the par- lor, the breast will have to be placed nearly perpendicular over the parlor breast, so as to receive the flues within the jambs of the fire-place. As it is desirable to have the chim- ney-breast as near the middle of the room as possible, it may be placed a few inches towards that point from over the breast below. So in arranging those of the stories above, always make provision for the flues from below. A 50.—Arranging the Stairs and Windows.—There should be at least as much room in the passage at the side of the stairs as upon them ; and in regard to the length of the pas- sage in the second story, there must be room for the doors which open from each of the principal rooms into the hall, and more if the stairs require it. Having assigned a posi- tion for the stairs of the second story, now generally placed | in the centre of the depth of the house, let the winders of the other stories be placed perpendicularly over and under them; and be careful to provide for head-room. To ascer- tain this, when it is doubtful, it is well to draw a vertical section of the whole stairs; but in ordinary cases this is not

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ЕКОХТ ОЕ A CITY HOUSE. 43 x J x i == к=. mI = == еее еже, 4: c dh cnl. e: жет шен. sr SS © е зө Хә T ай Fie. 21. ' 7 ELEVATION. City DWELLING. : ы Fic. 20. SECTION.

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44. ARCHITECTURE. necessary. To dispose the windows properly, the middle window of each story should be exactly in the middle of the front; but the pier between the two windows which light the parlor should be in the centre of that room; because when chandeliers or any similar ornaments hang m the centre-pieces of the parlor ceilings, it is important, in order to give the better effect, that the pier-glasses at the front and rear be in a range with them. If both these objects cannot be attained, an approximation to each must be at- tempted. The piers should in no case be less in width than the window openings, else the blinds or shutters, when thrown open, will interfere with one another; in general practice, it is well to make the outside piers 2 of the width of one of the middle piers. When this is desirable, deduct the amount of the three openings from the width of the front, and the remainder will be the amount of the width of all the piers; divide this by то, and the product will be 4 of a middle pier ; and then, if the parlor arrangements do Mt interfere, give twice this amount to each corner pier, and three times the same amount to each of the middle piers. 51.— Principles of Architecture.— To build well requires close attention and much experience. The science of build- ing is the result of centuries of study. Its progress towards perfection must have been exceedingly slow. Іп the con- struction of the first frail and rude habitations of men, the primary object was, doubtless, utility—a mere shelter from sun and rain. Dut as successive storms shattered his poor tenement, man was taught by experience the necessity of building with an idea to durability. Апа as the symmetry, proportion, and beauty of nature met his admiring gaze, contrasting so strangely with the misshapen and dispropor- . tioned work of his own hands, he was led to make gradual changes, till his abode was rendered not only commodious and durable, but pleasant in its appearance; and building became a fine art, having utility for its basis. 52.—Arrangement.—[n all designs for buildings of im- portance, utility, durability, and beauty, the first great prin- ciples, should be pre-eminent. In order that the edifice be

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'G3OSNWVHO AO NVALVHO | e " | | | А | | | WS | Am 3 | Il

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! ESSENTIAL REQUIREMENTS OF A BUILDING. 45 useful, commodious, and comfortable, the arrangement of the apartments should be such as to fit them for their several destinations ; for publio assemblies, oratory, state, visitors, retiring, eating, reading, sleeping, bathing, dressing, etc.— these should each have its own peculiar form and situation. To accomplish this, and at the same time to make their relative situation agreeable and pleasant, producing regu- larity and harmony, require'in some instances much skill and sound judgment. Convenience and regularity are very important, and each should have due attention; yet when both cannot be obtained, the latter should in most cases give place to the former. А building that 1s neither con- venient nor regular, whatever other good qualities it may possess, will be sure of disapprobation. 53.—Ventilation.— Attention should be given to such arrangements as are calculated to promote health: among these, ventilation is by no means the least. For this pur- pose, the ceilings of the apartments should have a respect- able height; and the sky-light, or any part of the roof that can be made movable, should be arranged with cord and pullies, so as to be easily raised and lowered. Small open- ings near the ceiling, that may be closed at pleasure, should be made in the partitions that separate the rooms from the passages — especially for those rooms which are used for sleeping apartments. All the apartments should be so ar- ranged as to secure their being easily kept dry and clean. In dwellings, suitable apartments should be fitted up for bathing with all the necessary apparatus for conveying water. 54..—Sstability.—Ilo secure this, an edifice should be de- signed upon well-known geometrical principles: such as science has demonstráted to be necessary and sufficient for firmness and durability. Jt is well, also, that it have the. appearance of stability as well as the reality, for should it - seem tottering and unsafe, the sensation of fear, rather than thośe of admiration and pleasure, will be excited in the be- holder. To secure certainty and accuracy in the applica- tion of those principles, a knowledge of the strength and |

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[ 46 ARCHITECTURE. other properties of the materials used is indispensable ; and in order that the whole design be so made as to be capable of execution, a practical knowledge of the requisite mechan- ical operations is quite important. 55.—Decoration.—The elegance of a design, although chiefly depending upon a just proportion and harmony of the parts, will be promoted by the introduction of orna- ments, provided this be judiciously performed ; for enrich- ments should not only be ot a proper character to suit the style of the building, but should also have their true posi- tion, and be bestowed in proper quantity. The most com- mon fault, and one which is prominent in Roman architec- ture, is an excess of enrichment: an error which is carefully to be guarded against. But those who take the Grecian models for their standard will not be liable to go to that extreme. Іп ornamenting a cornice, or any other assem- blage of mouldings, at least every alternate member should be left plain ; and those that are near the eye should be more finished than those which are distant. Although the charac- teristics of good architecture are utility and elegance, in connection with durability, yet some buildings are designed expressly for use, and others again for ornament: in the former, utility, and in the latter, beauty, should be the gov- erning principle. 56.— Elementary Parts of a Building. — The builder should be acquainted with the principles upon which the essential, elementary parts of a building are founded. A scientific knowledge of these will insure certainty and secu- rity, and enable the mechanic to erect the most extensive and lofty edifices with confidence. Тһе more important parts are the foundation, the column, the wall, the lintel, the arch, the vault, the dome, and the roof. А separate description of the peculiarities of each would seem to be necessary, and cannot perhaps be better expressed than in the following language of a modern writer on this subject, slightly modified :

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STRUCTURAL FEATURES OF A BUILDING. 47 57.—' The Foundation: of a building should be begun at a certain depth in the earth, to secure a solid basis, below the reach of frost and common accidents. The most solid basis is rock, or gravel which has not been moved. Next to these are clay and sand, provided no other excavations have been made in the immediate neighborhood. From this basis a stone wall is carried up to the surface of the ground, and constitutes the foundation. Where it is in- tended that the superstructure shall press unequally, as at its piers, chimneys, or columns, it is sometimes of use to occupy the space between the points of pressure by an inverted arch. This distributes the pressure equally, and prevents the foundation from springing between the differ- ent points. In loose or muddy situations, it is always un- safe to build, unless we can reach the solid bottom below. In salt marshes and flats, this is done by depositing timbers, or driving wooden piles into the earth, and raising walls upon them. Тһе preservative quality of the salt will keep these timbers unimpaired for a great length of time, and makes the foundation equally secure with one of brick or stone. 58.—The Column, or Pillar: is the simplest member in any building, though by no means an essential one to all. This is a perpendicular part, commonly of equal breadth and thickness, not intended for the purpose of enclosure, but simply for the support of some part of the superstruc- ture. The principal force which a column has to resist is that of perpendicular pressure. In its shape, the shaft of a column ‘should not be exactly cylindrical, but, since the lower part must support the weight of the superior part, in addition to the weight which presses equally on the whole column, the thickness should gradually decrease from bot- tom to top. The outline of columns should be a little curved, so as to represent a portion of a very long spheroid, or paraboloid, rather than of a cone. This figure is the joint result of two calculations, independent of beauty of appear- ance. One of these is, that the form best adapted for sta- bility of base is that of a cone; the other is, that the figure,

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48 ARCHITECTURE. which would be of equal strength throughout for support- ing a superincumbent weight, would be generated by the revolution of two parabolas round the axis of the column, the vertices of the curves being at its extremities. Тһе swell of the shafts of columns was called the ez/aszs by the ancients. It has been lately found that the columns of the Parthenon, at Athens, which have been commonly supposed straight, deviate about an inch from a straight line, and that their greatest swell is at about one third of their height. Columns in the antique orders are usually made to diminish one sixth or one seventh of their diameter, and sometimes even one fourth. The Gothic pillar is commonly of equal thickness throughout. 59.—The Wall: another elementary part of a building, may be considered as the lateral continuation of the column, answering the purpose both of enclosure and support. A wall must diminish as it rises, for the same reasons, and in the same proportion, as the column. It must diminish still more rapidly if it extends through several stories, support- ing weights at different heights. А wall, to possess the greatest strength, must also consist of pieces, the upper and lower surfaces of which are horizontal and regular, not rounded nor oblique. Тһе walls of most of the ancient structures which have stood to the present time are con- structed in this manner, and frequently have their stones bound together with bolts and clamps of iron. The same method is adopted in such modern structures as are intended to possess great strength and durability, and, in some cases, the stones are even dovetailed together, as in the light- houses at Eddystone and Bell Rock. But many of our modern stone walls, for the sake of cheapness, have only one face of the stones squared, the inner half of the wall being completed with brick; so that they can, in reality, be con- sidered only as brick walls faced with stone. Such walls are said to be liable to become convex outwardly, from the dif- ference in the shrinking of the cement. Rubble walls are made of rough, irregular stones, laid in mortar. The stones should be broken, if possible, so as to produce horizontal

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VARIOUS METHODS OF ERECTING WALLS. 49 surfaces. The coffer walls of the ancient Romans were made by enclosing successive portions of the intended wall in a box, and filing it with stones, sand, and mortar promis- cuously. This kind of structure must have been extremely insecure. The Pantheon and various other Roman build- ings are surrounded with a double brick wall, having its vacancy filled up with loose bricks and cement. The whole has gradually consolidated into a mass of great firmness. 60.—The Reticulated Walls: of the Romans— com- posed of bricks with oblique surfaces—would, at the present day, be thought highly unphilosophical. Indeed,they could not long have stood, had it not been for the great strength of their cement. Modern brick walls are laid with great precision, and depend for firmness more upon their position than upon the strength of their cement. The bricks being laid in horizontal courses, and continually overlaying each other, or breaking joints, the whole mass is strongly inter- woven, and bound together. Wooden walls, composed of timbers covered with boards, are a common but more per- ishable kind. They require to be constantly covered with а coating of a foreign substance, as paint or plaster, to pre- serve them from spontaneous decomposition. In some parts of France, and elsewhere, a kind of wall is made of earth, rendered compact by ramming it in moulds or cases. This method is called building in 2256, and is much more durable than the nature of the material would lead us to suppose. Walls of all kinds are greatly strengthened by angles and curves, also by projections, such as pilasters, chimneys, and buttresses. These projections serve to increase the breadth of the foundation, and are always to be made use of in large buildings, and in walls of considerable length. 61.— The Lintel, or Beam: extends in a right line over a vacant space, from one column or wall to another. The strength of the lintel will be greater in proportion as its transverse vertical diameter exceeds the horizontal, the strength being always as the square of the depth. The floor is the lateral continuation or connection of beams by means of a covering of boards.

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50 ARCHITECTU RE. 62.—' The Arch: is a transverse member of a building, answering the same purpose as the lintel, but vastly exceed- ing it in strength. Тһе arch, unlike the lintel, may consist of any number of constituent pieces, without impairing its strength. Itis, however, necessary that all the pieces should possess a uniform shape,—the shape of a portion of a wedge,—and that the joints, formed by the contact of their surfaces, should point towardsa common centre. Іп this case, no one portion of the arch can be displaced or forced inward ; and thé arch cannot be broken by any force which is not sufficient to crush the materials of which it is made. In arches made of common bricks, the sides of which are parallel, any oze of the bricks might be forced inward, were it not for the adhesion of the cement. Any zwo of the bricks, however, by the disposition of their mortar, cannot collect- ively be forced inward. An arch of the proper form, when complete, is rendered stronger, instead of weaker, by the pressure of a considerable weight, provided this pressure be uniform. While building, however, it requires to be sup- ported by a centring of the shape of its internal surface, until itis complete. The upper stone of an arch is called the keystone, but is not more essential than any other. In regard to the shape of the arch, its most simple form is that of the semicircle. ` It is, however, very frequently a smaller arc of a circle, or a portion of an ellipse. 63.—Hooke’s Theory of an Arch.—The simplest theory of an arch supporting itself only is that of Dr. Hooke. The arch, when it has only its own weight to bear, may be considered as the inversion of a chain, suspended at each end. The chain hangs in such a form that the weight of each link or portion is held in equilibrium by the result of two forces acting at its extremities; and these forces, or tensions, are produced, the one by the weight of the portion of the chain below the link, the other by the same weight increased by that of the link itself, both of them acting ori- ginally in a vertical direction. Now, supposing the chain inverted, so as to constitute an arch of the same form and weight, the relative situations of the forces will be the same, .

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VIADUCT AT CHAUMONT.

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PECULIARITIES OF THE ARCH. 5I only they will act in contrary directions, so that they are compounded in a similar manner, and balance each other on the same conditions. | | The arch thus formed is denominated a catenary arch. In common cases, it differs but little from a circular arch of the extent of about one third of a whole circle, and rising from the abutments with an obliquity of about 30 degrees from a perpendicular. But though the catenary arch is the best form for supporting its own weight, and also all addi- tional weight which presses in a vertical direction, it is not the best form to resist lateral pressure, or pressure like that of fluids, acting equally in all directions. Thus the arches of bridges and similar structures, when covered with loose stones and earth, are pressed sideways, as well as vertically, in the same manner as if they supported a weight of fluid. In this case, it is necessary that the arch should arise more perpendicularly from the abutment, and that its general fig- ure should be that of the longitudinal segment of an ellipse. In small arches, in common buildings, where the disturbing force is not great, it is of little consequence what is the shape of the curve. The outlines may even be perfectly straight, as in the tier of bricks which we frequently see over a window. This is, strictly speaking, a real arch, pro- vided the surfaces of the bricks tend toward a common centre. It is the weakest kind of arch, and a part of it is necessarily superfluous, since no greater portion can act in supporting a weight above it than can be included be- tween two curved or arched lines. 64.— Gothic Arches, — Besides these arches, various others are in use. Тһе acute or lancet arch, much used in Gothic architecture, is described usually from two cen- tres outside the arch. It is a strong arch for supporting vertical pressure. The rampant arch is one in which the two ends spring from unequal heights. Тһе horseshoe or Moorish arch is described from one or more centres placed-above the baseline. In this arch, the lower parts are in danger of being forced inward. The ogee arch is concavo-convex, and therefore fit only for ornament.

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52 ARCHITECTU RE. 65.—Arch: Definitions; Principles. — The upper sur- face is called the exzrados, and the inner, the zzrados. The spring is where the intrados meets the abutments. The span is the distance between the abutments. Тһе wedge- shaped stones which form an arch are sometimes called voussoirs, the uppermost being the keystone. The part ofa pier from which an arch springs is called the 27205, and the curve formed by the under side of the voussoirs, the archz- volt. Itis necessary that the walls, abutments, and piers on which arches are supported should be so firm asto resist the lateral ¢hrust, as well as vertical pressure, of the arch. It will at once be seen that the lateral or sideway pressure of an arch is very considerable, when we recollect that every stone, or portion of the arch, is a wedge, a part of whose force acts to separate the abutments. For want of attention to this circumstance, important mistakes have been commit- ted, the strength of buildings materially impaired, and their ruin accelerated. In some cases, the want of lateral firmness in the walls is compensated by a bar of iron stretched across the span of the arch, and connecting the abutments, like the tie-beam of a roof. This is thecase in the cathedral of Milan and some other Gothic buildings. 66.—An Arcade: or continuation of arches, needs only that the outer supports of the terminal arches should be strong enough to resist horizontal pressure. Іп the inter- mediate arches, the lateral force of each arch is counter- acted by the opposing lateral force of the one contiguous to it. In bridges, however, where individual arches are liable to be destroyed by accident, it is desirable that each of the piers should possess sufficient horizontal strength to resist . the lateral pressure of the adjoining arches. 67.— The Vault: is the lateral continuation of an arch, serving to cover an area or passage, and bearing the same relation to the arch that the wall does to the column. A simple vault is constructed on the principles of the arch, and distributes its pressure equally along the walls or abutments, A complex or groined vault is made by two vaults intersect- ing each other, in which case the pressure is thrown upon

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VARIOUS CONSTRUCTIONS OF THE DOME. 53 - springing points, and is greatly increased at those points. The groined vault is common in Gothic architecture. 68.—The Dome: sometimes called cupola, is a concave covering to a building, or part of it, and may be either a segment of a sphere, of a spheroid, or of any similar figure. When built of stone, itis a very strong. kind of structure, even more so than the arch, since the tendency of each part to fall is counteracted, not only by those above and below it, but also by those on each side. It is only necessary that the constituent pieces should have a common form, and that . this form should be somewhat like the frustum of a pyra- mid, so that, when placed in its situation, its four angles may point toward the centre, or axis, of the dome. During the erection of a dome, it is not necessary that it should be sup- ported by a centring, until complete, as is done in the arch. Each circle of stones, when laid, is capable of supporting itself without aid from those above it. It follows that the dome may be left open at top, without a keystone, and yet be perfectly secure in this respect, being the reverse of the arch. The dome of the Pantheon, at Rome, has been always open at top, and yet has stood unimpaired for nearly 2000 years. The upper circle of stones, though apparently the weakest, is nevertheless often made to support the addi- tional weight of a lantern or tower above it. Іп several of the largest cathedrals, there are two domes, one within the other, which contribute their joint support to the lantern, which rests upon the top. In these buildings, the dome rests upon a circular wall, which is supported, in its turn, by arches upon massive pillars or piers. This construction is called building upon fendentives, and gives open space апа room for passage beneath the dome. The remarks which have been made in regard to the abutments of the arch apply equally to the walls immediately supporting a dome. They must be of sufficient thickness and solidity to resist the lateral pressure of the dome, which is very great. The walls of the Roman Pantheon are of great depth and solid- ity. In order that a dome in itself should be perfectly secure, its lower parts must not be too nearly vertical, since,

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54 ARCHITECTURE. in this case, they- partake of the nature of perpendicular walls, and are acted upon by the spreading force of the parts above them. Тһе dome of St. Paul's Church, in Lon- don, and some others of similar coristruction, are bound with chains or hoops of iron, to prevent them from spreading at bottom. Domes which are made of wood depend, in part, for their strength on their internal carpentry. The Halle du Bled, in Paris, had originally a wooden dome more than 200 feet in diameter, and only one foot in thickness. This has since been replaced by a dome of iron. (See Art. 235.) 69. : is the most common and cheap method of covering buildings, to protect them from rain and other effects of the weather. It is sometimes flat, but more fre- quently oblique, in its shape. Тһе flat or platform roof is the least advantageous for shedding rain, and is seldom used in northern countries. The ent roof, consisting of two oblique sides meeting at top, is the most common form. These roofs are made steepest in cold climates, where they are liable to be loaded with snow. Where the four sides of the roof are all oblique, it is denominated а hipped roof, and where there are two portions to the roof, of different ob- liquity, it is a curb, or mansard roof. Іп modern times, roofs are made almost exclusively of wood, though frequently covered with incombustible materials. Тһе internal struc- ture or carpentry of roofs is a subject of considerable me- chanical contrivance. Тһе roof is supported by rafters, which abut on the walls on each side, like the extremities of an arch. If no other timbers existed except the rafters, they would exert a strong lateral pressure on the walls, tending to separate and overthrow them. To counteract this lateral force, а ftze-beam, as it is called, extends across, receiving the ends of the rafters, and protecting the wall from their horizontal thrust. To prevent the tie-beam from sagging, or bending-downward with its own weight, a $ing- post is erected from this beam, to the upper angle of the rafters, serving to connect the whole, and to suspend the weight of the beam. This is called ¢russing. Queen-posts.

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MANNER OF CONSTRUCTING ROOFS. 55 are sometimes added, parallel to the king-post, in large roofs: also various other connecting timbers. In Gothic buildings, where the vaults do not admit of the use of a tie-beam, the rafters are prevented from spreading, as in an arch, by the strength of the buttresses. In comparing the lateral pressure of a high roof with that of a low one, the length of the tie-beam being the same, it will be seen that a high roof, from its containing most materials, may produce the greatest pressure, as far as weight 1s concerned. Оп the other hand, if the weight of. both be equal, then the low roof will exert the greater press- ure; and this will increase in proportion to the distance of the point at which perpendiculars, drawn from the end of each rafter, would meet. In roofs, as well as in wooden domes and bridges, the materials are subjected to an inter- nal strain, to resist which the cohesive strength of the ma- terial is relied on. On this account, beams should, when possible, be of one piece. Where this cannot be effected, two or more beams are connected together by splicing. Spliced beams are never so strong as whole ones, yet they may be made to approach the same strength, by affixing lat- eral pieces, or by making the ends overlay each other, and connecting them with bolts and straps of iron. The ten- dency to separate is also resisted, by letting the two pieces into each other by the process called scarfing. Mortices, in- tended to ¢rvuss or suspend one piece by another, should be formed upon similar. principles. Roofs in the United States, after being boarded, receive a secondary covering of shingles. When intended to be incombustible, they are covered with slates or earthen tiles, or with sheets of lead, copper, or tinned iron. Slates are preferable to tiles, being lighter, and absorbing less moisture. Metallic sheets are chiefly used for flat roofs, wooden domes, and curved and angular surfaces, which require a flexible material to cover them, or have not a sufficient pitch to shed the rain from slates or shingles. Various artificial composi- tions are occasionally used to cover roofs, the most common of which are mixtures of tar with lime, and sometimes with sand and gravel.—Zzcy. Am. (See Art. 202.)

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SECTION IL—CONSTRUCTION. ART. 70.— Construction Esseniial.—Construction is that part of the Science of Building which treats of the Laws of Pressure and the strength of materials. To the architect and builder a knowledge of it is absolutely essential. It de- serves a larger place in a volume of this kind than is gene- rally allotted to it. Something, indeed, has been said upon the styles and principles, by which the best arrangements тау be ascertained; yet, besides this, there is much to be learned. For however precise or workmanlike the several parts may be made, what will it avail, should the system of framing, from deficient material, or an erroneous position of its timbers, fail to sustain even its own weight? Hence the necessity for a knowledge of the laws of pressure and the strength of materials. These being once understood, we can with confidence determine the best position and dimen- sions for the several pieces which compose a floor or a roof, a partition or a bridge. Ав systems of framing are more or less exposed to heavy weights and strains, and, in case of failure, cause not only a loss of labor and material, but frc- quently that of life itself, it is very important that the mate- rials employed be of the proper quantity and quality to serve their destination. And, on the other hand, any superfluous material is not only useless, but a positive injury, as it is an unnecessary load upon the points of support. It is neces- sary, therefore, to know the /cas¢ quantity of material that will suffice for strength. Not the least common fault in framing is that of using an excess of material. Economy, at least, would seem to require that this evil be abated. - Before proceeding to consider the principles upon which a system of framing should be constructed, let us attend to a few of the elementary laws in Mechanics, which will be found to be of great value in determining those principles.

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mS! T INTERIOR OF THE CATHEDRAL, SIENNA.

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DIRECT AND OBLIQUE SUPPORTS. 57 71.—LLaws of Pressure.—(1.) A heavy body always ex- erts a pressure equal to its own weight in a vertical direc- tion. Example: Suppose an iron ball weighing 100 lbs. be supported upon the top of a perpendicular post (Fig. 22-4); then the pressure exerted upon that post will be equal to the weight of the ball, viz., 100 165. (2.) But if two inclined posts (Fig. 22-8). be substituted for the perpendicular sup- port, the united pressures upon these posts will be more than equal to the weight, and will be in proportion to their position. The farther apart their fect are spread the greater will be the pressure, and тсе versa. Hence tremendous strains may be exerted by a comparatively small weight. And it follows, therefore, that a piece of timber intended for a strut or post should be so placed that its axis may coincide, as nearly as possible, with the diréction of the pressure. The direction of the pressure of the weight W (Fig. 22-В) is in the vertical line 0 d; and the weight W would fall in that line if the two posts were removed ; hence the best position for a support for the weight would be in a 116.422. that line. But as it rarely occurs in systems of framing that weights can be supported by any single resistance, they requiring generally two or more supports (as in the case of a roof supported by its rafters), it becomes important, there- fore, to know the exact amount of pressure any certain weight is capable of exerting upon oblique supports. Now, it has been ascertained that the three lines of a triangle, drawn parallel with the direction of three concurring forces in equilibrium, are in proportion respectively to these

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58 CONSTRUCTION. forces. For example, in /zg. 22-2, we have a representation of three forces concurring in a point, which forces are in equilibrium and at rest; thus, the weight W is one force, and the resistances exerted by the two pieces of timber are the other two forces. Тһе direction in which the first force acts is vertical —downwards; the direction of the two other forces is in the axis of each piece of timber respectively. These three forces all tend towards the point å. Draw the axes а 6 and бс of the two supports; make 2 d vertical, and from d draw de and d f parallel with the axes дс and 6 а repectively. Then the triangle 0 4 e has its lines parallel respectively with the direction of the three forces; thus, 24 is in the direction of the weight W, d e paral- lel with the axis of the timber D, and e? is in the direction of the timber C. In accordance with the principle above stated, the lengths of the sides of the triangle ? Ze are in propor- tion respectively to the three forces aforesaid ; thus— As the length of the line bd 0 Is to the number of pounds in the weight W, So is the length of the line де To the number of pounds’ pressure resisted by the timber C. Again-- Ав the length of the line 24 Is to the number of pounds in the weight JV, So is the length of the line Ze | Tothe number of pounds' pressure resisted by the timber D. And again— As the length of the line Фе Is to the pounds' pressure resisted by C, So is the length of the line d e To the pounds' pressure resisted by 2. These proportions are more briefly stated thus— 7.” Ce a =. 08: Pf, P being used as a symbol to represent the number of pounds’ pressure resisted by the timber C.

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PARALLELOGRAM OF FORCES. 59 24. bd:W::de:oQ ow) О representing the number of pounds’ pressure resisted by the timber D. 34. WOE Pe @ 72.—Parallelogram of Forces.—This relation between lines and pressures is applicable in ascertaining the pres- sures induced by known weights throughout any system of framing. The parallelogram ^? e Z f is called the Parallelo- gram of Forces; the two lines д е-апа ? f being called the components, and the line 2 d the resultant. Where it is re- quired to find the components from a given resultant (Fig. 22-В), the fourth line 4 f need not be drawn, for the triangle bde gives the desired result. But when the resultant is to be ascertained from given components (Zzg. 28), it is more convenient to draw the fourth line. 73.—The Resolution of Forces: is the finding of two or more forces which, acting in different directions, shail exactly balance the pressure of any given single force. To make a practical application of this, let it be required to ascertain the oblique pressure іп 226. 22-8. In this figure the line 2 4 measures half an inch (0-5 inch), and the line фе three tenths of an inch (0-3 inch). Now if the weight W be supposed to be 1200 pounds, then the first stated propor- tion above, bd:W::be:P, becomes 0-5: 1200::0-3:f. And since the product of the means divided by one of the extremes gives the other extreme, this proportion may be put in the form of an eguation, thus— 1200 X 0-3. p 0:5 Performing the arithmetical operation here indicated—that is, multiplying together the two quantities above the line, and dividing the product by the quantity under the line—the

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60 CONSTRUCTION. quotient will be equal to the quantity represented by 2, viz., the pressure resisted by the timber C. Thus— 1200 0:3 === о-5)360-0 720 = The strain upon the timber C is, therefore, equal to 720 pounds; and since, in this case (the two timbers being in- clined equally from the vertical), the line e Zis equal to the line 2 e, therefore the strain upon the other timber 2 is also 720 pounds. Fic. 23. 74.—1Inclination of Supports Unequal.—In Fig. 23 the pressures in the two supports are unequal. The supports are also unequal in length. The length of the supports, however, does not alter the amount of pressure from the concentrated load Supported; but generally long timbers are not so capable of resistance as shorter ones. For, not being so stiff, they bend more readily, and, since the com- pression is in proportion to the length, they therefore shorten more. To ascertain the pressures in Fig. 23, let the weight suspended from 2 4 be equal to two and three quarter tons (2.75 tons) Тһе line 2 4 measures five and a half tenths of an inch (0-55 inch), and the line 2 e half an inch (0:5 inch) Therefore, the proportion ба: W::be:P becomes 0-55: 2-75::0-5:P,

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STRAIN IN PROPORTION TO INCLINATION.— ——~61 алаған BRA DN 7 at а sk THE Ё Ry i ы and i dat hg томын ,--4 0-55 UNIVERSITY C „ OF “б amt, тү SLIFORYY 0-55)1-375(2-5 = Р, I IO 275 275 The strain upon the timber 4 is, therefore, equal to two and a half tons. Again, the line e 4 measures four tenths of an inch (0-4 inch); therefore, the proportion ba:W::ed:@Q becomes O-55:2+75::0-4:Q, 2-75 X 0-4 _ 0-55 " 2-75 O-4 0:55) 1- I00 (2 = Q. I IO 3 and The strain upon the timber 2 is, therefore, equal to two tons. 75.—The Strains Exceed the Weights.—Thus the united pressures upon the two inclined supports aiways exceed the weight. In the last case, 2% tons exert a pressure of 2% and 2 tons, equal together to 44 tons; and in the former case, 1200 pounds exert a pressure of twice 720 pounds, equal to 1440 pounds. The smaller the angle of inclination to the horizontal, the greater will be the pressure upon the sup- | ports. So, in the frame of a roof, the strain upon the rafters decreases gradually with the increase of the angle of incli- nation to the horizon, the length of the rafter remaining the same.

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62 CONSTRUCTION. This is true in comparing one system of framing with another; but in a system where the concentrated weight to be supported is not in the middle (see Fig. 23), and, in con- sequence, the supports are not inclined equally, the strain will be greatest upon that support which has the greatest inclination to the horizon. | 76.—Minimum Thrust of Rafters.—Ordinarily, as in roofs, the load is not concentrated, it being that of the fram- ing itself. Here the amount of the load will be in proportion to the length of the rafter, and this will increase with the increase of the angle of inclination, the span remaining the same. Soit isseen that in enlarging the angle of th tion to the horizon in order to lessen the oblique thrust, the load is increased in consequence of the elongation of the refter, thus increasing the oblique thrust. Hence there is an economical angle of inclination. А rafter-will have the least oblique thrust when its angle of inclination to the horizon is 35? 16” nearly. This angle is attained very nearly when the rafter rises 84 inches per foot, or when the height В C (Fig. 32) is to the base A C as 84 is to 12, or as o- 7071 is to 1-0. 77.—Practical Method of Determining Strains.— A com- parison of pressures in timbers, according to their position, may be readily made by drawing various designs of framing and estimating the several strains in accordance with the parallelogram of forces, always drawing the triangle 2 Ze so that the three lines shall be parallel with the three forces or pressures respectively. Тһе /глр// of the lines forming this triangle is unimportant, but it will be found more con- venient if the line drawn parallel with the known force is made to contain as many inches as the known force contains pounds, or as many tenths of an inch as pounds, or as many inches as 1015, or tenths of an inch as tons; or, in general, as many divisions of any convenient scale as there are units of weight or pressure in the known force. 1f drawn in this manner, then the number of divisions of the same scale found in the other two lines of the triangle will equal the units of pressure or weight of the other two forces respect- `

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HORIZONTAL THRUST OF RAFTERS. 63 ively, and the pressures sought will be ascertained simply by applying the scale to the lines of the triangle. For example, іп 276. 23, the vertical line 4 а, of the tri- angle, measures fifty-five hundredths of an inch (0-55 inch); the line 2 e, fifty hundredths (0-50 inch); and the line ға, forty (o-40 inch). Now, if it be supposed that the vertical pressure, or the weight suspended below 2 4, is equal to 55 pounds, then the pressure on A will equal 50 pounds, and that on Z will equal 40 pounds; for, by the proportion above stated, Od: ед ех Р, 55:55::50:50; апа so of the other pressure. If а scale cannot be had of equal proportions with the forces, the arithmetical process will be shortened somewhat by making the line of the triangle that represents the known weight equal to unity of a decimally divided scale, then the other lines will be measured in tenths or hundredths; and in the numerical statement of the proportions between the lines and forces, the first term being unity, the fourth term will be ascertained simply by multiplying the second and third terms together. For example, if the three lines are r, o-7, and 1-3, and the known weight is 6 tons, then bd: W::be: P becomes Wn 6 {% Oz ri Pom equals four and two tenths tons. Again— bd: W::ed: Q becomes Por; ul MF = /-0) equals seven and eight tenths tons. 78.—Horizontal Thrust. T Fig. 24, the weight W presses the struts in the direction of their length; their feet, 7 7, therefore, tend to move in the direction ло, and would so move were they not opposed by a sufficient resistance from the blocks, 4 and A. If a piece of each block be cut off at

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64 CONSTRUCTION. the horizontal line, ал, the feet of the struts would slide away from each other along that line, in the direction za; but if, instead of these, two pieces were cut off at the verti- cal line, 70, then the struts would descend vertically. To estimate the horizontal and the vertical pressures exerted by the struts, let 20 be made equal (upon any scale of equal parts) to the number of tons with which the strut is pressed ; Fic. 24. construct the parallelogram of forces by drawing oe parallel to ал, and of parallel to бл; then xf (by the same scale) shows the number of tons pressure that is exerted by the strut in the direction л a, and л e shows the amount exerted in the direction z 4. Ву constructing designs similar to this, giving various and dissimilar positions to the struts, and then estimating the pressures, it will be found in every case that the horizontal pressure of one strut is exactly equal to that of the other, however much one strut may be inclined more than the other; and also, that the united vertical pressure of the two struts is exactly equal to the weight W. (In this calculation the weight of the timbers has not been taken into consideration, simply to avoid complication to the learner. In practice it 1s requisite to include the weight of the framing with the load upon the framing.) Suppose that the twó struts, P and B (Fig. 24), were rafters of a roof, and that instead of the blocks, А and A, the walls of a building were the supports: then, to prevent

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TIES DESIRABLE IN ROOFS. 65 the walls from being thrown over by the thrust of B and B, it would be desirable to remove the horizontal pressure. 1115 may be done by uniting the feet of the rafters with a Fic. 25. rope, iron rod, or piece of timber, as in Fig. 25. This figure is similar to the truss of a roof. The horizontal strains on the tie-beam,-tending to pull it asunder in the direction of its length, may be measured at the foot of the rafter, as was shown at Fig. 24; but.it can be more readily and as accu- rately measured by drawing from fand e horizontal lines to the vertical line, dd, meeting it in o and о; then fo will be the horizontal thrust at 2, and eo at А; these will be found to equal one another. When the rafters of a roof are thus connected, all tendency to thrust out the walls horizontally is removed, the only pressure on them is in a vertical direc- tion, being equal to the weight of the roof and whatever it has to support. This pressure is beneficial rather than otherwise, as a roof having trusses thus formed, and the trusses well braced to each other, tends to steady the walls. 79.—Position of Supports.—/igs. 26 and 27 exhibit two methods of supporting the equal weights, Wand W. Let it be required to measure and compare the strains produced on the pieces, 4 Band AC. Construct the parallelogram of forces, e? fd, according to Art. 71. Then 2/ will show the

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665 ` CONSTRUCTION. strain on A В, and бе the strain on A С. By comparing the figures, ^ d being equal in each, it will be seen that the strains in б. 26 are about three times as great as those in Fic, 26. Fic. 27. Fig. 27; the position of the pieces, A В and А C, in Fig. 27, is therefore far preferable. ТІ Wh, " Fic. 28. 80.— The Composition of Forces: consists in ascertain- ing the direction and amount of оле force which shall be just capable of balancing two or more given forces, acting in different directions. This is only the reverse of the resolu-

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STRAINS INVOLVED IN CRANE. 67 tion of forces; and the two are founded on one and the . same principle, and may be solved in the same manner. For . example, let A and В (fig. 28) be two pieces of timber, pressed in the direction of their length towards 2—4 bv a force equal to 6 tons weight, and B о tons. То find the . direction. and amount of pressure they would unitedly exert, draw the lines ġe and 4/ in a line with the axes of the timbers, and make 22 equal to the pressure exerted by В, viz., 9; also make д f equal to the pressure on А, viz., 6, and complete the parallelogram of forces ebfd, then ба, the diagonal of the parallelogram, will be the direction, and its length, 9-25, will be the amount, of the united pressures of A and of P. The line 2415 termed the resultant of the two forces bf and фе. If А and P are to be supported by one post, C, the best position for that post will be in the direc- tion of the diagonal 4d; and it will require to be sufficiently strong to support the united pressures of 4 and of B, which are equal to 9-25 or 91 tons. UE Еіс. 29. 8{.—Another Example.—Let Fig. 29 represent a piece of f.aming commonly called a crane, which is used for hoist- ing heavy weights by means of the rope, B 4/, which passes overapulley ағд. This, though similar to Zzes. 26 and 27, is, however, still materially different. In those figures, the strain is in one direction only, viz., from 4 to 4; but in this there are two strains, from A to B and from A to W. The strain in the direction А В is evidently equal to that in thé

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68 CONSTRUCTION. direction А IW. То ascertain the best position for the strut A C, make бе equal to f, and complete the parallelogram of forces e? fd; then draw the diagonal 24, and it will be the position required. Should the foot, C, of the strut be placed either higher or lower, the strain on AC would be in- creased. Іп constructing cranes, it is advisable, in order that the piece B А may be under a gentle pressure, to place the foot of the.strut a trifle lower than where the diagonal ба would indicate, but never higher. a a FIG. 30. 82.—Ties and Struts.—Timbers in a state of tension are called ¢ies, while such as are in a state of compression are termed struts. This subject can be illustrated in the follow- ing manner : | Let A and B (Fig. 30) represent beams of timber sup- porting the weights W, W, and И; A having but one sup- port, which is in the middle of its length, and 2 two, one at each end. To show the nature of the strains, let each beam Бе sawed in the middle from ato 4. The effects are obvious: the cut in the beam A will open, whereas that in В will close. If the weights are heavy enough, the beam 4 will break at 6; while the cut in Б will be closed perfectly tight at a, and the beam be very little injured by it. But if, on the other hand, the cuts be made in the bottom edge of the timbers, from с to 4, B will be seriously injured, while A will scarcely be affected. . Dy this it appears evident that, in a piece of timber subject to a pressure across the direction of its length, the fibres are exposed to contrary strains. If the timber 15 supported at both ends, as at В, those from the top edge down to the middle are compressed in the direction

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TIES AND STRUTS. 69 of their length, while those from the middle to the bottom edge are in'a state of tension ; but if the beam is supported as at A, the contrary effect is produced; while the fibres at. the middle of either beam are not at all strained. The strains in a framed truss are of the same nature as those in a single beam. The truss for a roof, being supported at each end, has its tie-beam in a state of tension, while its rafters are compressed in the direction of their length. By this, it appears highly important that pieces in a state of tension should be distinguished from such as are compressed, in order that the former may be preserved continuous. A strut may be constructed. of two or more pieces ; yet, where there are many joints, it will not resist compression so well. 83.—To Distinguish Ties from Struts.—This may be done by the following rule. In Fig. 22-2, the timbers C and D are the sustaining forces, and the weight W 15 the %ғай іле force; and if the support be removed, the straining force would move from the point of support 4 towards а. Let it be required to ascertain whether the sustaining forces are stretched or pressed by the straining force. Rule: Upon the direction of the straining force 0 4, as a diagonal, construct a parallelogram e? fd whose sides shall be parallel with the direction of the sustaining forces C and D; through the point 0 draw a line parallel to the diagonal ¢/; this may then be called the dividing line between ties and struts. Because all those supports which are on that side of the dividing line which the straining force would occupy if unresisted are compressed, while those on the other side of the dividing line are stretched. In Fig. 22-В, the supports are both compressed, being оп that side of the dividing line which the straining force would occupy if unresisted. Іп 265. 26 and 27, in which А B and А С are the sustaining forces, А C is compressed, whereas А B is in a state of tension; 4 C being on that side of the line 22 which the straining force would occupy if unresisted, and A В on the opposite side. The place of the latter might be supplied by a chain or rope. In Zzg. 25, the foot of the ‘rafter at A is sustained by two forces, the wall and the tie-

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70 CONSTRUCTION. beam, one perpendicular and the other horizontal: the direction of the straining force is indicated by the line ѓа. The dividing line 22, ascertained by the rule, shows that the wall is pressed and the tie-beam stretched. 84.—Another Examp!c.—Let £ A B F (Fig. 31) represent a gate, supported by hinges at Æ and Æ. In this case, the straining force is the weight of the materials, and the direc- tion of course vertical. Ascertain the dividing line at the several points, G, В, J, Y, H, and F. It will then appear that the force at С is sustained by А С and СЕ, and the dividing line shows that the former is stretched and the latter com- pressed. The force at H is supported by A H and H E—the former stretched and the latter compressed. The force at 5 is opposed by А B and A В, ове pressed, the other stretched. The force at F is sustained by G F and FE, GF being stretched and FE pressed. Ву this it appears that A B is in a state of tension, and Æ F of compression; also, that 4 М and С F are stretched, while B H and СЕ are compressed : which shows the necessity of having А H and С F each in one whole length, while 2 H and G E may be, as they are shown, each in two pieces. The force at 715 sustained by G F апа 9 H, the former stretched and the latter compressed. The piece C D is neither stretched nor pressed, and could be dispensed with if the joinings at Y and / could be made.

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TO FIND THE CENTRE OF GRAVITY. | JI as effectually without it. In case 4 B should fail, then C D would be in a state of tension. 85.—Centre of Gravity.—The centre of gravity of a uni- form prism or cylinder is in its axis, at the middle of its length; that of a triangle is in a line drawn from one angle to Ke 524 of the opposite side, and at one third of the length of the line from that side; that of a right-angled tri- angle, at a point distant from the [NE ats equal to one third of the base, and distant from the base equal to one third of the perpendicular; that of а pyramid or cone, in the axis and at one quarter of the height from the base. The centre of gravity of a trapezoid (a four-sided.figure having only two of its sides parallel) is in a line joining the centres of the two parallel sides, and at a distance from the longest of the parallel sides equal to the product of the length in the sum of twice the shorter added to the longer of the parallel sides, divided by three times the sum of the two parallel sides. Algebraically thus: _1(2а+0) 3 (ab) where 4 equals the distance from the longest of the parallel sides, / the length of the line joining the two parallel sides, and a the shorter and 4 the longer of the parallel sides. Example.—A rafter 25 feet long has the larger end 14 inches wide, and the smaller end 10 inches wide: how far from the larger end is the centre of gravity located ? Неге / = 25, а = 1$, and = 1$, 22а-д 25(2x1$ +42) 25х1% _ 25 X 34 _ 3(«4-5 3+1) 3x12 hence Z — = = 11-8 = гї feet 95 inches nearly. In irregular bodies with plain sides, the centre of gravity may be found by balancing them upon the edge of a prism —upon the edge of a table—in two positions, making a line each time upon the body in a line with the edge of the prism, and the intersection of those lines will indicate the point re-

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72 CONSTRUCTION. quired. Or suspend the article by a cord or thread attached to one corner or edge ; also from the same point of suspen- sion hang a plumb-line, and mark ifs position on the face of the article; again, suspend the article from another corner or side (nearly at right angles to its former position), and mark the position of the plumb-line upon its face; then the intersection of the two lines will be the centre of gravity. Fic. 32. 86.— Effect of the Weight of Inclined Beams.—An in- clined post or strut supporting some heavy pressure applied at its upper end, as at 270. 25, exerts a pressure at its foot in the direction of its length, or nearly so. But when such a beam is loaded uniformly over its whole length, as the rafter of a roof, the pressure at its foot varies considerably from the direction of its length. For example, let 4 B (Fig. 32) be a beam leaning against the wall 2 с, and supported at its foot by the abutment А, in the beam А с, and let о be the centre of gravity of the beam. Through о draw the verti- cal line 6 4, and from В draw the horizontal line B 4, cutting bdin 6; join 6 and A, and 2 A will be the direction of the thrust. To prevent the beam from loosing its footing, the joint at A should be made at right angles to 6A. The amount of pressure will be found thus: Let 2 d (by any scale of equal parts) equal the number of tons upon the beam А В; draw d e parallel to В 6; then 2 «(Ьу the same scale) equals the pressure in the direction д А; and e 4 the pres- sure against the wall at 2--апа also the horizontal thrust at A, as these are always equal in a construction of this kind. The horizontal thrust of an inclined beam (Fig. 32)—the effect of its own weight—may be calculated thus: Rule.—Multiply the weight of the beam in pounds by.

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THRUST OF INCLINED BEAMS. 73 its base, AC, in feet, and by the distance in feet of its centre of gravity, o (see Art. 85), from the lower end, at A, and divide this product by the product of the length, A B, into the height, 7 С, and the quotient will be the horizontal d b «v | p 7 where 4 equals the distance of the centre of gravity, o, from the lower епа; б equals the base, A С; w equals the weight of the beam; % equals the height, B C; 7 equals the length of the beam ; and Z equals the horizontal thrust. Example.—A. beam 20 feet long weighs 300 pounds; its centre of gravity 15 at 9 feet from its lower end ; it is so in- clined that its base is 16 feet and its height 12 feet: what is the horizontal thrust? thrust in pounds. This may be stated thus: H = Here 24% becomes 2.5 9: 10. 890 ae vide A Al I2 x 20 5 = 180 = Н = the horizontal thrust. (m 9 х 4X5 This rule is for cases where the centre of gravity does not occur at the middle of the length of the beam, although it is applicable when it does occur at the middle; yet a shorter rule will suffice in this case, and it is thus: Rulce.—Multiply the weight of the rafter in pounds by the base, 4 C (Fig. 32), in feet, and divide the product by twice the height, Б C, in fect, and the quotient will be the horizontal thrust, when the centre of gravity occurs at the middle of the beam. If the inclined beam is loaded with an equally Ары load, add this load to the weight of the beam, and use this total weight in the rule instead of the weight ‘of the beam. And generally, if the centre of gravity of the combined weights of the beam and load does not occur at the centre of the length of the beam, then the former rule is to be used. In Zzg. 33, two equal beams are supported at their feet by the abutments in the tie-beam. This case is similar to the last: for it is obvious that each beam is in precisely the position of the beam іп 270. 32. The horizontal pressures at B, being equal and opposite, balance one another; and their horizontal thrusts at the tie-beam are also equal. (See 2177.

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74. CONSTRUCTION. 78—Fig. 25.) When the height of a roof (Fig. 33) is one fourth of the span, or of a shed (Fig. 32) is one halt the span, the horizontal thrust of a rafter, whose centre of grav- Fic. 33. ity is at the middle of its length, is exactly equal to the weight distributed uniformly over its surface. In shed or /ал-20 roofs, as Fig. 32, the horizontal pressure will be entirely removed if the bearings of the rafters, as 4 and B (Fig. 34), are made horizontal—provided, however, B A |. that the rafters and other framing do not bend between the points of support. Ifa beam or rafter have a natural curve, the convex or rounding edge should be laid uppermost. Pic. 34. 87.—Effect of Load on Beam.—The strain ina uniformly loaded beam, supported at each end, is greatest at the middle of its length. Hence mortices, large knots, and other defects should be kept as far as possible from that point; and in resting a load upona beam, asa partition upon a floor-beam, the weight should be so adjusted, if possible, that it will bear at or near the ends. Twice the weight that will break a beam, acting at the centre of its length, is required to break 1t when equally.

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VARYING PRESSURE ON BEARINGS. 75° distributed over its length; and precisely the same deflec- tion or sag will be produced on a beam by a load equally distributed that five eighths of the load will produce if act- ing at the centre of its length. 88.— Effect on Bearings.—When а uniformly loaded beam is supported at each end on level bearings (the beam itself being either horizontal or inclined), the amount of pressure caused by the load on each point of support is equal to one half the load ; and this is also the ase when the load is concentrated at the middle of the beam, or has its centre of gravity at the middle of the beam; but when the load is unequally distributed, or concentrated so that its centre of gravity occurs at some other point than the middle of the beam, then the amount of pressure caused by the load on one of the points of support is unequal to that on the other. The precise amount on each may be ascertained by the following rule. Rule.— Multiply the weight W (Fig. 35) by its distance, C D, from its nearest point of support, 5, and divide the рго-, duct by the length, 4 В, of the beam, and the quotient will FIG 35. be the amount of pressure on the remote point of support, 4. Again, deduct this amount from the weight W, and the re- mainder will be the amount of pressure on the near point of support, B; or, multiply the weight IV by its distance, А С, from the remote point of support, A, and divide the pro- duct by the length, A B, and the quotient will be the amount of pressure on the zear point of support, 2. When / equals the length between the bearings 4 and 4, n= АС, m= C В, апа W = the load; then

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"ERE ea CONSTRUCTION, - | W mm mM the amount of pressure at А, and E = B = the amount of pressure at В. Example.—A beam 20 feet long between the bearings has a load of 100 pounds concentrated at 3 feet from one of the bearings: what is the portion of this weight sustained by each bearing ? Here W = тоо; #, 17; m, 3; and 2, 20. W m тоох Hence А = 2 = 199*3 = 15. Load on A = 15 pounds. Load on B = 85 pounds. Total weight = 100 pounds. RESISTANCE OF MATERIALS. 89.— Weight—Strength.— Preliminary to designing a roof- truss or other piece of framing, a knowledge of two subjects is essential: one is, the effect of gravity acting upon the various parts of the intended structure ; the other, the power of resistance possessed by the materials of which the framing is to be constructed. Тһе former subject having been treated of in the preceding pages, it remains now to call at- tention to the latter. 90.— Quality of Materials, —Materials used іп construc- tion are constituted in their structure either of fibres (threads) or of grains, and are termed, the former fibrous, the latter granular. АП woods and wrought metals аге fibrous, while cast iron, stone, glass, etc., are granular. The strength of a granular material lies in the power of attrac- tion acting among the grains of matter of which the mate- rial is composed, by which it resists any attempt to separate its grains or particles of matter. А fibre of wood or ot:

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THE THREE KINDS OF STRAINS: C4 L IF A wrought metal has a strength by which it resists being com- pressed or shortened, and finally crushed; also a strength by which it resists being extended or made longer, and finally sundered. There is another kind of strength in a fibrous material: it is the adhesion of one fibre to another along their sides, or the lateral adhesion of the fibres. 9|, —Manner of Resisting.—In the strain applied to a post supporting a weight imposed upon it (272 36), we have an instance of an essay to shorten the fibres of which the timber is composed. Тһе strength of the timber in this case is termed the resistance to compression. Іп the strain ona piece of timber like a king-post or suspending piece (A, Fig. 37), we have an instance of an essay to extend or lengthen the fibres of the material. The strength here ex- hibited is termed the resistance to tension. When a piece of timber is strained like a floor-beam or any horizontal piece Fic. 38. carrying a load (Fig. 38). we have an instance in which the two strains of compression and tension are both brought into action ; the fibres of the upper portion of the beam be- ing compressed, and those of the under part being stretched.

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78 CONSTRUCTION. This kind of strength of timber is termed resistance to cross- strains. In each of these three kinds of strain to which tim- ber is subjected, the power of resistance is in a measure due to the /ateral adhesion of the fibres, not so much perhaps in the simple tensile strain, yet to a considerable degree in the compressive and cross strains. But.the power of timber, by which it resists.a pressure acting compressively in the direction of the length of the fibres, tending to separate the timber by splitting off a part, as in the case of the end of a tie-beam, against which the foot of the rafter presses, is wholly due to the lateral adhesion of the fibres. 92.— strength and Stiffness.—The strength of materials is their power to resist fracture, while the s¢zffness of mate- rials is their capability to resist deflection or sagging. A knowledge of their strength 1s useful, in order to determine their limits of size to sustain given weights safely; but a knowledge of their s¢zffness is more important, as in almost all constructions it is desirable not only that the load be safely sustained, but that no appearance of weakness be manifested by any sensible deflection or sagging. 93.— Experiments: Constants.— In the investigation of the laws applicable to the resistance of materials, it is found that the dimensions—length, breadth, and thickness—bear certain relations to the weight or pressure to which the piece is subjected. These relations are general; they exist quite independently of the peculiarities of any specific piece of material. These proportions between the dimensions and the load are found to exist alike in wood, metal, stone, and glass, or other material. One law applies alike to all materials ; but the capability of materials to resist differs in accordance with the compactness and cohesion of particles, and the tenacity and adhesion of fibres, those qualities upon which depends the superiority of one kind of material over another. The capability of each particular kind of material is ascertained by experiments, made upon several specimens, and an average of the results thus obtained is taken as an index of the capability of that material, and is introduced in. the rules as a constant number, each specific kind of ma-

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VALUES OF WOODS FOR COMPRESSION. 79 terial having its own special constant, obtained by ех- perimenting on specimens of that peculiar material. The results of experiments made to test the resistance of various materials useful іп construction—their capability to resist the three strains before named—will now be introduced. 94.— Resistance to Compression.—The following table exhibits the results of experiments made to test the resist- ance to compression of such woods as are in common use in this country for the purposes of construction. TABLE L—RESISTANCE TO COMPRESSION. е Ca H. à &, Z: z е. E 4% %% = б > а CE же E 55 | às | p3 | ЁЁ MATERIAL. ? 5 3 Е Б a = x Е E cd pp | 45 4 a S 9.5 RU veg an e E 5% 2443 B > Pounds Pounds Pounds | per inch. | per inch. | per inch. Сео аР ше ......ж...... 7. | Ө-біЗ 9500 840 2250 900 ЫС... TOME, К... : 0:762 11700 1160 2800 1120 Bosna... 2. “жас. aed. +» 0-774 8000 1250 2650 1060 Spruce k "m m * "EEME 0-369 7850 540 650 260 оне, аа... 0:388 6650 480 800 320 Hemlocke raa. aa —— зы» 0:423 5700 370 800 320 Vm УУ ООЧ......... Җ...... 0:397 3400 т. 800 320 Сазы. i cu. PALO Lee O-49I 6700 b. on 1250 500 P а. қа acrem ж-- 0-517 5850 wm 3100 I240 ЕГЕТ а... 0-574 8450 и 2700 1080 Hiokory . wire sss . macies. YT 05823» | 13750 |... .4100 1640 а T op Ж. +» 0:494 9050 .... 2500 IOOO Black? Walnut ............. O-421I 7800 „Ж. 2100 840 Mahogany (St. Domingo)..... | 0:837 | 11600 ж, 5700 2280 ы (Bay Wood) ....... 0-439 4900 ies 1700 680 мысты dx ELA SERRA „| 0:916 | 11100 5: 6800 2720 ое Е op 7T CTELIES | 1:282 | 12100 4254 7700 3080 Тһе resistance of timber of the same name varies much ; depending as it obviously must on the soil in which it grew on its age before and after cutting, on the time of year when cut, and on the manner in which it has been kept since it was cut. And of wood from the same tree much depends upon its location, whether at the butt or towards the limbs, and whether at the heart or of the sap, or at a point mid- way from the centre to the circumference of the tree. The

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80 CONSTRUCTION. pieces submitted to experiment were of ordinary good quality, such as would be deemed proper to be used in framing. The prisms crushed were generally small, about 2 inches long, and from 1 inch to 14 inches square; some were wider one way than the other, but all containing in area of cross section from 1 to 2 inches. The weight given in the table is the average weight per superficial inch. Of the first six woods named, there were nine specimens of each tested ; of the others, generally three specimens. The results for the first six woods named are taken from the author's work on Transverse Strains, published by John Wiley & Sons, New York. The results for these six woods, as well as those for all the others named in the table, were obtained by experiments carefully made by the author. The first six woods named were tested in 1874 and 1876, and upon a testing machine, in which the power is transmitted to the pieces tested, by levers acting upon knife-edges. For a description of this machine, see Transverse Strats, Art. 704. The woods named in the table, other than the first six, were tested some twenty years since, and upon a hydraulic press, which, owing to friction, gave results too low. . The results, as thus ascertained, were given to the public in the 7th edition of this work, in 1857. Іп the present edi- tion, the figures in Table I., for these woods, are those which have resulted by adding to the results given by the hydraulic press a certain quantity thought to be requisite to compensate for the loss by friction. Thus corrected, the figures in the table may be taken as sufficiently near approx- imations for use in the rules, —although not so trustworthy as the results given for the first six woods named, as these were obtained upon a superior testing machine, as Күрке stated. In the preceding table, the second column contains the specific gravity of the several kinds of wood, showing their comparative density. The weight in pounds of a cubic foot of any kind of wood or other material is equal to its specific gravity multiplied by 62.5, this number being the weight in pounds of a cubic foot of water. The third column

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EXPLANATION OF TABLE I. 8r contains the weight in pounds required to crush a prism having a base of one inch square; the pressure applied to the fibres longitudinally. In practice, it is usual never to load material exposed to compression with more than one fourth of the crushing weight, and generally with from one sixth to one tenth only. The fourth column contains the weight in pounds which, applied in line with the length of the fibres, is required to force off a part of the piece, causing the fibres to separate by sliding, the surface separated being one inch square. Thefifth column contains the weight in pounds required to crush the piece when the pressure is applied to the fibres transverselv, the piece being one inch thick, and the surface crushed being one inch square, and depressed one twentieth of an inch deep. The sixth column contains the value of P in the rules; P being the weight in pounds, ap- plied to the fibres transversely, which is required to make a sensible impression one inch square on the side of the piece, this being the greatest weight that would be proper fora post to be loaded with per inch surface of bearing, resting on the side of the kind of wood set opposite in the table. :А greater weight would, in proportion to the excess, crush the side of the wood under the post, and proportionably derange the framing, if not cause a total failure. lt will be observed that the measure of this resistance is useful in limiting the load on a post according to the kind of material contained, not in the post, but in the ZZmber upon which the post presses. 95. Resistance to Tension.—The resistance of materials to the force of stretching, as exemplified in the case of a rope from which a weight is suspended, is termed the resist- ance to tension. In fibrous materials, this force will be differ- ent in the same specimen, in accordance with the direction ‚ in which the force acts, whether in the direction of the length of the fibres orat right angles to the direction of their length. It has béen found that, in hard woods, the resistance in the former direction is about eight to ten times what it is in the latter; and in soft woods, straight grained, such as white pine, the resistance is from sixteen to twenty times. A knowledge of the resistance in the direction of the fibres 1s the most useful in practice.

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82 И CONSTRUCTION. In the following table are recorded the results of ex- periments made to test this resistance in some of the woods in common use, and also in iron, cast and wrought. Each specimen of the woods was turned cylindrical, апа about 2 inches diameter, and then the middle part reduced to about $ of an inch diameter, at the middle of the reduced part, and thence gradually increased toward each end, where it was considerably larger at its junction with the enlarged епа.. The results, in the case of the iron and of the first six woods named, are taken from the author's work, Trans- verse Strains, Table XX. Experiments were made upon the other three woods named by a hydraulic press, some twenty years since, and the results were first published in the 7th edition of this work, in 1857. These results, owing to friction, were too low. Adding to them what is supposed to be the loss by friction of the machine, the results thus corrected are what are given for these three woods in the following table, and may be taken as fair approximations, but are not so trustworthy as the figures given for the other six woods and for the metals. TABLE II.—RESISTANCE TO TENSION. 1 Specific Pounds re- MATERIAL. Gravity. quired to rup- ture one inch square. GscRa те... -................... — Дым 7 0:65 160co иф. Лб ИШ ОККО Ier ЕИ NEM 0:794 248co p 210 Acc TET S 0-774 19500 EMEN. uc mus cue Уы 0:432 19500 Т Ы ТҰН "б, 0-458 120С0 Memlock .......... фе «жар Дахон: E 0:402 8700 BENI. a. ose ore REDI e «aevo О.751 2босо КЕНЕ... Т. РОТ. жи шү» 0:694 20000 Di лм TOT RANMI TE ee Bis dado M a Г 0-608 15009 Gast Iron, American ) ..................... from 6-944 27000 jj BERT |... ee СГУ to 7:584 ' 17000 Wrought Iron, American | T TOR V T from 7:600 60000 ы ағы МЫ кт to 7:792 50000 The figures in the table denote the «mae capability of a bar one inch square, or the weight in pounds required to

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VALUES OF MATERIALS FOR CROSS-STRAINS. 83 produce rupture. Just what portion of this should be taken as the safe capability will depend upon the nature of the strain to which the material is to-be exposed. In practice it is found that, through defects in workmanship, the attach- ments zzay be so made as to cause the strain to act along oze side of the piece, instead of through its axis; and that in this case fracture will be produced with one third of the strain that can be sustained through the axis. Due to this and other contingencies, it is usual to subject materials exposed to tensile strain with only from one sixth to one tenth of the breaking weight. 96. In the follow- ing table are recorded the results of experiments made to test the capability of the various materials named to resist the effects of transverse strain. The figures are taken from - the author's work, Transverse Strains, before referred to. TABLE III.—TRANSVERSE STRAINS. B. F. е. а. MATERIAL. комната Mensis ' s^ ai "үм or rm M F авы в Fibres. Safety. Georgia Pine. ә...ыә.......... 850 м 5900 ma “00109. 1-84 Шоешс ©. oA oc." TELE 1200 5050 “0015 2-20 White мит EVA Lus... 650 | 3100 | “00086 3:39 SES . eese. dre GTC YT 550 3500 *00098 2-23 MENe Pile ^. Teller 500 2900 -OO0I4 I:7I ТОСЕ ....9.......... Mes 450 2800 *00095 2:35 White Wood -e.e aa: s а... 5 : боо 3450 -00096 2:52 Vh... 480 2550 “00103 2:54 Dh eee oom isl... 900 4000 *OOIII 2.82 | touc ST o ea ER IIOO 5150 -OOI4 2-12 пек mE... TOSO - 2850 “0013 2:01” Ghgfry«. Бани ewe owe cae 650 2850 -001563 | 2-03 Black Walnut a. y -e nnn 750 3900 *OO104 2°57 Mahogany (St. Domingo)...... 650 3600 .00116 2:16 a (Bay Wieod)......... 850 4750 ` 00109 . 2:28 Cast Iron, American cos ee 2500 . 50000 "n Bach 20... 2100 40000 Wrought Iron, American ...... . . 2600 62000 á English 22...... 1900 60000 Steelin BIS A. 6...2... 6000 70000 t Blue Stone Flagging ......... 200 Sahd тен 96. eee 59 Brick, епок .5:%........ 33 с | "P 37 Marble, East Chester ......... I47 b... — Mapa ЛН “|. oS —— ———

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84 CONSTRUCTION. “Тһе figures іп the second column, headed д, denote the weight in pounds required to break a «zz? of the material named when suspended from the middle, the piece being supported at each end. Тһе uz? of materialis a bar one inch square and one foot long between the bearings. The third column, headed /, contains the values of the several mate- rials named as to their resistance to flexure, as explained in Arts. 302-305, Transverse Strains. These values of F, as constants, are used іп the rules. The fourth column, headed е, contains the values of the several materials named, denot- ing the elasticity of the fibres, as explained in Art. 312, Transverse Strains. These values of e, as constants, are used in the rules. The fifth column, headed а, contains for the T ma- terials named the ratio of the resistance to flexure as com- pared with that to rupture, and which, as constants used in the rules, indicate the margin of safety to be given for each kind of material. The figures given in each case show the smallest possible value that may be safely given to a, the factor of safety. . Іп practice it is generally taken higher than the amount given in the table. For example, in the table the value of В, the constant for rupture by transverse strain for spruce, is 550. Now, if the dimensions of a spruce beam to carry a given weight be computed by the rules, using the constant 2, at 550, the beam will be of such a size that the given weight will just break it. But if, in the computation, instead of taking the full value of В, only a part of it be taken, then the beam will not break immediately; and if the part taken be so small that its effect upon the fibres shall not be sufficient to strain them beyond their limit of elasticity, the beam will be capable of sustaining the weight for an indefinite period; in this case the beam will be loaded by what is termed the safe weight. Or, since the value of a for spruce is 2-23 in the table, if, in- stead of taking B at 550, its full value, only the quotient arising from a division of B by a be taken—or 550 divided by 2-23, which equals 246-6—then the beam will be of suffi- ' cient size to carry the load safely. Therefore, while the con-

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THE VARIOUS CLASSES OF PRESSURES. 85. stant B is to be ащ for a breaking weight, for а safe load the quotient of — 15 to be used. But, again, if a be taken at its value as Б. in the table, the computed beam will be loaded up to its limit of safety. So loaded that, if the load be increased only in a small degree, the iimit of safety will be passed, and the beam liable, in time, to fail by rupture. Therefore, as the values of a, in the table, are the smallest possible, it is prudent in practice always to take a larger than the table value. For example, the table value of & for spruce is 2-23, but in practice let it be taken at 3 Or 4. 97.—Resistance to Compression.—T he resistance of ma- terials to the force of compression may be considered in several ways. Posts having their heights less than ten times their least sides will crush before bending; these belong to one class: another class is that which com- prises all posts the height of which is equal to ten times their least sides, or more than ten times; these will bend before crushing. Then there remains to be considered the manner in which the pressure is applied: whether in line with the fibres, or transversely to them; and, again, whether the pressure tends to crush the fibres, or simply to force off a part of the piece by splitting. The various pressures may be comprised in the four classes following, namely: Ist.—When the pressure is applied to the fibres trans- versely. 2d.— When the pressure is applied to the fibres longi- tudinally, and so as to split off the part pressed against, causing the fibres to separate by sliding. 3d.—When the pressure is applied to the fibres longi- tudinally, and on short pieces. 4th.— When the pressure is applied to the fibres longi- tudinally, and on long pieces. These four classes will now be considered in their reg- ular order.

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86 CONSTRUCTION. 98.— Compression Transversely to the Fibres.—In this first class of compression, experiment has shown that the resistance is in proportion to the number of fibres pressed, that is, in proportion to the area. For example, if 5000 pounds is required to crush a prism with a base 1 inch square, it will require 20,000 pounds to crush a prism having a base of 2 by 2 inches, equal to 4 inches area; because 4 times 5000 equals 20000. Therefore, if any given surface pressed be multiplied by the pressure per inch which the kind of material pressed may be safely trusted with, the product will be the total pressure which may with safety be put upon the given sur- face. Now, the capability for this kind of resistance is given in column P, in Table I., for each kind of material named in the table. Therefore, to find the limit of weight, proceed as follows: 99.—The Limit of Weight.—To ascertain what weight a post may be loaded with, so as not to crush the surface against which it presses, we have— Rule 1.—Multiply the area of the post in inches by the value of P, Table І. and the product will be the weight re- quired in pounds ; or— y SA: (f) Example.—A post, 8 by to inches, stands upon a white- pine girder; the area equals 8 x 10 — 80 inches. This being multiplied by 320, the value of P, Table I., set opposite white pine, the product, 25600, is the required weight in pounds. (00.—Aren of Post. — To ascertain what area a post must have in order to prevent the post, loaded with a given weight, from crushing the surface against which it presses, we have— | Rule TT.— Divide the given weight in pounds by the value of P, Table I., and the quotient will be the area required in ‘inches ; or—- > = H, (2).

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RESISTANCE. TO RUPTURE BY SLIDING. 87 - Example.—A post standing on а Georgia-pine girder is loaded with 100,000 pounds: what must be its area? Тһе weight, 100000, divided by goo, the value of Р, Table I., set opposite Georgia pine, the quotient, 111-11, is the required area in inches. Тһе post may be шо by 114, or to by 11 inches; or if square each side will be 10-54 inches, or 1139; inches diameter if round. (01. — &upture by Stiding.—In this the second class of rupture by compression, it has been ascertained by ex- periment that the resistance is in proportion to the area of the surface separated without regard to the form of the sur- face. Now, in Table I., column М, we have the ultimate resistance to this strain of the several materials named. But to obtain the safe load: per inch, the ultimate resist- ance of the table is to be divided by a factor of safety, of such value as circumstances may seem to require. Gener- ally this factor may be taken at 3. Then to obtain the safe load for any given case, we have but to multiply the given surface by the ultimate resistance, and divide by the factor of safety; therefore, proceed as follows:: . (02.— The Limit of Weight.—To ascertain what weight may be sustained safely by the resistance of a given area of surface, when the weight tends to split off the part pressed against by causing, in case of fracture, one surface to slide on the other, we have— | Rule III.—Multiply the area of the surface by the value of Z, in Table I. divide by the factor of safety, and the quotient will be the weight required in pounds; or— m 4H (3-) а Example.—The foot of a rafter is framed into the end of its tie-beam, so that the uncut substance of the tie-beam 1s 15 inches long from the' end of the tie-beam to the joint о! the rafter; the tie-beam is of white pine, and is 6 inches thick: what amount of horizontal thrust will this end of the tic-beain sustain, without danger of having the end of

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88 CONSTRUCTION. the tie-beam »split off?’ Here theearea of surface that-sus- : tains the pressure is 6 by 15 inches, equal to оо inches. This multiplied by 480, the value of H, set opposite to white pine, Table І., and divided by 3, as a factor of safety, gives a quotient of 14400, and this is the required weight in pounds. 103.—Area of Surface.— To ascertain the area of surface that 15 required to sustain a given weight safely, when the weight tends to split off the part pressed against, by causing, in case of fracture, one 'surface to slide on the other, we have— Rule IV.—Divide the given weight in pounds by the value of 77, Table I. ; multiply the quotient by the factor of safety, and the product will be the required area in inches ; or— = E (4-) Example.—The load on a rafter causes a horizontal thrust at its foot of 40,000 pounds, tending to split off the end of the tie-beam : what must be the length of the tie-beam be- yond the line where the foot of the rafter is framed into it, the tie-beam being of Georgia pine, and 9 inches thick? The weight, or horizontal thrust, 40000, divided by 840, the value of H, Table I., set opposite Georgia pine, gives a quo- tient of 47-619, and this multiplied by 3, as a factor of safety, gives a product of 142.857. This, the area of surface in inches, divided by о, the breadth of the surface strained (equal to the thickness of the tie-beam), the quotient, 15.87, 15 the length in inches from the end of the tie-beam to the rafter joint, say 16 inches. 104.—Tenons and Splices.—A knowledge of this kind of resistance of materials is useful, also, in ascertaining the length of framed tenons, so as to prevent the pin, or key, with which they are fastened from tearing out; and, also, in cases where tie-beams, or other timber under a tensile strain,

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CRUSHING STRENGTH OF STOUT POSTS. 80: аге spliced, this rule gives the length of the joggle at each end of the splice. 105.—Stout PPosts,—These comprise the third class of ob- jects subject to compression (A77. 97), and include all posts. which are less than ten diameters high. The resistance to compression, in this class, is ascertained to be directly in pro- portion to the area of cross-section of the post. Now in Table І., column C, the ultimate resistance to crushing is given for the several kinds of materials named ; from which the safe resistance per inch may be obtained by dividing it by a proper factor of safety. Having the safe resistance per inch, the resistance of any given post may be determined by multiplying it by the area of the cross-section of the post. Therefore, proceed as follows: (06.—Tnhe Limit of Weight.—To find the weight that can be safely sustained by a post, when the height of the post is less than ten times the diameter if round, or ten times the thickness if rectangular, and the direction of the pressure coinciding with the axis, we have— Rule V.—Multiply the area of the cross-section of the post in inches by the value of C,in Table I.; divide the pro- duct by the factor of safety, and the quotient will be the re- quired weight in pounds ; or— ш-4% = (5.) а Example—A Georgia-pine post is 6 feet high, and in cross-section 8 x 12 inches: what weight will it safely sus- tain? The height of this post, 12 x 6 — 72 inches, which is less than 10x 8 (the size of the narrowest side) = 80 inches ; it therefore belongs to the class coming under this rule. The area = 8 x 12 — 96 inches; this multiplied by 9500, the value of C,in the table, set opposite Georgia pine, and divided by б, as a factor of safety, the quotient, 152000, is the weight 1n pounds required. It will be observed that the weight would be the same for a Georgia-pine post of any height less than

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9o CONSTRUCTION. IO times 8 inches — 80 inches — 6 feet 8 inches, provided its breadth and thickness remain the same, 12 and 8 inches. (07.—Area of Post.—To find the area of the cross-sec- tion of a post to sustain a given weight safely, the height of the post being less than ten times the diameter if round, or ten times the least side if rectangular, the pressure coinciding with the axis, we have— Rule V1.—Divide the given weight in pounds by the value of C,in Table I.; multiply the quotient by the factor of safety, and the product will be the required area in inches ; or— |. Wa A= жәй (6.) Example—A weight of 40,000 pounds is to be sustained by a white-pine post 4 feet high: what must be its area of section to sustain the weight safely? Here, 40000 divided by 6650, the value of C, in Table I., set opposite white pine, and the quotient multiplied by 6, as a factor of safety, the pro- duct is 36; this, therefore, is the required area, and such a post may be 6x 66 inches. То find the least side, so that it shall not be less than one tenth of the height, divide the height, reduced to inches, by.ro, and make the least side to exceed this quotient. Тһе area divided by the least side so determined will give the wide side. If, however, by this process, the first side found should prove to be the greatest, then the size of the post is to be found by Rule IX., X., or XI. | (08.--Агеа of Round Posts.—In case the post is to be round, its diameter may be found by reference to the Table of Circles in the Appendix, in the column of diameters, op- posite to the area of the: post to be found in the column of areas, or opposite to the next nearest area. For example, suppose the required area, as just found by the example under Rule VIL, is 36: by reference to the column of areas, 35:78 is the nearest to 36, and the diameter set opposite is

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CRUSHING STRENGTH OF SLENDER POSTS. 01 6.75, which is а trifle too small. The post may therefore be, say, 6$ inches diameter. 109.—-Slender Posts.—When the height of a post is less than ten times its diameter, the resistance of the post to crushing is approximately in proportion to its area of cross- section. But when the height is equal to or more than ten diameters, the resistance per square inch is diminished. The resistance diminishes as the height is increased, the diameter remaining the same (Transverse Strains, Art. 643). Тһе strength of a slender post consists in a combination of the resistances of the material to bending and to crushing, and is represented in the following rule: 110.—The Limit of Weight, — То ascertain the weight that can be sustained safely by a post the height of which is at least ten times its least side if rectangular, or ten times its diameter if round, the direction of the pressure coincid- ing with the axis, we have— Rule V \1.—Divide the height of the post in inches by the diameter, or least side, in inches; multiply the quotient by itself, or take its square; multiply the square by the value of e, in Table III., set opposite the kind of .material of which the post is made; to the product add the half of itself; to the sum add unity (or one); multiply this sum by the factor of safety, and reserve the product for use, as below. Now multiply the area of cross-section of the post in inches by the value of C, in Table L, set opposite the material of the post, and divide the product by the above reserved product; the quotient will be the required weight in pounds; or— ACC. Ш а eB 7.) Example: A Round Post.—What weight may be safely placed upon а post'of Georgia pine 10 inches diameter and 10 feet high, the pressure coinciding with the axis of the post? The height of the post, (10 x 12 —) 120 inches, divided by 10, its diameter, gives a quotient of 12; this multiplied:

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92 CONSTRUCTION. by itself gives 144, its square; and this by -ooroo, the value of е for Georgia pine, in Table III., gives «15696; to which adding its half, the sum is 0-23544; to which adding unity, tlie sum is 1.23544; and this multiplied by 7, аз а factor of safety, the product is 8-648, the reserved divisor. Now the area of the post is (see Table of Areas of Circles, in the Ap- pendix, opposite its diameter, 10) 78-54; this multiplied by 9500, the value of C for Georgia pine, in Table I., gives a product of 746130; which divided by 8-648, the above re- served divisor, gives a quotient of 86278, the required weight in pounds. Another Example: A Rectangular Post.—W hat weight may be safely placed upon a white-pine post 10 x 12 inches, and 15 feet high, the pressure coinciding with the axis of the post? Proceeding according to the rule, we find the height of the post to be 180 inches, which divided by то, the least side of the post, gives 18; this multiplied by itself gives 324» its square; which multiplied by .o014, the value of e for white pine, in Table III., gives -4536; to which adding its half, the sum is -6804; to which adding unity, the sum is 1 -6804 ; and this multiplied by 8, asa factor of safety, the pro- duct is 13-4432, the reserved divisor. Now the area of the post, (10 x 12 —) 120 inches, multiplied by 6650, the value of C for white pine, in Table I., gives a product of 798,000, and this dividéd by 13-4432, the above reserved divisor, the quo- tient, 59360, is the required weight in pounds. ({1.—Diameter of the Post: when Round.— To ascertain the size of a round post to sustain safely a given weight, when the height of the post is at least ten times the diameter; the direction of the pressure coinciding with the axis of the post; we have— Rule VITI.—Multiply the given weight by the factor of safety, and divide the product by 1-5708 times the value of C for the material of the post, found in Table I. ; reserve the quotient, calling its value 6. Now multiply 432 times the value of е for the material of the post, found in Table IIL. by the square of the height in feet, and by the above quo- tient С; to the product add the square of G: extract the.

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SIZE ОЕ POST FOR GIVEN WEIGHT. 93 square root of the sum, and to it add the value of G; then the square root of this sum will be the required diameter; о 4 W a абы 1.5708 C” (5) 2-0 VAR бтк +G | (9.) Example.—What should be the diameter of a locust post IO feet high to sustain safely 40,000 pounds, the pressure coinciding with the axis? Proceeding by the rule, the given weight multiplied by 6, taken as а factor of safety, equals 240000. Dividing this by 1-5708 times 11700, the value of C for locust, in Table I., the quotient, 13-06, is the value of G, the square of which is 170-53. Now, the value of e for locust, in Table III., is -oo15. This multiplied by 432, by 100, the square of the height, and by the above value of G, gives a product of 846.2; which added to 170-53, the above square of G, gives the sum of 1016-73. То 31-90, the square root of this, add the above value of G; then 6-7, the square root of this sum, is the required diameter of the post. The post therefore requires to be 6-7, say 6% inches diameter. 112.—-Side of the Post: when Square.— To ascertain the side of a square post to sustain safely a given weight, when the height of the post is at least ten times the side ; the pres- sure coinciding with the axis ; we have— Rule 1X.—Multiply the given weight by the factor of safety, and divide the product by twice the value of C for the material of the post, found in Table I.; reserve the quo- tient, calling its value G.. Now multiply 432 times the value of e for the material of the post, found in Table IHI., by the square of the height in feet, and by the above quotient С; to the product add the square of G; extract the square root of the sum, and to it add the value of G; then the square 'root of this sum will be the required side of the post ; or— Wa - +e 10.) . E T (10.)

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94 CONSTRUCTION. s-4/ үз Gel*4 С + С. (11.) Example.—W hat should be the side of a Georgia-pine square post 15 feet high to sustain safely 50,000 pounds, the : pressure coinciding with the axis of the post? Proceeding by the rule, 50,000 pounds multiplied by 6, as a factor of safety, gives 300000; this divided by 2x9500 (the value of C) — 19000, the quotient, 15-780, is the value of 6. The value of e for Georgia pine is -00109; the square of the height is 225; then, 432 times -oo109 by 225 and by 15-780 (the above value of G) gives a product of 1672-86; the square of G equals 249-31; this added to 1672.86 gives a sum of 1022.17, the square root of which 15 43-843; which added.to 15-789, the value of C, gives 59-632, the square root of which is 7-722, the required side of the post. Тһе post, therefore, requires to be, say, 72 inches square. | ([3.—' Thickness of a Rectangular Post.—This may be definitely ascertained when the proportion which the thick- ness shall bear to the breadth shall have been previously determined. For example, when the proportion is as 6 to 8, then 14 times 6 equals 8, and the proportion is as 1 to 14; again, when the proportion is as 8 to то, then 13 times 8 equals 10, and in this case the раррацыов isas I to 14. Let the latter figure of the ratio 1 to 14, 1 to 14, etc., be called n, ог so that the proportion shall le as I to z, then— To ascertain the thickness of a post to sustain safely a given weight, when the height is at least ten times the thick- ness; the cM. of the weight coinciding with the axis; we have— Rule X.—Multiply the given weight by the factor of safety, and divide the product by twice the value of C for the material of the post, found in Table L, multiplied by л, as above explained ; reserve the quotient, zm itG. Now multiply 432 times the value of ¢ for the material of the post, . found in Table III., by the square of the height in feet, and by the above quotient G; to the product add the square of G ; extract the square root of the sum, and to it add thé value.

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BREADTH OF POST FOR GIVEN THICKNESS. 95 of G; then the square root of this sum will be the required thickness of the post; or— | | W a ТІГІ (22) "TL Се +С +С. (13.) Example. —What should be the thickness of a white-pine rectangular post 20 feet high to sustain safely 30,000 pounds, the pressure coinciding with the axis, and the proportion between the thickness and breadth to be as 10 to 12, or as 1 to 1-2? Proceeding according to the rule, we have the pro- duct of 30000, the given weight, by 6, as a factor of safety, equals 180000; this divided by twice Cx z, or 2 x 6650 x 1-2, (—15960) gives a quotient of 11.278, the value of G. Then, we have e = -oo14, the square of the height equals 400; therefore, 432 x -0014 x 400 x 11:278 = 2728-43.. To thisadd- ing 127-2, the square of С, we have 2855 63, the square root of which is 53-438; and this added to G gives 64-716, the square root of which is 8-045, the required thickness of the post. Now, since the thickness is in proportion to the breadth as 1 to 1-2, therefore 8-045 x 1-2 = 9:654, the re- quired width. The post, therefore, may be made 8 x o£ inches. 3 114.—Breadth of a Rectangular Post.—When the thick- ness of a post is fixed, and the breadth required; then, to ascertain the breadth of a rectangular post to sustain safely a given weight, the direction of the pressure of which coin- cides with the axis of the post, we have— Rule X1.— Divide the height in inches by the given thick- ness, and multiply the quotient by itself, or take its square: . multiply this square by the value of e for the material of the post, found in Table III.; to the product add its half, and to the sum add unity ; multiply this sum by the given weight, and by the factor of safety; divide the product by the pro- duct of the given thickness multiplied by the value of C for

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96 CONSTRUCTION. the material of the post, found in Table I., and the quotient will be the required breadth ; or— h Wa(it+3er?) b= = аг рч" (14.) Example.—W hat should be the breadth of a spruce post 18 feet high and 6 inches thick to sustain safely 25,000 pounds, the pressure coinciding with the axis of the post? According to the rule, 216 (= 12 x 18), the height in inches, divided by 6, the given thickness, gives a quotient ot 36, the square of which is 1296; the value of e for spruce is -00098; this multiplied by 1296, the above square, equals 1-27; which increased by -635, its half, amounts to 1-905; this mcreased by unity, the sum is 2:905; which multiplied by the given weight, and by the factor of safety, gives a product of 435749; and this divided by 6 (the given thickness) times 7850 (the value of C for spruce) — 47100, gives a quotient of 9-2516, the required breadth of the post. The.post, therefore, re- quires to be 6 x ọ4 inches. Observe that when the breadth obtained by the rule. is less than the given thickness, the result shows that the con- ditions of the case are incompatible with the rule, and that a new computation must be made; taking now for the breadth what was before understood to be the thickness, and proceeding in this case, by Rule X., to find the thickness. 1{5.—Resistanee to Tension.—In 477. 95 are recorded the results of experiments made to test the resistance of vari- ous materials to tensile strain, showing in each case the ca- pability to such resistance per square inch of sectional area. The action of materials in resisting a tensile strain is quite simple ; their resistance is found to be directly as their sec- tional area. Hence— | 116.—The Limit of Weight.—To ascertain the weight ог pressure that may be safely applied to a beam or rod as a tensile strain, we have— Rule XII.—Multiply the area of the cross-section of the beam or rod in inches by the value of 7; Table IL; divide

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AREA OF BEAM FOR TENSILE STRAIN. 97 the product by the factor of safety, and the quotient will be the required weight in pounds; or— 444 W = 2 (15.) The cross-section here intended is that taken at the small. est part of the beam or rod. А beam, in framing, is usually cut with mortices; the area will probably be smallest at the severest cutting ; TM area used in the rule must be that of the uncut fibres only. Example.— The tie-beam of a roof-truss is of white pine, 6 x 10 inches; the cutting for the foot of the rafter reduces the uncut area to 40 inches: whatamount of horizontal thrust from the foot of the rafter will this tie-beam safely sustain? Here 40 times 12000, the value of 7, equals 480000; this divided by 6, as a factor of safety, gives 80000, the required weight in pounds. ({7.—Sectional Area.—To ascertain the sectional area of a beam or rod that will sustain a given weight safely, when applied as a tensile strain, we have— Rule XIII.—Multiply the given weight in pounds by the factor of safety ; divide the product by the value of 7, Table II., and the quotient will be the area required in inches; or— " a A = wp (16.) This is the area of uncut fibres. If the piece is to be cut for mortices, or for any other purpose, then for this an adequate addition is to be made to the result found by the rule. Example—A rafter produces a thrust horizontally of 80,000 pounds; the tie-beam is to be of oak: what must be the area of the cross-section of the tie-beam in order to sustain the rafter safely? Тһе given weight, 80000, multiplied by Io, as a factor of safety, gives 800000; this divided by 19500, the value of 7, Table IL, the quotient, 41, is the area of uncut fibres. This should have usually one half of its amount

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98 CONSTRUCTION. added. to it as an allowance for cutting; therefore, 41 + 21 = 62. Тһе tie-beam may be 6 x 10$ inches. ' Another Example. — A tierod of American refined wrought iron is required to sustain safely 36,000 pounds: what should be its area of cross-section? Taking 7 as the factor of safety, 7 x 36000 = 252000; and this divided by 60000, the value of 7, Table Il., gives a quotient of 4-2 inches, the required area of the rod. ' 118.— Weight of the Suspending Piece Included.—Pieces subjected to a tensile strain are frequently suspended verti- cally. In this case, at the upper end, the strain is due not only to the weight attached at the lower end, but also to the weight of the rod itself. Usually, in timber, this is small in comparison with the load, and may be neglected; although in very long timbers, and where accuracy is decid- edly essential, as, also, when the rod is of iron, it may form a part of the rule. Taking the effect of the weight of the beam into account, the relation existing between the weights and the beam requires that the rule for the weight should be as follows: ~ Rule XIV.—Divide the value of 7 for the material of the ` beam or rod, Table II., by the factor of safety; from the quotient subtract 0-434 times the specific gravity of the ma- terial in the beam or rod multiplied by the length of the beam or rod in feet; multiply the remainder by the area of cross-section in inches, and the product will be the required weight in pounds ; or— = A (= — 0.43415). ! ( 17.) N. B.— This rule is based upon the condition that the sus- pending piece be not cut by mortices or in any other way. Example.—What weight may be safely sustained by a white-pine rod 4 x 6 inches, 40 feet long, suspended verti- cally? For white pine the value of 7 is 12000; this divid- ed by 8, as a factor of safety, gives 1500; from which sub- tracting 0-434 times 0-458 (the specific gravity of white pine, Table II.) multiplied by до, the length in feet, the remainder

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RESISTANCE ТО TRANSVERSE STRAINS, 99 is 1492-049; which multiplied by 24 (= 4x6, the area of cross-section) equals 35,761 pounds, the required weight to be carried. The weight which the rule would give, neglecting the weight of the rod, would have been 36000; ordinarily, so slight a difference would be quite unimportant. (19.--Агеа of Suspending Piece.—To ascertain the area of a suspended rod to sustain safely a given. weight, when the weight of the suspending piece is regarded, we have— Rule XV.—Multiply 0-434 times the specific gravity of the suspending piece by the length in feet; deduct the pro- duct from. the quotient arising from a division of the value of T, Table II., by the factor of safety, and with the remain- der divide the given weight in pounds; the quotient will be the required area in inches ; or— 8 И/ | 3 20-43415 i A == N.B.— This rule is based upon the condition that the rod be not injured in any wise by cutting. | | Example.—W hat should be the area of a bar of English cast iron 20 feet long to sustain safely, suspended from its lower end, a weight of 5000 pounds? Taking the factor of safety at 7-0, and the specific gravity also at 7, and the value of 7, Table lI., at 17000, we have the product of 0-434 x 7-0 x 20 = 60-76; then 17000 divided by 7 gives а quotient of 2428-57; from which deducting the above 60-76, there remains 2367-81; dividing 5000, the given weight, by this remainder, we have the quotient, 2: 11, which is the required area in inches. | ! RESISTANCE TO TRANSVERSE STRAINS. 120.— Transverse Strains: Rupture.— А load placed upon a beam, laid horizontally or inclined, will bend it, and, if the weight be proportionally large, will break it. The power in the material that resists this bending or breaking is termed the resistance to cross-strains, or transverse strains.

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IOO CONSTRUCTION. While in posts or struts the material is compressed or short- ` ened, and in ties and suspending pieces it is extended or lengthened, in beams subjected to cross-strains the material is both compressed and extended. (See 472. 91. When the beam is bent the fibres on the concave side are compressed, while those on the convex side are extended. The line where these two portions of the beam meet—that 15, the portion compressed and the portion extended — the hori- zontal line of juncture, is termed the zeuzra/ line or plane. It is so called because at this line the fibres are neither com- pressed nor extended, and hence are under no strain what- ever. The location of this line or plane is not far from the middle of the depth of the beam, when the strain 1s not suf- ficient to injure the elasticity of the material; but it re- moves towards the concave or convex side of the- beam as the strain 15 increased, until, at the period of rupture, its distance from the top of the beam is in proportion to its dis- tance from the bottom of the beam as the tensile strength of the material is to its compressive strength. [21.—Location of Mortices.—In order that the diminution of the strength of a beam by framing be as small as possible, all mortices should be located at or near the middle of the depth. There is a prevalent idea with some, who are aware that the upper fibres of a beam are compressed when sub- ject to cross-strains, that it is not injurious to cut these top fibres, provided that the cutting be for the insertion of an- other piece of timber—as in the case of gaining the ends of beams into the side of a girder. They suppose that the piece filled in will as effectually resist the compression as the part removed would have done, had it not been taken out. Now, besides the effect of shrinkage, which of itself is quite suf- ficient to prevent the proper resistance to the strain, there is the mechanical difficulty of fitting the joints perfectly throughout; and, also, a great loss in the power of resist- ance, as the material is so much less capable of resistance when pressed at right angles to the direction of the fibres than when directly with them, as the results of the experi- ments in the tables show.

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STRENGTH OF. BEAMS FOR CROSS-STRAINS. IOI 122.—' Transverse Strains: Relaátion of Weight to Di- mensions.—lhe strength of various materials, in their re- sistance to cross-strains, is given in Table III., 477. 96. The second column of the table contains the results of experi- ments made to test their resistance to rupture. In the case of each material, the figures given and represented by В indicate the pounds at the middle required to break a unit of the material, ora piece 1 inch square and 1 foot long between the bearings upon which the piece rests. To be able to use these indices of strength, in the computation of the strength of large beams, it 15 requisite, first, to establish the relation between the unit of material and the larger beam. Now,it may be easily comprehended that the strength of beams will be in proportion to their breadth; thàt is, when the length and depth remain the same, the strength will be directly as the breadth ; tor it is evident that a beam 2 inches broad will bear twice as much as one which is only т inch broad, or that one which is 6 inches broad will bear three times as much as one which is 2 inches broad. This establishes the relation of the weight to the breadth. With the depth, however, the relation is different ; the strength 15 greater than simply in proportion to the depth. If the boards cut from a squared piece of timber be piled up in the order in which they came from the timber, and be loaded with a heavy weight at the middle, the boards will deflect or sag much more than they would have done in the timber before sawing. The greater strength of the material when in a solid piece of timber is due to the cohesion of the fibres at the line of separation, by which the several boards, before sawing, are prevented from sliding upon each other, and thus the resistance to compression and tension is made to contribute to the strength. This resistance is found to be in proportion to the depth. Thus the strength due to the depth is, first, that which arises from the quantity of the material (the greater the depth, the more the material), which is in proportion to the depth; then, that which en- sues from the cohesion of the fibres in such a manner as to prevent sliding; this is also as the depth. Combining the two, we have, as the total result, the resistance in proportion

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IO2 * CONSTRUCTION. to the square of the depth. The relation between the weight and the length is such that the longer the beam is,- the less it will resist; a beam which is 20 feet long will sus- tain only half as much as one which is 10 feet long; the breadth and depth each being the same in the two beams. From this it results that the resistance is zzversely int pro- portion to the length. Го Obtain, therefore, the relation between the strength of the unit of material and that of a larger beam, we have these facts, namely: the strength of the unit is the value of В, as recorded in Table III. ; and the strength of the larger beam, represented.by W, the weight required to break it, is the product of the breadth into the square of the depth, divided by the length; or, while for the unit we have the гайо-- iB > i we have for the larger beam the ratio— баз W: ——. / Therefore, putting these ratios in an expressed proportion, we have— 5 y we From which (the product of the means equalling the pro- duct of the extremes; see 477. 373) we have— E Bod" Gio.) In which W represents the pounds required to break a beam, when acting at the middle between the two supports upon which the beam is laid ; of which beam 0 represents the breadth and Z the depth, both in inches, and Z the length: in feet between the supports ; and В is from Table IIl., and represents the pounds required to break a unit of material like that contained in the larger beam. .

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LIMIT OF WEIGHT AT MIDDLE. [23.— Safe Weight: Load at Middle.— The relation established, in the last article, between the weight and the dimensions is that which exists at the moment of rupture. The rule (19.) derived therefrom is not, therefore, directly practicable for computing the dimensions of beams for buildings. From it, however, one may readily be deduced which shall be practicable. In the fifth column of Table III. . are given the least values of a, the factor of safety, explained in Art. 96. Now, if in place of В, the symbol for the break- ing weight, the guotient of B divided by a be substituted, then the rule at once becomes practicable ; the results now being in consonance with the requirements for materials used in buildings. Thus, with this modification, we have— Dog" W = . . T (20.) Therefore, to ascertain the weight which a beam may be safely loaded with at the centre, we have— Rule XV\.—Multiply the value of B, Table IJI., for the kind of material in the beam by the breadth and by the square of the depth of the beam in inches; divide the pro- duct by the product of the factor of safety into the length of the beam between bearings in feet, and the quotient will be the weight in pounds that the beam will safely sustain at the middle of its length. | Example.— What weight in pounds сап be suspended safely from the middle of a Georgia-pine beam 4x 10 inches, and 20feet long between the bearings? For Georgia pine the value of B, in Table IIL, is 850, and the least value of a is 1.84. For reasons given in 477. 96, let a be taken as high as 4; then, in this case, the value of 0 is 4, and that of d is то, while that of Z is 20. Therefore, proceeding by the rule, 850 x 4 x 10? = 340000; this divided by 4 x 20 (= 80) gives a quotient of 4250 pounds, the required weight. Observe that, had the value of а been taken at 3, instead of 4, the result by the rule would have been a load of 5667 pounds, instead of 425o, and the larger amount would be none too much for a safe load upon such a beam ; although, Гр ІҢ ome 97577 УХ а

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104 CONSTRUCTION. with it, the deflection would be one third greater than with the lesser load. The value of а should always be assigned higher than the figures of the table, which show it at its least value; but just how much higher must depend upon the firmness required and the conditions of each particular case. 124.—Breadth of Beam with Safe Load,.—By a simple transposition of the factors in equation (20.), we obtain— -. Ма без РІН (21. а rule for the breadth of the beam. Therefore, to ascertain what should be the breadth of a beam of given depth and length to safely sustain at the middle a given weight, we have— Rule XVII.—Multiply the given weight in pounds by the factor of safety, and by the length in feet, and divide the product by the square of the depth multiplied by the value of B for the material in the beam, in Table III. ; the quotient will be the required breadth. . Example.—W hat should be the breadth of a white-pine beam 8 inches deep and то feet long between bearings to sustain safely 2400 pounds at the middle? For white pine the value of Л, in Table ІП., is 500. Taking the value of a at 4, and proceeding by the rule, we have 2400 x4x 10 — 96000; this divided by (8' x 500 —) 32000 gives a quotient of 3, the required breadth of the beam. 125.—Depth of Beam with Safe Load.—A transposition of the factors in equation (21.), and marking it for extraction of the square root, gives— Wal Pis : plora (22.) a rule for the depth of a beam. Therefore, to ascertain what should be the depth of a beam of given breadth and length to safely sustain a given weight at the middle, we have— ,

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WEIGHT AT ANY POINT. 105; Rule XVII.—Multiply the given weight by the factor of safety, and by the length in feet; divide the product by the product of the breadth into the value of B for the kind of wood, Table III.; then the square root of the quotient will be the required depth. Example.—What should be the depth of a spruce beam 5 inches broad and 10feet long between bearings to sustain safely, at middle, 4500 pounds? The value of P from the table is 550; taking а at 4, and proceeding by the rule, we have 4500 x 4x I5 = 2700С0; this divided by (550 x 5 =) 2750 gives a quotient ої 98-18, the square root of which is 9-909, the required depth of the beam. The beam should be 5 x 10 inches. 126.—Safe Load at any Point.—W hen the load is at the middle of a beam it exerts the greatest possible strain ; at any other point the strain would be less.. The strain de- creases gradually as it approaches one of the bearings, and when arrived at the bearing its effect upon the beam as a cross-strain is zero. Тһе effect of a weight upon a beam is in proportion to its distance from one of the bearings, mul- tiplied by the portion of the load borne by that bearing. The load upon a beam is divided upon the two bearings, as shown at 477. 88. Тһе weight which is required to rup- ture a beam is in proportion to the breadth and square of the depth, 2 2°, as before shown, and also in. proportion to the length divided by 4 times the rectangle of the two parts А Amn This, when the load is at the middle, may be put as / (see Fig. 35). into which the load divides the length, or 1, а result coinciding with the relation before ах47х%/ Ее given in 477. 122, viz.: “ The resistance is inversely in рго- portion to the length." The total resistance, thercfore, put- „ме ' al ting the two statements together, is 1n proportion о =< ; — pm There are, therefore, these two ratios, viz., W: A" " В : т; from which we have the proportion—

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106 CONSTRUCTION. d Ла Ж : 4 nau from which we have— Bbd'l (23-) 4ш т А А ». This is the relation at the point of rupture, and when — 18 used instead of B, the expression gives the safe weight. Therefore— | .B6d'I дати W (24-) is an expression for the safe weight. Now, to ascertain the weight which may be safely borne by a beam at any point in its length, we have— Rule XIX.-—Multiply the breadth by the square of the depth, by the length in feet, and by the value of В for the material of the beam, in Table III. ; divide the product by the product of four times the factor of safety into the rec- tangle of the two parts into which the centre of gravity of the weight divides the beam, and the quotient will be the required weight in pounds. Example.—W hat weight may be safely sustained at 3 feet from one end of a Georgia-pine beam which is 4 x 10 inches, and 20 feet long? The value of P for Georgia pine, in Malle IM., 1s 850; therefore, by the tule; 2 Xe X Go 850 6800000. Taking the factor of safety at 4, we have 4X4X3x17- 816. Using this as a divisor with which to divide the former product, we have as a quotient 8333 pounds, the required weight. 127.—Вгеаа or Depth: Load at any Point.—By a proper transposition of the factors of (24.) we obtain— ра 4Wamn = 397 , (25.) an expression showing the product of the breadth into the square of the depth; hence, to ascertain the breadth or

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DISTRIBUTED WEIGHT. 107 depth of a beam to sustain safely a given weight located at any point on the beam, we have— Rule X X.—Multiply four times the given weight by the factor of safety, and by the rectangle of the two parts into which the load divides the length; divide the product by the product of the length into the value of 7 for the mate- rial of the beam, found in Table III., and the quotient will be equal to the product of the breadth into the square of the depth. Now, to obtain the breadth, divide this product by the square of the depth, and the quotient will be the required breadth. But if, instead of the breadth, the depth be de- sired, divide the said product by the breadth; then the square root of the quotient will be the required depth. Example.—W hat should be the breadth (the depth being 8) of a white-pine beam 12 feet long to safely custain 3500 pounds at 3 feet from one end? Also, what should ре its depth when the breadth is 3 inches? By the rule, taking fhe МС oL Gaiety at 4, 4 х 350054 х3 о ="T5 DOO. The value of В for white pine, in Table 111., is 500; there- fore, 500 x 12 — 6000; with this as divisor, dividing 1512000, the quotient is 252. Now, to obtain the breadth when the depth is 8, 252 divided by (8x 8 —) 64 gives a quotient of 3:9375, the required breadth ; or the beam may be, say, 4 x 8. Again, when the breadth is 3 inches, we have for the quotient of 252 divided by 3 = 84, and the square root of 84 is 9:165, ог оф inches. For this case, therefore, the beam should be, say, 3 x gt inches. [28.— Weight Uniformly Distributed.— When the load is spread out uniformly over the length of a beam, the beam will require just twice the weight to break it that would be required if the weight were concentrated at the centre. Therefore, we һауе W = А where U represents the dis- tributed load. Substituting this value of W in equation (20.), we have— im e OA Bp ai or— ya 2 bee (26.) `

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108 CONSTRUCTION. Therefore, to ascertain the weight which may be safely sus- tained, when uniformly ап енені over the length of a beam, we havc— Rule XXI.—Multiply twice the breadth by the square of the depth, and by the value of B for the material of the beam, in Table II., and divide the product by the product of the length in feet by the factor of safety, and the quotient will be the required weight in pounds. Example.—What weight uniformly distributed may be safely sustained upon а hemlock beam 4 хо inches, and 20 feet long? The value of B for hemlock, in Table II., is 450; therefore, by the rule, 2 x 4 x 9° х 450 = 201600. Tak- ing the factor of safety at 4, we have 4 x 20 = 80, the pro- duct by which the former product is to be divided. This division produces a quotient of 3645, the required weight. 129.__ Breadth or Depth: Load Uniformly Distributed.— By a proper transposition of factors in (26.), we obtain— 4 U al bide dk BS (278) an expression giving the value of the breadth into the square of the depth. From this, therefore, to ascertain the breadth or the depth of a beam to sustain safely a given weight uni- formly distributed over the length of a beam, we have— Rule XXII.—Multiply the given weight by the factor of safety, and by the length; divide the product by the pro- duct of twice the value of B for the material of the beam, in Table ITI., and the quotient will be equal to the breadth into the square of the depth. Now, to find the breadth, divide the said quotient by the square of the depth; but if, instead of the breadth, the depth be required, then divide said quotient by the breadth, and the square root of this quotient will be the required depth. Example.—W hat should be the size ofa white-pine beam 20 feet long to sustain safely 10,000 pounds uniformly distributed over its length? The value of Z for white pine, in Table III., is 500. Let the factor of safety be taken at 4. Then, by the rule, 10000 x 4 x 20 = 800000; this divided by (2 х 5со--)

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WEIGHT PER BEAM IN FLOORS. 109 1000 gives a quotient of 800.: Now, if the depth be fixed at 12, then the said quotient, 800, divided by (12 x 12 =) 144 gives 53, the required breadth of beam; and the. beam may be, say, 5$ x12. Again, if the breadth is fixed, say, at 6, and the depth is required, then the said quotient, 8co, divided by 6 gives 133%, the square root of which, 11-55, isthe required depth. Тһе beam in this case should therefore be, say, 6 x 11$ inches. [30.—ILLoad per Foot Superficial.— When several beams are laid in a tier, placed at equal distances apart, as in a tier of floor-beams, it is desirable to know what should be their size in order to sustain a load equally distributed over the floor. If the distance apart at which they are placed, measured from the centres of the beams, be multiplied by the length of the beams between bearings, the product will equal the area of the floor sustained by one beam ; and if this area be multiplied by the weight upon a superficial foot of the floor, the product will equal the total load uniformly distributed over the length of the beam ; or, if c be put to represent the distance apart between the centres of the beams in feet, and / represent the length in feet of the beam between bearings, and f equal the pounds per superficial foot on the floor, then the product of these, or c f 7, will represent the uni- formly distributed load on a beam; but this load was before represented by U (Art. 128); therefore, we have c f 7 — U, and they may be substituted for it in (26.) and (27.). Thus wehave— 0 Bo uad 0.24 2B^ ог— bat = EET. (28.) в Therefore, to ascertain the size of floor-beams to sustain safely a given load per superficial foot, we have— - | Rule XXIII.—Multipiy the given weight per superficial foot by the factor of safety, by the distance between the `

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Ito CONSTRUCTION. centres of the beams in feet, and by the square of the length in feet; divide the product by twice the value of В for the material of the beams, in Table III., and the quotient will be equal to the breadth into the square of the depth. Now, to obtain the breadth, divide said quotient by the square of the depth, and this quotient will be the required breadth. But if, instead of the breadth, the depth be required, divide the aforesaid quotient by the breadth; then the square root of this quotient will be the required depth. Example.—What should be the size of white-pine floor- beams 20 feet long, placed 16 inches from centres, to sustain safely 90 pounds per superficial foot, including the weight of the materials of construction—the beams, flooring, plas- tering, etc.? The value of B for white pine is 500; the factor of safety may be put at 5. Then, by the rule, we have 90x 5 x48 x 20° = 240000. This divided by (2 x 500 =) 1000 gives 240. Now, for the breadth, if the depth be fixed at о inches, then 240 divided by (9’=) 81 gives а quotient of 2-963. The beams therefore should be, say, 3x9. Butif the breadth be fixed, say, at 2-5 inches, then 240 divided by 2-5 gives a quotient of 96, the square root of which is 9-8 nearly. The beams in this case would require therefore to be, say, 24 x 10 inches. N. B.—It is well to observe that the question decided by Rule XXII. is simply that of strength only. Floor-beams computed by it will be quite safe against rupture, but they will in most cases deflect much more than would be consist- ent with their good appearance. Floor-beams should be computed by the rules which include the effect of deflection. (Gee Arr. 152.) (31. —Levers: Load at One End.—The beams so far con- sidered as being exposed to transverse strains have been supposed to be supported at each end. When a piece is held firmly at one end only, and loaded at the other, it is termed a lever; and the load which a piece so held. апа loaded will sustain is equal to one fourth that which the same piece would sustain if it were supported at each end and loaded at the middle. Or, the strain in a beam sup-

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LEVERS TO SUSTAIN GIVEN тир III ported at each end caused by a given weight located at the middle is equal to that in a lever of the same breadth and depth, when the length of the latter is equal to oné half that of the beam, and the load at its end is equal to one half of that at the middle of the beam. Ог, when P represents the load at the end of the lever, and z its length, then W=2 P, апа /= 2и. Substituting these values of Wand / in equa- tion (20.), we have— aub d қ ^ 2an’ from which— Бра? 4an’ Pe (29.) Hence, to ascertain the weight which may be safely sus- tained at the end of a lever, we have— Rule XXIV.—Multiply the breadth of the lever by the square of its depth, and by the value of B for the material of the lever, in Table III. ; divide the product by the pro- duct of four times the length in feet into the factor of safety, and the quotient will be the required weight in pounds. Example—What weight can be safely sustained at the end of a maple lever of which the breadth is 2 inches, the depth is 4 inches, and the length is 6 feet? Тһе value of 2 for.maple, in Table IIL, is 1100; therefore, by the rule, 2 X4" x 1100 = 35200. Апа, taking the factor of safety at 5, 4X5x 6 = 120, and 35200 divided by 120 gives a quotient of 293-33, ог 2934 pounds. N. B.—When a lever is loaded with a weight uniformly distributed over its length, it will sustain just twice the load which can be sustained at the end. 132.—-Levers: Breadth or Depth.—By a proper trans- position of the factors in (29.), we obtain— T = i E (30.) Hence, to ascertain the breadth or depth of a lever to sus- tain safely a given weight, we have—

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ГІ2 CONSTRUCTION. Rule XXV.—Multiply four times the given weight by the length of the lever, and by the factor of safety; divide the product by the value of В for the material of the lever, in Table III., and the quotient will be equal to the breadth multiplied by the square of the depth. Now, if the breadth be required, divide said quotient by the square of the depth, and this quotient will be the required breadth; but if, instead of the breadth, the depth be required, divide the said quotient by the breadth ; then the square root of this quotient will be the required depth. Example.—What should be the size of a cherry lever 5 feet long to sustain safely 250 pounds at its end? Proceed- ing by the rule, taking the factor of safety at 5, we have 4x250x5$x5- 25000. The value ot 0 for cherry, in Table IIL, is 650; and 25000 divided by 650 gives a quotient of 38-46. Now,if the depth be fixed at 4, then 38-46 divided by (4x4 —) 16 gives a quotient of 2-4, the required breadth. But if the breadth be fixed at 2, then 38-46 divided by 2 gives a quotient of 19-23, the square root of which is 4-38, the required depth. Therefore, the lever may be 2.4 x 4, or 2 x 4% inches. (33.—Deflection: Relation to Weight.—When a load is placed upon a beam supported at each end, the beam bends more or less; the distance that the beam descends under the operation of the load, measured at the middle of .its length, is termed its deflection. In an investigation of the laws of deflection it has been demonstrated, and experiments have confirmed it, that while the elasticity of the material remains uninjured by the pressure, or is injured in but a small degree, the amount of deflection is directly in propor- tion to the weight producing it ; for example, if 1000 pounds laid upon a beam is found to cause it to deflect or descend at the middle a quarter of an inch, then 2000 pounds will cause it to deflect half an inch, 3000 pounds will deflect it three fourths of an inch, and so on. 134.—Deflection : Relation to Dimensions.—In Table IIT. are recorded the results of experiments made to test the -

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THE LAW OF DEFLECTION. 113. resistance of the materials named to deflection. The fig- ures in the third column designated by the letter Z (for flex- ure) show the number of pounds required to deflect a unit of material one inch. This is an extreme state of the case, for in most kinds of material this amount of depression would exceed the limits of elasticity ; and hence the rule would here fail to give the correct relation as between the dimensions and pressure. For the law of deflection asabove stated (the deflections being in proportion to the weights) is true only while the depressions are small in comparison with the length. Nothing useful is, therefore, derived from this position of the question, except to give an idea of the nature of the quantity represented by the constant 7; it being in reality an index of the stiffness of the kind of mate- rial used in comparing one material with another. Whatever be the dimensions of the beam, F will always be the same quantity for the same material; but among various materials F will vary according to the flexibility or stiffness of each particular material For example, Æ will be much greater for iron than for wood ; and again, among the various kinds of wood, it will be larger for the stiff woods than for those that are flexible. The value of 2, therefore, is the weight which would deflect the unit of material one inch, upon the supposition that the deflections, from zero to the depth of one inch, continue regularly in proportion to the increments of weight producing the deflections, or, for each deflection— ZR p 0; from which we have— W= 0 оу, Fra, in which 4 represents the deflection in inches corresponding to W, the weight producing it. This is for the unit of ma- terial. For beams of larger dimensions, investigations have shown (Transverse Strains, Chapters XIII. and XIV.) that the power of a beam to resist deflection by a weight at mid- dle is in proportion to its breadth and the cube of its depth, and it is inversely in proportion to the cube of the length ;

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114 CONSTRUCTION. or, when the resistance of the unit of material is measured, as above, by T: we have the relation between it and a larger beam of— ~ W ba’ т” Putting this ratio in a proportion with that of the unit of mieria, we have— -- W ba’ ЖР с Pa e T Lad which gives— W Ры? S xS from which we have— T ime — (31.) (35.— Deflection: Weight when at Middle.—In equation (31.) we have a rule by which to ascertain what weight is required to deflect a given beam to a given depth of deflec- tion ; this, in words at length, is— | Rule ХХУ1.—Мийшрї1у the breadth of the beam by the cube of its depth, and by the given deflection, all in inches, and by the value of F for the material of the beam, in Table ПІ.; divide the product by the cube of the length in feet, and the quotient will be the required weight in pounds. Example—What weight is required at the middle of a 4x12 inch Georgia-pine beam 20 feet long to deflect it three quarters of an inch? The value of F for Georgia pine, in Table III., is 5900; therefore, by the rule, we have 4X 12! x 0-75 x 5900 — 30585600, which divided by (20x20: x 20 =) 8000 gives a quotient of 3823-2, the required weight in pounds. 136.—Deflection: Breadth or Depth, Weight at Middle. --Ву a transposition of equation (31.), we obtain— Wr 6d = ту (32) .

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SIZE FOR A GIVEN DEFLECTION. 115 Lf a rule by which may be found the breadth or depth of a beam, with a given load at middle and with a given deflec- tion; this, in words at length, is— з Rule XXVII.—Multiply the given load by the cube of the length in feet, and divide the product by the product of the deflection into the value of F for the material of the beam, in Table III.; then the quotient will be equal to the breadth of the beam multiplied by the cube of its depth, both in inches. - Now, to obtain the breadth, divide the said quotient by the cube of the depth, and this quotient will be the required breadth. But if, instead of the breadth, the depth be re- quired, then divide the said quotient by the breadth, and the cube root of this quotient will be the required depth. But if neither breadth nor depth be previously fixed, but it be required that they bear a certain proportion to each Other; чий that Z: 0::1:», r being a decimal, then = 72 and 22° —rd*; then, to find the depth, divide the aforesaid quotient by the decimal z, and the fourth root (or the square root of the square root) will be the required depth, and this multiplied by the decimal will give the breadth. Example.—Whhat should be the size of a spruce beam 20 feet long between bearings, sustaining 2000 pounds at the middle, with a deflection of one inch? By the rule, the weight into the cube of the length is 2000 x 8000 — 16000000. The value of F for spruce, in Table III., is 3500; this by the deflection — 1 gives 3500, which used as a divisor in divid- ing the above 16000000 gives a quotient of 4571-43. Now, if the breadth be required, the depth being fixed, say, at 10, then 4571-43 divided by (10 x 10x 10 =) 1000 gives 4-57, the required breadth. The beam should be, say, 4% by 10 inches. But if the depth be required, the breadth being fixed, say, at 4, then 4571-43 divided by 4 gives 1142-86, the cube root of which is 10-46; so in this case, therefore, the beam is required to be 4x 10d inches. Again, if the breadth is to : bear a certain proportion to the depth, or that the ratio be- tween them is to Бе; вау,о-6 to 1, then let 7 = 0-6, and then 4571-43 = Оба”, and dividing by 0-6, we have 7619-05 = d*. This equals d’ xd’: therefore the square root of 7619

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116 CONSTRUCTION. is 87.29, and the square root of this is 9-343, the required depth in inches. Now 9:343 хо:6 equals the breadth, or 9:343 х0:6 = 5.6; therefore the beam is required to be 5-6 x9- 34 inches, or, say, 53 x оў inches. 137.—Deflection: when Weight is at Middle.—By a trans- position of the factors in (32.), we obtain— , ‚ (33.) a rule by which the deflection of any given beam may be as- certained, and which, in words at length, is— Rule XXVIII.—Multiply the given weight by the cube of the length in feet; divide the product by the product of the breadth into the cube of the depth in inches, multiplied by the value of / for the material of the beam, in Table III., ‚ and the quotient will be the required deflection in inches: Example.—To what depth will 1000 pounds deflect a 3x10 inch white-pine beam 20 feet long, the weight being at the middle of the beam? Ву the rule, we have 1000 x 20° — 8000000; then, since the value of F for white pine, in Table III., is 2900, we have 3 x 10° x 2900 = 8700000; using - this product as a divisor and by it dividing the former pro- duct, we obtain a.quotient of 0.9195, the required deflection in inches. | 138.—Defleetion : Load .Uniformly Distributed.—In two beams of equal capacity, suppose the one loaded at the middle, and the other with its load uniformly distributed over its length, and so loaded that the deflection in one beam shall equal that in the other ; then the weight at the middle of the former beam will be equal to five eighths of that on the latter. This proportion between the two has been de. monstrated by writers on the strength of materials. (See p. 484, Mechanics of Eng. and Arch., by Prof. Mosely, Am. ed. by Prof. Mahan, 1856.) Hence, when U is put to represent the uniformly distributed !oad, we have— А ЗЕ

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DEFLECTION FOR LOAD EQUALLY DISTRIBUTED. 117 or, when ап equally distributed load deflects a beam to a certain depth, five eighths. of that load, if concentrated at the middle, would cause an equal deflection. This value of W may therefore be substituted for it in equation (31.), and give— . 5 2 — Fóóqd'ó p 99 from which we obtain— l.6Fód?ó6 P= Tr тр (34-) a rule for a uniformly distributed load. 139.—Deflection: Weight when Uniformly Distributed. --Іп equation (34.) we have a rule by which we may ascertain what weight is required to deflect to a given depth any given beam. This, in words at length, is— Rule XXIX.—Multiply 1-6 times the deflection by the breadth of the beam, and by the cube of its depth, all in inches, and by the value of F for the material of the beam, in Table III.; divide the product by the cube of the length in feet, and the quotient will be the required weight in pounds. | | Example—What weight, uniformly distributed over the length of a spruce beam, will be required to deflect it to the . depth of 0-5 of an inch, the beam being 3 x 10 inches and Io feet long? Тһе value of Ғ for spruce, in Table IIL, 15 3500. Therefore, by the rule, we have 1-6x0-5 x3 x 10? x 3500— 8400000, and this divided by (10x 10x 10 =) 1000 gives 8400, the required weight in pounds. (40.—Deflection: Breadth or Depth, Load Uniformly Distributed.—By transposition of the factors in equation (34.), we obtain— TR dp a rule for the dimensions, which, in words at length, is—

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118 CONSTRUCTION. Rule XX X.—Multiply the given weight by the cube of the length of the beam; divide the product by 1-6 times the given deflection in inches, multiplied by the value of F for the material of the beam, in Table III., and the quotient will equal the breadth into the cube of the depth. Now, to ob- tain the breadth, divide this quotient by the cube of the depth, and the resulting quotient will be the required breadth in inches. But if, instead of the breadth, the depth be required, then divide the aforesaid quotient by the breadth, and the cube root of the resulting quotient will be the required depth іп inches. Again,if neither breadth nor depth be previously determined, but to be in proportion to each other at a given ratio, as 7 to 1, 7 being a decimal fixed at pleasure, then di- vide the aforesaid quotient by the value of 7, and take the square root of the quotient; then the square root of this square root will be the required depth in inches. The breadth will equal the depth multiplied by the value of the deci- mal ғ. | Example.—What should be the size of a locust beam 10 feet long which is to be loaded with 6000 pounds equally distributed over the length, and with which the beam is to be deflected $ of an inch? The value of F for locust, in Table III.,is 5050. By therule, we have 6000x(10 x 10 x 10 =) 1000 = 6000000, which is to be divided by (1.6 xo-75x 5050 =) бобо, giving a quotient of 990-1. Now, if the depth be, say, 6 inches, then 99o- 1 divided by (6x 6x 6 =) 216 gives a quo- tient of 4-584, the required breadth in inches, say 4&. But if the breadth be assumed at 4 inches, then 990-1 divided by 4.gives a quotient of 247-525, the cube root of which is 6-279, the required depth in inches, or, say, 61. And, again, if the ratio between the breadth and depth be as 0-7 to т, then 990-1 divided by o-7 gives a quotient of 1414-43, the square root of which is 37-609, of which the square root is 6.1326, the required depth in inches, or, say, 64; and then 6.1326 x 0.7 = 4:293, the required breadth in inches; or, the beam should be 443 x 64 inches. I41.—Deflection: when Weight is Uniformly Distributed. —By a transposition of the factors of equation (35.), we ob- tain— -

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DEFLECTION OF LEVERS AND BEAMS. 110 [7/3 анлы тета ae! a result nearly the same as that in equation (33.), which isa rule for deflection by a weight at middle, and which by slight modifications may be used for deflection by an equally distributed load. Thus by— Rule XX XI.—Proceed as directed in Rule XXVIII. (A77. 137), using the equally distributed weight instead of a con- centrated weight, and then divide the result there obtained for deflection by 1-6; then the quotient will be the required deflection in inches. Example.—TlTaking the example given under Rule XXVIII., in A77. 137, and assuming that the 1000 pounds load with which the beam is loaded be equally distributed, then O-9195, the result for deflection as there found, divided by 1-6, as by the above rule, gives 0.5747, the required deflection. This result is just five eighths of o-9195, the deflection by the load at middle. | N.B.—The deflection by a uniformly distributed load is just five eighths of that produced by the same load when concentrated at the middle of the beam; therefore, five eighths of the deflection obtained by Rule XXVIII. will be the deflection of the same beam when the same weight is uniformly distributed. 14-2.— Deflection of Levers.—The deflection of a lever is the same as that of a beam of the same breadth and depth, but of twice the length, and loaded at the middle with a load equal to twice that which is at the end of the lever. There- fore, if P represents the weight at the end of a lever, and > the length of the lever in feet, then 2 P= W and 2 » = /, and if these values of И/ апа / be substituted for those in equa- tion (33.), we obtain— "s 2 Px a" Wut viti which reduces to— m moe (37) Ға”

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120 CONSTRUCTION. a result 16 times that in equation (33.), which is the deflection in a beam. Therefore, when a beam and a lever equal in sectional area and in length be loaded by equal weights, the one at the middle, the other at one end, the deflection of the lever will be 16 times that of the beam. This proportion is based upon the condition that neither the beam nor the lever shall be deflected beyond the limits of elasticity. 143.—Defiection of a Lever: Load at End.—Equation (37.), in words at length, is— Rule X XXII —Multiply 16 times the given weight by the cube of the length in feet; divide the product by the product of the кейде into "^ cube of the depth multiplied by the value of / for the material of the lever, in Table HI., and the quotient will be the required deflection. Example.—What would be the deflection of a bar of American wrought iron one inch broad, two inches deep, loaded with 150 pounds at a point 5 feet distant from the wall in which the bar is imbedded? The value of / for American wrought iron, in Table ПІ., is 62000. Therefore, by ‘the rule, 16x ī50ọ x 5§^= 300000. wins divided by (1 x 2° x 62000 =) 496000 gives 0-6048, the required deflec- tion—nearly $ of an inch. [4-4.. —Dceflection of a Lever: Weight when at End.—By a transposition of the factors in equation (37.), we obtain— FUPZUS ES = 8. БЛ (38.) This result is equal to one sixteenth of that shown in equa- tion (31.), a rule for Ie weight at the middle. "Therefore, for— Rule XXXIII ӛскен as directed in Rule XXVII.; divide the quotient there obtained by 16, and the ее же quotient will be the required weight in pounds. Example.— What weight is required at the end of a 4 x 12 inch Georgia-pine lever 20 feet long to deflect it three quarters of an inch? Proceeding by Rule XXVII., we ob- tain a quotient of 3823.2; this divided by 16 gives 238:05, say 239, the required Sikha i in pounds.

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DEFLECTION OF LEVERS WITH UNIFORM LOAD. 121 (4-5.—Deflection of a Lever: Breadth or Depth, Load at End.—A transposition of the factors of equation (38.) gives— . | 6 Pa? ba*—! j І е FS (39.) а rule by which to obtain the sectional area of the lever. By comparison with equation (32.) it is seen that the result in (39.) 15 16 times that found by (32.). Therefore, the dimen- sions for a lever loaded at the end may be found by— Rule XX XIV.—Multiply by 16 the first. quotient found by Rule XXVII., and then proceed as farther directed in Rule XXVII., using the product of 16 times the quotient, instead of the said quotient. Example.—W hat should be the size of a spruce lever 20 feet long, between weight and wall, to sustain 2000 pounds at the end with a deflection of 1.inch? Proceeding by Rule XXVIL, we obtain a first quotient of 4571-43. By Rule XXXIV., 4571-43 x 16 = 73144-88. Now, if the depth be fixed, say, at 20, then 73144:88 divided by (20 x 20 x 20 —) 8000 gives 9-143, the required breadth. But to obtain the depth, fixing the breadth, say, at 9, we have for 73144-88 di- vided by 9 = 8127-21, thé cube root of which is 20-1055, the required depth. Again, if the breadth and depth are to be in proportion, say, as 0-7 to 1-0, then 73144-88 divided by 0-7 gives 104492. 7, the square root of which is 323-254, of which the square root is 17-98, the required depth in inches; and 17-98 x о-7 = 12-586, the required breadth in inches. The lever, therefore, should be, say, 12$ x 18 inches. 146.—Deflection of Levers: Weight Uniformly Distrib- uted.— A comparison of the effects of loads upon levers shows (Transverse Strains, Art. 347) that the deflection by a uniformly distributed load is equal to that which would be produced by three eighths of that load if suspended from the end of the lever. Or, Р = 2 U. Substituting this value of P, in equation (37.), gives— | бх Un’ O PES ^ which reduces to— f м4 6 Un (40.)

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P22 CONSTRUCTION. а rule for the deflection of levers loaded with an equally dis- tributed load. (47.— Deflection of Levers with Uniformly Distributed Load.—The deflection shown in equation (40.) is just six times that shown in equation (33.). The result by (33.) mul- tiplied by 6 will equal the result by (40); therefore, we have— Rule XXXV.—Proceed as directed in Rule XXVIII.; the result thereby obtained multiplied by 6 will give the required deflection. Example.—To what depth will 500 pounds deflect a 3 x 10 inch white-pine lever 10 feet long, the weight uniformly distributed over the lever? Here, by Rule XXVIII., we obtain the result 0-05747; this multiplied by 6 gives o- 3448, the required deflection. (48.—Defleetion of Levers: Weight when Uniformly Distributed.— By a transposition of factors in (40.), we ob- tain— | j^ qub | Я U = Com? ` (41.) This is equal to one sixth that of equation (31.); therefore, we have— Rule XXXVI.—Proceed as directed in Rule XXVI.; the quotient thereby obtained divide by 6, and the quotient thus obtained will be the required weight. Example.—What weight will be required to deflect a 4x 5 inch spruce lever 1 inch, the weight uniformly dis- tributed over its length? Proceeding as directed in Rule XXVI., the result thereby obtained is 1750; this divided Бу 6 gives 2912, the required weight in pounds. 149.—-Deflection of Levers: Breadth or Depth, Load Uniformly Distributed,—A transposition of factors in equa- tion (41.) gives— [2 Ж (42) 4

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SIMPLICITY IN CONSTRUCTION. I23 This result is just six times that of equation (32.); we, there- fore, have— Rule XXXV II.—Proceed as directed іп Rule XXVII. ; multiply the first quotient thereby obtained by 6; then in the subsequent directions use this multiplied quotient in- stead of the said first quotient, to obtain the required breadth and depth. Example.—W hat should be the size of a spruce lever то feet long, sustaining 26662 pounds, uniformly distributed over its length, with a deflection of 1 inch? Proceeding by Rule XXVII., the first quotient obtained is 761-905 ; this multiplied by 6 gives 4571-43, the multiplied quotient which is to be used in place of the said first quotient. Now, to obtain the breadth, the depth being fixed, say, at 10; 4571-43 divided by (cube of 10 =) 1000, the quotient, 4-57, is the required breadth. But if the breadth be fixed, say, at 4, then, to obtain the depth, 4571-43 divided by 4 gives 1142-86, the cube root of which is 10-46, the required depth. Again, if the breadth and depth are to be in proportion, say; as o-6 to 1-0, then 4571-43 divided by 0-6 gives 7619-05, the square root of which is 87.27, of which the square root is 9-343, the required depth in inches ; and 9-343 x 0-6 equals 5-6, the required breadth in inches; or, the lever may be, say, 5$ x 9} inches. CONSTRUCTION IN GENERAL. 150.— Construction: Object Clearly Defined. — In the various parts of timber construction, known as floors, par- titions, roofs, bridges, etc., each has a specific object, and in all designs for such constructions this object should be kept clearly in view, the various parts being so disposed as to serve the design with the least quantity of material. The simplest form is the best, not only because it is the most economical, but for many other reasons. The great number of joints, in a complex design, render the construction liable to derangement by multiplied compressions, shrinkage, and, in consequence, highly increased oblique strains ; by which its stability and durability are greatly lessened.

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124 CONSTRUCTION. FLOORS. [51.—Floors Described.—Floors are most generally con. structed szzg/c, that is, simply a series of parallel beams, each Fic. 39. spanning the width of the building, as seen at Fig. 39. Ос: casionally floors are constructed double, as at Fig. 40; and sometimes framed, as at Fig. 41; but these methods аге:

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RULES APPLIED TO FLOORS. 125 seldom practised, inasmuch as either of these requires more timber than the single floor. Where lathing and plastering is attached to the floor-beams to form a ceiling. below, the springing of the beams, by customary use, is liable.to' crack the plastering. Toobviate this in good dwellings, the double ` and framed floors have been resorted to, but more in former times than now, as the cross-furring (a series of narrow strips of board or plank nailed transversely to the underside of the beams to receive the lathing for the plastering) serves a like purpose very nearly as well. - [52.—Floor-Beams.— The size of floor-beams can be as- certained by the preceding rules for the s¢¢f/ness of materials. These rules give the required dimensions for the various kinds of material in common use. The rules may be some- what abridged for ordinary use, if some of the quantities represented in the formula be made constant within certain limits. For example, if the load per foot superficial upon the floor be fixed, and the deflection, then these, together with the constant represented by F, may be reduced to one

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126 CONSTRUCTION. constant. For dwellings, the load per foot may be taken at 7o pounds, the weight proper to be allowed for a crowd of people on their feet. (Transverse Strains, Art. 114) To this add 20 for the weight of the material of which the floor is composed, and the sum, 90, is the value of f, or the weight per foot superficial for dwellings. Then c f l= U (Art. 130). The rate of deflection allowable for this load may be fixed at 0-03 inch per foot of the length, or ô = 0-03 /. Substitut- ing these values in equation (35.), we obtain— es NA goc | _ 1875 67% 1:6 F030 Lix ЕНІ «Mie > or— D sm — re (43.) Putting 7 to represent 2 , we have— b qucm. (44-) А | 1875 Р i Now, by reducing—7——-, for the six woods in common use, the value of 7 for each is found as follows: Georgia Pine |. ЖИ К ОБ ОО. J "де «xocust ...... JEEP “М”... y" 0-37 White Oak... ла. ТЕ. J = оф Spruce .„;..... 0 ЖЖ. LO 7 = ogy Mite. Pines... . t ey eee J = 0-65 "а conce j = 0:67 Equation (44.) is a rule for the floor-beams of dwellings ; it may be used also to obtain the dimensions of beams for stores for all ordinary business: for it will require from 3 to 5 times the weight used in this rule, or from 200 to 400 (average 300) pounds to increase the deflection to the limit of elasticity in beams of fhe usual depths and lengths. For light stores, therefore, loaded, say, to 150 pounds per foot, the beams would be safe, but the deflection would be in-

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CONSTANTS FOR USE IN TÍIE RULES. 127 creascd to 0-06 per foot. When SO EN a Қ 25 this. would not be objectionable to the eye, then this rule (44). will serve for the beams of light stores. But for first-class stores, taking the rate of deflection at -o4 per foot, and ' fixing the weight per superficial foot at 275 pounds, includ- _ ing the weight of the material of which the floor is con- | structed, and letting £ represent the constant, then— . { LL DIC | (ag) and for— WV CLOUD E а... ра UY n Londi. — ss ME = 50285 "nass T TT ——— Да ЛЗ ТЕСЕ И. RR k = lq8 BOSNIA. deeper greet магт T. =з, | REuntc ICE k= Wig 153,—Floor-Beams for Dwellings.-—To find the dimen- sions of floor-beams for dwellings, when the rate of deflection” is 0-03 inch per foot, or for ordinary stores when the load is about 150 pounds per foot, and the deflection caused by this , weight is within the limits of the elasticity of the material, we have the following rule: | Rule XXXVIII.—Multiply the cube of the ЕТ” by the distance apart between the beams (from centres), both in (веб, and multiply the product by the value of 7 (2472. 7152) for the material of the beam; and the product will equal thc product of the breadth into the cube of the depth. Now, to find the breadth, divide this product by the cube of the depth in inches, and the quotient will be the breadth in ' inches. But if the depth is sought, divide the said product | by the breadth in inches, and the cube root of the quotient will be the depth in inches; or if the breadth and depth are | to be in proportion ав ғ is to unity, 7 representing any re- | quired decimal, then divide the aforesaid product by the value of 7, and extract the square root of the quotient, and | the square root of this square root will be the depth re- quired in inches, and the depth multiplied by the value of 7 will be the breadth in inches.

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128 CONSTRUCTION. Example. —In a dwelling Ог ordinary stOre, what must be the breadth of the beams, when placed 15 inches from centres, to support a floor covering a span of 16 feet, the depth being 11 inches, the beams of white oak? By the rule, 4096, the cube of the length, by 11, the distance from centres, and by о.б, the value of 7 s white oak, equals 3072. 'This divided by 1331, the cube of the depth, equals 2-31 inches, or 245; inches, the required breadth. But if, in- stead of the breadth, the depth be required, the breadth being fixed at 3 inches, then the product, 3072, as above, di- vided by 3, the breadth, equals 1024; the cube root of this is 10-08, ог, say, 10 inches nearly. But if the breadth and depth are to be in proportion, say, as 0-3 to 1-0, then the aforesaid product, 3072, divided by 0-3, the value of rv, equals 10240, the square root of which is 101-2, and the square root of this is 10-06, the required depth. This multiplied by о-3, the value of 7, equals 3.02, the re- quired breadth; the beam is therefore {о be, say, 3 x 10 inches. (54.—Floor-Beams for First-Class Stores.—T o find the breadth and depth of the beams for a floor of a first-class store sufficient to sustain 250 pounds per foot superficial (exclusive of the weight of the material in the floor), with a deflection of 0-04 inch per foot of the length, we have— Rule XXXIX.—The same as XXXVIII., with the ex- ception that the value of £ (Art. 152) is to bé used instead of the value of 7. Example.—YThe beams of the floor of a first-class store are to be of Georgia pine, with a clear bearing between the walls of 18 feet, and placed 14 inches from centres: what must be the breadth when the depth is 11 inches? By the rule, 5832, the cube of the length, and 11, the distance from centres, and 0-73, the value of £ for Georgia pine, all multi- plied together equal 4066-02; and this product divided by 1331, the cube of the depth, equals 3.732, the required breadth, or 3$ inches. But if, instead of the breadth, the depth be required: what must be the depth when the breadth is з inches:

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DISTANCE APART OF FLOOR-BEAMS. 129 The said product, 4966-92, divided by 3, the breadth, equals 1655-64, and the cube root of this, 11.83, or, say, 12 inches, is the depth required. But if the breadth and depth are to be in a given pro- portion, say 0:35 to т.о, the 4966-92 aforesaid divided by O-35, the value of 7, equals 14191, the square root of which is 119-13, and the square root of this square root is 10-01, Or, Say, ІІ inches, the required depth. And 10-91 multiplied by 0-35, the value of 7, equals 3-82, the required breadth, say 3% inches. 155.—Floor-Beams: Distance from Centres. — It is sometimes desirable, when the breadth and depth of the beams are fixed, or when the beams have been sawed and are now ready for use, to know the distance from cen- tres at which such beams should be placed in order that the floor be sufficiently stiff. By a transposition of the factors in equation (44.), we obtain— .. іссе 778 5 (46.) In like manner, equation (45.) produces— 04% C= are (47.) These, in words at length, are as follows: | Rule XL.—Multiply the cube of the depth by the breadth, both in inches, and divide the product by the cube of the length in feet multiplied by the value of 7, for dwellings and for ordinary stores, or by 2, for first-class stores, and the quotient will be the distance apart from centres in feet. Example—A span of 17 feet, in a dwelling, is to be cov- ered by white-pine beams 3x12 inches: at what distance apart from centres should they be placed? Ву the rule, 1728, the cube of the depth, multiplied by 3, the breadth, equals 5184. The cube of 17 is 4913; this by 0-65, the value of 7 for white pine, equals 3193-45. Тһе aforesaid 5184 divided by this 3193-45 equals 1-6233 feet, or, say, 20 inches.

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130 CONSTRUCTION. `- (56. — Framed Openings for Chimneys and Stairs. — ‘ Where chimneys, flues, stairs, etc., occur to interrupt the bearing, the beams are framed into a piece, 0 (27. 42), called a header. The beams, а а, into which the header is framed are called trimmers ог carriage-beams. These framed beams require to be made thicker than the common beams. Тһе? header must be strong enough to sustain one half of the ` weight that is sustained upon the Zaz-beams, с с (the wall at the opposite end or another header there sustaining the other | half) and the trimmers must each sustain one half of the : weight sustained by the header in addition to the weight it supports as acommon beam. It is usual in practice to make these framed beams one inch thicker than the common beams for dwellings, and two inches thicker for heavy stores. This . practice in ordinary cases answers very well, but in extreme cases these dimensions are not proper. Rules applicable generally must be deduced from the conditions of the case— the load to be sustained and the strength of the material. 157.—Breadth of Headers, —The load sustained by a header is equally distributed, and is equal to the superficial area of the floor supported by the header multiplied by the load on every superficial foot of the floor. This is equal to the length of the header multiplied by half the length of the tail-beams, and by the load per superficial foot. Putting g`

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DIMENSIONS OF HEADERS. aisg 'for the length of the header, z for the length of the tail- ' beams, and f for the load per superficial foot; U, the uni- ‘formly distributed load carried by the header, will equal 4 fng. By substituting for U, in equation (35.), this value of it, we obtain— Ww neis (= 1-6 Е6” The symbols g апа / here both represent the same thing, the length of the header; combining these, and for ó putting its value zz, we obtain— gp INIT af U*2Fr P To allow for the weakening of the header by the mor- tices for the tail-beams (which should be cut as near the middle of the depth of the header as practicable), the depth should be taken at, say, one inch less than the actual depth. With this modification, we obtain— wes 8 (аа “ ) If f be taken at go, and ғас 0-03, we have, by reducing— ‚кд оЗ b= ee те | (49.) which is а rule for the breadth of headers for dwellings and for ordinary stores. This, in words, is as follows: РЕГ Rule XLI.—Multiply 937:5 times the length of the tail- beams by the cube of the'length of the header, both in feet. The product divided by the cube of one less than the depth multiplied by the value of F, Table HI., will equal the breadth of the header in inches for dwellings or ordinary stores. | Example-—A header of white pine, for a dwelling, 1s 10 feet long, and sustains tail-beams 20 feet long; its depth 1s 12 inches: what must be its breadth? Ву the rule, 937-5 x 20x 10818750000. This divided by (12- tx 2900—

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132 CONSTRUCTION. 3850900, equals 4-858, say 5 inches, the required breadth. For first-class stores, f should be taken at 275, and r at 0-04. With these values the constants in equation (48.) reduce to 2148-4375, or, say, 2150. This gives— 2150 2 g? im Du (50.) a rule for the breadth of a header for first-class stores. It is the same as that for dwellings, except that the constant 2150 is to be used in place of 937-5. Taking the same ex- ample, and using the constant 2150 instead of 937-5, we obtain 11-14 аѕ the required breadth of the header for a first- class store. Modifying the question by using Georgia pine instead of white pine, we obtain 5-476 as the required thick- ness, say 5% inches. (58.—Breadth of Carriage-Beams.—4A carriage-beam or trimmer, in addition to its load as a common beam, carries one half of the load on the header, which, as has been seen in the last article, is equal to one half of the superficial area of the floor supported by the tail-beams multiplied by the weight per superficial foot of the load upon the floor ; therefore, when the length of the header in feet is repre- sented by g, and the length of the tail-beams by z, w equals “хх equals 1 fg z.* For a load not at middle, we have (25.)— pq 4Wamn ай” m * The load from the header, instead of being + / гл, is, more accurately, 3/тл(в-- с): because the surface of floor carried by the header is only that which occurs between the surfaces carried by the carriage-beams, each of which carries so much of the floor as extends half way to the first tail-beam Р Я с A Е from it, or the distance T therefore, the width of the surface carried equals the length of the header less (2 x 2 =) c,org—c. When, however, it is con- sidered that the carriage-beam is liable to receive some weight from a stairs or other article in the well-hole, the small additional load above referred to is not only not objectionable, but is really quite necessary to be included in the calculation. .

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THICKNESS OF CARRIAGE-BEAMS. 133 This is a rule based upon resistance to rupture. By substi- қ "AM. tuting for a, the factor of safety, Fa; its value in terms of resistance to flexure (Transverse Strains, (154.)), we have— ЖЕ 4WBimn 4Wmn, Blk dr 72722 4Wmn l оге де eM ә ^ Fro In this expression, И/ is a concentrated weight at the dis- tances 77 and > from the two ends of the beam. Taking the load upon a carriage-beam due to the load from the header, as above found, and substituting it for W, we obtain— „єх 41х47, Axifgnam — 3 Тәж Teal ar in poll a This is the expression required for the concentrated load. To this is to be added the uniformly distributed load upon the carriage-beam; this is given in equation (35.). Substi- tuting for U of this equation its value, fcl, gives— yc EE 47 == L6Fó Ёғ. Combining these two equations, we have for the total load— figmn'c&cI" D (51.) Ша ке 770 If, in this equation, / be taken at go, and 7 at 0-03, these reduce to 3000 ; therefore, with this value of 7 we have— 3000 (gm u* - $c ) ы. = a (52-) This rule for the breadth of carriage-beams with one header, for dwellings and for ordinary stores, is put in words as follows :

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134 CONSTRUCTION, 2 Rule XLII.—Multiply the length of the framed opening by its breadth, and by the square of the length of the tail- beams; to this product add 5 of the cube of site length into the distance of the common aes from centres—all in feet ; divide 3000 times the sum by the cube of the depth in inches multiplied by the value of Ffor the material of the beam, in Table ПІ, and the quotient will be the breadth in inches. Example.—In a tier of 3 хто inch beams, placed 14 inches from centres, what should be the breadth of a Georgia-pinc carriage-beam 20 feet long, carrying a header 12 feet long, having tail-beams 15 feet long? Неге the framed opening is 5.x1241eet. Therefore,accordingidto£he,rüle, 12x5 x 15°= 13500; to which add (5 x 20? x 44 =) 58334; the sum is 193338, and this by 3000 — 58000000. The value of. F for Georgia pine, in Table IIL, is 5900; the cube of the depth is 1000; the product of these two is 5900000; therefore, dividing the above 58000000 by 5900000 gives a quotient of 9.83, the required breadth in inches. ЇЇ, in equation (51.), f be taken at 275, and ғ at 0-04, then = becomes 6875, and the ‘equation. becomes— à Ae (53-) 6875 (gm nit с a қ ғ... ғ a rule for the breadth of carriage-beams for first-class stores ; the same as that for dwellings, except that the constant is 6875 instead of 3000. (59._Breadth of Carriage-Beams Carrying Two Sets of Tai}-Beams.—A rule for this is the same as that for a car- rage-beam carrying one set of tail-beams, if to it there be added the effect of the second set of tail-beams. | Equation (51.) with the addition named becomes— is eve qu in which z is the length of one set of tail-beams, and s the length of the other set; and m+n = /.

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CARRIAGE-BEAMS WITH TWO HEADERS. -135 { If f be taken at 90, and > at 0-03, these two reduce to` 3000, and we have— е b= ee (55.) a rule for the breadth of a tám carrying two sets ' of headers, for dwellings and for ordinary stores. It may be stated in words as follows: Rule XLIII.—Multiply the length of the longer. set of tail-beams by the difference between this length and the length of the carriage-beam, and to the тте add the square of the length of the shorter set of tail-beams ; mul- tiply the sum by the length of the longer set of. tail-beams, and by the length of the header; to this product add 5 of the product of the cube of the length of the carriage- beam into the distance apart from centres of the common beams; multiply this sum by 3000; divide this product by the product of the cube of the depth in inches into the value of F for the material of the carriage-beam, in Table II., and the quotient will be the required breadth. Example.—1n a tier of 3 x 12 inch beams, placed 14 inches from centres, what should be the breadth of a spruce car- riage-beam 20 feet long in the clear of the bearings, carry- ing two sets of tail-beams, one of them о feet long, the other 5 feet ; the headers being 15 feet long? The difference between the longer set of tail-beams and the Gasinqé-t dm is(20— 9 =) 11 feet. Therefore, by the rule, 9x 11-5! = 124; then (124x9x 15 =) 16740 - (5 x 20° x 1$ =) 5833$ = 225731; then 22573 х 3000 = 67720000. Nom "iie value of F for spruce, Table III., is 3500; this by 12°, the cube of the depth, equals 6048000; by this dividing the aforesaid 67720000, we obtain a quotient of 11-197, the required breadth of the carriage-beam. If, in equation (54.), f be taken at 275, and ғ at 0-04, these reduce to 6875, and we obtain— — 6875 [gn (m zu la tal 6. pacino 7 6) а rule for the breadth of carriage-beams carrying two sets

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136 CONSTRUCTION. of tail-beams, in the floors of first-class stores. .This is like the rule for dwellings, except that the constant is 6875 in- stead of 3000. 160. — Breadth o£ Carriage -Beam with Well-Hole at Middle.— When the framed opening between the two sets of tail-beams occurs at the middle, or when the lengths of the ` two sets of tail-beams are equal, then equation ($4.) reduces to 215 2 «АРА ER 52.2), (57) and if f be taken at 9o, and ғ at 0-03, these reduce to 3000, and we have— o 5 of? | pm SOOM (Pw хал) (58.) а rule for the breadth of a carriage-beam carrying two sets of tail-beams of equal length, in the floor of a dwelling or of ап ordinary storey; and which in words is as follows: Rule XLIV.—Multiply the length of the header by the square of the length of the tail-beams, and to the product add 5 of the product of the square of the length of the car- riage-beam by the distance apart from centres of the com- mon beams; multiply the sum by 3000 times the length of the carriage-beam; divide the product by the product of the cube of the depth into the value of F for the material of the carriage-beam, in Table II., and the quotient will be the required breadth. Example.—ln a tier of 3 x 12 inch beams, placed 12 inches from centres, what must be the thickness of a hemlock car- riage-beam 20 feet long, carrying two sets of tail-beams, each 8 feet long, with headers ro feet long? By the rule, тох 8+5 x I x 207 = 890; 890 x 3000 x 20 = 53400000. Now, the value of 7, in Table III., for hemlock is 2800; this by the cube of the depth, 1728, equals 4838400; by this dividing the former product, 53400000, and the quotient, 11. 0367, i: is the required breadth of the carriage-beam.

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CROSS-BRIDGING. TA If, in equation (57.), f be taken at 275, and > at 0.04, these will reduce to 6875, and we shall have— _ 6875 f(g f +¥c 1’) b= i э эы ж. ( 59.) а result the same as in equation (58.), except that the constant is 6875 instead of 3000. Equation (59.) is a rule for the breadth of carriage-beams carrying two sets of tail-beams of equal length, in the floor of a first-class store. In words at length, it is the same as Rule XLIV., except that the con- stant 6875 is to be used in place of 3000. A 161.—Cross-Bridging, or Herring-Bone Bridgirg.—The diagonal struts set between floor-beams, as in Fig. 43, are known as cross-bridging, or herring- bone bridging. By connecting the beams thus at intervals, say, of from 5 to 8 feet, the stiffness of the floor is greatly increased. The absolute strength of a tier of beams to resist a weight uniformly distributed over the whole tier is augmented but lit- tle by cross-bridging ; but the power of any one beam in the tier to re- sist a concentrated load upon it, as a heavy article of fur- , hiture or an iron safe, is greatly increased by the cross- bridging; for this device, by connecting the loaded beam with the adjacent beams on each side, causes these beams to assist in carrying the load. То secure the full benefit of the diagonal struts, it is very important that the beams be well secured from separating laterally, by having strips, such as cross-furring, firmly nailed to the under edges of the beams. The tie thus made, together with that of the floor-plank on the top edges, will prevent the thrust of the struts from sep- arating the beams. TIG. 43. 162.—Bridging: Value to Resist Concentrated Loads.— A rule for determining the additional load which any one beam connected by bridging will be capable of sustaining, by the assistance derived from the other beams, through the

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138 . CONSTRUCTION. bridging, may be found in Chapter XVIII., Transverse Strains. This rule may be stated thus: | А Я = L zu T 2° + 3°+ 4"+ ete.) ; (бо.) іп which A is the increased resistance, equal to the addi- tional load which may be put upon the loaded beam; cis the distance from centres in feet at which the beams in the tier are placed: fis the load in pounds per superficial foot upon the floor; Zis the length of the beams in feet; and 4 is the depth of the beams in inches. The squares within the bracket are to be extended to as many places as there are beams on each side which contribute assistance through the bridging. The rule given in the work referred to, for ascer- . taining the number of spaces between the beams, is— ; (61.) nN or, the depth of the beam in inches divided by the square of the distance from centres, in teet, at which the beams are placed will give the number of spaces between the beams which contribute on each side in sustaining the concentrated load. The nearest whole number, minus unity, will equal the required number of beams. The value of c for beams in floors of dwellings is given in equation (46.), and for those in first-class stores in equation (47). By a modification of equation (34.), putting c f/ for U, we have— Ci. -- ES апа-- C= E p 3 (62.) or— c= Е E - (63.) These equations give general rules for the value of c. е

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INCREASED LOAD BY CROSS-BRIDGING. 139. Now, the rule, in words at length, for the resistance offered by the adjoining beams to a weight concentrated upon one of the beams sustained by cross-bridging to the others, is— Rule XLV.—Divide the depth of the beam in inches by the square of the distance: apart from centres in feet at which the floor-beams are placed ; from the quotient deduct unity, and call the wholé number nearest to the remainder the First Result. Take the sum of the squares of the con- secutive numbers from unity to as many places as shall equal the above first result; multiply this sum by 5 times the - length in feet, by the load per foot superficial upon the floor, and by the fifth power of the distance apart from centres in feet at which the beams are placed; divide the product by 4 times the square of the depth in inches, and the quotient | will be the weight in pounds required. . Example. — Іп a tier of 3 x 12 inch floor-beams 20 feet long, placed in a dwelling 16 inches from centres and well bridged, what load may be uniformly distributed upon one of the beams, additional to the load which that beam is capable of sustaining safely when unassisted by bridging? Here, according to the rule, 12 divided by (14+1$ =) 1} equals Ge: б йе — 52, the nearest whole number to which is 6, the first ҮП” The sum of the square of the first 6 numbers equals (1-2% + 3* -- 4* 4-5! +6 =) 1 +4+9 + 16 25 +36 — 9r. Therefore, 91 x 5 x 20x go x (4) = 3451266.* Тһе square of the depth (12x12 =) 144 x4 = 576; by this dividing the above 3451266, we have the quotient 5001.78, say 5992 peunds, the required weight. This is the additional load which may be placed upon the beam. At 90 pounds per superficial foot, the common load on each beam, we have * The value of c, 16 inches, equals $ feet. The fifth power of this, or (3% is obtained by involving both numerator and denominator to the fifth power, and dividing the fifth power of the former by the fifth power of the latter; for 5 Gy = ШЕ. For the numerator we have 4x 4x 4x4x4=1024, and for the de- 3 nominator 3x3x3x3x3—243. Тһе former divided by the latter gives as a quotient 4-214, the value of ($4). The process of involving a number to a high power, or the reverse operation of extracting high roots, may be performed by logarithms with great facility. (See 477. 427.)

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140 CONSTRUCTION. до x 20 x $ = 2400 as the common load. To this add 5002, the load sustained through the bridging by the other beams, and the sum, 8392 pounds, will be the total load which may be safely sustained, uniformly distributed, upon one beam— nearly 34 times the common load. 163.—Girders,—W hen the distance between the walls of а building is greater than that which would be the limit for the length of ordinary single beams, it becomes requisite to introduce one or.more additional supports. Where sup- ports are needed for a floor and partitions are not desirable, it is usual to use a large piece of timber.called a girder, sus- tained by posts set at intervals of from 8to 15 feet; or, when posts are objectionable, a framed construction called a framed girder (Art. 196); or an iron box called a tubular iron girder (477. 182). When a simple timber girder is used it is advisable, if it be large, to divide it vertically from end to end and reverse the two pieces, exposing the heart of the timber to the air in order that it may dry quickly, and also to detect decay at the heart. "When the halves are bolted together, thin slips of wood should be inserted between them at the several points at which they are bolted, in order to leave sufficient space for the air to circulate freely in the space thus formed between them. This tends to prevent decay, which will be found first at such parts as are not exactly tight, nor yet far enough apart to permit the escape of moisture. When girders are required for a long bear- ing, it is usual to truss them; that is, to insert between the halves two pieces of oak which are inclined towards each other, and which meet at the centre of the length of the girder like the rafters of a roof-truss, though nearly if not quite concealed within the girder. This and many similar methods, though extensively practised, are generally worse than useless; since it has been ascertained that, in nearly all such cases, the operation has positively weakened the girder. A girder may be strengthened by mechanical contrivance, when its depth is required to be greater than any one piece of timber will allow. Fig. 44 shows a very simple yet invalu- able method of doing this. The two pieces of which the gir-.

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CONSTRUCTION OF GIRDERS. 141 der is composed are bolted or pinned together, having keys inserted between to prevent the pieces from sliding. Тһе keys should be of hard wood, well seasoned. The two pieces should be. about equal in depth, in order that the joint between them may be in the neutral line. (See Arts. 120, 121.) The thickness of the keys should be about half their breadth, and the amount of their united thickness should be equal to a trifle over the depth and one third of the depth of the girder. Instead of bolts or pins, iron hoops. are sometimes used ; and when they can be procured, they are far preferable. In this case, the girder is diminished at the ends, and the hoops driven from each end towards the middle. А girder may be spliced if timber of a sufficient length cannot be obtained ; though not at or near the mid- Fic. 44. dle, if it can be avoided. (See. Arz. 87.) Girders should rest from 9 to 12 inches on each wall, and a space should be left for the air to circulate around the ends, that the damp- ness may evaporate. (64.. —Girders: Dimensions.—The size of a girder, for any special case, may be determined by equations (21.), (22.), | (25.), (27.), and (28.), to resist rupture ; and to resist deflection, by equations (32.) and (35.). For girders in dwellings, equa- tion (44.) may be used. Іп this case, the value of c 1s to be taken equal to the width of floor supported by the girder, which is equal to the sum of the distances half way to the . wall or next bearing on each side. When there is but one

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142 CONSTRUCTION. girder between the two walls, the value of c is equal to half the distance between the walls. The rule for girders for dwellings, іп words, is— — Rule XLVI.—Multiply the cube of the length of the gir- der by the sum of the distances from the girder half way to the next bearing on each side, and by the value of 7 for the material of the girder, in 477. 152; the product will equal -the product of the breadth of the girder into the cube of the depth. To obtain the breadth, divide this product by the cube of the depth; the quotient will be the breadth. To obtain the depth, divide the said product by the breadth ; the cube root of the quotient will be the depth. If the breadth and depth are to be in a given proportion, say as r г i-o, then divide the aforesaid quotient by the value of ғ; take the square root of the quotient; then the square root of this square root will be the depth, and the depth multi- plied by the value of 7 will be the breadth. Example.—In the floor of a dwelling, what should be the size of а Georgia-pine girder 14 feet long between posts, placed at 1o feet from one wall and 20 feet from the other? The value of c here is. £9. - 20. = $9 — 15. The value of 7 for Georgia pine (Art. 152) is 0-32. Ву the rule, 14^ x 15 x 0-32 = I3171:2. Now, to find the breadth when the depth 15 12 inches; 13171-2 divided by the cube of 12, or by 1728, gives a quotient of 7.622, or 7$, the required breadth. Again, to find the depth, when the breadth is 8 inches: 13171-2 divided by 8 gives 1646-4, the cube root of which is 11-808, or, say, r1$ inches, the required depth. But if neither breadth nor depth have been previously determined, except as to their proportion, say as 0-7 to 1-0, then 13171:2 divided by о-7 gives 18816, of which the square root is 137-171, and of this the square root is 11-712, or, say, 112 inches, the required depth. For the breadth, we have 11.712 by 0.7 equals 8-198, or, say, 8}, the required breadth. Thus the girder is required to be 75 х 12, 8 x 114, or 81 х11# inches. This example is one in a dwelling or ordinary store; for first-class stores the rule for girders is the same as the last, except that the value of £ is to be taken instead of 7,.in Art. 152. :

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FIRE-PROOF TIMBER FLOORS. 143 165.—Solid Timber Floors.—Floors constructed with rolled-iron beams and brick arches are proof against fire only to a limited degree; for experience has shown that the heat, in an extensive conflagration, is sufficiently intense to deprive the iron of its rigidity, and consequently of its strength. Singular as it may seem, it is nevertheless true that wood, under certain circumstances, has a greater fire- resisting quality than iron. Floors of timber constructed, as is usual, with the beams set apart, have but little power to resist fire, but if the spaces between the beams be filled up solid with other beams, which thus close the openings against the passage of the flames, and the under surface be coated with plastering mortar containing a large. portion of plaster of Paris, and finished smooth, then this wooden floor will resist the action of fire longer than a floor of iron beams and brick arches. The wooden beams should be se- cured to each other by dowels or spikes. 166. Dwellings and Assem- bly-Rooms.—From Transverse Strains, Art. 702, we have— _ (82+ y d)/^ "0-576 F [ which may be modified so as to take this form: mE | ee ny a (64.) 0:576 F which is a rule for the depth or thickness of solid timber floors for dwellings, assembly-rooms, or office buildings, and in which y and / are constants depending upon the mate- rial; thus, for— Cough ЕШЕ................ y = aud 4 — 0-314 Cou mec A a gis HO 305 oe за TTC y = 24, “ л = 0-389 КЕШСЕ, MES. y == = 0-39 The rule may be stated in words thus: Rule XLVII.—Multiply the length by the value of 2

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144. CONSTRUCTION. and by the value of Ж, as above given; to the product add 82; multiply the sum by the cube of the length ; divide this product by 0-576 times the value of ~F, in Table III.; then the cube root of the quotient will. be the required depth i in inches. Example—What depth is required for a solid Georgia- pine floor to cover a span of 20 feet? For Georgia pine Е = 5900; у, ав above given, equals 4, and 7 equals 0-314; therefore, by the rule— pat E (82 +4 0-314 x 20) 4 314x20)7 — _ 4/ 856960 _„ O-576x5900 3398-4 or, the depth required is, say, 6-32 or 6,5, inches. 167. Class Stores.—The equation given for first-class stores, in Transverse Strains, (263 + y d) /* З. bet .768 F which may be changed to this form: \ / (263 +y k L) P? =) ое а sq in which y is as before, and £ for— Geongla Pine equals "Wee. c... о-4 Sbnmase сопа... РЕ... 0:472 севе ЕМЕ. ИЕ ЖКК... 4%. O- 502 Dremlogk'*egeae. "T 7 aae ж”... O- 506 This rule may be put in words the same as Rulé XLVII., except as to the constants, which require that 263 beused in place of 82, that 2 be used in place of %, and that 0-768 be used in place of 0-576. Table XXI. of Transverse Strains contains the results of computation showing the depths of solid timber floors for dwellings and assembly-rooms and for first-class stores, in floors of spans varying from 8 to 30 feet, and for the four kinds of timber before named. j

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IRON FLOOR-BEAMS. 145 168. — Rolled - Iron..Beams.—The dimensions of iron beams, whether wrought or cast, are to be ascertained by the rules already given, when the beams are of rectangular form in their cross-section; these rules are applicable alike to wood and iron (477. 93), and may be used for any mate- rial, provided the constant appropriate to the given mate- rial be used. But when the form of cross- . section 15 such as that which is usual for rolled-iron beams (Fig. 45) the rules _ need modifying. Without’ attempting 6 to explain these modifications (referring for this to Transverse Strains, Art. 457 and following article), it may Бе re- marked that the elements of resistance to flexure in a beam constitute what is , termed the Moment of Inertia. This, in ™ a beam of rectangular cross-section, is equal to 44 of the breadth into the cube of the depth; or— T= dy bd’. (66.) Fic. 45. This would be appropriate to rolled-iron beams if the hollow on each side were filled with metal, so as to complete the form of cross-section into a rectangle. The proper ex- pression for them may be obtained by taking first the moment for the beam as if it were a solid rectangle, and from this deducting the moment for the part which on each side is wanting, or for the rectangles of the hollows. Іп accordance with this view of the case, we have— I= т (bd — 0,42); (67) in which 0 is the breadth of the beam or width of the flanges; д, is the breadth of the two hollows, or is equal tod less the thickness of the web or stem; dis the depth includ- ing top and bottom flanges; and Z, is the depth in the clear between the top and bottom flanges. Now, if equation (32.) be divided by 12, we shall have—

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146 CONSTRUCTION. and since +, 2 4% represents the moment of inertia, we have— И/ 2 Ба туы (68.) This gives the value of / for а beam of апу form in cross- section loaded at the middle. Ву this equation the values of 7 have been computed for rolled-iron beams of many sizes, and the results recorded in Table XVII., Transverse Strains. A few of these are included in Table IV., as follows: TABLE IV.—ROLLED-IRON BEAMS. NAME. Depth. Esc J | NAME. Depth. | i cr] = É——— в | 5 ! -— om 2 Trenton.....| 4 30 2:84. ||Buffalo......| 9 90 109.117 Paterson....| 5 30 12-082 |Рһсепіх....! 9 150 190.63 Phoenix... | 5 36 14:317 || Buffalo. ....| 10} 90 151:436. Trenton....| ‘6 40 23:501. |Вийїа1о......| Tol 105 175-645 Phenix....| 7 Be 42:43 |Trenton..... тоў I35 | 241-473 Trenton....| 7 60 46-012 ||Buffalo..... 12} 125 286-019 Buffalo...... 8 65 64-526 ||Paterson....| 12} 125 292-05 Paterson....| 8 80 84-735 |Paterson.....| 124 170 398.935 Dhosnix.....| 9 70 2-207 ||Buffalo......| 124 180 418-945 Рһсепіх....| 0 84 107-793 |[Trenton.....| 15% 150 528.22 | | | 169. — Rolled - Iron Beams: Dimensions; Weight ai Middie.—If, in equation (68.), there be substituted for / its value for wrought iron, as in Table III., we shall have— — W 7 ^ 12x 62000 0’ or — W 28 | = eee (69.) This is a rule by which to ascertain the size of a rolled-iron beam to sustain a given weight .at middle with a given de- flection, and, in words at length, is as follows: Rule XLVIII.—Multiply the weight in pounds by the cube of the length in feet; divide the product by 744000 times the deflection in inches, and the quotient will be the

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DEFLECTION OF IRON BEAMS. 147 moment of inertia of the required beam, and may be found, or the next nearest number, in Table IV. in column headed J.. Opposite to the number thus found, to the left, will be found the name,'depth, and weight per yard of the required beam. Example.—W hich of the beams of Table IV. would be proper to carry 10,000 pounds at the middle with a deflection of one inch, the length between bearings being 20 feet? Here we have, substituting for the symbols their values— W 18 тоооох 20 80000000. = 7440000 — 744000x 1 .— 744000 110715203 or, the moment of inertia of the required beam is 107-527, the nearest to which, in the table, is 107-793, pertaining to the Phoenix g-inch, 84-pound beam. This, then, is the required beam. = |70.—Rolled-Iron Beams: Deflection when Weight is at Middle.—Dy a transposition of symbols in equation (69.), we have— po eee T 744000 Г (7o) > ora rule for the deflection of rolled-iron beams when the weight is at the middle. This, in words, is— Rule XLIX.—Multiply the weight in pounds by the cube of the length in feet; divide the product by 744000 times the value of 7 for the given beam, and the quotient will be the required deflection in inches. Example.—What will be the deflection of a Phoenix 9- inch, 70-pound beam 20 feet long, loaded at the middle with 7500 pounds? The value of 7 for this beam, in Table IV., is 92-207; therefore, substituting for the symbols their val- ues, and proceeding by the rule, we have— E = „(б . 7800x20" _ _ 087461; 744000 / 744000 x 92-207 or, the deflection will be, say, 2 of an inch.

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148 CONSTRUCTION. !71.—RoHed-Iron Beams: Weight when at Middle.— A transposition of factors in equation (7о.) gives— 744000 / 0 4 б f PUE e (71.) This is a rule for the weight at middlc, and, in words, is— Rule L.—Multiply 744000 times the value of / by the deflection in inches; divide the product by the cube of the length, and the quotient will be the required weight in pounds. Example-—What weight at the middle of a Buffalo g-inch, go-pound beam will deflect it one inch, the length between bearings being 20 feet? The value of / for this» beam, in Table IV., is 109-117; therefore— W = 744000 Г 8 __ 744000 x "eU ПО т = 75 | 29 — = ТО 55. or, the required weight 1s, say, 10,148 pounds. 172.— Rolled-Iron Beams: Weight at any Point.—The equation for а load at any point is ( Transverse Strains, Art. 485)— 186000 16 | {элл ’ (72.) in which mand > represent the two parts in feet into which the point where the load rests divides the length. This, in words, 15 as follows: Rule LI.—Multiply 186000 times the value of / by the deflection in inches; divide the product by the product of the length into the rectangle formed by the two parts into .which the point where the load rests divides the length ; the quotient will be the required weight in pounds. Example.— What weight is required, located at то feet from one end, to deflect 14 inches a Paterson 121-inch, r25- pound beam 25 feet long between bearings? The value of Г for this beam, in Table IV., is 292-05; m = 10, and л = 7 — m= 25 —10~ 15; therefore—

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VARYING WEIGHTS ON IRON BEAMS. I49 18600076 186000 x 292-05 x 1:5 pm = -- {жп ‚В 28 X 10X15 == 20729). 52%; or, the required weight 15, бау, 21,730 pounds. 173. КоПей-Ігоп Beams: Dimensions; Weight at any Point.—By transposition of factors in equation (72.), we ob- tain— T= И {шт — 71860000” (73.) This may be expressed in words as follows: Rule LII.—Multiply the weight by the length, and by the rectangle of the two parts into which the point where the weight rests divides the length ; divide the product by 186000 times the deflection, and the quotient will be the value of 7, which (or its next nearest number) may be found in Table IV., opposite to which will be found the required beam. Example.—What beam то feet long will be required to carry 5000 pounds at 3 feet from one end with a deflection of o-4 inch? Неге we have m equal 3, and z equal 7; therefore— = W imn $85000x10x3x7 _ — = 71860006 ^ 18600 xo.4 4l The value of 7 is 14-113, the nearest number to which in the table, is 14-317, the moment of inertia of the Phoenix 5- inch, 36-pound beam ; this, therefore, is the beam required. 174.—Rolled-Iron Beams: Dimensions; Weight Uniform- ly Distributed.—Since 5 (7 = W (Art. 138), equation (69.) may be modified by the substitution of this value of W, when we’ obtain— which reduces to— JL d с-з) ет cest 4-) ^A 1190400 Ô ` К

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150 CONSTRUCTION. a rule for the dimensions of a beam for a uniformly distrib- uted load, which, in words, 15 as follows: Rule LIII.—Multiply the uniformly distributed load by the cube of the length; divide the product by 1190400 times the deflection, and the quotient will be the value of 7, corre- sponding to which, or toits next nearest number will be found in Table IV. the required beam. Example.—What beam то feet long is required to sus- tain an equally distributed load of 14,000 pounds with a de- flection of half an inch? For this we have— 14000 x 10° “N 190400 x 0:5 = 29-52. This is the moment of inertia of the required beam ; nearly. the same as 23-761, in Table IV., the value of Z fora Tren- ton 6-inch, 40-pound beam, which will serve as the re- quired beam. 175.—Roltled-Iron Beams: Deflection; Weight Uniformly Distributed.—A transposition of the factors in equation (74.) gives— ы - 78 ^. 1190400 7” (75-) a rule for the deflection of a uniformly loaded beam, and which may be put in these words, namely : . Rule LIV.—Multiply the uniformly distributed load by the cube of the length ; divide the product by 1190400 times the value of /, Table IV., and the quotient will be the re- . quired deflection. Example.—To what depth will 14,000 pounds, uniformly distributed, deflect a Buffalo 10$-inch, go-pound beam 20 feet long? Тһе value of / for this beam, as per the p^ is 151-436; therefore— 14000 x 20? — 1190400 X 151-436 А” а: 2, or, the required deflection is, say, 5 of an inch.

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IRON FLOOR-BEAMS FOR DWELLINGS. I5I .176.—Rotled - Iron Beams: Weight when Uniformly Distributed.—Equation (75.), by a transposition of factors, gives— 1190400 J ô U = 11994 (76.) p - a rule for the weight uniformly distributed, and which may be worded thus: Rule LV.—Multiply 1190400 times the value of /, Table IV., by the deflection; divide the product by the cube of the length, and the quotient will be the required weight. Example—What weight uniformly distributed upon a Buffalo 104 -іпсһ, 105- pound beam 25 feet long between bearings will deflect it $ of an inch? The value of Z for this beam, as per Table IV., is 175-645; therefore— | A 3 гу — 1190400 2 645 x — or, the required weight is, say, 10,036 pounds. 177.—RoHed-Iron Beams: Floors of Dwellings or As- sembly-Rooms.— From Transverse Strains, Art. 500, we have— | те ЈР е (77.) a rule for the distance from centres of rolled-iron beams іп floors of dwellings, assembly-rooms, or offices, where the spaces between the beams are filled in with brick arches and . concrete. In the equation, c is the distance apart from сеп- tres in feet, and y is the weight per yard of the beam, This, in words, is thus expressed : Rule LV\.—Divide 255 times the value of Z by the cube of the length; from the quotient deduct one 420th part of the weight of the beam per yard, and the remainder will be the required distance apart from centres. , Example.--What should be the distance apart from сеп.

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152 CONSTRUCTION. tres of Buffalo 12}-inch, 125-pound beams 25 -feet long between bearings, in. the floor. of an.assembly-room? For these beams, in Table IV., 7 equals 286-019, and y — 125; therefore— _ Tas x 286.019 _ 125. A 25? 420” — 729348 . = 4. 668 — - T" 37; or, the required distance from centres is, say, 4 feet 44 inches. {78.—Rolled-Iron Beams: Floors of First-Class Stores. —From 7; ransverse Strains, Art. 504, we have— 148-8 a Y, =i Ey am -. 960’ (78.) a rule for the distance from centres of rolled-iron beams in the floor of a first-class store; the spaces between the beams being filled with brick arches and concrete. This rule may be put in words as follows: Rule LVII.—Divide 148-8 times the value of 7 by the cube of the length; from the quotient deduct one обо part of the weight of the beam per yard, and the remainder will be the distance apart of the beams from centres in feet. Example. — What should be the distance apart from cen- tres of Buffalo 121-inch, 180-pound beams 20 feet long between bearings, А the floor of a first-class store? For these beams the value of 7, Table IV., 15 418-945, and the value of y 1s 180; therefore— = 148. 8 x 418-945 180 — or, the required distance from centres is, say, 7 feet 71 inches.

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TIE-RODS FOR IRON BEAMS. 153 ' :[79.—Floor-Arehes: General Considerations.— In fill- ing the spaces between the iron beams of a floor, the arches should be: constructed with hard whole brick of good shape, laid upon the supporting: centre in contact with each other, and the joints thoroughly filled with cement grout, and keyed with slate. Made in this manner, the arches need not be‘over four inches thick at the crown for spans extending іо ‘у or 8 feet, and 8 inches thick at the springing, where they should be started upon:a proper skew-back. Тһе rise of the arch should not be less than 14 inches for each foot of the span. (80. — Floor - Arches ; Tie-Rods: Dwellings. — From Transverse Strains, Art. 507, we have— d = УО-0108 c s, | (79.) which is a rule for the diameter in inches of a tie-rod for an.arch in the floor of a bank, office building, or assembly- room ; in which Z 1s the diameter іп inches of the rod, 5 15 the span of the arch, and c is the distance apart between the rods (s and c both in feet): This rule requires that the arch rise 14 inches per foot of the span, and that the brick-work and the superimposed load each weigh 7o pounds, or to- gether 140 pounds. This rule, in words, is-as follows: Rule LVIIL—Multiply the span of the arch by the dis- tance apart at which the rods are placed, and by the decimal о-0108; the square root of the product will be the diameter of the required rod. "Example.—W hat should be the diameter. of the wrought- iron ties of brick arches of 5 feet span, in a bank or hall of assembly, where the ties are 8 feet apart? For this we have— | d == Vo-o198 х8 x 5 =V -792 = 0-89; or, the diameter of the required rods should be, say, 2 of an inch. ^ 181.—Floor-Arches ; Tie-Rods: First-Class Stores.—From the same source as in last article, we have— d = V0-04527 € s, . (80)

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154 CONSTRUCTION. which is a rule for the size of tie-rods for the brick arches of the floors of first-class stores, where the arches have a- rise of ІЗ inches for each foot of the span, and where the weight of the brick arch and concrete is not over 7o pounds per superficial foot of the floor, and the loading does not exceed 250 pounds per superficial foot. Ав the rule is the same as the one in the preceding article, except the deci- mal, a recital of the rule, in words, is not here needed. To Obtain the required diameter, proceed as directed in Rule LVIIL, using the decimal 0-04527 instead of the one there given. TUBULAR IRON GIRDERS. (82.—' Tubular Iron Girders: Description.—The use of wooden beams for floors is limited to spans of about 25 feet. When greater spans than this are to be covered, some expe- dient must be resorted to by which ) intermediate bearings for the floor- beams may be provided. Wooden girders may be used, but these need to be supported by posts at intervals of from 10 to r5 feet, unless the girders are trussed, or ; made upof topand bottom chords, 7 struts, and ties. And even this is objectionable, owing to the height such a piece of framing requires, and *which encumbers the otherwise free space of the hall. А substitute for the framed girder has been found in the tubular i iron girder, as in Fig. 46, made of rolled plate iron and angle irons, riveted. They require to be stiffened by an occasional upright 7 iron along eachside, and а cross-head at least at each bearing. (83.—'"Tubular Iron Girders: Area of Flanges; Load at Middte.—In wrought-iron tubular girders it is usual to make the top and bottom flanges of equal thickness. From Transverse Strains, Art. 551, we have—

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TUBULAR IRON GIRDERS. 155 Жы а EU (81.) a rule for the area of the bottom flange; in which а” equals the area of the flange in inches, W the weight in pounds at the middle, / the length and Z the depth of the girder, both in feet, and £ the safe load in pounds per inch with which the metal may be loaded, and which is usually taken at 9000. The rule may be stated thus: Rule LIX.—Multiply the weight by the length; divide the product by 4 times the depth into the value of 2, and the quotient will be the required area of the bottom flange. Example—In a girder до feet long and 3 feet high, to carry 75,000 pounds at the middle, what area of metal is required in the bottom flange, putting $ at gooo? For this we have, by the rule— mis /.— £5000 < MO PLAT m a ac? ГАРЕТ жы or, the area required is 272 inches. This is the amount of uncut metal. An allowance is required for that which will be cut by rivet-holes. This is usually an addition of one sixth. |84.-таһршаг Iron Girders: Area of Flanges; Load at any Point.—The equation suitable for this (Transverse Strains, Art. 553) is— , mn | adapto s l du in which m and z are the distances respectively from the lo- cation of the load to the two ends of the girder. The other symbols are the same as in the last article. This rule may be thus stated : Rule LX.—Multiply the weight by the values of % and of л; divide the product by the product of the depth into the length and into the value of 2, and the quotient will be the required area of the bottom flange. | Example.—1n a girder 50 feet long between bearings and

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156 ‚ 1 CONSTRUCTION, 34 feet high, what area of metal is required in the bottom flange to sustain 50,000 poundsat 20 feet from one end, when k equals gooo? Ву the rule, we have— m n: ОООО x 20 x 30 - Li W Uoc оге дай, 39 = 19:05; = diui c: 3% x 9000 x 50 or, each flange requires 19 inches of solid metal uncut for rivets. (85.—'Tubular Iron Girders: Area of Flanges; Load Uniformly Distributed.—The equation „appropriate. here is (Transverse Strains, Art..555)—. mene do (83.) This is a rule by which to obtain the area of cross-sec- tion of the bottom flange at any point in the length of the girder, the load uniformly distributed ; # and x being the respective distances from the point measured to the two ends of the girder, and (7 representing the uniformly dis- tributed load in pounds. This, in words, is described as follows : . Rule LXI.—Divide the weight by the product of twice the depth into the length and into the value of 2; then the quotient multiplied by the values of 72 and of z will be the required area of the bottom flange at the point measured, the distance of which from the ends equals m and л. Example.—ln a girder 50 feet long and 3} feet high, to carry a uniformly distributed load of 120,000 pounds, what area of cross-section is required in the bottom flange, at the middle and at intervals of 5 feet thence, to each support ; £ being taken at 9o00? Here we have, first-- mn 120000 777 Г ж. — M = ------------------- 00. ал UT Me «тео = 0-038095 72 n. Now, when m = 2 = 25, we have the middle point; then— a’ = 0-038095 m n = 0-038095 x 25 x 25 = 23-81;

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SHEARING STRAIN. i 57. М 97. e AA à ~ “. i 2 \\ 4 ог, the area of the bottom. flange at mid-length 15 23-81 inches.: . | When m = 20, then z — зо, and— a’ = 0-038095 x 20 x 30 = 22-86; or, the required area, at 5 feet either way from the middle, 15 221 inches. - When zz = 15, then z = 35, and— 8 = 0:088095 X 15 X 35,29 20«0 ; or, at 10 feet either way from the middle, the required area is 20 inches. When zz = 10, then z = 40, and— а” = 0-038095 x IOX 40 = 15:24; ог, at 15 feet either way from the middle, the required area is 154 inches. When m= 5, then # = 45, and— a’ = 0-038095 x 5x45 = 8-57 ; ог, at 20 feet each side of the middle, the required area 15 82 inches. The area of cross-section found in every case is that of the uncut fibres; to this is to be added as much as will be cut by the rivets. This is usually about one sixth of the area given by the rule. The top flange is to be made equal in area to the bottom flange. The flanges are unvarying in: width from end to end, the variation of area being obtained. by varying the thickness of the flanges, and this being at- tained by building the flange in lamina, or plates; but these should not be less than a quarter of an inch thick. There should be added to the length of the girder, in the clear, about one tenth of its length for supports on the walls: thus, a girder 30 feet long requires 3 feet added for supports, or 18 inches on each wall. [86.—' rubular Iron Girders: Shearing Strain.—The top and bottom flanges are provided of sufficient size to resist

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158 CONSTRUCTION. the transverse strain; the two upright plates, technically termed the wed, need, therefore, to be thick enough to resist only the shearing strain. This, upon a beam uniformly loaded, is at the middle theoretically nothing, but from thence it increases regularly towards each support, where it equals half the whole weight. For example, the girder of Art. 185, 50 feet long between supports, carries 120,000 pounds uniformly distributed over its length. Іп this case the shearing strain at the wall at each end is the half of 120,000 pounds, or 60,000 pounds ; at 5 feet from the wall it is 3; or 4 less, ог 48,000 pounds; at то feet from the wall it is £ less, or 36,000 pounds; at 15 feet it is 24,000; at 20 feet it is 12,000; and at 25 feet or the middle, it is nothing. [87.— Tubular Iron Girders: Thickness оғ Web.—The equation appropriate for this 15-- G t= PLE (84.) in which 2 1s the thickness of the web (equal to the sum of the thicknesses of the two side plates), 4 15 the height of the plate (¢ and d both in inches), С is the shearing strain, and >. is the effective resistance of wrought iron to shearing per inch of cross-section. This may be put in words as follows : Rule LXII.—Divide the shearing strain by the product of the depth in inches into the value of Z', and the quotient will be the thickness of the web, or of the two side plates taken together. Example.—What is the required thickness of web in а girder 50 feet between bearings, side plates 38 inches high between top and bottom flanges, and to carry 120,000 pounds, uniformly distributed? Here, putting the shearing resistance of the plates at 7000 pounds per inch, we have— ката - V. Т ak’ 7 38х yooo 26босо” The shearing strain at the supports, as in last article, is 60000 ; therefore, we have for this point—

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LIGHT IRON GIRDERS. 159 mm 60000 __ = "é65o i 9:225, When G = 48000, then— , bes toe ae еі. m P and when G = 36000, then— | ele Ун = 266000 — °° 135 Those nearer the middle of the girder are still less than these ; and these are all below the practicable thickness, which is half an inch for the two plates. The plates ought not in practice ever to be made less than a quarter of an inch thick. (88.—' rubular Iron Girders, for Floors of Dwellings, Assembly-Rooms, and Office Buildings.—When the floors of these buildings are constructed with rolled-iron beams and brick arches, then the following (Art. 568, Transverse Strains) is the appropriate equation for the area of cross-section of the bottom flange of the girder: 7 c mn . a! = (140+ oe T x T: (85.) in which a’ is in inches, and с, с, d, Z, m, and z are in feet. Also, a’ is the area required; y is the weight per yard of the rolled-iron beam of the floor ; с, their distances from centres ; c', the distance from centres at which the tubular girders are placed, or the breadth of floor carried by one girder; d, the depth of the girder; 2, the effective resistance of the metal per inch in the flanges of the girder; and z and z are the distances respectively from the two ends of the girder to the point at which the area of cross-section of the bottom flange is required. The rule may be thus described : Rule LXIII.—Divide the weight per yard of the rolled- iron beams by 3 times their distance from centres; to the quotient add 140 and reserve the sum ; deduct the length in feet from 7оо, and with the remainder as a divisor divide 700; multiply the quotient by {һе above reserved sum, and

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160 CONSTRUCTION. by the value of c’; divide the product by the product ot twice the depth into the value of 2, and the quotient multi- plied by the values of # and of z will be the required area of cross-section of the bottom flange at the point in the length distant from the two ends equal to m and x respec- tively. Example.—\n a floor of g-inch, 70-pound beams, 4 feet from centres, what ought to Бе the area of the bottom flange of atubular girder 40 feet long between bearings, 2 feet 8 inches deep, and placed 17 feet from the walls or from other girders; the area of the flange to be ascertained at:every 5 feet of the length; the value of £to be put at дооо? Неге y — 70, c=4, с = 17, L= 40, and Z—2$. Therefore, by the rule— a! = ( 1404 49.) - 796 зе. “еее st 3x4/ 700-40 2х2%х%000 ў a’ = 145-84 x 1.0606 x 0-0003541% xmn; a’ = 0-05478 m n. The values of 7z and м are— ӘШЕСіппасе...,............: Dn m = 20; k= 20 BP Ner fromoanddle.......... Ne. CR m= 1: 2E neo * ni ЕС oT. om с т. ж — ION 39 NS ^ ай саға бы аа m= 5; t= 35 These give— ЖЕГЕ middle. esce oae a’ = 0-05478 x 20 x 20 = 21.01 EN feet from middle... so’ = 0-05478 x 15 x 25 = 20-54 i ii к СЕО LES ох рю = 10-48 EE — s .a' = О-05478х 5x35 = 9:59 These are the areas of uncut fibres at the points named, in the lower flange ; the upper flange requires the same sizes. 189.—Tubular Iron Girders, for Floors of First-Class Stores.—The equation proper for this is (Transverse Strains, Art. 570)—

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С CAST IRON COMPARED WITH WROUGHT. 161 / А EE 204 2) Же. cmn 3 Mali ome 755 Wu 22 arule the same in form as that of the previous article ; hence it needs no particular exemplification. Rule LXIII. of last article may be used for this case, simply by using the constant 320 in place of that of 140. CAST-IRON GIRDERS. [90.—Cast-IYron Girders: Inferior.—Rolled-iron beams have been so extensively introduced within a few years as to have superseded almost entirely the formerly much used cast-iron beam or girder. Тһе tensile strength of cast iron' is far inferior to that of wrought iron. This inferiority and the contingencies to which the metal is subject in casting render it very untrustworthy ; it should not be used where rolled-iron beams can be procured. А very substantial gir- der to carry a brick wall 15 made by placing two or more rolled-iron beams side by side, and securing them together by bolts at mid-height of the web; placing thimbles or sep- arators at each bolt. Ав there may be cases, however, in which cast-iron girders will be used, a few rules for them will here be given. 191.—Cast-Iron Girder: Load at Middle.—The form of cross-section given to this girder usually is as shown in ETE AUS In the cross-section, the bottom flange is made to contain in area four times as much as the top flange. The strength will be in proportion to the area of the bottom flange, and to the height or depth of the girder at middle. Hence, to obtain the greater strength from a given amount of material, it is requisite to make the upright part, or the web, rather thin; yet, in order to prevent injurious strains in the casting while it is cooling, the parts should be nearly equal in thickness. The thickness of the three parts—web,

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162 CONSTRUCTION. top flange, and bottom flange—may be made in proportion as 5, 6, and 8. For a weight at middle, the form of the web should be that of a triangle; the top flange forming two straight lines declining from the centre each way to the bottom flange at the ends, like the rafters of а roof to its tie-beam. From Transverse Strains, Art. 583, we have— p uil a a= 4850 d” (87.) which is a rule for the area in inches of the bottom flange, for a load at middle; the area of the top flange is to be equal to one fourth of that of the bottom flange. To secure this, make the width of the top flange equal to one third of the width of the bottom flange; the thickness of the former, as before directed, being made equal to $ or $ of the lat- ter. The weight W is in pounds; the length / is in feet; and the depth dis ininches. The factor of safety a should be taken at not less than 3; better at 4 or 5. The equation in words may be as follows: Rule LXIV.—Multiply the weight by the length, and by the factor of safety; divide the product by 4850 times the depth at middle, and the quotient will be the area in inches of the bottom flange; divide this area by the width of the bottom flange, and the quotient will be its thickness. Ofthe top flange make its width equal one third that of the bot- tom flange, and its thickness equal to three quarters that of the latter. Make the thickness of the web equal to 5 that of the bottom flange. Example.—W hat should be the dimensions of the cross- section of a cast-iron girder 20 feet long between bearings, and 24 inches high at middle, where 30,000 pounds is to be carried; the factor of safety being put at 5? | Неге we һауе W = 30000; a = 5; /- 20; апа d= 24; therefore, by the rule— ‚ _ 30000 x 5 x 20 ~ 4850 x 24 = 25:773.

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THE BOWSTRING GIRDER. 163 This is the area of the bottom flange. If the width of this flange be 12 inches, then 25-773 divided by 12 gives 2-15, or 24 full, as the thickness. Опе third of 12 equals 4, equals the width of the top flange ; and 2 of 2-15 equals 1-61, or 15— its thickness. The thickness of the web equals £ x 2-15 = 1-34 Or 14 inches. |92. -Сая-Егоп Girder: Load Uniformly Distributed.- The equation suitable to this is— " Ual = g7oo d (88.) a rule of like form with that of the last article; therefore, Rule LXIV. may be used for this case, simply by substitut- ing 9700 for 4850. 193.—Cast - Iron Bowstring Girder.—An arched girder, such as that in /%g. 48, is technically termed a “bowstring . girder.” The curved part is a cast-iron beam of T form in section, and the horizon- tal line is a wrought-iron tie-rod attached to the ends of the arch. This girder has but little to commend it, and is by: no means worthy the confi- dence placed in it by aE builders, with many of whom it is quite popular. The brick arch usually turned over it is adequate to sustain the entire compressive force induced from the load (the brick wall built above it), and it thereby supersedes the necessity for the tron arch, which is a useless expense. The tie-rod is the only useful part of the bowstring girder, but it is usually made too small, and not infrequently is seriously injured by the needless strain to which it is subjected when it is “shrunk in” to the sockets in the ends of the arch. The bow- string girder, therefore, should never be used. |94. substitute for the Bowstring Girder.—As the cast- iron arch of a bowstring girder serves only to resist com- ә

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164 CONSTRUCTION. pression, its place can as well be filled by an arch of brick, footed on a pair of cast-iron skew-backs; and these held in position by a pair of wrought-iron tie-rods, as shown in Fig. 49. This system of construction is preferable to the bowstring girder, in that the tie-rods are not liable to injury by “shrinking in," and the cost is less. From Transverse Strains, Art. 596, we have— DV ША (89.) 9425 d' an equation in which 2 15 the diameter in inches of each of the two tie-rods of the brick arch; U is the load in pounds uniformly distributed over the arch; / is the span of the arch in feet; and d, in inches, is its versed sine, or its height measured from the centre of the tie-rod to the centre of the . thickness or height of the arch at middle. This equation may be put in words as follows: Rule LXV.—Multiply the weight by the length ; divide the product by 9425 times the depth, and the square root of the quotient will be the diameter of each rod. Example.—W hat should be the diameter of each of the pair of tie-rods required to sustain a brick arch 20 feet span from centres, with a versed sine or height at middle of 30 inches, to carry a brick wall 12 inches thick and 30 feet high, weighing 100 pounds per cubic foot? The load upon this arch will be for so much of the wall as will occur over the opening, which will be about one foot less than the span of the arch, or 20 — 1 = 10 feet. Therefore, the load will equal 19x 30x1 x100 = 57,000 pounds; and hence, U = 57000, / -х/20, d = 30, and, by the rule— | FIG. 49. РЕ”, Ж = V 4-0318 = 2-008; or, the diameter of each rod is required to be 2 inches.

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STRAINS REPRESENTED GRAPHICALLY. 165 FRAMED GIRDERS. [95.— Graphic Representation of Strains. —In the first part of this section, commencing at Art. 71, the method was developed of ascertaining the strains in the various parts of а frame by the parallelogram or triangle of forces. Тһе method, so far as there explained, is adequate to solve sim-: ple cases ; but when more than three pieces of a frame con- verge in one point, the task bv that method becomes difficult. This difficulty, however, disappears when recourse is had to the method known as that of * Reciprocal Figures, Frames, and Diagrams of Forces," proposed by Professor I. Clerk Maxwell in 1867. This is an extension of the method by the triangle of forces, and may be illustrated as follows: B Fic. 50. FIG. 51. Let the lines in Fig. 50 represent, in direction and amount, four converging forces in equilibrium in any frame, as, for example, the truss of a roof; let the lines іп Fig. 51 be drawn parallel to those in 270. 50, in the manner fol- lowing, namely: Let the line А B be drawn parallel with the line of Fig. 50 which is between the corresponding letters А and 5, and let it be of corresponding length; from В draw the line B C parallel with the line of Fig. 50 which is be- tween the letters B and С, апа of corresponding length; then from C draw C D, and from А draw A D, respectively parallel with the lines of 7zg. 50 designated by the corre- sponding letters, and extend them till they intersect at D. The lengths of these two lines, the last two drawn, are de- termined by the point D where they intersect; their lengths, therefore, need not be previously known. The lengths of the lines in Fig. 51 are respectively in proportion to the

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166 CONSTRUCTION. several strains іп /zg. 50, provided these strains are in equilibrium. 270: 51 is termed a ciosed polygon of forces. A system of such polygons, one for each point, in the frame where forces converge, so constructed that no line repre- senting a force shall be repeated, is termed a diagram of forces. This diagram of forces is а reciprocal of the frame from which it is drawn, its lines and angles being the same. The facility of tracing the forces in the diagram of forces depends materially upon the system of lettering here shown, and which was proposed by Mr. Bow, in his excellent work on the Economics of Construction. Їп this system each line of the frame is designated by the two letters which it separates ; thus the line between А and 2 is called line AB; that between C and 2 is called line CD; and so of others; and in the diagram the corresponding lines are called by the same letters, but here the letters designating the line are, as ` usual, at the ends of the line. Any point in a frame where forces converge is designated by the several letters which cluster around it; as, for example, іп Fig. 5o, the point of convergence there shown is designated as point 4 5 C D. This invaluable method of defining graphically the strains in the various pieces composing a frame, such as a girder or roof-truss, is remarkably simple, and is of general application. Its utility will now be exemplified in its appli- cation to framed girders, and afterwards to roof-trusses. 196.—Framed Girders.—Girders of solid timber are use- ful for the support of floors. only where posts are admissible as supports, at intervals of from 8 to 15 feet. For unob- structed long spans it becomes requisite to construct a frame to serve as a girder (47/5. 163, 182). А frame of this kind requires two horizontal pieces, a top and a bottom chord, | and a system of struts and suspension-pieces by which the top and bottom chords are held in position, and the strains from the load are transmitted to the bearings at the ends of the girders. Various methods of arranging these struts and ties have been proposed. One of the most simple and effective is shown in Zzg. 52, forming a series of isos- celes triangles. The proportion between the length and height of a girder is important as an element of economy

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RULES FOR FRAMED, GIRDERS. 167 both of space and cost. ‘When circumstances do not control in limiting the height, it may be determined by this equation from Transverse Strains, Art. 624— ‘ - 17542). ^. 2400 ' (90.) in which Zis the depth or height between the axes of the top and bottom chords, and 71s the length between the cen- tres of bearings at the supports (4 and / both in feet). This equation in words is as follows: Rule LXV I.—To the length add 175; multiply the sum by the length; divide the product by 2400, and the quotient will be the required height between the axes of the top and bottom chords. Example-—What should be the depth of a girder which is 40 feet long between the centres of action atthe supports? For this the rule gives— а 075-40) х40_ m 2400 "m 3:581; or, the proper depth for economy of material is 3 feet and 7 inches. i The number of bays, panels, or triangles into which the bottom chord may be divided is a matter of some considera- tion. Usually girders from— 20 to 59 feet long should have 5 bays. 50 “ 95 “ “ “ ‘ 6 : 85 «€ 107 “ “ “ “ 7 “ “ “ “ “ 9 “ 107 127 iy “ 146 T: “ “ “ 9 Do 197. Framed Girder and Diagram of Forces.—Let 722. 52 represent a framed girder of six bays of, say, 11 feet each, or of a total length of 66 feet. The lines shown are the axial lines, or the imaginary lines passing through the axes of the several pieces composing the frame. The six arrows indicate the six pressures into which the equally distributed load is supposed to be divided. Each of these is at the apex of a triangle, the base of which lies along the lower chord.

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168 ' CONSTRUCTION. Pire spaces between the arrows are lettered ; so, also, the space between the last arrow at either end and the point of support has a letter, and so has each triangle, and there is one for the space beneath the lower chord. These letters are to be used in describing the diagram of forces, as was explained in Art. 195. The diagram of forces (Fig. 53) for this girder-frame is drawn as follows, namely: Upon a verti- distances, to represent the six equal vertical pressures indi- cated by the arrows in Fig. 52. The equal distances A О, O P, etc., may be made of any convenient size; but it will Fic. 535. - serve to facilitate the measurement of the forces in the dia- gram if they are made bya scale of equal parts, and the number of parts given to each division be made equal to the number of tons of 2000 pounds each which is contained in the pressure indicated by each arrow. On this vertical line ‚ the distance A О represents the load at the apex of the tri- angle B, or the point A OCB (Art. 195); the distance OP

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GRAPHICAL DIAGRAMS OF FORCES. 169 represents the weight at the second arrow, or at the point ОРЕ РС, and so of the rest. If the weights upon the points in the upper chord had been unequal, then the divi- sion of the vertical line А M would have had to be corre- ` spondingly unequal, each division being laid off by the scale, to accord with the weight represented by each. Тһе line of loads, A JV, being adjusted, the other lines are drawn from it (Art. 195), so as to make a closed polygon for the forces converging at each point of the frame, Fig. 52—commenc- ing with the point A 27, Fig. 52, where there are three forces, namely, the force acting through the inclined strut A B, the horizontal force іп В T, and the vertical reaction А Tat the point of support. This last is equal to half the entire load, or equal to the pressure indicated by the three arrows, 4 О, O P, апа РО, and is represented in 27. 53 by A Qor А 7. From the point Q draw a horizontal line О В; this is parallel with the force B T of Fig. 52, in the lower chord. From the point А draw А P parallel with the strut A B of Fig. 52. This line intersects the line P T in В and closes the polygon A B 7A; the point P defines the length of the lines 4 Band 2 7, and these lines measured by the scale by which the line of loads was constructed give the required pressures in the corresponding lines, A Вапа 2 7, Gt РИ bot Taking next the point А B СО, where four forces meet, of which we already have two, namely, the force in the strut A B and the load А O—from the point О draw the hori- zontalline ОС; this is parallel to the horizontal force ОС of Fig. 52. Now from В draw B С parallel with the suspen- sion-piece ВС of Fig. 52. This line intersects OC in C, and the point C limits the lines О C and В C and closes the poly- gon А B CO A, the four sides of which are respectively in proportion to the four forces converging at the point A 5 CO of Fig. 52, and when measured by the scale by which the line of loads was constructed give the required strains ге- spectively in each. Taking next the point BC D T, where four forces converge, of which we already have two, "VT and В 7—from В extend the horizontal line 77 to D; from C draw C D parallel with C D of Fig. 52, and extend it to 1n- tersect TD in D, and thus close the polygon TB C D T.

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170 CONSTRUCTION, The lines in a part of this polygon coincide—those from B to T; this is because the two strains В Гапа D 7, Fig. 52, lie in the same horizontal line. Again, taking the point OCD ЕР, where five forces meet, three of which, ОР ОС, and C D, we already have—draw from D the line D £ parallel with D E of Fig. 52, and from Р the line PZ horizontally or parallel with РЕ of Fig. 52. These two lines intersect at Ж and close the polygon PO C D E P, the sides of which meas- ure the forces converging in the point PO C D E, Fig. 52. Next in order is the point D Е F T, Fig. 52, where four forces meet, two of which, 72D and D E, are known. From Ж draw E Е parallel with E Fin 226. 52; and from 7, TF parallel with TF in Fig. 52; these two lines meet in F and close the polygon TD E FT, the sides of which measure the required strains in the lines converging at the point DE FT, Fig. 52. Taking next the point PE FG Q, Fig. 52, where five forces meet, of which we already have three, ОР, РЕ, апа E F— trom F draw a line parallel with F G of Fig. 52, and from Q a line parallel with Q G of Fig. 52. These two intersect at C and complete the polygon QP E FG О, the lines of which measure the forces converging at PE FG Q in Fig. 52. In this last polygon, а peculiarity seems to indicate an error: the line FG has no length; it begins and ends at the same point ; or, rather, the polygon is complete without it. This is easily understood when it is considered that the two lines FG and G H do not contribute any strength towards sustaining the loads PQ and ОЁ, and in so far as these weights are concerned they might be dispensed with, and the space occupied by the three triangles F, С, and H left free, and be designated by only one letter instead of three. Thus it appears that there are only four instead of five forces at the point PE FG О, and that the four are represented by the lines of the polygon QP £ FQ. The peculiarity above explained arises from considering loads only on the top chord: the analysis of the case is cor- rect as worked from the premises given; but in practice there is always more or less load on the bottom chord at the middle, which should be considered. This will be included іп а case proposed in the next article. One half of the dia-

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P d LOAD ON BOTH CHORDS, 171 gram of forces is now complete. The other half being ex- actly the same, except.that it is in reversed order, need not here be drawn. (98.—Framed Girders: Load on Both Chords.—Let Pig. 54 represent the axial lines of a girder carrying an Fic. 54. equally distributed load on each chord, represented by the arrows and balls shown in the figure. Let each bay measure IO feet, or the length ot the girder be 50 feet, and its height Fic. 55. be 44 feet. The diagram of forces (Fig. 55) for this girder is obtained thus: | - The plan of the girder, 272. 54, requires to be lettere as shown; having one letter within each panel and outside the frame, and one between every two weights or strains. Then, in 20, 55, mark the vertical line K V at L, M, N, О,

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172 х CONSTRUCTION. and P, dividing it by scale into equal parts, corresponding "with the weights on the top chord represented by the ar- rows. For example, if the load at each arrow equals 6% tons, make K L, L M, M N, etc., each equal to 63 parts of.the scale. Then A P will equal the total load on the top flange. Make the distance P V equal to the sum of the loads on the bottom chord. Then AK V equals the total load on the gir- der. Bisect К Vin U; then K U or U V equals half the total load ; consequently, equals the reaction of the bearing at X or Р of Fag 5a: Now, to obtain the polygon of forces converging at K А U, Fig. 5а, we have one of these forces, X U, or the re- action of the bearing at KA U, equal to K U, Fig. 55. From U draw UA parallel with UA of Fig. 54, and from А draw KA parallel with the strut X A, Fig. 54, and intersecting the line U A at A,a point which marks the limitof X A and U A, and closes the polygon ЖА U K, the sides of which аге in proportion respectively to the three strains which converge at the point A UK, Fig. 54. For example, since the line K U by scale measures the vertical reaction, K U, of the bearing at A СК, Fig. 54, therefore the line ХА of the diagram of forces by the same scale measures the strain in the strut А A, Fig. 54, and the line А U of the diagram by the same scale measures the strain in the bottom chord at A U, Fig. 54. For the strains converging at KA B L, Fig. 54, of which two, К А and K L, are already known, we draw from A the line А B parallel with the line А В, Fig. 54, and from Z draw L B parallel with Z В, Fig. 54, meeting A P at B, a point which limits the two lines and closes the polygon КА B Т, K, the lines of which are in proportion respectively to the strains converging at the point KA B L, Pzg. 54, as before explained. Of the five strains converging at U A BC T, we already have three— Г (7, U A, and A В; to obtain the other two, make UQ equal to PV, equal to the total load upon the lower flange; divide U Q into four equal parts, Q R, RS, S 7, and T U, corresponding with the four weights on the lower chord, and represented, by the four balls, Fig.: 54. Now, from 7, the point marking the first of these divisions, draw T C parallel with TC, Fig. 54, and from B draw 7 C paral-

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VARIOUS STRAINS IN FRAMED GIRDERS, 173 lel with the strut BC, Fig. 54, meeting TC in C, а point which limits the lines BC and TC and closes the polygon TUA BC T,thesidesof which are in proportion respectively to the strains converging in the point TUA B C T, Fig. 54. Of the five forces converging at M L B C D, we already have three— 74 L, L B, and BC; to obtain the other two, from M draw MD parallel with M D, Fig. 54, and from С draw CD parallel with CD, Fig. 54, meeting M D at D, a point limit- ing the lines MD and CD and closing the polygon MLBCDM, the sides of which are in proportion to the strains converging at the point M L B CD, Fig. 54. Of the five forces converging at the point S TCD Е, three—S 7, T C, and C D—are known; to obtain the other two, from S draw SZ parallel with SZ, Fig. 54, апа from D draw D E, parallel with the strut D E, Fig. 54, meeting the line SZ in Е, a point limiting the two lines 5 E and D Е and closing the polygon S TCD E S, the sides of which are in proportion to the strains converging at S TCD E, Fig. 54. One half of the strains in Fig. 54 are now shown in its diagram of forces, 2. 55; and since the two halves of the girder are symmetrical, the forces in one half corresponding to those in the other, hence the lines of the diagram for one half of the forces may be used for the corresponding forces of the other half. [99. —Framed Girders; Dimensions of Parts. — The parts of a framed girder are the two horizontal chords (top and bottom) and the diagonals—the struts and ties. The top chord is in a state of compression, while the bottom chord experiences a tensile strain. Those of the diagonal pieces which have a direction from the top to the bottom chord, and from the middle towards one of the bearings of the girder, as K A, BC, or D E, Fig. 54, are struts, and aré subj jected to compression. The diagonal pieces which have a direction from the bottom to the top chord, and from the middle towards one of the supports, as 4 B or C D, Fig. 54. are fies, and are subjected to extension, (477. 83. Тһе amount of strain in each piece іп a framed girder having been ascertained in a diagram of forces, as shown in 47/5. 197 and 108, the dimensions of each piece may be obtained:

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I74 CONSTRUCTION. by rules already given. The dimensions of the pieces in a state of compression are to be ascertained by the rules for posts in Arts. 107 to 114, and those in a state of tension by 248. 117 to 119 (see Arts. 226 to 229). Саге is required, in obtaining the size of the lower chord, to allow for the joints which necessarily occur in long ties, for the reason that tim- ber is not readily obtained sufficiently long without splicing. Usually, in cases where the length of the girder is too great to obtain a bottom chord in one piece, the chord is made up of vertical lamina, and in as long lengths as practicable, and secured with bolts. A chord thus made will usually require about twice the material; or, its sectional area of cross-sec- tion will require to be twice the size of a chord which is in one whole piece; and in this chord it is usual to put the fac- tor of safety at from 8 to Io. The diagonal ties are usually made of wrought iron, and it is well to secure the struts, especially the end ones, with iron stirrups and bolts. And, to prevent the evil effects of shrinkage, it is well to provide iron bearings extending through the depth of each chord, so shaped that the struts and rods may have their bearings upon it, instead of проп the wood. PARTITIONS. 200.—Partitions.—Such partitions as are required for the divisions in ordinary houses are usuaily formed by tim- ber of small size, termed studs or joists. These are placed upright at 12 or 16 inches from centres, and well nailed. Upon these studs lath are nailed, and these are covered with plastering. The strength of the plastering depends in a great measure upon the clinch formed by the mortar which has been pressed through between the lath. That this clinch may be interfered with in the least possible degree, it is proper that the edges of the partition-joists which are presented to receive the lath should be as narrow as prac- ticable; those which are necessarily large should be reduced by chamfering the corners. The derangements in floors, plastering, and doors which too frequently disfigure the interior of pretentious houses with gaping cracks in the

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FRAMED PARTITIONS, 175 plastering and in the door-casings are due in nearly all cases to defective partitions, and to the shrinkage of floor-timbers. A plastered partition is too heavy to be trusted upon an ordi- nary tier of beams, unless so braced as to prevent its weight from pressing upon the beams. This precaution becomes es- pecially important when, in addition to its own weight, the partition serves as a girder to carry the weight of the, floor- beams next above it. In order to reduce to the smallest practicabie degree the derangements named, it is important that the studs in a partition should be ¢russed or braced so as to throw the weight upon firmly sustained points in the construction beneath, and that the timber in both partitions and floors should be well seasoned and carefully framed. To avoid the settlement due to the shrinkage of a tier of beams, it is important, ina partition standing over one in the story below or over a girder, that the studs pass between the beams to the plate of the lower partition, or to the girder ; and, to be able to do this, it is also important to ar- range the partitions of the several stories vertically over each other. All principal partitions should be of brick, especially such as are required to assist in sustaining the floors of the building. Р Fic. 56. ` 20|.— Examples of Partitions.—/72. 56 represents a par- tition having a door in the middle. Its construction 15 simple but effective. Fig. 57 shows the manner of constructing а

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176 CONSTRUCTION. partition having doors near the ends. The truss is formed above the door-heads, and the lower parts are suspended from іё. · The posts æ and ё аге halved, and nailed to the tie cd and the sill ef The braces in а trussed partition H | : E n b р Lodi FIG. 57. should be placed so as to form, as near as possible, an angle of до degrees with the horizon. Тһе braces in a partition should be so placed as to discharge the weight upon the E. l ШЕ [| Aw | Fic. 58. points of support. All oblique pieces that fail to do this should be omitted. When the principal timbers of a partition require to be large for the purpose of greater strength, it is a good plan

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WEIGHT UPON PARTITIONS. 177 to omit the upright filling-in pieces, and in their stead to place a few horizontal pieces, as іп Fig. 58, in order that upon these and the principal timbers upright battens may be nailed at the proper distances for lathing. А partition thus constructed requires a little. more space than others; but it has the advantage of insuring greater stability to the plas- tering, and also of preventing to a good degree the conver- sation of one room from being overheard in the adjoining one. Ordinary partitions are constructed with 3x4, 3x5, or 4x6 inch joists, for the principal pieces, and with 2 x 4, 2x5, or 2x6 filling-in studs, well strutted at intervals of about 5 feet. When a partition is required to support, in addition to its own weight, that of a floor or some other burden resting upon it, the dimensions of the timbers should beascertained, by applying the principles which regulate the laws of pressure and those of the resistance of timber, as explained in the first part of this section, and іп Arts. 196 to 199 for framed girders. The following data may assist in calculating the amount of pressure upon partitions: White-pine timber weighs from 22 to 32 pounds per cubic foot, varying in accordance with the amount of seasoning it has had. Assuming it to weigh 30 pounds, the weight of the beams and floor-plank in every superficial foot of the flooring will be— 6 pounds when the beamsare 5x 8 inches, and placed 20 inches from centres. 73 “ “ “ 3 х IO «t © 18 «€ [1] 9 “ “ “ 3 х 12 “ “ 16 [11 “ “JI «€ [1] “ 5 x I2 “4% “ 12 “ «€ I3 “ “ 66 4 х I2 “ «6 I2 “ “ 13 €€ 64 “ 4 х 14 “ “ 14 “ 4€ In addition to the beams and plank, there is generally the plastering of the ceiling of the apartments beneath, and some- times the deafening. Plastering may be assumed to weigh 9. pounds per superficial foot, and deafening 11 pounds. Hemlock weighs about the same as white pine. А parti- tion of 3x4 joists of hemlock, set 12 inches from centres, therefore, will weigh about 24 pounds per foot superficial and when plastered on both sides, 20% pounds. ;

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178 7 CONSTRUCTION. ROOFS. 202. In ancient Norman and Gothic buildings, the walls and buttresses were erected so massive and firm : that it was customary to construct their roofs without a tie- beam, the walls being abundantly capable of resisting the lateral pressure exerted by the rafters. But in modern buildings, usually the walls are so slightly built as to be in-. capable of resisting much if any oblique pressure; hence the necessity of care in constructing the roof so as to avoid oblique and lateral strains. The roof so constructed, instead of tending to separate the walls, will bind and steady them, Fic. 59. Fic. 60. Ес. 61. Fic. 62. Fic. 65. . 66. | Fic. 67. 203. s.—Designs for roof- trusses, illustrating various principles of roof construction, are herewith presented. The designs at 7725. 59 to 63 are distinguished from нењ at Pigs. 64 to 67 by having a horizontal tie-beam. Іп the latter group, and in all designs similarly destitute of the horizontal tie at the foot of the rafters; the strains are much - greater than in those having the tie, unless the truss be pro-

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VARIOUS FORMS OF ROOF-TRUSSES. 179 tected by exterior resistance, such as may be afforded by competent buttresses. ‘To the uninitiated it may appear preferable, іп Fig. 64, to extend the inclined ties to the rafters, as shown by the dotted lines. But this would not be beneficial; on the con- trary, it would be injurious. The point of the rafter where the tie would be attached is near the middle of its length, and consequently is a point the least capable of resisting transverse strains. The weight of the roofing itself tends to bend the rafter ; and the inclined tie, were it attached to the ‘rafter, would, by its tension, have a tendency to increase this bending. As a necessary consequence, the feet of the rafters would separate, and the ridge descend. — '/ _ In Fig. 65 the inclined ties are extended to the rafters; but here the horizontal strut or straining beam, located at the points of contact between the ties and rafters, counteracts the bending tendency of the rafters and renders these points stable. In this design, therefore, and only in such designs, it is permissible to extend the ties through to the rafters. Even here it is not advisable to do so, because of the in- creased strain produced. (See Figs. 77 and 79.) The design in Fig. 64, 66, or 67 is to be preferred to that in 272. 65. 204..—Force Diagram: Load upon Each Support.—By а comparison of the force diagrams hereinafter given, of each of the foregoing designs, we may see that the strains in the trusses without horizontal tie-beams at the feet of the rafters are greatly in excess of those havingthetie. In constructing these diagrams, the first step is to ascertain the reaction of, or load carried by, each of the supports at the ends of the truss. Іп symmetrically loaded trusses, the weight upon each support is always just one half of the whole load. 205.—Force Diagram for Truss in Fig. 59.—To obtain the force diagram appropriate to the design in fig. 59, first letter the figure as directed in 477. 195, and as in Fig. 68. Then draw a vertical line, E F (Fig. 69), equal to the weight W at the apex of roof ; or (which is the same thing in effect) equal to the sum of the two loads of the roof, one extending on each side of W half-way to the foot of the rafter. Di-

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180 CONSTRUCTION. vide Æ F into two equal parts at C. Make GC and G D eacn equal to one half of the weight У. Now, since Е G is equal to one half of the upper load, and G D toone half of the low- er load, therefore their sum, E G+ G D = ED, 15 equal to one half of the total load, or to the reaction of each support, E or F. From D draw D A parallel with DA of Fig. 68, and from Е draw Е А parallel with ЕА of Fig. 68. The three lines of the triangle 4 E D represent the strains, respectively, іп the three lines converging at the point А DE of 222. 68. Draw the other lines of the diagram parallel with the lines of E B с А р Fic. 69. Fig. 68, and as directed in Arts. 195 and 197. Тһе various lines of Fig. бо will represent the forces in the corresponding lines of 270. 68; bearing in mind (472. 195.) that while a line . 1n the force diagram is designated in the usual manner by the letters at the two ends of it, aline of the frame diagram 1s designated by the two letters between which it passes. Thus, the horizontal lines 4 D, the vertical lines A В, and the in- clined lines A Е have these letters at their ends in Zzg. бо, while they pass between these letters in Fzg. 68. 206.—Force Diagram for Truss in 7/77. 60.—For this truss we have, іп Fig. 70, a like design, repeated and lettered as required. We here have one load on the tie-beam, and three loads above the truss: one on each rafter and one at the ridge. In the force diagram, Fig. 71, make GH, H F, and 9 К, by any convenient scale, equal respectively to the weights GH, H F, and F K of Fig. 7o. Divide G K into two equal parts at Z. Make Г Е and LF each equal to one half the weight ЕР (Fig. 70). Then С Fis equal to one half the

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FORCE DIAGRAMS OF TRUSSES. 181 total load, or to the load upon the support С (Art. 205). Complete the diagram by drawing its several lines parallel with the lines of Fig. 70, as indicated by the letters (see 477. 205), commencing with G F, the load on the support G (Fig. 70. Draw from F and G the two lines FA and GA paral- lel with these lines in Fig. 7o. Their point of intersection defines the point 4. From this the several points Z, C, and D are developed, and the figure completed. Then the lines in Z7. 71 will represent the forces in the corresponding lines of Fg. 70, as indicated by the lettering. (See Art. 195.) FIG. 71. 207.—Force Diagram for Truss in 77g 61.—For this truss we have, in fig. 72, a similar design, properly prepared by weights and lettering ; and in ig. 73 the force diagram ap- propriate to it. : In the construction of this diagram, proceed as directed in the previous example, by first constructing N 5, the ver- tical line of weights; in which line VO, OP, РО, QR, and RS are made respectively equal to the several weights above the trussin 270. 72. Then divide VS into two equal parts at 7. Make 7X and TZ each equal to the half of the weight KZ. Make 7 K and Г M equal tothe weights F K and Z M of Fig. 72. Now, since M М is equal to one half of the weights above the truss plus one half of the weights below the truss, or half of the whole weight, it is therefore the weight upon the support N (Fig. 72), and represents the reaction of that support. А horizontal line drawn. from 77 will meet the inclined line drawn from JV, parallel with the rafter A JV (Fig. 72), in the

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182 - CONSTRUCTION. : point A, and the three sides of the triangle А M N, Fig. 73, will give the strains in the three corresponding lines meeting at. the point 4 M N, Fig. 72. The sidesof the triangle H F S, Fig. Fic. 72. 73, give likewise the strains in the three corresponding lines meeting at the point H F S, Fig. 72. Continuing the con- struction, draw all the other lines of the force diagram parallel N Fic. 73. with the corresponding lines of Fig. 72, and as directed in Art. 195. The completed diagram will measure the strains in all the lines of Fig. 72.

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FORCE DIAGRAMS CONTINUED. 4 183 208,—Force Diagram for Truss in Fig. 63.— The roof truss indicated at Fig. 63 is repeated in Fig. 74, with the ad c Fic. 74. Pici 75. dition of the lettering required for the construction of the force diagram, Fig. 75.

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184 | CONSTRUCTION. In this case there are seven weights, or loads, above the truss, and three below. Divide the vertical line OV at W into two equal parts, and place the lower loads in two equal parts on each side of W. Owing to the middle one of these loads not being on the tie-beam with the other two, but on the upper tie-beam, the line G М, its representative in the force diagram, has to be removed to the vertical 5 7, and the letter M is duplicated. The line VO equals half the whole weight of the truss, or 34 of the upper loads, plus one of the lower loads, plus half of the load at the upper tie- beam. It is, therefore, the true reaction of the support N О, and A X is the horizontal strain іп the beam there. It will be observed also that while 77 7 and GM (Fig. 75), which are equal lines, show the strain in the lower tie-beam at the middle of the truss, the lines CH and FG, also equal’ but considerab]y shorter lines, show the strains in the upper | tie-beam. Ordinarily, іп a truss of this design, the strain in the upper beam would be equal to that in the lower one, which becomes true when the rafters and braces above the upper beam are omitted. Іп ће present case, the thrusts of the upper rafters produce tension in the upper beam equal to CM or FM of Fig. 75, and thus, by counteracting the compression in the beam, reduce it to CH or FG of the force diagram, as shown. 209.— Force Diagram for Truss in 77g. 64.—The force diagram for the rooftruss at fig. 64 is given in Fig. 77, while Р/с. 78 is the truss reproduced, with the lettering requisite for the construction of Fig. 77. The vertical E F (Fig. 77) represents the load at the ridge. Divide this equally at W, and place half the lower weight each side of W, so that C D equals the lower weight. Then ÆD is equal to half the whole load, and equal to the reaction of the support Е (Fig. 76). The lines in the triangle A D Е give the strains іп the corresponding . lines converging at the point A D E of Fig. 76. The other lines, according to the lettering, give the strains in the cor- responding lines of the truss. (See 477. 195.) |

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FORCE DIAGRAMS CONTINUED. 185 210. reproduced i in Zzg. 78, with the letters X for use in the force diagram, Fig. 79. 2-7 г 4 З “а My / ^ j Y Р Fic. 76. Fic. 77. Fic. 79. Here the vertical G K, containing the three upper loads GH, Н%, and FK, is divided equally at W, and the lower

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186 М CONSTRUCTION. load Е F is placed half on each side of W, and extends from Ж to F. Then FG represents one half of the whole load of the truss, and therefore the reaction of the support G (Fig. 78). Drawing the several lines of 7zg. 79 parallel with the corre- sponding lines of Fig. 78, the force diagram is complete, and the strains in the several lines of 78 are measured by 5 cor- responding lines of 79. (See 474. 195.) | A comparison of the force diagram of the truss іп Fig. 76 with that of the truss іп Fig. 78 shows much greater strains in the latter, and we thus see that Y 76 or 64 is the ‘more economical form. Fic. 80. 211: —Еогсе Diagram for Truss in 777. 66.—This truss is reproduced and prepared by proper lettering in zz. 80, and its force diagram is given in 7zg. 81. Here the vertical YM contains the three upper loads ФК, КІ, and LM. Divide ¥ M into two equal parts at б, and make FG and G H respectively equal to the two loads FG and G H of Fig. 80. "Then H F represents one half of the whole weight of the truss, and therefore the reac- tion of the support ¥ From М and % draw lines par- allel with 4 H and A ¥ of Fig. 80, and the sides of the tri- angle A H 5 will give the strains in the three lines concen- trating in the point А H 9 (Fig. 80). The other lines of 2%

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EFFECT OF ELEVATING THE TIE-BEAM, 187 8; аге all drawn parallel with their corresponding lines in Fig. 80, as indicated by the lettering. (See 47. 195.) D Тс. 81. 212.—Roof-Truss: Effect of Elevating the Tie-Beam.— From Arts. 670, 671, Transverse Strains, it appears that the Fic. 82 — effect of substituting inclined ties for the horizontal tie at feet of rafters is— `

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188 CONSTRUCTION. 7-2. (91.) in which P represents half the weight of the whole truss and the load upon it; а+б = height of the truss at middle above a horizontal line drawn at the feet of the rafters; « equals the height from this line to the point where the two inclined ties meet; J, the height thence to the top of the truss; and V, the additional vertical strain at the middle of the truss due to elevating the tie from a horizontal line. Examples are given to show that when the elevation of the tie equals + of the whole height, the vertical strain there- by induced is equal to a weight which equals $ of half the whole load; and that when the elevation equals half the whole height, the vertical strain is equal to half the whole load. This is the strain in the vertical rod at middle. The strains in the rafters and inclined ties are proportionately increased. 213.—Planning a Roof.—In designing a roof for a build- ing, the first point requiring attention is the location of the trusses. These should be so placed as to secure solid bear- ings upon the walls; care being taken not to place either of the trusses over an opening, such as those for windows or doors, in the wall below. Ordinarily, trusses are placed so as to be centrally over the piers between the windows ; the number of windows consequently ruling in determining the number of trusses and their distances from centres. This distance should be from ten to twenty feet ; fifteen feet apart being a suitable medium distance. Тһе farther apart the trusses are placed, the more they will have to carry; not only in having a larger surface to support, but also in that the roof-timbers will be heavier; for the size and weight of the roof-beams will increase with the span over which they have to reach. In the roof-covering itself, the roof-planking may be laid upon jack-rafters, carried by purlins supported by the trusses ; or upon roof-beams laid directly upon the back of the principal rafters in the trusses. In either case, proper

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LOAD. UPON ROOFS. 189 struts should be provided, and set at proper intervals to re- sist the bending of the rafter. .In case purlinsare used, one of these struts should be placed at the location of each purlin. The number of these points of support rules largely in determining the design for the truss, thus: JA Fora short span, where the rafter will not require sup- port at an intermediate point, Fzg. 59 or 64 will be proper. For a span in which the rafter requires supporting at one intermediate point, take 77g. 60, 65, or 66. For a span with two intermediate points of support for the rafter, take 777. 61 or 67. For a span with three intermediate points, take 272. 63. Generally, it is found convenient to locate these points of support at nine to twelve feet apart. Тһеу should be sufh- ciently close to make it certain that the rafter will not be sub- ject to the possibility of bending. 214. — Load upon Roof-Truss.—In constructing the force diagram for any truss, it is requisite to determine the points of the truss which are to serve as points of support (see Figs. 70, 72, etc.), and to ascertain the amount of strain, ог. loading, which will occur at every such point. The points of support along the rafters will be required to sustain the roofing timbers, the planking, the slating, the snow, and the force of the wind. Тһе points along the tie- beam will have to sustain the weight of the ceiling and the flooring of a loft within the roof, if there be one, together with the loading upon this floor. The weight of the truss itself must be added to the weight of roof and ceiling. 215. Load on Roof per Superficial Foot.—ln any im- portant work, each of the items in 477. 214 should. be care- fully estimated, in making up the load to be carried. For ordinary roofs, the weights may be taken per foot superficial, as follows: Slate, about 7-0 pounds. Roof-plank, 2.7 " Roof-beams ог jack-rafters, “ 2.3 ранни Іп all, | 12 . , pounds.

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190 - CONSTRUCTION." This is for the superficial foot of the inclined roof. For the foot horizontal, the augmentation of load due to the angie of the roof will be in proportion to its steepness. In ordinary cases, the twelve pounds of the inclined surface will not be far from fifteen pounds upon the horizontal foot. For the roof-load we may take as follows: Roofing, about 15 pounds. Roof-truss, т 5 5; Snow, % 20 н Wind, Д IO А Total on roof, 50 pounds per square foot horizontal. This estimate is for a roof of moderate inclination, say one in which the height does not exceed } of the span. Upon a steeper roof the snow would not gather so heavily, but the wind, on the contrary, would exert a greater force. Again, the wind acting on one side of a roof may drift the snow from that side, and perhaps add it to that already lodged upon the opposite side. These two, the wind and the snow, are compensating forces. The action of the snow is vertical: that of the wind is horizontal, or nearly so. The power of the wind in this latitude is not more than thirty pounds upon a superficial foot of a vertical surface; except, perhaps, on elevated places, as mountain-tops for example, where it should be taken as high as fifty pounds per foot of vertical surface. 216.—Load upon Tie-Beam.—The load upon the tie- beam must of course be estimated according to the require- ments of each case. If the timber is to be exposed to view, the load to be carried will be that only of the tie-beam and the timber struts resting upon it. If there is to bea ceiling attached to the tie-beam, the weight to be added will be in accordance with the material composing the ceiling. If of wood, it need not weigh more than-two or three pounds per foot. -If of lath and plaster, it will weigh about nine pounds ; and if of iron, from ten to fifteen pounds, according to the

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WEIGHT UPON ROOFS, IN DETAIL. 191 thickness of the metal. Again, if there is to be a loft in the roof, the requisite flooring may be taken at five pounds, and the load upon the floor at from twenty-five to seventy pounds, according to the purpose for which it is to be used. 217.—Roof-Weights in Detail. —The load to be sustained by a roof-truss has been referred to in the previous three articles in general terms. It will now be treated more in detail. But first a few words regarding the slope of the roof. In a severe climate, roofs ought to be constructed steeper than in a milder one, in order that snow may have a tendency to slide off before it becomes of sufficient weight to endanger the safety of the roof. In selecting the material with which:the roof is to be covered, regard should be had to the requirements of the inclination: slate and shingles cannot be used safely on roofs of small rise. The smallest inclination of the various kinds of covering is here given, together with the weight per superficial foot of each. MATERIAL. Least Inclination. Weight upon a square oot. yr e VERE S: Don TET C CE TOT Rise r inch to a foot. Ф to r2 Ibs. COPPEN ы Иш .---.............. жара ue dM I {от} “ тт a ce cee sas Des ““ eanches * =" d 405 01 UE con Wesco no QUO TITO CREETETETERT ese | eo 20 0 Short рїйе'єїїш|е5................... Cp WC I} to 2 y Long cypress shingles... ............ Сыс оор. 2103“ COT? ume TELE жаланы, ағы С The weight of the covering as here estimated includes the weight of. whatever is used to fix it in place, such as nails, etc. The weight of that which the covering is laid upon, such as plank, boards, or lath, is not included. The weight of plank is about 3 pounds per foot superficial; of boards, 2 pounds ; and lath, about half a pound. pe Generally, for a slate roof, the weight of the covering, 1n- cluding plank and jack-rafters, amounts to about 12 pounds, as stated in A477. 215; but in every case, the weight of each article of the covering should be estimated, and the full load ascertained by summing up these weights.

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192 CONSTRUCTION. 218.—Load per Foot Horizontal.—The weight ot the covering as referred to in the last article is the weight per foot on the zzclined surface ; but it is desirable to know how much per foot, measured Zorzzezta/Zy, this is equal to. The horizontal measure of one foot of the inclined surface is equal to the cosine of the angle of inclination. Then, to ob- tain the inclined measure corresponding to one foot horizon- tal, we have— | where ? represents the pressure on a foot of the inclined surface, and C the weight of so much of the inclined cover- ing as corresponds to one foot horizontal. The cosine of an angle is equal to the base of the right-angled triangle divided by the hypothenuse (see Trigonometrical Terms, 477. 474), which in this case is half the span divided by the length of the rafter, or, , where s is the span, and / the length of the rafter. Hence, the load per foot horizontal equals— m c = (92.) 27 or, twice the pressure рег foot of inclined surface multiplied by the length of the rafter and divided by the span, both in feet, will give the weight per foot measured horizontally. 219.— Weight of Truss.—The weight of the framed truss will be in proportion to the load and to the span. This, for the weight upon a foot horizontal, will about equal— T = 0-077 Cs; which equals the weight in pounds per foot horizontal to be allowed for a wooden truss with iron suspension-rods and a horizontal tie-beam, near enough for the requirements of our present purpose; where s equals the length or span ‘of the

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EFFECT OF WIND ON ROOFS. 103 truss, and C the weight per foot horizontal of the roof cover- ing, as in equation (92.). Substituting for C its value, as in (92.), we have— Te 0:0077 s 2/2 ; $ Or— ' М T = 0-0154 12; (93.) which equals the weight in pounds per foot horizontal to be allowed for the truss. 220.— Weight of Snow оп Roofs.—The weight of snow will be in proportion to the depth it acquires, which will be in proportion to the rigor of the climate of the place where the building is to be erected. Upon roofs of ordinary incli- nation, snow, if deposited in the absence of wind, will not slide off; at least until after it has acquired some depth, and then the tendency to slide will be in proportion to the angle of inclination. The weight of snow may be taken, therefore, at its weight per cubic foot (8 pounds) multiplied by the depth it is usual for it to acquire. This, іп the latitude of New York, may be taken at about 24 feet. Its. weight would, therefore, be 20 pounds per foot superficial, meas- ured horizontally. i 221.—Effect of Wind on Roofs.—The direction of wind is horizontal, or nearly so, when unobstructed. Precipitous mountains or tall buildings deflect the wind considerably from its usual horizontal direction. Its power usually does not exceed зо pounds рег superficial foot except on ele- vated places, where it sometimes reaches 50 pounds or more. This is the pressure upon a vertical surface ; roofs, however, generally present to the wind an inclined surface. The ef- fect of a horizontal force on an inclined surface is in pro- portion to the sine of the angle of inclination; the direction of this effect being at right angles to the inclined surface, The force thus acting may be resolved into forces acting 1n two directions—the one horizontal, the other vertical; the former tending, in the case of a roof, to thrust aside the wa'ls

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104 CONSTRUCTION. on which the roof rests, and the latter acting directly on the materials of which the roof :15 constructed—this latter force being in proportion to the sine of the angle of inclination multiplied by the cosine. This will be made clear by the D followingexplanation. Re- ferring to 777.83,let DK E be the angle of inclination ofthe roof, D Е being equal toone foot. Bisect DK at А; draw A L parallel with Fic, 83. EK; make A L equal to the horizontal pressure of the wind upon one foot superficial of a vertical plane. Draw AC perpendicular to DK, and LF parallel with 4 C from F draw FC parallel with Е А; draw А B parallel with DZ. Тһе sides of the triangle Z A F rep- resent the three several forces in equilibrium: ZA is the force of the wind; Z F is the pressure upon the roof; and A F is the force with which the wind moves on up the roof towards D. Now,to find the relation of the force of the wind to the strain produced by it in the direction А C, we have— radi "sim C: adi. FC-—LA;therefore— СТЕР рім = WA Sin. ; A Gu ёш. or, the strain perpendicular to the surface of the roof equals the force of the wind multiplied by the sine of the angle of inclination. When AC represents this strain, then, of the two forces referred to above, В C represents the horizontal force, and А P the vertical force. То obtain this last force, we have— Гай’: cos. ;& AC: ND. Putting for A C its value as above, we have— rad. cost :х Faime A EF sim. cosy; V = ¥ Sint cose:

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FORCE OF THE WIND. 195 or, the vertical effect is equal to the product of the force of the wind upon a superficial foot into the sine and the cosine of the angle of inclination. This result is that which is due to the pressure of the wind upon so much of the inclined surface as is covered by one square foot of a vertical sur- face. The wind,acting horizontally through one foot super- ficial of vertical section, acts on an area of inclined surface equal to the reciprocal of the sine of inclination, and the horizontal: measurement of this inclined surface is equal to the cosine of the angle of inclination divided by the sine. This may be illustrated from /70. 83, thus— sin. аша DA: О D E equals 1 foot ; therefore— ч 5йп.:гай,::1:2К =——; sin. or, the surface acted upon by one square foot of: sectional area equals the reciprocal of the sine of the angle of incli- nation. Again, the horizontal measure of this inclined sur- face may be obtained thus— sin.: cos.:: DE: KE =<; sin. or, KE, the horizontal measurement, equals the cosine of the angle of inclination divided by the sine. In the figure, make A G equal to one foot; then we have— | СЗ ЕСЕ Е іп which V, as above, represents the vertical pressure due to the wind. acting upon the surface KD, and W the vertical pressure due to the wind acting upon the surface А27, or so much as covers К G, one foot horizontal. | : COS. Now we have, as above, K Е equal to cm K С = 1,and

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196 CONSTRUCTION. V = F sin. cos. Substituting these values, we have, instead of the above proportion— COS. 7 t= Cp eS. сов. ЖУ: sin. from which— F sin. cos. , W = — = F sin.’ ; | m (94-) sin. or, the vertical effect of the wind upon so much of the roof as covers each square foot horizontal, is equal to the pro- duct of the force of the wind per square foot into the square of thesine of the angle of inclination. Example.—W hen the force of the wind upon a square foot of vertical surface is 30 pounds, what will be the verti- cal effect per square foot horizontal upon a roof the inclina- tion of which is 26? 33”, or 6 inches to the foot? Неге we have F = 30, and the sine of 26° 33’ is 0-44608 ; therefore— W = 30x 0.44698" = 5.9937. This is conveniently solved by logarithms; thus— lap. зо = Тә ?/І2ІЗ 0-44698 = 9:6502868 О-44608 = 9-6502868 · 5:0037 = 0-7776949 or, the vertical effect is (5-9937, or) 6 pounds. The form of equation (94.) may be changed; for, in a right- angled triangle, the sine of the angle at the base is equal to the perpendicular divided by the hypothenuse; which, in the case of a roof, is the height divided by the length of the rafter; or—

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LOAD UPON ROOFS, 197 Therefore, equation (04.) may be changed to— = d = P (95.) or, the vertical effect upon each square foot of a roof is equal tothe product of the force of the wind per foot into the square of the height of the roof at the ridge, divided by the square of the length of the rafter (the height and "с" both in feet.) Example.—When the force of the wind is 30 бош the height of the roof 10 feet, and the length of the rafter 22. 36 feet, what will be the vertical effect of the wind? Неге we Hove 5-40, # = ro, and / — 22-26; апф— IO* W = 30x oe 6. 222, al Load per Foot Horizontal.—The various items comprising the total load upon a roof are the cover- ing, the truss, the wind, snow, the plastering or other kind of ceiling, and the load which may be deposited upon a floor formed in the roof; or, the total load— Же CLEP = srr. The value per foot horizontal for these has been found as Мб ca 2125 Pah ok: w= т For Sur 5 value must Бе taken according to circumstances, as in 477. 220. So, also, the value of Р and Г are to be assigned as required for each particular case, as in 477.216. Тһе total load, therefore, with these substitutions, will be— a = Si Cy 0154/9 FA S PL; which reduces to— = м= 1 (2 +0: o154) F^ +S+P+L; (90)

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198 _ CONSTRUCTION. in which Zis the length of the rafter; 2 is the weight of the covering per foot superficial, including the roof boards or slats, the jack-rafters, etc.; sis the span of the roof; 1s the vertical height above a horizontal line passing through the feet of the rafters; F is the force of the wind per square foot against a vertical surface; S is the weight of snow per square foot horizontal; P is the weight per superficial foot of the ceiling at the tie Баа: and Z, the load per superficial foot in the roof, including weight of flooring and floor- timbers. The dimensions, 5, /, and #, are each in feet; the | weight of 2, F, S, P, and Z are each in pounds. The value of 2 is for a square foot of the zzcZined surface. 223.— Strains in Roof-Timbers Computed. —The graphic method of obtaining the strains, as shown in 4725. 205 to 211, is, for its conciseness and simplicity, to be preferred to any other method; yet, on some accounts, the method of obtain- ing the strains by the parallelogram of forces and by arith- metical computations will be found useful, and will now be referred to. By the parallelogram of forces, the weight of the roof is in proportion to the oblique thrust or pressure in the axis of the rafter as twice the height of the roof 1s to the length of the rafter; or— JOY : асы or, transposing— R A 20: J: RS са (97-) l where Y equals the pressure in the axis of the rafter, and К the weight of one truss and its load. Again, the weight of the roof is in proportion to tke horizontal thrust in the tie- beam as twice the height of the roof is to half the span; or— Ae ES :24:34 2 ог, transposing— 5 Rs au Libe 7-3 | (98.)

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THI STRAINS SHOWN GEOMETRICALLY. 199 i where Æ equals the horizontal thrust in the tie-beam. To obtain А, the weight of the roof, multiply M, the load per foot, as in equation (96.), by s, the span, and by с, the dis- tance from centres at which the trusses are placed ; or— Ж = с 5. With this value of & substituted for it, we have— | Mes. Y= Fi (99.) and— Mast JEL к= 42 , (100.) in which Y equals the strain in the axis of the rafter, and А the strain in the tie-beam. These are the greatest strains in the rafter and tie-beam. At certain parts of these pieces the strains are less, as will be shown in the next article. 224.—Strains in Roof-Timbers Shown Geometrically, The pressure in each timber may be obtained as shown in Fig. 84, where A В represents the axis of the tie-beam, А С the axis of the rafter, D Е and FB the axes of the braces, and DG, F E, and C P the axes of the suspension-rods. Іп this design for a truss, the distance 4 В is divided intothree equal parts, and the rods located at the two points of division, G and E. Bythisarrangement the rafter 4 Cis supported at equidistant points, D and 7. The point D supports the rafter for a distance extending half-way to А and half-way to /, and the point / sustains half-way to D and half-way to С. Also, the point C sustains half-way to Р, and, on the other rafter, half-way to the corresponding point to F. And because these | points of support are located at equal distances apart, there- fore the load on each is the same in amount. On 2 С make Da equal by any decimally divided scale to the number of hundreds of pounds in the load on J, and draw the parallel- ogram а%/ Ос. Then, by the same scale, Dd represents (Art. 71) the pressure in the axis of the rafter by the load at

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200 CONSTRUCTION, 2 Fic. 84. hj

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STRAINS IN А TRUSS. 20I D; also, Dcthe pressure in the brace DE. Draw cd hori- zontal; then D d is the vertical pressure exerted by the brace DEat E. The point F sustains, besides the common load represented by 2a, also the vertical pressure exerted by the brace DE; therefore, make Fe equal to the sum of Da and D d, and draw the parallelogram F gef. Then Fg, meas- ured by the scale, is the pressure in the axis of the rafter caused by the load at F, and F f is the load in the axis of the brace ҒВ. Draw fA horizontal; then FA is the vertical pressure exerted by the brace F Bat P. The point C, besides the common load represented by D z, sustains the vertical pressure Fh caused Буе brace FB, and a like amount from the corresponding brace on the opposite side. There- fore, make Су equal to the sum of Da and twice FA, and draw 70 parallel to the opposite rafter. Then CZ is the pressure in the axis of the rafter at C. This is not the only pressure. in tbe rafter, although it is the total pressure at its head C. At the point ~F, besides the pressure С 2, there is Fg. Atthe point D, besides these two pressures, there is the pressure D 0. At the foot, at A, there is still an addi- tional pressure; for while the point D sustains the load half- way, to F and half-way to A, the point А sustains the load half-way to D. This load is, in this case, just half the load at D. Therefore draw A m vertical, and equal, by the scale, to half of Da. Extend CA 104; draw mZ horizontal. Then A/ is the pressure in the rafter at A caused by the weight of the roof from A half-way to D. Now the total of the pressures in the rafter is equal to the sum of A /+ D 2+ Fg added to C£. Therefore make £z equal to the sum of Al+D6+Fg,and draw zo parallel with the opposite raf- ter, and z7 horizontal. Then Co, measured by the same scale, will be found equal to the total weight of the roof on both sides of CB. Since Da represents s, the portion of the weight borne by the point D, therefore Co, representing the whole weight of the roof, should equal six times 2a, as it does, because D supports just one sixth of the whole load. Since C z is the total oblique thrust in the axis of the rafter at its foot, therefore z 7 is the horizontal thrust in the tie- beam at А.

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202 CONSTRUCTION. 225.—Application of the Geometrical System of Strains.— The strains in a roof-truss can be ascertained geometrically, as shown in Art. 224. To make a practical application of the results, in any particular case, it is requisite first to as- certain the load at the head of each brace, as represented by the line Da, Fig. 84. The load corresponding to any part of the roof is equal to the product of the superficial area of that. particular part (measured horizontally) multiplied by the weight per square foot of the roof. Or, when M equals the weight per square foot, с the distance from centres at which the trusses are placed, and z the horizontal distance between the heads of the braces, then the total load at the head of a brace is represented by— N = Meg. (101.) The value of M is given in general terms in equation (96.). To show its actual value, let it be required to find the weight per square foot upon a root 52 feet spanand 13 feet high at middle; or (Fig. 84) where A В equals half the space, or 26 feet, and СВ 13 feet, then A C, the length ofthe rafter, will be 26-069, nearly. And where the weight of covering per square foot, on the inclination, is 12 pounds, the force of the wind against a vertical plane is 30 pounds; the weight of snow per foot horizontal is 20 pounds; the weight of the plastering forming the ceiling at the tie-beam is 9 pounds; and the load in the roof is nothing ;—with these quantities substituted, equation (96.) becomes— ШИ = ‚об 2 Ed . 13 : 29-009 x I G 0:0154) +305 +2049 +0 = (29-069 x 12 x 0-05386) + (30 x 0-2) + 20 - 9; M = 18.788 +6 + 29 = 53-788; or, say, 53-8 pounds. Then if с, the distance from centres between trusses, is IO feet, and z, the distance between braces, is one third of А P, Fig. 84, or % = 82, the total load at the head of a brace will be, as per equation (101.)- N —$3:8x lox 8E = 4665;

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. TABLE OF STRAINS, 203 or, say, 4650 pounds. Now, by any decimally divided scale, make D a, Fug. 84, equal to 464 parts of the scale; this being the number of hundreds of pounds contained in the weight at D, as above. Then, by the same scale, the several lines in the figure drawn as before shown will be found to repre- sent respectively the weights here set opposite to them, as follows : Dues ao =| г == 234, and represents 23215 ронан айас b f= Fal E 4650 s Diese d$ ss. Fg 52 a 5200 d fec —DageDd = бой j 6975 P Ff = 65% “обу; ©“ C7 = 3Da= 139% "o вво“ CK ге 3E 156 “ — 15600 as Cn Ck+Fg+Db+Al=312 “ 31200 u МЕ Ох 320 0 = 6.Db—2CÉ = мг Ұ 31260 x i 7 = Си: 6 Da= 6 хабі-- 279/* | 27600 9 It should be observed here that the equality of the lines z7 and Co is a coincidence dependent upon the relation which in this particular case the line C B happens to bear tothe line А В; A В being equal to twice СВ. Апа so of some other lines in the figure. If the inclination of the roof were made greater or less, the equality of the lines referred to would disappear. It should also be observed that the strains above found are not quite exact; they are, however, correct to within a fraction of a hundred pounds, which is a sufh- ciently near approximation for the purpose intended. From the results obtained above, we ascertain that the strain in the rafter, from F to C, is represented by СА, and is equal to 15,600 pounds; while the strain at the foot of the rafter, from А to 2, is represented by Сл, and equals 31,200 pounds, or double that which is at the head of the rafter. Weascer- tain, also, that the maximum strain in the tie-beam, repre- sented by zj, is 27,900 pounds; that that in the brace DZ, represented by Рс, is 5200 pounds; and that that in the Brace, represented by 7 Ж is 6575 pounds. The strain .

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204. CONSTRUCTION, in the vertical rod DG is theoretically nothing. There is, however, a small strain in it, for it has to carry a part of the tie-beam and so much of the ceiling as depends for support upon that part. But the manner of locating the weights, adopted in this article, does not recognize any load located at the point G. This is an objection to this system, but 1t is not material. For a recognition of weights at the tie-beam, see Arts. 205 to 211. The load at G may be found by obtaining the product of the surface carried into the weight per foot of the ceiling; or, say, IO cz = 10x 10x 82 = 867 pounds. The load to be carried by the rod F Z is shown at D d = Ae, which above is found to be 2325 pounds. To this is to be added 867 pounds for the ceiling at Æ, as before found for the ceiling at С; or, together, 3192 pounds. The central rod СВ has to carry the two loads brought to В by the two braces footed there; and also the weight of the ceiling sup- ported by >. The vertical strain from the brace F P is rep- resented at F%, and equals 4650 pounds; therefore, the total load on СВ is 4650 + 4650 + 867 — 10,167 pounds. 226.—Roof-Timbers: the Tie-Beam.— The roof-timbers comprised in the truss shown in Zig. 84 are the rafters, tie-beam, two braces, and three rods. Of these, taking first the zze-beam, we have a piece subject to tension and some- times to cross-strain (see Art. 682, Transverse Strains). In this case the tensile strain: only need be considered. For this a rule is given in 472. 117. Іп this rule, if the factor of safety be taken at 20, the result will be sufficiently large to allow for necessary cuttings at the joints. Therefore, if the beam be of Georgia pine, equation (16.), 477. 117, becomes— 27900 x 20 A L— M m ЖІ Қ 16000 346; or, say, 35 inches. This is ample to resist the tensile strain ; but, to resist the transverse strains to which such а long piece of timber is subjected in the hands of the workman, it would be proper to make it, say, 6 x 9.

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STRAIN UPON THE RAFTER. 205 227.— The Rafter.—A rafter, like a post, is subject to а compressive force, and is liable to fail in three ways, name- ly : by flexure, by being crushed, or by crushing the material against which it presses. To render it entirely safe, there- fore, it is requisite to ascertain the requirements for resisting failure in each of these three ways. Of these it will be convenient to consider, first, that of the liability to being crushed. The rule for this is found in Art. 107. Let the rafter be of Georgia pine, then the value of C, Table I., will be 9500. The strain in the rafter (Art. 225) is 31,200 pounds. Now, taking the value of a, the fac- tor of safety, at то, we have, by Rule VI. (Art. 107.)— 31200 x 10 A= cen ME 1 E or, 33 inches area of cross-section. This is the size of the rafter at its smallest section; for example, at any one of the joints where it is customary to reduce the area by cutting for the struts and rods. Again: Let the liability of the rafter to flexure be now considered. For this we have a rule in 477. 114. The length of the rafter between unsupported points is nearly 9$ feet, or 92 x 12 = 116 inches. Let the thickness of the rafter be taken at 6 inches. Then,by Rule XI. (474. 114), we have— AD Walter) _ 31200 x 10 (1+ ё х -OO0109 х^"), Ser 9500 x 6 7 116 к. v = 19 ; 1021 = 373:8. Then, $ x 00109 x 373-8 = О-б11127 adding unity = 1. 1-612127 Substituting this, we have— _ 31200 x IO x 1.611127 __:502671 -624 “9500 x 6 Ыру “=;

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206 CONSTRUCTION. or, to resist flexure the breadth is required to be 8.82, or, say, 9 inches; or, the rafter is to'be б xo inches at the foot. The strain in the rafter at the upper end is only half that at the foot; the area. of cross-section, therefore. at the head need not be more than half that which is required at the foot ; but itis usual to make it there about 2 of the size at the foot. In this case it would be, therefore, 6 x 6 inches at the upper end. | Lastly, the requirement to resist crushing the surfaces against which the rafter presses is to be considered. The fibres of timber yield much more readily when pressed together by a force acting at right angles to the di- rection of their length than when it acts zz a /ine with their length. The value of timber subjected to pressure in these two ways is shown in 47/5. 94,98. In Table I., the value per square inch of the first stated resistance is expressed by P, and the ultimate resistance of the other by 2 The value of timber per square inch to safely resist crushing may be ex- pressed by = in which @ is the factor of safety. Timber pressed in an oblique direction will resist a force exceeding that expressed by Р, апа less than that expressed by En When the angle of inclination at which the force acts is just. 45^, then the force will be an average between 2 and x And for any angle of inclination, the force will vary inverse- ly as the angle; approaching P as the angle is enlarged, but approaching A as the angle is diminished. It will be equal to - when the angle becomes zero, and equal P ` when the angle becomes 9o?. Тһе resistance of timber per square inch to an oblique force is therefore expressed by— М-Р- e с — Py (102.)

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RESISTANCE OF SURFACES. 207 where 4% equals the complement of the angle of inclination. In a roof, 4° is the acute angle formed by the rafter with a vertical line.. If no convenient instrument be at hand to measure the angle, describe an arc upon the plan of the truss—thus: with C В (Fig. 84) for radius, describe the arc В g, and get the length of this arc in feet by stepping it off witha pair of dividers. Then— where 2 equals the length of the arc, and 2 equals В C, the height of the roof. Therefore— M = P+ 0-633 е -P) (тоз.) A \a equals the value of timber per square inch in a tie-beam, C and P being obtained from Table I., 477. 94. When С for the kind of wood in the tie-beam exceeds C set opposite the kind of wood in the rafter, then the latter is to be used in the rules instead of the former. ; The value of M, equation (103.),,is the resistance per square inch of the surface pressed at the foot of the rafter. The resistance of the entire surface will therefore be M A, = where A equals the area of the joint. Then, when the re- sistance equals the strain, we will have— Wan = S л EET СЕРІ); from which we have— Е ASO 4 E С Ы Prosa (aP) a (104.) in which S is the strain to be resisted. Now, the end of the rafter must be of sufficient size to afford a joint the area of which will not be less than that expressed by А in equation (104.). For example, the strain to which the rafter, Fig. 84, 1S subject at its foot is ascertained to be (477. 225) 31,200 pounds. For Georgia pine, the material of the tie-beam, P = 900 (Art. 94, Table Г), and C = 9500.

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208 CONSTRUCTION. The length of the arc Bg is about 14-4 feet; the height В Cis 13 feet. Let a, the factor of safety, be taken at то, then we have (104.)— 31200 , доо + (0-633 х 444) (2500. — ооо) | $1200 = 960 (o: 705 х50) | A= = 33-36; or, the етеін! area of the bearing at the joint required to prevent crushing the tie-beam is 334 inches. The results of the computations show that the rafter is required to be 6 inches thick, 9 inches wideat the foot, and 6 inches wide at the top. It 1s also ascertained that, in cut- ting for the bearing for the struts and boring for the sus- pension-rods, it is required that there shall be at least 33 inches area of cross-section left intact; and, farther, that the area of the surface of the joint against the tie-beam should not be less than 334 inches. 228.—The Braces.—Each brace is subject to compres- sion, and is liable to fail if too small, in the same manner as the rafter. Its size is to be ascertained, therefore, in the manner described for the rafter; which need not be here repeated, except, perhaps, as to the liability to fail by flexure ; for inthis case we have the breadth given, and need to find the thick- ness. The breadth of the brace is fixed by the thickness of the rafter, for 1t is usual to have the two pieces flush with each other. Rule XI. (477. 114) is to be used, but with this difference, namely : instead of the thickness, use the breadth as one of the factors in the divisor. Thus— Wa(r-4$er | ү == күр p. | (105.) In working this rule, it is required, in order to get the value of 7, the ratio between the height and thickness, to assume the thickness before it is ascertained ; and after com- putation, if the result shows that the assumed value was not a near approximation, a second trial will have to be made. Usually the first trial will be sufficient.

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STRAIN UPON BRACES. 209 For example, the brace D £ is about 9$ feet or 116 inches long. As the strain in it is only 5200 pounds, the thickness will probably be not over 3 inches. Assuming it at this, we / i have. г=++®= 38$; the square of whicl 45 UNIVE RSITY) Therefore, we have— $ x0-oo0109 x 1495 = 2. * a 3 3-4445 т=з The equation reduces, therefore, to this— 5200 X IOX 3:4445 . = 9500x6 — = 3:1424; ` or, the required: thickness of the brace is 34 inches, or the brace should be, say, 31 x 6 inches. Іп this case the result is so near the assumed value, a second trial.is not needed. For the second brace, we have the length equal to about 121 feet ог 147 inches; and the strain equal to 6575 pounds (Art. 225). The ratio, therefore, may be obtained’ by assum- ing the thickness, say, at 4. With this, we have— dii 2 141-- 36.75; the square of which is 13507. With this value of 7"— $ x «00109 x 13509; == 2.2081 add unity = I. 3-208I Then— . 6575 x 10 x 3.2081 ШИТ: Comparing this result with the assumed value of 7 = 4, we find the difference so great as to require a second trial. As the value of ғ was taken too low, the result obtained, is correspondingly low. The true value is somewhere between . 3:7and 4. Assume it now, say,at 3-0. With this value, we bave— E ——— —3- 7006. T -24 = 37.692; the square.of which is 1420 7.

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210 CONSTRUCTION. With this value of 7?— $ x ООІООх 1420-7 = 2.32282 add unity = т. 3 -32282 Then— 52 6575 x 10 X 3+32282 . D ọ500x6 _ = 3:833. This result is a trifle less than the assumed value, 3-9. The true value is between these, and probably is about 3-86. This is quite near enough for use. This brace, therefore, is required to be 3-86 x 6 inches, or, say, 4 x 6 inches. ` 229.—The Suspension-Rods.— These are usually made of wrought iron. This metal, when of excellent quality, may be safely trusted with 12,000 pounds per inch sectional area. But it is usual, for good work, to compute the area at only 9000 pounds per inch, and, as ordinarily made, these rods ought not to be loaded with more than 7000 pounds. The strain divided by this value per inch of the metal will give the sectional area of cross-section. For example, the strain in the rod DG, Fig. 84, 15 867 pounds (477. 225); therefore— E a ЖЕБЕ кома ог, the sectional area required is only an eighth of ап inch. By reference to the table of areas of circles in the Appen- dix, the diameter of а rod containing the required area, as above, will be found to be a little less than half an inch. A rod half an inch in diameter will therefore be of ample strength. For appearance's sake, however, no rod in a truss should be less than 3 of an inch in diameter. “Тһе rod ҒЕ hasto resist a strain of 3192 pounds. For this, then, we have— 3192 4 E J 0:456. А reference to the table of areas shows that a гоа contain-

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ROOF-BEAMS. 211 ing this area would be a little more ап $ of an inch in di- ameter; it would be of ample strength, зау, at £ of an inch in diameter. The rod C B, at the centre, has to carry a strain of 10,167 pounds. For this, then, we have— A reference to the table of areas shows that this rod should be r£ inches in diameter. 230.—Roof-Beams, Jack-Rafters, and Purlins.—T hese timbers are subject to loads nearly uniformly distributed, and their dimensions may be obtained by Rule ХХХ. equa- tion (35.), 477. 140. In this equation, U =c fl (Art. 152). Substituting this value for (7, and 7 Z for ô, equation (35.) be- comes— жү 2 ба = L6Fr and putting for > the rate of deflection, .04, we have— p.17 | Odri str ys (106.) a formula convenient for roof-timbers. Example.—In a roof where the roofing is to be supported оп white-pine roof-beams то feet long, placed 2} feet from centres, and where the load per foot superficial is to be 40 pounds, including wind and snow: what should be the di- mensions of the roof-beams? Ву equation (106.)— 40 x 2$ x то" 0-064 х 2900 — Mg == = 538.8. Now if 2, the breadth, be fixed, say, at 3, then— 538-8 3 d — 5-64 nearly. d? = = 1796;

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212 CONSTRUCTION. The roof-beams, therefore, require to be 3 x 52, or, say, 3 x 6. АП pieces of timber subject to cross-strains will sustain safely much greater strains when extended in one piece over two, three, or more distances between bearings; therefore, roof-beams, jack-rafters, and purlins should, if possible, be made in as long lengths as practicable; the roof-beams and purlins laid on, not framed into, the principal rafters, and extended over at least two spaces, the joints alternating on , the trusses ; and likewise the jack-rafters laid on the purlins in long lengths. 231.—Five Examples of Roofs: are shown at Figs. 85, 86, 87, 88, and 89. In 722. 85, а 15 an iron suspension-rod, 0, 2 are braces. In Fig. 86, a, а, and b are iron rods, and d,d, с, с are braces. In 7zg. 87, а, б are iron rods, 4, 4 braces, and c the straining beam. In Fic. 85. Fig. 88,а,а, b, b are iron rods, £, €, d, d are braces, and с is а straining.beam. In 270. 89, pur- SEU =. 7E alijs 7o NN eT - 0 Шын ызы о. Fic. 87. lins are located at P P, etc.; the inclined beam that lies upon them is the jack-rafter; the post at the ridge is the king-

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TRUSS WITH BUILT-RIB. | 213 post, the others аге queen-posts.. In this design the tie-beam is increased in height along the middle by a strengthening piece (Art. 163), for the purpose of sustaining additional weight placed in the room form- ed in the truss (477. 216). Fig. до shows a method of constructing а truss having a butlt-rib in the place of prin- cipal rafters. The proper form for the curve is that of the par- abola (Art. 560). This curve, when as flat as is described in the figure, approximates so close- ly to that of the circle that the latter may be used іп its stead. | The height, 2d, is just half of ac, the curve to pass through the middle of the rib. The rib is composed of two series of abutting pieces, bolted together. . These pieces should be as long ' as the dimensions of the timber will admit, in order that there may be but few joints. The sus- pending pieces are in halves, notched and bolted to the tie- beam and rib, and a purlin is framed upon the upper end of each. А truss of this construc- tion needs, for ordinary roofs, no diagonal braces between the suspending pieces, but if extra strength is required the braces may be added. The best place for the suspending pieces is at the joints of the rib. A rib of this kind will be sufficiently strong if the area of its section contain about one fourth more timber than is required for that of a rafter for a roof of the same size. The proportion of the depth to the thick- ness should be about as 10 to 7.

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214 CONSTRUCTION. 232.—Roof-Truss with Elevated Tie-Beam.—Designs such as are shown in 270. 91 have the tie elevated for the ac- commodation of an arch in the ceiling. This and all similar designs are seriously objectionable, and should always be 37 Jeet. Fic. 89. avoided; as the smail height gained by the omission of the tie-beam can never compensate for the powerful lateral strains which are exerted by the oblique position of the 38 feel. FIG. 90. » supports, tending to separate the walls. Where an arch is required in the ceiling, the best plan is to carry up the walls as hizh as the top of the arch. Then, by using a horizontal tie-beam, the oblique strains will be entirely re-

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HIP-ROOFS. 215 moved. It is well known that many a public building has been all but ruined by the settling of the roof, consequent upon a defective plan in the formation of the truss in this respect. It is very necessary, therefore, that the horizontal Етс. от. tie-beam be used, except where the walls are made so strong and firm by buttresses, or other support, as to prevent a possibility of their separating. (See Art. 212.) A a des l > | b |^ 7 f | Ih OX ee A i 5. z d Я Fig. 92. 233.—Hip-Roofs: Lines and Bevils.—The lines a б and bc, in Fig. 92, represent the walls at the angle of a building ; b e is the seat of the hip-rafter, and g f of a jack or cripple rafter. Draw e Aat right angles to ġe, and make it equal

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216 CONSTRUCTION. to the rise of the roof; join 0 and #, and 20 will be the length of the hip-rafter. Through e draw dz at right angles to ^c; upon 2, with the radius 24, describe the arc Z i, cutting dz in 7; join 0 and 2, and extend g f to meet 47 in у; then gj will be the length of the jack-rafter. The length of each jack-rafter is found in the same manner—by extend- ing its seat to cut the line 47, From /draw /£ at right angles to f g, also f/ at right angles to ġe; make /2 equal to f 7 by the arc 22, or make g £ equal to gj by the arc 77; then the angle at 7 will be the sop-devil of the jack-rafters, and the oneat £ will be the down-bevil.* 234.—The Backing of the Hip-Rafter.—At any con- venient place in ġe (Frig. 92), as o, draw zz м at right angles to бе; from o, tangical to 6 4, describe a semicircle, cutting 0 е in 5; join zz and s and ғ and е; then these lines will form at s the proper angle for bevilling the top of the.hip-rafter. DOMES.t 235.—Domes.—The usual form for domes is that of the sphere; the base circular. When the interior dome does not Fic. 93. rise too high, a horizontal tie may be thrown across, by which any degree of strength required may be obtained. * The lengths and bevils of rafters for roof-va//eys can also be found by the above process. 1 See also 474. 68.

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CONSTRUCTION OF DOMES. 217 Fig. 93 shows a section, and Fig. 94 the plan, of a dome of this kind, аб being the tie-beam in both. Two trusses of this kind (727. 93), parallel to: each other, are to be placed one on each side of the opening in the top of the dome. Upon these the whole framework 15 to depend for support, and their strength must be calculated accordingly. (See . Arts. 70 to 80 and 214 to 222.) If the dome is large and of importance, two other trusses may be introduced at right angles to the foregoing, the tie-beams being preserved in Fic. 95. one continuous length by framing them high enough to pass over the others. 236. —Ribbed Dome, When the interior must be kept free, then the framing may be composed of a succession of ribs standing upon a continuous circular curb of timber, as

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218 CONSTRUCTION. seen at Figs. 95 and g6—the latter being a plan and the former asection. This curb must be well secured, as it serves iu the place of a tie-beam to resist the lateral thrust of the ribs. In small domes these ribs may be easily cut from wide plank; but where an extensive structure is required, they must be built in two thicknesses so as to break joints, in the same manner as is described for a roof at Art. 231. They should be placed at about two feet apart at the base, and strutted as at a in 77g. 95. FIG. .96. The scantling of each thickness of the rib may be as fol- lows: For domes of 24 feet-diameter, 1 x 8 inches. “ “ [T1 36 “” “ IX x IO “ “ “ “ бо “ “ 2 х 13 “ “ “ “ 90 “ “ 24 x 13 “ “ é€ [T4 108 [T1 “ 3 x 13 [11 237.—Dome: Curve of Equilibrium.— The surfaces of а dome may be finished to any curve that may be desired, but the framing should be constructed of such form that the curve of equilibrium shall be sure to pass through the middle of the depth of the framing. The nature of this curve is such that, if an arch or dome be constructed in accordance with it, no one part of the structure will be less capable than another of resisting the strains and pressures to which the whole fabric may be exposed. The curve of equilibrium for an arched vault or a roof, where the load is equally diffused

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CURVE OF EQUILIBRIUM. 219 over the whole surface, is that of a. parabola (472. 460); for a dome having no /antern, tower, or cupola above it, a cubic parabola (Fig. 97); and for one having a tower, etc., above it, a curve approaching that of an hyperbola must be adopted, as the greatest strength is required at its upper parts. If the curve of a dome be circular (as in the vertical section, Fig.95), the pressure will have a tendency to burst the dome outwards at about one third of its height. Therefore, when this form is used in the construction of an extensive dome, an iron band should be placed around the framework at that height; and whatever may be the form of the curve, a’ band or tie of some kind is necessary around or across the base. FIG. 97. If the framing be of a form less convex than the curve of equilibrium, the weight will have a tendency to crush the ribs inwards, but this pressure may be effectually overcome by strutting between the ribs; and hence it is important that the struts be so placed as to form «бийке horizontal circles. 238,—Cubie Parabola Computed.—Let a 2 (Fig. 97) be the base, and бс the height. Bisect a д at d, and divide ad into roo equal parts; of these give de 26, ef 184, fg 144, A 124, 4 10$, 77 94, and the balance, 83, toja; divide фе into 8 equal parts, and from the points of division draw lines parallel to a 2, to meet perpendiculars from the several points

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220 CONSTRUCTION. * of division in а 0, at the points о,о,о, etc. Then a curve traced through these points will be the one required. .. 239.—Small Domes over Stairways: are frequently made elliptical in both plan and section; and as no two of the ribs Fic. 98. in one quarter of the dome are alike in form, a method for obtaining the curves may be useful. Fic. 99. To find the curves for the ribs of an elliptical dome, let abcd (Fig. 98) be the plan of a dome, and e the seat of one of the ribs. Then take e f for the transverse axis and twice the rise, o g, of the dome for the conjugate, and de-

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COVERING OF DOMES. 221 scribe (according to Arts. 548, 549, etc.) the semi-ellipse ес which will be the curve required for the rib egf. The other ribs are found in the same manner. 240.—Covering for а Spherica! Dome.—To find the shape, let А (Fig 99) be the plan, and 2 the section, of a given dome. From а draw ас at right angles to аб; find the .stretch-out (477. 524) of оф, and make dc equal to it; divide the arc o ? and the line d c each into a like number of equal parts, as 5 (a large number will insure greater accuracy than a small опе); upon с, through the several points of division in c d, describe the arcs odo, 1e 1, 2 f 2, etc. ; make do equal to half the width of one of the boards, and draw os parallel | to ac; join sand a, and from the points of division in the arc ob drop perpendiculars, meeting а sin 7 7 kl; from these points draw 7 4, 7 3, etc., parallel to ас; make 4 o, e 1, etc., on the lower side of ac, equal to da, e 1, etc., on the upper side; trace a curve through the points o, I, 2, 3, 4, c, ОП each side of dc; then oco will be the proper shape for the board. By dividing the circumference of the base А into equal parts, and making the bottom, o 4 o, of the board of a size equal to one of those parts, every board may be made of the same size. In the same manner as the above, the shape of the covering for sections of another form may be found, such as an ogee, cove, etc. | , To find the curve of the boards when laid in horizontal courses, let A В C (Fig. тоо) be the section of a given dome,

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222 CONSTRUCTION. and D Б its axis. Divide BC into as many parts as there are to be courses of boards, in the points 1, 2,3, etc. ; through 1 and 2 draw a line to meet the axis extended at a; then а will be the centre for describing the edges of the board, F. Through 3 and 2 draw 3 2; then 4 will be the centre for de- T ————= H M. D 4 с a did Fic. ror. scribing 7. Through 4 and 3 draw 44; then d will be the centre for G. Bis the centre for the arc то. If this method is taken to find the centres for the boards at the base of the dome, they would occur so distant as to make it impracti- cable; the following method is preferable for this purpose : G being the last board obtained by the above method, ex- tend the curve of its inner edge until it meets the axis, D В, Fic. 102. in e; from 3, through е, draw 3 f, meeting the arc A B in f; join f and 4, f and 5, and fand 6, cutting the axis, D D, in s, я, апа m ; from 4, 5, and 6 draw lines parallel to А C and cutting the axis in с, 2, andr; make c 4 (Fig. 101) equal toc 4in the pre- vious figure, and c s equal to cs also in the previous figure ; then describe the inner edge of the board H, according to Art. 516; the outer edge can be obtained by gauging from the inner edge. In like manner proceed to obtain the next

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DESIGNS OF BRIDGES. 223 board—taking 2 5 for half the chord, and рл for the height of the segment. Should the segment be too large to be de- scribed easily, reduce it by finding intermediate points in the curve, as at 477. 515. 24|.—Polygona! Dome: Form of Angle-Rib.—To ob- tain the shape of this rib, let A G H (Fig. 102) be the plan of a given dome, and C D a.vertical section taken at the line ef. From 1, 2, 5, etc., in the arc C D draw ordinates, paral- lel to А D, to meet fG; from the points of intersection on fG draw ordinates at right angles to fG; make s 1 equal to o 1, 52 equal too2, etc.; then G f B, obtained in this way, will be the angle-rib required. Тһе best position for the sheathing-boards for a dome of this kind is horizontal, but if they are required to be bent from the base to the vertex, their shape may be found in a similar manner to that shown at Fig. 99. BRIDGES. \ AT = EN ‘si FIG. 103 24.2.—Bridges.—Of plans for the construction of bridges, perhaps the following are the most useful. Fig. 103 shows а method of constructing wooden bridges where the banks of the river are high enough to permit the use of the tie- beam, ab. The upright pieces, cd, are notched and bolted on in pairs, for the support of the tie-beam. A bridge of this construction exerts no lateral pressure upon the abut- : ments. This method may be employed even where the banks of the river are low, by letting the timbers for the roadway rest immediately upon the tie-beam. In this case the 1rame- work above will serve the purpose of a railing.

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224 CONSTRUCTION. Fig. 104 exhibits a wooden bridge without a tie-beam. ‘Where staunch buttresses can be obtained this method may be recommended ; but if there is any doubt ot their stability, it should not be attempted, as it is evident that such a sys- tem of framing is capable of a tremendous lateral thrust. . Fic. 104. 243.—Bridges: Built-Rib.— 777. 105 represents a bridge with a duilt-rib (see Art. 231) as a chief support. The curve of equilibrium will not differ much from that of a parabola; this, therefore, may be used—especially if the rib is made Fic. 105. gradually a' little stronger as it approaches the buttresses. As itis desirable that a bridge be kept low, the following table is given to show the least rise that may be given to the rib. ! Span іп Feet.|Least Risein Feet| Span in Feet. [Least Rise in Feet| Span in Feet. |Least Rise in Feet ! 30 . 0.5 120 7 280 24 40 o: 8 140 8 200 28 50 І:4 160 IO 320 32 60 2 180 II 350 39 70 24 200 I2 380 47 80 3 220 14 400° 53 go 4 240 17 IOO 5 260 20 | j———————»——————M—Á———— ————————————

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2 ERSITY DIMENSIONS OF THE BUILT-RI The rise should never be made less than this, but in all - cases greater if practicable ; as a small rise requires a greater quantity of timber to make the bridge equally strong. The greatest uniform weight with which a bridge is likely to be loaded is, probably, that of a dense crowd of people. This may be estimated at 70 pounds per square foot, and the fram- ing and gravelled roadway at 230 pounds more; which amounts to 300 pounds on a square foot. The following rule, based upon this estimate, may be useful in determining the area of the ribs. Rule LXVIT.—Multiply the width of the bridge by the square of-half the span, both in feet, and divide this pro- duct by the rise' in feet multiplied by the number of ribs ; the quotient multiplied by the decimal o-oor1 will give the area of each rib in feet. When the roadway is only planked, use the decimal 0.0007 instead of o-oo11. | Example.—W hat should be the area of the ribs for a bridge of 200 feet span, to rise 15 feet and be 30 feet wide, with three curved ribs? Тһе half of the span is 100, and its square is 10000; this multiplied by 30 gives 300000, and 15 multiplied by 3 gives 45; then 300000 divided by 45 gives 66662, which multiplied by 0-oo11 gives 7-333 feet or 1056 inches for the area of each rib. Such а rib may be 24 inches thick by 44 inches deep, and composed of 6 pieces, 2 in width and з in depth. The above rule gives the area of a rib that would be requisite to support the greatest possible «zzform load. But in large bridges, а variable load, such as a heavy wagon, is capable of exerting much greater strains; in such cases, therefore, the rib should be made larger.* In constructing these ribs, if the span be not over 5o feet, each rib may be made in two or three thicknesses of timber (three thicknesses is preferable), of convenient lengths bolted together; but in larger spans, where the rib will be such as to render it difficult to procure timber of sufficient breadth, they may be constructed by bending the pieces to the proper curve and bolting them together. In this case, where tim- * See Tredgold's Carpentry by Hurst, Arts. 174 to 177.

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226 CONSTRUCTION. ber of sufficient length to span the opening cannot be ob- tained, and scarfing is necessary. such joints must be made as will resist both tension and compression (see 270. 114). To ascertain the greatest depth for the pieces which compose the rib, so that the process of bending may not injure their elasticity, multiply the radius of curvature in feet by the decimal 0-05, and the product will be the depth in. inches. Example.—Suppose the curve of the rib to be described with a radius of roofeet, then what should bethe depth? Тһе radius in feet, 100, multiplied by 0-03 gives a product of 5 inches. White pine or oak timber 5 inches thick would freely bend to the above curve ; and if the required depth of such a rib be 20 inches, it would have to be composed of at least 4 pieces. Pitch pine is not quite so elastic as white pine or oak—its thickness may be found by using the deci- mal 0-046 instead of 0-05. MEE P Wy = С ? XX ХАТЫ SS LT | -- — Fic. 106. 244.—Bridges: Framed Rib.—In spans of over 250 feet, a framed rib, as in Fig. 106, would be preferable to the foregoing. Of this, the upper and the lower edges are formed as just described, by bending the timber to the proper curve. The pieces that tend to the centre of the curve, called radials, are notched and bolted on in pairs, and the cross-braces are halved together in the middle, and abut end to end between the radials. The distance between the ribs of a bridge should not exceed about 8 feet. The roadway should be supported by vertical standards bolted to the ribs

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THE ROADWAY AND ABUTMENTS. 227 at about every то to 15 feet. At the place where they rest on the ribs, a double, horizontal tie should bé notched and bolted on the back of the ribs, and also another. on the un- derside ; and diagonal braces should be framed between the standards, over the space between the ribs, to prevent lat- eral motion. Тһе timbers for the roadway may be as light as their situation will admit, as all useless timber is only an unnecessary load upon the arch. 245.—Bridges: Roadway.—I{ a roadway be 18 feet wide, two carriages can pass without inconyenience. Its width, therefore, should be either 9, 18, 27, or 36 feet, ac- cording to the amount of travel. The width of the foot- path should be two feet for every person. When a stream of. water has a rapid current, as few piers as practicable should be allowed to obstruct its course; otherwise the bridge will be liable to be swept away by freshets. When the span is not over 300 feet, and the banks of the river are of sufficient height to admit of it, only one arch should be employed. Тһе rise of the arch is limited by the form of the roadway, and by the height of the banks of the river (see Art. 243). The rise of the road way should not exceed one in 24 feet, but as the framing settles about one in 72, the - roadway should be framed to rise one in 18, that it may be one in 24 after settling. The commencement of the arch at the abutments—the spring, as it is termed—should not be below high-water mark ; and the bridge should be placed at right angles with the course of the current. | 246. Bridges: Abutments.—The best material for the abutments and piers of a bridge is stone; and no other should be used. The following rule is to determine the ex- tent of the abutments, they being rectangular, and built with stone weighing 120 pounds to a cubic foot. Rule LXVIII —Multiply the square of the height of the abutment by 160, and divide this product by the weight ofa square foot of the arch, and by the rise of the arch ; add unity to the quotient, and extract the square root. Dimin- ish the square root by unity, and multiply the root so dimin-

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228 CONSTRUCTION. ished by half the span of the arch, and by the weight of a square foot of the arch. Divide the last product by 120 times the height of the abutment, and the quotient will be the thickness of the abutment. Example.—Let the height of the abutment from the base to the springing of the arch be 20 feet, half the span 100 feet, the weight of a square foot of the arch, including the great- est possible load upon it, 300 pounds, and the rise of the arch 18 feet: what should be its thickness? The square of the height of the abutment, 400, multiplied by 160 gives 64000, апа зоо by 18 gives 5400; 64000 divided by 5400 gives a quotient of 11-852; one added to this makes 12-852, the square root of which is 3-6; this, less one is 2-6; this mul- tiplied by 100 gives 260, and this again by 300 gives 78000; this divided by 120 times the height of the abutment, 2400, gives 32 feet 6 inches, the thickness required. The dimensions of a pier will be found by the same enis; for, although the thrust of an arch may be balanced by an adjoining arch when the bridge is finished, and while it re- mains uninjured, yet, during the erection, and in the event of one arch being destroyed, the pier should be capable of sustaining the entire thrust of the other. Piers are sometimes constructed of timber their princi- pal strength depending on piles driven into the earth; but such piene should never be adopted where it is possible to avoid them ; for, being alternately wet and dry, they decay much sooner than the upper parts of the bridge. Spruce and elm are considered good for piles. Where the height from the bottom of the river to the roadway 15 great, it isa good plan to cut them off at a little below low-water mark, cap them with a horizontal tie, and upon this erect the posts for the support of the roadway. This method cuts off the part that is continually wet from that which is only occa- sionally so, and thus affords an opportunity for replacing the upper part. The pieces which are immersed will last a great length of time, especially when of elm; for it.is a well-established fact that timber is less durable when subject to alternate dryness and moisture than when it 1s either con- tinually wet or continually dry. lt has been ascertained that \ .

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CENTRING FOR BRIDGES. 229 the piles under London Bridge, after having been driven about 600 years, were not materially decayed. These piles are chiefly of elm, and wholly immersed. 24.7.— Centres for Stone Bridges.—/ig, 107 is a design for a centre for a stone bridge where intermediate supports, as piles driven into the bed of the river, are practicable. Its timbers are so distributed as to sustain the weight of the arch-stones as they are being laid, without destroying the original form of the centre ; and also to prevent its destruc- tion or settlement, should any of the piles be swept away. The most usualerror in badly-constructed centres is that the timbers are disposed so as to cause the framing to rise at the crown during the laying of the arch-stones up Fic. 107. the sides. To remedy this evil, some have loaded the crown with heavy stones; but a centre properly constructed will need no such precaution. Experiments have shown that an arch-stone does not press upon the centring until its bed is inclined to the horizon at an angle of from 30 to 45 degrees, according to the hardness 'of the stone, and whether it is laid in mortar or not. For general purposes, the point at which the pressure com- mences may be considered to be at that joint which forms an angle of 32 degrees with the horizon. At this point the pressure is inconsiderable, but gradually increases towards the crown. The following table gives the portion of the weight of the arch-stones that presses upon the framing at the various angles of inclination formed by the bed of the

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230 CONSTRUCTION. . stone with the horizon. The pressure perpendicular to the curve is equal to the weight of the arch-stone multiplied by the decimal— “О, when the angle of inclination is 32 degrees. :04 К y iE ж < -08 “ “ “ 36 66. 212 ““ “ T: 38 “ -17 à Ы sf 40 % “21 “ 4 ч 42 и .25 “ « “ 44 “ .20 44 44 “ 46 с в T HR í "I -37 ч i 5 брит “ 4 “ “ “ 52 “ 44 E - » 54 - -48 < Z : 56 “ -52 “ 4€ Tm 58 T 54 “ 4€ “ бо Ce From this it is seen that at the inclination of 44 degrees the pressure equals one quarter the weight of the stone; at 57 degrees, half'the weight; and when a vertical line, as аф (Fig. 108), passing through the centre of gravity of the arch-stone, does not fall within its bed, c Z, the pressure may be con- sidered equal to the whole weight of the stone. This will be the case at about бо degrees, when the depth of the stcne is double its breadth. Тһе direction of these pressures is considered in a line with the radius of the curve. The weight upon a centre being known, the pressure may be estimated and the timber calculated accordingly. But it must be remembered that the whole weight is never placed upon the framing at once—as scems to have been the idea had in view by the designers of some centres. Іп building the arch, it should be commenced at each buttress at the same time (as is generally the case) and each side should progress equally towards the crown.. In designing the fram- Fic. 108.

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CENTRE FOR A STONE BRIDGE. 231 ing, the effect produced by each successive layer of stone should be considered. The pressure of the stones upon one ‚side should, by the arrangement of the struts, be counter- poised by that of the stones upon the other side. Over a river whose stream 15 rapid, or where it is neces- sary to preserve an uninterrupted passage for the purposes of navigation, the centre must be constructed without in- termediate supports, and without a continued horizontal tie at the base; such a centre is shown at Fig. 100. In laying the stones from the base up to æ and с, the pieces dd and ба act as ties to prevent any rising at 2. After this, while the stones are being laid from а and from с to 4, they act as struts; the piece fg is added for additional security. Upon this plan, with some variation to suit circumstances, FIG. 109. centres may be constructed for any span usual in stone- bridge building. ? In bridge centres, the principal timbers should abut, ап not be intercepted by a suspension or radial piece between. These should be in halves, notched on each side and bolted. The timbers should intersect as little as possible, for the more joints the greater is the settling ; and halving them together is a bad practice, as it destroys nearly one half the strength of the timber. Ties should be introduced across, especially where many timbers' meet; and as the centre 15 to serve but a temporary purpose, the whole should be de- ' signed with a view to employ the timber afterwards for other uses. For this reason, all unnecessary cutting should be avoided. |

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232 CONSTRUCTION, Centres should be sufficiently strong to preserve a staunch and steady form during the whole process of .build- ing; for any shaking or trembling will have a tendency to prevent the mortar or cement from setting. For this pur- pose, also, the centre should be lowered a trifle immedi- ately after the key-stone 15 laid, in order that the stones may take their bearing before the mortar is set; otherwise the joints will open on the underside. "The trusses, in centring, are placed at the distance of from 4 to 6 feet apart, accord- ing to their strength and the weight of the arch. Between every two trusses diagonal braces should be introduced to prevent lateral motion. In order that the centre may be easily lowered, the frames, or trusses, should be placed upon wedge-formed sills, as is shown at d (Fig. 109). These are contrived so as to admit of the settling of the frame by driving the wedge d with a maul, or, in large centres, with a piece of timber: mounted as a battering-ram. The operation of lowering a centre should be very slowly performed, in order that the parts of the arch may take their bearing uniformly. Тһе wedge pieces, instead of being placed parallel with the truss, are sometimes made sufficiently long and laid through the arch, in a direction at right angles to that shown at 272. 109. This method obviates the necessity of stationing men beneath the arch during the: process of lowering ; and was originally adopted with success soon after the occurrence of an accident, in lowering a centre, by which nine men were killed. To give some idea of the manner of estimating the pres- sures, 1n order to select timber of the proper scantling, cal- culate the pressure (Art. 247) of the arch-stones from z to д (Fig. 109), and suppose half this pressure concentrated at a, and acting in the direction a f. Then, by the parallelogram of forces (Art. 71), the strain in the several pieces compos- ing the frame da may be computed. Again, calculate the pressure of that portion of the arch included between 4 and с, and consider half of it collected at б, and acting in a vertical direction; then, by the parallelogram of forces, the pressure оп the beams 04 and 24 may be found.. Add the, e

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JOINTS OF THE ARCH-STONES. 233 pressure of that portion of the arch which is included be- tween гапа 0 to half the weight of the centre, and consider this amount concentrated at 4, and acting in a vertical direc- Поп; then, by constructing the parallelogram of forces, the pressure upon d j may be ascertained. The strains having been obtained, the dimensions of the Several pieces in the frames dad and bcd may be found by computation, as directed in the case of roof-trusses, from Arts. 226 to 229. The tie-beams dd, dd, if made of suffi- cient size to resist the compressive strain acting upon them from the load at 2, will be more than large enough to resist the tensile strain upon them during the laying of the first part of the arch-stones below a and c. i 248.—Arch-Stones : Joints.— Їп an arch, the arch-stones are so shaped that the joints between them are perpendicu- lar to the curve of the arch, or to its tangent at the point at which the joint intersects the curve. Іп a circular arch, the FIG. 110. joints tend toward {һе centre of the circle; in an elliptical arch, the joints may be found by the following process: To find the direction of the joints for an elliptical arch ; Bic. 111. a joint being wanted at a (Fig. 110), draw lines from that point to the foci, / and f; bisect the angle f a f with the line ad; then a ? will be the direction of the joint.

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234 CONSTRUCTION. To find the direction of the joints for a parabolic arch: a joint being wanted at а (Fig. 111), draw ae at right angles to the axis eg; make cg equal to ce, and join а and g; draw ak at right angles to ас; then a% will be the direction of the joint. Тһе direction of the joint from 0 is found in the same manner. The linesa g and ё f are tan- gents to the curve at those points respectively; and any number of joints in the curve may be obtained by first ascertaining the tangents, and then drawing lines at right angles to them. (See 472. 462.) | JOINTS. 249,—Timber Joints.—The joint shown in Fig. 112 is simple and strong; but the strength consists wholly in the bolts, and in the friction of the parts produced by screwing the pieces firmly together. Should the timber shrink to і | % 2 n LÁ 2 Fic. 112. even a small degree, the strength would depend altogether on the bolts. It would be made much stronger by indent- ing the pieces together, as at the upper edge of the tie-beam а... зі 9 T-— _ FIG. 113. in Fig. 113, or by placing keys in the joints, as at the lower : edge in the same figure. This process, however, weakens the beam in proportion to the depth of the indents. V = eee ^ 28 i с--с--с-- 9 Fic. 114 Fig.114 shows а method of scarfing, or splicing, a tie- beam without bolts. The keys are to be of well-seasoned, . hard wood, and, if possible, very cross-grained. The addi-

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THE SPLICING OF TIMBER. 235- tion of bolts would make this a very strong splice, or even white-oak pins would add materially to its strength. Fig. 115 shows about as strong a splice, perhaps, as can well be made. It is to be Fecontmerided for its simplicity ; as, on account of there being no oblique joints in it, it can be readily and accurately eect баа A complicated joint is the worst that can be adopted; still, some have proposed joints that seem to have little else besides complication to recommend them. In proportioning the parts of these scarfs, the depths of Fic. 115. all the indents taken. together should be equal to one third of the depth of the beam. In oak, ash or elm, the whole length of the scarf should be six times the depth, or thick- ness, of the beam, when there are no bolts; but, if bolts in- stead of indents are used, then three times the breadth ; and when both methods are combined, twice the depth of the beam. The length of the scarf in pine and similar soft woods, depending wholly on indents, should be about 12 times the thickness, or depth, of the beam ; when depend- ing wholly on bolts, 6 times the breadth; and when both methods are combined, 4 times the depth. Fic. 116. Sometimes beams have to be pieced that are required to resist cros&-strains—such as a girder, or the tie-beam of a roof when supporting the ceiling. In such beams, the fibres of the wood in the upper part are compressed ; and therefore a simple butt joint at that place (as in 270. 116) is far preferable to any other. In such case, an oblique joint is the very worst. The under side of the beam being in a state of tension, it must be indented or bolted, or both ; and an iron plate under the heads of the bolts gives a great addition of strength.

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236 : . CONSTRUCTION. Scarfing requires accuracy and care, as allthe'indents should bear equally; otherwise, one being strained more than another, there would be a tendency to splinter off the parts. Hence the simplest form that will attain the object is by far.the best. In all beams that are compressed end- wise, abutting joints, formed at right angles to the direction of their length, areat once the simplest and the best. Fora temporary purpose, 777. 112 would do very well; it would be improved, however, by having a piece bolted on all four sides. zg. 113, and indeed each of the others, since they have no oblique joints, would resist compression well. In framing one beam into another for bearing purposes, such as a floor-beam into a trimmer, the best place to make the mortise in the trimmer is in the neutral line (447/5. 120, 121), which is in the middle of its depth. Some have thought that, as the fibres of the upper edge are compressed, Fic. 117. a mortise might be made there, and the tenon driven in tight enough to make the parts as capable of resisting the compression as they would be without it; and they have therefore concluded аё plan to be the best. This could not be the case, even if the tenon would not shrink; fora joint between two pieces cannot possibly be made to resist compression so well as a solid piece without joints. The proper place, therefore, for the mortise is at the middle of the depth of the beam; but the best place for the tenon, in the floor-beam, is at its bottom edge. For the nearer this is placed to the upper edge, the greater is the liability for it to splinter off; if the joint is formed, therefore, as at 77g. 117, it will combine all the advantages that can be obtained. Dou- ble tenons are objectionable, because the piece framed into is needlessly weakened, and the tenons are seldom so accu- rately made as to bear equally. For this reason, unless the tusk

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THE FRAMING IN A ROOF-TRUSS. 237 ata in the figure fits exactly, so as to bear equally with the tenon, it had better be omitted. And in sawing the shoulders care should be taken not to saw into the tenon in the least, as it would wound the beam in the place least able to bear it. Thus it will be seen that framing weakens both pieces, more orless. It should, therefore, те avoided as much as possible , and where it is practicable one piece should rest upon the other, rather than be framed into it. This re- mark applies to the bearing of floor-beams on a girder, to the purlins and jack-rafters of a roof, etc. In a framed truss for a roof, bridge, partition, etc., the joints should be so constructed as to direct the pressures through the axes of the several pieces, and also to avoid every tendency of the parts to slide. To attain this object, ^ Fic. 118. FIG. 119. FIG. 120. the abutting surface on the end of a strut should be at right angles to the direction of the pressure ; as at the joint shown in Fig. 118 for the foot of a rafter (see 477. 86), in 7g. 119 for the head of a rafter, and in Fig. 120 for the foot of a strut or brace. The joint at Fig. 118 is not cut completely across the tie-beam, but a narrow lipis left standing in the middle, and a corresponding indent is made in the rafter, to prevent the parts from separating sideways. The abutting surface should be made as large as the attainment of other necessary objects will admit. The iron strap is added to prevent the rafter sliding out, should the end of the tie- beam, by decay or otherwise, splinter off. In making the joint shown аё 270. 119, it should be left a little open at a, so as to bring the parts to a fair bearing at the settling of the truss, which must. necessarily take place from the shrink- ing of the king-post and other parts. If the joint is made fair at first, when the truss settles it will cause it to open at

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238 CONSTRUCTION. the under side of the rafter, thus throwing the whole pres- sure upon the sharp edge at a. This will cause an indenta- tion in the king-post, by which the truss will be made to settle further; and this pressure not being in the axis of the rafter, it wil be greatly increased, thereby rendering the rafter liable to split and break. Fic. 121. Fic. 122. Fic. 189. If the rafters and struts were made to abut end to end, as in Figs. 121, 122 and 123, and the king or queen post notched on in halves апа bolted, the ill effects of shrinking would be avoided. This method has been practised with success in some of the most celebrated bridges and roofs in Europe ; and, were its use adopted in this country, the un- seemly sight of a dogged ridge would seldom be met with. Fic. 124. Fic, 125. A plate of cast-iron between the abutting surfaces will equalize the pressure. Fig. 124 1s a proper joint for a collar-beam in a sal roof: the principle shown here should characterize all tie- joints. The dovetail joint, although extensively practised in the above and similar cases, is the very worst that can be employed. Тһе shrinking of the timber, if only to a small

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WHITE-OAK PINS AND IRON STRAPS. 239 degree, permits the tie to withdraw—as is shown at Fig. 125. Тһе dotted line shows the position of the tie after it has shrunk. Locust and white-oak pins are great additions to the strength of a joint. In many cases they would supply the place of iron bolts ; and, on account of their small cost, they should be used in preference wherever the strength ef iron is not requisite. In small framing, good cut nails are of great service at the joints; but they should not be trusted to bear any considerable pressure, as they are apt to be brittle. Iron straps are seldom necessary, as all the joinings in carpentry may be made without them. They can be used to advantage, however, at the foot of suspending-pieces, and for the rafter at the end of the tie-beam. Іп roofs for ordinary purposes, the iron straps for suspending-pieces may be as follows: When the longest unsupported part of the tie-beam is— ` : 10 feet, the strap may Бе 1 inch wide by $ fr thick. 15 “ “ “ 14 [T4 ү “ “ i 20 € “ : “ 2 “ “ 2. € In fastening a strap, its hold on the suspending-piece will be much increased by turning its ends into the wood. Ітоп straps should be protected from rust; for thin plates of iron decay very soon, especially when exposed to dampness. For this purpose, as soon as the strap is made let it be heated to about a blue heat, and, while it is hot, pour over its entire surface raw linseed oil, or rub it with beeswax. Either of these will give it a coating which HARE: will not penetrate.

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SECTION III.—STAIRS. 250.— Stairs: General Itequirements. The STAIRS is that commodious arrangement of steps in a building by which access is obtained from one story to another. Their position, form, and finish, when determined with discrimi- nating taste, add greatly to the comfort and elegance of a structure. As regards their position, the first object should be to have them near the middle of the building, іп order that they may afford an equally easy access to all the rooms and passages. Nextin importance is light; to obtain which they would seem to be best situated near an outer wall, in which windows might be constructed for the purpose ; yet a skylight, or opening in the roof, would not only provide light, and so secure a central position for the stairs, but may be made, also, to assist materially as an ornament to the building, and, what 1s of more importance, afford an oppor- tunity for better ventilation. All stairs, especially those of the most important build- ings, should be erected of s/oze or some equally durable and fire-resisting material, that the means of egress from a burn- ing building may not be too rapidly destroyed. Winding stairs, or those in which the direction is gradu- ally changed by means of winders, or steps which taper in width, are interesting by reason of the greater skill required in their construction; but are objectionable, for the reason that children are exposed to accident by their liability to fall when passing over the narrow ends of the steps. Stairs of this kind should be tolerated only where there is not sufh- cient space for those with /Zyezs, or steps of parallel width. Stairs in one long continuous flight are also objection- able. Platforms or landings should be introduced at inter- vals, so that any one flight may not contain more than about twelve or fifteen steps. The width of stairs should be in accordance with the im-

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* OM Na as n 20: ~ Pt S KHORSABAD.—ASSYRIAN TEMPLE, RESTORED.

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THE GRADE OF STAIRS. 241 portance of the building in which they are placed, varying Írom 3 to 12 feet. Where two persons are expected to pass each other conveniently the least width admissible is 3 feet. Still, in crowded cities, where land is valuable, the space allowed for passages is correspondingly small, and in these stairs are sometimes made as narrow as 24 feet. From 3 to 4 feet is a suitable width for a good dwelling ; while 5 feet will be found ample for stairs in buildings occu- pied by many people; and from 8 to 12 feet is sufficient for the width of stairs in halls of assembly. To avoid tripping or stumbling, care should be exer- cised, in the planning of a stairs, to secure an even grade. To this end, the zoszzg, or outer edge, of each step should be exactly in line with all the other nosings. In stairs com- posed of both flyers and winders, precaution in this regard is especially needed. Іп such stairs, the steps—flyers and winders alike—should be of one width on the line along which a person would naturally walk when having his hand upon the rail. This ¢vead-line, consequently, would be paral- lel with the hand-rail, and is usually taken at a distance of from 18 to 20 inches from the centre of it. In the plan of the stairs this tread-line should be drawn and divided into equal parts, each part being the ¢read, or width of a flyer from the face of one riser to the face of the next. 251.- Тһе Grade of Stairs.—The extra exertion required in ascending a staircase over that for walking on level ground is due to the weight which a person at each step is required to lift; that is, the weight of his own body. Hence the difficulty of ascent will be in proportion to the height of each step, or to the vise, as it is termed. To facilitate the operation of going up stairs, therefore, the risers should be low. The grade of a stairs, or its angle of ascent, depends not only upon the height of the riser, but also upon the width of the step; and this has a certain relation to the riser; for the width of a step should be in proportion to the smallness of the angle of ascent. The distance from the top of one riser to the top of the next is the distance travelled at each step taken, and this dis-

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242 STAIRS. tance should vary as the grade of the stairs; for a person who in climbing a ladder, or a nearly vertical stairs, can travel only r2 inches, or less, at a step, will be able with equal or greater facility to travel at least twice this distance on level ground. Тһе distance travelled, therefore, should be in proportion inversely to the angle of ascent ; or, the di- mensions of riser and step should be reciprocal: a low rise should have a wide step, and a high rise а narrow step. 252.—Pitch-Board : Relation of Rise to Tread .— Among the various devices for determining the relation of the rise to the ¢read, or net width of step, опе is to make the sum of the two equal to 18 inches. For example, for a rise of 6 inches the tread should be 12, for 7 inches the tread should be 11; or— б + 12% WB ӨШ LONE 64 + 114 = 18 | 85 + 95 = 18 pm uw = 71s алт Іс 7$ + 10$ = 18 OE T a = IS This rule 1s simple, but the results in extreme cases are not satisfactory. If the ascent of a stairs be gradual and easy, the length from the top of one rise to that of another, or the hypothenuse of the pitch-board, may be proportionally long; but if the stairs be steep, the length must be shorter. There is a French method, introduced by Blondel in his Cours а” Architecture. Itis referred to in Gwilt’s Encyclo- pedia, Art. 2813. - This method is based upon the assumed distance of 24 inches as being a convenient step upon level ground, and upon 12 inches as the most convenient height to rise when the ascent is vertical. These are French inches, old system. The 24 inches French equals about 257; inches English. With these distances as base and perpendicular, a right- angled triangle is formed, which is used as a scale upon which the proportions of a pitch-board are found. For example, let a line be drawn from any point in the hypothe- nuse of this triangle to the right angle of the triangle; then this line will equal the length of the pitch-board, along the

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PROPORTIONS OF THE PITCH-BOARD. 243. rake, for a stairs having a grade equal to the angle formed by this line and the base-line of the scale. In the absence of the triangular scale, the lengths of the pitch-boards, as found by this rule, may be computed by this expression— W = 2575 —2h; (107.) in which W equals the tread, or base of the pitch-board, and й the riser, or its perpendicular height. For example, let 2 = 6; then— W= 2545 —2x6- 1345. This result is greater than would be proper in some cases. The length of the hypothenuse of the pitch-board should be proportional not only to the angle of ascent (477. 251), but also to the strength and height of the class of people who are to use the stairs. Tall and strong persons will take longer steps than short and feeble people. The hypothe- nuse of the pitch-board should be made in proportion to the distance taken at a step on level ground by the persons who are to use the stairs. If people are divided into two classes, one composed of robust workmen and the other of delicate women and іп- firm men, then there may be two scales formed for the pitch- boards of stairs—one to be used for shops and factories, and the other for dwellings. The distance on level ground trav- elled per step, by men, varies from about 26 to 32 inches, or on an average 28 inches. The height to which men are accustomed to rise on ladders is from 12 to 16 inches at each step, or on the average 14 inches. With these dimensions, therefore, of 14 and 28 inches, a scale may be formed for pitch-boards for stairs, in buildings to be used exclusively by robust workmen. And with 12 and 24 inches another scale may be formed for pitch-boards for stairs, in buildings to be used by women and feeble people. These two scales are both shown in 777. 126. They are made thus: Let C A B bea right angle. Make 4 B equal to 28 inches, and A C equal to 14; then join В and

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244 STAIRS. C. At right angles to C В, from A, draw А F; then with A Е for radius describe the arc F G. Then a line, as 4 K or А L, drawn from А at any angle with А B and limited by the line G F P will give the length of the hypothenuse of the pitch-board, for shop stairs of a grade equal to the angle which said line makes with A 5. From X, perpendicular to А B, draw K N; then К.М will be the proper riser for a pitch-board of which A Nis the tread. So, likewise, L M will be the appropriate riser for the tread АМ. The arc FG is introduced to limit the rake-line of pitch-boards occur- ring between F and C, in order to avoid making them longer than the one at / The scale for the stairs for dwellings is made in the same manner; 4 D — 24 inches being the base, A Е = 12 inches the rise, and H D the line limiting the - rake-lines of pitch-boards. CS MIC? А M N D B Fic. 126. To compute the length of risers and treads, we have for the scale for shops, for those occurring. between F and 5- r—i(28—1): (108.) i= к. (тоо.): and for those between Fand б, we have— >= 4156.8 — 7°; - (то8, А.) t= *is6.8— 5; (109, А.) For the scale for dwellings, we have, for those occurring between А and D— "= { (24—17); (108, B.) — 1 =104— 27; (109, В.)

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STAIRS FOR SHOPS AND FOR DWELLINGS. 245 and for those between 77 and % we have— (opt ry (108, C.) м а= фато", (тоо, С.) where, in each equation, ғ represents the riser, and 2 the tread, or net step. By these formule, the following tables have been com- puted : | STAIRS FOR SHOPS. Tread. Вабо Rise to Tread. | Rise. Tread. Ratio—Rise to Tread. Rise 2: 24: т {о I2. 7:40 I3:20 I to тё 3: 22. Tm 4848 + 4-60 | 12.80 p o ' 1:08 2:50 21: у“ 6: 7:80 12.40 I “ 1-50 4° 20: i ge 8. I2: т 7977-15 4:50 19: І 4 4-28 8.20 II:6 I '* NM 5. 18. I “ 3:60 8.50 ІІ Іі.“ I-29 5:4 I7:20 I. '* 3949 8.80 IO:40 I ' NH 5:7 16-60 I “ Or 9: 29: | г “жым 6. 16. т '* 2467 9-30 9:40 Jb ос 6-25 15:50 I “ 225 9:60 8-80 І "002 6:50 15. 90: 10: 8. Т.“ 0-80 6-70 14°60 т “ane 10:50 7 I “ 0-67 6:90 14:20 I '* 2.06 II; 6. I “ 0:55 m I4: mM qu n x59 |е 4595 I “ 0:43 7:20 I3:60 nm “ 50-1 12. 2:58 14 0:30 STAIRS FOR DWELLINGS. Rise Tread. |Ratio—Rise to Tread. Rise. Tread. |Ratio—Rise to Tread. 2: 20: I to 10. 7:40 9:20 I to I-24 3: 18. p ow 7:50 9- I “ I-2b 3:50 17: I “ 4.86 7-60 8.80 І MESS fuo 4: 16. p TAX 7:70 8.бо i "ре 4°50 15: re **45 5.53 7:80 8-40 Т 4 гой 5: 14: 4 2-50 7:90 8.20 y '* I-Oq 5-40 13:20 Г-747 4.41 8. 8. мы 5-70 12-60 мй 64 «© оу 8.10 7:80 I “ 0o96 6. 12: ИЕ" 8-30 7°40 I ‘* 0-89 6-25 II+50 IN 1 Oy 8.50 7 I “4: 6-08 6-50 ІІ: lei qe69 8-75 6-50 г.“ QA 6-75 10:50 a ^^ 66 9: 6. 1797 OU 7. IO: м. y dag 9:30 5.40 І i» 0-58 7°10 9*80 p ' 4.38 9:60 4:80 L ОЙ 7:20 9-60 Hag IO: 3:90 I » 0:39 7:30 9:40 Ia 1:29 10:50 2:20 І О.21 These tables will be useful in determining questions in-

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246 STAIRS. volving the proportion between the rise and tread of a pitch-board. For stairs in which the run is limited, to determine the number of risers which would give an easy ascent: Divide the run by the Летейї, and find in the proper table, above, the ratio nearest to the quotient, and in a line with this ratio, in the second column to the left, will be found the corre- sponding riser. With this divide the zzse in inches; the guo- tent, or the nearest whole number thereto, will be the required number of risers in the stairs. | Example.—For the stairs in a dwelling, let the rise be 12’ 8” ог 1524nches. Let the run between the extreme risers be 17 2". To this, for the purpose of obtaining the correct angle of ascent, by having an equal number of risers and treads, add, for one more tread, say 10 inches, its probable width; thus making the total run 18 feet, or 216 inches. Thus we have for the run 216, and for the rise 152. Divid- ing the former by the latter gives 1-42 nearly. In the table of stairs for dwellings, the ratio nearest to this is 1-43, and in the line to the left, in the second column, is 7, the approxi- mate size of riser appropriate to this case. Dividing the rise, 152 inches, by this 7, we have 215 as the quotient. This is nearer to 22 than to 21 ; therefore, the number of risers required is 22. When the number of risers 15 determined, then the rise divided by this number will give the height of each riser ; thus, in the above case, the rise is 152 inches. This divided by 22 gives 6-909 inches for the height of the riser. When the height of the riser is known, then, if the run is unlimited, the widtb of tread will be found in the proper table above. For example, if the riser is 7 inches or nearly that, then in the table of stairs for dwellings, in the next column to the right, and opposite 7 in the column of risers, is found 10, the approximate width of tread. By the use of equation (109, B.), the width may be had exactly according to the scale. For example, equation (109, B.) with 6-91 for the riser, becomes— _ # = 24 — 2 x 6:91 = 10:18, or about 10,8, inches.

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TO CONSTRUCT THE PITCH-BOARD. 247 When the run is limited and the number of risers is known, then the width of tread is obtained by dividing the run by the number of treads. There are always of treads one 1655 than there are of risers, in each flight. i 253.—Dimensions of the Pitch-Board.—The first thing in commencing to build a stairs is to make the 22/с/-Ьоага; this is done in the following manner: Obtain very accurate- ly, in feet and inches, the rise, or perpendicular height, of the story in which the stairs are to be placed. This must be taken from the top of the lower floor to the top of the upper floor. Then, to obtain the number of rises and treads and their size, proceed as directed in 477. 252. Having obtained these, the pitch-board may be made in the following man- ner: Upon a piece of well-seasoned board about 5 of an inch thick, having one edge jointed straight and square, lay the corner БА of a steel square, as shown at Fig. 127. Make a equal to the riser, and dc equal to the tread; mark along the edges with a knife, and cut by the marks, making the edges of the pitch-board perfectly square. The grain of the wood should run in the direction indicated in the figure, because, in case of shrinkage, the rise and the tread will be equally affected by it. When a pitch-board is first made, the dimensions of the riser and tread should be preserved in figures, in order that, in case of shrinkage or damage otherwise, a second may be made. Fic, 127. 254.—The String of a Stairs. — The space required for timber and plastering under the steps is about 5 inches for ordinary stairs, or 6 inches if furred; set a gauge, there- fore, at 5 ог 6 inches, as the case requires, and run it on the lower edge of the plank, as a^ (Fig. 128). Commencing at one end, lay the longest side of the pitch-board against the gauge-mark, a ё, as at c, and draw by the edges the lines for the first rise and tread ; then place it successively as at d, e,

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248 . STAIRS. and f, until the required number of risers shall be laid down. To insure accuracy, it is well to ascertain the theoretical raking length of the pitch-board by computation, as in note. to Art. 536, by getting the square root of the sum óf the squares of the rise and run, and using this by which to divide the line аф into equal parts. с 4 е ҒА " i a Fic. 128. 255.—Step and Riser Connection.—/7e. 129 represents a section of step and riser, joined after the most approved method. In this, а represents the end of a block about 2 wy Fic. 120. inches long, two or three of which, in the length of the step, are glued in the corner. The cove at 015 planed up square, glued in, and stuck or moulded after the glue is set. PLATFORM STAIRS. 256.—Platform Stairs: the Cylinder.—A platform stairs ascends from one story to another in two or more flights, having platforms or landings between for resting and to change their direction. This kind of stairs, being simple, is

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CYLINDER OF PLATFORM STAIRS. 249 easily constructed, and at the same time is to be preferred to those with winders, for the convenience it affords in use (Art. 250). The cylinder may be of any diameter desirable, from a few inches to 3 or more feet, but it is generally small, about 6 inches. It may be worked out of one solid piece, but a better way is to glue together 3 pieces, as in 7g. 130; in which the pieces a, 0, and c compose the cylinder, and Z and e represent parts of the strings. The strings, after being glued to the cylinder, are secured with screws. The joining at о and о is the most proper for that kind of Joint, 5 те 4 FIG. 130. ж FIG. 121. 257. — Form of Lower Edge of Cylinder. — Find the stretch-out, de (Fig. 131), of the face of the cylinder, a £c,

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250 STAIRS. according to Art. 524; from 4 and e draw df and eg at right angles to de; draw 4g parallel to Ze, and make £f and gz each equal to one riser; from 7 and f draw 77 and fk parallel to 22; place the tread of the pitch-board at these last lines, and draw by the lower edge the lines £4 and 22; parallel to these draw mx and ор, at the requisite distance for the dimensions of the string; from 5, the centre of the plan, draw sq parallel to df; divide л and g g each into two equal parts, as at v and w; from v and w draw vz and wo parallel to fd; join z and o, cutting gs in r; then the angles unr and rot, being eased off according to Art. 521, will give the proper curve for the bottom edge of the cylinder. A centre may be found upon which to describe these curves, thus: from u draw их at right angles to mn; from r draw rx at right angles to zo; then x will be the dente for the curve zr. The centre for the curve r£ may be found ina similar manner. Centres from which to strike these curves are usually quite unnecessary; an experienced workman will readily form the curves guided alone by his practised eye. ; Fic. me. 258. the first baluster, д (Fig. 132), half its diameter from the face of the riser, cd, and one third its diameter from the end of the step, ed; and place the centre of the other baluster, a, half the tread from the centre of the first. А line through the centre of the rail will occur vertically over the centres of the balusters. Тһе usual length of the balusters is 2 feet 5 inches and 2 feet 9 inches respectively, for the short and long balusters. Their length may be greater than is here indicated, but, for safety, should never beless. The differ- ence in length between the short and long balusters is equal to one half the height of a riser.

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CONSTRUCTION OF WINDING STAIRS. 251 259.—Winding Stairs: have the steps narrower at one end than at the other. Іп some stairs there are steps of parallel width incorporated with the tapering steps; in this case the former are called flyers, and the latter winder-. 260.— Regular Winding Stairs. [n Fig. 133, abcd rep- resents the inner. surface of the wall enclosing the space allotted to the stairs, « e the length of the steps, and e fgh the cylinder, or face of the front-string. The linea eis given as the face of the first riser, and the point 7 for the limit of ` : JS a ПОЕНЕ j “ Tucs 139. the last. Make 27 equal to 18 inches, and upon o, with 02 for radius, describe the arc 77; obtain the number of risers and of treads required to ascend to the floor at 7, according to Art. 252, and divide the arc 77 into the same number of equal parts as there are to be treads: through the points of division, I, 2, 3, etc., and from the wall-string to the front- string, draw lines tending to the centre, о: then these lines will represent the face of each riser, and determine the form and width of the steps. Allow the necessary projection for the nosing beyond ae, which should be equal to the thick-

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252 STAIRS. ness of the step, and then a e/ £ will be the dimensions for each step. Make a pitch-board for the wall-string having a 2 for the tread, and the rise as previously ascertained; with this lay out on a thicknessed plank the several risers and treads, as at Fug, 128, gauging from the upper edge of the string for the line at which to set the pitch-board. Upon the back of the string, with a 14-inch dado plane, make a succession of grooves ті inches apart, and parallel with the lines for the risers on the face. These grooves must be cut along the whole length of the plank, and deep enough to admit of the plank's bending around the curve абса. Then construct a drum, or cylinder, of. any com- mon kind of stuff, made to fit a curve with a radius tbe thickness of the string less than да; upon this the string must be bent, and the grooves filled with strips of wood, called keys, which must be very nicely fitted and glued in. After it has dried, a board thin enough to bend around on the out- side of the string must be glued on from one end to the other, and nailed with clout-nails. In doing this, be careful not to nail into any place opposite to where a riser or step is to enter on the face. After the string has been on the drum a sufficient time for the glue to set, take it off, and cut the mortices for the steps and risers on the face at the lines previously made; which may be done by boring with a centre-bit half through the string, and nicely chiselling to the line. Тһе drum need not be made to extend over the whole space occupied by the stairs, but merely so far as requisite to receive one piece of the wall-string at a time; for it is evident that more than one will be required. Тһе front-string may be constructed in the same manner; taking e instead of a & for the tread of the pitch-board, dadoing it with a smaller dado plane, and bending it on a drum of the proper size. 261.— Winding Stairs: Shape and Position of Timbers.— The dotted lines in Fig. 133 show the position of the timbers as regards the plan; the shape of each is obtained as follows: In Fig. 134, the line 1 а is equal to a riser, less the thickness of the floor, and the lines 2 m, 3 n, 4 0, 5 , and 6 are each

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TIMBERS FOR WINDING STAIRS. 253 equal to one riser. The linea 2 is equal to a m in Fig. 133, the line m 3 to m n in that figure, etc. In drawing this figure, commence at a, and make the lines a 1 and a 2 of the length above specified, and draw them at right angles to each other; draw 2 » at right angles to a 2, and m3 at right angles to m 2, and make 2 апа m з of the lengths as above specified; and so proceed to the end. Then through the points т, 2,3, 4, 5, and 6trace the line 10; upon the points 1, 2, 3, 4, etc., with the size of the timber for radius, describe arcs as showin in the figure, and by these the lower line may be traced parallel to the upper. This will give the proper shape for the timber, a 4, in Fig. 133; and that of the others may be found in asimilar manner. In ordinary cases, the shape of one face of the timber will be sufficient, for a good workman can easily hew it to its proper level by that; but where great accuracy is desirable, a pattern for the other side may be found in the same man- a mn » o ё 2 ІШ ud > ue т. = a Fic. 134. ner as for the first. In many cases, the timbers beneath cir- cular stairs are put up after the stairs are erected, and with- out previously giving them the required form; the work- man in shaping them being guided by the form marked out by the lower edge of the risers. 262. — Winding Stairs with Flyers: Grade of Front- String.—In stairs of this kind, if the winders are confined to the quarter circle, the transition from the winders to the flyers is too abrupt for convenience, as well as in appear- ance. To remove this unsightly bend in the rail and string, it is usual to take in among the winders one or more of the flyers, and thus graduate the width of the winders to that of the flyers. But this is not always done so as to secure the best results. By the method now to be shown, both rail and strings will be gracefully graded. Іп 27. 135, а б repre- sents the line of the facia along the floor of the upper story,

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254 STAIRS. бес the face of the cylinder, and cd the face of the front- string. Make g^ equal to of the diameter of the baluster, and parallel to a 0, b ec, and cd draw the centre-line of the rail, fg, z hi, and Z7; make г and gZ each equal to half the width of the rail, and through & and 2, parallel to the centre- line, draw lines for the convex and the concave sides of the rail; tangical to the convex side of the rail, and parallel to km, draw по; obtain the stretch-out, g 7, of the semicircle, k p m, according to Art. 524; extend a £ to 2, and ёз to s; make c s equal to the length of the steps, and 2 и equal to 18 inches, and parallel to z p describe the arcs 57 and u 6; from 7 draw 2 w, tending to the centre of the cylinder; from 6, and on the line 6 z x, run off the regular tread, as at 5, 4, 3, 2, 1, and о; make и x equal to half the arc 4 6, and make the point of division nearest to x, as v, the limit of the par- allel steps, or flyers; make ғ о equal to mz; from о draw о a?* at right angles to z o, and equal to one riser; from a? draw a? 5 parallel to xz о, and equal to one tread; from s, through o, draw 5 07, Then from w draw w c? at right angles to л o, and set up on the line w c? the same number of risers that the floor, A, is above the first winder, P,as at 1, 2,3, 4, 5, and 6; through 5 (on the arc би) draw а? e?, tending to the centre of the cylinder; (гот e? draw e? f? at right angles to жо, and through 5 (on the line ж c?) draw g? f° parallel to л o; through 6 (on the line тес?) and f? draw the line 2% 0°; make 6 c? equal to half a riser, and from c? and 6 draw с? г? айа 6 7* parallel to ж о; make £*.z*vequal to &* /*; from 2° draw 7? 2°? at right angles to 7? 27, and from f? draw f? 27 at right angles to f? 22%; upon $°, with Ё? f°? for radius, de- scribe the arc /° 2°; make 2? /? equal to 2? f°, and ease oft the angle at 2? by the curve f? /?, In the figure, the curve is described from a centre, but as this might be imprac- ticable in a full-size plan, the curve may be obtained accord- —— * In the references a°, 07, etc., а new form is introduced for the first time. During the time taken to refer to the figure, the memory of the form of these may pass from the mind, while that of the sound alone remains; they may then be mistaken fora2,22,etc. This can be avoided in reading by giving them a sound corresponding to their meaning, which is a second, б second, etc.

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MOULDS FOR QUARTER-CIRCLE STAIRS. 255 ing to Art. 521. Then from 1; 2, 3, and 4 (on the line w c?) draw lines parallel to z o, meeting the curve in 27°, z?, o?, and p*; from these points draw lines at right angles to z o, . and meets! Шектеу” z?, 5°, and 7*; from x? and >*% атаў end ы 23 43 Lv шы” J Fic. 135. lines tending to «*, and meeting the convex side of the rail in y? and 2°; make mv? equal tor 53, and m w? equal to 7 1*; from y? z?,v?, and зо, through 4, 3, 2, and т, draw lines meeting the line of the wall-string in а, 05, c°, and d°; from

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256 , STAIRS. e?, where the centre-line of the rail crosses the line of the floor, draw e? f? at right angles to z o, and from f°, through б, draw f?g*;-then the heavy lines f? g?, 222°, yea 2? 53, v?*c?, qo? d?, and zy will be the lines for the risers, which, being extended to the line of the front-string, 2 e c d, will give the dimensions of the winders and the grading of the front-string, as was required. HAND-RAILING. 263.—Hand-Railing for Stairs.—A piece of hand-rail- ‘ing intended for the curved part of a stairs, when properly shaped, has a twisted form, deviating widely from plane sur- faces. If laid upon a table it may easily be rocked to and fro, and can be made to coincide with the surface of the table in only three points. And yet it is usual to cut such twisted pieces from ordinary parallel-faced plank; and to cut the plank in form according to a /ace-mould, previously formed from given dimensions obtained from the plan of the stairs. The shape of the finished wreath differs so widely from the piece when first cut from the plank as to make it appear to a novice a matter of exceeding difficulty, if not an impossibility, to design a face-mould which shall cover accu- rately the form of the completed wreath. But he will find, as he progresses in a study of the subject, that it is not only а possibility, but that the science has been reduced to such a system that all necessary moulds may be obtained with great facility. To attain to this proficiency, however, re- quires close attention and continued persistent study, yet no more than this important science deserves. The young car- penter may entertain a less worthy ambition than that of desiring to be able to form from planks of black-walnut or mahogany those pieces of hand-railing which, when secured together with rail-screws, shall, on applying them over the stairs for which they are intended, be found to fit their places exactly, and to form graceful curves at the cylinders. That railing which requires to be placed upon the stairs before cutting the joints, or which requires the curves or butt-joints to be refitted after leaving the shop, is discredit-

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PRINCIPLES OF HAND-RAILING. 297 able to the workman who makes it. No true mechanic will be content until he shall be proved able to form the curves and cut the joints in the shop, and so accurately that no altera- tion shall be needed when the railing is brought to its place on the stairs. The science of hand-railing requires some knowledge of descriptive geometry—that branch of geornetry which has for its object the solution of problems involving three dimensions by means of intersecting planes. The method of obtaining the lengths and bevils of hip and valley rafters, etc., as in Art. 233, is a practical example of descrip- tive geometry. The lines and angles to be developed in problems of hand-railing are to be obtained by methods dependent upon like principles. 264..-—Hand-Railing: Definitions; Planes and Solids, —Preliminary to an exposition of the method for drawing the face-moulds of a hand-rail wreath, certain terms used in descriptive geometry need to be defined. Among the tools used by a carpenter are those well-known implements called planes, such as the jack-plane, fore-plane, smoothing-plane, etc. These enable the workman to straighten and smooth the faces of boards and plank, and to dress them out of wind, or so that their surfaces shall be true and unwinding. The term plane, as used in descriptive geometry, however, refers not to the implement aforesaid, but to the unwinding surface formed by these implements. A plane in geometry is defined to be such a surface that if any two points in it be joined by a straight line, this line will be in contact with the surface at every point in its length. With like results lines may be drawn in all possible directions upon such a sur- face. This can be done only upon an unwinding surface; therefore, a plane is an unwinding surface. Planes are understood to be unlimited in their extent, and to pass freely through other planes encountered. The science of stair-building has to do with prisms and cylinders, examples of which are shown in 2205. І 36, 137, and 138. A right prism (igs. 136 and 137) is a solid standing upon a horizontal plane, and with faces each of which is a plane. Two of these faces—top and bottom—are horizontal

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258. Be STAIRS. and are equal polygons, having their corresponding sides parallel. | The other faces of the prism are parallelograms, each of which:is a vertical plane. When the vertical sides of a prism are of equal width, and in number increased indefi- nitely, the two polygonal faces of the prism do not differ essentially from circles, and thence the prism becomes a cylinder. Thus a right cylinder may: be defined to be a prism, with circles for the horizontal faces (Fig. 138). Rig. 136: PIG 971 Fic. 138. 265.— Hand- Railing: Preliminary Considerations, — If within the zwed/-hole, or stair-opening, of a circular stairs a solid cylinder be constructed of such diameter as shall fill the well-hole completely, touching the hand-railing at all points, and then if the top of this cylinder be cut off on a line with the top of the hand-railing, the upper end of the cylinder would present a winding surface. But if, instead of cutting the cylinder as suggested, it be cut by several planes, each of which shall extend so as to cover only one of the wreaths of the railing, and be so inclined as to touch its top in three points, then the form of each of these planes, at its intersec- tion with the vertical sides of the cylinder, would present the shape of the concave edge of the face-mould for that particular piece of hand-railing covered by the plane. Again, if a hollow cylinder be constructed so as to be in contact with the outer edge of the hand-railing throughout its length, and this cylinder be also cut by the aforesaid

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FACE-MOULDS FOR HAND-RAILS. 259 planes, then each of said planes at its intersection with this latter cylinder would present the form of the convex edge of the said face-mould. ` A plank of proper thickness may now have marked upon 1t the shape of this face-mould, and the piece covered by the face-mould, when cut from the plank, will evidently contain a wreath like that over which the face-mould was formed, and which, by cutting away the surplus material above and below, may be gradually wrought into the graceful form of the required wreath. ` By the considerations here presented some gengral idea may be had of the method pursued, by which the form of a face-mould for hand-railing is obtained. А little reflection upon what has been advanced will show that the problem to be solved is to pass a plane obliquely through a cylinder at certain given points, and find its shape at its intersection with the vertical surface of the cylinder. Peter Nicholson was the first to show how this might be done, and fer the invention was rewarded, by a scientific society of London, with a gold medal. Other writers have suggested some slight improvements on Nicholson's methods. The method to which preference is now given, for its simplicity ot work- ing and certainty of results, is that which deals with the tangents to the curves, instead of with the curves themselves; so we do not pass a plane through a cylinder, but through a prism the vertical sides of which are tangent to the cylinder, and contain the controlling tangents of the face-moulds. The task, therefore, is confined principally to finding the tangents upon the face-mould. This accomplished, the rest is easy, as will be seen. The method by which is found the form of the top of a prism cut by an oblique plane will now be shown. 266.—A Prism Cut by an Oblique Plane.— А prism is shown in perspective at Fig. 139, cut by an oblique plane. | The points abcd are the angles of the horizontal base, and abg, bcf, сасу, and adeg are the vertical sides; while , e fbg is the top, the form of which is to be shown. 267. -—Form of Top of Prism.—In 2/2. 139 the form of the top of the prism is shown as it appears in perspective,

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260 STAIRS. not in its read shape; this is now to be developed. In Fig. 140, let the. sauare aécd represent by scale the actual form a FIG. 139. and size of the base, adcd, of the prism shown in 270. 139. Make c c, and Z4, respectively equal to the actual heights at UU æ a == Fic. 140. c f and de, Fig. 139; the lines dd, and сс, being set up per- pendicular to the line dc. Extend the lines dc and 4, с, until

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ILLUSTRATION BY PLANES. 261 they meet in 4; join ^ and Л. Now this line 44 is the inter- section of two planes: one,the base, or horizontal plane upon which the prism stands; the other, the cuz/izg plane, or the plane which, passing obliquely through the prism, cuts it so as to produce, by intersecting the vertical sides of the prism, the form Fep F 130. To show that 04 is the line of intersection of these two planes, let the paper on which the triangle ZZ d, is drawn (designated by the letter 5) be lifted by the point Z, and revolved on the line ZZ until d, stands vertically over а, and c, over c; then Б will be a plane standing on the line Z7, vertical to the base-plane A. Тһе point / being in the line cd extended, and the line cd being in the base-plane 4, there- fore % is іп the base-plane A. Now the line Z c, represents the line e f of Fig. 139, and is therefore in the cutting plane ; consequently the point 2, being also іп the line d,c, ex- tended, is also in the cutting plane. By reference to Fig. 139 it will be seen that the point å is in both the cutting am base planes; we must therefore conclude that, since the two points J and 2 are in both the cutting and base planes, a line joining these two points must be the intersection of these two planes. The determination of the line of intersection of the base and cutting planes is very important, as it is a control- ling line; as will be seen in defining the lines upon which the form of the face-mould depends. Care should therefore be taken that the method of obtaining it be clearly under- stood. It will be observed that the intersecting line 24, being in the horizontal plane 4, is therefore a horizontal line. Also, that this horizontal line 0 4 being a line in the cutting plane, therefore all lines upon the cutting plane which are drawn parallel to ? 4 must also be horizontal lines. The import- ance of this will shortly be seen. Through a, perpendicular to oh, draw the line ё, @,, and parallel with this line draw dd on d as centre describe the arc 4,4,,,; draw 4, 4, 1? parallel with dd, „апа extend the latter to 4,,; on 4, as centre describe the arc 4,4,,; join 9, and Z,. We now have three vertical planes which are to be brought into position around the base-plane А, as follows: Revolve B

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262 | STAIRS. upon ZZ, E upon ZZ,, and C upon 2, d, each until it stands U 202 perpendicular to the plane. 4. Then the points Z, and а, 444 will coincide and be vertically over d; the points Z,, and 4, will coincide and stand vertically over Z, ; and с, will cover с. 4” These vertical planes will enclose а wedge-shaped figure, lying with one face, 2,, d,d h4, horizontal and coincident with the base-plane A, and three vertical faces, ?,, d, Z,,,, d d,, d, d. and Add. By drawing the figure upon a piece of stout paper, cutting it out at the outer edges, making creases in the lines hd, dd,, 4,0,, then folding the three planes В, Ж, and C at right angles to А, the relation of the lines will be readily seen. Now, toobtain the form of the top or cover to the wedge-shaped figure, perpendicular to 2, 2,,, draw 0,, 4 4/44 7471 4 and Z,,e; on ё, as centre describe the arc 2й,; make Z,,e 43 4/4 equal to d,d; join e and 2. Now the form of the top of the . wedge-shaped figure is shown within the bounds d, b, с. - By revolving this plane 2 on the line 2,,Z,, until it is at a right angle to the plane C, and this while the latter is supposed to be vertical to the plane A, it will be perceived that this movement will place the plane D on top of the wedge-shaped figure, and in such a manner as that the point e will coincide with 2, Z,, and the point Z, will fall upon and be coincident with the point Z4, and the lines of the cover will coincide with the corresponding lines of the top edges of the sides of the figure; for example, the line 4,d,,, is common to the top and the side С; the line d,,,e equals d, d, which equals а,4,,; therefore, the line Z,,e will comcide with d,d, of the side Æ; the line eZ, will coincide with Z,Z of the side B; and the line 5, 4, will coincide with the line b,h. Thus the figure D bounded by 2, ей, will exactly fit as a cover to the wedge-shaped figure. Upon this cover we may now develop the form of the top of the prism. Preliminary thereto, however, it will be observed, as was before remarked, that lines upon the cutting plane which are parallel to the intersecting line 2, £4, are horizontal; and each, therefore, must be of the same length as the line in the base-plane A vertically beneath it. For example, the line Z,,e, is a line in the cutting plane D, parallel with the line д,,7, in the same plane, and this line 4,,4, will (when the

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EXPLANATION OF THE DIAGRAMS. 263 cutting plane D is revolved into its proper position) be co- incident with the intersecting line 2, 2; therefore, the line а „е is а line in the cutting plane 2, drawn parallel with the intersecting line 2,0. Now this line Z,, e, when in position, will be coincident with the line Z,,,, dy, which lies vertically over the line Z, Z-of the base-plane 4; its length; therefore, is equal to that of the latter. In like manner it may be shown that the length of any line on the plane 2 parallel to 5, À,is equal in length to the corresponding line upon the plane A vertically beneath it. x Therefore, to obtain the form of the top of the prism, we proceed as follows: Perpendicular to 2,4, draw сс, and аа,,; perpendicular 0,4, draw с, and equal toc, с; on ё, as centre describe the arc 25,; join д,а,,, Ё, f, and «,, e. Now we have here in plane D the form of the top of the prism, as shown in the figure bounded by the lines a,b, f e. This will be readily seen when the plane D is revolved into position. Then the point a,, will be vertically over а; the point e coincident with 4, Z,,,, and vertically over 2; the point f coincident with c, and vertically over c; while 2, will coin- cide with 4 of the base-plane A. The figure a,, 2, fe, therefore, represents correctly both in form and size the top of the prism as it is shown in per- spective at Ф f eg, Fig. 139. The line e f, Fig. 140, is equal to the line Z, c, and so of the other lines bounding the edges of the figure. The cutting plane 6 feg, Fig. 139, may be taken to repre- sent the surface of the plank from which the wreath of hand- railing is to be cut; the wreath curving around from 9 to e, as shown in Fig. 141, the lines dg and ge being tangent to the curve in the cutting plane; while ад and а 4 are tan- gents to the curve on the base plane, or plane of the cylin- der. The location of the cutting plane, however, is usually not at the upper surface of the plank, but midway between the upper and under surfaces. The tangents in the plane are found to be more, conveniently located here for deter- mining the position of the butt-joints. Fora moulded rail two curved lines, éach witb a pair of tangents, are required upon the cutting plane, one for the outer edge of the rail,

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264 STAIRS. and the other for the inner edge; but for a round rail only one curve with its tangents is required, as that from д to e іп 276. 141, which is taken to represent the curved ‘line run- ning through the centre of the cross-section of the rail. As an easy application of the principles regarding the prism, just developed, an example will now be given. e Fic. 141. 268.—Face-Mould for Hand-Railing of Platform Stairs. —Let 7 £ and / m, Fig. 142, represent the central or axial lines of the hand-rails of the two flights, one above, the other be- low the platform; and let the semicircle / 2 / be the central line of the rail around the cylinder at the platform, the risers at the platform being located at у and 4 Vertically over the platform risers draw gg,; make gr, equal to a riser of the lower flight, and х g, and ss, each equal to a riser of the upper flight. Draw 2,5 and g&, horizontal and equal each to a tread of each flight respectively. Through r, draw Ёа, and through g, draw 5,7. Vertically over d draw a, 7, Horizontally draw a,,a,,,, and 7, 7,,. It 16 usual to extend the wreath of the cylinder so as to include a part of the straight rail—such a part as convenience may require. Let the straight part here to be included ex- tend from / to ё on the plan. Vertically over д draw 5, c,,,,, and horizontally draw 0, w,,; at any point on 4, w, locate w and make w, w, equal to 77, and bisect it in о; erect the perpendiculars w,a,,,,, W d,,,, and w, v ; join ¢,,and «,,,, ; from d,,, horizontally draw 4,, d,,; parallel with r,£, draw v 7 P. We now have the plan and elevations of the prism, 44? M PPP

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FACE-MOULD FOR PLATFORM STAIRS. 265 containing at its angles the tangents required for the wreath extending from 2 to d on the plan. The elevation / isa view of the cylinder looking in the direction dc. Comparing Fig. 142 with Fig. 141, the line 6, w, is the trace, upon a vertical plane, of the horizontal plane abcd

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266 STAIRS. ` of Fig. 141, or is the ground-line from which the heights of the prism are to be taken. The triangle a, ġa, 15 represented іп с, 141 at а bg, and the inclined line д, а, is the tangent of the rail of the lower flight, and is represented mee wat at bg, while æ, 2, 15 the tangent of the railing around the cylinder, and the E of it is represented in 276. 141 at ge. The height £,c,,, 1 shown in Zzg. 141 at c f, while the height жа,,, or a, ЖЕ shown in Fig. 141 at d e. { The vertical planes P Е C may now ‚ be constructed about the prism asin 276. 140, proceeding thus: Make cz, equal to b, ^e, and dd, equal tow Ф; through c, draw d, 2 ; through б draw /02,,; perpendicular to % 6, through a draw б, d,; from d parallel with 2,, d, draw d 4;,, ; on 4 as centre describe thearc d,d dara d,, also d d, ; parallel with Z5, ,; didi ; 4444 оп а), as centre describe the arc d, Z,,; join d, to 2,. Раг- allel Eo b, draw from each important point of the plan, as shown, an ordinate extending to the line 2, Z,,, and thence across plane D draw ordinates perpendicular to b, d, and make them respectively equal to. the corresponding ordinates of the plane 4, measured from the line 2, d,; join e to ла, 100.2. BO 2, v 6, to f; also join / to z. Then «,,,?, is the tangent standing over a4, and a,,eis the tangent standing over ad. The line 2,7, is the part of the tangent which stands over 2Z, the portion of the wreath which is straight. The curve ez, 5,7, is the trace upon the cutting plane of the quarter circle d z pi, traced through the points л,2, and as many more as desirable, found by ordinates as any other point in the plane 4. Thus we have complete the line b, 1, n,e, the central line of the wreath extending from 2 to d in the plan. Thisis the essential part of the face-mould, which is now to be drawn as follows: At 270. 143 repeat the par- allelogram а,,,0, fe of Fig. 142, and, with a radius equal to half the diameter of the rail, describe, from centres taken on the central line, the several circles shown; and tangent to these circles draw.the outer and inner edges of the rail. The joint at 4, is to be drawn perpendicular to the tangent б.а, while that at eis to be perpendicular to the tangent «a,, This completes the face-mould for the wreath over

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WREATHS FOR A ROUND RAIL. 267 b Ind of the plan. If the pitch-board of the upper flight be the same as that of the lower flight, the face-mould at Fig. 143 will, reversed, serve also for the wreath over the other half of the cylinder. In using this face-mould, place it upon a plank equal in thickness to the diameter of the rail, mark its form upon the plank, and saw square through; then chamfer the wreath to an octagonal form, after which carefully remove the angles so as to produce the required round form. Тһе joints, as well as the curved edges, are to be cut square through the plank. Many more lines have been used in obtaining this face- mould than were really necessary for so simple a case, but no more than was deemed advisable in order properly to eluci- date the general principles involved. А very simple method Fic. 143 for face-moulds of platform stairs with small cylinders will now be shown. 269.—More Simple Method for Mand-Rail to Platform Stairs.—In Fig. 144, 7g e represents a pitch-board of the first flight, and 4 and 7 the pitch-board of the second flight ofa plat- form stairs, the line e f being the top of the platform; and abe is the plan of a line passing through the centre of the rail around the cylinder. Through? апа 4 draw 22, and through 7 and e draw jÅ, from £ draw £7 parallel to fe; from 6 draw » parallel to gd; from / draw /r parallel to Ej; from x draw zt at right angles to j$, on the line од make o7 equal to 72; join c and 7 on the line jc, Fig. 145, make ec equal to ez at Fig. 144; from с draw с? at right angles to 7c, and make cż equal to c7 at Fzg. 144; through 2 draw 2 / parallel to jc, and make ¢/ equal to ¢/at 272. 144; join 7 and с, and complete the parallelogram ¢c/s; find the points о, о, о, according to 477. 551; upon е, 0, 0, 0, and 4

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208 STAIRS. S successively, with a radius equal to half the width of tlie rail, describe the circles shown in the figure; then a curve traced on both sides of these circles, and just touching them, Fic. 144. will give the proper form for the mould. Тһе joint at 7 is drawn at right angles to c Z Fiod. 145. This simple method for obtaining the face-moulds for the hand-rail of a platform stairs appeared first in the early edi- tions of this work. It was invented by a Mr. Kells, an

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HAND-RAIL TO PLATFORM STAIRS. 269 eminent stair-builder of this city. А comparison with Fig. 142 will explain the use of the few lines introduced. For a full comprehension of it reference is made to Fig. 146, in which the cylinder, for this purpose, is made rectangular -з--,-7-7----. m _ oc Ј / instead of circular. ,The figure gives a perspective view of a part of the upper and of the lower flights, and a part of the platform about the cylinder. The heavy lines, 277, » с, and су, show the direction of the rail, and are supposed to pass through the centre of it. Assuming that the rake of

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270 STAIRS. the second flight is the same as that of the first, as is gener- ally the case, the: face-mould for the lower twist will, when reversed, do for the upper flight; that part of the rail, there- fore, which passes from e to c, and from c to 7, is all that will need explanation. Suppose, then, that the parallelogram eaoc represent а plane lying perpendicularly over еад/, being inclined in the direction ec, and level in the direction co; suppose this 4 Fic. 147. plane ёа ос be revolved on ес as an axis, in the manner indi- cated by the arcs ол and ал, until it coincides with the plane ez£c; the line «o will then be represented by the line an; then add the parallelogram x 777, and the triangle cz /, deducting the triangle evs; then the edges of the plane es dc, inclined in the direction ec, and also in the direction с/, will lie perpendicularly over the plane ea?/. From this we gather that the line co, being at right angles to ес, must, in

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HAND-RAIL FOR LARGE CYLINDER. 271 order to reach the point Z be lengthened the distance zz, and the right angle ec? be made obtuse by the addition to it of the angle zc 2. By reference to Fig. 144, it will be seen that this lengthening is performed by forming the right- angled triangle cof, corresponding to the triangle co? in бар. 146. The line c£ is then transferred to Fig. 145, and placed at right angles to'ec; this angle есі is then increased by adding the angle ¢c/, corresponding to /с/, Fig. 146. Thus the point / is reached, and the proper position and length of the lines ec and cZóbtained. То obtain the face- mould for a rail over a cylindrical well-hole, the same process is necessary to be followed until the length and position of these lines are found; then, by forming the parallelogram ecis, and describing a quarter of an ellipse therein, the proper form will be given. 270.—Wiand-Railing for a Larger Cylinder.— 77. 147 represents a plan and a vertical section of a line passing through the centre of the rail as before. From 6 draw 0 2 parallel to cd; extend the lines zd and 7e until they meet £4 in £ and f; from > draw z/ parallel to o^; through / draw 27 parallel to 72; from £ draw 22 at right angles to 74; оп the line o? make o: equal to £z. Make ec (Fig. 148) equal to e£ at Fig. 147; from с draw сё at right angles to ec, and equal to c7 at Fig. 147; from 2 draw £f parallel to ce, and | make 7/ equal to ¢/ at Ре. 147; complete the parallelogram ecl s, and find the points о, o, o, as before ; then describe the circles and complete the mould as in /zg. 145. The difference between this and Case 1 is that the line c7, instead of being raised and thrown out, is lowered and drawn in. A method of planning a cylinder so as to avoid the necessity of cant- ing the plank, either up or down, will now be shown.

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272 STAIRS. 27\.—Face-Mould without Canting the Plank.—Instead of placing the platform-risers at the spring of the cylinder, a . more easy and graceful appearance may be given to the rail, and the necessity of canting either of the twists entirely obviated, by fixing the place of the above risers at a certain distance within the cylinder, as shown in 77g. 149—the lines indicating the face of the risers cutting the cylinder at # and l, instead of at р and g, the spring of the cylinder. То ascertain the position of the risers, let абс be the pitch- board of the lower flight, and cde that of the upper flight, these being placed so that фе and cd shall forma right line. Extend ас to cut dein f; draw fg parallel ‘to 40, and of indefinite length; draw go at right angles to fg, and equal in length to the radius of the circle formed by the centre of the rail in passing around the cylinder ; on o as centre describe the semi- circle 777; through о draw 25 par- allel to 46; make of equal to the . radius of the cylinder, and describe on о the face of the cylinder 27; then extend 20 across the cylinder, cutting it in 7 and £—giving the position of &he face of the risers, as required’ То find the Таде- mould for the twists is simple and obvious: it being merely a quarter of an ellipse, having o7 for semi- minor axis, and sf for the semi-major axis; or, at 772g. 151, let “с? bea right angle; make cz equal to оу, Fig. 149, and dc equal to sf, Fig. 149; then draw do parallel to ci, апа com- plete the curve as before. | Fic. 149. 272. meets the Level.—In Fig. 150, абс is the plan of a line pass- ing through the centre of the rail around the cylinder as before, and 7e is a vertical section of two steps starting ^ from the floor, 4g. Bisect eZ in а, and through d draw df

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HAND-RAIL AT RAKE AND LEVEL. 27/3 parallel to Ag; bisect fx in 2 and from / draw 22 parallel to пу; from z draw мі at right angles to ул; on the line o? make of equal to zz. Then, to obtain a mould for the twist going up the flight, proceed as at Fig. 145 ; making ec in that figure equal to ez in Fig. 150, and the other lines of а length and. position such as is indicated by the letters of reference in each figure. То obtain the mould for the level FIG. 150. rail, extend фо (Fig. 150) to 2; make o; equal to //, and join iand с; make cz (fig. 151) equal to ciat Fig. 150; through - 0 4 j d 0 Еіс. 151. c draw са at right angles to c; make de equal to df at Fig. 150, and complete the parallelogram 04с2; then pro- ceed as in the previous cases to find the mould. ‚ 273.—Application of Faec-Moulds to Plank, — АП the moulds obtained by the preceding examples have been for round rails. For these, the mould may be applied to a plank of the same thickness as the rail is intended to be, and the

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274 STAIRS. plank sawed square through, the joints being cut square from the face of the plank. А twist thus cut and truly rounded will hang in a proper position over the plan, and present a perfect and graceful wreath. 274.—Face-Moulds for Moulded Rails upon Platform Stairs.—In Fig. 152, а бес is the plan of a line passing through ' ! | D і | ! Fic. 152. the centre of the rail around the cylinder, as before, and the lines above it are a vertical section of steps, risers, and plat- form, with the lines for the rail obtained as in Fig. 144. Set half the width of the rail from 0 to f and from ё to 7, and from f and 7 draw. fe and rd parallel to ca. At Fug. 153 the centre-lines of the rail 7c and с/ are obtained as in the previous examples, making jc equal ju of Fig. 152, c£

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FACE-MOULD APPLIED TO PLANK. 275 equal с2 of Fag. 152, and 7/ equal 5/ of Fug. 152. Make cz and c£ each equal to cz at Fig. 152, and draw the lines zm and £g parallel to с/; make Ze and /r equal to ze and zd at Fig. 152, and draw dz and eg parallel to Zc; also, through 7 draw og parallel to Zc; then, in the parallelograms muro and goeg, find the elliptic curves, dm and eg, according to Art. 551, and they will define the curves. The line 42, being drawn through / perpendicular to /c, defines the joint which is to be cut square through the plank. 275.—Application of Face-Moulds to Plank.—In Fig. 152 make a drawing, from 4 to 4, of the cross-section of the hand-rail, and tangent to the lower corner draw the line g Z. The distance between the lines 7e and g Z is the thickness of the plank from which the rail is to be cut. Lay the face- mould upon the plank, mark its shape upon the plank, and Fic. 153. saw it square through. To proceed strictly in accordance with the requirements of the principles upon which the face- mould is formed, the cutting ought to be made vertically through the plank, the latter being in the position which it would occupy when upon the stairs. Formerly it was the custom to cut it thus, with its long raking lines. But, owing to the great labor and inconvenience of this method, efforts were made to secure an easier process. By investigation it was found that it was possible, without change in the face- mould, to cut the plank square through and still obtain the correct figure for the railing, and this method is the one now usually pursued. Not only is the labor of sawing much re- duced by this change; but to the workman it is an entire re- lief, as he now, after marking the form of the wreath upon the . plank, sends it to a steam saw-mill, and, at a small cost, has it

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276 STAIRS. cut out with an upright scroll.saw. When thus cut out in the square, the upper surface of the plank is to be faced up true and unwinding, and the outer edge jointed straight and square from the face. Then а figure of the cross-section of the hand-railing is to be carefully drawn on the ends of the squared block as shown in /igs.154 and 155, and which are regulated so as to be correctly in position, as follows. First, as to the end 2 of the straight part 27: In Fag. 154, let 20 c d bean end view of the squared block, of which.a e fd is the shape of the end of the straight part. Let the point g be the centre of this end of the straight part; through g draw upon the end ae fd the line 77, so that the angle 2/2 shall be equal to the angle 22с, Fig: 152. This is the angle at which the plank is required to be canted, revolving it on Fic. 155. the axis of the straight part of the rail. Through g draw the line z 2 parallel with ад. Upon a thin sheet of metal (zinc.is preferable) mark carefully the exact figure of the cross-section of the rail, drawing a vertical line through its centre, cut away the surplus metal, then, with this template as a pattern, mark upon the end a e fd, Fig. 154, the figure of the rail as shown, the vertical line upon the template being made to coincide with the line / k. From z and # draw the vertical lines 7 »; and 7 z parallel with 7 4. Now, as to the other end of the square block: Let ^c fe, Fig. 155, represent the block, of which cvn is the form of the end at the curved part, and о its centre. Through o draw fg, so that the angle e pq shall be equal to the angle jub, Fig. 152. Also, through о draw d £ parallel with e 0;

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CUTTING THE TWIST-RAIL. 277 from 4 and 2 draw the vertical lines 4 r and ds parallel with pq. Place the template on cv z, the end of the block, so that the vertical line through its centre shall coincide with q; mark its form, then from y, at mid-thickness, draw w y parallel with 2 4. In applying the mould, let /zg. 156 represent the upper face of the squared block, а i x with the face-mould lying TUM upon it. With the distance 4 а 4, Fig. 154, and by the edge a x, mark a gauge-line upon the upper face of the Fic. 156. squared block. Set the outer edge of the face-mould to coin- cide with this gaugé-line. Let the end of the face-mould be set at w, e w being equal to e w, Fig. 155; then mark the block by the edge of the face-mould. Now turn the block over and apply the face-mould to the underside, as in /zg. 157. ` With the distance 4 2, Fig. 154, and by the outer edge of 28" the block, mark а gauge- f T. lite róm т, yis. Gert? / the inner edge of the face- mould to this gauge.line, and slide it end wise till the ІІІ 1 distance ez shall equal ew, Fig. 155, then mark the block by the edges of the face-mould. The over wood may now be re- moved as indicated by the vertical lines at the sides of the cross-section marked on each end of the block (see also 276: 167): the direction of the cutting at the curves must be verti- cal; the inner curve will require a round-faced plane. А com- parison of the several figures referred to, with the directions given, together with a little reflection, will manifest the reasons for the method here given for applying the face- mould. Especially so when it is remembered that the face- mould was obtained not for the 7o? of the rail, but for the rail at the mid-thickness of the block. So, therefore, in the application to the upper surface of the block, the face-mould is slid up the rake far enough to put the mould in position vertically over its true position at mid-thickness ; and on the “т -” -- чч = — — —— че ы e € er A

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298- STAIRS. contrary, in applying the face-mould to the underside of the plank, it is slid down until it is vertically beneath its true position at the mid-thickness of the block. When the vertical faces are completed, the over wood above and below the wreath 1s to be removed. In doing this, the form at the ends, as given by the template, is a suf- ficient guide there. Between these the upper and under surfaces are to be warped from one end to the other, so as to form a graceful curve. With a little practice an intelli- gent ПОШИ ТС will be able to work these surfaces with facility. Тһе form of cross-section produced by this opera- tion is that of a parallelogram, tangent to the top, bottom, and two sides of the rail; and which at and near the ends of the block is not quite full. The next operation is that of. working the moulding at the sides and on top, first by re- bates at the sides, then chamfering, and finally moulding the curves. Templates to fit the rail, one.at the sides, another . on top, are useful as checks hr ume cutting away too much óf the wood. The joints are all to be worked square through the plank in the line drawn perpendicular to the tangent, as shown in Fig. 153. 276.—Wand-Railing for Circular Stairs.—Let it be re- quired:to furnish the face-moulds for a circular stairs similar to that shown in Zzg. 133. Preliminary to making the face-moulds it is requisite to make a plan, or horizontal projection of the stairs, and on this to locate the projections of the tangents and develop their vertical projections. For this purpose let cde fg, Fig.158,be the horizontal projection of the centre of the rail, and the lines numbered from 1 to 10 be the risers. At any point, а, оп an extension of the line of the first riser locate the centre of the newel. Опа asa centre describe the two circles; the larger one equal in diameter to the diame- ter of the newel-cap, the inner one distant from the outer one equal to half the width of the rail, Let the first joint in the hand-rail be located at 4, at the fourth riser ; through 2 draw 220 tangent to the circle. Select a point, 4, on this

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PLAN OF CIRCULAR STAIRS. 279. tangent which shall be equally distant from 0 and from the inner circle of the newel-cap, measured on a line tending to а; join £ and а, and from a point, q, on the line 2o describe pe 11° Fic. 158. the curve from å to the point of the mitre of the newel-cap, the curve being tangent, at this point, to the line ah. Select positions for the other joints in the hand-rail as at c, d, e, and f.

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280 STAIRS. Through these draw lines tangent to the circle.* Then the ` horizontal projection of the tangents will be the lines 22, 27, Lm, m n, and np. Now, if a vertical plane stand upon each of these lines, these planes would form a prism not quite complete standing upon the base-plane, 4. Upon these ver- tical planes, C, D, Е, F, G, and ÆA, lines may be drawn which at each joint shall be tangent to the central line of the rail. These are the tangents now to be sought. ане to the tangents at 5, с, d, etc., draw the lines 64, 4,44, ff, ge, andthe), hk, Rk, Д0 “ etc. Ав015а4 т. x riser, and the height is eoutited from the top of the first riser, make 0 0, equal to three risers. (To avoid extending the drawing to inconvenient dimensions, the heights in it are made only half their actual size. As this is done uniformly throughout the drawing, this reduction will lead to no error in the desired results.) As с is on the eighth riser, therefore make сс, equal to seven risers, and so, in like manner, make the heights ZZ,ee,, and ҒҒ each of a height to correspond with the number of the riser at which it is placed, deduct- ing one riser. These heights fix the location of each tangent at its point of contact with the central line of the rail. But each tangent is yet free to revolve on this point of contact, up or down, as may be required to bring the ends of each pair of tangents in contact; or, to make equal in height the edges of gach pair of vertical planes, which coincide after they are revolved on their base-lines into a vertical position ; as, for example: the edges 22, and 2 £,, of the planes C and D must be of equal height; so, also, the edges 7/7, and 77, of the planes D and £ must be of equal height. Тһе method of establishing these heights will now be shown. To this end let it be observed, that of the horizontal pro- jection of any pair of intersecting tangents, their lengths, from the point of intersection to the points of contact with the circle, are equal; for example: of the two tangents 2 £ and ZZ, the distances from 2, their point of intersection, to 0 and с, their points of contact with the circle, are equal ; and so also c/ equals 42, dm equals em, etc. It will be observed * A tangent is a line perpendicular to the radius, drawn from the point of : contact.

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THE FALLING-MOULD FOR THE RAIL. 281 that this equality is not dependent on 0,с, 4, etc., the points of contact, being disposed at equal distances; for, in this example, they are placed at unequal distances, some being at ` three treads apart and others at four; and yet while this un- equal distribution of the points д, c, d, etc., has the effect of. causing the point of contact, as 0, с, or e, to divide each whole tangent into two unequal parts, it does not disturb the. equality of the two adjoining parts of any two adjacent tan- gents. Now, because of this equality of the two adjoining parts of a pair of tangents, the height to be overcome in passing from one point of contact to the next must be divided equally between the two; each tangent takes half the distance. Therefore, for stairs of this kind, the arrange-. ment: being symmetrical, we have this rule by which to fix the height of the ends of any two adjoining tangents, namely : To the height at the lower point of contact add half the dif- ference between the heights at the two points of contact ; the sum will be the required height of the two adjoining ends of tangents. For example: the heights at 0 and с, two adjacent points of contact, are respectively three and seven risers; the difference is four risers; half this added to three, the height of the lower rise, gives five risers as the height of ££, kk „the height at the adjoining ends of the. tangents 206 and Z$. Again, the heights at c and 4 are re- spectively seven and ten risers; their difference is three; half of which, or one and а half-risers, added to seven, the height at the lower point of contact, makes nine and a half risers as the heights //,//,, at the ends of the adjoining tangents £/ and m/. In a similar manner are established the heights of the tangents at zz, л, and 5. | The rule for finding the heights of tangents as just given is applicable to circular stairs in which the treads are di- vided equally at the front-string, as in 77g. 158. Stairs of irregular plan require to have drawn an elevation of the rail, stretched out into a plane, upon which the tangents can be located. This will be shown farther on. The locations of the joints c, 4, e, in this example, were disposed at unequal distances merely to show the effect on the tangents as before noticed. Іп practice it is proper to

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282 STAIRS. locate them at equal distances, for then one face-mould in such a stairs will serve for each wreath. When the tangent at G has been drawn, the level tangent for the landing mày be obtained in this manner: As the joint fis located at the eighteenth riser, one riser below the landing, draw a horizontal line at s, one riser above the point f, and at half a riser above this draw the level line at 2, ; then this line is the level tangent, and 2 its point of intersection with the raking tangent. Draw the vertical line 2, 2, and from ? draw the tangent 2 g, which is the horizontal projec- tion of the tangent 2, ғ, on plane H (which, to avoid undue enlargement of the drawing, is reduced in height), where f p,, equals 20. To obtain the horizontal tangent 7 и at the newel, pro- ceed thus: Fix the point 7, in the tangent 7 £, at a height above 27 equal to the elevation of the centre of the newel above the height of a short baluster—for example, from 5 to 8 inches—and draw a line through x parallel to 27; this is a horizontal line through the middle of the height of the newel-cap, and upon which and the rake the easement to the newel is formed. Perpendicular to 62 draw 72, and join £ and и; then / is the horizontal tangent. : 277.--Ғасс-Мош148 for Circular Stairs.— At Fig. 159 the plan of the newel and the adjacent hand-rail are repeated, but upon an enlarged scale; and in which 22, is the reduced height of the point 4, or is equal to 00, less £r, Fig. 158, and the angle 00, 2 equals the angle 207 of Pzg. 158. In this plan the actual heights must now be taken. Join 7 and и; then Zz is the level tangent, as also the line of intersection of the cutting plane C and the horizontal plane 4. Регреп- dicular to 7 u, at a point 2 or anywhere above it, draw ж, 0,,. Parallel with 7 4 draw 64,,; make 0,0, equal to 64,; join 5,, and и; then the angle 22,, v, is the angle which the plank in position makes with a vertical line, or what is usually termed the /umó-bevil. Perpendicular to 5,, и, draw и, и, and 6,,6,,,; make 5,, 0,,, equal to 25,,; make и, £, equal to и, Z, and u, u, tou, 4 ; join 2,,,, and 2,; then 5,,,, 7, is the tangent in the cutting plane, the horizontal projection of

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FACE-MOULD FOR FIRST SECTION. 283 which is ^7. The butt-joint at 0,,, is drawn square to the tangent бф, 7. Parallel to the intersecting line 2 и, draw ordinates across the plane 4 from as many points as desir- able, and extend them to the rake-line и, 0,,; through the points of their intersection with this line, and perpendicular to it, draw corresponding ordinates across the plane C. Make а,а,, equal to d, d, and so in like manner, for all other points, Еіс. 159. obtain in the plane C for each point in the horizontal plane A its corresponding point in the plane C: in each case taking the distance to the point іп the plane 4 from the line hb, and applying it in the plane C from the rake-line и, Ж For the curves bend a flexible strip to coincide with the several points obtained, and draw the curve by the side of the strip. The point of the mitre is at @,,,, the mitre-joint is

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284 STAIRS. shown at 4 2,, and Z,,c,. The line fc, 15 drawn through с, the most projecting point of the mitre, and parallel to the rake-line z,06,,. Additional wood is left attached, extending from Z to f; this is an allowance to cover the mitre, which has to be cut vertically ; the butt-joint at 4, and the face at f «,, are both to be cut square through the plank. The face Хе,» because it is parallel to the rake-line z, 0,,, is a vertical face, as well as being perpendicular to the surface of the plank. On it, therefore, lines drawn according to the rake, or like the angle и,0,, ?,, will be vertical and will give the direction of the mitre-faces. We now have at C the face- mould for the railing over the plan from 6 tod in А. The mould thus found is that made upon a cutting plane C, passed through the plank, parallel to its face, but at the middle of its thickness. To put it in position, let the plane C be lifted by its upper edge с, and revolved upon the line z, ?,,, until it stands perpendicular to the plane В. Now revolve both C and P (kept in this relative position during the revolution) upon the line z, 0, until the plane 2 stands perpendicular to the plane 4. Then every point upon plane C will be verti- cally over its corresponding point in the. plane 4. For ex- ample, the point 2,, will be vertically over 2, 2, over 2, and so of all other points. To show the application of the face-mould to the plank, make 4, б, equal to half the thick- ness of the plank; parallel to z, 4,, draw 2, с, а line which represents the upper surface of the plank, for the line x, ё is at the middle of the thickness. Through 5,,,, and parallel with £,,, u, draw the line c, à,,,, and extend it across the face- mould; make 2, c, equal to 6, с; through c, and parallel with b,,,, t, draw с,е. Now, m no, pis an end view of the plank, showing the face view of the butt-joint at 0,,. Through ғ, the centre, draw a line parallel with the sides. Then д, rep- resents the point 0,,,; make 4, e, equal to 2,,,, e; through ғ, the centre, draw e, > across the face of the joint; then e, 7 is a vertical line (see 477. 284), parallel and perpendicular to which the four sides of the squared-up wreath are to be drawnas shown. Inapplying the face-mould to the plank at first, for the purpose of marking by its edges the form of the face-mould, it will be observed that the face-mould is under- stood to have the position indicated by the line и,0,, or at 47

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FACE-MOULDS FOR CIRCULAR STAIRS. 285 the middle of the thickness of the plank. By this marking the rail-piece is cut square through the plank, and this cut- ting gives the correct form of the wreath, but oz/y at the middle of the thickness of the plank. After itis cut square through the plank, then, to obtain the form at the upper and under surfaces, the face-mould is required to be moved end- wise, but parallel with the auxiliary plane 4,-and so far as to bring the face-mould into a position vertically over or under its true position at the middle of the thickness of the plank. For example, the point д,,, if the mould were placed at the middle of the thickness of the plank, would be at the height of the point 2,,; but when upon the top of the plank, the point 2,,, would have to be at the height of the point с, therefore the mould must be so moved that the point 2, shall pass from 4, to с; consequently ё, с is the distance the mould must be moved, or, as it 15 technically termed, the sliding distance; hence 0,,, с,, which is equal to 4, c, is the distance the mould is to be moved: up when on top, and down when underneath. This is more fully explained in Art. 284. 278.—Face-Monlds for Circular Stairs.— At 77e. 160 so much of the horizontal projection of the hand-railing of stairs in 7Zg. 158 is repeated as extends from the joint 0 to that at d, but at an enlarged scale. Upon the tangent c£ set up the heights as given in 27. 158; for example, make kk, equal to £,, Ё, of Fig. 158, and сс, equal to c,,c, of Fig: 158. Join с, and 4, and extend the line to meet c 2, extended, in а. Join а and 0; then ад is the line of intersection of the cutting and horizontal planes; it is therefore a horizon- tal line, parallel to which the ordinates are to be drawn. Perpendicular to аф draw 2,c,,. Parallel to ад draw cc, and ££, ; join 6, and с„; the angle cc, ё, is the plumb-bevil ; perpendicular to 2, c, draw 2, ,,, Ё, Ё,» and, c, ; make b,b, equal to 4 ġ, and so of the other two points, £,,, and ¢,,,, make them respectively equal to their horizontal projections upon the plane A. Join с,, and £,,; also, £,, and 2,,; then 5, £,, and Ё, с з are the tangents. From c,, draw the line c, 6, 44744 Р : parallel to 4,c,,; this is the slide-line. In this example, this

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286 STAIRS. line passes through the point 4,,; the slide-line does not always pass through the ends of the two tangents; it is not required to pass through both, but it 1s indispensable that it be drawn parallel with the.rake-line 2, с,. Тһе lines for the joints at each end are drawn square to the tangent lines. Points in the curves, as many as are desirable, are now to be found by ordinates as. shown in the figure, and as before d, Fic. 160. explained for the points in the tangents. The curves are made by drawing a line against the side of a flexible strip bent to coincide with the points. The face-mould may be put in position by revolving the planes C and B, as explained in the last article, for the rail at the newel. The face-mould for the rail over the plan from c to Z is to

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FACE-MOULDS CONTINUED. 287 be obtained in a similar manner, taking the heights from Fg. 158. Por — make dd, equal to аа, of Fig. 158, and 4 equal to 2,2, of Fig. 158 (taking the heights at their - actual measurement now). Join d, апа Z, and extend the line to meet the line 4/ extended in ғ; join 7 and с; then rc is the line of intersection, and parallel to which the dette are to be drawn. The points in the face-mould may now be Obtained as in the previous cases, giving attention first to the tangent and slide-line; drawing the lines for the joints perpendicular to the tangents. It may be remarked here that the chord-line 2c is parallel with the measuring line 2,2,,, and that the line oZ bisects the chord-line; so, also, the line o7 bisects the chord-line c4. This coincidence is not accidental; it will always occur in a regular circular stairs. Hence in cases of this kind it is not necessary to go through the preliminaries by which to obtain the intersect- ing line ад, but draw it at once parallel to the line o7, bisecting the chord бс and passing through the point of intersection of the two tangents: For the distance to slide the. mould in its after-application, the lines are given at c, and 4), and their use is explained in the last article, and 4 more fully in 477. 284. 279.--Ғасе-Мош48 for Circular Stairs, again.—At 70. 161 so much of the plan of the hand-railing of the stairs of Fig. 158 is repeated as is required to show the rail from f£ to g, but drawn at a larger scale. To prepare for the face- moulds, perpendicular to 2 / draw 22,, and make 22, equal to ~,,p,, of Fig. 158 (taking this height now at its actual measurement); join 2, and f; then /2,15 the tangent of the vertical plane C, and f is a point in the cutting plane at its intersection with the base-plane 4. Now since rs, the tan- gent over pg, is horizontal and is in the cutting plane, therefore from f draw fa parallel with rs ог pg; then fa is the line of intersection of the cutting and horizontal planes, and gives direction to the ordinates. Draw /,5,, perpendicular to fa; make 2,,2,, equal to pf, ; join 2, and Ж then the angle 22,718 the plumb-bevil ; perpendicular

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288 ` STAIRS. to Ж, 7; draw pon and ІЛЕ таКе Pui Pui equal to 5,,£; 2..4 equal to 2,2; join d and 7,; then df, and 45,,, are the tangents. Make 2, е equal to half the thickness of the plank; draw f,a parallel with f 5, ; make f, a equal to ec; draw ac, parallel with the tangent f,d; through /,, per- pendicular to 7, 4, draw the line for the butt-joint ; then f,c, is the distance required to determine the vertical line on the face of the joint at /,, as shown at 4. Through 5,,,, per- Fic. 161. pendicular to the tangent 2,,,, 4, draw the line for the butt- joint; make 7,,,,d equal to ec; then 5,,, ^ is the distance required for determining the vertical line on the face of the joint at 2,,,, as shown at B (see Art. 284). The curved lines are obtained by drawing a line against the edges of a flexi- ble rod bent to as many points as desirable, obtained by measuring the ordinates of the plan at А and transferring them to the face-mould by the corresponding ordinates, as before explained.

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RAILING FOR QUARTER-CIRCLE STAIRS. 289 280.—Hand-Railing for Winding Stairs. — The term winding is applied more particularly to a stairs having steps of parallel width compounded with those which taper in width, as in 77g. 135, and as is here shown in Fig. 162, in which fa bc represents the central line of the rail around the cylinder, and the quadrant Ze, distant from the first quadrant 20 inches, is the tread-line, upon which from 4, a point taken. at pleasure, the treads are run off. Through e, perpendicu- Frio. 162. lar to af, draw ae (the occurrence here of one of the points of division on the tread-line perpendicularly opposite a, the spring of the circle, is only an accidental coincidence) ; make aa, equal to two risers; join а, апа f. With the diameter ac, оп д as a centre, describe the arc at 2, crossing ac extended; through 0 draw gé,; then a, is the stretch- out, or development of the quadrant аб. Through 4 draw 22, tending toward the centre of the

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290 STAIRS. cylinder; make 0,2, equal to 02; perpendicular to fò, draw 0,0, апа 227. As there are four risers from e to Z, make a, a, equal to four risers, and draw a, t, parallel with fa; through 2, draw 2,2,,; by intersecting lines, or in any con- venient maker; ease off to any extent the angle fa,z, Through 7, a point in this curve (chosen so as to be perpen- dicularly over m, a point between а апа 7, nearer to а), draw £/,a tangent to the. curve. Perpendicularly to this tangent, through 7, draw the line for a butt-joint; also through 2,, and perpendicularly to а,0,, draw the line for the joint at the centre of the half circle. On the line aa, set up points of division for the riser heights, and through these points of division draw horizontal lines to the line 5, f- From these points of contact drop perpendiculars to the line fad, and transfer such.of them as require itto the circle at, by drawing lines tending to g. Through these points of intersection with the central line of the rail, and through the points of division оп the tread-line, draw the riser-lines me, an,etc. At half a riser above the floor-line, on top of the upper riser draw a horizontal line, and ease off the angle as shown; the intersection of the floor-line with this curve gives the position of the top riser at the centre of the rail. This completes the plan of the steps and the elevation of the rail —requisite preliminaries for the face-moulds. The gradu- ation of the treads from flyers to winders obviates an abrupt angle at their junction in the rail and front-string. The objection to the graduation, that it interferes with the regularity of stepping at the tread-line, is not realized in practice. ing Stairs.—At zg. 163 so much of the plan at Fig. 162 is repeated as is required for the face-moulds, but for perspicuity at twice the size. The hori- zontal projection of the tangents for the first wreath are ad and 46 drawn at right angles to each other, tangent to the circle аға and 2. Let those tangents be extended beyond’ d; through m, the lower end of the wreath, draw 77 d,, mak- ing an angle with 7247 equal to that in Fig. 162, between the

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FACE-MOULDS FOR THE TWISTS. 29I line af and af; or let the angle 4744, equal a fa, of Fig. 162. Make 44, equal to 44, Make 22, equal to 4,4, of Fig. 162; join Z, and 6, and extend the line to e,,; make 6,6, equal to 2,,0,,, of Fig. 162, and draw 5, е, parallel with е, u^ йс NN Ё ' SN s | ЗАТЫ | AYE Ж CRZ DAN BA 74/255 N 00 Fic. 163. de. From e, draw e,,e parallel with 0,0; through e and f draw e f tangent to the circle at f; then фе and e f are the horizontal projections of the tangents for the upper wreath. Then if the plane В be revolved on ad, the plane C on de,

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292 STAIRS. and the plane D on ef until they:each stand vertical to the plane A, the lines жа, 4, ё,» and e,, f, will constitute the tangents of the two wreaths i in position. This arrangement locates the upper joint of the upper wreath at. /, leaving fc, а part of the circle, to be worked as a part of the long level rail on the landing. As the tangent over ef is level, the raking part of the rail will all be included in the wreath 2 f, so that at the joint f the rail terminates on the level. The portion fc, therefore, is a level rail requiring no canting, and it requires no other face-mould than that afforded by the plan from f to c. For the face-mould for the rail over m a б, let the line A, a), be extended to z4, a point in the base-line 4 m,; then zz, 4 А point in the base-plane 4,'às well as in the cutting plane Æ; therefore the line mm, 74 15 the zxztersecting line parallel to which all the ordinates on plane A are to be drawn. Per- pendicular to this intersecting line zz, m, at any convenient place draw m, 6,; make 2, 6,, parallel to 77, т and equal to 05, ; connect 2) P 74, à point at the intersection of the lines zz, mand 6, m,; then the angle 00, 74,15 the plumb- bevil. Through É parallel to m, m, draw 427%; from the three points 72, 4,, and 5,, draw lines perpendicular to т,б: make zz, 77, equal to m,m; make 0,0, equal to 2,0. Since the measuring base-line 7%, ё, passes through 4, the point of the angle formed by the two tangents, g, is the point of this angle in the cutting plane Æ; therefore join zz,, and Z, , also d, and 0,,,,; then 5,,, d,,, ља а A», andthe two tangents at right angles to which the joints at z,, and b, аге drawn. The curves of the face-mould are now found as usual, by transferring the distances by ordinates, as shown, from the plane А to the plane Ж, making the distance from the rake-line zz, б, to each point in plane Ж equal to the dis- tance from the corresponding point in the plane 4 to the measuring base-line 7,4. Now, to obtain the sliding distance and the vertical line upon the butt-joints, make 0,, д, equal - to half the thickness of the plank; parallel with zz, 2,, draw б, бл; also, à,,, & and ж, m,,; make 5,,, бу and 7, m, each equal to 4, 24; through б and zz,,, w parallel to the respective tangents, draw би, 0, and m,,, m,,,,; then ё, and

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FACE-MOULDS FOR WINDING STAIRS. 293 74,,, are the points from which, through the centre of the butt-joints, a line is to be drawn which will be vertical when the wreath is in position. (See Art. 284.) For the face-mould for the upper quarter, through 4, Fig. 163, draw 6 e, parallel with d, е; make e е, equal to ee; draw е, f, parallel with e f. Now, since е, f, is а horizon- tal line and is in the cutting plane 2, therefore, parallel with e,,, f, and through 0, draw ġ n; then 0 715 the required in- tersecting hne. Extend e f to f,; make f f, equal to ff; join f, and z; then the angle f f, nis the plumb-bevil. Per- pendicular to z f, draw f, f, апа z л,, and make these lines respectively equal to e f and бл; join f, and /,,,; also f, and z,; then /, f, and f,,z, are the required tangents. The butt-joints at f, and z, are drawn perpendicular to their respective tangents. То get the slide distance and vertical lines on the butt-joints, make /, f, equal to half the thickness of the plank; parallel with л f, through f, draw /, f; also, through z, draw л, z,,; make ж, »;, equal to f, f; through л, parallel with x, f, draw и, z,,,; then z,,, 15 the. point through which a line is to be drawn to the centre of the butt-joint, and this line will be in the vertical plane contain- ing the tangent. So, also, parallel with the tangent f, fn» and through f,,, draw f,,,, fa; then / is the point through which a line is to be drawn to the centre of the butt-joint (see Art. 284). Тһе curve is now to be obtained by the ordinates, as before explained. 282. FFnce-Moulds for Winding Stairs, again.—In the last article, in getting the face-moulds fora winding stairs, the two wreaths are found to be very dissimilar in length. This dissimilarity may be obviated by a judicious location of the butt-joint connecting the two wreaths, as shown in fg. 164. Instead of locating the joint precisely at the middle of the half circle, as was done in Fig. 163, place it farther down, - say at z, which is at z in Fig. 162, two risers down from the top, or at any other point at will. Then through z in the plan draw m,s tangent to the circle at л; and perpendicu- lar to this tangent draw x z,,, and d d,,; make x z, equal to n, n of Fig. 162; from d erect d а, perpendicular to m 2; / €

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294 STAIRS. make the angle d m d, equal to that of ё, 7 / of Fig. 162. Make Z Z,, equal to d Z,; join Z,, and z,, and extend the line to m,, a point of intersection with the base-line л z,; then z, is a point in the base-plane, as also in the cutting plane; Sy SOC Жора Г PARES NY ФУ \ Fic. 164. therefore zz, 7% is the intersecting line parallel to which ail the ordinates of the plan are to be drawn, and perpendicular to which эл, z,, the measuring base-line, is drawn. Make 2, n, equal to 2 z,,; connect m,, and z,,,, and then transfer

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CARE REQUIRED IN DRAWING. 295 by the ordinates to the cutting plane z d and n the three points of the plan at the ends of the tangents, as before de- scribed, as also such points in the curve as may be required to mark the curve upon the face-mould, all as shown in previ- ous examples. Forthe face-mould of the upper wreath, make п,т,, equal to ли, of 72g. 162. From ғ, draw z,,, s,, par- allel with zz, s; extend the line Z, z, to intersect л, s, in s,,; parallel with z,,, z draw s,,s; from s draw s > tangent to the circle at 7 (s z equals s7); through 7, tending to the centre of the cylinder, draw the butt-joint; then 7 sand s z are the horizontal projections of the tangents for the upper wreath- piece, the tangent s7 being level and, consequently, parallel to the intersecting line drawn through xz. Perpendicular to rs draw 7, 2; parallel with л, s,, draw %5,; make 7,7,, equal toss,; joinz, and р. From this line and the measuring base- line 7, 2, the points for the tangents are first to be obtained and then the points in the curve, all as before described. The part of the circle from 7 to ¢ is on the level, as before shown, and may be worked upon the end of the long level rail, its form being just what is shown in the plan from c tor. 283.—Face-Moulds: Test of Aecuraey.— The methods which have been advanced for obtaining face-moulds are based upon principles of such undoubted correctness that there can be no question as to the results, when the methods given are thoroughly followed. Апа yet, notwitbstanding the correctness of the system and its thorough comprehen- sion by the stair-builder, he will fail of success unless he exercises the greatest care in getting his dimensions, his per- pendiculars, and his angles. Тһе slightest deviation in a perpendicular terminated by an oblique line will result in a magnified error at the oblique line. То secure the greatest possible. degree of accuracy, care must be exercised in the choice of the instruments by which the drawings are to be made: care to know that a straight-edge is what it purports to be; that a square, or right-angle, is truly a right-angle; that the compasses or dividers be well made, the joint per- fect, and the ends neatly ground to a point. Then let the drawing-board be carefully planed to a true surface ; and,

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296 STAIRS. if possible, let the drawing, full size, be made upon large, stout roll-paper rather than upon the drawing-board itself, as then the points forthe face-mould may be pricked through upon the board out of which the face-mould is to be cut, and thus a correct transfer be made. For long straight lines it is better to use a fine chalk-line than the edge of a wooden straight-edge. The line is more trustworthy. Perpendicu- lars, especially when long, are better obtained by measure- ment or by calculation (477. 503) than by. a. square. The pencil used should be of fine quality—rather hard, in order that its point may be kept fine. With these precautions in regard to the instruments used, and with due care in the manipulations, the face-moulds may be correctly drawn, accurate in size and form. Ав a test of the.accuracy of the work, it will be well to observe in regard to the tangents, that the length of a tangent, as found upon the face-mould, should always equal its length as shown upon the vertical plane. .For example, іп 77g. 160, the tangent £, с,,, on the face-mould should be equal to Å, с,, the tangent on the vertical plane В; and in cases like this, where the stairs are quite regular, with equal trezds at the front-string, the two tan- gents of a face-mould are-equal to each other, or £,,c,,, equals PEE) k, 6; and in this case, the line 0,,е,, should equal the rake- кте б, с 4? 44744, асы Again, as another example, in 77g. 161, d f, the tangent upon the face-mould, should be equal to /р,, the tangent of the vertical plane C; while d 5,,,, the other tangent on the face-mould, should be equal to 7 s, the tangent of the vertical plane D. But the more important test is in the length of the chord-line joining the ends of the two tangents ; as, for ex- ample, the chord z,, 0,,, of Fzg. 163, the horizontal projec- tion of which is the chord 7z ё in plane A. Perpendicular to mb draw 6g; make g equal to 46,, and join g and зм; then m, b, the chord of the face-mould, should be equal to 72 g. After fully testing the accuracy of the drawing for the face- mould, choose a well-seasoned thin piece of white-wood, or any other wood not liable to split, and plane it to an even thickness throughout ; mark upon it the curves, joints, tan- gents, and slide-line, and cut the edges true to the curve-

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THE FACE-MOULD APPLIED TO PLANK. 297 lines and joints square through the board ; then square over such marks as are required to draw each tangent and the slide-line also upon the reverse side of the board. This completes the face-mould. 284.——Applieation of the Face-Mould. In order that a more comprehensive idea of the lines given for applying a face-mould may be had, let 4, Fig. 165, represent one end of a wreath-piece as it appears when first cut from a plank, and when held up in the position it is to occupy at completion over the stairs. Also, let Б represent the corresponding face-mould, laid upon the wreath-piece A in the position which it should have after sliding. And, for the purpose of a clearer illustration, let it be supposed that the two pieces, А апа В, are transparent. Then let a, а ò dc, e, represent a solid of wedge form, having a triangular level base, а 0 4, upon the three lines of which stand these three vertical planes, namely: on the line a ё the plane а, а 2с, upon the line а d the plane a, a d e, and on the line Z2 the plane 6d e,c,; the top of the solid is an inclined plane, a, с, e, and the vertical line а, a is the edge of the wedge. Now, it will be observed that the point a in the base of the solid is identical with а, the centre of the butt-joint, and the point а, (at the intersection of two vertical planes and the inclined plane of the solid) is vertically over а, and is identical with а„ a point in the upper surface of the plank. Also, the inclined plane e, c, а,, which forms the top of the solid, coincides with the upper surface of the plank A; from which the wreath-piece has been squared ; and the line c, а, (at the angle formed by the inclined plane e, v, а, and the vertical plane а, a 2 с,) coin- cides with f g, the slide-line drawn upon the top of the plank; also, the line e, а, (at the angle formed by the in- clined plane e, c, a, and the vertical plane а, a 4 е) coincides with а, 2, the tangent line upon the underside of the face- mould after it has been slid to its new position, vertically over its true position at the middle. of the thickness of the plank. From а the line a с 15 drawn parallel with а; с,; so, also, the line a eis drawn parallel with a, ¢,; consequently the line ec is parallel with e, с,; and the planee cæ is parallel

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298 STAIRS. with the plane е, с, а,, and coincides with a plane passing through the middle of the thickness of the plank, and, conse- quently, is the cutting plane referred to in previous articles, upon which the lines are drawn which give shape to the v Ж rie: 155. face-mould. When the face-mould is first laid upon the plank. the line 7, 7, coincides with 2, 7,, and ‘when in that position, its form marked upon the plank is the form by which the plank is sawed square through; but this gives the form of

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THE SLIDING OF THE FACE-MOULD. 200 the wreath, not as it is at the surface of the plank, but as it is at the middle of the thickness of the plank, or upon the plane ace; so that, for example, the line 7, 7,, represents the line 2/ drawn through a, the centre of the butt-joint ; and when the mould 2 is slid to the position shown in the figure, the line 2, 7, comes into a position vertically over Z7; hence the three lines 7, 7, а, а, and 7,7 are each vertical and in а vertical plane, zz, 7, у. By these considerations it will be seen that the face-mould Æ, located as shown in the figure, is in its true position for the second marking, by which the addi- tional cutting is now to be performed vertically. This being , established, it will now be shown how to get upon the butt- joint a line in the vertical plane containing the tangent. If the top and bottom lines of the vertical plane a, a бс, be ex- tended, they will meet in the point /, and will extend the plane into a triangle /02 c, cutting the upper edge, of the butt-joint in f, the end of the tangent, and the point in which the point a, of the underside of the face-mould was located when the mould was first applied to the plank. The line fa on the butt-joint is perpendicular to z j or z, 7,, Again, if the top and bottom lines of the plane 2, а d e, be extended, they will meet in 2, and will extend the plane into the tri- angle 2 d e, cutting the edge of the butt-joint in 4, a point from which, if a line be drawn upon the butt-joint to a, its centre, this line will be in the vertical plane 2 4 e, which plane contains the tangent perpendicular to which the butt- joint is drawn; consequently lines upon the butt-joint par- allel to л а will each be in a vertical plane parallel to the vertical tangent plane, and lines drawn upon the butt-joint perpendicular to these lines will be horizontal lines; hence the line 4 a is the required line by which to square the wreath at the butt-joint. Now, it will be observed that the triangle a f a, is like that given in the various figures for ob- - taining face-moulds, to regulate the sliding of the face-mould and the squaring at the butt-joint. For example, іп 27. 163, the right-angled triangle 4,,, 9,0, is the one referred to. This triangle is in a vertical plane parallel to one containing the slide-line; its longer side is a vertical line; one of the sides containing the right angle is equal to half the thickness

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300 STAIRS. TUS of the. plank, while the other, drawn parallel to the face of: the plank, is the distance the face-mould is required to slide. Precisely like this, the triangle æ fa, of Fig. 165 is in the vertical plane 2 2 c, containing fg, the slide-line; its longer side, а, a, is a vertical line; fa, one of the sides containing the right angle, is equal to half the thickness of the plank, while the other side, drawn coincident with the surface of. fh T 4 Pe Fic. 166. the plank, 15 the distance to slide the face-mould. "Therefore the triangle а, fa of Fig. 165 gives the required lines by which to regulate the application of the face-moulds. The relative position of the more important of these lines is geo- metrically shown in /7g. 166, in which A and P are upon the horizontal plane of the paper, С is in a vertical plane stand- ing on the ground-line ? а, and D is a plan of the butt-joint, revolved upon the line 2, j, into the horizontal plane, and

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BLOCKING-OUT OF THE RAIL. 301 then perpendicularly removed to the distance f f. The let- tering corresponds with that in zg. 165. Тһе shaded part of D shows the end of the squared wreath. When the blocked piece has been marked by the face-mould in its second application, its edges are to be trimmed vertically as shown in Zzg. 167, after which the top and bottom surfaces of the wreath are to be formed from the shape marked on the butt-joints. Fic. 167. 285. — Face-Mould Curves are Elliptical.—The curves of the face-mould for the hand-railing of any stairs of circular plan are elliptical, and may be drawn by a: trammel, or in any other convenient manner. The trouble, however, attending the process of obtaining the axes, so as to be able to employ the trammel in describing the curves, is, in many cases, greater than it would be to obtain the curves through points found by ordinates, in the usual manner. But as

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302 STAIRS. this method for certain reasons may be preferred by some, an example 15 here given in which the curves are drawn by a trammel, and in which the method of obtaining the axes is shown. Let /zg. 168 represent the plan of a hand-rail around part DX Fic. 165, of a cylinder and with the heights set up, the intersection line obtained, the measuring base-line drawn, the rake-line established, and the tangents on the face-mould located—all in the usual manner as hereinbefore described. Then, to prepare for the trammel, from o, the centre of the cylinder, draw о д, parallel with the intersecting line, and 2,o, perpen-

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FACE-MOULDS FOR ROUND RAILS. 303 dicular to 2, f, the rake-line; make 4,0, equal to бо, and o,a, . equal to оа; through o, draw o, parallel with 6 f. From o draw oe perpendicular to оё,; continue the central circular line of the rail around to e; parallel with ob, draw ef, and from f,the point of intersection of ef with 0, f, and perpendicular to 2, f, draw f,e,; make fe, equal to fe; then o, is the centre of the ellipse, and о,а, the semi-conju- gate diameter and o,e, the semi-transverse diameter of an ellipse drawn through the centre of the face-mould. To get the diameters for the edges of the face-mould, make a,c, and а,а, each equal to half the width of the rail, as at cad; par- allel to a line drawn from а, to e, and through с, draw the line c, 2; also, parallel with a line drawn from a, to e, draw d, (see Art. 559); then for the curve at the inner edge of the face-mould, o, is the semi-transverse diameter, and о, с the semi-conjugate; while for the curve at the outer edge o, is the semi-transverse diameter, and о, 4, the semi-conju- gate. So much of the edges of the face-mould as are straight are parallel with the tangent. Now, placing the trammel at the centre, as shown in the figure, and making the distance on the rod from the pencil to the first pin equal to the semi-conjugate diameter, and the distance to the second pin equal to the semi-transverse diameter, each curve may be drawn as shown. (See Art. 549.) | 286. —Face-Moulds for Round Rails.—The previous ex- amples given for finding face-moulds are intended for moulded rails. For round rails the same process is to be followed, with this difference: that instead of finding curves on the face-mould for the sides of the rail, find one for a centre-line and describe circles upon it, as at Fig. 145; then trace the sides of the mould by thé points so found. The thickness of stuff for the twists of a round rail is the same as for the straight part. The twists are to be sawed square through. 287.— Position of the Butt-Joint.—When a block for the wreath of a hand-rail is sawed square through the plank, the joint, in all cases, is to be laid on the face-mould square to the tangent and cut square through the plank.

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304. STAIRS. Managed. in.this way, the butt-joint is in a plane pierced perpendicularly by the tangent. But if the block be not. sawed square through, but vertically from the edges of the Fic. 169. face-mould, then, especially, care 15 required in locating the joint. The method of sawing square through is attended with so many advantages that it is now generally followed ; yet, as it is possible that for certain reasons some may prefer,

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POSITION OF THE BUTT-JOINT. . ini some cases, to saw vertically, it is proper that керу of finding the position of the joint for that purpose should be given. Therefore, let A, Fig. 169, be the plan of the rail, and B the elevation, showing its side; in which £z is the direction of the butt-joint. From £ draw 20 parallel to Zo, and ġe at right'angles to £4; from 2 draw 6 f, tending to the centre of the plan, and from f draw fe parallel to 22; from /, through e, draw 22, and from 2 draw 24 parallel to ef; join d and 4, and db will be the proper direction for the joint on the plan. The direction of the joint on the other side, ас, сап be found by transferring the distances x? and od to xa and ос. Then the allowance for over wood to cover the butt-joint 1s shown as that which is included be- tween the.lines ox and 40. The face-mould must be so drawn as to cover the plan to the line 4d for the wreath at the left, and to the line ас for that at the right. By some the direction of the joint is made to radiate toward the centre of the cylinder; indeed, even Mr. Nicholson, in his Carpenter's Guide, so advised. That this is an error may be shown as follows: In Fig. 170, 4777 is the plan of a part of the rail about the joint, sz is the stretch-out of az, and ср is the helinet, or vertical projection of the plan arji. This, is found by drawing a horizontal line from the height set upon each perpendicular standing upon the stretch-out line s. The lines upon the plan arji are drawn radiating to the centre of the cylinder, and therefore correspond to the horizontal lines of the helinet drawn upon its upper and under surfaces. Bisect ғ2 on the ordinate drawn from the centre of the plan, and through the middle draw c? at right angles to gv; from 6 and c draw cd and бе at right angles to s»; from d and e draw lines radiating toward the centre of Т” plan ; then do and em will be the direction of the joint on the plan, according to Nicholson, and c its direction on the falling-mould. It must be admitted that all the lines on the upper or the lower side of the rail which radiate toward the centre of the cylinder, as do, em, ог 27, are level; for instance, the level line wv, on the top of the rail in the helinet, is a true representation of the radiating line 77 on

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306 STAIRS. the plan. The line 02, therefore, on the top of the rail in the helinet, is a true representation of ez; on the plan, and | ёс оп the bottom of the rail truly represents Zo. From 2 draw £/ parallel to có, and from % draw 2 f parallel to dc; FIG. 170. join 7 and 4, also c and f; then c £7? will be a true repre- sentation of the end of the lower piece, P, and cf 4àó of the end of the upper piece, 4; and / or 22 will show how much the joint is open on the inner, or concave, side of the rail.

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CORRECT LINES FOR BUTT-JOINT. 307 To show that the process followed in 477. 287 is correct, let Zo and em (Fig. 171) be the direction of the butt-joint found as at 70. 169. Now, to project, on the top of the rail in the helinet, a line that does not radiate toward the centre of the cylinder, as jk, draw vertical lines from 7 and £ to 10 and #, and join w and 0; then it will be evident that wh is a true representation in the helinet of 72 on the plan, it being in the same plane as 7 £, and also in the same winding surface as wv. The line Zz, also, is a true representation en

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308 - STAIRS. the bottom of the helinet of the line f% in the plan. The line of the joint e zz, therefore, is projected in the same way, and truly, by 206 on the top of the helinet, and the line Zo by ca on the bottom. Join а and 2, and then it will be seen that the lines ca, az, and 76 exactly coincide with c2, the line of the joint on the convex side of the rail; thus proving the | lower end of the upper piece, 4, and the upper end of the lower piece, В, to be in one and the same plane, and that the direction of the joint on the plan is the true one. Ву refer- ence to Fig. 169 it will be seen that the line 72 corresponds to xz in Zzg. 171; and that e£ in that figure is а representa- tion of fb, and т^ of dù. 288.—Serolls for Hand-Rails: General Rule for Size and Position of the Regulating Square. — The breadth which the scroll is to occupy, the number of its revolutions, and the relative size of the regulating square to the eye of the scroll being given, multiply the number of revolutions by 4, and to the product add the number of times a side of the square is contained іп the diameter of the eye, and the sum will be the number of equal parts into which the breadth is to be divided. Make a side of the regulating square equal to one of these parts. To the breadth of the scroll add one of the parts thus found, and half the sum will be the length of the longest ordinate. Fic. 172. 289.— Centres in Regulating Square.—Let a 2 1 6 (Fig. 172) be the size of a regulating square, found according to the previous rule, the required number of revolutions being

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SCROLLS AT NEWEL. 309 [] 1$. Divide two adjacent sides, as a 2 and 2 1, into as many equal parts as there are quarters in the number of revolu- tions, as seven ; from those points of division draw lines across the square at right angles to the lines divided ; then 1 being the first centre, 2, 3, 4, 5, 6, and 7 are the centres for the other quarters, and 8 is the centre for the eye ; the heavy lines that determine these centres being each one part less in length than its preceding line. 2 =. - е-е е... pne E “қылы. am € Bo m a nay i a VQ, matics y 55 ч 1 \ ‘ 9 E ees 4 ` l : 1 | { / ЫР” же ч» __ Оса Н 62 Fic. 173. 290.— Scroll for Hand-Rail Over Curtail Step. — Let a b (Fig. 173) be the given breadth, 17 the given number of revolutions, and let the relative size of the regulating square to the eye be 4 of the diameter of the eye. Then, by the rule, 12 multipled by 4 gives 7, and 3, the number of times a side of the square is contained in the eye, being added, the sum is 10. Divide a4, therefore, into 10 equal parts, and set one from 0 to c; bisect a c in e; then ae will be the length of the longest ordinate (1 Z огт е). From a draw a d, from e draw e 1, and from ё draw å f, all at right angles to ab; make ет equal to а, and through r draw т d parallel

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310 STAIRS. to аф; set dc from 1.to 2, and upon 12 complete the regu lating square; divide this square as at Fig. 172; then de- scribe the arcs that compose the scroll, as follows: upon 1 describe 4 e, upon 2 describe e /, upon 3 describe f g, upon 4 describe ғ Z, etc.; make d Z equal to the width of the rail, and upon 1 describe / »;, upon 2 describe mz, etc.; de- scribe the eye upon 8, and the scroll is completed. ' 291.—Scroll for Curtail Step.—Bisect d / (Fig. 173) in o, and make o v equal to 4 of the diameter of a baluster ; make vw equal to the projection of the nosing, and e x equal to 4; upon І describe w y, and upon 2 describe yz; also, upon 2 describe xz, upon 3 describe 7 7, and so around to 2; and the scroll for the step will be completed. 292.— Position of Balusters Under Scroll, —Bisect d Z (Fig. 173) in o, and upon 1, with 1 o for radius, describe the circle o zz; set the baluster at ? fair with the face of the second riser, c?, and from 2, with half the tread in the divi- ders, space off as at 0,4,7, 5,2, u, etc., as far as 47; upon 2, 3, 4, and 5 describe the centre-line of the rail around to the eye of the scroll; from the points of division in the circle o ru draw lines to the centre-line of the rail, tending to 8, the centre of the eye; then the intersection of these radiating lines with the centre-line of the rail will determine the posi- tion of the balusters, as shown in the figure. 293.—Falling-Mould for Raking Part of Sceroll.— Tangi- cal to the rail at % (Fig. 173) draw 7 Ё parallel toda; then ka? will be the joint between the twist and the other part of the scroll. Make de? equal to the stretch-out of d e, and upon d e? find the position of the point £, as at £^; at Fig. 174, make ed equal to e? d in Fig. 173, and dc equal to dc? in that figure; from с draw са at right angles to ec, and equal to one rise; make c? equal to one tread, and from 2, through a, draw £ 7; bisect a c in / and through / draw 7% 4 parallel to e£; mg is the height of the level part of a scroll, which should always be about 34 feet from the floor ; ease off the angle m fj, according to 472. 521, and draw

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FACE-MOULD FOR THE SCROLL. 3IL gwn parallel to x 7, and.at a distance equal to the thick- ness of the rail; at a convenient place for the joint, as 2, draw 29 at right angles to 07; through z draw уй at right angles to 2%; make 42 equal to d £? in Fig. 173, апа from 2 draw £o at right angles to ей; at Pig. 173, make а л? equal to dh in Fog. 174, and draw A? 2? at right angles to 429; Fic. 174. then £ a’ and 220% will be the position of the joints on the plan, and, at Fig. 174, o P and zz their position on the falling- mould; and po zz (PZg.174) will be the required falling- mould which is to be bent upon the vertical surface from 02 to & (Fag. 173). FIG. 175. 294.. —Face-Mould for the Scroll.—At Fig. 173, from 4 draw 7° at right angles to r* ad: at Pig. 172, make 4 4 equal to A? 7? іп Fig. 173, and from 7 draw rsat right angles tov; from the intersection of z s with «s level line : m q, through 2, draw s 7; at Fig. 173, make й? 6° equal to 02 in Fig. 172, and join 6° and 7°; from a*, and from as many

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312 STAIRS. other points in the arcs, а? / and Ға, ав is thought neces- sary, draw ordinates to 7? 7 at right angles to the latter; make x 2 (Fig. 175) equal in its length and in its divisions to the line 7? 6? in Fig. 173; from 7, n,o, №, 9, and 2 draw the lines 7 £2 d,oa, f e, д f, and іс at right angles to 7 4, and equal to 7° $, d? 5°, f° a?, etc., in Fig. 173; through the. points thus found trace the curves £/ and ас, and complete the face-mould, as shown in the figure. This mould is to be ap- plied to a square-edged plank, with the edge, / ? parallel to the edge of the plank. The rake-lines upon the edge of the plank are to be made to correspond to the angle s 7 in Fig. 174. The thickness of stuff required for this mould is shown at с. 174, between the lines s гапа и v—z v being drawn parallel to 5 2. Fic. 176. 295. Form of Newcl-Cap from a Section of the Rail. ‚ — Draw а 0 (Fig. 176) through the widest part of the given section, and parallel to с d; bisect а біп e, and through a, e, and 2 draw 22, f g,and £ 7 at right angles to а 4; at a con-

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BORING FOR BALUSTERS. 313. venient place on the line fg, аз o, with a radius equal to half the width of the cap, describe the circle 2/2; make 7 / equal to e ġ or ea; join Zand 7, also Zand 2; from the curve f 6 to the line 2 7 draw as many ordinates as is thought necessary parallel to fg; from the points at which these ordinates meet the line / у, and upon the centre, o, describe arcs in continuation to meet ор; from %2 x, etc., draw z s, £u, etc., parallel to fg; make z s, ¢ и, etc., equal to e f, : ә, etc.; make T y Әс. сапа! to zæ, etc; make 22,03, etc., equal to on, o t, etc.; make 2 4 equal to z s, and in this way find the length of the lines crossing 07%; through the points thus found describe the section of the newel-cap as shown in the figure. Fic. 177. 296.—Boring for Balusters in a Round Rail before it is Rounded.—Make the angle o c 7 (Fig. 177) equal to the angle o c ¢ at Fig. 144; upon с describe a circle with a radius equal to half the thickness of the rail; draw the tan- gent 6d parallel to ¢c, and complete the rectangle e b d f, having sides tangical to the circle; from с draw са at right angles to o c; then, ё 4 being the bottom of the rail, set a gauge from 4 to a, and run it the whole length of the stuff ; in boring, place the centre of the bit in the gauge-mark at a, and bore in the direction а с. To do this easily, make chucks as represented in the figure, the bottom edge, g 2, being par- allel to o c, and having a place sawed out, as e f, to receive the rail. These being nailed to the bench, the rail will be held steadily in its proper place for boring vertically. The distance apart that the balusters require to be, on the under- side of the rail, is one half the length of the za£e-side of the pitch-board.

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314 STAIRS. SPLAYED WORK. 297.—The Bevels іп Splayed Work.—The principles employed in finding the lines in stairs are nearly allied to those required to find the bevels for splayed work—such as hoppers, bread-trays, etc. А method by which these may be Fic. 178. obtained will, therefore, here be shown. Іп Fig. 178, аф c is the angle at which the work is splayed, and 0 d, on the upper edge of the board, is at right angles to а 6; make the angle f 2.7 equal to a ðc, and from f draw f% parallel to са; from ё draw 6 oat right angles to а 6; through о draw Te parallel to c б, and join e and 4; then the angle a ed will be the proper bevil for the ends from the inside, or 2 Ze from the outside. If a mitre-joint is required, set f g, the thickness of the stuff on the level, from e to 77, and join zz and d; then $d m will be the proper bevil for a mitre-joint. If the upper edge of the splayed work is to be bevelled, so as to be horizontal when the work is placed in its proper position, then / 7, the same as ад e, will be the proper bevel for that purpose. Suppose, therefore, that a piece in- dicated by the lines 4 g, g f, and fh were taken off; then a line drawn upon the bevelled surface from 4 at right angles to d would show the true position of the joint, because it would be in the direction of the board for the other side ; but a line so drawn would pass through the point o, thus proving the principle correct. So,if a line were drawn upon the bevelled surface from 4 at an angle of 45 degrees to £ d, it would pass through the point z. ` “

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ў ге Т ji үү iu | о F mes reatum ere EI, a ---% E m $ Tn . — ғ. p a re CU eol VIEW IN THE ALHAMBRA.

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SECTION IV.—DOORS AND WINDOWS. DOORS. 298.— General Requirements, — Among the architectural arrangements of an edifice, the door is by no means the least in importance; and if properly constructed, it is not only an article of use, but also of ornament, adding materially to the regularity and elegance of the apartments. The dimen- sions and style of finish of а door should be in accordance with the size and style of the building, or the apartment for which it is designed. Ав regards the utility of doors, the principal door to a public building should be of suffi- cient width to admit of a free passage for a crowd of people; while that of a private apartment will be wide enough if it permit one person to pass without being incommoded. Ех- perience has determined that the least width allowable for this is 2 feet 8 inches; although .doors leading to inferior and unimportant rooms may, if circumstances require it, be as narrow as 2 feet 6 inches; and doors for closets, where an entrance is seldom required, may be but 2 feet wide. The width of the principal door to a public building may be from 6 to 12 feet, according to the size of the building ; and the width of doors for a dwelling may be from 2 feet 8 inches to 3 feet 6 inches. If the importance of an apart- , ment in a dwelling be such as to require a door of greater width than 3 feet 6 inches, the opening should be closed with two doors, or a door in two folds; generally, in such cases, where the opening is from 5 to 8 feet, folding or slid- ing doors are adopted. As toéhe height of a door, it should in no case be less than about 6 feet 3 inches; and generally not less than 6 feet 8 inches. 299. ——Fhe Proportion between Width and Height: of single doors, for a dwelling, should be as 2 is to 5; and, for, -entrance-doors to public buildings, as 1 is to 2. If the width is given and the height required of a door for a

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316 DOORS AND WINDOWS. dwelling, multiply the width by 5, and divide the product by 2; but if the height is given and the width required, divide by 5 and multiply by 2. Where two or more doors of different widths show in the same room, it is well to pro- portion the dimerisions of the more important by the above rule, and make the narrower doors of the same height as the wider ones; as all the doors in a suit of apartments, except the folding or sliding doors, have the best appear- ance when of one height. The proportions for folding or sliding doors should be such that the width may be equal to 4 of the height; yet this rule needs some qualification ; for if the width of the opening be greater than one half the width of the room, there will not be a sufficient space left Еіс. 179. for opening the doors; also, the height should be about one tenth greater than that of the adjacent single doors. 300.—Panels.—Where doors have but two panels in width, let the stiles and muntins be each 4 of the width; or, whatever number of panels there may be, let the united widths of the stiles and the muntins, or the whole width of the solid, be equal to 3 of the width of the door. Thus: in a door 35 inches wide, containing two panels in width, the stiles should be 5 inches wide; and in a door 3 feet 6 inches wide, the stiles should be 6 inches. If a door 3 feet 6 inches wide is to have 3 panels in width, the stiles and muntins should be each 44 inches wide, each panel being 8 inches. The bottom rail and the lock-rail ought to be each equal in width to +; of the height of the door; and’ the top

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% TRIMMINGS FOR DOORS. 317 rail, and all others, of the same width as the stiles. The moulding on the panel should be equal in width to 1 of the width of the stile. 301.—' Trimmings.—777. 179 shows a method of trimming doors: а is the door-stud; 0, the lath and plaster; c, the ground, d, the jamb; e, the stop ; f and g, architrave casings; and Z, the door-stile. It is customary in ordinary work to form the stop for the door by rebating the jamb. But when the door is thick and heavy,a better plan is to nail on a piece as at e in the figure. This piece can be fitted to the door and put on after the door 1s hung ; so, should the door be a trifle winding, this will correct the evil, and the door be made to shut solid. Fig. 180 is an elevation of a door and trimmings suitable for the best rooms of adwelling. (For trimmings generally, see Sect. V.) The number of panels into which a door should be divided may be fixed at pleasure; yet the present style of finishing requires that the number be as small as a proper regard for strength will admit. In some of our best dwellings, doors have been made having only two upright panels. A few years' experience, however, has proved that the omission of the lock-rail is at the expense of the strength and durability of the door; a four-panel door, therefore, is the best that can be made. 302.—Hanging Doors.—Doors should all be hung so as to open into the principal rooms; and, in general, no door should be hung to open into the hall, or passage. As to the proper edge of the door on which to affix the hinges, по general rule can be assigned. | WINDOWS. 303.— Requirements for Light.—A window should be of such dimensions, and in such a position, as to admit a sufficiency of light to that part of the apartment for which it is designed. Хо definite rule for the size can well be given that will answer in all cases; yet, as an арргохіта-

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* 318 DOORS AND WINDOWS. tion, the following has been used for general purposes. Multiply together the length and the breadth in feet of the apartment to be lighted, and the product by the height in Fic. 180. | feet; then the square root of this product will show the required number of square feet of glass. 304.— Window-Frames,—For the size of window-frames, add 44 inches to the width of the glass for their width, and

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WIDTH OF INSIDE SHUTTERS. 319 63 inches to the height of the glass for their height. -These give the dimensions, in the clear, of ordinary frames for 12- light windows; the height being taken at the inside edge of the sill. In a brick wall, the width of the opening is 8 inches more than the width of the glass—44 for the stiles of the sash, and 33 for hanging stiles—and the height between the stone sill and lintel is about 104 inches more than the height of the glass, it being varied ageing to the thick- ness of the sill of the frame. boxes require to have the Боер about one inch wider than the flap, in order that the flap may not interfere when both are folded into the box. The usual margin shown be- tween the face of the shutter when folded into the box and the quirk of the stop-bead, or edge of the casing, is half an inch; and, in the usual method of letting the whole of the thickness of the butt hinge into the edge of the box-shutter, it is necessary to make allowance for the z%row of the hinge. This may, in general, be estimated at 4 of an inch at each hinging ; which being added to the margin, the entire width of the shutters will be 14 inches more than the width of the frame in the clear. Then, to ascertain the width of the box- shutter, add 14 inches to the width of the frame in the clear, between the pulley-stiles; divide this product by 4, and add half an inch to the quotient, and the last product will be the required width. For example, suppose the window to have 3 lights in width, 11 inches each. Then, 3 times 11 is 33, and 44 added for the wood of the sash gives 37%; 37% and 14 is 39, and 39 divided by 4 gives 92; to which add half an inch, and the result will be той inches, the width required for the box-shutter. 306.—Proportion: Width and Height.—In disposing and locating windows in the walls of a building, the rules of architectural taste require that they be of different heights in different stories, but generally of the same width. The windows of the upper stories should all range perpendicu- larly over those of the first, or principal, story ;` and they

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320 DOORS AND WINDOWS. should be disposed so as to exhibit a balance of parts throughout the front of the building. То aid in this it is always proper to place the front door in the middle of the front of the building ; and, where the size of the house will admit of it, this plan should be adopted. (See the latter part of Art. бо) The proportion that the height should bear to the width may be, in accordance with general usage, as follows: The height of basement windows, 14 of the width. s 5 principal-story * 2i T * ‘i second-story “ m ^ à s third-story D uU ii A fourth-story “ m Ы “ Jj attic-story “ the same as the width. But, in determining the height of the windows for the several stories, it is necessary to take into consideration the height of the story in which the window is to be placed. For, in addition to the height irom the floor, which is gen- erally required to be from 28 to 30 inches, room is wanted above the head of the window for the window-trimming and the cornice of the room, besides some respectable space which there ought to be between these. 307. Doors and windows usually ter- minate in a horizontal line at top. "These. require no special directions for their trimmings. But circular-headed doors and windows are more difficult of execution, and require some attention. If the jambs of a door or window be placed at right angles to the face of the wall, the edges of the soffit, or surface of the head, would be straight, and its length be found by getting the stretch-out of the circle (477. 524); but when the jambs are placed obliquely to the face of the wall, occasioned by the demand for light in an oblique direction, the form of the soffit will be obtained by the fol- lowing article; as also when the face of the wall is circular, as shown in the succeeding figure.

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OBLIQUE SOFFITS OF. WINDOWS. : 321 308.—Form of Soffit for Circular Window-Heads.— When the light is received in an oblique direction, let abcd (Fig. 181) be the ground-plan of a given window, and e faa vertical section taken at right angles to the face of the jambs. Fic. 181. From a, through e, draw ag at right angles to a4; obtain the stretch-out of e fa, and make eg equal to it; divide eg and e fa each into a like number of equal parts, and drop perpendiculars from the points of division in each; from the points of intersection, I, 2, 3, etc., in the line ad, Fic. 182. draw horizontal lines to meet corresponding perpendicu- lars from eg; then those points of intersection will give the curve line dg, which will be the one required for the edge of the soffit. The other edge, cA, is found in the same manner.

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322 DOORS AND WINDOWS. For the form of the soffit for circular window-heads, when the face of the wall is curved, let a bc d (Fig. 182) be the ground-plan of a given window, and e fa a vertical sec- tion of the head taken at right angles to the face of the jambs. Proceed as in the foregoing article to obtain the line Z7; then that will be the curve required for the edge of the soffit, the other edge being found in the same manner. If the given vertical section be taken in a line with the face of the wall, instead of at right angles to the face of the jambs, place it upon the line cé (Fig. 181), and, having drawn ordinates at right angles to c2, transfer them to e fa; in this way a section at right angles to the jambs can be obtained.

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He 77. Ч. “а РРР «х (wg INTERIOR OF ST. PETER'S, ROME. * LA Е ў saci 6 e DAZ А IERI IE FUR JT LOIS PS IES IIE RP EUN м ^ = ys ed

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b st | BRA pS К UNIVERSITY] SECTION V.—MOULDINGS AND CORNICES. MOULDINGS. 309.—Mouldings: are so called because they are of the same determinate shape throughout their length, as though the whole had been cast in the same mould or form. The regular mouldings, as found in remains of classic architec- ture, are eight in number, and are known by the following names: | ar 14, Fic. 18. Annulet, band, cincture, fillet, listel or square. Fic. 184. Astragal or bead. ғғ Torus or tore. Fic. 185. Bic, M46 Scotia, trochilus or mouth. nas Ovolo, quarter-round or echinus. Fic. 187 cr Cavetto, cove or hollow. Fic. 188.

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324. MOULDINGS AND CORNICES. Cymatium, or cyma-recta. PIG. 150. Mr M үк e кес, Ogee. Inverted cymatium, or cyma-reversa. FIG. 190. Some of the terms are derived thus: Fillet, from the French word fil, thread. Astragal, from astragalos, a bone of the heel—or. the curvature of the heel. Bead, because this moulding, when properly carved, resembles a string of beads. Torus, or tore, the Greek for rope, which it resembles when on the base of a column. Scotia, from sofia, darkness, be- cause of the strong shadow which its depth produces, and which is increased by the projection of the torus above it. Ovolo, from ovum, ап egg, which this member resembles, when carved, as in the Ionic capital. Cavetto, from cavus, hollow. Cymatium, from kumaton, a wave. 310.—Characteristies of Mouldings.—Neither of these ` mouldings is peculiar to any one of the orders of architect- ure; and although each has its appropriate use, yet it is by no means confined to any certain position in an assemblage of mouldings. The use of the fillet is to bind the parts, as also that of- the astragal and torus, which resemble ropes. The ovolo and cyma-reversa are strong at their upper ex- tremities, and are therefore used to support projecting parts above them. The cyma-recta and cavetto, being weak at their upper extremities, are not used as supporters, but are placed uppermost to cover and shelter the other parts. The scotia is introduced in the base of a column to separate the upper and lower torus, and to produce a pleasing variety and relief. Тһе form of the bead and that of the torus is the same; the reasons for giving distinct names to them are that the torus, in every order, is always considerably larger than the bead, and is placed among the base mouldings,

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GRECIAN MOULDINGS. 325 whereas the bead is never placed there, but on the capital or entablature; the torus, also, is seldom carved, whereas the bead is; and while the torus among the Greeks is frequently elliptical in its form, the bead retains its circular shape. While the scotia is the reverse of the torus, the cavetto is the re- verse of the ovolo, and the cyma-recta and cyma-reversa are combinations of the ovolo and cavetto. = FIG. 191. The curves of mouldings, in Roman architecture, were most generally composed of parts of circles; while those of the Greeks were almost always elliptical, or of some one of the conic sections, but rarely circular, except in the case of the bead, which was always, among both Greeks and Ro- mans, of the form of a semicircle. Sections of the cone af- ford a greater variety of forms than those of the sphere; and perhaps this is one reason why the Grecian architecture so

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326 MOULDINGS AND CORNICES. much excels the Roman. Тһе quick turnings of the ovolo and cyma-reversa, in particular, when exposed to a bright sun, cause those narrow, well-defined streaks of light which give life and splendor to the whole. 311.—А Profile: is an assemblage of essential parts and mouldings. That profile produces the happiest effect which FIG. 192. Етс. 193. is composed of but few members, varied in form and size, and arranged so that the plane and the curved suríaces suc- ceed each other alternately. 312.—' Тһе Grecian Torus and Seotia.—]oin the extremi- ties a and 2 (Fig. 191), and from / the given projection of the moulding, draw fo at right angles to the fillets; from д 4 Fic. 194. — Fic. 195. draw 2/ at right angles to аб; bisect a^ inc; join f and с, and upon с, with the radius c f, describe the arc /2, cutting біп: й; through с draw Ze parallel with the fillets; make dc and ce each equal to 64; then Ze and аб will be conju- gate diameters of the required ellipse. To describe the curve by intersection of lines, proceed as directed at 472.

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THE GRECIAN ECHINUS. 327 551 and ме; by a trammel, see 77. 549; and to find the foci, in order to describe it with a string, see 477. 548. 313.—The Grecian Echinus.— 7773. 192 to 199 exhibit, va- riously modified, the Grecian ovolo, orechinus. gs. 192 to 196 are elliptical, 2d and 2c being given tangents to the curve; parallel to which the semi-conjugate diameters, ad and de, Fic. 196. | FIG. 197. are drawn. In Figs. 192 and 193 the lines ad and dc аге semi- axes, the tangents, ad and бс, being at right angles to each other. To draw the curve, see 472. 551. In Fig. 197 the curve is parabolical, and is drawn according to 477. 560. In Figs. 198 and 199 the curve is hyperbolical, being described according to Art. 561. Thelength of the transverse ax's, ад, Fic. 198. FIG. 199. being taken at pleasure in order to flatten the curve, ад should be made short in proportion to ac. 314.--Тһе Grecian Cavetto.—In order to describe this, Figs. 200 and 201, having the height and projection тШ see Art. 551. 315.—The Grecian Cyma-Rcecta.—When the projection is more than the height, as at 7zg. 202, make а 6 equal to the

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328 MOULDINGS AND CORNICES. height, and divide абс into four equal parallelograms ; then proceed as directed in note to 477. 551. When the projec- | tion is less than the height, draw da (Fig. 203) at right angles ГЕ. Fic. 201. Ес. 200. to аб; complete the rectangle, abcd; divide this into four equal rectangles, and proceed according to Artz. 551. 316.—The Grecian Cyma-Reversa.— When the projection Ғіс. 202. is more than the height, as at 77g. 204, proceed as directed for the last figure; the curve being the same as that, the position only being changed. When the projection is less Fic. 204. d Fic. 205. than the height, draw a d (Fig. 205)at right angles to the fillet; make ad equal to the projection of the moulding; then - proceed as directed for Fig. 202.

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FORMS OF ROMAN MOULDINGS. 329 3i7.—Roman Mouldings: are composed of parts of circles, and have, therefore, less beauty of form than the Grecian. The bead and torus are of the form of the semicircle, and the scotia, also, in some instances ; but the latteris often composed of two quadrants, having different radii, as at Figs. 206 and 207, which resemble the elliptical curve. The ovolo and ca- an Fic. 206. HIG. 207. vetto are generally a quadrant, but often less. When they are less, as at 77g. 210, the centre is found thus: join the extrem- ities, а and 2, and bisect að in с; from с, and at right angles to «à, draw cd, cutting a level line drawn from a in 4; then d will be the centre. This moulding projects less than its height. When the projection is more than the height, as at Fig. 212, extend the line from c until it cuts a perpendicular Y 2 Fic. 208. Fic. drawn from a, as at 4; and that will be the centre of the curve. Ina similar manner, the centres are found for the mouldings at Figs. 207, 211, 213, 216, 217, 218, and 219. The centres m the curves at xs. 220 and 221 are found thus: bisect the linea 2 at c; upona,c and 2 successively, with ac or сё for radius, describe arcs intersecting at 4 and d; then those intersections will be the centres.

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330 MOULDINGS AND CORNICES. FIG. 212. : Fic. 212. Fic. 216. Fic. 217.

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FORMS OF MODERN MOULDINGS. 331 318.—Modern Mouldings: are represented in Figs. 222 to 229. They have been quite extensively and successfully used in inside finishing. Fig. 222 is appropriate for a bed- moulding under a low projecting shelf, and is frequently used under mantel-shelves. The tangent 2 2 is found thus: bisect the line аб at c, and фе at d; from d draw de at right angles to ей; from 0 draw 6 f parallel to ed; upon 4, Fic, 218. FIG. 219. with 6d for radius, describe the arc df; divide this arc into 7 equal parts, and set one of the parts from s, the limit of the projection, too; make о 2 equal to ое; from 4, through c, draw the tangent 22; divide 22, hc, с?, and za each into a like number of equal parts, and draw the intersecting lines as directed at 477. 521. If a bolder form is desired, draw the tangent, 2 2, nearer horizontal, and describe an elliptic FIG. 220. FIG. 221. curve as shown in Figs. 191 and 224. Fig. 223 is much used on base, or skirting, of rooms, and in deep panelling. The curve is found in the same manner as that of Fig. 222. Іп this case, however, where the moulding has so little projec- tion in comparison with its height, the point 2 being found as in the last figure, 25 may be made equal to s e, instead of де as in the last figure. Fig. 224 is appropriate for a crown

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. Pie. 222. FIG, 224.

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.PLAIN MOULDINGS. | 333 moulding of a cornice. In this figure the height and pro- jection are given; the direction of the diameter, 22, drawn Fic. 225. - Fic. 226. through the middle of the diagonal, e /, is taken at pleasure; and dc is parallel to ағ. To find the length of dc, draw 0 2 PIG. 227. Fic. 228. FIG. 229. at right angles to 202; upon o, with o f for radius, describe the arc, fh, cutting ёй in й; then make oc and od each

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% \ 334 MOULDINGS AND CORNICES. equal to 54.* To draw the curve, see note to 477. 551. Figs. 225 to 229 are peculiarly distinct from ancient mouldings, being composed principally of straight lines; the few curves they possess are quite short and quick. Pigs. 230 and 23r are designs for antz caps. The di- ameter of the anta is divided into 20 equal parts, and the height and projection of the members are regulated in ac- cordance with those parts, as denoted under H and P, height and projection. Тһе projection is measured from the mid- dle of the antæ. These will be found appropriate for por- ticos, doorways, mantelpieces, door and window trimmings, Н.Р. Кіс. 221. etc. Тһе height of the antz for mantelpieces should be from 5 to 6 diameters, having an entablature of from 2 to 2} diameters. This is a good proportion, it being similar to the Doric order. But fora portico these proportions are * The manner of ascertaining the length of the conjugate diameter, dc, in \ this figure, and also in Figs. 191, 241, and 242 is new, and is important in this application. It is founded upon well-known mathematical principles, viz.: All the parallelograms that may be circumscribed about an ellipsis are equal to one another, and consequently any one is equal to the rectangle of the two axes. And again: The sum of the squares of every pair of conjugate diame- ters is equal to the sum of the squares of the two axes.

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EAVE CORNICES. 385 much too heavy: an antz 15 diameters high and an entab- lature of 3 diameters will have a better appearance. CORNICES. 319.— Designs for Cornices.—/igs. 232 to 240 are designs for eave cornices, and 2/5. 241 and 242 are for stucco cor- nices for the inside finish of rooms. Іп some of these the projection of the uppermost member from the facia is divided into twenty equal parts, and the various members — — ~m «ншы» же» ғыз — өле» Pic. 242. are proportioned according to those parts, as figured under Ab and Р. | 320.—Eave Cornices Proportioned to Height of Huild- ing.—Draw the line ac (Fig. 243), and make бс and ба each equal to 36 inches; from b draw да at right angles to ac, and equal in length to $of ас; bisect dd in е, and from а, through e, draw af; upon a, with ac for radius, describe the arc cf, and upon e, with e for radius, describe the arc fd; divide the curve dfc, into 7 equal parts, as at 10, 20, 30, etc, and from these points of division draw lines to dc

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336 MOULDINGS AND CORNICES. Fic. 233. Si LL TOZ DIII II ТІТІСІШІЛІІІІ ТІГІ Fic. 234.

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“ХАМРІ,Е5 OF CORNICES. 337 FIG. 235. Fic. 226.

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338 MOULDINGS AND CORNICES. FIG. 237. Fic. 238.

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VARIOUS DESIGNS OF CORNICES. 339 FIG. 240. -

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340 MOULDINGS AND CORNICES - FiG. 241. F1c. 242.

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PROPORTION OF CORNICES. : 341 parallel to Z5; then the distance фт is the projection of a cornice for a building то feet high; 22, the projection at 20 feet high; 23, the projection at 30 feet, etc. If the projec- tion of a cornice for a building 34 feet high is required, divide the arc between зо and до into то equal parts, and а i б I 234 C Fic. 243. from the fourth point from 30 draw a line to the base, £c, parallel with 04; then the distance of the point at which that line cuts the base from 0 will be the projection re- quired. So proceed fora cornice of any height within 70 feet. The above is based on the supposition that 36 inches FIG. 244. is the proper projection for a cornice 7o feet high. This, for general purposes, will be found correct ; still, the length of the line фе may be varied to suit the judgment of those who think differently. Having obtained the projection of a cornice, divide it into 20 equal parts, and apportion the several members

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342 | MOULDINGS AND CORNICES. according to its destination—as is shown at Figs. 238, 230, and 240. 321.--бөгпісе Proportioned to a given Cornice.—Let the cornice at Fig. 244 be the given cornice. Upon апу point in the lowest line of the lowest member, as at 2, with the height of the required cornice for radius, describe an intersect- ing arc across the uppermost line, as at 0; join а and 0; then 2 1 will be the perpendicular height of the upper fillet for the proposed cornice, 1 2 the height of the crown mould- ing—and so of all the members requiring to be enlarged to the sizes indicated on this line. For the projection of the 1 Fic. 245. proposed cornice, draw ad at right angles to ад, and c d at right angles to 2c; parallel with cd draw lines from each projection of the given cornice to the line ad; then ed will be the required projection for the proposed cornice, and the perpendicular lines falling upon e Z will indicate the proper projection for the members. To proportion a cornice according to a larger given cor- . nice, let A (Fig. 245) be the given cornice. Extend ао to å, and draw са at right angles to аб; extend the horizontal lines of the cornice, A, until they touch od; place the height of the proposed cornice from о to e, and join fand e; upon o, with the projection of the given cornice, oa, for radius,

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TO FIND THE ANGLE BRACKET. 243 describe the quadrant ad; from d draw då parallel to fe; upon o, with оф for radius, describe the quadrant ёс; then oc wil be the proper projection for the proposed cornice. Join а and с; draw lines from the projection of the different members of the given cornice to ао parallel to od; from these divisions on the line ао draw lines to the line oc parallel to ас; from the divisions on the line o f draw lines to the line oe parallel to the line fe; then the divisions on the lines o e and oc will indicate the proper height and pro- jection for the different members of the proposed cornice. In this process, we have assumed the height, o e, of the pro- posed cornice to be given; but if the projection, ос, alone Fic. 246. Fic. 247. be given, we can obtain the same result by a different pro- cess. Thus: upon o, with oc for radius, describe the quad- rant có; upon o, with оа for radius, describe the quadrant ad; join d and 6; from f draw fe parallel ќо 202; then oe will be the proper height for the proposed cornice, and the height and projection of the different members can be obtained by the above directions. By this problem, a cor- nice can be proportioned according to a smaller given one as well as to a darger ; but the method described in the pre- vious article is much more simple for that purpose. 322.—Angle Bracket in a Built Cornice—Let A (Fig. 246) be the wall of the building, and B the given bracket,

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344 : OULDINGS AND CORNICES. which, for the present purpose, is turned down horizontally. The angle-bracket, C, is obtained thus: through the ex- tremity, а, and parallel with the wall, fd, draw the line ad; make ec equal af, and through с draw c? parallel with ed; join 4 and 4, and from the several angular points in B draw ordinates to cut Z ? in т, 2, and 3; at those points erect lines perpendicular to Z5; from й draw 4g parallel to fa; take j CT ‘ Fic. 248. the ordinates, т o, 20, etc., at P, and transfer them to C, and the angle-bracket, C, will be defined. In the same manner, the angle-bracket for an internal cornice, or the angle-rib of a coved ceiling, or of groins, as at Fig. 247, can be found. 323.—Raking Mouldings matched with Level Returns.— Let A (Fig. 248) be the given moulding, and A 6 the rake of

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CROWN MOULDING ON THE RAKE. 345 the roof. Divide the curve of the given moulding into any number of parts, equal or unequal, as at 1, 2, and 3; from these points draw horizontal lines to a perpendicular erected from c; at any convenient place on the rake, аз ағ В, draw acat right angles to A 0; also from ? draw the horizontal line ба; place the thickness, da, of the moulding at A from 6 to a, and from а draw the perpendicular line ae; from the points 1, 2, 3, at A, draw lines to C parallel to 42; make a I, a 2, and a 3, at B, and at C, equal to аы, etc., at Ae through the points, 1, 2, and 3, at В, trace the curve—this will be the proper form for the raking moulding, From 1, 2, and 3,at C, drop perpendiculars to the corresponding ordinates from 1, 2, and 3, at 4; through the points of inter- section, trace the curve—this will bethe proper form for the return at the top.

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ІЛЕ У- =i] sjaju |

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i 4 PART II. SECTION VI —GEOMETRY. 324.— Mathematics Essential.—In this and the following Sections, which will constitute Part II., there are treated of certain matters which may be considered as elementary. They are all very necessary to be understood and acquired by the builder, and are here compactly presented in a shape which, it is believed, will aid him in his studies, and at the same.time prove to be а great convenience as a matter of reference. The many geometrical forms which enter into the composition of a building suggest a knowledge of Elemen- tary Geometry as essential to an intelligent comprehension of its plan and purpose. One of the prime requisites of a building is stability, a quality which depends upon a proper distribution of the material of which the building is con- structed; hence a knowledge of the laws of pressure and the strength of materials is essential; and as these are based upon the laws of proportion and are expressed more con- cisely in algebraic language, a knowledge of Proportion and of Algebra are likewise indispensable to a comprehensive understanding of the subject. There will be found in this work, however, only so much of these parts of mathematics as have been deemed of the most obvious utility in the Science of Building. Fora more exhaustive treatment of the subjects named, the reader is referred to the many able works, readily accessible, which make these subjects their specialties. 325,.-_Elementary Geometry.—In all reasoning defini- tions are necessary, in order to insure in the minds of the

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348 | | GEOMETRY. proponent and respondent identity of ideas. А соғо ағу is an inference deduced from a previous course of reasoning. ` An axiom is a proposition evident at first sight. In the fol- lowing demonstrations there are many axioms taken for granted (such as, things equal to the same thing are equal to one another, etc.); these it was thought not necessary to introduce in form. 326.—Definition.—If a straight line, as A 2 (Fig. 249), stand upon another straight line, as C D, so that the two A ЖЕ Fic. 249. angles made at the point B are equal—4 B C to A BD (Art. 499, obtuse angle) —then each of the two angles is called a right angle. 327.—Definition. — The circumference of every circle is supposed to be divided into 360 equal parts, called degrees; hence a semicircle contains 180 degrees, a quad- rant 9o, etc. b ———. 4 SA 328.—Definition.—_ The measure of an angle is the num- ber of degrees contained between its two sides, using the angular point as a centre upon which to describe the arc. Thus the arc C E (Fig. 250) 15 the measure of the angle СВЕ, E A of the angle E P A, and A D of the angle 4 В D. 329.— Corollary.—As the two angles at B (Fig. 249) are right angles, and as the semicircle, C A D, contains 180 de- grees (Art. 327), the measure of two right angles, therefore, 15

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UNIVERSITY G RIGHT ANGLES AND OBLIQUE ANG 540 1 10:4 3524 180 degrees; of one right angle, 90 degrees; of halfa right. angle, 45; of one third of a right angle, 30, etc. 330.— Definition.—In measuring an angle (472. 328), no regard 15 to be had to the length of its sides, but only to the degree of their inclination. Hence egual angles are such as have the same degree of inclination, without regard to the length of their sides. 331.—Axiom.—If two straight lines parallel to one another, as 4А В and C D (Fig. 251), stand upon another. straight line, as Г F, the angles 4 BF and C D F are equal, and the angle А B Е is equal to the angle C D Ж. 332. Definition.—If a straight line, as 4 B (fig. 250), stand obliquely upon another straight line, as C D, then one КЄ 251. of the angles, as A BC, is called az obtuse angle, and the other, as 4 BD, an acute angle. 333.— Axiom.—The two angles A В Dand А BC (Fig. 250) are together equal to two right angles (4775. 326, 329); also, the three angles 4 BD, EBA, and CB E are together equal to two right angles.: 334.--СоғоШағу.-Непсе all the angles that can be made upon one side of a line, meeting in a point in that line, are together eua to two right angles. 335.—Corollary. =e all the angles that can be made on both sides of a line, at a point in that line, or all the angles that can be made about a point, are together equal to four right angles.

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350 GEOMETRY. / 336.—Proposition.—If to each of two equal angles a third angle be added, their sums will be equal. Let A BC and D Е F (fig. 252) be equal angles, and the angle Z F K the one to be added. Make the angles С В А and Б.Е D each equal to the given angle J F K; then the angle G 5 C will be equal to the angle H E F; for if A B C and DEF be angles Fic. 252. of 9o degrees, and 7 7 К 30, then the angles GBC and. HEF wil be each equal to go and 30 added, viz., 120 degrees. 337.—Proposition.—Triangles that have two of their sides and the angle contained between them respectively equal, have also their third sides and the two remaining A D c Ж A Ғіс. 253. angles equal; and consequently one triangle will every way equal the other. Let А BC (Fig. 253) and DEF be two given triangles, having the angle at А equal to the angle at Г, Ње side A В equal to the side D Е, and the side ЯС equal to the side D F; then the third side of one, 2 C, is equal to the third side of the other, £ F; the angle at B is equal to the angle at 2, and the angle at C is equal to the angle at

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\ EQUAL TRIANGLES IN PARALLELOGRAMS. 351 F. For if one triangle be applied to the other, the three points Л, А, C, coinciding with the three points Б, D, F, the line B must coincide with the line ЕР; the angle at д with the angle at Е; the angle at C with the angle at 2; and the triangle B А C be every way equal to the triangle BD 338. Proposition.—The two angles at the base of an isosceles triangle are equal. Let ABC (Fig. 254) be an FIG. 254. isosceles triangle, of which the sides, A Band А C, are equal. Bisect the angle (Art. 506) В A C by the line А D. Then, the line В A being equal to the line A C, the line A D of the triangle Æ being equal to the line A D of the triangle F (being common to each), the angle B A D being equal to the angle D А C,—the line 5 D must, according to Art. 337, be E F F E С р D е Fic. 255. equal to the line D C, and the angle at B must be equal to the angle at C. 339. Proposition, —A diagonal crossing a parallelogram divides it into two equal triangles. Let CD EF (Fig. 255) be a given parallelogram, and C F a line crossing it diag- onally. Then, as ÆC is equal to FD, and £F to C D, the angle at Æ to the angle at D, the triangle A must, according to Art. 337, be equal to the triangle 2.

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352 GEOMETRY. 340.—Proposition.—Let 2 А L M (Fig. 256) be a given parallelogram, and A Z a diagonal. At any distance between 3 K and L M draw УР parallel to FK; through the point G, the intersection of the lines KZ and МР, draw H1 parallel to X M. In every parallelogram thus divided, the parallelogram A is equal to the parallelogram В. For, ac- cording to Art. 339, the triangle # K-Z is equal to the tri- angle K L M, the triangle C to the triangle D, and Е to F; Fic, 256. this being the case, take D and F from the triangle K LM, and C and £ trom the triangle Y K L, and what remains in one must be equal to what remains in the other; therefore, the parallelogram 4 is equal to the parallelogram В. ^ 341,—Proposition.—Parallelograms standing upon the same base and between the same parallels are equal. Let ABCD and EFCD (fig. 257) be given parallelograms A B E F С D Fic, 257. standing upon the same base, C D, and between the same parallels, 4 F and C D. Then А В and ЕР, being equal to C D, are equal to one another; Z being added to both A B and ЕР, А Е equals BF; the line А С being equal to В D, and A Е to В F, and the angle C A Е being equal (Art. 331) to the angle D B F, the triangle А Е C must be equal (Art. 337) to the triangle В F D; these two triangles being equal, take the same amount, the triangle Z Е G, from each,

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TRIANGLE EQUAL TO QUADRANGLE. 353 and what remains in one, 4 BGC, must be equal to what remains in the other, E DG; these two quadrangles being equal, add the same amount, the triangle C G D, to each, and they must- still be equal; therefore, the parallelogram A B C D is equal to the parallelogram £ F C D. 342.—Corollary.— Hence, if a parallelogram and triangle . Stand upon the same base and between the same parallels, F G pe H i D Fic. 258. the parallelogram will be equal to double the triangle. Thus, the parallelogram A D (Fig. 257) is double (477. 339) Ше triangle C E D. 343.—Proposition.—Let FG H D (Fig. 258) bea given quadrangle with the diagonal FD. From С draw G E Fic. 259. parallel to FD; extend HD to Е; join Fand Æ; then the triangle FE H will be equal in area to the quadrangle FGHD. For since the triangles FD G and FD E stand upon the same base, FD, and between the same parallels, FD and С E, they are therefore equal (Arts. 341, 342); and since the triangle C is common to both, the remaining tri-

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354 GEOMETRY. angles, 4 and В, are therefore equal; then, В being equal to A, the triangle FE H is equal to the quadrangle F G H D. 34.4.. —Proposition.—1f two straight lines cut each other, as FG and H F (Fig. 259), the vertical, or opposite angles, А апа С, аге equal. Thus, FE, standing upon A 9, forms the angles В and C, which together amount (477. 333) to two right angles ; in the same manner, the angles А and В form two right angles; since the angles A and В are equal to В and C, take the same amount, the angle В, from each pair, and what remains of one pair is equal to what remains of the other; therefore, the angle 4 is equal to the angle C. The same can be proved of the opposite angles В and D. F E B С b Fic. 260. 34 5.— Proposition.—The three angles of any triangle аге equal to two right angles. Let 4 BC (Fig. 260) be a given triangle, with its sides extended to F, Е and 2, and the line С С drawn parallel to ВЕ. Ав С Cis parallel. to Е В, the angle at H is equal (477. 331) to the angle at 2; as the lines ЕС and РЕ cut one another at A, the opposite angles at M and X are equal (477. 334); as the angle at Vis equal (477. 331) to the angle at F, the angle at 715 equal to the angle at М; therefore, the three angles meeting at C are equaltothe - three angles of the triangle 4 ВС; and since the three angles at C are equal (472. 333) to two right angles, the three angles of the triangle 4 BC must likewise be equal to two right angles. Any triangle can be subjected to the same proof. 34-6.—Corollary.— Hence, if one angle of a triangle be a right angle, the other two angles amount to just one right angle.

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RIGHT ANGLE IN SEMICIRCLE. 355 34-7.—Corollary.—If one angle of a triangle be a ‘right ángle and the two remaining angles are equal to one another, these are each equal to half a right angle. 348.—Corollary-—If any two angles of a triangle amount to a right angle, the remaining one is a right angle. 349.—Corollary.—If any two angles of a triangle are to- gether equal to the remaining angle, that remaining angle is a right angle. 350.—Corollary.—If апу two angles of a triangle are each equal to two thirds of a right angle, the remaining angle is also equal to two thirds of a right angle. ЗБІ.-СогоПағу.--Непсе, the angles of an equilateral triangle are each equal to two thirds of a right angle. Fic. 261. 352.—Proposition.—If from the extremities of the di- ameter of a semicircle two straight lines be drawn to any point in the circumference, the angle formed by them at that point will bea right angle. Let А BC (Fzg. 261) bea given semicircle; and 4 Band AC lines drawn from the extrem- ities of the diameter AC to the given point 2; the angle formed at that point by these lines is a right angle. Join the point В апа the centre D; the lines DA, D В, and DE, being radii of the same circle, are equal; the angle at 15, therefore, equal (472. 338) to the angle at Е; also, the angle: at C is, for the same reason, equal to the angle at F; the angle A BC, being equal to the angles at A and C taken to- gether, must, therefore (477. 349), be a right angle. 353. — Proposition.—The square on the hypothenuse of a right-angled triangle is equal to the squares on the two re-

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356° GEOMETRY. maining sides. Let A 5 C (Fig. 262) be a given right-angled triangle, having a square formed on each of its sides; then the square B Æ 15 equal to the squares А C and С В taken together. Thiscan be proved by showing that the parallelo- gram В L is equal to the square G В; and that the parallelo- gram CL is equal to the square НС. The angle C 5-D"is a right angle, and the angle A В F is aright angle ; add to each of these the angle A BC; then the angle FB С will evidently be equal (477. 336) to the angle A BD; the triangle FBC and the square С В, being both upon the same base, F B, and between the same parallels, F В and С C, the square С P is equal (Art. 342)to twice the triangle “BC; the triangle A B D and the parallelogram В L, being both upon the same FIG. 262. base, 2 D, and between the same parallels, 5 D and A L, the parallelogram 2 / is equal to twice the triangle ALD; the triangles, “BC and А BD, being equal to one another (Art. 337), the square GB is equal to the parallelogram 2 2, either being equal to twice the triangle Сог А BD. The method of proving H C equal to C L is exactly similar—thus proving the square B Æ equal to the squares 77 C and GB, taken together. This problem, which is the 47th of the First Book of Euclid, is said to have been demonstrated first by Pythago- ras. It is stated (but the story is of doubtful authority) that as a thank-offering for its discovery he sacrificed a hun- dred oxen to the gods. From this circumstance it is some- times called the hecatomb problem. It is of great value in

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* DIAGONAL OF SQUARE FORMING OCTAGON. 252 the exact sciences, more especially in Mensuration and As- tronomy, in which many otherwise intricate calculations are by it made easy of solution. 354-. —Proposition.—In an equilateral octagon the semi- diagonal of a circumscribed square, having its sides coinci- dent with four of the sides of the octagon, equals the dis- tance along a side of the square from its corner to the more remote angle of the octagon occurring on that side of the square. Let Fig. 263 represent the square referred to; in which O is the centre of each; then A O equals A D. To prove this, it need only be shown that the triangle 4 O D is an isosceles triangle having its sides A Oand A D equal. The МЕ С В X A | D ^ Fic. 263. octagon being equilateral, it is also equiangular, therefore the angles BCO, ECO, ADO, etc., are all equal. Of the right-angled triangle FEC, FC and FE being equal, the two angles FE C and FCE, are equal (Art. 338), and are therefore (Art. 347) each equal to half a right angle. In like manner it may be shown that FA B and FB A are also each equal to half a right angle. And since F E C and FA P are equal angles, therefore the lines E C and 4 P are parallel (Art. 331,) and hence the angles Е СО and А OD are equal. These being equal, and the angles E CO and A D O being equal by construction, as before shown, therefore the angles AOD and A D О are equal, and consequently the lines 4 О and А D are equal. (Art. 338.)

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358 GEOMETRY. 355.—Proposition.— An angle at the circumference of a circle is measured by half the arc that subtends it; that is, the angle A B C (Fig. 264) is equal to half the angle A DC. Through the centre D draw the diameter BE. The tri- angle A BD isan isosceles triangle, A Dand P D being ra- dii, and therefore equal; hence, the two angles at F and G are equal (472. 338), and the sum of these two angles is equal to the angle at Æ (Art. 345), and therefore one of them, С, is equal to the half of М. The angles at H and at G (or АВЕ) are both subtended by the arc А Е. Now, since the angle B, E Fic. 264. at Н is measured by the arc A E, which subtends it, there- fore the half of the angle at Æ would be measured by the half of the arc 4 £ ; and since G is equal to the half of H, therefore С or А B E is measured by the half of the arc 4 Е. It may be shown in like manner that the angle ЕВ С 15 measured by half the arc £ C, and hence it follows that the angle A BC is measured by half the arc, 4 C, that sub- tends it. 356.— Proposition.—In a circle all the inscribed angles, A, B, and С (Fig. 265), which stand upon the same side of the

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EQUAL ANGLES IN CIRCLES. 359 : chord D E аге equal. For each angle is measured by half the arc DF E (Art. 355). Hence the angles are all equal. 357.—Corollary.—Equal chords, in the same circle, sub- tend equal angles. Fic. 265. 358.— Proposition.—The angle formed by a chord and tangent is equal to any inscribed angle in the opposite seg- | Етіс. 266. ment of the circle; that is, the angle D (Fig. 266) equals the angle A. Let HF be the chord, and ZG the tangent ; draw the diameter 2 H; then YH G is a right angle, also FFH 18

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360 GEOMETRY. а right angle. (477.352.) The angles А and В together equal а right angle (477. 346); also the angles P and D together equal a right angle (equal to the angle ¥ HG); therefore, the sum of А and B EE the sum of 7 and D. From each of these two equals, taking the like quantity В, the remainders А and Р are equal. Thus, it is proved for the angle at 4; it 1s also true for any other angle; for, since all other in- scribed angles on that side of the chord line HF equal the angle A (Art. 356), therefore the angle formed by a chord and tangent equals any angle in the opposite segment of the circle. This being proved for the acute angle D, it is also true for the obtuse angle E AF; for, from any point, K (Fig. 267) in the arc A K F, draw lines to 9, F and М; now, if itcan Fic. 267. be proved that the angle E H F equals the angle FK H, the entire proposition is proved, for the angle FK H equals any of all the inscribed angles that can be drawn on that side of the chord. (Art. 356.) To prove, then, that E H F equals H KF: the angle EHF equals the sum of the angles 4 and Б; also the angle HX F equals the sum of the angles C and D. Тһе angles P and D, being inscribed angles on the same chord, F F, are equal. The angles C and А, being right angles (Art. 352), are likewise equal. Now, since A equals C and B equals 2, therefore the sum of А and В equals the sum of C and D—or the angle £ H F equals the angle HK F. 359. — Proposition. — The areas of parallelograms of equal altitude are to each other as the bases of the parallelo- -

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PARALLELOGRAMS PROPORTIONATE TO BASES. 361 grams. In Fig. 268 the areas of the rectangles A B C D and BE DF are to each other as the bases CD and DF. For, putting the two bases in form of a fraction and reducing this: fraction to its lowest terms, then the numerator and denomi- nator of the reduced fraction will be the numbers of equal parts into which the two bases respectively may be divided. For example, let the two given bases be 12 and 9 feet respect- ively, then 12 = 4, and this gives four parts for the larger base and three parts for the smaller one. So, in 272. 268, divide the base C D into four equal parts, and the base D F into three equal parts; then the length of any one of the . parts in C D will equal the length of any one of the parts in DF. Now, parallel with A C, draw lines from each point of division to the line A Е. These lines will evidently divide the whole figure into seven equal parts, four of them occupy- A B с D е “ Fies 265. Е ing the area A BC D, and three of. them occupying the area BEDF. Now it is evident that the areas of the two rect- angles are in proportion as the number of parts respectively into which the base-lines are divided, or that-—. , APCD PEO: CD MDF. The areas in this particular case are as 4 to 3.. But in gen- eral the proportion will be as the lengths of the bases. ` Thus the proposition is proved in regard to rectangles, but it has been shown (Art. 341) that all parallelograms of equal base and altitude are equal. Therefore the proposition is proved in regard to parallelograms generally, including rect- angles. 360.— Proposition.—Triangles of equal altitude are to each other as their bases. It has been shown (A77. 359) that

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362 GEOMETRY. parallelograms of equal altitude are in proportion as their bases, and it has also been shown (477. 342) that of a triangle and parallelogram, when of equal Баѕе and altitude, the parallelogram is equal to double the triangle. Therefore triangles of equal altitude are to each other as their bases. 361.— Proposition. —Homologous triangles have their corresponding sides in proportion. Let the line C D (Fig. 269) be drawn parallel with. 4 B. Then the angles E C D and Æ A B are equal (А72. 331), also the angles Е D С and ЕВ А are equal. Therefore the triangles E C D and E AB are homologous, or have their corresponding angles equal. E Fic. 269. For, join C to В, and A to D, then the triangles 4 C D and BCD, standing on the same base, C D, and between the same parallels, C D and А В, are equal in area. To each of these equals join the common area C D Е, and the sums ADE and ВСЕ will be equal. The triangles C D Е and А РЕ, having the same altitude, are to each other as their bases СЕ and A Ж (Art. 360), or— CUE TO Che: MES Also the triangles C D Ё and B СЕ, having the same alti- tude, are to each other as their bases D Е and B Е, or— CD EL LUC ie oo D BUE.

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CHORDS GIVING EQUAL RECTANGLES. 363 And, since the triangles 4 D E and Б СЕ are equal, as before shown, therefore, substituting in the last proportion A DE - for В C E, we have— БИЕ ЛАРЕ: : РЕ: ВЕ The first two factors here being identical with the first two in the first proportion above, we have, comparing the two proportions— Ch: AH >: DE: D. е or, we have the corresponding sides of one triangle, C D Ж, in proportion to the corresponding sides of the other, 4 Б Е. “Ете. 270. 362. g. 279), intersecting, the parallelogram or rectangle formed by the two parts of one is equal to the rectangle formed by the two parts of the other. That is, the product of C G multi- plied by С 2 is equal to the product of E G multiplied by GF. Тһе triangle А is similar to the triangle В, because it has corresponding angles. The angle H equals the angle G (Art. 344); the ang at F equals the angle at А, because they stand upon the same chord, D F (Art. si for the same

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364 GEOMETRY. reason the angle M equals the angle 2, for each stands upon the same chord, E C. Therefore, the triangle A having the same angles as the triangle B, the length of the sides of one are in like proportion as the length of the sides in the other (Art. 361). So— QE С : 7-6 C D. Hence, as the product of the means equals the product of the extremes (Art. 373), E G multiplied by GF is equal to C С multiplied by GD. 363.— Proposition.—If the sides of a quadrangle are bisected, and lines drawn joining the points of bisection in the adjacent sides, these lines will form a parallelogram. С ы М Pic. 271. Draw the diagonals 4 B and C D (Fig. 271). It will here be perceived that the two triangles 4 £ О and A C D are homol- ogous, having like angles and proportionate sides. Two of the sides of one triangle lie coincident with the two corres- ponding sides of the other triangle, therefore the contained angles between these sides in each triangle are identical. By construction, these corresponding sides are proportion- ate; AC being equal to twice A Е, and A D being equal to twice A О; therefore the remaining sides are proportionate, C D being equal to twice Æ О, hence the remaining corres- ponding angles are equal. Since, then, the angles 4 £O and А C D are equal, therefore the line Æ Q is parallel. with

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PARALLELOGRAM IN QUADRANGLE. 365 the diagonal C D—so, likewise, the line M Љ is parallel to the same diagonal, C D. If, therefore, these two lines, Æ О and M N, are parallel to the same line, C D, they must be parallel to each other. In the same manner the lines O N and ЕМ are proved parallel to the diagonal А В, and to each other ; therefore the inscribed figure ME OW isa parallelogram. It may be remarked, also, that the parallelogram so formed will contain just one half the area of the circumscribing quadrangle.

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SECTION VII.—RATIO, OR PROPORTION. 364.—Merchandise.—A carpenter buys 9 pounds of nails for 45 cents. Не afterwards buys 87 pounds at the same rate. How much did he pay for them? An answer to this question is readily found by multiply- ing the 87 pounds by 45 cents, the price of the 9 pounds, and dividing the product, 3915, by 9; thequotient, 435 cents, is the answer to the question. 365.—The “Rule of Three.”—The process by which this problem is solved is known as the Rule of Three, or Proportion. | Та cases of this kind there are three quantities given, to find a fourth. Previous to working the question it is usual to make a statement, placing the three given quantities in such order that the quantity which is of like kind with the answer shall occupy the second place; the quantity upon which this depends for its value 15 put in the first place, and the remaining quantity, which is of like kind with that in the, first place, is assigned to the third place. When thus arranged, the second and third quantities are multiplied together and the product is divided by the first quantity; the quotient, the answer to the question, is a fourth quantity. These four quantities are related to each other in this manner, namely: the first is in proportion to the second as the third is to the fourth; or, taking the quantities of the given example, and putting them in a for- mal statement with the customary marks between them, we have— 9 : 45 1: : 87 : 435, which is read: 9 is to 45 ав 87 is to 435; ог, 9 is in propor- tion to 45 as 87 is.to 435; or, 9 bears the same relation to 45 as 87 does to 435..

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EQUALITY OF RATIOS. 367 366.—Couples: Antecedent, Consequent.—These four quantities are termed Proportionals, and may be divided into two couples ; the first and second quantities forming one couple, and.the third and fourth the other couple. Of each couple the first quantity is termed the antecedent, and the last the consequent. Thus 9 is an antecedent and 45 its con- sequent ; so, also, 87 is an antecedent and 435 its consequent. 367.—Equal Couples: an Equation.— These four quan- tities may be put. in form thus: Hy... 185 9 87 Each couple is here stated as a fraction: each has its ante- cedent beneath its consequent, and the two couples are separated by a sign, two short parallel lines, signifying equality. This is an equation, and is read thus: 45 divided by 9 is equal to 435 divided by 87; or, as ordinary fractions: 45 ninths are equal to 435 eighty-sevenths. 368.— Equality of Ratios. — Each couple isalso termed a Ratio, and the two the Equality of Ratios. Thus the ratio = equal to the ratio с If the division indicated in these two ratios be actually performed, the equality between the two will at once be apparent, for the quotient in each case is 5. Тһе resolution of each couple into its simplest form by actual division is shown thus: These are read: 45 divided by 9 equals 5; and 455 divided by 87 equals 5. 369.— Equals Multiplied by Equals Give Equals.—If two equal quantities be each multiplied by a given quantity, the

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368 RATIO, OR PROPORTION. | two products will be equal. For example, the fractions 2 апа 3 are each equal to 4, and are therefore equal to each other. If these two equal quantities be each multiplied by any given number, say, for example, by 4, we shall have 4 times 2 equals 8, and 4 times $ equals 12; these products, % and 12 are each equal to 2, and therefore equal to each other. 370.—Multiplying an Equation.—The quantity on each side of the sign = is called a member of.the equation. If each member be multiplied by the same quantity, the equality of the two members is not thereby disturbed (477. 369); therefore, if the two members of the equation 45 _ 9 thus: x (Art. 367) be each multiplied by 87, or be modified 45x87. 435 X87. a. $7’ in which x, the sign for multiplication, indicates that the quantities between which it is placed are to be multiplied together; this ddition to each member of the equation does not destroy the equality; the members are still equal, though considerably enlarged. The equality may be easily tested by performing the operations indicated in the equa- tion. For example: for the first member, we have 45 times 87 equals 3915, and this divided by 9 equals 435. Again, for the second member we have 435 times 87 equals 37845, and this divided by 87 equals 435, the same result as that for the first member. Thus the multiplication has not interfered with the equality of the members. 371.—Multiplying and Dividing one Member of an Equation: Caneelling.—If a quantity be multiplied by a given number, and the product be divided by the same given number, the quotient will equal the original quantity. For example: if 8 be multiplied by 3, the product will be 24; then if this product be divided by 3, the quotient will be 8, the original quantity. Thus the value of a quantity is not

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TRANSFERRING FACTORS IN EQUATIONS. 369 changed by multiplying it by a number, provided it be also divided by the same number. From this, also, we learn that the value of a queat which is required to be multiplied and divided by the same number will not be changed if the multiplication and divis- ion be both omitted ; one cancels the other. Therefore the number 87, appearing in the second member of the equation in the last article both as a multiplier and a divisor, may be omitted without destroying the equality of the two mem- bers. The equation thus treated will be reduced to— 45 x 87 9 == 435. This expression is read: the product of 45 times 87 divided by 9 equals 435. It will be observed that we have here the four terms of the problem in 72. 365, three of them in the first member, and the fourth, the answer to the problem, in the second member. 372.—Transferring a Factor.—Each of the four quan- tities in the aforesaid equation is termed a factor. Compar- ing the equation of the last article with that of Art. 43, it appears that the two are alike excepting that the factor 87 has been transferred from one member of the equation to the other, and that, whereas it was before a divisor, it has now become a multiplier. From this we learn that a factor may be transferred from one member of an equation to the other, provided that in the transfer its relative position to the horizontal line above or below it be also changed ; that is, if, before the transfer, it be below the line, it must be put above the line in the other member; or, if above the line, it must be put below, in the other member. For example: in the equation of the last article let the factor 9 be removed to the second member of the equation. It stands as a divi- sor in the first member ; therefore, by the rule, it must appear as a multiplier after the transfer ; or— 45 x 87 = 9 x 435;

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370 RATIO, OR PROPORTION. which is read, 45 times 87 equals 9 times 435. Бу actually : performing the operations here indicated, we find that each member gives the same product, 3915; thus proving that the equality of the two members was not interfered with by the transfer. 373. : Means and Extremes.— Іп Art. 366, the four factors are put in the usual form of four proportionals. А comparison of these with the four factors as they appear in the equation in the last article, shows that the first member contains the second and third of the four proportionals, and the second member contains the first and the fourth; or, the first contains what are termed the means, and the second, the extremes: From this we learn that in any set of four proportionals, the product of the means equals the product of the extremes. Ав for example, 6 ; , 2 I$; so, also, = 14, an equality of ratios: hence the four factors, 2, 3, 4, 6, are four proportionals, and may be put thus: Extreme, mean, mean, extreme. Jio Ааа dm and, as above stated, the product of the means (3x4 —) 12, equals the product of the extremes (2 x 6 —) 12. Triangles Proportionate. — The TR of the subject of Ratios has thus far been con- fined to its relations with the mercantile problem of Artz. 364. The rules of proportion or the equality of ratios apply equally to questions other than those of a mercantile character. They apply alike to all questions in which quan- tities of any kind are comparable. For example, in geome- try, lines, surfaces, and solids bear a certain fixed relation to one another, and are, therefore, fit subjects for the rules of proportion. It is shown, in 472. 361, that the correspond- ing sides of homologous triangles are in proportion to one another. Hence, when, of two similar triangles, two corres- ponding sides and one other side are given, then by the equality of ratios the side corresponding to this other side

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RATIOS APPLIED TO TRIANGLES. 371 тау Бе computed. For example: in two triangles, such as ECD and E АВ (Fig. 269), having their corresponding angles equal, let the side Е C, in the triangle £ C D, equal 12 feet, and the corresponding side ХЕ A, in the triangle E A В, equal 16 feet, and the side E D, of triangle E C D, equal 14 feet. Now, having these three sides given, how can we find the fourth? Putting them in proportion, we have, as in Art. 361— | URL : 4: : DE: BE; and, substituting for the known sides, their dimensions, we have— I277 T0 7 $ Inda о and, by Art. 373— | 12x BE = 16x14. Dividing each member by 12, gives— IÓ x I pw 12 Performing the multiplication and division indicated, we have— Thus we have the fourth side equal to 18$ feet. 375.—The Steelyard.—An example of tour proportion- als may also be found in the relation existing between the arms of a lever and the weights suspended at theirends. А familiar example of a lever is seen in the common szee/yard used by merchants in weighing goods. Thisisa bar, 4 В, of steel, arranged as in Fig. 272, with hooks and links, and a suspended platform to carry ХА, the article to be weighed ; and with a weight P, suspended by a link at B, from the bar A B, along which the weight P is movable. The entire load is sustained by links attached to the ful- crum, or point of suspension C. The apparatus is in equi- librium without Æ and P. In weighing any article, X, the

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372 RATIO, OR PROPORTION. weight P is moved along the bar BC until the weight just balances the load, or until the bar A P will remain in а hori- zontal position. If the weight P be too far from the fulcrum C the end of the bar 5 will fall, but if it be too near it will . rise. 376.—The Lever Exemplified by the Steelyard. — То exemplify the principle of the lever, let the bar А 5 (Fig. 272) be balanced accurately with the scale platform, but without the weights R and P. Then, placing the article № upon the platform, move the weight P along the beam until there is an equilibrium. Suppose the distances 4 C and BC are found to be 2 and 40 inches respectively, and suppose Fre. 272, the weight P to equal 5 pounds, what at this point will be the weight of R? By trial we shall find that Æ = 100 pounds. Again, if a portion of R be removed, then the weight P would һауе to be moved along the bar BC to produce an equilibrium ; suppose it be moved until its distance from C be found to be 20 inches, then the weight of Æ would be found to be 5o pounds, or— R = 50 pounds. Again, suppose a part of the weight taken from A be re- stored, and the weight P,on being moved to a point re- quired for equilibrium, be found to measure зо inches from C, then we shall find that— К = 75 pounds.

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RATIOS OF THE LEVER. 373 Thus when— S aA) R= тоо; on, — co 40 ШІ? зе 20, R= 50; ог, и showing an equality of ratios; or, in general, BC is in pro- portion to №, or— DC ote If, instead of moving P along В C, its position be permanent, and the weight P be reduced as needed to produce equilib- rium with the various articles, №, which in turn may be put upon the scale; then we shall find that if when the weight P equals 5 pounds the article Æ equals 100, and there is an equilibrium, then when— ER 5, i 9. o P= o m E R will equal ig X 100 = 90; 8 8 | = то х5 = 4 & will equal — х тоо = ӨШ; ЕНІ = __ i ШЕ — _ P= (o*573:5 R will equal io Х 100 = 70; and so on for other proportions, and in every case we shall "m have the ratio > equal 20, thus— Ae. m roo A T EL R80, DT» ыш “ Ё 70 = = ст = 20. +64

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374 - RATIO, OR PROPORTION. Thus we have an equality of ratios in comparing the weights. Again, if the weight P and the article A be permanent in. weight, and the distances 4 C, B C be made to vary, then if there be an equilibrium when AC is 2 and BC is 40, we shall find that when— 8 ы 8 LEN 16Х2:1-6; ВС will equal ig X 40=32, ! 6 ACE o x2 = 1:2; BC will equal -5 х Ag mc и EM USE Б га Or A = io X 2 = 0:8; BC will equal — x 40 =16 ; and so on for other proportions, and in every case we shall " ^ have the ratio TE = 203 thus— Bi er AC 8 злым I» ENS XM EY тыт BC 16 — те = vg ^ 20: producing thus an equality of ratios in comparing the arms of the lever. From these experiments we have found, in comparing the article weighed with an arm of the lever, the constant ratio ВС: №, апа when comparing the weights we have found the constant ratio Р: R. Again, in com- paring the arms of the lever, we find the constant ratio AC: BC. Putting two of these couples in proportion, we have— лт. : DECRE Hence (Art. 373) we have— AC x@= BC х Р.

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PRINCIPLE OF THE LEVER DEMONSTRATED. 375 Dividing both members by А C, we have— ах P : "Ne In a steelyard the short arm, А C, and the weight, or poise, P, are unvarying ; therefore we have— P R= BCx жы; Ld P or, when AG is constant, we have— Ж:%ФС 377.—The Lever Principle Demonstrated.—The rela- tion between the weights and their arms of leverage may be demonstrated as follows : * Let A B G H, Fig.273, represent a beam of homogeneous material, of equal sectional area throughout, and suspended upon an axle or pin at C, its centre. This beam is evidently in a state of equilibrium. Of the part of the beam A DG K, let Е be the centre of gravity ; and of the remaining part, D. B КН, let F be the centre of gravity. If the weight of the material in A D G K be concentrated at Е, its centre of gravity, and the weight of the material in * The principle upon which this demonstration is based may be found in an article written by the author and published in the Mathematical Monthly, Cam- bridge, U. S., forer858, p. 77.

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376 RATIO, OR PROPORTION. DBKH be concentrated in F, its centre of gravity, the state of equilibrium will not be interfered with. "Therefore let the ball R be equal in weight to the part A D G K, and the ball P equal to the weight of the part D BK A; and let these two balls be connected by the rod EF. Then these two balls and rod, supported at C, will evidently be in a state of equilibrium (the rod Æ F being supposed to be with- out weight). Now, it is proposed to show that A is to P as C F is to CE. This can be proved; for, since Ё equals the area A4 D G K and P equals the area D B КН, therefore R 15 in proportion to A D, as P is to DB (Art. 359); or, taking the halves of these lines, № is in proportion to A % as Р is to EE. Also, F L equals half the length of the beam; for FD is the half of A D, and DZ is the half of DB; thus these two parts (7 D+ DL) equal the half of the two parts (4 D+D В); or, 7 L equals the half of А В; or, we have— Adding these two equations together, we have— AD DEB СО OS а xL юш: 2 f. Now, 97» О Т = 71, and AD+DAS=AS. therefore, y= Thus we have А М = FL. From each of these equals take FM, common to both, then the remainders, A F and МІ, will be equal; therefore, A F = СЕ. We have also Л P = XL. From each of these equals take ML, common to both, and the remainders, YM and LB, wil be equal; therefore, Г В = ЕС. As was above shown— R: AF FP ш "

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TO FIND A FOURTH PROPORTIONAL. ^ Substituting for A F aa. L В their values, as just found, we have— FOSC. ЕС; \ from which we have (Art. 373)— E OB T. EC Thus it 1s demonstrated that the product of one weight into its arm of leverage, is equal to the product of the other weight into its arm of leverage: а proposition which is known as the law of the lever. 378 .—Any One of Four Proportionals may be Found, —Any three of four proportionals being given, the fourth may be found ; for either one of the four factors may be made to stand alone; thus, taking the equation of the last article, if we divide both members by CF (Art. 371), we have— PERC. m mx me xum З 6 In the first member C F, in both numerator and denominator, cancel each other (477. 371), therefore— REC A > so likewise we may obtain— T PN ND a” АС Rx G Fag = Ta

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SECTION VIIL—FRACTIONS. 379.—4A Fraction Defined.—As a fracture is a break or division into parts, so a fraction is literally a piece broken off; a part of the whole. The figures which are generally used to express a frac- tion show what portion of the whole, or of an integer, the fraction is: for example, let the line A В, (Fig. 274), be divided into five equal parts, then the line A C, containing three of those parts, will be three fifths of the whole line 4 В, and % may be expressed by the figures 3 and 5, placed thus, 2, which is known as a fraction and is read, three fifths. The number 5 below the line denotes the number of parts into which an integer ог unit, 4 B, is supposed to be divided; it i aE aes a ШИ | А ВЕ € B Bac. 274. is therefore called the denominator, and expresses th? denom- ination or kind, whether fifths, sixths, ninths, or any number, into which a unit is supposed to be divided. The number 3 above the line, denoting the number of parts contained in the fraction, is termed the xumerator, and expresses the number of parts taken, as 2, 3, 4, or any other number. 380.— Graphical Representation of Fractions: Effect of Multiplication.—In /7g. 275, let the line А В be di- vided into three equal parts; the line C D into six equal parts; the line E F into nine equal parts; the line С H into twelve equal parts, and the line 7 Ж into fifteen equal parts. The lines 4 B, CD, E F, G H, and 7 K, being all of equal length.

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FRACTIONS ILLUSTRATED. 379 Then the parts of these lines, 4 Z, C M, EN, etc., may be expressed respectively by the fractions, 2, Ч 4 and а In each case the figure below the line, as, 3, 6, 9, 12, or 15, expresses the number of parts into which the whole is di- vided, and the figure above the line, as 1, 2, 3, 4, or 5, the Fic, 275. number of the parts taken; and, as the lines A L, C M, EN, etc., are all equal to each other, therefore these fractions are all equal to each other. If the numerator and denominator of the first fraction be each multiplied by 2, the products will equal the numerator and denominator of the second fraction; thus— | зле 3 X — 2. so, also, З 375 S and hee BE ah 3*4 -— 12 Я Іх = 5 d = e a же 16 Thus it is shown that when the numerator and denomi- nator of a fraction are each multiplied by the same factor, the product forms a new fraction which is of equal value with the original. Е 23 4 5 n | In like manner we have, =~, =, —, --, etc., each equal to 8 12 16 20 one fourth; and which may be found by multiplying’ the à I ; numerator and denominator of = successively by 2, 3, 4, 5, etc.

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380 FRACTIONS. -. 381.— Form of Fraction Changed by Division.—By an operation the reverse of that in the last article, we may re- duce several equal fractions to оле of equal value. Thus, if in each we divide the numerator and denominator by the same number, we reduce it to a fraction of equal value, but with smaller factors. | | For example, taking the fractions of the last article, 2, 4 45» 455, let each be divided by a number which will divide both numerator and denominator without a remainder.* Thus, 2+2=1 3—3 =1 жале Se 9—3 = 3 A4t4-i 555-і 124-4 7 D5-59.$ As these fractions are shown (477. 380) to be equal, and as the operation of dividing each factor by a common num- ber produces quotients which in each case form the same fraction, 4, we therefore conclude that the numerator and denominator of a fraction may be divided by a common number without changing the value of the fraction. 382.— improper Fractions.—The fractions $, 1, 2, etc., all fractions which have the numerator larger than the de- nominator are termed proper fractions. They are not im- proper arithraetically, but they are so named because it isan improper use of language to call that a фаг? which is greater than the whole. | As expressions of this kind, however, are subject to the same rules as those which are fractions proper, it is custom- ағу to include them all under the technical term of fractions. Expressions like these—all expressions in which one number is separated by a horizontal line from another number below it, or one set of numbers is thus separated from another set below it—may be called fractions, and are always to be un- derstood as indicating division, or that the quantity above the line is to be divided by the quantity below the line. * Division is indicated by this sign ~~, which is read '' divided by.”

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IMPROPER FRACTIONS. 381 9 17 24 3x8x4 17x82 ! Thus, Жарасады сты etc., are all fractions, tech nically, although each may be greater than unity. And it is understood in each case that the operation of division is re- 2 8 T quired. Thus, = “г , ET = 83 po 4. When the divis- 7” 3 2.х I2 LY ion cannot be made without a remainder, then the fraction, by cutting the numerator into two, may be separated into two parts, one of which may be exactly divided, and the other will be a fraction proper. Thus, the fraction id is equal to 15 ts ? (for I5 +2 = 17); and since = equals 3, therefore, 5 Y MEN = ке * So, likewise, the fraction 79805 1:504 1355 ло. ж 18 — NN 125 125 125 1425 125 125 383.— Reduction of Mixed Numbers to Fractions.—By an operation the reverse ot that in the last article, a given mixed number (a whole number and fraction) can be put into the form of an improper fraction. This is done by multiplying the whole number by the de- nominator of the fraction, the product being the numerator of a fraction equal in value to the whole number ; the de- nominator of this fraction being the same as that of the given. fraction. The numerator of this fraction being added to the ‘numerator of the given fraction, the sum will be the numera- tor of the required improper fraction, the denominator of which is the sameas that of the givenfraction. Forexample, the required numerator for— 21.45 2L So = 24,18 2х4 T — 9. d jur 3%, is 3x 5+2 = 17. So 3$ ро. 384. 5 120. 5 7%. tion.—Factors placed in the form of a fraction as >

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382 FRACTIONS. Уолт” X S a indicate division (4477. 382); the denominator (the fac- tor below the line) being the divisor, and the numerator (the factor above the line) the dividend, while the value of : : : : 2 the fraction is the quotient. Thus of the fraction, E e 2 41 15 the divisor, 820 the dividend, and 20 the quotient. From this we learn that division may always be indicated by placing the factors in the form of a fraction, so that the divisor shall form the denominator and the dividend the nu- merator. . | 385.— Addition of Fractions having Like Denomina- Р А : I 2 tors.—Let it be required to add the fractions z and A By referring to Art. 379 we see that A4 D (Fzg. 274), 1s one of the five parts into which the whole line A B is divided; it is, therefore, = We also see that D C contains two of the five parts; it is, therefore, С. Wealsoseethat 4 D--DC-— AC, which contains three of the five parts, or 4 C — Н of A B. UN ; — We therefore conclude that 24.5 = 3, In this operation it Б мү is seen that the denominator is not changed, and that the resultant fraction has for a numerator a number equal to the sum of the numerators of the fractions which were required : to be added. Ву thisit is shown that to add fractions we simply take the sum of the numerators for the new numerator, making the denominator of the resultant fraction the same as that of the fractions to be added. For example: What is the sum of the fractions 2 and | Here we have 1+3+4=8 for the numerator, therefore—

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SUBTRACTION AND ADDITION OF FRACTIONS. 383 386.—Subtraction of Fractions of Like Denominators.— Subtraction is the reverse of addition; therefore, to sub- tract fractions a reverse operation is required to that had in the process of addition; or simply to subtract instead of adding. For example, if 2 be required to b ‘ted if , =, 1 if p 5 4 о be sup 3 0169 ТР we have— QS onanii \ (UNIVERSITY I m т \ \£4 LIFOR NE | By reference to Fig. 274 ап exemplification this-will be 3 Un | t9 seen where we have 4 С = 3 Ep = = ane! с = = and 5 5 5 we have— fel — A A= ЕС СО 2m еу We therefore have this rule for the substraction of frac- tions: Subtract the less from the greater numerator ; the remain- der will be the numerator of the required fraction. The denom- znator to be the same as that of the given fractions. 387.——Dissimilar Denominators Equalized.— The rules just given for the addition and subtraction of fractions re- quire that the given fractions have like denominators. When the denominators are unlike it 1s required, before add- ing or substracting, that the fractions be modified so as to make the denominators equal. For example: Let it be re- quired to find the sum of ; and 5 By reference to Fm 6 ; 275, we find that Son line 4 B is equal to E on line И 6 These being equal, we may therefore substitute a for - ж "Then we have—

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384 FRACTIONS. 9 Now, it will be seen that the fraction С may be had by mul- —Á numerator and denominator of the given. Їтас- х з= 6 he es e we have seen (477. 380) that this operation does not change the value of the fraction. From this we learn that the denominators may be made equal by multiplying the smaller denominator and its numerator by әр number which will effect such a result. tion = z By 3, fo I For example: == = 9,“ =; 2 and La = ae es Sa ы ыс) “же” а p = I UT Въ аф. + ЕС. 12 апа 4 12 "16 No ИТЕ 6, 4 In this example the second fraction is changed by multiply- ing by 1% 388.— Reduction of Fractions to their Lowest Terms.— The process resorted to in the last article to equalize the denominators, is not always successful. What is needed for а common denominator is to find the smallest number which shall be divisible by each of the given denominators. Before seeking this number, let each given fraction be reduced to its lowest terms, by dividing each factor by a common number. For example: i may, by dividing by 5, be reduced to е which is its equivalent. So, also, = by di- viding by 7,15 reduced to 2 its lowest terms. 389.—Least Common Denominator.— 70 find the least common denominator place the several fractions in the order of their dietis ее: increasing toward the right. If the largest denominator be not divisible by each of the others, double it; if the division cannot now be performed, treble

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LEAST COMMON DENOMINATOR. 385 it, and so proceed until it is multiplied by some number which will make it divisible by each of the other denomina- tors. This number multiplied by the largest denominator «vill be the least common denominator. То raise the denominator of each fraction to this, divide the common denominator by the de- nominator of one of the fractions, the quotient will be the number by which that fraction is to be multiplied, both numerator and denominator, and so proceed with each frac- tion. For example: What is the sum of the fractions п d One of these, =, may be reduced, by divid- ing by 2, to 2. Therefore, the series is = * 2, ко 4, and ME and for the fourth, 24--8 = 3, and 593 = 21 x —~-. Thus the fractions in their reduced form аге: ло йы dO ES Ры 24 — DIAS UL Fae. 24° Ж О WO ба 24 390.—Least Common Denominator Again.— When the denominators are not divisible by one another, then to ob- tain a common denominator, it is requisite to multiply to- gether all of the denominators which will uot divide any of the other denominators. For example: What is the sum of the l г 3 3 4 fract —, —, ~, and -? ractions drain n 7

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386 FRACTIONS. In this case the first denominator will divide the last, but the others are prime to each other. Therefore, for the common denominator, multiply, together all but the first ; or— 5 x 7x9 = 315 the common denominator ; and— | 315 + 3 = 105, common factor for the first fraction; 315 +5 = 63, common factor for the second fraction; 318 = 7 = 45, common factor for the third; 315-+9 = 35, common factor for the fourth. And, then— 105 = 105, 2x 63 =126 3x 45 = 135 4x 35 «BE: 5 02 «S15. 77 5 45 x 915 IO Fo T ы о IOS 126 135 140 506 |19! =s c ДЕ === кыы сш 4 Mm. 315 315 315 315 315 315 391.— Fractions Multiplied Graphically.—Let 4 BCD (Fig. 276) be a rectangle of equal sides, or 4 B equal 4 С and each equal one foot. Then А B multiplied by 4 C will A B Fic. 276. equal the area A BCD, or 1 x = 1 square foot. Let the line Е F be parallel with А В, апа midway between 4 В and CD. Then АВ х AE equals half the areaof A 5 CD, or 1x4= 4. Again; let С H be parallel with Æ C, and mid- way between Е Cand FD. Then EC x EC = $ x $ equals the area E С СН, which is equal to a quarter of the area

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MULTIPLICATION OF FRACTIONS. 397 ABCD;orkix$-i; which is a quarter of the superficial area. | - The product here obtained is less than either of the factors producing it. It.must be remembered, however, that while the factors represent Zizes, the product represents superficial area. The correctness of the result may be recognized by an inspection of the diagram. | 392.—Fractions Multiplied Graphically. — In Fig. 277 ` let А В equal 8 feet and А C equal 5 feet; then the rect- PIG. 277 angle A BCD contains 5 х8 = 40 feet. The interior lines divide the space included within A BCD into до equal Squares of опе foot each. Let A Е equal 3 feet or = di 4 С. Let А С equal 7 feet or а Of АВ. Then the rectangle Е ҒА С contains ix 7 = + or twenty-one fortieths of the whole area A B CD. Thus, while the factor fractions i and 1 A | Ne 5. represent Zizes, it is shown that the product fraction РТ] гер- resents surface. Thus E is а fraction, Е FA G, of the whole surface, CD A B. 393.— Rule for Mutiplication of Fractions, and Exam- ple.—In the example given in the last article it will be ob-

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388 FRACTIONS. served that the product of the denominators of the two given fractions equals the area of the whole figure (А В C D), while the product of the.numerators equals the area of the rectangle (Æ F A G), the sides of which are equal respec- tively to the given fractions. From this we obtain for the product of fractions this— RULE.—Multiply together the denominators for the new de- ` nominator, and the numerators for a new numerator. > For example: what is the product of = and 2; Неге we have 20x21 =420 for the new denominator, апа 7X13 = 91 for the new numerator; therefore the product of— A DE ср 214 29 420” or, of a rectangular area divided one way into 20 parts апа the other way into 21 parts, thus containing 420 rectangles, I l the product of the two fractions E and £ is equal to 91 of these rectangles, or 2". of the whole. 420 394.—Fractions Divided Graphically.— Division is the reverse of multiplication; or, while multiplication requires the product of two given factors, division requires one of the factors when the other and the product are'given. Or (referring to Fig. 277) in division we have the area of the rectangle, E FA G, and one side, Е A; given, to find the other side, 4 G. Now itis required to:find the number of times E A is contained in E FAG. By inspection of the figure we per- ceive the answer to be, A С times; for EA x A4G— EFA С, the given area. Or, when E A FG is given as £ апа £ A as 5, we have as the given problem—

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DIVISION OF FRACTIONS. 389 Since division is the reverse of multiplication, instead of multiplying we divide the factors, and have— o 7 4025 8 Thus, to divide one fraction by another, for the numerator of the required factor, divide the numerator of the product by the numerator of the given factor, and for the denominator of the required factor divide the denominator of the product by the denominator of the given factor. For example: Divide = by 2 Answer, 2. Піуіде a by aa Answer, а с 9 3 395.-Еше for Division of Fractions.—The rule just given does not work well when the factors are not commen- 5 surable. For example, if it be required to divide 7 by b we have by the above rule— Producing fractional numerators and denominators for the resulting fraction, which require modification in order to reach those composed only of whole numbers. ІҒ the nu- merators, 5 and 7, of this compound fraction be multiplied by 9 (the denominator of the denominator fraction), or the compound fraction by 9, we shall have— % 5х9 2 "тек E 750 9 9

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390 FRACTIONS. And, if these be again multiplied by 2 (the denominator of the numerator fraction), we shall have— 5х9 5хох2 2 Mx 2 = ——€— 7X9 7х9х 2 9 9 Like figures above and below in each fraction cancel each other (477. 371), therefore, the result reduces to— 5x9 y o in which we find the factors of the two original fractions. In one fraction 2 we have the factors in position as given, ) = кк "қ, but in the other © they are inverted. The fraction in which. the factors are inverted 1s the divisor. Hence, for division of fractions, we have this— RULE.—Jnvert the factors of the divisor, and then, as in multiplication, multiply the numerators together for the numera- tor of the required fraction, and the denominators for the de- nominator of the required fraction. Thus, as before, if = is required to be divided by о үге һауе-- 29 — 5 JRZ 14. And, to divide eh by L we have—. 47 E) 23x9 207 4x7 329 Again, to divide E by 5, we have— 25x04 7254 5 Ё

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CANCELLING IN ALGEBRA. 301 This last example has two factors, 9 and 45, опе of which „у л ү measures the other ; also, the first fraction 25 is not 1n its lowest terms; when reduced it is T The question, there- i fone, may be stated thus: for the two g’s cancel each other.

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SECTION IX.—ALGEBRA. 396.—Algebra Defined.—It occurs sometimes that a student familiar only with computation by numerals is needlessly puzzled, in approaching the subject of Algebra, to comprehend how it 15 possible to multiply Zetters together, or to divide them. То remove this difficulty, it may be suf- ficient for them to learn that their perplexity arises from a misunderstanding in supposing the /e¢ters themselves аге ever multiplied or divided. It is true that in treatises on the subject it is usual to speak as though these operations were actually performed upon the letters. It is always un- derstood, however, that it is not the letters, but the guan- tities represented by the letters, which are to be multiplied or divided. For example, іп 477. 361 it is shown, in comparing similar sides of homologous triangles, that the bases of the two tri- angles are to each other as the corresponding sides, or, ШО огге to Гг. 260, we have CE: E:: DE: ВЕ. Now, let the two bases C E and A £ be represented respec- tively by а and 0, and the two corresponding sides 2 Ж and B E by c and а respectively; or, for— | CET ЛИЕ. : De гав 5, риф = qq c 0b and, by Art. 373, we have— | bx ¢ =a хб) which may be written— рс ak; for x, the sign for multiplication, is not needed between let- ters, as it is between numeral factors. The operation of

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APPLICATION OF ALGEBRA. 393 multiplication is always understood when letters are placed side by side. Now, here we have an equation in which, as usually read, we have (һе product of 4 and c equal to the product of a and а. But the meaning is that the product of the quantities represented Бу д and с is equal to the product of the guan- tities represented by а and 4, and that this equation is in- tended to represent the relation subsisting between the four proportionals, C Æ, А E, D E, and P Е, of Fig. 269. In order to secure greater conciseness and clearness, the four small letters are: substituted for the four pair of capital letters, which are used to indicate the lines of the figures referred to. 397.— Example: Application.—It was shown in the last article that the four letters а, 0, c, and d represent the cor- responding sides of the two triangles of Fzg..269, and that— besad. Now, let each member of this equation be divided by a, then (Art. 371)— — = d. If now the dimensions of the three sides represented by a, б, and с are known, and it is required to ascertain from these the length of the side represented by 4, let the three given dimensions be severally substituted іог the letters герге. senting them. For example, let a= 40 feet; 6 = 52 feet, and c= 45 feet; then— óc 52х45 2340 am c. с d= —55-£ (сес. The quantities being here substituted ‘or tne letters ; we have but to perform the arithmetical processes indicated to obtain the arithmetical value of d. Frem this example it is seen that before any practical use can be made of an algebraical formula in computing dimensions, it is requisite to substitute numerals for the letters and actually perform arithmetically such operations as are only indicated by the letters.

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394 | ALGEBRA. 398.—Algebra Useful in Constructing Rules.—lIn all problems to be solved there are certain conditions or quan- tities given, by means of which an unknown quantity is to be evolved. For example, in the problem in 2. 397, there were three certain lines given'to find a fourth, based upon the condition that the four lines were four proportionals. Now, it has been found that the relation between quantities and the conditions of a question can better be stated by let- ters than by numerals; and it is the office of algebra to present by letters a concise statement of a question, and by certain processes of comparison, substitution and elimina- tion, to condense the statement to its smallest compass, and at last to present it in a formula or rule, which exhibits the known quantities on one side as equal to the unknown on the other side. Here algebra ends, at the completion of the rule. To ие the rule is the office of arithmetic. For, in using the rule, each quantity in numerals must be substi- tuted for the letter representing it, and the arithmetical processes indicated performed, as was done in Art. 397. 399.—Algebraic Rules are General.—One advantage derived from algebra is that the rules made are general in their application, For example, the rule of Art. 397, - — d,is applicable to all cases of homologous triangles, however they may differ in size or shape from those given in Fig. 269—and not only this, but it is also applicable in all cases where four quantities are in proportion so as to con- stitute four proportionals. For example, the case of the four proportionals constituting the arms of a lever and the weights attached (Arts. 375-378). For, taking the rela- tion as expressed in 477. 377— БАС ue Сх ЫС, we may substitute for C F the letter л, and for Æ C the letter m, then m will represent the arm of the lever Е С (21е. 262), and z “ав arm of the lever FC. Then we have— Pn = Кї,

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SYMBOLS CHOSEN AT PLEASURE. 395 and from this, dividing by z (477. 372), we have— mo HER m (110.) or, dividing by m, we have— R= =. (1ай s which is a rule for computing the weight of R, when P and the two arms of leverage, m and л, аге known. For example, let the weight represented by P be 1200 pounds, the length of the arm 7 be 4 feet, and that of z be 8 feet, then we have— Ет ж 1200х8 | " => = Р == 2400 pounds. Pn , ae | This rule, R=, is precisely like that in 477. 397— bc ; ге din which three quantities are given to find a fourth, the four constituting a set of four proportionals. 400.—Symbols Chosen at Pleasure.—The particular letter assigned to represent a particular quantity is a matter of no consequence. Any letter at will may be taken; but when taken, it must be firmly adhered to to represent that par- ticular quantity, throughout all the modifications which may be requisite in condensing the statement into which it enters into a formula for use. For example, the two rules named in Art. 399 are precisely alike—three quantities given to find a fourth—yct they are represented by different letters. In опе, Rand P represent the two weights, and # and л the arms of leverage at which they act; while in the other the letters а, b,c, and d represent severally the four lines which constitute two similar sides of two homologous triangles. The two rules are alike in working, and they might have been con- stituted with the same letters. And instead of the letters chosen any others might have been taken, which con- venience or mere caprice might have dictated. Іп some 4 vom

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396 ALGEBRA. questions it is usual to put the first letters, as a, 4, c, etc., to. represent known quantities, and the last letters, as x, y, z, for the quantities sought. In works on the strength of materials it is customary to represent weights by capital letters, as P, №, U, W, etc., and lines or linear dimensions by the small letters, as 2, d, 7, for the breadth, depth, and length, respectiv ely, of a beam. Any other letters may be put to represent these quantities, although the initial letter of the word serves to assist the memory in recognizing the partic- ular dimensions intended. 40l.— Arithmetical Processes Indicated by Signs,—In algebra, the four processes of addition, subtraction, multi- plication, and division, are frequently required; and when the required process cannot be actually performed upon the letters themselves, a certain method has been adopted by which the process is indicated. For example, in additon, when it is required to add а to 4, the two letters cannot be intermingled as numerals may be, and their sum presented ; but the process of addition is simply indicated by placing between the two letters this sign, +, which is called plus, meaning added to; therefore, to add a to 6 we have— a+b, which is read а plus 2, or the sum of а and 2. When the quantities represented by а and 6 are substituted for them— and not till tHen—they can be condensed into one sum. For example, let а equal 4 and 0 equal 3, then for— a+o we have— 4+ 3; and we may at once write their sum 7, instead of 4+ 3. So, likewise, in the process of subtraction, one letter can- not be taken from another letter so as to show how much of this other letter there will be left as a remainder; but the process of subtraction can be indicated by a sign, as this, —, which is called minus, less, meaning subtracted from. For

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ALGEBRAICAL SIGNS. 397 example, let it be required to subtract ё from а. To do this we have— | а-- 6; which is read а minus 0, апа when the values of a and баге substituted for them, we һауе, when a equals 4; and 6 equals 3— a— 6, or— dn and now, instead of 4 — 3, we may put the value of the two, which is unity, or 1. The algebraic signs most frequently used are as follows: +, plus, signifies addition, and that the two quantities be- tween which it stands are to be added together; as а + ё, read а added to 0. —, minus, signifies subtraction, or that of the two quantities between which it occurs, the latter is to be subtracted from the former; as a — д, read a minus 2. . X, multiplied бу, or the sign of multiplication. It denotes that the two quantities between which it occurs are to be multiplied together; as a x 4, read a multiplied by å, “or a times №. This sign is usually omitted between symbols or letters, and is then understood, as a6. This has the same meaning as сх ё. It is never omitted between arithmetical numbers; as 9x 5, read nine times five. --, divided бу, or the sign of division, and denotes that of the two quantities between which it occurs, the former is to be divided by the latter; as а--д, read а divided by б. Division is also represented thus: Z in the form of a fraction. This signifies that а is to be 4 divided by 2. When more than one symbol occurs anr cm that the product of the symbols above the line is to be divided by the product of those below the line. above or below the line, or both, as ; it депо 4

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398 ALGEBRA. =, 25 equal to, or sign of equality, and denotes that the quantity or quantities on its left are equal to those on its right; as а — 2 = с, read a minus 0 is equal to с, or equals с; ог, 9— 5 = 4, read nine minus five equals four. This sign, together with the symbols on each side of it, when spoken of as a whole, is called an equation. | а? denotes a squared, or a multiplied by а, or the second power of a, and а denotes а cubed, or a multiplied by a and TET multi- plied by а, or the third power of a. The small figure, 2, 3, Or 4, etc., is termed the index or exponent of the power. It indicates how many times the symbol is to betaken. hus, @ =a, d =n ам, a WES y is the radical sign, and denotes that the square root of the quantity following it is to be extracted, and y denotes that the сибе root of the quantity following it is to be extracted. Thus, уо = 3, and 427 = 3. The extraction of roots is also denoted by a fractional in- dex or exponent, thus— ' а% denotes the square root of a, аз denotes the cube root of a, a% denotes the cube root of the square of a, etc. 4-02.— Example in Addition and Subtraction: Cancel- ling.—Let there be some question which requires a state- ment to represent it, like this— 4 4 d = с — Ё, which indicates that if the quantity represented by о. be added to the quantity represented by 4, the sum will be equal to the quantity represented by c, after there has been subtracted from it the quantity represented by 0; or, as it is usually read, а plus d equals c minus 4; or the sum of а and d equals the difference between с and б. For illustration, take in place of these four letters, in the order they stand, the numerals 4, 2, 9, 3, and we shall have by substitution — a+d= c — 6, 4+2=9 — 3, or adding and subtracting— 6 96;

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TRANSFERRING SYMBOLS. 399 If it be required to add to each member of the equation the quantity represented by д, this will not interfere with the equality of the members. For a+d are equal to c — d, and if to each of these two equals a common quantity be added, the sums must be equal ; therefore— a+d+b=c— b+, or by numerals— м. | 4+2+3=9— 3+3, or— ! 959 It will be observed that the right hand member contains the quantity — ё and +. This shows that the quantity £ is to be subtracted and then added. Now, if 3 be subtracted from о, the remainder will be 6, and then if 3 be added, the sum will be 9, the original quantity: Thus it is seen that when in the same member of an equation a symbol appears as a minus quantity and also аза plus quantity, the two cancel each other, and may be omitted. Therefore, the expression— . at+d+b=c—b+6 becomes— a+d+b=e. 403.—Transferring a Symbol to the Opposite Member. —In comparing, in the last article, the first equation with the last, it will be seen that the same symbols are contained in each, but differently arranged: ‘that while in the first equa- tion 0 appears in the right hand member and with a minus or negative sign, in the last equation it appears in the left hand member and with a plus or positive sign. Thus it is . seen that in the operation performed 4 has been made to pass from one member to the other, but in its passage it has been changed. A similar change may be made with another of the symbols. For example, from the last equation, let Z be subtracted, or this process indicated, thus— at+d+b—d=c—d.

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400 ALGEBRA. The plus and minus d, in the left hand member cancel each other, therefore— ~~ акб-с--а, “ ог, by numerals— · Ере ore Reducing— 7 = EN By this we learn that алу quantity (connected by + or —) may be passed from one member of the equation to the other, pro- vided the sign be changed. 404.—Signs of Symbols to be Changed when they are to be Subtracted,—As ап example in subtraction, let the quantities represented by + 0 — a — f-- с, be taken from the quantities represented .by +а +02 – с – 7. This may be written— | (5-ағб-с--:У)--(-0--а--/-ке), an expression showing that the quantities enclosed within the second pair of parentheses are to be subtracted from those included within the first pair. Let the quantities represent- ed in the first pair of parentheses for convenience be repre- sented by A, or, a+b —c —f = A. Now, by theterms of the | problem, we are required to subtract from А the quantities enclosed within the second pair of parentheses. To do this take first the positive quantity, 0, and subtract it or indicate the subtraction, thus— А --6; we will then subtract the positive quantity c, or-indicate the subtraction, thus— 7 A — b -— с. ` We have yet to subtract — а and — f, two negative quanti- ties. | The method by which this can be accomplished may be discovered by considering the requirements of the problem. The plus quantities 0 and c, before being subtracted from 4, were required to have the two negative quantities а and f de-

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THE OPERATION TESTED. 401 ducted from them. It is evident, therefore, that in subtract- ing ? and с, before this deduction was made, too much has been taken from А, and that the excess taken is equal to the sum of а and f. То correct the error, therefore, it is neces- sary to add just the amount of the excess, or to add the sum of a and f, or annex them by the plus sign, thus— A—b—ct+atf. To test the correctness of the operation as here performed, let numerals be substituted for the symbols; let a= 2,0 = 3, c— 1, f= $; then the given quantities to be subtracted,— (+d—a—f+o), become— (+ og $+ 1), (4 =з 2%) zd Thus the quantity to be substracted equals 13. Applying the numerals to the above expression— which reduces to— A—b+a+f—c becomes-- A—3+24+4—-1=A—4+24=A—14. A correct result; itis the same as before. Restoring now the symbols represented by 4, we have for the whole ex- pression— +at+b—c—f—b+a+f—e, which, by cancelling (427. доз) апа by adding like symbols with like signs, reduces to— - ШҚ o 2 6; To test this result, let the quantity which was represented by А have the proper numerals substituted, thus : +at+tb—c—f, ғ253-І-4ӛл%- 18 = 34.

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402 ALGEBRA. Гһе sum of the given quantity required to be subtracted was before found to amount to 12, therefore— — I$ becomes— 34 — 1$ = 2. And the result by the symbols as above was— 2a-—2c, which becomes— Qu 122—201, or— i a ә; a result the same as before, proving the work correct. An examination of the signs in the above expression, which de- notes the problem performed, will show that the sign of each symbol which was required to be subtracted has been changed in the operation of subtraction. Before subtract- ing they were— (46—a—ft+e); after subtraction they are— (— d+a+f—c). By this result we learn, that zo subtract a quantity we have but to change its sign and annex it to the quantity from which it was required to be subtracted. Example: Subtract a — 6 fromc +d. Answer, "7 a+. If numerals be substituted, say a=7,6=4,c= 5, and d — 9, then— | c+d becomes 5+9= 14, а= Ф E 7--4-- 3, c+d — (a — b) = 14 — 3-11. So, also, — À Сте = ао becomes— 5+90-—7/+4ж 11.

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FRACTIONS ADDED AND SUBTRACTED. 403 405.—Algebraic Fractions: Added and Subtracted.— When algebraic fractions of like denominators аге to be% added or subtracted, the same rules (42775. 385 and 386) are to be observed as in the addition or subtraction of numeri- cal fractions—namely, add or subtract the numerators for a new numerator, and place beneath the sum or difference the common denominator. ? ~A For example, what is the sum of 4 % For this we have— а-с-ғ-а Жо. à б Subtract 5 from 7 For this we have— б--с J What is the algebraical sum of— p c n r = 2s — p For these we have— ф+с—п—7” 4 To exemplify this;let 0 represent greo = 9, 22 WEE and d — 12. Then, for the algebraic sum, we have— 038—239... 12 — =I 12 Do Now, taking the positive and negative fractions sep- arately, we have— Lm hg sae 12” and— mA a — 5

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404 м | ALGEBRA Together— - 17 — 5 12 as before. ~e 406.—The Least Common Denominator.—When the denominators of algebraic fractions differ it is necessary be- fore addition or subtraction can be performed to harmonize them, as in the reduction of the denominators of numerical fractions са 388-390) For example, add together the 4 jS buc that they collectively contain the letters а, 0 and с, and по others. It will be requisite, therefore, that each of the frac- tions be modified so that its denominator shall have these three factors. То effect this it will be seen that it 15 neces- sary to multiply each fraction by that one of these letters which is lacking in its denominator. Thus, in the first, а is lacking, therefore (477. 380) a s E] T In the second a a fractions D. In these denominators we perceive and с are lacking, therefore -- хас а and in the third бхасшадбс 7х? = Т? жа EE Placing them now 6 is lacking, therefore together we have— аа+асе+ ёт а е * FS dH + —, ave. be m ПФ The factor а а may be represented thus 2”, which means that а occurs twice, the small figure at the top indicating the number of times the letter occurs; 2° is called а squared, a a а == а”, and is called а cubed. - In order to show that the above fraction, resulting as the sum of the three given fractions, is correct, let a = 2,0 = 3, ¢=4,¢=5,and7=6. Then the three given fractions аге қ. 4 Тік... 5 l ec + 9 477A © 5

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FRACTIONS SUBTRACTED. 405 In equalizing these denominators we multiply the second fraction by 2, and the third by 14, which will give— 5x2=10 gxib dd Хы 6) дах 6? then— + Жын р _ 7 В em у cee? и = заа f Now the sum of the fractions is— a*+ace+br abe 4 - 5 же ана 1 4! 28x 3 X4 Or, з 18, ODay pel te а) TAN 24 24 24 ull the same result as before, thus showing that the reduction was rightly made. 4.07.—Algebraic Fractions Subtracted.—To exemplify the subtraction of fractions, let it be required to find the ; б : algebraic sum of - pem 5 These denominators all dif- fer. Тһе fractions, therefore, require to be niodified, so that each denominator shall contain them all. To accom- plish this, the first fraction will need to be thus treated : axdf-adf. оох бн са f’ the second— de me AE | dxcf= cdf’ the third— i ехса- cde ~ fxed= саў The sum of these is— adf—bcf—cde саў 1

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406 ALGEBRA. That this is a correct answer, let the result be proved by hgures; thus, for a put 15; 0, 25 285; - ЖАТАР Г we shall have— — e --- .. --- == л ым-- .-- --- ` It will be observed that these denominators may be equal- ized by multiplying the first fraction by 2, and the second by 14, therefore we have— To make the required subtraction we are to deduct from 30 (the numerator of the positive fraction), first 5, then 5 ; or, the sum of the numerators of the negative fractions; or for the numerator of the new fraction we have 30 — 8 — 22. The required result, therefore, is— 22:0 ТІ To apply this test to the algebraic sum we have— adf —bcf —cde _ KEZELE FEFELE FETE DU, T "Oe aom which by multiplication reduces to— збо — 36 — бо 264 22 II a result the same as before, proving the work шай An- other example: a P. с | From -- — — take —, — and —; 2 n Wt М 24 A -o — oie — == --- — 2... --

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DENOMINATORS HARMONIZED. 407 The fractions which have the same denominator may be grouped together thus: oss * — ль <--е — | — m i$ ÁÀÓÀÀÀÀ— ———= and— То harmonize these two denominators, m and м, the first fraction must be multiplied by ж and the last by л, or— m (a -- с — е) Қ n(b—d) mí(a—c—^e)tn(b—d) mn mn — mn d In the polynomialfactor within the parentheses (a — c — e) we have the positive quantity a, from which is to be taken the two negatives c and e, or their sum is to be taken from а, or (а — (с+е)). With this modification we have for the alge- braic sum of the five given fractions— m (a — (c--e))--n(b—d) mn ` ў To test the accuracy of this result, let the value of the sev- eral letters respectively be as follows: а = 11,6 —9,c— 3, d = 4, е = 5, m = 10, and я = 8. Then the sum is— то (11 —(3+5)) +8(9—4) _70_7 ох 8 80 8” Now, taking the fractions separately, we have— NC 2 = (24 ==, 5-21 n mn - 8 78 ні 8? 2. ваа. aa " m 10 i5 A a" or, together we have, as the sum of these two results— 5 е 3 g T IO

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408 ` ALGEBRA. To harmonize these denominators we may multiply the first fraction by 5, and the second by 4, thus: — LM — . Дохи дол. 40. 8’ the same result as before, thus the accuracy of the work 15 established. 4.08,— Graphical Representation of Multiplication.— In Fig. 278, let A BC D, a rectangle, have its sides 4 Б and Fic. 278. A C divided into equal parts. Then the area of the figure will be obtained by multiplying one side by the other, or putting а for the side A В, and 6 for the side А С, then the area will be ох 2, orab. This will be the correct area of the figure, whatever the length of the sides may be. If, as shown, the area be divided into 4 x 7 = 28 equal rectangles, then а would equal 7, and 0 equal 4, and ad = 7 хд = 28, the area, “li AS equal 28 and AC equal 16, then will a = 28, and"? = 16, апа ex = 29 Тб = 449 mtliegüra. . 409.— Graphical Multiplication: Three Factors.—Let ABCDEFG (fig. 279) represent. a rectangular solid which may be supposed divided into numerous small cubes as shown. Now,if a be put for the edge A В, 0 for the edge A C, and с for the edge C D, then the cubical solidity of the

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MULTIPLICATION OF A BINOMIAL. 409. whole figure will be represented bye xóxc-aóbc. Ifthe edge A В measures 6, the edge А C 3, and the edge C D 4, then a6c=6x3x4=72= the cubic contents of the figure, or the number of small cubes contained in it. ` FIG. 279. - 410.— Graphic Representation: Two and Three Fac- tors.—/725. 278 and 279 serve to illustrate the algebraic ex- pressions аф and абс. In the former it is shown that the multiplication of two lines produces a rectangular suríace, or that if а and 0 represent lines, then ад may represent a rectangular surface (2760. 278) having sides respectively equal to а апа 0. Апа so if a, 0, and c represent three sev- eral lines, then абс may represent a rectangular solid (Fig. 279) having edges respectively equal to а, 0, and с. Fic. 280. 411. — Graphical Multiplication of a Binomial, — Let A B C D (Fig. 280) be a rectangular surface, and ВЕ D F an- other rectangular surface, adjoining the first. The area of the whole figure is evidently equal to— (AB BE)x4A C.

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410 ALGEBRA. The area is also equal to— ZU AVC В E хар; or, since A C = B D, the area equals— | ра C Bie с; or, if symbols be put to represent the lines, say a for А В, б for B E, and с for А C, then the two representatives of the area, as above shown, become: The first— (24.0) с zen t and the last— (gx д 4. (xc) = area. Hence we have— (a+ ё) с=ас+ё с. This result exemplifies the algebraic multiplication of a bi- nomial, which is performed thus: Let a + 2 be multiplied by c. The problem is stated thus: (a + Ру. To perform the multiplication indicated we proceed thus: | 2 WT с ackbc multiplying each of the factors of the multiplicand sepa- rately and annexing them by the sign foraddition. Putting the two together, or showing the problem and its answer in an equation, we have— (а-біс--ас--бе, producing the same result, above shown, as derived from the graphic representation. 412.— Graphical Squaring of a Binomial.—Let EGC 9 (Fig. 281) be a rectangle of equal sides, and within it draw

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SQUARING OF A BINOMIAL. 41I the two lines, A H and F D, parallel with the lines of the rectangle, and at such a distance from them that the sides, АВ and P D, of the rectangle, A B C D, shall be of equal length. We then have in this figure the three squares, EGCF, ABCD, and FG B Н, also the two equal rect- angles, EF A B and BHD 9. Let Е F be represented by а and FG by 0, then the area of ABCD will be axa=a’; the area of FG B H will be охо = 6°; the area of E FA B will be a xb = a 0, апа” thas Fic. 281. of BHD 9 will be the same. Putting these areas together thus— a*+2ab+06’, the sum equals the area of the whole figure—equals the prod- uct of EG x E C—equals the product— (a+) x (а +8). So, therefore, we have— (a+b) (a+b) = а 4 2ab-6*; (112.) or, in general, the square of a binomial equals the square of the first, plus twice the first бу the second, plus the square of the second. This result is obtained graphically. The same re- sult may be obtained by algebraic multiplication, combining

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412 ALGEBRA. each factor of the multiplier with each factor of the multi- plicand and adding the products, thus— а+ ё aid a*+ab ab+* a + 2.020 G 5. The same result as above shown by graphical representa- tion. 413.—Graphical Squaring of the Difference of Two Factors.—Let the line E С (Fig. 281) be represented by с, and the line A Æ and A C as before respectively by 4 and а, then— Е С ДЕ А С. c—b=a. From this, squaring both sides, we have— (c — 2) = а", The area of the square 4 P C D тау be obtained thus: From the square Е С C 9 take the rectangle E G x Е А and the rectangle FG x D 7, minus the. square FG B H, or from с? take the rectangle с 0, and the rectangle c 0, minus the square, 0%, and the remainder will be the square, 2°; or, in proper form— с cb —chbrb'-a' In deducting from с? the rectangle c? twice, we have taken away the small square twice; therefore, to correct this error, we have to add the small square, or ?*, Then, when reduced, the expression becomes— с — 220 +b’ = а" = (с — by. Thisresult 15 obtained graphically. Тһе result by algebraic

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PRODUCT OF THE SUM AND DIFFERENCE. 413 process will now be sought. : The square of a quantity. may be obtained by multiplying the quantity by itself, or— - c — b c— b с? — бс Е 5 (c — bP = с — 20с+ 0". (113.) In this process, as before, each factor of the multiplier is combined with each factor of the multiplicand and the sev- eral products annexed with their proper signs (Art. 415), and thus, by algebraic process, a result is obtained precisely like that obtained graphically. This result 15 the square of the difference of c and 0; and since с and д may represent any quantities whatever, we have this general— RULE.— The square of the difference of two quantities ts equal to the sum of the squares of the two quantities, minus twice their product. Fic. 282. 4(4.—€Graphical Product of the Sum and Difference of Two Quantities.—Let the rectangle 4 B C D (Fig. 282) have its sides each equal toa. Let the line Æ F be parallel with 4 B and at. the distance 0 from it, also, the line FG made parallel with 2 D, and at the distance ? from it. Then the line Е F equals a + 2, and the line Ё С equals а- Ё. Therefore the area of the rectangle E F C С equals n+ 2,

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414 i ALGEBRA. multiplied by a — 2. From the figure, for the area of this rectangle, we have— ABCO-—-MBEH-HmFDG-ZEFCG; or, by substitution of the symbols, а —abab(a— б). Multiply the last quantity T м а--2. б ab — b’ = 0 (а — ё). Substituting this in the above we have— a*—ab+ah—b*=(ard)x(a—D). Two of these like quantities, having contrary signs, cancel each other and disappear, reducing the expression to this— а? — 5? = (a+ b) x (a -“й). The correctness of this result is made manifest by апіпврес- tion of the figure, in which it is seen that the rectangle E FCG is equal to the square 4 5 CD minus the square B FHF. For ABEZ equals B¥DG. Now, if from the square ABCD we take away A BE H, and place it so as to cover BS DG, we shall have the rectangle Е CG plus the square BYHF; showing that the square A В.С, is equal to the rectangle E FCG plus the square BY ҒҒ; or— a* —(a- b) x (a— b) +0". The last quantity may be transferred to the first member of the equation by changing its sign (477. 403). Therefore— а — b? = (a + b) x (a — b), as was before shown.

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MULTIPLICATION—PLUS AND MINUS. 415 The result here obtained is derived from the geometrical figure, or graphically. * Precisely the same result may be obtained algebraically ; thus— а +ò a—b a’*+ab LAC Y (a+ 6) х (2 —– 6) = а?.... — b? (114.) Here the two like quantities, having unlike signs, cancel each other and disappear, leaving as the result du the dif- ference of the squares. The result here obtained is general; hence we have this— RULE.— The product of the sum and difference of two quan- tities equals the difference of their squares. Fic. 283. 415.—Plus and Minus Signs in Muitiplication.—In pre- vious articles the signs in multiplication have been given to products in accordance with this rule, namely: Zīke signs give plus ; unlike signs, minus. This rule may be illustrated graphically, thus: In the rectangular Fig. 283, let it be re- quired to show the area of the rectangle A G CH, in terms of the several parts of the whole figure. Thus the area of AGES equal ABEF—GBF¥F and the area of EFCH equals EFCD — YFH D. And the areas of AGE T ЕСН equals the area of А С CH. Therefore the sum of the two former expressions equals 4 G CH. Thus A BE F— GByYF.--EFCD—SyYEFHDEZJIGCH. Lect the several lines now be represented by algebraic symbols; for example,

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416 ALGEBRA. let daB = E F uude Ca zi Је ФЕ D; and E C = ЎН = d, and let these symbols be substituted for the lines they represent, thus ABE F—GBZF-EFCD-— ЕЗГІ D'E АСТАН” -ас--бс-аа--ба-(а- 6) х (с+ а). An inspection of the figure shows this to be a correct result. It will now be shown that an algebraical multiplica- tion of.the two binomials, allotting the signs in accordance with the rule given, will produce a like result. For example— а= 0 cd ac—bc+ad—bd. 4.(6.—Equality of Squares on Hypothenuse and Sides of Right-Angled Triangle.—The truth of this proposition has been proved geometrically in 2477. 353. It will now be shown graphically and proved algebraically. à Let A BCD (Fig. 284) be a rectangle of equal sides, and BE D the right-angled triangle, the squares upon the sides of which, it is proposed to consider. Extend the side БЕ to F; parallel with BF draw DG, CK, and A. Parallel with Æ D draw А X and LG. These lines produce triangles, AHB, ACS, ALC, CK D, and CGD, each equal to the given triangle BED (Art. 337). Now, if from the square

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SQUARES ON RIGHT-ANGLED TRIANGLE. 417 ABCD wetake A BA and place it at CDG; and if we take BED and place it at AL C we will modify the square ABCD, зо as to produce the figure LG DE HA L, which is made up of two squares, namely, the square DE FG and the square AL ЕМ, and these two squares are evidently equal to the square A BCD. Now, the square D Е FG 15 the square upon Е D, the base of the given right-angled triangle, and the square A LFH is the square upon A H = B Е, the perpendicular of the given right-angled triangle, while the square 4 B C D is the square upon & D, the hypothenuse of the given right-angled triangle. Thus, graphically, it is shown that the square upon the hypothenuse of a right-angled triangle ts equal to the sum of the squares upon the remaining two sides. To show this algebraically, let B Е, the perpendicular of the given right-angled triangle, be represented. by a; & D, the base, by 2, and B D, the hypothenuse, by c., Then it is required to show that— Є qm, Now, since D K = BE = a, therefore, E К = ED-— D K = b — a, and the square £ K 9 H equals (6 — a)’, which (Art. 413) equals 63--2аб ға” This is the value of the square Е KF H which, with the four triangles surrounding it, make up the area of the square ABCD. Placing the triangle A B H of this square outside of it at C D C, and the triangle B E D at A L C, we have the four triangles, grouped two and two, and thus forming the two rectangles C G D К and AL CX. Each of these rect- angles has its shorter side (4 Z, CG) equal to BZ =a, and its longer side LC, GD, equal to E D = 6; and the sum of the two rectangles is ab -- a6 — 2a. This represents the area of the two rectangles, which are equal to the four tri- angles, which, together with the quaa E K X Н, equal the square A BCD; or— ABCD-EKSHACGDKAALCY,

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7 . of a — b by a — b, or the square of a — b. = 418 ALGEBRA. or— = (6 — аў аб + а 0, or— = (b — аў + 2аб. Then, substituting for (^ — ay, its equivalent as above, we have— | с®— б®—2аф+а*+ 2а. Remove the two like aes with unlike signs (Art. 402), and we have— c —b5 r.a! (115.) which was to be proved. 4.7.— Division the Reverse of Multiplication.—4As di- vision is the reverse of multiplication, so to divide one quan- tity by another is but to retrace the steps taken in multipli- cation. If we have the area a? (Fig. 278), and one of the factors a given to find the other, we have but to remove from a ё the factor a, and write the answer б. If we have the cubic contents of a solid adc (Fig. 270). and one of the factors a given to find the area represented by the other two, we have but to remove a, and write the others, 2c, as the answer. If there be given the area represented by a (b+c) (see Art. 411), and one ofthe factors а to find the other, we have but to remove aand write the answer ó--c. Sometimes, how- ever, а (b + c) is written 20 + ас. Then the given factor 15. to be removed from each monomial and the answer written д + с. | If there be given the area represented by а%--2 > ab + 6° to find the factors, then we know by Art. 412 that this area’ 15 that of a square the sides of which measure a+ 6, and that the area is the product of a + ? by a + б; or, that а +% is the square root оға? +2020 + 8%. If there be given the area a? — 220 + b° to find its fac- tors, then we know by Art. 413 that this: area is that of a square whose sides measure a — 0, or that it is the product

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PROCESSES IN DIVISION. 419 If there be given the difference, of the squares of two quantities, or the area represented by a? — 2°, to find its fac- tors, then we know by Art. 414 that this is the area pro- duced by the multiplication of a — ё by a+ б. 4.(8.— Division : Statement of Quotient.—In any case of division the requirement may be represented as a fraction; thus: To divide c + Z — f by а – ё we write the quotient thus— Са у ЖТ - For example, to illustrate by numerals, let a = 7, 6 = 3, с= 4, d= 5, апа f= 6. Then the above becomes— 445-0 3 jr ы: 419.—Division ; Reduction.—When each monomial in either the numerator or denominator contains a common quantity, that quantity may be removed and placed outside of parentheses containing the monomials from which it was taken; thus, in— : 2а6-дас--8 ed f we have 2 and a factors common to each monomial of the numerator. Therefore the expression may be reduced to 2a(5: 2с —44d) f • To test this arithmetically we willl put « = 9, 0 = 7,6 = 5, d — 4, and f — 6. Then for the first expression we have— 2x9x74A4X9x5—8x9x4 OS 257 `+ s which equals— 126 + 180 — 288 =н zo.

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420 ALGEBRA. And for the second expression— 2х9(7 + 2х5 — 4х4) which equals— 18 (17 -- Lnd iom i + 5 1 2: the same result as before. It will be observed that in this process of removing all common factors algebra furnishes the means of performing the work arithmetically with many less figures. The reduction is greater when the common factors are found in both numerator and denominator. For example, in the expression— Зап+ оби — 15 сл "12 dn — 18 fn we have 3 7 a factor common to each monomial in the nu- merator and denominator; therefore the expression reduces to 3п(а+30— 5c) 3n(4d—6f) ` And now, since 3 л isa factor common to both numerator and denominator, these cancel each other; therefore (Art. 371) the expression reduces to— 24-30 — 5с LA AT To test these reductions arithmetically, let а = о, 6 = 8, ст 4.4- 6, f= 3, апа я,= s. Then the first expression becomes— 3X9X5--9x8x5 — I5 X4x5 I2x 6x5— I8x3x5 which equals— 135 4-360 — 300 1099,1, 360 — 270 7 90 76"

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FORMULA OF THE LEVER. 421 and the second expression becomes— 9:-3x8 —5x4. 4x6—6x3' which equals— The same result, but with many less figures. 420.—Proportionals: Analysis.—In the formula of the lever (Art. 377), PX CF ZR х EC. Let abe put юг Ше arm of leverage CF and m for E С. Then we have— Рн = Rm, from which by division (Art. 372) we have (Art. 399)— a (110.) = жа R= БЕ (111.) Suppose there bea case in which neither & nor P severally are known, but that their sum is known; and it is required from this and the # and z to ind Rand P. Let— VW= RES then— | И—К=Р. (See Art. 403.) The value of P was above found to be— 77. P R.. n Since P = Ж Z and also equals W — R, therefore— › Жж Be n

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422 ALGEBRA. Transferring R to the opposite member (A77. 403) we have— We * ИИ == RtR— Here A appears as a common factor and may be separated by division (477. 419); thus— w= А (1 +). By division the factor (1 E *) may be transferred to the opposite member (4477. 371). Thus we have— W Aoc - Iz ; (116.) by which we find the value of R developed. Asan example, let W = 1000 pounds, # = 3 feet and z = у feet; then— IOOO IOOO тм ee ee 128 9 Multiplying the numerator and denominator by 7, we get— Жа” See. 10 Since— е + P= сө, and— R = “700, then— , І p = amo: But a process similar to the above Gone eps an expression for the value of P, which is— We. I1 Е m (117.) f= Putting this to the test of figures, we have—

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NEGATIVE EXPONENTS. 423 421.— Raising a Quantity to any Power.— When a quantity is required to be multiplied by its equal, the prod- uct is called the square of the quantity. Thus a xa = e (Art. 412). If the square be multiplied by the original quantity the result is a cube; or, a’ x а = а"; or, generally, for— ауаауааа,аааа,ааааа, ме put— in which the small number at the upper right-hand cor- ner indicates the number of times the quantity occurs in the expression. Thus, if w= 2, them си, 9 = 4 х 2 = 8, а = 8 х2 = 16, а — үб «2 Са, m NER in the series of powers may be found by multiplying the preceding one by а, or by dividing the succeeding one by а. 5 a Whus.a' xa ze, andy— = a’. a 4.22.— Quantities with Negative Exponents,—The series of powers, by division, may be extended backward. Thus, 5 4 3 2 А е = 42 а SQOO альн ME if we divide --а; ——4; -=@,—=¢@ ; -—@, —- 7. a a a a a In this series we have 2 =a’. Butaquantity divided by о . . l a its equal gives unity for quotient, or 3—1. Therefore, EE апа 2° = 1. This result is remarkable, and holds good re- gardless of the value of a. From this and the preceding negative exponents we de- rive the following :

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424 ALGEBRA. a к= т-а a ы. І x JNEXL— —, a “ ET Rr IU "P ———m a aa a 59 а І І a= = — = — etc a aa a LJ Showing that а quantity with a negative exponent may have substituted for it the same quantity with a positive exponent, but used as a denominator to a fraction having unity for the numerator. 423.—Addition and Subtraction of Exponential Quan- tities. —Equal quantities raised to the same power may be added or subtracted; as, а? + 2 а? = 34^; but expressions іп which the powers differ cannot be reduced ; thus, +a — a cannot be condensed. 424.— Multiplication of Exponential Quantities. — It will be observed in 472. 421 that in the series of powers, the index or exponent increases by unity; thus, a’, a’, a’, a’, etc.; and that this increase is effected by multiplying by the root, or original quantity. From this we learn that zo multiply two quantities having equal roots we simply add their exponents. Thus the product of a, a’, and а? 15 a’ x a’ ха? = а". The product of 2°, а”, and е? is a° x à? x а = a’. The exponents here, are: —2 +3 +5 = 8 — 2 = 6. 425.—Division of Exponential Quantities.— As division 15 the reverse of multiplication, 70 divide equal quantities raised to various powers, we need simply to subtract the expo- nent of the divisor from that of the dividend. Thus, to divide a’ by а? we have a" = a°. That this is correct is manifest ; for the two factors, a’ x a’, in their product, а”, produce the dividend. 5 "E i. Е ° . I To divide a’ by а, we have a** = a°, which is equal to 2

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EXPLANATION OF LOGARITHMS., 425 (see Art. 422). The same result may be had by stating the question in the usual form. Thus, to divide a’ by а” we have 2 а ; "o. 1 . "IE fraction which is not in the lowest terms, for it may be а? a — ч. aa a numerator and denominator the quantity a’, which cancel put thus, , by which it is seen that it has in both its each other (472. 371). Therefore, A ЕТ the same result as before. 4.26.—Extraction of Radicals.—We have seen that the square ofa isa x а-а”; of 241524х 2a = да“; in each case the square is obtained by doubling the exponent. To obtain the square root the converse follows, namely. take half of the exponent. Thus the square root of a* 15 a’, of a’ is а, of а° 15 a’. The same rule, when the exponent is an odd number, gives a fractional exponent, thus: the square root of a’ is a? ; or, of а”, is ai. So, also, the square root of a, or a’, is at. Therefore, we have at = Va, equals the square root of a, and the cube root of æ = a = Va. 427.—Logarithms,-—We have seen in the last article the nature of fractional exponents. Thus the square root of а” equals a3, which may be put a. In this way we may have an exponent of any: fraction whatever, as a. Between the exponents 2 and 3, we may have any number of fractional exponents all less than 3 and more than 2. So, also, the same between 3 and 4, ог any other two consecutive num- bers. The consideration of fractional exponents or indices has led to the making of a series of decimal numbers called logarithms, which are treated in the manner in which expo- nents are treated; namely— To multiply numbers add their logarithms. To divide numbers, subtract the logarithm of the divisor from the logarithm of the dividend. -

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426 ALGEBRA. To raise any number to а given power, multiply its agat by the exponent of that power. To obtain the root of any power, divide the logarithm of the given number by the exponent of the given power. As an example by which to exemplify the use of loga- rithms: What is the product of 25 by 375? We first make this statement : Log Tof 25. = 1 NONE Putting at the left of the decimal point the integer char- acteristic, or whole number of the logarithm at one less than the number of figures in the given uber at the left of zzs decimal point. To find the decimal part of the required logarithm we seek in a book of Logarithms (such as that of Law's, in Weale's Series, London) in the column of numbers for the given number 25, or 250 (which is the same as to the mman- tissa) and opposite to this and in the next column we find 7940 and a place for two other figures, which a few lines above are seen to be 39; annex these and the whole number is 0-397940. These we place as below: Log. of 25- = 1: 397940. Now, to find the logarithm of 375, the other factor, we turn to 375 in the column of numbers and find the figures opposite to it, 4031, which are to be preceded by 57, the two figures found a few lines above, making the whole, - 574031, which are placed as bélow, and added together. Log. of 25. = 1-397940 "375+ = 2:574031 The sum = 3*97T971 This sum is the logarithm of the product. To find the product, we seek in the column of logarithms, headed o., for -971971, the decimal part. We find first оу, the first two

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EXAMPLES OF LOGARITHMS. 427 figures, and a little below seeking for 1071, the remaining four figures, we find 1740, those which are the next less, and opposite these, to the left, we find 7, and above 93, or together, 937; these are the first three figures of the required product. | For the fourth figure we seek in the horizontal column opposite 7 and 1740 for 1971, the remaining four figures of the logarithm, and find them in the column headed 5. This figure 5 is the fourth of the product and completes it, as there are only four figures required when the integer number of the logarithm is 3. The completed statement therefore is— Log. of 25: = 1-397940, “ " $375: = 2-574031; "^ ЗБ 3 ЭОЕ Another example in the use of logarithms. What is the product of 3957 by 94360? The preliminary statement, as explained in last article, is— Log. 3957 — 3. “ 94360 = 4. In the book of logarithms.seek in the column of numbers for 3957. In the first column we find only 395, and opposite to this, in the next column, we find a blank for two figures, above which are found 59. Take these two figures as the first two of the mantissa, or decimal part of the required logarithm, thus, 0-59. Again, opposite 395 and in the col- umn headed by 7 (the fourth figure of the given number), we have the four figures 7366. These are to be annexed to (о· 59) the first two obtained. Тһе decimal part of the loga- rithm, therefore, is 0-597366. To obtain the logarithm for 94360, the other given num- ber, we proceed in a similar manner, and, opposite 943, we find 0-97; then, opposite 943 and in column headed б, we find 4788, or, together, the logarithm 15 0.974788. Тһе whole is now stated thus— `

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428 ALGEBRA. Log. of . 3957 = 3:597366 . wil 94360 = 4-974788 —2 wi 273282000 = 8-572154. = sum of logs. The two logarithms are here added together, and their sum is the logarithm of the product оҒ the two given factors. The number corresponding to the above resultant logarithm may be found thus: Look in the column headed o for 57, the first two numbers of the mantissa, íhen in. the same column, farther down, seek 2154, the other four figures of the mantissa; or, the four (1709) which are the next less than the four sought, and opposite these to the left, in the column of numbers, will be found 373, the first three figures of the product ; opposite these, to the right, seek the four figures next less than 2154, the other four figures of the man- tissa. These аге found in the column headed 3 and are 2058. Тһе 3 at the head of the column is the fourth figure in the product. From 2154, the last four figures of the man- tissa, deduct the above 2058, or— 2154, 2058, Remainder, 06. At the bottom of the page, opposite the next less number (3727) to that contained in 3733, the answer already found, seek the number next less to the above remainder, 96. This is 02.8, and is in the column headed 8. Then 8 is the next number in the product. From 96 deduct 92.8, and multi- ply it by 10, or— - 06 92-8 3:2 10 AR 387 Then, in the same horizontal column, seek for 32 or its next less number. This is 23-2, found in column 2. This 2 isthe next figure in the product. Additional figures may be ob- tained by the table of proportional parts, but they cannot be

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THE SQUARE OF A BINOMIAL. 429 depended upon for accuracy beyond two or three figures. We therefore arrest the process here. The product requires one more figure than the integer of the logarithm indicates; as the integer 1s 8, there must be nine figures in the product. We have already six; to make the requisite number nine we annex three ciphers, giving the completed product— 3957 x 94360 = 373382000. By actual multiplication we find that the true product in the last article is 373382520. Іп a book of logarithms, carried to seven places, the required result is found to be 373382500, which is more nearly exact. The utility of logarithms is more apparent when there are more than two factors to be multiplied, as, in that case, the operation is performed all in one statement. Thus: What is the product of 3-75, 432-95, 1712, and 0-0327? The statement is as follows : Log. 3-75 = 0-574031 432-95 = 2-636438 І712- = 3:233504 0327 edit Product = 00801. = 4-958521 16 5. Explanations of working are given more in detail in most of the books of logarithms. 428.—Completing the Square of a Binomial.—We have seen in Art. 412 that the square of a binomial (a + 0) equals a? + 2 ab + 6*—a trinomial—the first and last terms of which are each the square of one of the two quantities, while the second term contains the second quantity multi- plied by twice the first quantity— In analytical investigations it frequently occurs that an expression will е obtained which may be reduced to this form :

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430 2 ALGEBRA. oo Ww aw, (T18) in which zz is the coefficient of the second term, and a and 6 are two quantities represented by а and 0 or any other two symbols. A comparison of this expression with the square of a bi- nomial (112.) contained in 477. 412, shows that the member at the left comprises two out of the three terms of the square of a binomial; as thus— Gur ep IU but with a coefficient ж instead of 2. It is desirable, as will be seen, to ascertain a proper third term for the given ex- pression; or, as it is termed, ‘to complete the square." The method by which this is done will now be shown. А consideration of the above trinomial shows that the third term is equal to the square of the quotient obtained by dividing the second term by twice the square root of the first; or— К > ЕЕ . 2 Now a third term tothe above binomial, equation (118.), may “һе obtained by this same rule. For. example— maby? 2029): 2 The rule for the third term then is: Divide the second term by twice the square root of the first, and square the quotient. As an example, let i& be required to find the third term required to complete the square in the expression— ALTE =F, in which z and f are known quantities and x unknown. Putting it in this form— | an ОЙТ, and dividing by 4, we have—

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BINOMIALS CONTINUED. 431 e vá x i лы which reduces to— 3 Ж 2 I — “a x м шек Now applying the above rule for finding the third term, we и (2) (3,4) which is the required third term. To complete the square we add this third term to both members of the above re- duced expression, and have— x +} Л + (ё )=2+ (às). The member of this expression at the left is the completed square of a binomial, the two quantities constituting which are the square roots of the first and third terms respectively ; or x and 3 л, and we therefore have— ARTE ТҮРЕ? „)==+% 2 and now taking the square root of both sides of the expres- sion, we have— t and, by transferring the second quantity to the right mem- ber, we have— ж=ү + (2) a: an expression in which т, the unknown quantity, is made to stand alone and equal to known quantities. The process of completing the square is useful, as has been shown, in developing the value of an unknown quan-

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432 ' ALGEBRA. tity where it enters into an expression in two forms, one as the square of the other. As an example to test the above result, let f = 256 and n = 8. Then we have by the last expression for the value of x— бы y 23°, (is ) — 3: 8, = v8 36 — 6, = 100 — 6, X — IO — 6 = 4. Now this value of x may be tested in the original expres- sion— ow xai y, for which we have— 6x8x4+4x 4 =f, LO2. 4064 = f, 256 — f; the correct value as above. PROGRESSION. 429.—Arithmetieal Progression.—In a series of num- bers, as I, 3, 5, 7, О, etc., proceeding in regular order, in- creasing by a common difference, the series is called an arithmetical progression ; the quantity by which one num- beris increased beyond the preceding one is termed the difference. If Z represent the difference and а the first term, then the progression may be stated thus— Terms— i, 2, 5. 4, 5, a apg тет 4 бақа ча д, ctc. The coefficient of Z is equal to the number of terms preced- ing the one in which it occupies a place. Thus the fifth term is a + 4d, in which the coefficient 4 equals = number of the preceding terms. From this we learn the rule by which at once to desig-

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ARITHMETICAL PROGRESSION. 433 nate any term without finding all the preceding terms. For the one hundredth term we should have a + 99 Z, or, if the number of terms be represented by 7, then the last term would be represented by— l= a + (п — 1) 4. (119.) For examiple, in a progression where а; the first term; equals 1, d the difference, 2, and z, the number of terms, go, the last term will be— eese deii ae p Therefore, to find the last term: To the first term add the ym of the common difference into the number of terms less one. * By a transposition of the terms in the above expression, so as to give it this form— а= 1 -– (n – 1) 4, (120.) we have a rule by which to find the first term, which, in words, 15-- Multiply the number of terms less one by the common difer- ence, and deduct the product from the last term, the remainder will be the first term. By a transposition of the terms of the former expression to this form— /-а-ш-(б(т - rd, апа dividing both members by (2 — 1), we һауе-- l—a d= талан Я (121.) which is а rule for the common difference, and which, in words, 15-- Subtract the first term from the last, and divide the remain- der by the number of terms less one ; the quotient will be the com- mon difference. Multiplying both members of the equation (121.) by (n — 1) and dividing by а, we obtain—

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434 ALGEBRA. A п—1= —5 Transferring 1 to the second member, we have— p # = SET (1225) which is a rule for finding the number of terms, and which, in words, is— Divide the difference between the first and last terms by the common diference; to the quotient add unity, and the sum will be the number of terms. Thus it has been shown, in equations (119,) (120), (121), and (122), that when, of the four quantities in arithmetical progression, any three are given, the fourth may be found. The sum of the terms of an arithmetical progression may be ascertained by adding them; but it may also be had bya shorter process. If the terms are written in order in a hori- zontal line, and then repeated in another horizontal line be- neath the first, but in reversed order, as follows: I, 3, 5, 7, 9, II, 13, I5, 15, I3, II, О, 75 5 3, I, 16, 16, 16, 16, 16, 16, 16, 16, and the vertical columns added, the sums will be equal. In this case the sum of each vertical couple is 16,and there are 8 couples; hence the sum of these 8 couples is 8 x 16 = 128. And in general the sum will be the product of one of the couples into (һе number of couples. It will be observed that the first couple contains the first and last terms, 1 and 15; therefore the sum of the double series is equal to the product of the sum of the first and last terms into the number of terms. Or if S be put to represent the sum of the series, we shall have— 2X8 = (о + J), and, dividing both sides by = S ign Wak (123.)

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GEOMETRICAL PROGRESSION EXPLAINED. 435 Or, іп words: The sum of an arithmetical series equals the prod- uct of the sum of the first and last terms, into Moe! the number of terms. 430.—Geometrical Progression.—A series of numbers, such as I, 2, 4, 8, 16, 32, 64, 128, 256, etc., in which any one of the terms is obtained by multiplying the preceding one by a constant quantity, is termed a Geometrical Progression. The constant quantity is termed the common Ха, and is equal to any term divided by the preceding one. Thus in 1$ 8 the above example g or 3 or t = 2, equals the common ra- tio of the above series. In the series, I, 3, 9, 27, etc., we have for the ratio— which is the common ratio of this series. А geometrical series may be put thus: Terms: . Ln 2 S 4; Progress. : ILL 1х3, 1I*3x3, ХЗ К or thus— Terms: 0, 2, 2 Progress. : “тха Lx 3! in which the common ratio, in this case 3, appears in each term and with an exponent which 15 equal to the number at terms preceding that in which it occupies a place. If the first term be represented by a and the common ra- tio by 7, then the following will represent any geometrical progression— а, ar, ar*, a r5, ат”, etc. (124.) For example, let a = 2 and 7 = 4; then the progression will be— 2, 8, 32, 128, $12, etc!

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436 » ALGEBRA. If > = unity, then when a = 2 the progression becomes— о Ка, 2, 2, SHC. If 7 be less than unity, then the progression will be a de- creasing one. - For example, let а = 2 and > = 4. Then we have for the progression— Cd ж-е- Кы I [BM CI AE im те If the number of terms be represented by z, and the last by Athen the last term will be— 7e ажет" "omes For example, let >” equal 6, then the progression will be— Terms: I, Әу 3, 4, iis 6; Progress.: T TA ERE 0 TD a in which .the exponent of the last term equals z — І = 6 — І = 5. If 5 be put for the sum of a geometrical progression, we will have— =a er er +... ee! ee |, Multiply each member by 7, then— i rex Uo аға rib up ett eq m E Subtract the upper line from the lower; then— 57 та GPa TU TT MNT а: ое", 5 ЖАР а РТ... ТЕ”; 77 4 ani Sr— s= =a ** * * Т А Sr — s= — Gs @r, | Ял ы ағымда ar"—a O= тағу Ч

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GEOMETRICAL PROGRESSION CONTINUED. 437 The last term (equation (125.)) equals / = az"^*', and since ar" -—rxar""' —ri,therefore— rl—a S ЖЕТ. (126.) Thus, to find the sum of a geometrical progression: Multi- ply the last term бу the ratio; from the product deduct the first term, and divide the remainder by the ratio less unity. For example, the sum of the geometrical progression— S = I +3 9 + 27 + 81 + 243 + 729 = 1093 by actual addition. | To obtain it by the above rule— i the correct result. If there Бе а decreasing geometrical progression, as I, 4, 3» зт, etc., in which the ratio equals 4, the sum will be— IM I I I | я S=1 крш у er etc., to infinity. Multiply this by 3, and subtract the first from the last— I I I I 5 4 35—341«5 459 5 ТТР i м І I = lt РЕЗЕ er 25 = 3701S ve + to infinity. In a decreasing progression let 7, the common ratio, be represented by = (б less than с), and the first term by а, then the sum will be— 2 3 S = 0.46 JE etc., to infinity.

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438 ALGEBRA. Multiply this by =. and subtract the product from the above— b b e p T - 5--а-жад-аз 4- etc., to infinity. Ее б p б infini = а 4а ыы» 0 Ra nity. b b p б? ағы Ss --а-ү ша іс = + tomhnity.. [^ с (67 [^ сабы? eui ы и с b Or— 5(1--)-а, с а dcum (127.) А For example, let the first term of а geometrical progression equal 2, and the ratio equal 4, then the sum will be— = M at йә 791 |77 From this, therefore, we have this rule for the sum of an in- finite geometrical progression, namely : Divide the first term by unity less the ratio. |

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SECTION X.—POLYGONS. 431. — Relation of Sum and Difference of Two Lines.— Let AB and CD (Fig. 285) be two given lines; make ЕМ == В Fic. 285. equal to 4 В, апа AG equal to CD; then ЕС equals the sum of the two lines. Make FG equal to А В, which is equal to £ H. Bisect Æ біп 9; then, also, 7 bisects HF; for— BD cu, and— EH=FG. Subtract the latter from the former ; then— EY LiFe Pes but— EY BA FF. and— YyG—FG--—ZF; therefore— . В H F = FF. Now, Е 7 is half the sum of the two lines, and Æ $ is half the difference; and— Е9-Н?7--ЕН- АВ. Or: Half the sum of two quantities, minus half their dif- ference, equals the smaller of the two quantities.

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440 POLYGONS. Let the shorter line be designated by а, and the longer by 2; then the proposition is expressed by— (128.) We also have £ F + FF = EF CD; or, half the sum of two quantities, plus half their difference, equals the larger quantity. 432.—Perpendicular, in Triangle of Known Sides.— Let A BC (Fig. 286) be the given triangle, and С a perpen- dicular let fall upon A B, the base. Let the several lines of Fic. 286. the figure be represented by the symbols a, 0, c, d, g, and f, as shown. Then, since 4 Е C and BEC are right-angied triangles, we have (477. 416) the following two equations, and, by subtracting one from the other, the third— SK, tad) = a’, е* + qq == ф°%, he ai— 63, Then (Art. 414), by substitution, we have— FT гу у= (4 + б) (а — 0). By division we obtain— нэ д Дні Даве 7—8 yx p

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RULE FOR TRIGONS. 441 + According to Art. 431, equation (128.), we have— g-Lti- Ems In this expression let the. value. of f — E as above, be. substituted, then we will have— = (а + b) (а — 6) 2 2(f +8) Multiply the first fraction by (f + g), then join the two fractions, when we will have— - EE = 5) (an) re eR e e = ГЕ m) The lines f and g, in the figure, together equal the line c; therefore, by substitution— | c — (a + 6) (a — b) m (129.) d = This is the value of the line p It may be expressed in words, thus: The shorter of the two parts into which the base of a triangle is divided by a perpendicular let fall from the apex upon the base, eguads the quotient arising from a division by twice the base, of the differ- ence between the square of the base and the product of the sum and difference of the two inclined lines. | As an example to show the application of this rule, let а == 9, ф = б, and c — 12; then equation (129.) becomes— I2? — (о + 6 — 6 T ix Оо бу 2 x I2 _ 144 — 15 x 3 = à 122488 hy” ^8

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442 ' POLYGONS. Now, to obtain the length of Z, the perpendicular, by the figure, we have— LED SS, -апа, — the square root— d MESE ua (130.) or, in words: The altitude of a triangle equals the square root of the difference of the squares of one of the inclined sides and tts base. As an example, take the same dimensions.as before, then ©. equation (130.) becomes— = ve Fr. The square of 6 = 6. “ “ “ 41 = 17-015625 6* — 43°= 18:084375, the square root of which 15 4.44234; therefore— d = V6! — 4Ẹ = 4-44234. This may be tested by applying the rule to the other in- clined side and its base— C == 2 = 44 f= 7% Then, p бы” — or, g^ = 81. pm 62-015625 9' — 74° = 18:984375.

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TRIGON—RADIUS OF CIRCLES. 443 The same result as before, producing for its square root the same, 4-44234, the value of 4; therefore— d = Y 9* —7$* = 4-44234. 433.—Trigon: Radius of Circumscribed and Inscribed Circles: Area.—Let A B C (Fig. 287) be a given trigon or triangle with its circumscribed and inscribed circles. Draw the lines 4 DF, РВ and D C. The three triangles, А B D, A C D, and В D C, have their apexes converging at D, and form there the three angles, САРВ, ADC, and BDC. These three angles together form Ес; 287. four right angles (477. 335), and each of them, therefore, equals $ of a right angle. | The angles of the triangle BDC together equal two right angles (477. 345). Ав above, the angle BDC equals 4 of a right angle, hence 2 — $ = $5* = $ of a right angle, equals the sum of the two remaining angles at P and С. The triangle BDC is isoceles (A77. 338) ; for the two sides B D and D C, being radii, are equal; therefore the two angles at the base B and C are equal, and as their sum, as above, ' equals $ of a right angle, therefore each angle equals $ of a right angle. Draw the two lines FC and FB. Now, be- і

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444. : POLYGONS. cause DC and.) F are radii, they are equal, hence DFC is an isoceles triangle. It was before shown that the angle 5 D C equals 4 of a right angle; now, since the diameter 4 F bisects the chord B C, the angles ВР Е and Е DC are equal, and each equals the half of the angle BDC; ог, 4 of 4 of a right angle equals %of aright angle. Deducting this from two right angles (the sum’ of the three ‘angles of the triangle), or:2— $ =-~ I$ = 4 of a right angle equals the sum of the angles at F and C; hence each equals the half of 4; or 2 of a right angle; therefore the triangle DFC is equilateral. The triangles DBFand DFC are equal. The angles В О Сапа P FC are equal; the line BC is perpendicular to DF and bisects it, making DE and £F equal; hence DE equals half DF, or D B, radii of the circumscribing circle. Therefore, putting R to represent B D, the radius of the circumscribing circle, and 6= BC, a side of the triangle A BC, by Art. 416, we bave— BD = bn? FWA, r= G (8) Transferring and reducing— д Л le д "WW Abr утын, 4 4 3 р? Los 4 d АҒ Tan 52514 3 a b RE сая; Lo. үз (131.) Or, The Radtus of the circumscribing circle of а regular trigon or equilateral triangle, equals a side of the triangle divided by the square root of 3.

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AREA OF EQUILATERAL TRIANGLE. 445 By reference to Fig. 287 it will be observed, as was | DF BP above shown, that Dk = Же е ДЕ DE venas dius of the inscribed circle, equals half the radius of the circumscribed circle; or, again, dividing equation (131.) by 2, we have— CES. 2: «2 Mrz È and, putting > for the radius of the inscribed circle, we have— б == e 132. B (132-) Or: The radius of the inscribed circle of a regular trigon equals the half of a side of the trigon divided by the square root of 3. To obtain the area of a trigon or equilateral triangle; we have (Art. 408) the area of a parallelogram by multiplying its base into its height; and (4725. 341 and 342) the area of a triangle is equal to half that of a parallelogram of equal base and height, therefore, the area of the triangle Б D C (Fig. 287) is obtained by multiplying Z С, the base, into the half of E D, its height. Or, when JV is put for the area— 3 M CU E, R ^ Lm ый substituting for А its value (131.)— б N =6 ==; "4 V3 à? М-- . гиз This is the area of the triangle BDC. The triangle A В С is compounded of three equal tri- angles, one of which is the triangle BDC; therefore the area of the triangle 4 В С equals three times the area of the triangle B DC; or, when A represents the area—

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446. POLYGONS. 112 3° (133.) Or: Тһе area of a regular 7000 or equilateral triangle equals three fourths of the square of a side of the triangle di- vided by the square root of 3. 4.34.—Tetragon; Radius of Circumscribed and En- scribed Circles: Area.—Let A B CD (Fig. 288) be a given tetragon or square, with its circumscribed and inscribed Fic. 288. circles, of which A Æ is the radiusof the former and Е F that of the latter. The point F bisects А P, the side of the square. А F equals E F and equals half А В, a side of the square. Putting A for the radius of the circumscribed circle and 0 for A В, we have (Art. 416)— A ESA be, ӘЛБЕТТЕ R=, a34) Or: The vadius of the circumscribed circle of a regular tetra- gon equals а side of the square divided by the square root of 2.

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SIDE AND AREA OF HEXAGON. * 447. By referring to the figure it will be seen that the radius of the zzserzbed circle equals half a side of the square— í (135.) = Тһе.ағеа of the square equals the square of a side— A= O°, (136.) 435.— Hexagon: Radius of Circumscribed and In- scribed Circles: Area.—Let A BC DE F (Fig. 289) be an equi- lateral hexagon with its circumscribed and inscribed circles, of which £G is the radius of the former, and С Z that of the latter. The three lines, A4 D, B E, and CF, divide the ! Fic. 289. hexagon into six equal triangles with their apexes converg- ing at G. The six angles thus formed at G are equal, and since their sum about the point G amounts to four right angles (Arf. 335), therefore each angle equals $ ог? of a right angle. The sides of the six triangles radiating from G are the radii of the circle, hence they are equal; therefore; each of the triangles is isosceles (471. 338), having equal angles at the base. In the triangle E G D, the sum of the three angles being equal to two right angles (477. 345), and the angle at С being,as above shown, equal to $ of a right angle, therefore the sum of the two angles at Æ and D equals 2 — $ = 4 of aright angle; and, since they equal cach other,

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448 - POLYGONS. therefore each equals $ of a right angleand equals the angle at С; therefore E G D is an equilateral . triangle.. . Hence E D, a side of a hexagon, equals E G, the radius of the circum- scribing circle— К- 6. | (137) As to the radius of the inscribed circle; represented by G H, a perpendicular from the centre upon £ D, the base; the point 77 bisects E D. Therefore, Е H equals half of a side of the. hexagon, equals half the radius of the circumscribing circle. Let R = this radius, and r the radius of the inscribed circle, while б = a side of the hexagon; then we have (4775. 353 and 416) GH?= ЕСЕН? і 2, P Now, R = 2, therefore— Waa, r=py = TIN 3A, p ae шш o қамты | (138.) Or: The radius of the inscribed circle of a regular hexagon equals the half of a side of the hexagon, multiplied by the square root of 3. | As to the area of the hexagon, it will be MM that the six triangles, А BG, В С C, etc., converging at С, the centre, are together equal to the area of the hexagon. The area of E G D, one of these triangles, is equal to the product of £ D, the base, into the half of G H, the perpendicular; or, when N is put to equal the area— TERT GH

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SIDE AND AREA OF OCTAGON. 449 ІУ--дх T 2 and, since 7, as above, equals V 3 к. 4 2 b үз: N=bx Eb i 4 VERTE 4 This is the area of one of the six.equal triangles; therefore, when 4 is put to represent the area of the hexagon, we have— A = 6x Ve uem 4 Mi в А = ¥3 2. (139) Or: The area of a regular hexagon equals three halves of the square of a side multiplied by the square root of 3. Fic. 290. 436.—Octagon: Radius of Cireumscribed and In scribed Circles: Aren.—Let CE D B F (Fig. 290) represent a quarter of a regular octagon, in which Fis the centre, E D a side, and СЕ and DB each half a side, while СХ and B Fare radii of the inscribed circle, and £/ and ‘DF are radii of the circumscribed circle.

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450 POLYGONS. Let R represent the latter, and > the former; also let 6 represent £ D, one of the sides, and n be put for А D, and for А Е. Then we have— AD+DB= CF. M 2 3 ё e z= 7 — —, Or 2 Since 4 D E is a right-angled triangle (477. 416), we have— ИЕ Д0 EDX Placing the value of z, equal to the value before found we have— * D T ж”. r= 4.50 This coefficient may be reduced Бу multiplying the first fraction by V2, thus—

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RULES FOR OCTAGON. 451 therefore— C (Rip ES ж =(V¥2 +1) £ (140.) Or: Тһе radius of the inscribed circle of a regular octagon equals half а side of the octagon multiplied by the sum of unity plus the square root of 2. In regard to the radius of the cir- cumscribed circle, by 477. 416 we have— DE = Bw? + DB’, wars (2) In this expression substituting for 7’, its value as above, we have— waves y (4) + (4). Ga [( V2.-- ту «rA The square of the coefficient ( V2 + 1) by 422. 412 equals 2 +2V2 +1 = 2 #2 + 3, then— KR? = [(2 V2 + 3) +1] wey R? = (2 Уз + 4)(—)- R—a/2 Va ka S. (141.) Or: The radius of the circumscribed circle of a regular octagon equals half a side of the octagon multiplied by the square root of the sum of twice the square root of 2 plus 4. In regard to the area of the octagon, the figure shows that one eighth of it is contained in the triangle D ЕР,

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452 * = POLYGONS. The area of D E F, putting it equal to JV, is— um ET. xí 3 2 pe 2 2d. E "t No TUES TEA 2 2 N= (Via 2. This is the area of one eighth of the octagon; the whole area, therefore, is— i A = (2.4.1) ian A 4 % A4 -x(V2-1) 2. (142.) Or: The area of a regular octagon equals twice the square of a side, multiplied by the sum of the square root of 2 added to unity. When a side of the enclosing square, or diameter of the inscribed circle,is given, a side of the octagon may be found ; for from equation (140.), multiplying by two, we have— ay = (V2 + I) Ё. Dividing by 42 + 1, gives— ar. V2 41 (143.). The numerator, 27, equals the diameter of the inscribed circle, or a side of the enclosing square ; therefore: The side of à regular octagon, equals а side of the enclosing square divided by the sum of the square root of 2 added to unity. ” 437,—Dodecagon: Radius of Circumscribed and In- scribed Circles: Area,—Let A BC (Fig: 291) be an equilat-

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SIDE AND AREA OF DODECAGON. 453. eral triangle. Bisect A B 1р F; draw C FD; with radius А C describe the arc A DB. Join A and D, also D and В; bisect AD in E; with the radius Е С describe the arc E С: Then A D and D P are sides of a regular dodecagon, or twelve- sided polygon; of which A C, D C, and В C are radii of the circumscribing circle, while Æ C is a radius of the inscribed сеје. The line A В is the side of a regular hexagon (Art. 43 5). Putting R equal to A C the radius of the circumscribing cir- cle; ғ, = Е С, the radius of the inscribed circle ; д, = А Р, а side of the dodecagon, and n = DF. Then comparing the Fic. 201. homologous triangles, 4 DF and AEC (the angle ADF equals the angle Е A C, and the angles DFA and А Е C are right angles); therefore, the two remaining angles DAF and А СЕ must be equal, and the two triangles homologous (Art. 345)... Thus we have— DF: DH :: AB: AC, E .24

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454- "^ A POLYGONS. In Art. 435 it was shown that FC (Fig. 291), or GH of и À | І d Fig. 289, the radius of the inscribed hexagon, equals У з Эр and in which its? — R; Ес- 34: Now ( Fig. 291)— | Ту боа MEM ore; | | — № - Substituting this value of x,in the above expression, we have— a „м 7 gw — Be Bi - Multiplying by 2 and reducing, we have— к=ү 7 (144.) Or: The radius of the circumscribed circle of a regular dodec- agon, equals.a side of the dodecagon multiplied by the square root of a fraction, having unity for its numerator and for 115 denominator 2 minus the square root of 3. Comparing the same triangles, as above, we have— Ol) ы E G or— (145-)

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RULE FOR DODECAGON. 455 Or: The radius of the inscribed circle of a regular.dodecagon equals a side of the dodecagon divided by the difference between 4 and the square root of 3. The area of a dodecagon is equal to twelve times the area . of the triangle A DC (Fig. 291). The area of tliis triangle is equal to half the base by its perpendicular; or, dE x EC; or— b — X 7, 2 . or, where N equals the area— N= br. Or, for the area of the whole dodecagon— I2 М ="6 br, A = 67. Substituting for 7 its value as above, we have— 6 A= ----- b 146. "ELS | (146.) Or: The area of a regular dodecagon equals the square of a side of the dodecagon, multiplied by a fraction having 6 for tts numerator, and for tts denominator, 4 minus twice the square root of 3. 438.—Hecadecagon : Radius of Circumscribed and In- scribed Circles: Area.—Let A B СР (Fig. 292) be a square enclosing a quarter of a regular octagon CE FB, E F being one of its sides, and СЕ апа FB each half a side, while FD is the radius of the circumscribed circle, and # D the radius of the inscribed circle of the octagon. Draw the diagonal A D; with D F for radius, describe the circumscribed circle EGF; join G with F and with Е; then ЕС and СР will each be a side of a regular hecadecagon, or polygon of six- teen sides. An expression for FD, the radius of the circumscribed

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Ф 456 POLYGONS. circle, may be obtained: thus: Putting FD = R; HD=r; GF=6; GF =n; апа Е = > (Art. 416), we have— iue CRLF F*, C та Fic. 292. Comparing the two homologous (477. 361) triangles, G ¥ F and FH D (Art. 374), we have— CF diri © dur, iD і ұбт Ф "Qs 2 di. к= m ЖЕ -— 23 asi t Putting this value of м? in an equation against the - former value, we have— E ir al dm (5 ). In 2472. 436, the value of F D, as the radius of the cir- cumscribed circle of a A octagon, is given in equation (141.) as— KS % V2+4 2

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SIDE AND AREA OF HECADECAGON. 457 in which 0 represents:a side of the octagon, or ЖР, for which we have put s. Substituting s for б and putting the numerical. coefficient under the radical, equal to В, we have— | | К = УВ; Squaring each member gives— е-ә6). From which, by transposition, we have— Substituting in the above expression for (7, this value of it, gives— ашасы = «ыша ZR. PUN Transposing, we have— E Р 4 DX gre MO Multiplying the first term by В, and the second by 4 Ж”, we have— : | Bo* 4%" 2 No ы е, y Р ADR* 4ВА? ВФ‘ + 4R' рз ABR* E , B + gk ты Б.С” 0". Transposing, we have— ARA—RBR' bz — Bo.

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458 POLYGONS. To complete the square (477. 428) we proceed thus— OR'LBRUCSZ-ABÜ. R'-BRV+EBSY=(GBS'*Y— TBO". Taking the square root, we have— R°—4Bb'= ИГЕ Р, R'—wyiBuü'—iBb4t4Bb, A E DyXB —iLB+4B0",. R'—b(yYRB!—IB-34B) к= ОИЕ) ABO Restoring В to its value, 2 #2 + 4 as above, we have— 1B=4V2+1, B-1=2 ТЖ 3: multiply these— : 2 +2142, 34 3V2, к }+В(В 1) =5+#42, В = М2 + 2. Therefore— К" s eg V2 2 2), Rabi Vs*3V24 Vita. (147)

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RULES FOR HECADECAGON. 459. Or: Тһе radius of the circumscribed circle of а regular hecadecagon equals а side of the hecadecagon multiplied by the square root of the sum of two quantities, one of which is the square root of 2 added to 2, and the other is the square root of the sum of seven halves of the square root of 2 added to 5. To obtain the radius of the inscribed circle we have (Fig. 292)— | "ud Балара ГЕ", Substituting for R? its value as above, we have— r= 6"( VEB(B— 1) +48) - (2). 2 r? = 5*[( V4 B(B—1) +4 8) — (4)7, ry VEB(B— 32 4. қ The coefficient of 6 is the same as in the case above, ex- cept the — 1; therefore its numericak value will be 4 less, or— i r= by V5 +4 Ус. (148). Or: The radius of the inscribed circle of a regular hecadeca- gon equals a side of the hecadecagon multiplied by the square root of two quantities, one of which ts the square root of 2 added to 14, and the other is the square root of the sum of seven halves of the square root of 2 added to 5. To obtain the area of the hecadecagon it will be observed that the area of the triangle G FD (Fig. 292) equals HD х H F, апа that this is the ṣẹ part of the polygon; we there- _ fore have— A= V6 DS AY, А = 072 = 87 6.

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460 POLYGONS. The value of 7 is shown in (148.); therefore we have— A= 857 4/ "s +3V¥24¢V24+1% ^ — (149) Or: The area of a.regular hecadecagon equals eight times the square of its side, multiplied by the square root of two quan- tities, one of which is the square root of 2 added to 1$, and the other 15 the square root of the sum of seven halves of the square root of 2 added to 5. 439.—Polygons: Radius of Circumscribed and Inscribed Circles: Area.—In 47s. 433 to 438 the relation of the radii to a side in a trigon, tetragon, hexagon, octagon, dodeca- gon and hecadecagon have been shown by methods based Fic. 293. upon geometrical proportions. This relation in polygons of seven, nine, ten, eleven, thirteen, fourteen and fifteen. sides, cannot be so readily shown by geometry, but can Бе easily obtained by trigonometry—as also said. relation of the parts in a regular polygon of any number of sides. The na-. ture of trigonometrical tables is discussed in Arts. 473 and 474. So much as is required for the present purpose will here be stated. Let A BC (Fig. 293) represent one of the triangles into which any polygon may be divided, in which BC = 2? = а side of the polygon; 4 C R = the radius of the circum- scribed circle ; and 4 D =r = the radius of the inscribed circle.

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GENERAL RULES FOR POLYGONS. 461 Make ЕС equal unity ; on C asa centre describe the arc Е Е; draw FH and Е С perpendicular to В C, or parallel to АР; then for. the uses of trigonometry Æ С is called the tangent of с, or of the angle A C B, and FH 15 the sine, and H C the cosine of the same angle. These trigonometrical quantities for angles varying from zero up to ninety degrees have been computed and. are to be found in trigonometrical tables. . Referring now to Fig. 293 we Hen HUE WP AC, Cos. c : TL 2 NY. — 2 cos.c (150.) Again— 3 E Cu ЖА G zo D, y»: Damn. с == MEN E 7 E tanya. а (151.) These two equations give tlie: required. radii. of the cir- cumscribed and inscribed circles. They may be stated thus: The radius of the circumscribed circle of any regular poly- gon equals a side of the polygon divided by twice the. cosine of the angle formed by a q^ of the polygon and a radius from one end of the side. The radius of muie -circle of any regular polygon equals half of a side of the polygon multiplied by the tangent of the angle formed by a side of the polygon and a radius from- one end of the side. Тһе area of a polygon equals the area of the triangle A B C (Fig. 293), (of which ZC is one side of the polygon: and А is the centre), multiplied by the number of sides in the polygon; or, if z be put to represent the number of. the sides and A the area, then we have—

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462 POLYGONS. dl Bn, in which B equals the area of the triangle. The area of A B C (Fig. 293) is equal to A D x B D, or— | B=rx-. 2 For v substituting its value, as in equation (151), we have— pol v ELM WOMEN tan. c. 2 T re Therefore, by substitution— Az "LA n tan. с. (152.) Or: The area of a regular polygon equals the square of а side of the polygon, multiplied by one fourth of the number of tts sides, and by the tangent of the angle formed бу a side of the - polygon, and a radius from one end of the sides. 440.—Polygons: Their Angles.—Let a line be drawn from each angle of a regular polygon to its centre, then these lines form with each other angles at the centre, which taken together amount to four right angles, or to 360 de- vngesq(/17/5.327, 335). If this 360 degrees be didide by the number of the sides of the polygon, the quotient will equal the angle at the cen- tre of the polygon, of each triangle formed by a side and two radii drawn from the ends of the side. For example: if А B C (Fig. 293) be one of the triangles referred to, having ВС one of the sides of the polygon and the point 4 the cen- tre of the polygon, then the angle B A C will be equal to 360 degrees divided by the number of the sides of the polygon. If the polygon has six sides, then the angle B A C will contain E = бо degrees ; or if there be то sides, then the angle at 6 aL the centre, will contain - 36 degrees. The angle

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SIDE AND AREA OF PENTAGON. 463 В А D equals half the angle ВА C, or, when x equals the number of sides, the angle Б 4 C equals— | 360 n and the triangle BA pafa 2 equals— Now the angles 84 D- DB.A equal one right angle (Art. 346), or go degrees. Hence the angle D ВА = 9o? — D А D, or the angle с equals— 360". 2 Mak С 90° ж (1359 For example, if z equal 6, or the polygon have six sides, then— : c? = 90° — aS = 90 — 30 = 60°. Therefore, the angle c, contained in equations (150.), (151.), and (152.), equals до degrees, less the quotient derived from a di- vision of 360 by twice the number of sides to the polygon. 4.41. —Pentagon: Radius of the Cireumseribed and In- scribed Cireles: Area.— The rules for polygons developed in the two former articles will here be exemplified in their application to the case of a regular pentagon, or polygon of five sides. ! To obtain the angle c° (153), we һауе z = 5, and— o oso REESE, UO 2° = 90 apo 36 = 54°. For the radius of the circumscribed circle, we have (150.)— 6. ee , 2 COS. с

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464. POLYGONS. b 2 COS. 54° I d Ааа қ 2 COS. 54 Using a table of logarithmic sines and tangents (472. 427), we have— Log. 2 = о: 3010300 Cos. 54° = 0:7602187 Their sum = 0:0702487 — subtracted from Log. 1 = 0-0000000 О-85065 = 9:9297513 Therefore— R= о.85065 д. Or: The radius of the circumscribed circle of a regular penta- gon equals a side of the pentagon multiplied by the decimal о. 85065. For the radius of the inscribed circle, we have (151.)— = 2 tan. C, tan. 54° 2 Ж-- / For this we have— Log. tan. 54? = 0- 1387390 Log. 2 = о: 3010300 | 0-68819 = о: 8377090. Therefore— 7 = 0.68810 2. Or: The radius of the inscribed circle of a regular pentagon equals a side of the pentagon multiplied by the decimal 0-68819. For the area we have (152.)— | ASTO witan с, дже ЖБ tan. 524% И —Et SA? 7° І

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TABLE FOR REGULAR POLYGONS, 465 For this we have— Log. 5. = 0-6989700 Log. tan. 54° = 0- 1387390 | 0-8377090 Log. 4 = 0-6020600 1.72048 = 0-2356490 Therefore— : | А = 72048 %7. Or: The area of a regular pentagon equals the square of its side multiplied by 1.72048. 44.2.— Polygons: Table of Constant Multipliers, — To - obtain expressions for the radii of the circumscribed and in- scribed circles, and for the area for polygons of 7, 9, 10, 11, 13, I4, and 15 sides,a process would be needed such pre- cisely as that just shown in the last article for a pentagon, except in the value of z and с, which are the only factors which require change for each individual case. Мо useful purpose, therefore, can be subserved by ex- hibiting the details of the process required for these several polygons. Тһе values of the constants required for the radii and for the areas of these polygons have been com- puted, and the results, together with those for the polygons treated in former articles, gathered in the annexed Table of Regular Polygons. REGULAR POLYGONS. Ж ” А SipEs | Ti E rt пола... е... о. cm Eo | +57735 -28868 *43301 LE SA Ctrageme 2.2.2 M ‚70711 * 50000 І-ООООО Б. PENON. es еек огоо Е | - 85065 68810 I:72048 ОТОО. . cessere oe со |! 1.00000 86603 2.59808 2s Мер рөп............... ЛШ. | 1-15238 I 03826 3:63301 8. OCZOM. ................‹ о | 11-30656 I:207II 4°82843 о, М@парОп.......... canes cee | r*46190 1:37374 6- 18182 19, Л®е саф Ой. Locros suede de ENS. | 1-61803 1-53884 7:69421 Lr. ЗАСЕБАН. п. о. е... оин | 1:77473 1.70284 9:36564 . se. ПБОЦНЕОН............ 2222... | 1193185 1.86603 | ІІ-109615 ІЗ. “DPC Ol. -o iosi eee tie жс 2-08020 2-02858 | 13:18577 I4. Tetradecagon...-... ЖЕ”... 2.24698 | 2-190604 | I5-5345I 15. Pentadecagon. ..............5.. 2-40487 2:3522І 17:64236 16. Несайесароп...............Ж®Ш НЕЗ СООДО 2.51367 20- 10936

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466 POLYGONS. In this table R represents the radius of the circumscribed circle; > the radius of the inscribed circle; 4 one of the sides, and A the area of the polygon. Ву the aid of the constants of this table, №, the radius of the circumscribed circle of any of the polygons named, may be found when a side of the polygon is given. For this purpose, putting ж for any constant of the table, we have— = T (154.) As an example: let it be required to find Ж, for a penta- gon having each side equal to 5 feet; then the above expres- sion becomes— . 5 x 0-85065,. 4* 25325. The radius will be 4 feet 3 inches and a small fraction. In like manner the radius of the inscribed circle will be— r = Т; (155.) and for a pentagon with sides of 5 feet, we have— r= 5 x 0.68810, r = 3:44095. Or, the radius of the inscribed circle will be 3 ft. 45, and а small fraction. Or, multiplying the decimal by 12, 3 ft. 5 in. 2 and a small fraction. The area of any polygon of the table may be obtained by this expression— Lie P tow (156.) and, applying this to the pentagon as before, we have— A --/5"х 1-720468, A = 43-012.

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EXPLANATION OF THE TABLE. 467 Or, the area of a pentagon having its sides equal to 5 feet, is 43 feet апа 4124 of a foot. By the constants of the table a side of any of its poly- gons may be found, when either of the radii, or the area, are known. When Ё is known, we have— „л І 15 known, we have— ы =. (158.) 24 When the area is known, we have— б- V A, (159.)

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SECTION XI.—THE CIRCLE. 443.—Cireles: Diameter and Perpendicular: Mean Proportional.—Let 4 В C (Fig. 294) be a semicircle. From C, any point in the curve, draw a line to 4 and another to B; then ABC will be a right-angled triangle (477. 352). Draw the line CD perpendicular to the diameter АВ; then C D will divide the triangle А BC into two triangles, ACD and CBD, which are homologous. For, let the triangle CBD be revolved оп D asa centre until its line CD shall come to the position Ж D, and the line DB oc- cupy the position D F, each in a position at right angles Q А Е р В FIG 294. to its former position, the point 2 describing the curve БЕ, and the point C the curve CZ, and each forming а ` quadrant or angle of ninety degrees. Since these points have revolved ninety degrees, therefore the three lines of the triangle C В D have revolved into a position at right angles to that which they before occupied ; hence the line E Fis at right angles to СВ, and (from the fact that A C B is а right angle) parallel with А С. Since the triangle E FD equals the triangle СЛ D. and since the lines of E FD are parallel respectively to the corresponding lines of А C D, therefore the triangles 4 C D and СВ D are homologous. Comparing the lines of these triangles and putting а = А В, у= CD, апа х = DB, we have—

Page 507

RADIUS FROM CHORD AND VERSED:' SINE. 469 ЖАО АС: АР, ала a 2 ye = 40 (а — ж). (160.) Or, in a semicircle, a perpendicular to the diameter terminated by the diameter and the curve zs a geometric mean, or mean proportional, between the two parts into which the perpendicular divides the diameter. 444.— Circle: Radius from Given Chord and Versed Sine.—Let А B (Fig. 205) be a given chord line and C Da versed sine. Extend C D to the opposite side of the circle; it will passthrough 7, the centre. Join 4 and С, also Е and „ав Mu . 4 FIG. 205. E. The line AD, perpendicular to the diameter C Æ, is a mean proportional.between the two parts CD and DE (Art. 443) ; or, putting a = A D, b = C D, and ~ equal the radius Ғ.Ж, we have— CU AMD à DE. d. 07-7. а? = 6 (27 — 6), а = 2 рф —– 6’, © тб” СРО, а? 4 6? Wm (161.)

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470 ЗЫРТ тнв СНЕ. Or: The radius of a circle equals the sum of the squares of half the chord and the versed sine, divided by twice the*versed sine. Another expression for the radius may be obtained ; for the two triangles CBD and СЕ В (Fig. 295) are homologous (Art. 443) and their corresponding linesin proportion. Put- ting f for CB, we have— CORB: CBE ОЧЕ, or— (а ж ЖИЗ, or— Wy 29 and— E Ж-- Vii (162.) = РА) Or: The radius of a circle equals the square of the chord of half the arc divided by twice the versed sine. 44.5, — Circle: Segment from Ordinates.— When the curve of a segment of a circle is required for which the radius can- not be used, either by reason of its extreme length, or be- Fic. 296. cause the centre of the circle is inaccessible, it is desirable to obtain the curve without the use of the radius. This may be done by calculating ordinates, a rule for which will now be developed. Let DCB (fig. 206) be a right angle, and A DB a cir- cular arc described from C as a centre, with the radius BC.= GD =e Drew’ P, parallel wath sD C, "and АС parallel with СВ. Now, іп the segment А DG, we have given .4 С, its chord, and D Е, its versed sine, and it is re-

Page 509

RULE FOR ORDINATES. 471 quired to find an expression by which its ordinates, as РР, may be computed. From Art. 416, we һауе-- ` PM?=CP?— СИЗ; or, putting for these lines their usual symbols— у= ек5 МО — у, now we have— EO DC Xu = FM-DC—DE, ЕМ = + — 0. Then we have— РЕ-РМ--ҒМ; or, putting 7 for P F and substituting for P M and FM their values as above, we have— -ғ-у-(ғ-9, and for y, substituting its value as above, we have— t= yr OX? (xD. —— (163.) Or: The ordinate in the segment equals the square root of the difference of the squares of the radius and the abscissa minus the difference of the radius and the versed sine. For example: let the chord A С (Fig. 296) in a given case equal 20 feet, and the versed sine, 4, or the rise D Е, equal 4 feet; and let the ordinates be located at every 2 feet along the chord line, A G. In solving this problem we require first to find the radius. | This is obtained by means of equation (141.)— Saw son "m wis )

Page 510

472 - THE CIRCLE. For a, half the chord, we have 10 feet ; for 4, the versed sine, we have 4 feet; and, substituting these values, we have— ; IO?--4? . 116 > 4 — == r 2X4 8 The radius equals— ,14*5 The versed sine equals— 4-0 (r -- 0) == 10:5 The square of 14-5, the radius, equals 210-25. Now we have, substituting these values in equation (163.)— t= МОТО — Ёё — 10-5. The respective values of x, as above required, are о, 2, 4, 6, 8 and то Substituting successively for x one of these yalues, we shall have, when— | r= орїш= V 210-25 — 0? —10:5 = 4- Ж: 2; {= Wy 210-25 — 2? — 10.5 = 3.8614 x= 4) 1— V 210-25 — 4* — 10-5 = 3:4374 x= 6:98 —N es GENERE rete = 299694 2 =й = demo secar = 10-5 195954 x= 10; / — 210.25 — I0? — 10-5 — 0-0 Values for 7 may be taken at points as numerous as desira- ble for accuracy. In ordinary cases, however, they need not be nearer than in this example. | After the points are secured, let a flexible piece of wood be bent so as to coincide with at least four of the points at a time, and then draw the curve against the strip. 446.—Cirele: Relation of Diameter to Circumferenee, —In A77. 439 it is'shown that the area of a polygon equals the radius of the inscribed circle multiplied by half of a side of the polygon and by the number of the sides; or,

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TO FIND THE CIRCUMFERENCE. 473 HM =з} x z = CZ or, the area equals half the radius by a side into the number of sides; or, half the radius into the periphery of the polygon. Now, if a polygon have very small sides and many of them, its periphery will approxi- mate the circumference of the circle inscribed within it; in- deed when the number of sides becomes infinite, and conse- quently infinitely small, the periphery and circumference become equal. Consequently, for the area of the circle, we have— y 50 (164.) where c represents the circumference. By computing the area of a polygon inscribed within a given circle, and that of one circumscribed about the circle, the area of one will approximate the area of the other in proportion as the number of the sides of the polygon are increased. | For example: if polygons of 4 sides be inscribed within and circumscribed about a circle, the radius of which is 1, the areas will be: respectively 2 and 4. Ifthe polygons have 16 sides, the areas are each 3 and a fraction, the fractions being unlike; when they have 128 sides the areas are each 3-14 and with unlike fractions; when the sides are increased to 2048, the areas each equal 3-1415 and unlike fractions, and when the sides reach 32768 in number the areas are equal each to 3-1415026, having like decimals to seven places. Тһе computations have been continued to 127 places (Gregory's “ Math. for Practical Меп”), but for all possible uses in building operations seven places will befound to be sufficient. From this result we have (һе diameter in proportion to the circumference as I : 3: 1415926, ОГ as— 1 :$- 0591, I : 3: I4T59, І 3 Mi 6. Of these proportions, that one may be used which will give

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474 | THE CIRCLE. a result most nearly approximating the degree of accuracy required. For many purposes the last proportion will be sufficiently near the truth. | For ordinary purposes the proportion 7 : 22 is very use- ful, and 15 correct for two places of decimals; it fails in the third place. The proportion 113°: 355 is correct to six places of deci- mals. “Бог the’ quantity -3-1415926 putting the Greek letter т (called ру), and 27 = d for the diameter, we have— “с-та. (165.) To apply this: in a circle of 50 feet diameter, what is the:circumference ? € = 3-1416 x 50 £o 1577900 //. If the more accurate value of z be used, we have— С 78714159807x 50, € = 157-07963. The ‘difference between the two results 15 0.00037, which for all ordinary purposes, would be inappreciable. By the rule of 7 : 22, we have— € — 90x 42 — 157 -1428571, an excess over the more accurate result above, of 0-0632271, which is about # of an inch. By the rule of 113 : 355, we have— c = 50 x - 157-079646. This result gives an excess of only 0-000016 ; it is sufficiently near for any use required in building. From these results we have these rules, namely: To obtain the circumference of a circle, multiply its diameter by

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TO FIND THE AREA. 475 22, and divide the product бу.ў ; or, more accurately, multiply the diameter бу 355 and divide the product by, 113; ог, by mul- tiplication only, multiply the diameter by 3-1416; or, бу 3-141591; or, ду 3:1415926; according to the degree of accuracy required. And conversely: To obtain the diameter from the cir- cumference, multiply the circumference by 7 and divide the product бу 22; or, multiply by 113 and divide by 355; or, di- vide the circumference by 3-1416; or, бу 3-141591; or, by 3- 1415926. 447.—Cirele: Length of an Arc.—Considering the cir- cle divided into 360^, the length of an arc of one degree in a circle the diameter of which is unity may be thus found. The circumference for 360° is 3- 14159265 ; ; ae = 0-00872664625 ; - which equals an arc of one degree in a circle having unity as its diameter; or, for ordinary use the decimal 0-008727 ог 0-0087} may be taken; or putting а for the arc and ғ for the number of degrees, we have— а = 0:00872665 dg. . (166.) Wherefore: To obtain the length of an arc of a circle; multiply the diameter of the circle by the number of degrees in the arc, and by the decimal о-00873, or, instead thereof, dy 0-008727. 448.-» 0r— A=4obur, where 6 equals a side of the polygon and z the number of sides; so that 2 z equals the perimeter of the polygon. Now, if for the perimeter of the polygon there. be sub-

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476 THE'CIRGLE. > stituted the circumference of the circle, we shall have, put- ting for the circumference 3- 1416 d, or, т d (Art. 446)— ades 3 mdr, in which v is the radius. Since 2 7 = d, the diameter, апа d r= Nue have— m А= ўл 2 = А = їлп 2°. And since— ! т = 3- 14159265, } л = o. 78539816, or— lz = 0.7854, nearly. Therefore— A dn. (167.) Or: The area of a circle equals the square of the diameter mul- tiplied by 0-7854. D Fic. 297. 2 As an example, the area of a circle 10 feet in diameter is found thus— ТО x ТӨТ. 100: 100 x 0.7854 = 78-54 feet. 449,— Cirele: Area of a Sector.—The area of A 5 CD (Fig. 297), a sector of a circle, is proportionate to that of the whole circle. For,as the circumference of the whole circle is to its area, so is the arc A BC to the area of A BCD.

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AREA OF SECTOR. 477 The circumference of a circle is (165.) С = та. The area of a circle is (167.) А = ·7854 2? For the arc A BC puta, and for the area of 4 B CD put s. Then we have from the above-named proportion— - mue: Том ага: то: 5, 78544” н Ж The coefficient 0.7854 is E (Art. 448). Therefore, multiplying the fraction by 4, we have— = л О Eg" t or— S =tdg=4 wa. (168.) Wherefore: To obtain the area of a sector of a circle, multiply a quarter of the diameter by the length of the are. Thus: let A D equal 10; also let АВ Ca, equal i12. Then the area of A B C D is— 9 = 60716702, S = 60. The length of the arc may be had by the rule іп 472. 447. article, A BCD (Fig. 297) 1s called the sector of a circle. Of this the portion included within A Е C B is a segment of a circle. The area of this equals the area of the sector minus the area of the triangle A DC; or, putting 77 for the area of the seg- ment, S for the area of the sector, and T for the area of "d triangle, then— М --.5-- 7. Putting c for 4 C (276 297) and A oY D E, then Т = > A. In the last article, s = 47a, in which a = the length of the

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478 THE CIRCLE. arc- A-B С. Substituting this value of s in the above, we have— ar—ckh. a с MT yh (169.) Or: When the length of the arc is known, also that of the chord and the perpendicular from the centre of the circle, then the area of the segment equals the difference between the product of half the arc into the radius, aud half the chord into 115 perpendicular to the centre of the circle. But ordinarily the length of the arc and of the chord are unknown.. If in this case the number of degrees contained between the two radii, DA, DC, are known, then the area of Fic. 298. the segment may be found by a rule which will now be de- veloped. In Fig. 298 (a repetition of Zzg. 297) upon D as a centre, and with D F = unity for a radius, describe the arc 77 F. Then С Fis the sive of the angle C D B, and DG is the co- sine; and we have— DIE ta Geb. Di. : EK, or— с ығ: 2. 2 Again— O usc бы. ОДЕ, Or — I: COS:: 7: # == > cos.

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RULE FOR AREA OF SEGMENT. 479 i By equation (166.) we have— a = 0-00872665 dg, | in which a is the length of the arc; g the number of degrees contained in the arc ; and Z is the diameter of the circle. Since d = 2 7, therefore— a = 0:0174533 7 8. Putting B for the decimal coefficient, we have— а == The expression (169.), by substitution of values as above, becomes— a с М = =r — =й, 2 2 Br Я ШИ = E y r sin, x 7 COS. М = 4 Врт? — sin. cos. r’ M = r'($.Bg — sin. cos.) M = r'(0-00872665 g — sin. cos.) (170.) Or: The area of a segment of a circle eguals the square of the radius into the difference between 0-00872665 times the number of degrees contained in the arcof the circle, and the product of the sine and cosine of half the arc. When the number of degrees subtended by the arc is unknown, or tables of sines and cosines are not accessible, then the area may be obtained by equation (169.), provided the chord and versed sine are known ; but before this equa- tion can be used.for this purpose, expressions giving their values in terms of the chord and versed sine must be ob- tained, for a, the arc, 7, the radius, and 2, the perpendicular to the chord from the centre of the circle. For the value of the arc we have (from * Penny Сусі.,” Art. Segment) as a close approximation— а = } (8 f—c).

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480 THE CIRCLE. By equation (162.) we have— T 2v Then— Же- 2 — or— | кы: = 2v Substituting these values in equation (169.) we have— м= |68 7 — Jia — — ‚(нь v) |. (aue) This rule is the rule (169.) expanded. The written rule for equation (169.) may be used, substi- tuting for “half the arc," one sixth of the difference between eight times the chord of half the arc and the chord (or т of 8 times A B, Fig. 298, minus A C, the chord). Also substitute for “ the radius,” the square of the chord of half the arc divided by twice the versed sine. Also, for.**zzs perpendicular to the centre of the circle,” substitute, the quotient of the square of the chord of half the arc divided by twice the versed sine, minus the versed sine. When the arc is small the curve approximates that of a parabola. In this case the equation for the area of the par- abola, which is quite simple, may be used. It is this— 2 М = –со. (172). 9 Or, in segments of circles where the versed sine is small in comparison with the chord, 7Ze area equals approximately two thirds of the chord into the versed sine.

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"SECTION XII.—THE ELLIPSE. 451. —Ellipse: Definitions.—Let two lines, P F, P F” (Fig. 299), be drawn from any point P to any two fixed points F F', and let the point P move in such a manner that the sum of the two lines, P F, РЕ”, shall remain a constant quantity ; then the curve PM KO GA DB P, traced by P, will be an Ellipse; the two fixed points 2) 2”, the Foci; the point C at FIG. 299. the middle of FF’, the centre; the line 4 M drawn through F F' and terminated by the curve, the Major or Transverse Axis; the line ВО, drawn through C and at right angles to A M, the Minor or Conjugate Axis; the line GP, drawn through P and C and terminated by the curve, the Diameter to the point P; the line D К drawn through C, parallel with the tangent P 7, and terminated by the curve, the diameter Conjugate to PG; the line E ZA drawn parallel with DK is a double ordinate to the abscissas G H and А P of the di- ameter GP (E H = БК); the line 7 Z drawn through Fata

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482 THE ELLIPSE. right angle to А M and terminated by the curve, the Param- eter, or Latus Rectum. When the point P reaches and coincides with В, the two lines P F and PF’ become equal. The proportion between the major and minor axes de- pends upon the relative position of F, F’, the foci; the nearer these are placed to the extremities of the major axis the smaller will the minor axis be in comparison with the major axis. The nearer F, F” approach C, the centre, the nearer will the minor axis approach the length of the major axis. When 2, 2 reach and coincide with the centre, the minor axis will equal the major axis, and the ellipse will become a P A С N B FIG. 300. Sinele. Themswe hye? A= PA P C= А С» From this we learn PF 4 PF'=2 А C= А М; also, when PF— РР”, tihemP P= BF=— AC From this we may, with given major and minor axes, find the position of “and F’. Todo this, оп В,ав a centre, with A C for radius, mark the major axis at F and Z”. 452.—Ellipse : Equations to the Curve.—An equation to a curve is an expression containing factors two of which, called co-ordinates, measure the distance to any point in the curve. For example: in a circle it has been shown (477. 443) that PN is a mean proportional to 4 Vand МВ. Or, puie a NY y= PN, and a 594 B, we Мауе- КЛ NUM t: Л Жа. or кк sy tg MS, or— y" =x (а — 24):

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EQUATION TO THE ELLIPSE. 483 This is the equation to the circle having the origin of x and y, the co-ordinates at A,the vertex of the curve: It wall be observed that the factors are of such nature in this equa- tion, that it may be employed to measure the distance, rect- angularly, to P, wherever іп the curve the point P may be located. Dy this equation the rectangular distance to any and every point in the curve may be measured ; or, having the curve and one of the lines x or y, the other may be com- puted, From this example, the nature and utility of an equation to any curve may be understood. The equation to the ellipse having the origin of co-ordinates at the vertex, is similar to that for the circle. In the form usually given by writers on Conic Sections, it is— ` 2 2 ф y --а(2ал-- x"), (173.) in which a= A C (Fig. 299); 2 = B C; x equals A №, and у = RN. If, as before suggested, the toci be drawn towards the cen- tre and finally made to coincide with it, the minor axis would then become equal to the major axis, changing the ellipse into а circle. [n this case, the factors e and біп the equation would 2 TC Р become equal; and the fraction == would equal 4, = and hence the equation would become— лаға - 37 or— у= ena — ж); precisely the same as in the equation to the circle above shown. The 2a of this equation is equivalent to a of the circle; for « in the ellipse represents only half the major axis; while in the equation to the circle а represents the diameter. The relation between the ellipse and the circle is thus shown: indeed, the circle has been said to be an ellipse in its extreme conditions.

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484 THE ELLIPSE. і 453.—Ellipse: Relation of Axis to Abscissas of Axes, — Multiplying equation (173.) by а" we have— C EN RN b? (2 ах — ж)) or— a y zm х (2 а — ж). 4 These four factors may be put in a proportion, thus— UTE bam = ey Ж representing — Жа” C ues VM Р, Or: The rectangle of the two parts into which the ordinate divides the axis majoris zz proportion to the square of the ordinate, as the square of the semi-axis major is to the square of the semt-axis minor. Fic. 301. It is shown by writers on Conic Sections that this rela- tion is found to subsist, not only with the axes and ordinate, but also between an ordinate to any diameter and the ab- scissas of that diameter ; for example, referring to Fig. 299— Ш СР CHUP: p If A B^ P'M (Fig. Зот) be a semi-circle, then (422. 443) PN UNPM. Substituting this value of A Vx N M in— AC: BC PN’: PN’, we have— AC:BC::P'N: PN:

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RELATION OF TANGENT TO AXIS. . 485 Or: The ordinate in the circle is in proportion to its correspond- ing ordinate in the ellipse, as the semi-axis major is to the semi- axis minor, or as the axis major is to the axis minor. 454.— Ellipse: Relation of Parameter and Axes.—The equation to the ellipse when the origin of the co-ordinates is at the centre is, as shown by writers on Conic Sections, thus— а” уы 22-4 +, (174.) or— aye В (a° —a If x’ equal С (Fig. 299) then the ordinate will be located at F F, and —À 2 «est ЫЗ; | —-e-p (UNIVERSITY Then— а — x'* = а? — (a° — b CALIFORNIA: | = а — а + 0°, a*—x'* = 6’, This is shown also by the figure. Substituting in the above this value of a° — 2/*, we have— Gy =) ОМ од. From dh taking the square root— a je or— LEGEN VOY Y Now y, located at F 7, is the semi-parameter; hence we have the semi-minor axis a third proportional to the semi- major axis and the semi-parameter. Or: The parameter is а third proportional to the two axes of an ellipse. 455.—Ellipse: Relation of Tangent to the Axes.—Let T T' (Fig. 301) be a tangent to Р, a point in the ellipse; then, as has been shown by writers on Conic Sections— Cle И, or— CM ЄЛ. ЭС: Ci.

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486 THE ELLIPSE. Or: The semi-major axis is a mean proportional between the ab- scissa C N and C T, the part of the axis intercepted between the centre and the tangent. This relation is found also to subsist between the similar parts of the minor axis; for— ПҚ С” = Ср. This relation affords an easy rule for finding the point 7, or 7"; tor from the above we have— | C or, putting 2 for C 7, we have— 9" | das 2 (175-) or— b? £! = —. 176.) - (176. Since the value of 7 is not dependent upon y nor upon 2, therefore 7 is constant for all ellipses which may be de- scribed upon the same major axis A M; and since the circle is an ellipse (Art. 452) with equal major and minor axes, therefore rule (175.) is applicable also to a circle, as shown іп Fig. 301. The equation (175.) gives the value of z = C7: From this deducting CNV = ж, we have N T, the subtangent, or— С ЕЛ == ус lonr a. or, substituting for 2 its value in (175.), we have— | B zd | (177.) Or: Тһе subtangent to an ellipse eguals the difference between the quotient of the square of the semi-major axis divided by the abscissa, and the abscissa ; the origin of the co-ordinates being at C, the centre. |

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AXES TO CONJUGATE DIAMETER. 497 456.—Ellipse : Relation of Tangent with the Foci.—Let the two lines from the foci to P (Fig. 302), any point in the ellipse, be extended beyond P. With the radius РЕ” de- scribe from Р the arc F’ G, and bisect it in М. Then the line P 7, drawn through Æ, will. be a tangent to the ellipse at 2. This has been shown by writers on Conic Sections. The construction here shown affordsa ready method of drawing a tangent. And from the principle here given we learn that a tangent makes equal angles with the lines from the tangential point to the two foci. | For, because G H = H F', we have the angle FP H = HPG. Тһе angles А PG and KPF are opposite, and hence (477. 344) are equal; and since the two triangles F'PHand KP Fare each equal to HP G, therefore F' PH and KPF are equal tó each other. Or: А ¢angent to an ellipse makes equal angles with the two lines drawn from the point of tangency to the two foct. | Experience shows that light shining from one focus is reflected from the ellipse into the other focus. It is for this reason that the two points Fand 2 are called /осг, the plu- | ral of focus, a fireplace. 457.— Ellipse : Relation of Axes ‘to Conjugate Diame- ters.—Parallel with X 7 (Fig. 302) let D Е be drawn through

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488 THE ELLIPSE. C, the centre, and Z Q through ¥, one end of the diameter from the point P. Parallel with this diameter P¥ draw L К and О А through the extremities of the diameter D Ж. Then D E is a diameter conjugate to the diameter P F, and КЕ, R'Q, О L, and "P" are tangents at the extremities of these conjugate diameters. | Now it is shown by writers on Conic Sections (Fig. 302) that— | ; Tt — v» ат b? = a 24 0%. Or: The sum of the sguares of the two axes equals the sum of the sguares of any two conjugate diameters. From this it is also shown that the area of the parallelo- gram А C equals the rectangle A C x BC; or, that a paral- lelogram formed by tangents at the extremities of any two conjugate diameters 1s equal to the rectangle of the axes. 458.—Ellipse; Area.—Let Æ equal the area of an ellipse; A the area of a circle, of which the radius 2 equals the semi- major axis of the ellipse, and let 0 equal the semi-minor axis. Then it has been shown that— БӘМ Du а б. Б? a The area of a circle (Art. 448) is— | A mdr weigh and when the radius equals a— E) cr This value of 4, substituted in the above equation, gives— И | = 7 a -, a Жі Po ab. (178.)

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PRACTICAL SUGGESTIONS, 489 Or: The area of an ellipse equals 3.141591 times the product of the semi-axes у or 0-7854 times the product of the axes. 459. ——Ellipse: Practical Suggestions.—In order to de- scribe the curve of an ellipse, it is essential to have the two axes; or, the major axis and the parameter ; or, the major axis and the focal distance. If the two axes are given, then with the semi-major axis for radius, from B (Fig. 299) as centre an arc may be made at Fand F’, the foci; and then the curve may be described by any of the various methods given at Arts. 548 to 552. If the major axis only and the parameter are given, then (Art. 454) since— 0% = ay, we have— b= Vay. (179) Or: The semtminor axis of an ellipse equals the square root of the product of the semi-major axis into the semt-parameter. Then, having both of the axes, proceed as before. If the major axis and the focal distance are given, or the location of the foci; then with the semi-major axis for ra- BAP CTW M Fic. 303. dius and from the focal points as centres, describe arcs cut- ting each other at B and О (Fig. 299). Тһе intersection of the arcs gives the limit to BO, the minor axis. With the two axes proceed as before. Points in the curve may be found by computing the length of the ordinates, and then the curve drawn by the side of a flexible rod bent to coin- cide with the several points., For example, let it be required to find points in the curve of an ellipse, the axes of which are 12 and 20 feet; or

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490 THE ELLIPSE. the semi-axes 6 апа 10 feet, or 6 x 12 = 72 inches, and Io x 12 = 120 mchés. Fix the positions of the points Л VV’, etc., along the semi- major axis C M (Fig. 303) at any distances apart desirable. It is better to so place them that the ordinates when drawn shall divide the curve В PM into parts approximately equal. If CM be divided into éight parts as shown, these parts measured from C will be well graded if made equal severally to the following decimals multiplied by С 27. In this case C M = 120; therefore— ~ am e — mS o ВЕ а 5 CN моеот Na СМ” = 120 x 0-625 = 75. — ж МӘ - сол” Moo. = I20 xX49-95 == 102- = x’ 1209/50992 г 1205009075. = m7 =r [201 -© -=o Ег. The equation of the ellipse having the origin of co-ordi- nates at the centre (Art. 454) 15— a? у = beue =), or, dividing by а%— "E ОЛЕНИ 9 /д\ o = PL —« ) | j/ 7% or— = PE (22-2 б or— "dan M а ee (180.) - in which а and 0 represent the semi-axes. Substituting for these their values in this case, we һауе-- ДО” =_= t cA S D /9 У =0-°6 414400 — v’,

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LENGTH OF ORDINATES. 49I Now, substituting in this equation the several values of x? successively, the values of the corresponding ordinates will be obtained. For example, taking 36, the first value of ж, as above, we have— y= 0-6 714460 — 36% y= 68-684 ; y = 0-6 Ут4400 — 57? у = 63:359; and so in like manner compute the others. The ordinates for this case are as follows, viz. : When xs = о, y = 72-6 4 ox = 36,09) = 68:054 “o a = 57, y = 63-359 “ X = 75, у = 56-205 “ x = 90, у = 47:624 * т 29 102, 7 = 377428 ao Е ru 27 “ X = 117, у = 15-999 “ Ur = 120 у = Sere The computation of these ordinates is accomplished eas ly by the help of a table of square roots and of logarithms. For example, the work for one ordinate is all comprised within the following, viz. : y = 0-6 V14400 — 36? = 68.684. 120° = 1469 je == ДЕ 13104 = 4:1174039 Half = 2-0587020 o. 6 7701613 68.684 = 1-8368533.. The logarithm of 13104 = 41174030. The half of this is - the logarithm of the square root of 13104. То the half log- arithm add the logarithm of 6-6; the sum is the logarithm cf 68-684 found in the table (see A77. 427).

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SECTION XIIL—THE PARABOLA. 460.—Parabola: Definitions.—The parabola is one of the most interesting of the curves derived from the sections of a cone. The several curves thus produced are as fol- lows: When cut parallel with its base the outline is a circle ; when the plane passes obliquely through the cone, it is an ellipse ; when the plane is parallel with the axis, but not in the axis, it is a hyperbola; while that which is produced by . Ес. 304. a plane cutting it parallel with one side of the cone is a parabola. Let the lines Z M and L N (Fig. 304) be at right angles ; draw CF B parallel with LM; make LQ = LF; draw ОВ parallel with LF; then FB = Б О. Now let the line AL move from FL, but remain parallel with it, and as it moves let it gradually increase in length in such manner that the point A shall constantly be equally distant from the line Z M and from the point F. Then A BP, the curve described by the point А, will be a semi-parabola. For example, the lines FB and В Q are equal; the lines F Pand PM are equal, , and so of lines similarly drawn from any point in the curve ABP. Let PN be drawn parallel with LM; then for the

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EQUATION TO THE CURVE. 493 point P, A Nis the abscissa and NP its ordinate (see Art. 452). The double ordinate C В drawn through F, the focus, is the Parameter. А Fisthe focal distance. А is the vertex of the curve. The line Z M is the directrix. 4.61.—Parabola : Equation to the Curve.—In Fig. 304 FPN is а right-angled triangle, therefore— NP’ = Рр FN’; but— FP = i PE апа-- FIV DM М Therefore— NP'—HACAEPT ANE NEN or— у= (2+0) r eT р being put for the distance Z F = FB (see Art. 452). Ав in Arts. 412 and 413, we have— (ж + 4 ) = 2" p44 2258 (x —4)*=2*—prtip? Lu 1.7 А (181.) by subtraction. This is the usual equation to the parabola, , in which we have the rule: The square of the ordinate equals the rectangle of the corresponding abscissa with the param- ever. From (181.) we have— ЖЕЛЕ 409 Os or: The parameter is a third proportional to the abscissa and its corresponding ordinate. 462.— Parabola : Tangent.—From M, any point in the directrix, draw a line to 25 the focus (Fig. 305); bisect M F in А, and through А draw UT perpendicular to M F, then the line ZU will be a tangent to the curve. For, draw M D

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494. THE PARABOLA. perpendicular to Z V, and from P, the point of its intersection with the line Г U, draw а line to 2, the focus; then, because R Pisa perpendicular from the middle of M F, M PF is an isosceles triangle, and therefore the lines 77 Pand FP are equal, or the point Pis equidistant from the focus and from the directrix, and therefore is a point in the curve. To show that the line 7 U touches the curve but does not pass through it, take U, any point in the line 7 0, other than FIG. 305. the point P; join U to Mand to F. Then, since VU is a point in the line 7 U, M U F, for reasons above given, is an isosce- les triangle ; from U draw U V perpendicular to Z V. Now, if the point Ube also in the curve, the lines VV and UF, by the law of the curve, must be equal; but UF, as before shown, is equal to UM, a line evidently longer than UV; therefore, it is evident that the point U is not in the curve. A similar absurd result will be reached if any other point than the point U in the line U T be assigned, excepting the

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RULE FOR THE TANGENT. І 495 point P. Therefore the line ГР touches ће curve in only one point, Р; hence it 1s a tangent. Parallel with Z V, from A, draw A S, the vertical tangent. Now A 5 bisects M F or intersects it in the point №. For the two right-angled triangles FZ M and FA R are homolo- gous; and because РА = А L, by construction, therefore тк = gM. Or: The vertical tangent bisects all lines which can be drawn from the focus to the directrix. The lines Pf and FT are equal; for the lines M P and NT being parallel, therefore the alternate angles M PT and V ТР are equal (A77. 345); and because the line P 7 bi- sects M F, the base of an isosceles triangle, therefore the angles M P Гапа FP Г are equal. We thus have the two angles V TP and FP T each equal to the angle ZAMAM; therefore the two angles № 7 P and FP T are equal to each other; hence the triangle PF Tis an isosceles triangle, hav- ing the points Гапа Р equidistant from 7, the focus. Also because the-line M F is perpendicular to P 7, there- fore the line MF bisects the tangent PT in the point 5. And because TR = RP, therefore, comparing triangles maf and ZPO, TF = A0. The opposite angles M P T and U PD made by the two in- tersecting lines U Гапа M D (Art. 344) are equal, and since the angles /Р7 and FP T are equal, as before shown, therefore the angles F P Тапа UPD are equal. Itis because these two angles are equal, that, in reflectors, rays of light and heat proceeding from. F, the focus, are re- flected from the parabolic surface in lines parallel with the axis. | For an equation expressing the value of the tangent, we have— ТР = Bias та (ere у", = V4 a? T^ (182.) Or: The tangent to a parabola equals the square root of the sum of four times the square of the abscissa added to the square of the ordinate.

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496 THE PARABOLA. 4.63.— Parabola: Subtangent.—The line T N (Fig. 305), the portion of the axis intercepted between 7, the point of intersection of the tangent, and JV, the foot of a perpendicu- lar to the axis from Р, the point of contact, is the subtangent. The subtangent is bisected by the vertex, or TA = AN. For, the two triangles TRA and TPN are homologous ; and, as shown in the last article, the line M F bisects P.T in x; or ТТТ Therefore, we have— gu Ted Uu MP uw N, (a xY NE TAr bee but— T Жаса ДЫР. therefore— РХ TV = TAK X T Tay TH Or: The subtangent of a parabola is bisected by the vertex ; or is equal to z«zce the abscissa. And because of the similarity of the two triangles TR A and ГР N, as above shown, we have— NW =e ЛИС mAAR. Or: The ordinate equals twice the vertical tangent. 464.—Parabola: Normal and Subnormal.—The line РО (Fig. 305) perpendicular to P T, is the normal and N O, the part of the axis intercepted between the normal and the ordinate, is the subnormal. For the normal, from similar triangles, we have— Ж», P. Тб КР О NP хл ie o noc pa 2%

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DIAMETER AND RECTANGLE OF BASE. 497 Or: The normal equals the rectangle of the ordinate and tan- gent, divided by twice the abscissa. The subnormal equals half the parameter. For (181.)— y —2pz, Or— | 22-72 22 AV. Dividing by 2 А № gives— | | а ^ mw wie | Ерс (А.) In the similar triangles (477. 443) OPN and PTN, we have— VO: VP T: NP: ae . ло US M As shown in the previous article, V 7 =2 A N; therefore— NOn NP. (B.) 2247 Comparing equations (А.) апа (B.), we have— NO = ЖБ. Or: The subnormal of a parabola equals half the parameter, a constant quantity for the subnormal to all points of the curve. . 465.—Parabola: Diameters.— In the parabola BAC (Fig. 306), P D, a diameter (a line parallel with the axis) to the point P, is in proportion to В Dx D C, the rectangle of the two parts into which the base of the parabola 15 divided by.the diameter. This may be shown in the following manner: - DP Æ EIME РИ = ID. (A.)

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498 THE PARABOLA. For Е А we have, taking the co-ordinates, for the point C, (181.)— y* —2fpz, | y (nc 22 — Xx Өр... ЕС: or— on = (В.) Ғе А В Е ‚р е Fic. 306. For VA we have, taking the co-ordinates to the point P, (181.)— | y'-—2psz, oe Е Or— 2) oop - NP? er 72) = NWA. (С) Using these values (B.) and (С.) іп (A.), we have— JD Р ЕЗГІ A, ' ac. vo Eme 2p 2p 2p If 7 be put for BC and z for DC, then— NP—-EC—DC-—-il—nm,

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TO FIND THE CONSTITUENT PARTS. 499 and— p ESSET GUY then (Art. 413)— ppatl BE lain, 2 2p or (Art. 415)— Hip Же жы an Ие 22 i 3 K рор aa 2 p : : рр-“4-Ф L 27. 3 рр Б СхХВВ 290 —— Now, since 2 2, the parameter, is constant, we have D P, the diameter, in proportion to D C x B D, the two parts of the base. | Putting Z for the diameter, we have— n (l— n) = CIEN (183.) Or: The diameter of a parabola equals the quotient obtained by dividing the rectangle-—formed by the two parts into which the diameter divides the base—by the parameter. It has been shown by writers on Conic Sections that a diameter, P ¥ (Fig. 307), to any point Pin a parabola bisects all chord lines, SG, D Е, etc., drawn parallel with the tan- gent to the point P; the diameter being parallel with the axis of the parabola. 4.66.—Parabola : Elements.—From any given parabola, to find the axis, tangent, directrix, parameter and focus, draw any two parallel lines or chords, SG and DE (Fig. 307), and bisect them in / and 7; through these points draw 9 P; then X P will be a diameter of the parabola—a

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500 THE PARABOLA. line parallel with the axis. Perpendicular to P 5 draw the double ordinate РО and bisect it in M; through / and par- allel with P5 draw TO, cutting the curve in A; then TO will be the axis. Make A T= А М, join Гапа P; then ТР will be the tangent to the point 2; from P draw PO per- pendicular to P 7; then PO will be the normal, and JV O the subnormal. With AWO for radius, from № as a centre, describe the quadrant OR; draw RC parallel with А O, cutting the curve Fic. 307. in C; from C draw СР perpendicular to A O, cutting A O in F; then F will be the focus and C B the parameter. Make АГ = AF; draw LM perpendicular to TO; then LM will be the directrix. Extend P7 to meet LM at М; join P and , F; then, if the work has been properly performed, FP will equal M P. 4.67.—Parabola ? Described Mechanically.— With VP (Ғғ. 308) а given base, and VA a given height, set perpen-

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THE CURVE DESCRIBED MECHANICALLY. SOI dicularly to the base, extend VA beyond A, and make 4 T equal to VA; join Zand P; from P perpendicularly to TP draw PO; bisect O N in R; make А Г and A F each equal to VR; through L, perpendicular to LO, draw D E, the di- rectrix. [ Let the ruler C DES be laid to the line DZ, then with ЎС Н, а set-square, the curve may be described іп the fol- lowing manner: Placing the square against the ruler and with its edge Fic. 308. ҰН coincident with the line MP, fasten to it a fine cord on the edge РЕ, and extend it from P to 7, the focus, and se- cure it to a pin fixed in F. The cord FP will equal the edge M P. . To describe the curve set the triangle 2 G H at M PE, slide it gently along the ruler towards 2, keeping the edge ¥G in contact with the ruler, and, as the square is moved, keep the cord stretched tight, holding for this pur- pose a pencil, as at А, against the cord. Thus held, as the square is moved the pencil will describe the curve. That this operation will produce.the true curve we have but to

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502 THE PARABOLA. consider that at all points the line FK will equal Ж 9, which is the law of the curve (A77. 460). 468.—Parabola: Described from Points.—With given base, V P (Fig. 309), and given height, А JV, to find the points D, F, M, etc., and describe the curve. Make A 7 equal to AN (Art. 462); join T and P; perpendicular to ГР draw Frc. 309. PO; make АВ equal to twice VO; take C, any point in the axis A О, and bisect P С in ¥; on ? asa centre describe the semi-circle BCG cutting А L, a perpendicular to BO in С; on 4 C and А С complete the rectangle AC DG. Then D is a point.in the curve. Take М, another point in the axis; bisect B Hin К; on K as a centre describe the semi-circle БЕН cutting A L in £; this by £E F and AF, gives F, an-

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THE CURVE DESCRIBED FROM ARCS. 503 other point in the curve; in like manner procure M, and as many other points as may be desired. This simple and accurate method of obtaining points in the curve depends . upon two well-established equations; one, the equation to the parabola, and the other, the equation to the circle. The line GD, an ordinate in the parabola, is equal to AC, an” ordinate in the circle ВСС; А С, the abscissa of the para- bola, is also the abscissa of the circle; in which we have (Art. 443)— AG $40: €: ME “ 425 з, > у= х(а — ж). For the parabola, we have (181.)— убори. Comparing these two е. we have— х(е- 2) 20 а--л-- 27, or— BG—AG=2p. By construction 4 В equals 2 VO, or twice the subnor- mal; the subnormal (477. 464) equals half the parameter. Hence, twice the subnormal equals the parameter—equals 25. Therefore, the method shown in / 22. 309 is correct. 469.— Parabola: Described from Ares.—Let VP (Fzg. зто) be the given base and A N the given height of the par- abola. Make A T (Art. 462) equal AN. Join Т to P; draw | РО perpendicular to PT; bisect V O in R; make AZ and АҒ each equal to VR; then Г М, drawn perpendicular to T O, will be the directrix. Parallel to Z M draw the lines DD C, etc., BM discretion. Then with the distance BZ for

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504. THE PARABOLA. radius, and on Fas a centre, mark the line B D with an arc; the intersection of the arc and the line will be a point in the curve (Art. 460). Again, with CZ for radius and on Fas a centre, mark the line CE with an arc; this gives another point in the curve. In like manner, mark each horizontal Fic. 310. line from Fas a centre by a radius equal to the perpendicu- lar distance between that line and Z M, the directrix. Then a curve traced through the points of intersection thus ob- tained will be the required parabola. 4770.—Parabola: Described from Ordinates, —With а given base, JV P (Fig. 311), and height, A N, a parabola may be drawn through points F, H, C, etc., which are the extrem- ities of the ordinates 2 7, CH, D G, etc.; the lengths of the ordinates being computed from the equation to the curve, (181.)— gea For any given parabola, in base and height, the value of *

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THE CURVE FORMED BY ORDINATES. 505 №2 may be had by dividing both members of the equation by 2 х; by which we have— PETIT me от 2427; (23 from which, N P and A N being known, 2 may be computed. With the value of 2, a constant quantity, determined, the equation is rendered practicable. For, taking the square root of each member of equation (181.), we have— - у= ГУ орх. (B.) which by computation will produce the value of y, for every assigned value of x, as A Б, A C, A D, etc. m Oo D» Fic. 371, As an example: let it be required to compute the ordi- nates in a parabola in which the base, VP, equals 8 feet, and the height, A N, equals то feet. With these values equation (A.) as above becomes— А — —==» EGO x. E. nei 8.24 £g 2AN 2x10 20 3:2; 2p шб.4.

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506 "HE PARABOLA. Then, with this value in (B.) as above, we have, for each or- dinate— У- y 6.42. In order to assign values to x, let А V be divided into any number of parts at 5, C, D, etc., say, for convenience in this example, in ten equal parts; then each part will equal one foot, and we shall have the consecutive values of x — 1, 2, 3, 4, etc., to 10, and the corresponding values of y will be as follows. When— r= 1, у= y6.4x I= б.д = 2-5297 = BF, r= 2, у= y6.Ax 2= ү 12.8 = 3:5777 = CH, z= Bey = E 3 = Ve 2 a е DG, r= 4 у= VO-4x 4= V 25.6 = 5:0596 = Е 5, r= 5, y= V6.4x 5 = V 32. = 5-6569 = (etc.), , У= үб.дх 6= V 38-4 = 6:1968 = ж = 7, у= VO-4x у= V 44-8 = 6:6933 = x= 8 у= V6-4x me Vg «Qut 554, == = ово ИБ 7 6805 = X —10, у= 46.4x107 464. =80 = МР, With these values of 7, respectively, set on the correspond- ing horizontal lines 2%, CH, DG, ES, etc., points in the curve 9 H, G, S, etc., are obtained, through which the curve | may be drawn. The decimals above shown are the decimals of a foot; they may be changed to inches and decimals of an inch by multiplying each by 12. For example: 12 х0:5297: = 6.3564 equals 6 inches and the decimal о: 3564 of an inch, which equals nearly $ of an inch. Near the top of the curve, owing to its rapid change in direction and to the approximation of the direction of the curve to a parallel with the direction of the ordinates, it is

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THE CURVE FOUND BY DIAMETERS. 507 desirable to obtain points in the curve more frequent than those obtained by dividing the axis into egual parts. Instead, therefore, of dividing the axis into equal parts, it 15 better to divide it into parts made gradually smaller toward the apex of the curve—or, to obtain points for this part of the curve as shown in the following article. 471.—Parabola: Described from Diameters.— Let EC (Fig. 312) be the given base and 4 Е the given height, placed perpendicularly to ЕС. Divide ЕС in several parts at pleasure, and from the points of division erect perpendicu- lars to ЕС. The problem is to compute the length of these diameters, as D P, and thereby obtain points in the curve, as at P. For this purpose we have equation (183.), which gives Fic. 312. the length of the diameters, and in which л equals DC (Fig. 312), Z equals twice Æ C, and 2 equals half the parameter of the curve. The value of p is given in equation (А.), (272. 470), in which y equals EC (Fig. 312), and x equals A Е, Substituting these symbols in equation (A.), we have— . MERE NI P= ye 2xAB 2h’ where ё = E C, the base, and 4 = A Е, the height. For 2, substituting this. its value, in equation (183.), we have— eX n (| — n) а n(l— т) (184.)

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508 : THE PARABOLA. As an example: let it be required in а parabola in which the base equals 12 feet and the height 8 feet, to compute the length of several diameters, and through their extremities describe the curve. Then / will equal 8, and 6 12. If the base be divided into 6 equal parts, as in 772. 312, each part will equal 2 feet. Then we have— m COR NEN p" cue" ча 18" and— a= LI п), b n (24— n) Ee 18 In this equation, substituting the consecutive values of z, we have, when— I8 2X22 i == 7. dau c 2084 n= 4, J = ша 18 6x18 d а <= 8 x 16 = 8, a= ТЕ n=10, d=—" 4 547 18 E a= 22 unum ^ 18 The several diameters, as P D, in Fig. 312, may now be made equal respectively to these computed values of Z, and the curve traced through their extremities.

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AREA EQUALS TWO THIRDS OF RECTANGLE. 509 472.—Parabola: Area.—From (181.), the equation to the parabola, and by the aid of the calculus, it has been shown that the area of a parabola is equal to two thirds of the circumscribing rectangle. For example: if the height, АЕ (Fig. 312), equals 8 feet, апа E C, the base, equals 12 feet, then the area of the part included within the figure APCE А equals $ of 8x 12 = x 6 = 64 feet; ог, it is equal to $ of the rectangle AB CE.

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SECTION XIV.—TRIGONOMETRY. 473.—Right-Angled Triangles: The Sides.—In right- angled triangles, when two sides are given, the third side may be found by the relation of equality which exists of the squares of the sides (4775. 353 and 416). For example, A 6 р В с а FIG. 312. if the sides а and 2 (Fig. 313) are given, с, the third side, may be computed from equation (115.)— 2 С an. Extracting the square root, we have— Cmm +a”. When the hypothenuse and one side are given, by transposi- tion of the factors in (115.), we have— a*=c’— 63: a= ус”--7% (А. O= о с — а): А (В.)

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TWO SIDES GIVEN TO FIND THE THIRD. SII Owing to the factors being involved to the second power in this expression, the labor of computation is greater than that in a more simple method, which will now be shown. In equation (A.) or (B.) the factors under the radical may be simplified. Ву equation (114.) we have— c* — b? = (c b) (с — б). Therefore, equation (A.) becomes— i a= с + 6) (с — 6), а form easy of solution. For exainple: let с equal 20-732 and 0 equal 13-216, then we have— 29:732 13-216 The sum = 42-948 The difference = 16-516 By the use of a table of logarithms ME 427) the problem may be easily solved ; thus— Log. 42-948 = 1-6329429 16-516 = 1-2179049 To get the square root 2) 2-8508478 а = 26.6539 =i -4254239 This method is applicable to the sides of a triangle, only ; for the hvpothenuse it will not serve. The length of the hypothenuse as well as that of either side may, however, be obtained by proportion; provided a triangle of known di- mensions and with like angles be also given. For example: in 7g. 314, in which the two sides а and б are known, let it be required to find с, the hypothenuse. Draw the line DZ parallel with А C, then the two trian- gles BDE and BAC are homologous; consequently their

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512 | TRIGONOMETRY. corresponding sides аге in proportion (477. 361). Hence, if d equals unity, we have— ELA жі с. from which, when а and f are known, с is obtained by sim- ple multiplication. 474.— Right-Angled Triangles: Trigonometrical Ta- bles.—To render the simple method last named available, the lengths of d, e and f (Fig. 314) have been computed for triangles of all possible angles, and the results arranged in . FIG, 314: tables, termed Trigonometrical Tables. The lines æ, e, and f, are known as sines, cosin’s, tangents, cotangents, etc., as shown in Fig. 315—where A B is the radius of the circle BCH. Drawaline A F, from A, through any point, C, of the arc BG. From C draw CD perpendicular to АВ; from В draw BE perpendicular to A B; and from G draw G F per- pendicular to А G. Then, for the angle FAB, when the radius A C equals unity, CD is the sine; AD the cosine; DB the versed sine ; BE the tangent; GF the cotangent ; А E the secant; and A F the cosecant. But if the angle be larger than one right angle, yet less than two right angles, as BA М, extend HA to K and E В to К, and from Z draw H F perpendicular to A 7.

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TRIGONOMETRICAL TABLES. | 513 Then, for the angle BA H, when the radius А H equals unity, 27 F is the sine; А F the cosine; BF the versed sine ; BK the tangent ; and А К the secant. When the number of degrees contained in a given angle is known, the value of the size, cosine, etc., corresponding to that angle, may be found in a table of Natural Sines, Co- 4 E/F Tangent. Cosine. Fic. 315. sines, etc. Or, the logarithms of the sines, cosines, etc., may be found in logarithmic tables. In the absence of such a table, and when the degrees contained in the given angle are unknown, the values of the sine, cosine, etc., may be found by computation, as fol- lows:—Let ABC (Fig. 316) be the given angle. At any distance from В draw 0 perpendicular to BC. Ву any scale of equal parts obtain the length of each of the three lines а, b,c. Then for the angle at B we have, by proportion—

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514. TRIGONOMETRY. быз @ єз © : seer 5 == : cosec. D = с си? I-O : sin ye с собака le Oni. GOS. В = <. b СС бұй то «tah. Le A a D NEGO ace. В =. @ a é pu Or, in any right-angled triangle, for the angle contained between the base and hypothenuse— When perp. dise by hyp., the quotient equals the size. * — base “ hp. £ “ “ cosine, “Шеш “ “ base, 7“ “ “ tangent. * — base D “ рер, '"'' i “ cotangent. 4 6G hyp. « * base, “ 2 “ secant. "* hyp. * “ perms, “ e * cosecant. To designate the angle to which a trigonometrical term applies, the letter at the intended angle is annexed to the Fic. 216. name of the trigonometrical term ; thus, in the above exam- ple, for the sine of А BC we write sin. Б; for the cosine, COS. В, etc.

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EQUATIONS TO RIGHT-ANGLED TRIANGLES. HT: By these proportions the two acute angles of a right- angled triangle may be computed, provided two of the sides are known. For when the perpendicular and hypoth- enuse are known, the sine and cosecant may be obtained. When the base and hypothenuse are known, the cosine and secant may be computed. Апа when the base and perpen- dicular are known, the tangent and cotangent may be com- puted. Either one of these, thus obtained, shows by the trigo- nometrical tables the number of degrees in the angle; and, deducting the angle thus found from 90°, the remainder will be the angle of the other acute angle of the triangle. For % А T P B С M D Fic. 317. example: in a right-angled triangle, of which the base is 8 feet and the perpendicular 6 feet, how many degrees are contained in each of the acute angles? Having, in this case, the base and perpendicular known: by referring to the above proportions we find that with these two sides we may obtain the tangent; therefore— ! O Tan. Pg с = OTI. a 8 Referring to the trigonometrical tables, we find that 0-75 is the tangent of 36? 52' 12", nearly ; therefore— The quadrant equals оо. о. o The angle В equals 36-52-12 The angle A equals 53-07-48

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516 TRIGONOMETRY. 475.—Right-Angled Triangles: Trigonometrical Value of Sides.—In the triangle A BC (Fig. 317), with B P= 1 for radius, and on 7 as a centre, describe the arc P D, and from its intersection with the lines A B and BC, draw PM and TD perpendicular to the line BC. Then from homologous triangles we have these proportions for the perpendicular— BD BET - BC: C2: " tan. ыш Базе c, perp:;; T pintan Ж Et амедее atau: (185.) Also— ы “ЖР БАШ» tato due 7 USB. 5 дһур.: perp., iisi 2 асв Л. (186.) For the base, we have— ВРе BM :: BUY c, >: cos. Б.” hyp. : Dase, I: Cos. Ё$ i reia ac COR E, (187.) Again— | J dus ОО iuc А tan. B : 7 :шрегр., base, Ex 2:17:55: (188.) ~ tan. B" For the hypothenuse, we have— DM ОТУ xvn: іп. Віт: perp. : hyp.,

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RULE FOR THE PERPENDICULAR, 517 STER T Н c 5. (189.) Again— ТАРАТА ТІ NIS NS >”: séc Ip": dase "hyns a I‘: sec. B:: @ ¢c=2eee. Бе =e. 190. cos. В (190.) This substitution of the cos. for the sec. 1s needed because tables of secants are not always accessible. That it is an equivalent is clear; for we have— BM:BP:: БЕП I COS. By these equations either side of a right-angled triangle may be computed, provided there are certain parts of the triangle given. As, for example: of the six parts of a tri- angle (the three sides and the three angles), 7Zree must be given, and at least one of these must be a side. As an example: let it be required to find two sides of a right-angled triangle of which the base is 100 feet, and the acute angle at the base is 35 degrees. Here we have given one side and two angles (the base, acute angle, and the right angle) to find the other two sides, the perpendicular and the hypothenuse. Among the above rules we have, in equation (185.), for the perpendicular— b = іал. Or: The perpendicular equals the product of the base into the tangent of the acute angle at the dase.

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518 TRIGONOMETRY. Then (Art. 427)— ы * The logarithmic tangent of В (= 35?) is 9-8452268 Log. of а (= 100) is 2:0000000 | Perpendicular, 0 (= 70-02075) = 1 - 8452268 And іог the hypothenuse, taking equation (190.) we have— | соза Or: Тһе hypothenuse equals the quotient of the dase divided by.the cosine of the acute angle at the dase. i For this we have— Log. of a (= 100) 15 2- 0000000 “ соз. B (= 35°) is 9-9133645 Hypothenuse с (= 122-0775) = 2.0866355 We thus find that a right-angled triangle, having an angle of 35 degrees at the base, has its three sides, the perpendic- Fic. 318. ular, base, and hypothenuse, respectively equal to 70-02075, 100, and 122-0775. N.B.—The angle at А (722.317) is obtained by deducting the angle at B from 9o? (Art. 346) Thus, 90 — 35 = 55; this is the angle at 4, in the above case. If the perpendicular be given, then for the base use equation (188.), and for the hypothenuse use equation (189.). ' If the hypothenuse be given, then for the base use equation (187.) and for the perpendicular use equation (186.). -

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USEFUL RULE FOR THE SIDES. 519 476. : Sines and Sides.—In the oblique-angled triangle 4 BC (Fig. 318) from C and per- pendicular to 4 В draw C D. "This line divides the oblique- angled triangle into two right-angled triangles, the lines and angles of which may be treated by the rules already given; but there is a still more simple method, as will now be shown. As shown in Art. 474: “When the perpendicular is di- vided by the hypothenuse the quotient d. the sine.' Applying this to 22. 318, we have— — sin. A = f á oE т” а sn A __ б sin. 8 d’ * S or, reducing, we have— 5ш. 4 Ж -—LG = or, putting the equation in the form of a proportion— sin. Б.Ш ОКИС or; the sizes are in proportion as the szdes, respectively ор- posite. Or, as commonly stated, the szzes are.in proportion as the sides which subtend them. This is a rule of great utility ; by it we obtain the follow- ing: Referring to Fig. 318, we have— in. A Eme 5 : Sin; ат”; qp (191.)

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520 TRIGONOMETRY. sin. A SUD. Са т РТЖ Oe C (192.) sib. A sine БЕ: : ра ЭНИ 2 (193.) sin. C 2 sin. с — = = (194.) sin: 21: (SIM С ta, С 2 E (195.) sn Bisin.C::bse= #96. (196) These expressions give the values of the three sides respec- tively; two expressions for each, one for each of the two remaining sides; that is to be used which contains the given side. From these expressions we derive the values of the sines; thus— 4 sin. A = sin. B. (97) sin. А = sin. c. (198.) 2 sin. 2 = Әт A (199.) sin. 2 «с-ш. (С 2 (200.) sin. C = sin. AĜ, (201.) sih. (С imm. B5. (202.) 477.- Oblique-Angled Triangles: First Class. — The problems arising in the treatment of oblique-angled trian- gles have been divided into four classes, one of which, the

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TO FIND THE TWO SIDES. | 521 first, will here be referred to. The problems of the first class are those in which a side and two angles are given, to find the remaining angle and sides. As to the required angle, since the three angles of every triangle amount to just two right angles (A77. 345), or 180°, the third angle may be found simply by deducting the sum of the two given angles from 180°. For example: referring to Fig. 318, if angle А = 18° and angle В = 42°, then their sum is 18 + 42 = бо, and 180 — бо = 120° = the angle ЖСБ. To find the two sides: if a be the given side, then to find the side 6 we have, equation (193.)— sin. 5. es же sin. A’ or, the side 4 equals the product of the side а into the quo- tient obtained by a division of the sine of the angle opposite 6 by the sine of the angle opposite a. For example: in a triangle (Fig. 318) in which the angle | A = 18°, the angle B = 42° (and, consequently (477. 345) the - angle C = 120°), and the given side « equals 43 feet; what are the lengths of the sides 2 and с? Equation (193.) gives— Performing the problem by logarithms (477. 427), we have— Log. а (== 43) = 1-6334685 біп. В (= 42°) = 9-8255109 1-4589794 Sin. A (= 18°) = 9- 4899824 Log. ^ (= 93: 1102) = 1:9689970. Thus the side 4 equals 93:1102 feet, or 93 feet 1 inch and nearly one third of ай inch.

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$22 TRIGONOMETRY. For the side с, we have, equation (195.)— g SP: Ga " sin. AY GI Log. а(= 43) = 1-6334685 | Sin. C (= 120°) = 9-9375306 1° -5709991 Sin. С 18 °) = о: 4899824 Log. с (= 120-508) = 2-0810167 or, the base c equals 120 feet 6 inches and one tenth of an. inch, nearly. But if instead of a the side 0 be given, then for а use equation (191.), and for c use equation (106.). And, lastly, if с be the given side, then for а use equation (192.), and for б use equation (194.). . 478.—Oblique-Angled Triangles: Second Class.— The problems which comprise the second class are those in which two sides and an angle opposite to one of them are given, to find the two remaining angles and the third side. The only requirement really needed here is to find a second angle; for, with this second angle found, the problem is reduced to one of the first class; and the third side may then be found under rules given in Art. 477. To fifid a second angle, use one of the equations (197.) to (2029. For example: in the triangle ABC (Fig. 318), let a (= 43) and 2 (= 93-11) be the two given sides, and A, the angle op- posite а, be the given angle (— 18?). Then to find the angle В, we have equation (199.)—(selecting that which in the right hand member contains the given angle and sides)— sin. Б = sin. A d a = біп, 4 23: 1H у | 43

Page 561

OBLIQUE-ANGLED TRIANGLES. 523 Ву logarithms (A77. 427), we have— Log. sin. A (= 18?) = 9-4899824 * 2 93: = 9689970 I -4589794 4 46 = 1+6334685 * sin |2562?) = 028252100 By reference to the log. tables, the last line of figures, as above, is found to be the sine of 42?; therefore, the required angle Bis 42°. Then 180? — (18? 4- 42?) = 120? = the angle C. With these angles, or with any two of them, the third side c may be found by rules given in 477. 477. Етс. 319. 479.—Oblique-Angled Triangles: Sum and Differenee of "Two Angles.—Preliminary to a consideration of prob- lems in the third class of triangles, it is requisite to show the relation between the sz»: and difference of two angles. In Fig. 319, let the angle AFM and the angle A # V be the two given angles; and let A FM be called angle А, and A ЖУ, angle В. Now the sum and difference of the angles may be ascertained by the use of the sum and difference of the sines of the angles, and by the sum and difference of the tangents. In the diagram, in which the radius 4 F equals

Page 562

524 ; TRIGONOMETRY. unity, we have MP, the sine of angle А (= A FM) and N Q = RP, the sine of angle B(= AFM). Тһеп— MP—RP-MR: equals the difference of the sines of the angles; and since PM’ = PM— PM'+RP=RM, equals the sun of the sines of the angles. With the radius # С describe the arc 7 D E, and tangent to this arc draw FH parallel with M M’, or perpendicular to 4 В. Then FD is the tangent of the angle M C IV, and D Н is the tangent of the angle VC’. Now since an angle at the circumference 15 equal to half the angle at the centre standing on the same arc (477. 355), therefore the measure of the angle M CN is the half. of M N, equals— | | 4(AM—AN)=}(4—B). Similarly, we have— k(AM'+ AN) ут + А), for the angle VC M". . Therefore we have for the tangent of the angle M C N— EOD ec (D and, for the tangent of the angle VC M'— DH = tan. (А + D). And, because FC and MCR are homologous triangles, as, also, DCH and RCM’, therefore— ZEE WERDE EM

Page 563

SUM AND DIFFERENCE OF TWO ANGLES. ` 525 sin. 4 + sin. В : sin. 4-- sin. B :: tan. 3(4 + B) : tan. 3(4 — В), from which we have— sin. 4 — sin. / tan. $ (A — D) ЕЕЕ sin. A + sin. tan. §(4 + D) 49.) То obtain а proper substitute for the first member of this expression we have, equation (195.)— sin. C = a-- at, ш or— E с sin. A = asin, (б (M.) We also have, equation (196.)— жазшы — sin. B’ or— с sin. В = 0 sin. C. (N.) These two equations, (M.).and (N.), added, give— c sin. А + c sin. B= asin. C+ 6 sin. C. or— c (sin. A + sin. Б) = sin. C (a + 0). (ES) But, if equation (N.) be subtracted from equation (M.), we have— c sin. A —csin. B=asin. C — Ё sin. C, c(sin. А — sin. B) = A sin. C (a — б). (R.) If equation (R.) be divided by equation (P.), we have— c(sin. А — sm. B) sin. C(a — 0) c(sin. A + sin. B) sin. C (a + b)’

Page 564

526 TRIGONOMETRY. which reduces to— Р sin. 4 — sin. В a — b SET + sm. 2 из. The first member of this equation is identical with the first member of the above equation (D.), and therefore its equal, the second member, may be substituted for it; thus— а-60 tan. 3(4 — B) a+b tan.a(A + B) From which we have— tan. $(4 — P) = (203.) We have (Art. 431) the proposition, that if half the differ- ence of two quantities be subtracted from half their sum, the remainder will equal the smaller quantity. For example: if A represent the larger quantity and В the smaller, then— }(4 + B) —4(4— B)— В; (204.) and, again, we also have (472. 431)— (А+ B) +5(4 — P) = A. (205.) 480.—Oblique-Angled Triangles: Third Class. — The third class of problems comprises all those cases in which two sides of a triangle and their included angle are given, to find the other side and angles. In this case, as in the problems of the second ыш, the only requirement here is to find a second angle; for then the problem becomes one belonging to the first class. But the finding of the second angle, in problems of the third class, is attended with more computation than it is in prob- lems of the second class. The process is as follows: Hav- ing one angle of a triangle, the swm of the two remaining

Page 565

OBLIQUE-ANGLED TRIANGLES. 527 angles is obtained by subtracting the given angle from 189? — the sum of the three angles. Then with equation (203.) the diference of the two angles is obtained. And then, having the sum and difference of the two angles, either may be found by one of the equations (204.) and (205.). For example: let Fig. 320 represent the triangle in which a (= 36 feet) and (= 27 feet) are the given sides; and С с FIG. 320. C (— 105?) the angle included between the given sides, а and б. The sum of the two angles А and В, therefore, will be— (A + B) = 180 — 105 = 75°, and the half of the sum of A and B is 45 = 37° 30’. The sum of the given sides is 36 + 27 = 63, and their dif- ference is 36 — 27 — 9. Then from equation (203.) we have— . tan. à (4 — В) = tan. 37? 309. Solving this by logs. (477. 427), we have— Log. tan. 37? 30' — 9-8849805 9 = 0:9542425 O: 8392230 63 = 1:7993405 tan. $(A — В) (= 6? 157 20.5”) = 90398825 Thus half the difference of A and B is 6? 15' 20- 5", nearly.

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528. TRIGONOMETRY. By equation (204.)— 37^ 30 69 15 20-5” The difference, 31° 17 SS = В, and by equation (205.)— 37-30 б.15.20:5 The sum, 43.45.20:5 = Л From above, 31.14.39:5 = В ГСһе илеп angle, 105. оо =C The three angles, 180. o. o Thus, by adding together the three angles, the work is tested and proved. Having the three angles, the third side may now be found by the rule for problems of the first class. 481.— Oblique-Angled Triangles: Fourth Class. — The fourth class comprises those problems in which the three sides of the triangle are given, to find the three angles. The method by which the problems of the fourth class are solved is to divide the triangle into two right-angled triangles; then, by the use of equation (129.), to find one side of one of these triangles, and then with this side to find one of the angles, then by rules for the second class prob- lems, obtain the second and third angles. Thus, from equation (129.), we have— fe еч. ш a 2¢ By the relation of sines to sides (A77. 9) \ we have (Fig. 321)— | дір: 8m. E a illoa

Page 567

TRIANGLES—FOURTH CLASS. 529. But the angle Е is a right angle, of which the sine is unity, therefore— ne: 1: sin, Fae, Substituting for g its value as above, we have— , зіп, Mi= ee 714225 мыз (206.) To illustrate: let а, 2, c (Fig. 321) be the three given sides FIG. 321. of the triangle A BC, respectively equal to 12, 8 and 16 feet. With these, equation (206.) becomes— dicm 1 10 — 2 4 Зоор 2x8 x 16 в. г __ 256 m Sine Жас $55 Solving this by logarithms (477. 427), we have— Log. 176 = 22455127 “ 256 = 2-4082400 . Log. sin. 43° 26’ = 98372727 оғ, the angle at F equals 43° 26’, nearly. Of the triangle

Page 568

530 TRIGONOMETRY. АСЕ (Ре-з29-Е 35 @ right angle, therefore the sum of F and A, the two remaining angles, equals 90° (Art. 346). Hence, for the angle at 4, we have— A = 90° — 43° 26' = 46° 34’. We now have two sides а and 0 and A, an angle opposite to one of them, to find 5, a second angle. For this, equa- tion (199.) is appropriate. Thus— sin. B = sin. A is а This may be solved as shown in 472. 478. And, when the second angle is obtained, the third angle is Tom by subtracting the sum of the first ави ѕесопа ап- gles from 180°. But to test the accuracy of the work, it is well to com- pute the angle'C from the angle А, and the sides a and c. For this, equation (201.) will be appropriate. 482.— Trigonometrie Formulx: Right-Angled Trian- gles. — For facility of reference the formule of previous п А с р В C & Fic. 322. articles are here presented in tabular form. The symbois referred to are those of Fig. 322.

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FORMUL IN TABULAR FORM. 531 RIGHT-ANGLED TRIANGLES. GIVEN. REQUIRED, FORMULE. bm с c = Маф b’. mue. b, b = (с + а) (c— а). а = ((с+ b) (с — 8). A, B, D, А. b, п, с, а, В, b, TERI а, p b 4.83. —Trigonometrical Formule: First Class, Oblique. С с FIG. 323. —The symbols of the formule of the following table indi- cate quantities represented in Fig. 323 by like symbols.

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532 TRIGONOMETRY. OBLIQUE-ANGLED TRIANGLES: FIRST CLASS. GIVEN, | REQUIRED. 4, 10, Є, А, С, | Je | D C, | A, A, 9, b, а, E. C, с, | a, 1, Б,а, бф, lee Ё | A, C, a, | C, | ы с, | 484. FORMULA, C = 180 — A + В. B= Bo- AEC. | Zo ш | | Trigonometrical Formulz : Second Class, Oblique. —The symbols in the formulz of the following table refer to quantities represented іп Fig. 323, by like symbols.

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FORMUL FOR TRIANGLES, SECOND, CLASS. 533 OBLIQUE-ANGLED TRIANGLES: SECOND CLASS. dc —s— s -————- да”, та | GIVEN. REQUIRED. FORMULA. | 19, а, b, 4, sin. 4 = Sim BT. | А КО, в, с, 4, sin. A = sin. ct. А, a, 6, D, sin. Б. sim. 4d E А : b ШЕ 5 с, D, . sm. В = sin CM EP |^. Е - EE С. sin. C = # = DAS с, C, sin. C = SI: Be. m. ©, A, А —180— В + С. A, C, B, В —180— А + C. Bw, a C — 180 — À Ж B. E —————— e, For— б, See Formule, First Class.

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534 i TRIGONOMETRY. 4.85. —TFrigonometrical Formule: Third Class, Oblique. —-The symbols in the formulz of the following table refer to quantities shown by like symbols in 77g. 323. OBLIQUE-ANGLED TRIANGLES: THIRD CLASS. GIVEN. | REQUIRED. FORMULA. A+B,| А + ВБ = 180 — С. шол шк, tan. $(4 — Б) = tan. &(4 + B) 27, — am с VI н C + В, C + B= 080 — А. AL b,c, | CMB, tan. 4(C — B) = tan. $(C+ 275. al сас 9 EES Ме By. CAI C+A = ое. ENS C — A, tan. 4 (C — A) = tan. $(C + Ау———. C, | С=ңс+лу++4(С— A). A, V 42C 4) 8 tC ee. For the remaining side consult formulz for the frst class. 4.86. Trigonometrical Formuke: Fourth Class, Ob- lique.—The symbols in the formule of the following table refer to quantities shown by like symbols in 777. 321.

Page 573

FORMUL/E FOR TRIANGLES, FOURTH CLASS. 535 OBLIQUE-ANGLED TRIANGLES: FOURTH CLASS. ————— TT Given z, 0, с топа A, B,C. c? — (a + бф) (a — б) 26с 4 A = go F. sin, F = sin. B = sin. A A a н , с sin. C = sin. 4 —. a C = 180 — (4 + B). -

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SECTION XV.—DRAWING. 487.—General Remarks.—A krowledge of the proper- ties and principles of Zizes can best be acquired by practice. Although the various diagrams throughout this work may be understood by inspection, yet they will be impressed upon the mind with much greater force, if they are actually drawn out with pencil and paper by the student. Science is acquired by study—art by practice; he, therefore, who would have anything more than a theoretical (which must of necessity be a superficial) knowledge of carpentry and geometry, will provide himself with the articles here speci- fied, and perform all the operations described in the fore- going and following pages. Many of the problems may appear, at the first reading, somewhat confused and intricate ; but by making one line at.a time, according to the explana- tions, the student will not only succeed in copying the fig- ures correctly, but by ordinary attention will learn the principles upon which they are based, and thus be able to make them available in any unexpected case to which they may apply. | 488.—4Articles Required. — The following articles are necessary for drawing, viz.: a drawing-board, paper, draw- ing-pins or mouth-glue, a sponge, a T-square, a set-square, two straight-edges, or flat rulers, а lead pencil, a piece of india-rubber, a cake of india-ink, a set of drawing-instru- ments, and a scale of equal parts. 489.— The Drawing-Board.— Тһе size of the drawing- board must be regulated according to the size of the draw- ings which are to be made upon it. Yet for ordinary prac- tice, in learning to draw, a board about fifteen by twenty inches, and one inch thick, will be found large enough, and

Page 575

PRAWING PAPER. 537 more convenient than a larger one. This board should be well seasoned, perfectly square at the corners, and without clamps on. the ends. А board is better without clamps, because the little service they are supposed to render by preventing the board from warping is overbalanced by the consideration that the shrinking of the panel leaves the ends of the clamps projecting beyond the edge of the board, and thus interfering with the proper working of the stock of the T-square. When the stuff is well-seasoned, the warp- ing of the board will be but trifling ; and by exposing the rounding side to the fire, or to the sun, it may be brought back to its proper shape. 490.—Drawing-Paper.— For mere line drawings, it is unnecessary to use the dest drawing-paper ; and since, where much is used, the expense will be considerable, it is desirable for economy to procure a paper of as low a price as will be suitable for the purpose. The best paper is made in Eng- land and water-marked “ Whatman.” This is a hand-made paper. There is also a machine-made paper at about half- price, and the manilla paper, of various tints of russet color, is still less in price. These papers are of the various sizes needed, and are quite sufficient for ordinary drawings. 491.—To Secure the Paper to the Board.—4A drawing- pin is a small brass button, having a steel pin projecting from the underside. By having one of these at each corner, the paper can be fixed to the board; but this can be done ina better manner with mouth.glue, The pins will prevent the paper from changing its position on the board; but, more than this, the glue keeps the paper perfectly tight and smooth, thus making it so much the more pleasant to work on. | To attach the paper with mouth-glue, lay it with the bottom side up, on the board ; and with a straight-edge and penknife cut off the rough and uneven edge. With a sponge moderately wet rub all the surface’ of the paper; except a strip around the edge about half an inch wide. As soon as the glistening of the water disappears turn the sheet

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539 DRAWING. * over and place it upon the board just where you wish it glued. Commence upon one of the longest sides, and pro- ceed thus: lay a flat ruler upon the paper, parallel to the edge, and within a quarter of an inch of it. With a knife, or anything similar, turn up. the edge of the paper against the edge of the ruler, and put one end of the cake of mouth- glue between your lips to dampen it. Then holding it up- right, rub it against and along the entire edge of the paper that is turned up against the ruler, bearing moderately against the edge of the ruler, which must be held firmly with the left hand. Moisten the glue as often as it becomes dry, until a sufficiency of it is rubbed on the edge of the paper. ‘Take away the ruler, restore the turned-up edge to the level of the board, and lay upon it a strip of pretty stiff paper. By rubbing upon this, not very hard but pretty rapidly, with the thumb-nail of the right hand, so as to cause a gentle friction and heat to be imparted to the glue that is on the edge of the paper, you will make it adhere to the board. The other edges in succession must be treated in the same manner. Some short distances along one or more of the edges may afterward be found loose; if so, the glue must again be applied, and the paper rubbed until it adheres. Тһе board must then be laid away in a warm or dry place; and in a short, time the surface of the paper will be drawn out, perfectly tight and smooth, and ready for use. The paper dries best when the board is laid level. When the drawing is finished lay a straight-edge upon the paper and cut it from the board, leaving the glued strip still attached. This may afterward be taken off by wetting it freely with the sponge, which will soak the glue and loosen the paper. Do this as soon as the drawing 1s taken off, in order that the board may be dry when it is wanted for use again. Care must be taken that, in applying the glue, the edge of the paper does not become damper than the rest; if it should, the paper must be laid aside to dry (to use at another time) and another sheet be used in its place. Sometimes, especially when the drawing-board is new, the paper will not stick very readily; but by persevering

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THE T-SQUARE. 539 this difficulty may be overcome. In the placé of the mouth- glue a strong solution of gum-arabic may be used, and on some accounts is to be preferred ; for the edges of the paper need not be kept dry; and it adheres more readily. Dissolve the gum in a sufficiency of warm water to make it of the consistency of linseed-oil. It must be applied to the paper with a brush, when the edge 15 turned up against the ruler, as was described for the mouth-glue. If two drawing-boards are used, one may be in use while the other is laid away to dry; and as*they may be cheaply made, it is advisable to have two. The drawing-board having a frame around it, commonly called a panel board, may afford rather more facility in attaching the paper when this is of the size to Fic. 324. suit; yet it has objections which overbalance that consid- eration. / 492.—' The Т-бапаге.-А 7-sguare of mahogany, at once’ simple in its construction and affording all necessary service, may be thus made: let the stock or handle be seven inches long, two and a quarter inches wide, and three eighths of an inch thick; the blade, twenty inches long (exclusive of the stock), two inches wide, and one eighth of an inch thick. In joining the blade to the stock, a very firm and simple joint may be made by dovetailing it—as shown at Fig. 324. 493.—' The Set-Square.— The sc-sguareé is in the form of a right-angled triangle ; and is commonly made of mahogany,

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540 DRAWING. one eighth of an inch in thickness. The size that is most convenient for general use is six inches and three inches respectively for the sides which contain the right angle, although a particular length for the sides 15 by no means necessary. Care should be taken to have the square corner exactly true. This, as also the T-square and rulers, should have a hole ,bored through them, by which to hang them upon a nail when not in use. 494.—' The Rulers, — One of the rulers may be about twenty inches long, and the other six inches. The pencil ought to be hard enough to retain a fine point, and yet not so hard as to leave ineffaceable marks. It should be used lightly, so that the extra marks that are not needed when the drawing is inked, may be easily rubbed off with the rubber. The best kind of zzdza-zu£ is that which will easily rub off upon the plate; and, when the cake is rubbed against. the teeth, will be free from grit. 495.--Тһе Instruments.—The drawing-instruments may be purchased of mathematical instrument makers at various prices; from one to one hundred dollars a set. In choosing a set, remember that the lowest price articles are not always the cheapest. A set, comprising a. sufficient number of instruments for ordinary use, well made and fitted in a ma- hogany box, may be purchased of the mathematical instru- ment makers in New York for four or five dollars. But for permanent use those which come at ten or twelve dollars will be found to be better. 496.—The Seale of Equal Parts.—The best scale of equal parts for carpenters’ use, is one that has one eighth, three sixteenths, one fourth, three eighths, one half, five eighths, three fourths, and seven eighths of an inch, and one inch, severally divided into ¢wedfths, instead of being divided, as they usually are, into tenths. Ву this, if it be required to proportion a drawing so that every foot of the object represented will upon the paper measure one fourth of an inch, use that part of the scale which is divided into one

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THE SET-SQUARE. 541 fourths of an inch, taking for every foot one of those divis- ions, and for every inch one of the subdivisions into twelfths; and proceed in like manner in proportioning a drawing to any of the other divisions of the scale. An instrument in the form of a semi-circle, called a protractor, and used for laying down and measuring angles, is of much service to surveyors, and occasionally to carpenters. 497.—The Use of the Set:Square.—In drawing parallel lines, when they are to be parallel to either side of the board, use the T-square; but when it is required to draw lines parallel to a line which is drawn in a direction oblique Fic. 325. to either side of the board, the set-square must be used. Let ab (Fig. 325) be a line, parallel to which it is desired to draw one or more lines. Place any edge, as cd, of the set- square even with said line; then place the ruler г against one of the other sides, as ce, and hold it firmly ; slide the set-square along the edge of the ruler as far as it is desired, as at f; anda line drawn by the edge ¿f will be parallel toa 2. | To draw a line, as $Z (Fig. 326), perpendicular to another, as аб, set the shortest edge of the set-square at the line аё; place the ruler against the longest side (the hypothenuse of the right-angled triangle); hold the ruler firmly, and slide the set-square along until the side eZ touches the point £;. then the line Z, drawn by it, will be perpendicular to аф.

Page 580

542 DRAWING. In like manner, the drawing of other problems may be facil- itated, as will be discovered in using the instruments. 498.— Directions for Drawing.—In drawing a problem, proceed, with the pencil sharpened to a point, to lay down the several lines until the whole figure is completed, ob- serving to let the lines cross each other at the several angles, instead of merely meeting. By this, the length of every line will be clearly defined. With a drop or two of water, rub one end of the cake of ink upon a plate or saucer, until a sufficiency adheres to it. Be careful to dry the cake of Ere 226 ink; because if it is left wet it will crack and crumble in pieces. With an inferior camel's-hair pencil add a little water to the ink that was rubbed on the plate, and mix it well. It should be diluted sufficiently to flow freely from the pen, and yet be thick enough to make a d/ack line. With the hair pencil place a little of the ink between the nibs of the drawing-pen, and screw the nibs together until the pen makes a fine line. Beginning with the curved lines, proceed to ink aZ the lines of the figure, being careful now to make every line of its requisite length. If they are a trifle too short or too long the drawing will have a ragged appear- ance; and this is opposed to that neatness and accuracy which is indispensable-to a good drawing. When the ink is dry efface the pencil-marks with the india-rubber. If the

Page 581

` PUTTING THE DRAWING IN INK. 543 pencil 15 used lightly they will all rub off, leaving those lines only that were inked. In problems all auxiliary lines are drawn light; while the lines given and those sought, in order to be distinguished at a glance, are made much heavier. The heavy lines are made so by passing over them a second time, having the nibs of the pen separated far enough to make the lines as heavy as desired. ІҒ the heavy lines are made before the drawing is cleaned with the rubber they will not appear so black and neat, because the india-rubber takes away part ofthe ink. If the drawing is a ground-plan or elevation of a house, the shade-lines, as they are termed, should not be put in until the drawing is shaded ; as there is danger of the heavy lines spreading when the brush, in shading or color- ing, passes over them. If the lines are inked with common: writing-ink they will, however fine they may be made, be subject to the same evil; for which reason india-ink is the only kind to be used.

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SECTION XVI.—PRACTICAL GEOMETRY. 499,—Definitions.— Geometry treats of the properties of magnitudes. А point has neither length, breadth, nor thickness. А line has length only. Superficies has length and breadth only. А plane is a surface, perfectly straight and even in every direction; as the face of a panel when not warped nor "winding. . , A solid has length, breadth, and thickness. A right, or straight, line is the shortest that can be drawn between two points. Parallel lines are equidistant throughout their length. a Tui Buc. 327- Fic. 328. FIG. 329. An angle is the inclination of two lines towards one an- other (Fig. 327). A right angle has one line perpendicular to the other (Fig. 328). An oblique angle is either greater or less than a right angle (Figs. 327 and 329). An acute angle is less than a right angle (Fig. 327). An obtuse angle is greater than a right angle (Fig. 329). When an angle is denoted by three letters, the middle one, in the order they stand, denotes the angular point, and the other two the sides containing the angle; thus, let а, 4, с (Fig. 327) be the angle, then 0 will be the angular point, and a b and дс will be the two sides containing that angle.

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TRIANGLES AND RECTANGLES. 545 А triangle is a superficies having three sides and angles (Figs. 330, 331, 332, and 333). Ап equilateral triangle has its three sides equal (А0. 330). Ап zsosceles triangle has only two sides equal (Fig. 331). DE oO Fic. 330. FIG 331. A scalene triangle has all its sides unequal (Fig: 332). А right-angled triangle has one right angle (72g. 333). An acute-angled triangle has all its angles acute (Figs. 330 and 331). FIG. 252: F1G. 333. An obtuse-angled triangle has one obtuse angle (Fig, 332). A quadrangle has four sides and four angles (Figs. 334 to 339). "E A parallelogram is a quadrangle having its opposite sides parallel (Figs. 334 to 337). FIG. 3345 Fic. 335. A rectangle is a parallelogram, its angles being right angles (Figs. 334 and 335). A square is a rectangle having equal sides (Fig: 334). A rhombus is an equilateral parallelogram having oblique angles (Fig. 336). | @8

Page 584

546 PRACTICAL GEOMETRY. А rhomboid is a parallelogram having oblique s (Fig. 337). А trapezoid is a quadrangle having only two of its sides parallel (Ае. 338). 7 / Еіс:95. . —FtGw597. A trapezium is a quadrangle which has no two of its sides parallel (Fig. 339). A polygon is a figure bounded by right lines. A regular polygon has its sides and angles equal. An irregular polygon has its sides and angles unequal. Fic, 338. Fic. 339. A frigon is a polygon of three sides (Fzgs. 330 to 333); а tetragon has four sides (Figs. 334 to 339); a pentagon has five (Fig. 340); a hexagon six (Fig. 341); a heptagon seven (fig. 342); an octagon eight (Fig. 343); a "onagon nine ; a decagon ten; an zzdecagon eleven; and a dodecagon twelve sides. EO OO F 1G. 340. Еіс. 341. Кіс. 342. | FIG. 343. A circle is a figure bounded by a curved line, called the circumference, which 18 everywhere equidistant from a cer- tain point within, called its centre. The circumference is also called the periphery, and some- times the circle.

Page 585

PARTS. OF THE CIRCLE. 547 The radius of a circle is a right line drawn. from the centre to any point in the circumference (ад, Fig. 334). All the ғай of a circle are equal. а The diameter is a right line passing through the centre, and terminating at two opposite points in the circumference. Hence it is twice the length of the radius (са, Fig. 344.) An are of a circle is a part of the circumference (сё, or bed, Fig. 244). A chord is a right line joining the extremities of an arc (b d, Fig. 344). | А segment is any part of а circle bounded by an arc and its chord (А, Fig. 344). Ғіс. 245. Fic. 346. A sector is any part of a circle bounded by an arc and two radii, drawn to its extremities (5, Fig. 344). А quadrant, or quarter of a circle, 15 a sector having a quarter of the circumference for its arc (C, Fig. 344). A tangent is a right line which, in passing a curve, touches, withont cutting it (fg, Fig. 344).

Page 586

548 PRACTICAL GEOMETRY. А cone is а solid figure standing upon a circular base di- minishing in straight lines to a point at the top, called its vertex (Fig. 345). The aris of a cone is a right line passing through it, from the vertex to the centre of the circle at the base. An ellipsis is described if a cone be cut by a plane, not parallel to its base, passing quite through the curved surface (ab, Fig. 346). A parabola is described if a cone be cut by a plane, par- allel to a plane touching the curved surface (са, Fig. 346— са being parallel to fg). An hyperbola is described if a cone be cut by a plane, Fic, guge parallel to any plane within the cone that passes through its vertex (eh, Fig. 346). Foci are the points at which the pins are placed іп de- scribing an ellipse (see 477. 548, and /, f, Fig. 347). The ¢ransverse axis is the longest diameter of the ellipsis (a b, Fig. 347). The conjugate axis 15 the shortest diameter of the ellipsis ; and is, therefore, at right TES to the transverse axis (са, Fig. 347). The parameter is a right line passing through the focus of an ellipsis, at right angles to the transverse axis, and ter- minated by the curve (gf and gt, Fig. 347).

Page 587

RIGHT LINES AND ANGLES. 549 * A diameter of an ellipsis is any right line passing through the centre, and terminated by the curve (£4 or m n, Fig. 347). A diameter is conjugate to another when it is parallel to a tangent drawn at the extremity of that other—thus, the di- ameter mn (Fig. 347) being parallel to the tangent of, is therefore conjugate to the diameter 22. А double ordinate is any right line, crossing a diameter of an ellipsis, and drawn parallel to a tangent at the extremity of that diameter (22, Fzg. 347). | А cylinder is а solid generated by the' revolution of a right-angled parallelogram, or rectangle, about one' of its Fic. 348. Fic. 349. sides; and consequently the ends of the cylinder are equal circles (Fig. 348). The ал? of a cylinder is a right line passing through it from the centres of the two circles which form the ends. A. segment of a cylinder is comprehended under three planes, and the curved surface of the cylinder. Two of these are segments of circles; the other plane is a parallelo- gram, called by way of distinction, the plane of the segment. The circular segments are called the ends of the cylinder (Fig. 349). PROBLEMS. RIGHT LINES AND ANGLES. 500.—To Bisect a Lite Upon the ends of the line ад (Fig. 350) as centres, with any distance for radius greater than half аб, describe arcs cutting each other in c and 4;

Page 588

550 PRACTICAL GEOMETRY. e draw the line са, and the point e, where it cuts аб, will be: the middle of the line a à. In practice, a line is generally divided with the com- passes, or dividers; but this problem is useful where it is с SSS nm dy V | а Ріс. 350. desired to draw, at the-middle of another line, one at right angles to it. (See Z7. 514.) 501.--То Erect a Perpendicular.—From the point а (Fig. 351) set off any distance, as а б, and the same distance from а to c; upon с, as a centre, with any distance for radius greater than ca, describe an arc at Z; upon 4, with the same d "d \ \ е а i b Fic. 351. radius, describe another at 7; join d and а, and the line da will be the perpendicular required. This, and the three following problems, are more easily performed by the use of the set-square (see 477. 493). Yet they are useful when the operation is so large that a set- square cannot be used.

Page 589

TO ERECT A PERPENDICULAR. 551 502.—'To let Fall a Perpendicular.—Let a (Fig. 352) be the point above the line бс from which the perpendicular is required to fall. Upon а, with any radius greater than ad, describe an arc, cutting ^c ateand /; upon the points e and f, with any radius greater than ed, describe arcs, cutting FIG. 352. each other at ғ; join а and g, and the line ad will be the perpendicular required. 503.— To Erect a Perpendicular at the End of a Line. —Let a (Fig. 353), at the end of the line ca, be the point at which the perpendicular is to be erected. Take any point, as д, above the line ca, and with the radius да describe the arc dae; through 4 and 2 draw the line de; join e and 2, then ea will be the perpendicular required. Fic. 353. The principle here made use of is a very important one, and is applied in many other cases (see 477. 510, 3d, and 477, 513. For proof of its correctness, see Art. 352). ^ A second method. Let 6 (Fig. 354), at the end of the line a à, be the point at which it is required to erect a perpendic- ular. Upon 4, with any radius less than да, describe the arc ced; upon c, with the same radius, describe the small arc at e,

Page 590

552 PRACTICAL GEOMETRY. and upon e, another at Z; upon e and Z, with the same or any other radius greater than half e а, describe arcs intersecting at f; join fand 0, and the line / will be the perpendicular required. This method of erecting a perpendicular, and that of the following article, depend for accuracy upon the (E b Fic. 354. fact that the side of a hexagon is equal to the radius of the circumscribing circle. A third method. Let 6 (fig. 355) be the given point at which it is required to erect a perpendicular. Upon 2, with any radius less than 2а, describe the quadrant Ze f; upon d, with the same radius, describe an arc at e, and upon e an- other at c; through 4 and e draw dc, cutting the arc in с; join с and 0, then c? will be the perpendicular required. Fic. 355. This problem сап be solved by the siv, eight and ten rule, as it 1s called, which is founded upon the same principle as the problems at Arts. 536, 537, and is applied as follows: let ad (Fig. 353) equal eight, and ae, six; then, if Ze equals ten, the angle cad isa right angle. Because the square of six and that of eight, added together, equal the square of

Page 591

EQUAL ANGLES. 553 ten, thus: 6 x6 — 36, ФИГ 5 X 8 = O4; 365 62 аб Ox то = 100. Any sizes, taken in the same proportion, as six, eight and ten, will produce the same effect; as 3, 4 and 5, or 12, 16-and 20. (See Art. 536.) By the process shown at 27. 353, the end of a board may be squared without a carpenters-square. All that is neces- sary is a pair of compasses and a ruler. Let ca be the edge of the board, and а the point at which it is required to be squared. Take the point 0 as near as possible at an angle of forty-five degrees, or on a mztre-line from а, and at about the middle of the board. This is not necessary to the work- ing of the problem, nor does it affect its accuracy, but the result is more easily obtained. Stretch the compasses from б to a, and then bring the leg аға around to ж; draw a line from d, through 0, out indefinitely; take the distance 40 and place it from 0 to e; join e and а; then ea will be at right angles to са. In squaring the foundation of a building, or laying out a garden, a rod and chalk-line may be used in- stead of compasses and ruler. 504..—To let Fall a Perpendicular near the End of a Line.—Let e (Fig. 353) be the point above the line са, from which the perpendicular is required to fall. From e draw any line, as ed, obliquely to the line са; bisect edat 2; upon б, with the radius фе, describe the arc ead; join eand a; then ea will be the perpendicular required. 505.—To Маке an Angle (as е d f, Fig. 356) Equal to a Given Angle (as ba c).—From the angular point a, with any РД ^ \ a ó d g 4 Fic. 356. radius, describe the arc ёс; and with the same radius, оп | the line Ze, and from the point 4, describe the arc fg; take the distance 4c, and upon g, describe the small arc at /;

Page 592

5 54 PRACTICAL GEOMETRY. join f and Z; and the angle ed f will be equal to the angle bac. If the given line upon which the angle is to be made is situated parallel to the, similar line of the given angle, this may be performed more readily with the set-square. (See Art. 497.) 506.—To Bisect an Angle.—Let abc (Fig. 357) be the angle to be bisected. Upon 2, with any radius, describe the FIG. 357. arc ac; upon a and с, with a radius greater than half ас, describe arcs cutting each other at 7; join 6 and d; and dd will bisect the angle abc, as Was required. This problem is frequently made use of in solving other problems; it should therefore be well impressed upon the memory. 507.—'"ro Triscet a Right Angle.—Upon а (fig. 358), with any radius, describe the arc дс; upon д and с, with the ё Fic. 358. same radius, describe arcs cutting the arc бс at 4 and e; from d and e draw lines to а, and thex will trisect the angle, as was required.

Page 593

«est IBRA S тт p> TO DIVIDE A' GIVEN LINE. The truth of this is made evident by е fo EG ation: divide a circle into quadrants; also, take the radius in the dividers, and space off the circumference. This will divide the circumference into just six parts. А semi-circum- ference, therefore, 15 equal to three, and a quadrant to one and а half of those parts. © The radius, therefore; is equal to two thirds of a quadrant ; and this is equal to a right angle. 508.— Through a Given Point, to Draw a Line Parallel to a Given Line.—Let a (Fig. 350) be the given point, and е а FIG. 359. бс the given line. Upon any point, as d, in the line дс, with the radius da, describe the arc ас; upon а, with the same radius, describe the arc de; make Ze equal to ac; through e and a draw the line ea, which will be the line required. This is upon the same principle as A77. 505. 509.— To Divide а Given Line into any Number of Equal Parts.—Let a? (Fig. 360) be the given line, and 5 the number of parts. Draw ac at any angle to ад; on ас, from Є, 3 Fic. 360. а, set off five equal parts of any length, as at 1, 2, 3, 4 and c; join с and 4; through the points т, 2, 3, and 4, draw re, 2 f, 3 g and 44, parallel to có; which will divide - line av) as was required.

Page 594

555 PRACTICAL GEOMETRY. The lines ад and ас are divided in the same proportion. (See Art. 542.) THE CIRCLE. 510.--То Find the Centre of а Cirele.— Draw any chord, as ab (Fig. di and bisect it with the perpendicular cd; bi- Fic. 361. sect cd with the line e f, as at g; then g is the centre, as was 'required. A second method. Upon any two points in the circumfer- ence nearly opposite, as а and 0 (Fig. 362), describe arcs cut- Fic. 362. ting each other at с and 4; take any other two points, as г and f, and describe arcs intersecting, as at g and 2; join g and лапа c and 4; the intersection o is the centre. This is upon the same principle as Art. 514. A third method. Draw any chord, as ад (Fig. 363), and from the point а draw ас at right angles to a? ; join c and b; bisect c? at 4—which will be the centre of the circle.

Page 595

A TANGENT AT A GIVEN POINT. 557 If a circle be not too large for the purpose, its centre may very readily be ascertained by the help of a carpenters’- square, thus: apply the corner of the square to any point in the circumference, as at а; by the edges of the square (which the lines ад and ас represent) draw lines cutting the 2 S. Fic. 363. circle, as at б and с; join ? and с; then if dc is bisected, as at d, the point d will be the centre. (See 477. 352.) 51{.—At a Given Point in a Circle to Draw a Tangent thereto, —Let а (Fig. 364) be the given point, and 2 the cen- а Гіс. 364. tre of the circle. Join а and 0; through the point 2, and at right angles to 24, draw cd; then cd is the tangent required. 5(2.—' Тһе Same, without making usc of the Centre of the Cirele.—Let a (Fig. 365) be the given point. From a set off any distance to 0, апа the same from 2 to c; join а and с; upon а, with a£ for radius, describe the arc dbe; make d 6 equal to фе; through а and 4 draw a line; this will be the tangent required.

Page 596

558 PRACTICAL GEOMETRY. The correctness of this method depends upon the fact that the angle formed by a chord and tangent is equal to any inscribed angle in the opposite segment of the circle (477. 358); ад being the chord, and dca the angle in the opposite segment of the circle. Now, the angles “аб and dca are equal, because the angles dad and фас are, by construction, Fic. 365. equal; and the angles бас and dca are equal, because the triangle а бс is an isosceles triangle, having its two sides, а and ёс, by construction equal; therefore the angles dad and бса are equal. 513.—4 Circle and a Tangent Given, to Find the Point of Contact.—From any point, as а (Fig. 366), in the tangent Fic. 366. бс, draw a line to the centre d; bisect ad at e; upon e, with ` the radius ea, describe the arc afd; f is the point of con- tact required. If f and Z were joined, the line would form right angles with the tangent dc. (See Art. 352.)

Page 597

A CIRCLE THROUGH GIVEN POINTS. 550 5[4..— Through any Three Points not in a Straight Line, to Draw a Circle, —Let а, ? and с (Fig. 367) be the three given points. Upon а and 4, with any radius greater than half а 0, describe arcs intersecting at 4 and e; upon д and с, with any radius greater than half бс, describe arcs intersect- ing at f and ғ; through 4 and e draw a right line, also 5 { Fic. 367. another through f and ғ; upon the intersection 7, with the radius да, describe the circle абс, and it will be the one re- quired. | 5[5.—' Three Points not іп a Straight Line being Given, to Find a Fourth that shall, with the Threc, Lic in the Circumference of a Circle.—Let абс (Fig. 368) be the given points. Connect them with right lines, forming the triangle Fic. 368. асб; bisect the angle cba (Art. 506) with the line 24; also bisect c a in e, and erect ed perpendicular to ас, cutting dd ind; then 4 is the fourth point required. A fifth point may be found, as at f, by assuming a, d and б, as the three given points, and proceeding as before. 50,

Page 598

560 PRACTICAL GEOMETRY. also, any number of points. may be found simply by using any three already found. .This problem will be serviceable in obtaining short pieces of very flat sweeps. (See 477. 240.) The proof of the correctness of this method is found in the fact that equal chords subtend equal angles (477. 3 57). Join 4 and с; then since ae and ес are, by construction, equal, therefore the chords ad and dc are equal; hence the angles they subtend, “ба and dbc, are equal. So, like- wise, chords drawn from а to f, and from f to d, are equal, and subtend the equal angles 40/7 and fba. Additional points деуола а or 6 may be obtained on the same principle. To obtain a point beyond а, оп 0, as a centre, describe with any radius the arc 207; make oz equal to oz; through ё and n draw 6g; оп а as centre and with a f for radius, describe the arc, cutting gb at g, then g is the point жишш 516.—'ro Describe a Segment of a Circle by a Set-Tri- angle.—Let a 2 (Fig. 369) be the chord, and cd the height [4 d e | | T Fic. 369. of the segment. 'Secure two straight-edges, or rulers, in the position c e and c f, by nailing them together at c, and affixing а brace from e to f; put in pins аға and д; move the angu- lar point с in the direction acd; keeping the edges of the triangle hard against the pins .« and д; a pencil held at c will describe the arc acb. A curve described by this process is accurately circular, and is not a mere approximation to a circular arc, as some may suppose. This method produces a circular curve, be- cause all inscribed angles on one side of a chord-line are equal (Art. 356). To obtain the radius from a chord and its versed sine, see Art. 444. If the angle formed by the rulers at c be a right angle,

Page 599

TO FIND THE VERSED SINE. 561 the segment described will be a semi-circle. This problem is useful in describing centres for brick arches, when they are required to be rather flat. Also, for the head hang- ing-stile of a window-frame, where a brick arch, instead of a stone lintel, is to be placed over it. 517.—To Find the Radius of an Arc of a Circle when the Chord and Versed Sine are Given.—The radius is equal to the sum of the squares of half the chord and of the versed sine, divided by twice the versed sine. This is expressed, QS algebraically, thus: 7 , where 7 is the radius, c the chord, and v the versed sine (477. 444). Example.—ln a given arc of a circle a chord of 12 feet has the rise at the middle, or the versed sine, equal to 2 feet, what is the radius ? Half the chord equals 6, the square of 6 is, 6 x 6 — 36 The square of the versed sine 15, 2 х 2 ек Their sum equals, 40 ‘ Twice the versed sine equals 4, and 40 divided by 4 equals то. Therefore the radius, in this case, is ro feet. This result is shown in less space and more neatly by using the above algebraical formula. For the letters substituting ey a 12ү a G42" becomes 7 = Gey + 2° 20 2 ха? and performing the arithmetical operations here indicated equals— their value, the formula 7 = 5[8.—' ro Find the Versed Sine of an Arc of a Circle when the Radius and Chord are Given.—The versed sine is equal to the radius, less the square root of the difference of the squares of the radius and half chord; expressed alge- braically thus: v = r — ¥r* — (6), where > is the radius, v the versed sine, and ¢ the chord. (Equation (161.) reduced.)

Page 600

562 PRACTICAL GEOMETRY. Example.—1ln an arc of a circle whose radius is 75 feet, what is the versed sine to a chord of 120feet? By the table in the Appendix it will be seen that— The square of the radius, 75, equals . * 5625 The square of half the chord, бо, equals . 5600 The difference is 4 i A Е 2025 The square root of this is . й ; ; "т This deducted from the radius . : . ai The remainder 15 the versed sine, E ЕЕ? This is expressed by the formula, thus— v—75— 075° — (280) = 75 — #5625 — 3600 = 75 — 45 = 30. 519.—Te Describe the Segment of a Circle by Intersee- tion of Lines.—Let a? (Fig. 370) be the chord, and cd the |o 01... (али height of the segment. Through с draw e f parallel to аф, draw 2 f at right angles to cb; make ce equal to cf; draw ag and b% at right angles to а0; divide ce, cf, da, db, ағ, and 24, each into a like number of equal parts, as four; draw the lines 1 1, 2 2, etc., and from the points o, o, and о, -draw lines to с; at the intersection of these lines trace the curve, acd, which will be the segment required. In very large work, or in laying out ornamental gar- dens, etc., this will be found useful; and where the centre of the proposed arc of a circle is inaccessible it will be inval- uable. (To trace the curve, see note at 477. 550.) The lines ea, са, and f), would, were they extended, meet in a point, and that point would be in the opposite side of the circumference of the circle of which ac? isa

Page 601

ORDINATES TO AN ARC. 22 563. segment. The lines 11,22, 3 3, would likewise, if extended, meet in the same point. The line cd, if extended to the op- posite side of the circle, would become a diameter. The line f 6 forms, by construction, a right angle with 2c, and hence the extension of fò would also form a right angle with £c, on the opposite side of фе; and this right angle would be the inscribed angle in the semi-circle; and since this is re- quired to be a right angle (Art. 352), therefore the construc- tion thus far is correct, and it will be found likewise that at each point in the curve formed by the intersection of the radiating lines, these intersecting lines are at right angles. 520. — Ordinates, — Points in the circumference of a circle may be obtained arithmetically, and positively accu- rate, by the calculation of ordinates, or the parallel lines o 1, Етс. 371. 02,03,04 (Fzg. 371). These ordinates are drawn at right angles to the chord-line ад, and they may be drawn at апу distance apart, either equally distant or unequally, and there may be as many of them as is desirable; the more there are the more points in the curve will be obtained. If they are located in pairs, equally distant from the versed sine са, calculation need be made only for those on one side of cd, as those on the opposite side will be of equal lengths, re- spectively ; for example: o t, on the left-hand side of cd, is equal to o1 on the right-hand side, 02 on the right equals 02 on the left, and in like manner for the others. The length of any ordinate is equal to the square root of the difference of the squares of the radius and abscissa, less the difference between the radius and versed sine (477. 445) The abscissa being the distance from the foot of the versed sine to the foot of the ordinate. Algebraically,

Page 602

564 PRACTICAL GEOMETRY. t= Vr? — x? — (r— 6), where 2 is put to represent the ordi- nate; x, the abscissa; 0, the versed sine; and 7, the radius. Example.—An arc of a circle has its chord ад (Fig. 371) 100 feet long, and its versed sine cd, 5 feet. It is required to ascertain the length of ordinates for a sufficient number of points through which to describe the curve. To this end it is requisite, first, to ascertain the radius. This is readily уь, done іп accordance with 477.517. For Dar becomes ; 2v ОЕ ; : 4 в = 252.5 = radius. Having the radius, the curve might at once be described without the ordinate points, but for the impracticability that usually occurs, in large, flat segments of the circle, of getting a location for the centre, the centre usually being inaccessible. The ordinates are, therefore, to be calculated. In Ато. 371 the ordinates are located equidistant, and are то feet apart. It will only be requisite, therefore, to calculate those on one side of the versed sine cd. For the first ordinate 01, the formula t= Vr? — x? — (r — b) becomes— t = ¥252-5" — 10* — (252-5 — 5). = /63756:33 — 100 — 247-5. = 252-3019 — 247-5. | = 4:8019 = the first ordinate, o 1. For the second— | 212 425209" —20 — АВЕ nM | 251-7066 — 247-5. = 4:2066 = the second ordinate, o2. For the third— f 22/252.5* — 30* = 247-5. = 250-7115 — 247: 5. E 3:2115 — the third ordinate, o 3.

Page 603

TO DESCRIBE A TANGED CURVE. 565 For the fourth— (t= 9252-5! — 40! — 247-5. = 249:31I5 = 247: 5. = 1 -8115 =the fourth ordinate, o 4. The results here obtained are in feet and decimals of a foot. To reduce these to feet, inches, and eighths of an inch, proceed as at Reduction of Decimals in the Appendix. If the two-feet rule, used by carpenters and others, were decimally divided, there would be no necessity of this re- duction, and it is to be hoped that the rule will yet be thus divided, as such a reform would much lessen the labor of computations, and insure more accurate measurements. Versed sine cd = ft.5-0 = Ё. 5-0 inches. Ordinates от = 4.8019 = 4-9% inches, nearly. << o2= 4.2000 = 442% inchesiearly. « O3: 32ҒІ5 с. 3-48 inches, Шу. “ O4- 1.8115 = 1-0) inches, nearly. 521.—In a Given Angle, to Describe a ranged Curve. —Let abc (Fig. 372) be the given angle, and г in the line aå, 5 | Fig. 272: and 5 in the line dc, the termination of the curve. Divide 1 б.апа 45 into а like number of equal parts, as аёт, 2, 3, 4, and 5; join 1 and 1,2 and 2, 3 and 3, etc.; and a regular curve will be formed that will be tangical to the line а, at the point 1, and to dc at 5. This is of much use in stair-building, in easing the angles formed between the wall-string and the base of the hall, also between the front string and level facia, and in many other instances. The curve is not circular, but of the form of the parabola (Zzg. 418); yet in large angles the difference

Page 604

566 PRACTICAL GEOMETRY. is not perceptible. This problem can be applied to describ- ing the curve for door-heads, window-heads, etc., to rather better advantage than 477. 516. For instance, let аб (Fig. 373) be the width of the opening, and cd the height of the FIG. 373; arc. Extend са, and make de equal to cd; join а and e, also e and 6; and proceed as directed above. . 522. a Circle within any Given Triangle, | so that the Sides of the Triangle shall be Tangical.— Let adc (Fig. 374) be the given triangle. Bisect the angles FIG. 374. же а and 6 according to Art. 506; upon 4, the point of intersec- tion of the bisecting lines, with the radius de, describe the required Сте 523.—About a Given Circle, to Describe an Equilateral Triangle.—-Let а4бс (Fig. 375): be the given circle. Draw the diameter cd; upon d, with the radius of the given circle, describe the arc aeb; join а and д; draw fg at right angles їо dc; make fc and cg each equal to аб; from f, through a, draw fh, also from g, through 2, draw gh; then fgh% will be the triangle required. | 524.-то Find a Right Line nearly Equal to the Cir- cumference of a Circle.—Let abcd (Fig. 376) be the given

Page 605

A RIGHT LINE EQUAL TO A CIRCUMFERENCE. 567 circle. Draw the diameter ac; on this erect an equilateral triangle а ес according to Art. 525; draw gf parallel to ac; extend ес to f, also ea to g; then gf will be nearly the * length of the semi-circle adc; and twice gf will nearly equal the circumference of the circle a бс d, as was required. Lines drawn from e, through any points in the circle, as о, о and o, to 2, рапа 2, will divide gf in the same way as the semi-circle а Zc 1s divided. So, апу portion of a circle may be transferred to a straight line. This is a very useful Еле. 376. problem, and should be well studied, as it is frequently used to solve problems on stairs, domes, etc. Another method. Let ab f c (Fig. 377) be the given circle. Draw the diameter а с; from Z, the centre, and at right an-

Page 606

568 l PRACTICAL GEOMETRY. gles to ac, draw 40; join 6 and с; bisect dc at e; from d, through e, draw df; then ef added to three times the di- ameter, will equal the circumference of the circle sufficiently near for many uses. The result is a trifle too large. If the Fre. 377- circumference found by this rule be divided by 648.22 the quotient will be the excess. Deduct this excess, and the remainder will be the true circumference. This problem is rather more curious than useful, as it is less labor to perform the operation arithmetically, simply multiplying the given diameter by 3- 1416, or, where a greater degree of accuracy is needed, by 31415926. (See Art. 446.) | POLYGONS, ETC. 525.—Upon a Given Line to Construct an Equilateral Triangle.—Let a b (Fig. 378) be the given line. Upon а and € “ b Fic. 399. б, with a 6 for radius, describe arcs, intersecting аё с; join a and с, also с and 6; then ac? will be the triangle required. 526.-То Describe an Equilateral Rectangle, or Square. —Let a? (Fig. 379) be the length of a side of the proposed

Page 607

POLYGONS IN CIRCUMSCRIBING CIRCLES. 569 square. Upon a and 0, with ад for radius, describe the arcs ad and бс; bisect the arc ae in f; upon e, with ef for ra- dius, describe the arc cfd; join а and c, c and d, d and 6; then acd? will be the square required. FIG. 379. 527.—Within a Given Circle, to Inscribe an Equilateral Triangle, Hexagon or Dodecagon.— Let abcd (Fig. 380) be the given circle. Draw the diameter 24; upon 0, with the radius of the given circle, describe the arc aec; join а and с, also a and d, and c and d—and the triangle is completed. For the hexagon: from a, also from c, through e, draw the lines af and cg; join а and 0, 6 and с, cand f, etc., and the hexagon is completed. The dodecagon may be formed by bisecting the sides of the hexagon. Each side of a regular hexagon is exactly equal to the radius of the circle that bp the figure. For the radius is equal to a chord of an arc of 60 degrees; and, as every circle is supposed to be divided into 360 degrees, there is just 6 times 60, or 6 arcs of 60 degrees, in the whole cir- cumference. А line drawn from each angle of the hexagon

Page 608

570 PRACTICAL GEOMETRY. to the centre (as in the figure) divides it into six equal, equi- lateral triangles. 528. | —Let abcd (Fig. 381) be the given square. Draw the diagonals Fic 381. ай and бс; upon a, 0, c, and а, with ae for radius, describe arcs cutting the sides of the square at т, 2, 3, 4, 5, 6, 7, and 8; join r and 2, 3 and 4, 5 and 6, etc., and the figure is com- pleted. In order ‘to eight-square a hand-rail, or any piece that is Fic. 382. to be afterwards rounded, draw the diagonals ad and дс upon the end of it, after it has been squared-up. Set a gauge to the distance ағ and run it upon the whole length of the stuff, from each corner both ways. This will show

Page 609

BUTTRESSED OCTAGON. 571 how much is to be chamfered off, in order to make the piece octagonal (Art. 354). 529.--Те Find the Side of a Buttressed Octagon.— Let ABCDE (Fig. 382) represent one quarter of an octagon structure, having a buttress ÆHF G F at each angle. The distance M H, between the buttresses, being given, as also F G, the width of a buttress; to find HC or C F, in order to obtain BC, the side of the octagon. Let BC, a side of the octagon, be represented by 2; or DC by 46. Let M H = а; От 7 П--ча; and 7 CCS Then we have— ¥D+¥C=CD, a+ x = ў 0, а+ 25 = 0. ГОРА С put p; ОРАСИ = Now £ D is the radius of an inscribed circle and, as per = б equation (140.), equals 7 = ( 2 + 1) эм Also, E C is the radius of a circumscribed circle, and, às per equation (141.), equals К = Уз 42 4 42: The two triangles, C X and C E D, are homologous; for the angles at C are common and the angles at X and D are right angles. Having thus two angles of one equal respectively to the two angles of the other, therefore (Art. 345) the remaining angles must be equal. Hence, the sides of the triangles are proportionate, or— гав ата СЕ R к: № iada х= 0. 'The value of the side, as above, is— b—artzr—actf

Page 610

572 PRACTICAL GEOMETRY. And taking the value of R and 7, as above, we have— р Чд” Y2V2 4, — —— ДИ is Км 7 iv 1 м 4 . . R Substituting this for “>, ме Һауе— · М2 #22 баты ab V2 + І The numerical coefficient m p reduces to 1-0823923 or 1.0824, nearly. Therefore we have— b = a + 1:0824 2. ^ (207.) Or: The szde of a buttressed octagon equals the distance be- tween the buttresses plus 1-0824 times the width of the face of the buttress. For example: let there be an octagon building, which measures between the buttresses, as at M H, 18 feet, and the face of the buttresses, as FG, equals 3 feet; what, in such a building, is the length of a side BC? For this, using equa- tion (207.), we have— бл 18 + 1.0824 x 3 = 18 + 3.2472 01 . 8272» Or: The side of the octagon 2 C equals 21 feet and nearly 3 inches. 530.— Within a Given Cirele to Inscribe any Regular Polygon.— [Let adc 2 (Figs. 383; 384, and 385) be given circles. Draw the diameter 2c; upon this erect an equilateral trian- gle aec, according to Art. 525; divide ac into as many equal parts as the polygon is to have sides, as at 1, 2, 3, 4, etc. ; from e, through each even number, as 2, 4, 6, etc., draw lines

Page 611

TO DESCRIBE ANY REGULAR POLYGON. 573 cutting {һе circle in the points 2, 4, etc.; from these points and at right angles to ас draw lines to the opposite part of the circle; this will give the remaining points for the polygon, as 0, f, etc. In forming a hexagon, the sides of the triangle erected upon ас (as at Fig. 384) mark the points 0 and f. This method of locating the angles of a polygon is an approxima- tion sufficiently near for many purposes; it is based upon the like principle with the method of obtaining a right line nearly equal to a circle (Art. 524). The method shown at Art. 531 1s accurate. [/1 ў "; Fic, 386. Fic. 388 53І.-Сроп a Given Line to Describe any Regular Polygon.—Let a? (Figs. 386, 387, апа 388) be given lines, equal to a side of the required figure. From 0 draw дс at right angles toad; upon а and 4, with аб for radius, describe the arcs acd and feb; divide ac into as many equal parts

Page 612

574 PRACTICAL GEOMETRY. as the polygon 15 to have sides, and extend those divisions from c towards 4; from the second point of division, count- ing from c towards а, as 3 (Fig. 386), 4 (Fig. 387), and 5 (Fig. 388), draw a line to 0; take the distance from said point of division to а, and set it from ? to e; join e and a; upon the intersection o with the ‘radius oa, describe the circle af Z5; then radiating lines, drawn from 4 through the even numbers on the arc аа, will cut the circle at the several angles of the required figure. In the hexagon (Fig. 387), the divisions on the arc ad are not necessary ; for the point o is at the intersection of the arcs ad and fb, the points f and 4 are determined by the intersection of those arcs with the circle, and the points above g and # can be found by drawing lines from a and 6 through the centre о. In polygons of a greater number of sides than the hexagon the intersection o comes above the arcs; in such case, therefore, the lines ae and 65 (Frig. 388) have to be extended before they will intersect. This method of describing polygons is founded on correct principles, and is therefore accurate. In the circle equal arcs subtend equal angles (4775. 357 and 515). Although this method is accurate, yet polygons may be described as accurately and more simply in the following manner. It will be observed that much of the process in this method is for the purpose of ascertaining the centre of a circle that will circumscribe the proposed polygon. By reference to the Table of Poly- gons in Art. 442 it will be seen how this centre may be ob- tained arithmetically. This is the rule: multiply the given side by the tabular radius for polygons of a like number of sides with the proposed figure, and the product will be the radius of the required circumscribing circle. Divide this- circle into as many equal parts as tbe polygon is to have sides, connect the points of division by straight lines, and the figure is complete. For example: It is desired to de- scribe a polygon of 7 sides, and 20 inches a side. The tabu- lar radius is 1-15238. This multiplied by 20, the product, 23-0476 is the required radius in inches. Тһе Rules for the Reduction of Decimals, in the Appendix, show how to change decimals to the fractions of a foot or an inch. From

Page 613

EQUAL FIGURES. 575 this, 23-0476 is equal to 23,4, inches, nearly. It is not needed to take all the decimals in the table, three or four of them will give a result sufficiently near for all ordinary practice. 532.—' To Construct a Triangle whose Sides shall be severally Equal to Three Given Lines.—Let a, 2 and c (Fig. 389) be the given lines. Draw the line de and make it equal f a O08 Fic, 389. c; upon e, with 6 for radius, describe an arc at f; upon d, with а for radius, describe an arc intersecting the other at /; join d and f, also f and e; then dfe will be the triangle required. 533.--То Construct a Figure Equal to a Given, Right- | lined Figure. — Let adcd (fig. 390) be the given figure. Make e f (Fig. 391) equal to cd; upon f, with Za for radius, 6 Гіс. 390. FIG. 301. describe an arc at g; upon e, with ca for radius, describe an arc intersecting the other at g; join сапа e; upon f and g, with 40 and аф for radius, describe arcs intersecting at д; join g and A, also £4 and f; then 270. 391 will every way equal Ғ?р. 390. So, right-lined figures of any number of sides may be copied, by first dividing them into triangles, and then pro-

Page 614

576 PRACTICAL GEOMETRY. ceeding as above. Тһе shape of the floor of any room, or of any piece of land, etc., may be accurately laid out by this problem, at a scale upon paper; and the contents in square feet be ascertained by the next. 534.—To Make a Parallelogram equal to a Given Triangle.—Let a dc (Fig. 392) be the given triangle. From а draw ad at right angles to dc; bisect ad in e; through e FIG. 392. draw fg parallel to ^c; from 0 and c draw £ f and c g par- allel to Ze; then ^ / gc will be a parallelogram containing a surface exactly equal to that of the triangle a дс. Unless the parallelogram is required to be a rectangle, the lines f and cg need not be drawn parallel to Ze. Ifa rhomboid is desired they may be drawn at an oblique angle, provided they be parallel to one another. То ascertain the area of a triangle, multiply the base dc by half the perpen- FIG. 393. dicular height Za. Іп doing this it matters not which side is taken for base. 535.—A Parallelogram being Given, to Construct An- other Equal to ít, and Having a Side Equal to a Given Linc. —Let A (Fig. 393) be the given parallelogram, and В the given line. Produce the sides of the parallelogram, as at

Page 615

SQUARE EQUAL TO TWO OR MORE SQUARES. 577 а, 6, c, and 2; make eZ: equal to Б; through d draw c f par- allel to £^; through e draw the diagonal са; from а draw af parallel to ed; then C will be equal to А. (See Art. 340.) 536.--То Make а Square Equal to two or more Given Squares,—Let 4 and B: (fig. 394) be two: given squares.. FIG. 394. Place them so as to form a right angle, as at a; join band с; then the square C, formed upon the line бс, wil! be eqval in extent to the squares A and P added together. Again: if a à (Fig. 395) be equal to the side of a given square, ca, placed at right angles to a£, be the side of another given square, Fic. 395. and са, placed at right angles to cå, be the side of a third given square, then the square A, formed upon the line 25, will be equal to the three given squares. (See Arz. 353.) The usefulness and importance of this problem are pro- verbial. То ascertain the length of braces and of rafters in

Page 616

573 PRACTICAL GEOMETRY. framing, the length of stair-strings, etc., are some of the pur- poses to which it may be applied in carpentry. (See note to Art. 503.) If the lengths of any two sides of a right- angled triangle are known, that of the third can be ascer- tained. Because the square of the hypothenuse is equal to the united squares of the two sides that contain the right angle. (r)—The two sides containing the right angle being known, to find the hypothenuse. Rule.— Square each given side, add the squares together, and from the product extract the square root ; this will be the answer. For instance, suppose it were required to find the length of a rafter for a house, 34 feet wide—the ridge of the roof to be о feet high, above the level of the wall-plates. Then 17 feet, half of the span, is one, and 9 feet, the height, is the other of the sides that contain the right augle. Proceed as directed by the rule: 17 D 17 29 119 81 — square of 9. 17 289 — square of 17. 289 — square of 17. 370 Product. I ) 370( 19-235 + = square root of 370; equal 19 feet 27 in., 1. I | nearly ; which would be the required length of the rafter. 3843 ) 13600 223 11529 38465): “207100 102325

Page 617

TO FIND THE LENGTH OF A RAFTER. 579 (By reference to the table of square roots in the Appen- dix, the root of almost any number may be found ready calculated ; also, to change the decimals of a foot to inches and parts, see Rules for the Reduction of Decimals in the Appendix.) | Again: suppose it be required, in a frame building, to find the length of a brace having a run of three feet each way from the point of the right angle. The length of the sides containing the right angle will be each 3 feet; then, as before— ЕТ .9— square of one side. @times 3 = о = square ot the other sige. 18 Product: the square root of which is 4- 2426+ ft., Or 4 feet 2 inches and 7 full. (2.)— The hypothenuse and one side being known, to find the other side. Rule.—Subtract the square of the given side from the square of the hypothenuse, and the square root of the prod- uct will be the answer. Suppose it were required to ascertain the greatest per- pendicular height a roof of a given span may have, when pieces of timber of a given length are to be used as rafters. Let the span be 20 feet, and the rafters of 3 x 4 hemlock joist. These come about 13 feet long. Тһе known hy- pothenuse, then, 15 13 feet, and the known side, то feet— that being half the span of the building. I3 13 39 13 169 — square of hypothenuse. 10 times IO = 100 = square of the given side. 69 Product: the square root of which is

Page 618

Ded т 9808 7” PRACTICAL GEOMETRY. 8.3066+ feet, or 8 feet з inches and 8 full. This will be the greatest perpendicular height, as required. Again: suppose that in a story of 8 feet, from floor to floor, a step-ladder is required, the strings of which are to be of plank 12 feet long, and it is desirable to know the greatest run such a length of string will afford. In this case, the two given sides are—hypothenuse 12, perpendicular 8 feet. 12 times 12 — 144 — square of hypothenuse, 8 times 8 = 64 = square of perpendicular. 80 Product: the square root of which is 8-9442+ feet, ог 8 feet 11 inches and 4$—the answer, as re- quired. ` Many other cases might be adduced to show the utility of this problem. A practical and ready method of ascer- taining the length of braces, rafters, etc., when not of a great length, is to apply a rule across the carpenters’- square. Suppose, for the length of a rafter, the base be 12 feet and the height 7. Apply the rule diagonally on the square, so that it touches 12 inches from the corner on one side, and 7 inches from the corner on the other. The number of inches on the rule which are intercepted by the sides of the square, 132, nearly, will be the length of the rafter in feet ; viz., 13 feet and 2 of a foot. . If the dimensions are large, as 30 feet and 20, take the half of each on the sides of the square, viz., 15 and то inches; then the length in inches across will be one half the number of feet the rafter is long. This method is just as accurate as the preceding ; but. when the length of a very long rafter is sought, it requires great care and pre- cision to ascertain the fractions. For the least variation on the square, or in the length taken on the rule, would make perhaps several inches difference in the length of the rafter. For shorter dimensions, however, the result will be true enough. 537.—To Make a Circle Equal to two Given Circles.— Let A and В (Fig. 306) be the given circles. In the right- angled triangle абс make a? equal to the diameter of the.

Page 619

АБ; Ы); S UNIVERSES НОД circle В, and cà equal to the diameter of {һе сіт the hypothenuse ac will be the diameter of a circle ie will be equal in area to the two circles 4 and В, added together. “ы SIMILAR FIGURES. Fic. 396. Any polygonal figure, as A (Fig. 397), formed on the hy- pothenuse of a right-angled triangle, will be equal to two similar figures,* as B and C, formed on the two legs of the triangle. d Етс. 397. 538.—To Construct a Square Equal to a Given Rect- angle.—Let А (Fig. 398) be the given rectangle. Extend the side ад and make ёс equal to ġe; bisect ac in f, and upon f, with the radius fa, describe the semi-circle асс; extend e? till it cuts the curve in g; then a square £g Ad, formed on the line 02, will be equal in area to the rectan- gle A. * Similar figures are such as have their several angles respectively equal, and their sides respectively proportionate.

Page 620

582 PRACTICAL GEOMETRY. Another method. Let А (Fig. 399) be the given rectangle. Extend the side ад and make ad equal to ac; bisect ad in e; upon e, with the radius ea, describe the semi-circle afd; - extend gé till it cuts the curve in f; join а and f; then Fic. 398. the square 2, formed on the line a f, will be equal in area to the rectangle 4. (See Arts. 352 and 353.) 539.—— To Form a Square Equal to a Given Triangle. — Let аб (Fig. 398) equal the base of the given triangle, and де equal half its perpendicular height (see Fig. 392); then pro- ceed as directed at Art. 538. 540.—'Two Right Lines being Given, to Find a Third Proportional Thereto.— Let А and B (Fig. 400) be the given lines. Make a? equal to 4; from a draw ас at any angle

Page 621

PROPORTIONATE DIVISIONS IN LINES. 583 with 20; make ас and ad each equal to B; join с and 08; from 4 drag de parallel to có; then ae will be the third proportional required. That i IS, a e bears the same propor- tion to В as B does to A. А ДУ: Fic. 400. 54..—' Three Right Lines being Given, to Find a Fourth Proportional Thereto.—Let А, В, and C (Fig. дог) be the given lines. Make аф equal to 4; from a draw ас at any суб е <a we. NC е ő Fic. 401. angle with аб; make ac equal to В and ae equal to С; join с and 2; from e draw е parallel to cd; then a f will be the fourth proportional required. That 15, af bears the same proportion to C as В does to A. To apply this problem, suppose the two axes of a given _ ellipsis and the longer axis of a proposed ellipsis are given. Then, by this problem, the length of the shorter axis to the proposed ellipsis can be found ; so that it will bear the same proportion to the longer axis as the shorter of the given ellipsis does to its longer. (See also A77. 550.) 542. g Given, to’ Divide Another, Longer or Shorter, Given Line in the Same Proportion.—Let А (Fig. 402) be the line to be di- vided, and Z the line with its divisions. Make ад equal to B with all its divisions, as at 1, 2, 3, etc.; from a draw ac at any angle with ад; make ac equal to 4; join c and д; from .

Page 622

584 PRACTICAL GEOMETRY. the points 1, 2, 3, etc., draw lines parallel to сб; then these will divide the line ec in the same proportion as 2 is divided —-as was required. This problem will be found useful in proportioning the A $$ $$$ [4 a 1 2 8 NN FIG. 402. members of a proposed cornice, in the same proportion as those of a given cornice of another size. (See 477.321.) So of a pilaster, architrave, etc. 543.—Between Two Given Right Lines, to Find a Mean Proportional.—Let 4 and В (Fig. 403) be the given lines. On the line ас make a? equal to А and bc equal to В; bisect ас in e; upon e, with ea for radius, describe the semi-circle adc; at ё erect Od at right angles to ас; then 731 ж а е ё с А 8 Етс. 403. ба will be the mean proportional between A and 5. That is, 20 15 to dd as bd is to ёс. This is usually stated thus: 20: 020: 04: bc, and since, the product of the means equals the product of the extremes, therefore, a 0 x dc = 24°. This is shown.geometrically at 477. 538. CONIC SECTIONS. 544.—Definitions. — ЇГ a cone, standing upon a base that is at right angles with its axis, be cut by a plane, per-

Page 623

AXIS AND BASE OF. PARABOLA. 585 pendicular to its base and passing through its axis, the sec- tion will.be an isosceles triangle (as абс, Fig. 404); and the base will be a semi-circle. If a cone be cut by a plane in the direction e f the section will be an elipsis; if in the direction mt, the section will be a parabola, and if in the direction 70, ап hyperbola. (See Art. 499.) If the cutting planes be at right angles with the plane абс, then— 545.—To Find the Axes of the Ellipsis: bisect г f (fig. 404) in g; through g draw Z parallel to аб; bisect #2 in 7; Fic. 404. upon /, with 74 for radius, describe the semi-circle 202; from g draw gk at right angles to 22; then twice g+ will be the conjugate axis and e f the transverse. 546.—To Find the Axis and Base of the Parabola.— Let э» / (Fig. 404), parallel to ac, be the direction of the cut- ting plane. From m draw md at right angles toad; then Zm will be the axis and height, and zz an ordinate and half the base, as at Figs. 417, 418. .547.— To Find the Height, Base, and Transverse Axis of an Hyperbola.—Let or (Fig. 404) be the direction of the

Page 624

586 PRACTICAL GEOMETRY. cutting plane. Extend оғ and ac till they meet at z; from о draw ор at right angles to ад; then ro will be the height, nr the transverse axis, and ор half the base; as at Fig. 419. 548. being Given, to Find the Foci, and to Describe an Ellipsis with a String.—Let ad (Fig. 405) and cd be the given axes. Upon c, with ae or бе for radius, de- scribe the arc ff; then f and /, the points at which the arc cuts the transverse axis, wil be the foci. At f and f place two pins, and another at c; tiea string about the three pins, so as to form the triangle ffc; remove the pin from c and place a pencil in its stead; keeping the string taut, Fic. 405. move the pencil in the direction c ga; it will then describe the required ellipsis. The lines fg and g/ show the posi- tion of the string when the pencil arrives at g. This method, when performed correctly, is perfectly ac- curate; but the string is liable to stretch, and is, therefore, not so good to use as the trammel. Іп making an ellipse by astring or twine, that kind should be used which has the least tendency to elasticity. For this reason, a cotton cord, such as chalk-lines are commonly made of, is not proper for the purpose; a linen or flaxen cord is much better. 549. to Deseribe an Ellipsis with a Trammel.—Let a? and cd (Fig. 406) be the given axes. Place the trammel so that a line passing through the centre of the grooves would coincide with the axes; make

Page 625

ELLIPSE BY TRAMMEL. . 587 the distance from the pencil e to the nut f equal to half c4; also, from the pencil e to the nut g equal to half ad; letting the pins under the nuts slide in the grooves, move the tram- mel eg in the direction cbd; then the pencil at e will de- scribe the required ellipse. | А trammel may be constructed thus: take two straight strips of board, and make a groove on their face, in the cen- tre of their width; join them together, in the middle of their length, at right angles to one another; as is seen at Fég. 406. А rod is then to be prepared, having two movable nuts made of wood, with a mortise through them of the size of the rod, and pins under them large enough to fill the grooves. Make a hole at one end of the rod, in which to Fic. 406. / place a pencil. In the absence of a regular trammel a tem- porary one may be made, which, for any short job, will an- swer every purpose. Fasten two straight-edges at right angles to one another. Lay them so as to coincide with the axes of the proposed ellipse, having the angular point at the centre. Then, іп a rod having a hole for the pencil at one end, place two brad-awls at the distances described at 477. 549. While the pencil is moved in the direction of the curve, keep the brad-awls hard against the straight-edges, as directed for using the trammel-rod, and one quarter of the ellipse will be drawn. Then, by shifting the straight- | edges, the other three quarters in succession may be drawn. If the required ellipse be not too large, a carpenters’-square may be made use of, in place of the straight-edges. An improved method of constructing the trammel is as

Page 626

588 PRACTICAL GEOMETRY. follows: make the sides of the grooves bevelling from the face of the stuff, or dove-tailing instead of square. | Prepare two slips of wood, each about two inches long, which shall be of a shape to just fill the groove when slipped in at the end. These, instead of pins, are to be attached one to each of the movable nuts'with a screw; loose enough for the nut to move freely about the screw as an axis. The advantage of this contrivance is, in preventing the nuts from slipping out of*their places during the operation of describing the curve. 550.—' Tro Describe an EHipsis by Ordinates.—Let 22 and cd (Fig. 407) be given axes. With ce or ed for radius Fic. 407, | describe the quadrant Үс; divide f4, ae, and e£, each into a like number of equal parts, as at 1, 2, апа 3; through these points draw ordinates parallel to cd апа: g; take the distance 12 and place it at 1 7, transfer 2 / to 277, and 3£ to 32; through the points a, 2; m, 2, and с, trace a curve, and the ellipsis will be completed. The greater the number of divisions on a, e, etc., in this and the following problem, the more points in the curve can be found, and the more accurate the ‘curve can be traced. If pins are placed in the points л, zz, 7, etc.; and a thin slip of wood bent around by them, the curve can be made quite ' correct. This method is mostly used in tracing face-moulds for stair hand-.railing. 55|.—'ro Describe an Ellipsis by Intersection of Lines. —Let a ? and ed (Fig. 408) be given axes. Through с, draw

Page 627

ELLIPSE BY INTERSECTION OF LINES. 589 Sg parallel to.2^; from а апа 6 draw af and ? g at right angles to ад; divide fa, gb, ae, and её, each into a like. number of equal parts, as at т, 2, 3, and о, о, о; from 1, 2, and 3, draw lines to с; through о, о, and o, draw lines from 4, intersecting those drawn to с; then a curve, traced through the points 2, 2, 2, will be that of an ellipsis. Where neither trammel nor string is at hand, this, per- . haps, is the most ready method of.drawing an ellipsis. The divisions should be small, where accuracy is desirable. By this method an ellipsis may be traced without the axes, pro- vided that a diameter and its conjugate be given. Thus, ad and cd (Fig. 409) are conjugate diameters: fg 15 drawn par- allel to ад, instead of being at right angles to cd; also, fa and g^ are drawn parallel to c4, instead of being at right angles to a 2.

Page 628

590 PRACTICAL GEOMETRY. 552.—' To Describe an Ellipsis by Intersecting Arcs.— Let аф and са (Fig. 410) be given axes. Between опе of the foci, / and /, and the centre e, mark any number of points, at random, as 1, 2, and 3; upon f and f, with 41 for radius, describe arcs at 2,2, с, and ғ; upon f and f, with a1 for А Ете. 410. radius, describe arcs intersecting the others at ¢, e; g, and ғ; then these points of intersection will be in the curve of the ellipsis. The other points, 2 and z, are found in like manner, viz.: Æ is found by taking 22 for one radius, and a2 for the other; zis found by taking 03 for one radius, and a 3 for the Fic. ШІ. other, always using the foci for centres. Then by tracing a curve through the points c, g, 2, 2, б, etc., the ellipse will be completed. This problem is founded upon the same principle as that of the string. This is obvious, when we reflect that the length of the string is equal to the transverse axis, added to

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TO DESCRIBE AN OVAL. 59I the distance- between the ioci. 'See' Fig. 405, in wite 77 equals аг, the half of the transverse axis. | 553.—' To Deseribe a Figure Nearly in the Shape of an Elipsis, by a Pair of Compasses.—Let ad and cd (Fig. 411) be given axes. From с draw ce parallel to 22; from а draw ae parallel to cd; join e and 4; bisect ea in f; join. f and c, intersecting ed in 2; bisect гс ino; from о draw og at right. angles to 2с, meeting cd extended to ғ; join гапа g, cutting the transverse axis іп ғ; make Z 7 equal to Zg, and 2 Ё equal to Zr; from 7, through 7 and 2, draw 7 m and ул; also, from g, through £, draw gZ; upon g and 7, with gc for radius, describe the arcs 2/ and mz; upon > and &, with ra for Fic. 412. radius, describe the arcs 2 and /z; this will complete the figure. 1 When the axes are proportioned to one another, as at 2 to 3, the extremities, с and d, of the shortest axis, will be the centres for describing the arcs 2/ and ma; and the inter- section of ed with the transverse axis will be the centre for describing the arc ж, 2, etc. Ав the elliptic curve 15 contin- ually changing its course from that of a circle, a true ellipsis cannot be described with a pair of compasses. The above, therefore, is only an approximation. 554.-То Draw an Oval in the Proportion Seven by Nine.—Let cd (Го. 412) be the given conjugate axis. Bisect

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592 PRACTICAL GEOMETRY. c d in o, and through o draw a? at right angles to cd; bisect coin e; upon o, with ое for radius, describe the circle e fg% ; from e, through Z and f, draw ej and et; also, from g, through # and f, draw gk and g/; upon g, with ge for radius, describe the arc 22; upon e, with ed for radius, de- scribe the arc 72; upon д and f, with / £ for radius, describe the arcs 7 and 77; this will complete the figure. Fic. 413. This is an approximation to an ellipsis; and perhaps no method can be found by which-a well-shaped oval can be drawn with greater facility. Ву a little variation in the process, ovals of different proportions may be obtained. If quarter of the transverse axis is taken for the radius of the circle e fg 4, one will be drawn in the proportion five by seven. Fic. 414. 555.—To Draw a Tangent to an Ellipsis.—Let abcd (Fig. 413) be the given ellipsis, and 4 the point of contact. Find the foci (472. 548) f and f, and from them, through 4, draw fe and fd; bisect the angle (472. 506) edo with the line sz; then sz will be the tangent required.

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TO FIND THE AXES OF AN ELLIPSE. 593 556. —An Ellipsis with a Tangent Given, to Detect the Point of Contact.—Let a gb f (Fig. 414) be the given ellip- sis and tangent. Through the centre e draw a? parallel to the tangent; anywhere between e and f draw cd parallel to ab; bisect cd in o; through о апа е draw.fg; then g will be the point of contact required. 557.—A Diameter of an Ellipsis Given, to Find its Conjugate,—Let aé (Fig. 414) be the given diameter. Find the line fg by the last problem; then fg will be the diam- eter required. 558. ——Any Diameter and its Conjugate being Given, to Ascertain the Two Axes, and thence to Describe the Ellipsis. —Let a à and са (Fig. 415) be the given diameters, conjugate x У e Nf TN Fic. 415. to one another. Through c draw c f parallel to аб; from c draw cg at right angles to ef; make cg equal to ай or 20; join g and 2; upon g, with gc for radius, describe the arc ikcj; upon 2, with the same radius, describe the arc 7s; through the intersections / and л draw 20, cutting the tan- gent ef in о; upon o, with og for radius, describe the semi- circle её: join e and ғ, also gand f, cutting the arc 7c 7 in £ and 7; from e, through 2, draw e m, also from f, through h, draw Ұр; from 0 and 2 draw £r and 7s parallel to gh

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594 PRACTICAL GEOMETRY. cutting em in r, and /р іп s; make 2 equal to 4r, and 2) equal to 25; then z; and sp will be the axes required, by which the ellipsis may be drawn in the usual way. 559.—To Describe an Ellipsis, whose Axes shall be Proportionate to the Axes of a Larger or Smaller Given One.—Let acbd (Fig. 416) be the given ellipsis and axes, and 7 d Fic. 416. 77 the transverse axis of a proposed smaller one. Join а and c; from 2 draw Ze parallel to ac; make o f equal to ve; then ef wil be the conjugate axis required, and will bear the same proportion to 2/ as cd does to ад. (See Art. 541.) 560.— To Describe a Parabola by Intersection of Lines. —Let m/ (Fig. 417) be the axis and height (see с. 404) and dd а double ordinate and base of the proposed parabola. Through / draw аа parallel to dd; through d and d draw da and da parallel to m/; divide ad and dm, each into a like number of equal parts; from each point of division in

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TO DESCRIBE AN HYPERBOLA. 595 dm draw the lines 11, 22, etc., parallel to 2/4; from each point of division in Za draw lines to /; then a curve traced through the points of intersection a, o, and o, will be that of a parabola. Another method. Let ml (fig. 418) be the axis and height, and dd the base. Extend m/ and make /a equal to mZ; join а and d, and а and ж; divide ad and ad, each into a like number of equal parts, as at т, 2, 3, etc. ; join 1 and 1,2 and 2, etc., and the parabola will be completed. (See Arts. 460 to 472.) | 561.—Тө Describe ап Hyperbola by Intersection of Lines.—Let 7 о (Fig. 419) be the height, рр the base, and xr the transverse axis. (See Fig. 404.) Through r draw aa Fic. 418. Fic. 419. parallel to 22; from р draw ар parallel to ro; divide ар and po, each into a like number of equal parts; from each of the points of division in the base, draw lines to z;. from each of the points of division in ар, draw lines to ғ; then a curve traced through the points of intersection o, o, etc., · will be that of an hyperbola. қ The parabola and hyperbola afford handsome curves for various mouldings. (See Pzgs. 191 to 205; 222 to 224; 241 and 242; also note to Art. 318.)

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SECTION XVII.—SHADOWS. 562.—The Art of Drawing consists in representing solids upon a plane surface, so that a curious and nice ad- justment of lines is made to present the same appearance to the eye as does the human figure, a tree, or a house. It is by the effects of light, in its reflection, shade, and shadow, that the presence of an object is made known to us; so upon paper it is necessary, in order that the delineation may appear real, to represent fully all the shades and shad- ows that would be seen upon the object itself. In this sec- tion I propose to illustrate, by a few plain examples, the simple elementary principles upon which shading, in archi- tectural subjects, is based. The necessary knowledge of drawing, preliminary to this subject, 15 treated of in Section XV., from Arts. 487 to 498. 563.—The Inclination of the Line of Shadow.— This is always, in architectural drawing, 45 degrees, both on the elevation and on the plan; and the sun is supposed to be behind the spectator, and over his left shoulder. This can be illustrated by reference to 270. 420,1n which А repre- sents a horizontal plane, and B and C two vertical planes placed at right angles to each other. A represents the plan, C the elevation, and В а vertical projection from the eleva- tion. In finding the shadow of the plane P, the line а 2 is drawn at an angle of 45 degrees with the horizon, and the line c? at the same angle with the vertical plane З. The plane В being a rectangle, this makes the true direction of the sun's rays to be in a course parallel to 40, which direc- tion has been proved to be at an angle of 35 degrees and 16 minutes with the horizon. It is convenient, in shading, to have a set-square with the two sides that contain the

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“LS ХО ILdAUO *“adOULna

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CONVENTIONAL PLANES OF SHADOW. 597 right angle of equal length; this will make the two acute angles each 45 degrees, and will give the requisite bevel when worked upon the edge of the T-square. One reason why this angle is chosen in preference to another is that when shadows are properly made upon the drawing by it, the depth of every recess is more readily known, since the breadth of shadow and the depth of the recess will be equal. To distinguish between the terms shade and shadow, it will be understood that all such parts of a body as are not ex- posed to the direct action of the sun's rays are in shade; Fic. 420. while those parts which are deprived of light by the inter- position of other bodies are in shadow. 564.— ro Find the Line of Shadow on Mouldings and other Horizontally Straight Projections.—/izgs. 421, 422, 423, and 424 represent various mouldings in elevation, re- turned at the left, іп the usual manner of mitering around а projection. А mere inspection of the figures is sufficient to see how the line of shadow is obtained, bearing in mind that the ray ад is drawn from the projections at an angle of 45

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598 SHADOWS. degrees. When there is no return at the end, it is neces- sary to draw a section, at any place in the length of the mouldings, and find the line of shadow from that. 565.— To Find the Line of Shadow Cast by а Shelf.—In Fig. 425, А is the plan and В is the elevation of a shelf attached to a wall. From а and с draw a 0 and са, accord- ing to the angle previously directed; from 6 erect a per- pendicular intersecting са at d; from d draw de parallel to Fic. 421. Fie. 422. Fic. 423. Pic. 424. the shelf; then the lines cd and Ze will define the shadow cast by the shelf. There is another method of finding the shadow, without the plan A. Extend the lower line of the shelf to /; and make c f equal to the projection of the shelf from the wall; from f draw fg at the customary angle, and from с drop the vertical line cg intersecting fg at g; from £ draw ge parallel to the shelf, and from с draw c4 at the usual angle; then the lines cd and de will determine the extent of the shadow as before.

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SHADOWS OF STRAIGHT AND OBLIQUE SHELVES. 500 ,566.—To Find the Shadow Cast by a Shelf whieh is Wider at one End than at the Other.—In Fig. 426, А is the plan, and Z the elevation. Find the point 4, as in the pre- Fic. 425. vious example, and from any other point in the front of the shelf, as а, erect the perpendicular ae; from а and e draw аф and ec,at the proper angle, and from 0 erect the perpendicu- lar бс, intersecting ec in с; from 4, through с, draw do: then the lines 24 and Zo will give the limit of the shadow cast by the shelf. e

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600 SHADOWS. 567.—To Find the Shadow of a Shelf having one End Acute or Obtuse Angled.—/7g. 427 shows the plan and ele- vation of an acute-angled shelf. Find the line eg as before ; from а erect the perpendicular a5; join ó and e; then de and ez will define the boundary of shadow. 568.— To Find the Shadow Cast by an Inclined Shelf.— In Fig. 428 the plan and elevation of such a shelf are shown, having also one end wider than the other. Proceed as di- Ғіс 428. rected for finding the shadows of Fig. 426, and find the points d and c; then ай and dc will be the shadow required. If the shelf had been parallel in width on the plan, then the line dc would have been parallel with the shelf а 6.

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SHADOWS OF INCLINED AND CURVED SIIELVES. бот 569.-То Find ihe Shadow Cast by a Shelf Inclined in its Vertical Section either Upward or Downward.—From а (figs. 429 and 430) draw а б at the usual angle, and from % draw дс parallel with the shelf; obtain the point e by draw- Fic. 429. Fic. 430. ing.a line from d at the usual angle. In 772. 429 join e and 2; then ze and ec will define the shadow. Іп Fig. 430, from о draw oz parallel with the shelf; join 2 and e; then ze and cc will be the shadow required. The projections in these several examples are bounded Fic? 352. by straight lines; but thc shadows of curved lines may be found in the same manner, by projecting shadows from sev- eral points in the curved line, and tracing the curve of shadow through these points. (Figs. 431 and 432.)

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602 SHADOWS. :570.—' ro Find the Shadow of a Shelf having its Front Edge, or End, Curved on the Plan,—In 2705. 431 and 432 А and А show an example of each kind. From several points, as a, a, in the plan, and from the corresponding points 0, 0 in the elevation, draw rays and perpendiculars intersect- FIG. 433. ing at е, с, еїс.; through these points oí intersection trace the curve, and it will define the shadow. 571.—To Find the Shadow of a Shelf Curved in the Ele- vation.—In 2/2. 433 find the points of intersection, e, c and Fico 424 , as in the last examples, and a curve traced сай them will define the shadow. The preceding examples show how to find shadows when cast upon a vertical plane ; shadows thrown upon carved sur- . faces are ascertained in a similar manner. (Fig. 434.) | -

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SHADOW UPON AN INCLINED WALL. 603 572.—To Find the Shadow Cast upon 2 Cylindrical Wall by a Projection of any Kind. — Dy an inspection of fig. 434, it will be seen that the only difference between this and the last examples is that the rays in the plan die against the circle a2, instead of a straight line. Ғіс. 435. 573.-То Find the Shadow Cast by a Shelf upon an In- clined Wail,—Cast the ray 22 (Fig. 435) from the end of the shelf to the face of the wall, and from 4 draw бс parallel to the shelf; cast the ray de from the end of the shelf; then the lines Ze and ec will define the shadow. Fic. 436. These examples might be multiplied, but enough has been given to illustrate the general principle by which shad- ows in all instances are found. Let us attend now to the application of this principle to such familiar objects as are likely to occur in practice.

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„ 624 SHADOWS. 574..—To Find the Shadow of a Projecting Horizontal Beam.—From the points a, a, etc. (Fig. 436), cast rays upon the wall; the intersections e, e, e of those rays with the per- pendiculars drawn from the plan will define the shadow. If the beam be inclined, either on the plan or elevation, at any angle other than a right angle, the difference in the manner ІШІ | | ШІ Ш of proceeding сап be seen by reference to the preceding examples of inclined shelves, etc. 575.--То Find the Shadow in a Reeess.—From the point a (Fig. 437) in the plan, and å in the elevation, draw the rays ac and бе; from с erect the perpendicular ce, and from e Fic. 438. draw the horizontal line eZ; then the lines ce.and ed will show the extent of the shadow. This applies only where the back of the recess is parallel with the face of the wall. ..576.—To Find the Shadow in a Recess, when the Face of the Wall is Inclined, and the Back of the Recess is Vertical.—In /7g. 438, А shows the section and В the eleva-

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tion of a recess of this kind. < From 4, and from any other SHADOW.IN А FIREPLACE. 605 point in the line ġa, as а, draw the rays бс and ае; from с, a, and e draw the horizontal lines cg, af, and ей; from Z ^ ^n | | | ІТ НІШ Гг | FIG. 439. 4 and f cast the rays ZZ and f/, from 2, through 4, draw 25; then sz and zg will define the shadow. 577. Yo Find the Shadow in a Fireplace.— From а and 6 (Fig. 439) cast the rays ас and бе, and from с erect the Кіс. 440. perpendicular ce; from e draw the horizontal line eo, and join о and 4; then ce, eo, and od will give the extent of the shadow, ,

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606 SHADOWS. 578.—To Find the Shadow of a Moulded Window-Lin- tel.—Cast rays from the projections а, o, etc., іп the plan (Fig. 440), and а, e, etc., in the elevation, and draw the usual perpendiculars intersecting the rays at 2,2, and 2; these in- tersections connected, and horizontal lines drawn from them, will define the shadow. Тһе shadow on the face of the lin- tel is found by casting a ray back from 2 to 5, and drawing the horizontal line s z. 579.—To Find the Shadow Cast by the Nosing of a Step. —From а (Fig. 441) and its corresponding point с, cast the E We FIG. 4q1. rays a 0 and cd, and from? erect the perpendicular Od: tàm- gical to the curve at e cast the ray e f, and from e drop the perpendicular e o, meeting the mitre-line a g in o; cast a ray from o to 2, and from z erect the perpendicular z/; from 2 draw the ray 206; from f to d and from d to £ trace the curve as shown in the figure; from £ and / draw the hori- zontal lines £z and / s; then the limit of the shadow will be completed. 580.—To Find the Shadow Thrown by a Pedestal upon Steps.—From a (Fig. 442) in the plan, and from c in the ele- vation, draw the rays а0 and ce; then ao will show the ex-

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SHADOWS ON STEPS AND COLUMNS. 607 tent of the shadow on the first riser, as at 4; fg will deter- mine the shadow on the second riser, as at B; cd gives the amount of shadow on the first tread, as at C, and 22 that on the second tread, as at D; which completes the shadow of Fic. 442. the left-hand pedestal, both on the plan and elevation. А mere inspection of the figure will be sufficient to show how the shadow of the right-hand pedestal is obtained. Fic. 443. FIG. 444. 581.—To Find the Shadow Thrown on a Column by a Square Abacus.—From a and 0 (Fig. 443) draw the rays ас and ġe, and from с erect the perpendicular се; tangical to the curve at d draw the ray Z/, and from 2, corresponding to fin the plan, draw the ray 40; take any point between а” and f as ғ, and from this, as also from a corresponding point

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608 SHADOWS. п, draw the rays Z7 апа xs; from т and from d erect the perpendiculars rs and do; through the points e,s, and о trace: the curve as shown in the figure; then the extent. of the shadow will be defined. . Fic. 445. 582.—' To Find the Shadow Thrown on a Column Әу a Circular Abacus.—This is so nearly like the last example that no explanation will be necessary, farther than a refer- ence to the preceding article.

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SHADOWS ON THE CAPITAL OF A COLUMN. 609 583.—To Find the Shadows on the Capital of a Column. -—This may be done according to the principles explained in the examples already given; a quicker way of doing it, however, is as follows: if we take into consideration one ray of light in connection with all those perpendicularly under and over it, it is evident that these several rays would form a vertical plane, standing at an angle of 45 degrees with the face of the elevation. Now we may suppose the column to be sliced, so to speak, with planes of this nature— cutting it in the lines a 0, с d, etc. (Fig. 445), and, in the ele- | |. | ' Fic. 446. vation, find by squaring up from the plan, the /ле5 of section which these planes would make thereupon. For instance: in finding upon the elevation the line-of section a2, the plane cuts the ovolo at e, and therefore f will be the correspond- ing point upon the elevation ; 2 corresponds with g, 2 with 7, o with s, and Z with 2. Now, to find the shadows upon this line of section, cast from m the ray mx, from 4 the ray ho, etc.; then that part of the section indicated by the letters | m fin,and that part also between 4 and o will be under

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610 " SHADOWS. shadow. By an inspection of the figure, it will be seen that the same process is applied to each line of section, and in that way the points ЖҰ) м, wer, ав also Т, 2,3, etes аге а Fic. 447. successively found, and the lines of shadow traced through them. ' Fig. 446 is an example of the same capital with all the shadows finished in accordance with the lines obtained on Fig. 445.

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SHADOW OF A COLUMN ON A WALL. біш 584.-То Find the Shadow Thrown on a Vertical Wall by a Column апа Entablature Standing іп Advanec of said Wall.—Cast rays from a and 6 (Fig. 447), and find the point c as in the previous examples; from 4 draw the ray de, and from e the horizontal line ef; tangical to the curve at g and д draw the rays gj and 22, and from 2 and 7 erect the per- pendiculars 2/ and 7£; from m and z draw the rays mf and nk, and trace the curve between £and f; cast a ray from oto р, а vertical line from 2 %о s, and through s draw the horizon- tai line 52; the shadow as required will then be completed. ІШІ | 1 М ! HI n | ІНІҢ | | ІШ ШЕ | 1 | ІН Fig. 448 is an example of the same kind as the last, with all the shadows filled in, according to the lines obtained in the preceding figure. 585.—Shadews on a Cornice.—/775. 449 and 450 are examples of the Tuscan cornice. Тһе manner of obtaining the shadows 1s evident. 586. Refleeted Light.—In shading, the finish and life of an object depend much on reflected light. This is seen to advantage іп Fig. 446, and on the column in 7g. 448. Re-

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612 SHADOWS. flected rays are thrown in a direction exactly the reverse of direct rays; therefore, on that part of an object which is subject to reflected light, the shadows are reversed. The = = L s | | | | | ІН. ІН СИ | = FIG. r fillet of the ovolo in Fig. 446 is an example of this. On the right hand side of the column, thesface of the fillet is much darker than the cove directly under it. The reason of this A ч "| Rd A Fig. 450. is, the face of the fillet is deprived both of direct and re- flected light, whereas the cove is subject tothe latter. Other instances of the effect of reflected light will be seen in the other examples.

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CONTENTS. — ----- PART I. SECTION I.—ARCHITECTURE. Art. 1. Building defined, p. 5.—9. Antique Builcings ; Tower of Babel, p. 5.—9. Ancient Cities and Monuments, p. 6.—4. Architecture in Greece, p. 6.-5. Architecture in Rome, р. 7.—6. Rome and Greece, p. 8.—7. Ar- chitecture debased, p. 9.—8, The Ostrogoths, p. 9.—9. The Lombards, р. 10. —10. The Byzantine Architects, p. 10.—11. The Moors, p. 10.—12. The Architecture of England, p. 11.-- ЕЗ, Architecture Progressive, p. 12.—114. Architecture in Italy, р. 12.—45. The Renaissance, p. 13.—106, Styles of Ar- chitecture, p. 13.—1 7. Orders, p. 14.--1%. The Stylobate, p. 14.—19. The Column, p. 14.—20. The Entablature, p. 14.--21. The Base, p. 14 —22. The Shaft, p. 15.—28%. The Capital, p. 15.--24. The Architrave, p. 15.—995. The Frieze, p. 15.—26. The Cornice, p. 15.-9%7, The Pediment, p. 15.—28. | The Tympanum, р: 15.—29. The Attic, p. 15.--30. Proportions in an Order, p. 15.--З1. Grecian Styles, p. 16.-39. The Doric Order, р. 16.—38. The Intercolumniation, p. 17.—84. The Doric Order, p. 19 —85. The ‘Ionic Order, p. 19.—36. The Intercolumniation, p. 20.—%. To Describe the Ionic Volute, p. 20.—88. The Corinthian Order, p. 23.—89. Persians and Carya- tides, p. 24.—40. Persians, p. 24—41. Caryatides, p. 26.—42. Roman Styles, p. 26. —43.. Grecian Orders modified by the Romans, p. 27.—44. The Tuscan Order, р. 30.—45. Egyptian Style, p. 30.—46. Building in General, p. 33.--47. Expression, p. 35.—48. Durability, p. 37.—49. Dwelling- Houses, р. 37.—80, Arranging the Stairs and Windows, р. 42.—91. Prin- ciples of Architecture, p. 44.—992. Arrangement, p. 44—93. Ventilation, p. 45.—94. Stability, p. 45.--55. Decoration, p. 46.—96. Elementary Parts of a Building, p. 46.—977, The Foundation, p. 47.—9$8. The Column, or Pillar, p. 47.—99. The Wall, p. 48.—60, The Reticulated Walls, р. 49.—61. The Lintel, or Beam, p. 49.—62. The Arch, p: 50.—63. Hooke's Theory of ап Arch, р. 50.—64. Gothic Arches, p. 51.—695, Arch: Definitions ; Principles, p. 52.-66. An Arcade, p. 52.—67. The Vault, p. 52.—68., The Dome, p. 53. . --69. The Roof, p. 54.

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614 CONTENTS. SECTION IL.—CONSTRUCTION. Art. 70. Construction Essential, p. 56.—*1. Laws of Pressure, p. 57.— 79. Parallelogram of Forces, р. 59.—73. The Resolution of Forces, p. 59.— 74. Inclination of Supports Unequal, р. б6о.-%Ә. The Strains Exceed the Weights, p. 6r.— 76. Minimum Thrust of Rafters, p. 62.--7%7. Practical Method of Determining Strains, p. 62.— 78. Horizontal Thrust, p. 63.--79. Position of Supports, p. 65.--80. Тһе Compositipn of Forces, p. 66.—$1. Another Example, p. 67.—82. Ties and Struts, p. 68.—83. To Distinguish Ties from Struts, p. 69.—84, Another Example, p. 70.—898. Centre of Gravity, p. 71.—86. Effect of the Weight of Inclined Beams, p. 72—87. Effect of Load on Beam, p. 74.—$8, Effect on Bearings, p. 75.—89. Weight-Strength, p. 76.—90. Quality of Materials, p. 76.—91. Manner of Resisting, p. 77.— 92. Strength and Stiffness, p. 78.—93. Experiments : Constants, p. 78.—94. Resistance to Compression, р. 79.—99. Resistance to Tension, p. 81.—96, Resistance to Transverse Strains, p. 83.—97. Resistance to Compression, p. 85.—98. Compression Transversely to the Fibres, p. 86.—99. The Limit of Weight, p. 86.—800. Area of Post, p. 86.—4301. Rupture by Sliding, p. 87. —1092. The Limit of Weight, р. 87.—103. Area of Surface, p. 88.—104. Tenons and Splices, p. 88.--105. Stout Posts, p. 89. 106. The Limit of Weight, p. 89.—4 0*7. Area of Post, p. до. —108, Area of Round Posts, p. до. —1Ф9, Slender Posts, р. 91.—1 10. The Limit of Weight, p. 91.—111. Diameter of the Post: when Round, p. 92.--119, Side of Post: when Square, p. 93.—8 13, Thickness of a Rectangular Post, р. 95.414. Breadth of a Rectangular Post, p. 95.—8 85. Resistance to Tension, p. 96.—116. “Тһе Limit of Weight, p. 96.—1117. Sectional Area, p. 97. —3 18. Weight of the Suspending Piece Included, p. 98.—889. Area of Suspending Piece, Р. 99. RESISTANCE TO TRANSVERSE STRAINS. Art. 120, Transverse Strains: Rupture, р. 99.—121. Location of Mor- tises, p. 190.—122, Transverse Strains: Relation of Weight to Dimensions, p. r01.—123. Safe Weight: Load at Middle, p. 103. 024. Breadth of Beam with Safe Load, p. 104.—1125. Depth of Beam with Safe Load, p. 104.—1926. Safe Load at any Point, p. 105.--197. Breadth or Depth : Load at any Point, p. 106.—1928, Weight Uniformly Distributed, p. 107.—14 299. Breadth ог Depth: Load Uniformly Distributed, p. 108.—830., Load per Foot Super- ficial, p. 100.--ЕЗІ. Levers: Load at one End, p. 110.—132. Levers: Breadth or Depth, p. 111.—133. Deflection: Relation to Weight, p. 112.—184. Пе- flection: Relation tó Dimensions, р. 112.—185. Deflection: Weight when at Middle, p. 114.—1:36. Deflection : Breadth or Depth, Weight at Middle, р. 114.--19?7. Deflection: When Weight is at Middle, p. 116.— 138., Deflection: Load Uniformly Distributed, p. 116.—139. Deflection: Weight when Uni- formly Distributed, р. 117.--140. Deflection: Breadth or Depth, Load Uni- formly Distributed, p. 117.—1418. Deflection: When Weight is Uniformly Distributed, p. 118.—142. Deflection of Lever, p. 119.—1483. Deflection of a Lever: Load at End, p. 120.—144. Deflection of a Lever: Weight when at End, p. 120.—145. Deflection of a Lever: Breadth or Depth, Load at End,

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CONTENTS. 615 p. 121.—146. Deflection of Levers: Weight Uniformly Distributed, p. 121. — 147. Deflection of Levers with Uniformly Distributed Load, p. 122.—148, Deflection of Levers: Weight when Uniformly Distributed, p. 122.—1449. Deflection of Levers: Breadth or Depth, Load Uniformly Distributed, p. 122. CONSTRUCTION IN GENERAL. Art. 850. Construction: Object Clearly Defined, p. 123.—151. Floors Described, р. 124.—159. Floor-Beams, p.125.—153. Floor-Beams for Dwell- ings, р. 127.—454. Floor-Beams for First-Class Stores, p. 128.—155. Floor- Beams: Distance from Centres, p. 129.—1356. Framed Openings for Chimneys and Stairs, p. 150.—157. Breadth of Headers, p. 130.—1958. Breadth of Carriage-Beams, р. 132.—4359. Breadth of Carriage-Beams Carrying Two Sets of Tail-Beams, р. 134.--А60. Breadth of Carriage-Beam with Well-Holo at Middle, р. 136.—161. Cross Bridging, or Herring- Bone Bridging, p. 137. —162. Bridging: Value to Resist Concentrated Loads, p. 137.—1 63. Gird- ers, p. 140.--І64. Girders: Dimensions, p. 141. FIRE-PROOF TIMBER FLOORS. Art. 165. Solid Timber Floors, p. 143.—166. Solid Timber Floors for Dwellings and Assembly-Rooms, p. 143.—1 67. Solid Timber Floors for First- Class Stores, p. 144.—86$8. Rolled-Iron Beams, р. 145.--169. Rolled-Iron Beams: Dimensions; Weight at Middle, р. 146.—1'70. Rolled-Iron Beams: Deflection when Weight is at Middle, p. 147.—4 71. Rolled-Iron Beams: Weight when at Middle, р. 148.—172. Rolled-Iron Beams: Weight at апу Point, p. 148.—12'78. Rolled-Iron Beams: Dimensions; Weight at any Point, р. 149---174. Rolled-Iron Beams: Dimensions ; Weight Uniformly Distrib- uted, р. 149.—8 75. Rolled-Iron Beams: Deflection ; Weight Uniformly Dis- tributed, p. 150. —176. Rolled-Iron Beams : Weight when Uniformly Distrib- uted, p. 151.—14 77. Rolled-Iron Beams: Floors of Dwellings or Assembly- Rooms, p. 151.—178. Rolled-Iron Beams: Floors of First-Class Stores, p. 152.—1 79. Floor-Arches: General Considerations, p. 153.—180. Floor- Arches: Tie-Rods; Dwellings, p. 153. — 481. Floor-Arches: Tie-Rods; First Class Stores, p. 153.. TUBULAR IRON GIRDERS. Art. 182. Tubular Iron Girders: Description, p. 154.—4 $83. Tubular Iron Girders : Area of Flanges ; Load at Middle, p. 154.—1 84. Tubular Iron Girders: Area of Flanges ; Load at any Point, p. 155.—1 85, Tubular Iron Girders: Area of Flanges; Load Uniformly Distributed, p. 156.—1$6. Tu- bular Iron Girders: Shearing Strain, p. 157.-187. Tubular Iron Girders : Thickness of Web, р. 158.—198%. Tubular Iron Girders for Floors of Dwell- ings, Assembly-Rooms, and Office Buildings, p. 1590.-189. Tubular Iron Girders for Floors of First-Class Stores, p. 160. CAST-IRON GIRDERS. Art. 190. Cast-Iron Girders: Inferior, p. 161.—191. Cast-Iron Girder: Load at Middle, р. 161.—1992. Cast-Iron Girder: Load Uniformly Distributed,

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616 CONTENTS. p. 163.—193. Cast-Iron Bowstring Girder, р. 163.--194. Substitute for the Bowstring Girder, p. 163. FRAMED GIRDERS. Art. 195. Graphic Representation of Strains, p. 165.—196, Framed Girders, р. 166.—197. Framed Girder and Diagram of Forces, p. 167.—198. Framed Girders : Load on Both Chords, p. 171.—199. Framed Girders: Di- mensions of Parts, p. 173. PARTITIONS. Art. 200. Partitions, p. 174.—901. Examples of Partitions, p. 175. ROOFS, Art. 202. Roofs, р. 178.—203. Comparison of Roof-Trusses, p. 178.— 204. Force Diagram: Load upon Each Support, p. 179.—205. Force Dia- gram for Truss in /72. 59, p. 179.--206. Force Diagram for Truss in Fig. бо, p. 180.—20%, Force Diagram for Truss in Fig. бі, p. 181.—208, Force Dia- gram for Truss in 77g. 63, p. 183.—209. Force Diagram for Truss in 270. 64, p. 184. —210. Force Diagram for Truss in Zig. 65, p. 185.2048. Force Dia- gram for Truss іп Fig. 66, р. 186.—9192. Roof-Truss : Effect of Elevating the Tie-Beam, р. 187.—213. Planning a Roof, р. 188.—2184. Load upon Roof- Truss, р. 189.—215. Load on Roof per Superficial Foot, p. 189. —216. Load upon Tie-Beam, р. І00.-217; Roof Weights in Detail, p. 191.—218. Load per Foot Horizontal, p. 192.--219. Weight of Truss, р. 1902.-290. Weight of Snow on Roofs, p. 193.--991. Effect of Wind on Roofs, p. 1903.-299, Total Load per Foot Horizontal, p. 197.--998. Strains іп Roof Timbers Computed, р: 108:--224, Strains in Roof Timbers Shown Geometrically, p. 109.--995. Application of the Geometrical. System of Strains, p. 202.—226, Roof Timbers: the Tie-Beam, p. 204.—92'7, The Rafter, p. 205.—929$, The Braces, p. 208.—229, The Suspension Rod, p. 210.—230. Roof-Beams, Jack-Rafters, and Purlins, p. 211.—931. Five Examples of Roofs, p. 212.— 232. Roof-Truss with Elevated Tie-Beam, p. 214.—233. Hip-Roofs : Lines and Bevels, р: 215.—9334. The Backing of the Hip-Rafter, p. 216. DOMES, Art. 235. Domes, p. 216.—236, Ribbed Dome, р. 217.—-237. Domes: Curve of Equilibrium, p. 218.—28%, Cubic Parabola Computed, p. 219.— 239. Small Domes over Stairways, p. 220.—240, Covering for a Spherical Dome, p. 221.—9241. Polygonal Dome: Form of Angle-Rib, p. 223. BRIDGES. Art. 242. Bridges, p. 223.—243. Bridges: Built-Rib, p. 224.--244. Bridges: Framed Rib, p. 226.—245, Bridges: Roadway, p. 227.--246. Bridges: Abutments, p. 227.— 247. Centres for Stone Bridges, p. 229.--248, Arch Stones: Joints, p. 223. JOINTS. Art. 949. Timber Joints, p. 234.

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CONTENTS. 617 SECTION III.—STAIRS. Art. 250, Stairs: General Requirements, р. 240.—951. The Grade of Stairs, p. 241.--252, Pitch-Board : Relation of Rise to Tread, p. 242.—958. Dimensions of the Pitch-Board, p. 247.—254. The String of a Stairs, p. 247. --255. Step and Riser Connection, p. 248. PLATFORM STAIRS Art. 956. Platform Stairs: the Cylinder, р. 248.—925'7. Form of Lower Edge of Cylinder, p. 249.—258. Position of the Balusters, p. 250.—259. Wind- ing Stairs, p. 251.—260. Regular Winding Stairs, p. 251.—261. Winding Stairs: Shape and Position of Timbers, p. 252.—969, Winding Stairs with Flyers: Grade of Front String, p. 253. HAND-RAILING. Art. 263. Hand-Railing for Stairs, p. 256.—964. Hand-Railing: Defini- tions ; Planes and Solids, p. 257. —2965. Hand-Railing » Preliminary Consider- ations, p. 258.—266. A Prism Cut by an Oblique Plane, p. 259. — 267. Form of Top of Prism, p. 259.—268. Face-Mould for Hand-Railing of Platform Stairs, p. 264.—269. More Simple Method for Hand-Rail to Platform Stairs, p. 267.-270. Hand-Railing for a Larger Cylinder, p. 271.—2'71. Face- Mould without Canting the Plank, p. 272.—9 72, Railing for Platform Stairs where the Rake meets the Level, р. 272.—9'78. Application of Face-Moulds to Plank, p. 273.-974. Face-Moulds for Moulded Rails upon Platform Stairs, p. 274.—92 75. Application of Face-Moulds to Plank, р. 275.—276. Hand-Railing for Circular Stairs, p. 278.--97%7. Face-Moulds for Circular Stairs, p. 282.—9' 78. Face-Moulds, for Circular Stairs, p. 285.— 9279. Face- Moulds for Circular Stairs, again, p. 287.—280. Hand-Railing for Winding Stairs, p. 289.288. Face-Moulds for Windjng Stairs, р. 290.—282. Face- Moulds for Winding Stairs, again, p. 293.—288. Face-Moulds: Test of Accu- гасу, p. 295.—284. Application of the Face-Mould, p. 297.— 285. Face-Mould Curves are Elliptical, p. 301.—9$6, Facc-Moulds for Round Rails, p. 303.— 287. Position of the Butt Joint, р. 303.—288. Scrolls for Hand-Rails: Gen- eral Rule for Size and Position of the Regulating Square, p. 308.—289. Cen- tres in Regulating Square, р. 308.—2990, Scroll for Hand-Rail Over Curtail Step, p. 300.--291. Scroll for Curtail Step, p. 310.--299, Position of Balus- ters Under Scroll, p. 310 —293. Falling-Mould for Raking Part of Scroll, p. 310.—29 4. Face-Mould for.the Scroll, p. 5311.-996. Form of Newel-Cap from a Section of the Rail, р. 3212.-906. Boring for Balusters in a Round Rail before it is Rounded, p. 313. SPLAYED WORK. Art. 297. Thc Bevels іп Splayed Work, p. 314. SECTION IV.—DOORS AND WINDOWS. DOORS. Art. 298, General Requirements, р. 315.—999. The Proportion between Width and Height, р. 315.—300. Panels, p. 316.--301. Trimmings, p. 317. --309. Hanging Doors, p. 317.

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618 CONTENTS. Y б WINDOWS. Art. 303. Requirements for Light, p. 317.—904. Window Frames, p. 318.—305. Inside Shutters, р. 3190.-306. Proportion: Width and Height, р. 3190.-207, Circular Heads, p. 320.—908, Form of Soffit for Circular Win- dow Heads, p. 321. 4 SECTION V.—MOULDINGS AND CORNICES. MOULDINGS. Art. 309. Mouldings, p. 323.--310. Characteristics of Mouldings, p. 324.--З311. A Profile, p. 5326.-219. The Grecian Torus and Scotia, p. 326.— ЗІЗ. The Grecian Echinus, р. 327,—814, The Grecian Cavetto, р. 327.— 885. The Grecian Cyma-Recta, p. 327.--216. The Grecian Cyma-Reversa, р. 3293.-217. Roman Mouldings, р. 329.—328. Modern Mouldings, p. 331. 7 CORNICES. Art. 219. Designs for Cornices, p. 335.—820. Eave Cornices Propor- tioned to Height of Building, р. 335.—3 21. Cornice Proportioned to a given Cornice, p. 342.—22, Angle Bracket in a Built Cornice, p. 343.—323. Rak- ing Mouldings Matched with Level Returns, p. 344. PAR ТТ SECTION VI.—GEOMETRY. Art. 394. Mathematics Essential, p. 5347.-395. Elementary Geometry, р. 347—326. Definition—Right Angles, p. 348.—32%. Definition—Degrees іп a Circle, р. 348.—828,. Definition—Measure of ап Angle, p. 348.—329. Corollary—Degrees іп а Right Angle, p. 348.—330. Definition—Equal Angles, p. 349—331. Axiom—Equal Angles, p. 349.—882. Definition— Obtuse and Acute Angles, p. 349.888. Axiom—Right Angles, р. 349.— 334. СогоПагу--Туо Right Angles, p. 2349.-225. Corollary—Four Right Angles, p. 3190.-826. Proposition—Equal Angles, p. 550.-28%. Propo- sition—Equal Triangles, p. 350.-928%8. Proposition—Angles in Isosceles Triangle, p. 351.—839. Proposition—Diagonal of Parallelogram, p. 351.— 340, Proposition—Equal Parallelograms, p. 352.—841. Proposition—Paral- lelograms Standing on the Same Base, p. 352.—842. Corollary— Parallelo- gram and Triangle, р. 353.—%4%. Proposition—Triangle Equal to Quadrangle, р. 353.—344. Proposition—Opposite Angles Equal, p. 354.— 34%. Proposi- tion—Three Angles of Triangle Equal to Two Right Angles, p. 354. —346. Corollary—Right Angle in Triangle, p. 354—347. Corollary—Half a Right

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CONTENTS. 619 Angle, p. 355.—848. Corollary—Right Angle:in a Triangle, p. 5355.-349. Corollary—Two Angles Equal to Right Angle, p. 355. —990. Corollary—T wo Thirds of a Right Angle, р. 3555.-951. Corollary—Equilateral Triangle, р. 355. --252. Proposition—Right Angle іп Semi-circle, p. 355.—858. Proposition— The Square of the Hypothenuse Equal to the Squares of the Sides, p. 355.— 354. Proposition—Equilateral Octagon, p. 357.—355. Proposition—Angle at the Circumference of a Circle, p. 355. —9396. Proposition—Equal Chords give Equal Angles, p. 358.—39 77. Corollary of Equal Chords, p. 3590.-358, Proposition—Angle Formed by a Chord. and Tangent, p. 359.—359. Propo- sition—Areas of Parallelograms, p. 360.--260. Proposition—Triangles ot Equal Altitude, p. 361. —3961. Proposition —Homologous Triangles, p. 362.— 362. Proposition—Parallelograms of Chords, p. 363.—868. Proposition— Sides of Quadrangle, p. 364. SECTION VII—RATIO, ОК PROPORTION. Art. 364, Merchandise, p. 366.—365. The Rule of Three, p. 366.— 366. Couples: Antecedent, Consequent, p. 367.—:267. Equal Couples: an Equation, p. 367.—368, Equality of Ratios, p. 367.—869. Equals Multiplied by Equals Give Equals, р. 367.-270. Multiplying an Equation, р. 3638.-371, Multiplying and Dividing one Member of an Equation: Cancelling, p. 368.— 372, Transferring a Factor, p. 3690.-328. Equality of Product: Means and Extremes, p. 369—8374. Homologous Triangles Proportionate, p. 370.— 325. Тһе Steelyard, p. 371. —376. The Lever Exemplified by the Steelyard, p. 372.—93 7*4. The Lever Principle Demonstrated, p. 375.—378. Any One ot Four Proportionals may be Found, p. 377. SECTION VIII.—FRACTIONS. Art. 379. A Fraction Defined, p. 378.—88%0. Graphical Representation of Fractions: Effect of Multiplication, p. 378.—981. Form of Fraction Changed by Division, p. 380.—382. Improper Fractions, p. 380.—383, Re- duction of Mixed Numbers to Fractions, р. 3981.-384. Division Indicated by the Factors put as a Fraction, p. 3981.-386. Addition of Fractions having Like Denoniinators, p. 382.--3%6. Subtraction of Fractions of Like Denominators, p. 383.—387. Dissimilar Denominators Equalized, р. 3583.-288. Reduction of Fractions to their Lowest Terms, p. 384.—389. Least Common Denomina- tor, p. 384.—390. Least Common Denominator Again, p. 385.—391. Frac- tions Multiplied Graphically, p. 386.—392. Fractions Multiplied Graphically Again, р. 387.—3938. Rule for Multiplication of Fractions, and Example, p. 387.--294. Fractions Divided Graphically, p. 388.—895. Rule for Division of Fractions, p. 389. SECTION IX.—ALGEBRA. Art. 396. Algebra Defined, p. 392.--297. Example: Application, р. 393.298. Algebra Useful in Constructing Rules, p. 394.—399. Algebraic Rules are General, p. 394—400. Symbols Chosen at Pleasure, p. 395.—401. Arithmetical Processes Indicated by Signs, p. 396.—402. Examples іп Addi-

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620 | CONTENTS. tion and Subtraction: Cancelling, p. 398.—408. Transferring a Symbol to the Opposite Member, p. 399. —404. Signs of Symbols to be Changed when they are to be Subtracted, p. 400.--405. Algebraic Fractions, Added and Sub- tracted, p. 403.—406. The Least Common Denominator, р. 404.—407. Alge- braic Fractions Subtracted, p. 405.—40%, Graphical Representation of Multi- plication, p. 408.—409. Graphical Multiplication: Three Factors, p. 408.— 410. Graphic Representation: Two and Three Factors, p. 409.—411. Graph- ical Multiplication of а Binomial, p. 409.—412. Graphical Squaring of a Binomial, p. 4І0.--418, Graphical Squaring of the Difference of Two Fac- tors, p. 412.—414. Graphical Product of the Sum and Difference of Two Quantities, p. 413.—415. Plus and Minus Signs in Multiplication, p. 415.— 416. Equality of Squares on Hypothenuse and Sides of Right-Angled Tri- angle, p. 416.--417. Division the Reverse of Multiplication, р. 418.—41 8. Division : Statement of Quotient, p. 419.—419. Division: Reduction, p. 419. —420. Proportionals: Analysis, p. 421.—421. Raising a Quantity to any Power, p. 423—422. Quantities with Negative Exponents, р. 423.—423. Addition and Subtraction of Exponential Quantities, p. 424.—424. Multipli- cation of Exponential Quantities, p. 424.—425. Division of Exponential Quantities, р. 424—426. Extraction of Radicals, p. 425.—42'7. Logarithms, р. 425.—428, Completing the Square of a Binomial, p. 429. PROGRESSION, Art. 429. Arithmetical Progression, p. 432.—430. Geometrical Progres sion, p. 435. SECTION X.+POLYGONS. Art. 433. Relation of Sum and Difference of Two Lines, p. 439.—432. Perpendicular, in Triangle of Known Sides, p. 440.483. Trigon: Radius of Circumscribed and Inscribed Circles : Area, P. 443.—434, Tetragon: Radius of Circumscribed and Inscribed Circles: Area, p. 446.--485. Hexagon: Ra- dius of Circumscribed and Inscribed Circles: Area, p. 447.-4386. Octagon: Radius of Circumscribed and Inscribed Circles: Area, р. 4490.-497. Dodec- agon: Radius of Circumscribed and Inscribed Circles: Area, p. 452. —A38., Hecadecagon : Radius of Circumscribed and Inscribed Circles: Area, p. 455. —439. Polygons: Radius of Circumscribed and Inscribed Circles: Area, p. 460.—440. Polygons: Their Angles, p. 462. —441. Pentagon: Radius of the Circumscribed and Inscribed Circles; Area, p. 463.—442. Polygons: Table of Constant Multipliers, p. 465. SECTION XI.—THE CIRCLE. Art. 443, Circles: Diameter and Perpendicular: Mean Proportional, p. 468.—444.. Circle: Radius from Given Chord and Versed Sine, p. 469.— 445. Circle: Segment from Ordinates, p. 470.—446. Circle: Relation of Diameter to Circumference, p. 472.—447. Circle: Length of an Arc, p. 475. --448, Circle: Area, p. 475.—449. Circle: Area of a Sector, p. 476.—450. Circle: Area of a Segment, p. 477.

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CONTENTS. 621 SECTION XII.—THE ELLIPSE. Art, 451. Ellipse: Definitions, р. 481.—452, Ellipse: Equations to the Curve, p. 482.—4953. Ellipse: Relation of Axis to Abscissas of Axes, p. 484. —4954. Ellipse: Relation of Parameter and Axes, p. 485.—4959. Ellipse: Relation of Tangent to the Axes, p. 485.—456, Ellipse: Relation of Tangent with the Foci, p. 487.--45%7. Ellipse: Relation of Axes to Conjugate Diam- eters, p. 487.—49 8, Ellipse: Area, р. 488.—459. Ellipse: Practical Sugges- tions, p. 489. SECTION ХПІ.-ТНЕ PARABOLA. . Art. 460. Parabola: Definitions, р. 492.—461. Parabola: Equation to the Curve, p. 493.—462. Parabola: Tangent, p. 493.--468. Parabola: Sub- tangent, p. 496.—464. Parabola: Normal and Subnormal, p. 496.—465. Parabola: Diameters, p. 497.—466. Parabola: Elements, p. 499.—467*. Parabola: Described Mechanically, p. 500.—468. Parabola : Described from Points, p. 502.--469. Parabola: Described from Arcs, р. 503.—470. Para- bola: Described from Ordinates, p. 504.--4%71. Parabola: Described from Diameters, р. 507.—472. Parabola: Area, p. 509. SECTION XIV.—TRIGONOMETRY. Art. 473. Right-Angled Triangles: The Sides, p. 510.—474. Right- Angled Triangles: Trigonometrical Tables, p. 512.—4*75, Right-Angled Triangles: Trigonometrical Value of Sides, p. 516.—446. Oblique-Angled Triangles: Sines and Sides, p. 519.—47 7. Oblique-Angled Triangles: First Class, p. 520.—478. Oblique-Angled Triangles: Second Class, p. 522.-- 479. Oblique-Angled Triangles: Sum and Difference of Two Angles, p. 523. —4$80. Oblique-Angled Triangles: Third Class, p. 526.—481. Oblique- Angled Triangles: Fourth Class, p. 528.—482, Trigonometrical Formule: Right-Angled Triangles, p. 530.—483. Trigonometrical Formule: First Class, Oblique, p. 531.—484. Trigonometrical Formule: Second Class, Oblique, p. 532.--485. Trigonometrical Formulz: Third Class, Oblique, p, 534.--486. Trigonometrical Formula: Fourth Class, Oblique, p. 534. SECTION XV.—DRAWING Art. 487. General Remarks, р. 536.—488. Articles Required, р. 536.— 489. The Drawing-Board, p. 556.—490. Drawing-Paper, р. 537.—491. То Secure the Paper to the Board, p. 537.—492. The T-Square, p. 539.—498. The Set-Square, p. 539.—494. The Rulers, р. 540.--496. The Instruments, p. 540.—496. The Scale of Equal Parts, p. 540.—497. The Use of the Set- Square, p.,541.—498. Directions for Drawing, p. 542.

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622 CONTENTS. SECTION XVI.—PRACTICAL GEOMETRY. Art. 499, Definitions of Various Terms, p. 544. PROBLEMS. RIGHT LINES AND ANGLES. Art. 500. To Bisect a Line, р. 5490.-501. To Erect a Perpendicular, p. 550.—902. To let Fall a Perpendicular, p. 551.—903. To Erect a Perpen- dicular at the End of a Line, p. 551.—504. To let Fall a Perpendicular near the End of a Line, р. 553.—905., To Make an Angle Equal to a Given Angle, p. 553.--206. To Bisect an Angle, р. 554.—907. To Trisect a Right Angle, р. 554.--50%, Through a Given Point to Draw a Line Parallel to a Given Line, p. 555.--509. To Divide a Given Line into any Number of Equal Parts, р. 555. THE CIRCLE. Art. 510. To Find the Centre of a Circle, p. 556.—911. At a Given Point in a Circle to Draw a Tangent thereto, p. 557.--519, The Same, with- out making use of the Centre of the Circle, p. 557.—913. A Circle and а Tangent Given, to Find the Point of Contact, p. 5598.-114. Through any Three Points not in a Straight Line to Draw a Circle, p. 559.—989. Three Points not in a Straight Line being Given, to Find a Fourth that Shall, with the Three, Lie іп the Circumference of a Circle, p. 559.—916. To Describe a Segment of a Circle by a Set-Triangle, p. 560.—51'7. To Find the Radius of an Arc of a Circle when the Chord and Versed Sine are Given, p. 561.—d18. To Find the Versed Sine of an Arc of a Circle when the Radius and Chord are Given, p. 561.—519. To Describe the Segment of a Circle by Intersection of Lines, p. 562.--590. Ordinates, p. 563.--5921. Іп a Given Angle to De- scribe a Tanged Curve, p. 565.--299, To Describe a Circle within any Given Triangle, so that the Sides of the Triangle shall be Tangical, p. 566.—9223. About a Given Circle to Describe an Equilateral Triangle, p. 566.—9924. То Find а Right Line nearly Equal to the Circumference of a Circle, p. 566. POLYGONS, ETC. Art, 525. Upon a Given Line to Construct ап Equilateral Triangle, p. 568.-5926. To Describe ап Equilateral Rectangle, or Square, p. 568.--597. Within a Given Circle to Inscribe an Equilateral Triangle, Hexagon, or Dodec- agon, p. 569.—528. Within a Square to Inscribe an Octagon, р. 570.—529. To Find the Side of a Buttressed Octagon, р. 571.--520. Within a Given Circle to Inscribe any Regular Polygon, p. 572.—991. Upon a Given Line to Describe any Regular Polygon, p. 573.-532. To Construct a Triangle whose Sides shall be severally Equal to Three Given Lines, p. 575.—388. To Con- struct a Figure Equal to a Given Right-lined Figure, р. 575.—934. То Make a Parallelogram Equal to a Given Triangle, р. 576.—585. A Parallelogram being Given, to Construct Another Equal to it, and Having a Side Equal to a

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CONTENTS. 623 Given Line, p. 576.—536, To Make a Square Equal to two or more Given Squares, p. 577.—587. To Make a Circle Equal to two Given Circles, p. 580. — 58%, To Construct a Square Equal to a Given Rectangle, p. 581.—539. То Form a Square Equal to a Given Triangle, p. 582.—540. Two Right Lines being Given, to Find a Third Proportional thereto, p. 582.—9541. Three Right Lines being Given, to Find a Fourth Proportional thereto, p. 583.—5492. А Line with Certain Divisions being Given, to Divide Another, Longer or Shorter, Given Line in the Same Proportion, p. 583.—548. Between Two Given Right Lines to Find a Mean Proportional, p. 584. CONIC SECTIONS, Art. 544, Definitions, p. 584.—545. To Find the Axes of the Ellipsis, р. 585.—9546. To Find the Axis and Base of the Parabola, р. 585.—547. To Find the Height, Base, and Transverse Axis of an Hyperbola, p. 585.—548, The Axes being Given, to Find the Foci, and to Describe an Ellipsis with a String, p. 586.—549. The Axes being Given, to Describe an Ellipsis with a Trammel,.p. 586.—590. To Describe an Ellipsis by Ordinates, p. 588. —551. To Describe an Ellipsis by Intersection of Lines, p. 588.—552. To Describe an Ellipsis by Intersecting Arcs, p. 590.—998. To Describe a Figure Nearly in the Shape of an Ellipsis by a Pair of Compasses, p. 591.—994. To Draw an Oval in the Proportion Seven by Nine, p. 591.—555. To Draw a Tangent to an Ellipsis, p. 502.--586. An Ellipsis with a Tangent Given, to Detect the Point of Contact, p. 593.—55%7. A Diameter of an Ellipsis Given, to Find its Conjugate, р. 593.--558, Any Diameter and its Conjugate being Given, to Ascertain the Two Axes, and thence to Describe the Ellipsis, р. 593.—999. To Describe an Ellipsis, whence Axes shall be Proportionate to the Axes of a Larger or Smaller Given One, p. 594.—860. To Describe a Parabola by Intersection of Lines, p. 594.—561. To Describe an Hyperbola by Intersec- tion of Lines, p. 595. SECTION XVII.—SHADOWS. Art. 569, The Art of Drawing, р. 5906.-263. The Inclination of the Line of Shadow, p. 596.—564. To Find the Line of Shadow on Mouldings and other Horizontally Straight Projections, p. 597.—9695. To Find the Line of Shadow Cast by a Shelf, p. 598.—966. To Find the Shadow Cast by a Shelf which is Wider at one End than at the Other, p. 599.—967. To Find the Shadow of a Shelf having one End Acute or Obtuse Angled, p. 600.—d68, To Find the Shadow Cast by an Inclined Shelf, p. 600.—969. To Find the Shadow Cast by a Shelf inclined in its Vertical Section either Upward or Downward, р. 601.—570, To Find the Shadow of a Shelf having its Front Edge or End Curved on the Plan, p. 602.—5 78. To Find the Shadow of a Shelf Curved in the Elevation, p. 602.—572, To Find the Shadow Cast upon a Cylindrical Wall by а Projection of any Kind, p. 603.—9 35. To Find the Shadow Cast by a Shelf upon an Inclined Wall, p. 603.—9'74. To Find the Shadow of a Projecting Horizontal Beam, p. 604.—6 76. To Find the Shadow

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624 CONTENTS. іп a Recess, p. 604.—9 76. To Find the Shadow іп a Recess, when the Face of the Wall is Inciined, and the Back of the Recess is Vertical, p. 604.—577, To Find the Shadow in a Fireplace, p. 605.—578. To Find the Shadow of a Moulded Window-Lintel, p. 606—579, To Find the Shadow Cast by the Nosing of a Step, p. 606.—580, To Find the Shadow Thrown by a Pedestal upon Steps, p. 6c6.—581. To Find the Shadow Thrown on a Column by a Square Abacus, p. 6о7.-5829, To Find the Shadow Thrown on a Column by a Circular Abacus, p. 608.—583.' To Find the Shadows on the Capital of a Column, p. 609.—584. To Find the Shadow Thrown on a Vertical Wall by à Column and Entablature Standing in Advance of said Wall, p. 611.—95$895, Shadows on a Cornice, p. 611.—386, Reflected Light, p. 611.

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AMERICAN HOUSE CARPENTER. APPENDIX. AO — «et ич UNIVERSITY, CALIFORNI”

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CONTENTS. i PAGE. GUOSssARY.. со сооо оо ОЛСО oo etea e err e heir EODD 000 TABLE or SQUARES, CUBES AND Коөтв................................ 688 RULES FOR THE REDUCTION OF ПЕсіМАІ5............................. 647 ЖЕМЕ or СИС ............. а. Pits ve sieeve a OB TABLE SHOWING THE CAPAC'TY OF WELLS, CISTERNS, ЕТС.............. 653 SABLE OF THE WEIGHTS OF MATERIAIS... о... ee rtr ht hs 654

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GLOSSARY. Terms not found here can be found in the lists of definitions in other parts of this book, or in common dictionaries. A bacus.—The uppermost member of a capital. Abattoir.—A slaughter-house. Abbey. —The residence of an abbot or abbess. Abutment.—That part of a pier from which the arch springs. Acanthus.—A plant called in English dear’s-breech. Its leaves are employed for decorating the Corinthian and the Composite capitals. Acropolis.—The highest part of a city ; generaily the citadel. Acroteria.—The small pedestals placed on the extremities and apex of a pediment, originally intended as a base for sculpture. Atsie.—Passage to and from the pews of a church. In Gothic architecture, the lean-to wings on the sides of the zave. Alcove.—Part of a chamber separated by an estrade, or partition of columns. Recess with seats, etc., in gardens. Altar.—A pedestal whereon sacrifice was offered. In modern churches, the area within the railing in front of the pulpit. Alto-relievo.—High relief ; sculpture projecting from a surface so as to appear nearly isolated. Amphitheatre.—A double theatre, employed by the ancients for the exhibi- tion of gladiatorial fights and other shows. Ancones.—Trusses employed as ап apparent support to a cornice upon the flanks of the architrave. Annulet-—A small square moulding used to separate others ; the fillets in the Doric capital under the ovolo, and those which separate the flutings of col- umns, are known by this term. Anta.—A pilaster attached to a wall. Apiary.—A place for keeping beehives. Arabesque.—A building after the Arabian style. Areostyle.—An intercolumniation of from four to five diameters. Arcade—A series of arches. Arch.—An arrangement of stones or other material in a curvilinear form, so as to perform the office of a lintel and carry superincumbent weights. Architrave.—That part of the entablature which rests upon the capital of a column, and is beneath (һе frieze. Тһе casing and mouldings about a door or window. . Archivolt.—The ceiling of a vault; the under surface of an arch. Area, —Superficial measurement. An open space, below the level of the ground, in front of basement windows.

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628 APPENDIX, Arsenal.—A. public establishment for the deposition of arms and warlike stores. Astragal.—A small moulding consisting of a half-round with а fillet on each side. Айіс.--А low story erected over an order of architecture. A low additional story immediately under the roof of a building. Aviary.—A place for keeping and breeding birds. Balcony.—An open gallery projecting from the front of a building. Baluster.—2A. small pillar or pilaster supporting a rail. Balustrade А series of balusters connected by a rail. Barge-course.—That part of the covering which projects over the gable of a building. Base.—The lowest part of a wall, column, etc. Basement-story.—That which is immediately under the principal story, and included within the foundation of the building. Basso-relievo.—Low relief ; sculptured figures projecting from a surface one half their thickness or less. See Alto-relievo, Battering.—See Talus. Battlement.—Indentations on the top of a wall or parapet. Bay-window.—A window projecting in two or more planes, and not form- ing the segment of a circle. Bazaar.—A species of mart or exchange for the sale of various articles of merchandise. Bead.—A circular moulding. Bed-mouldings.—Those mouldings which are between the corona and the frieze. Belfry.—That part of the steeple in which the bells are hung; anciently called campanile. Belvedere.—An ornamental turret or observatory commanding a pleasant prospect. у Bow-window.—A window projecting in curved lines. Bressummer—A beam or iron tie supporting a wall over a gateway or other opening. ы Brick-nogging.—The brickwork between studs of partitions. Buttress.—A. projection from a wall to give additional strength. Саб/е.--А cylindrical moulding placed in flutes at the lower part of the col- umn. Camber.—To give a convexity to the upper surface of a beam. Campanile.—A tower for the reception of bells, usually, in Italy, separated from the church. Canopy. —An ornamental covering over a seat of state. Cantalivers.—The ends of rafters under a projecting roof. Pieces of wood or stone supporting the eaves. | Capital, —The uppermost part of a column included between the shaft and the architrave. | Caravansera.—In the East, a large public building for the reception of trav- ellers by caravans in the desert.

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GLOSSARY. 629 Carpentry.—(From the Latin carpentum, carved wood.) That department of science and art which treats of the disposition, the construction, and the relative strength of timber. The first is called descriptive, the second con- structive, and the last mechanical carpentry. Caryatides.—Vigures of women used instead of columns to support an entablature. Сазіпо.--А small country-house. Castellated.—Built with battlements and turrets in imitation of ancient castles. Cast/e,.—A building fortified for military defence. А house with towers, usually encompassed with walls and moats, and having a donjon, or keep, in the centre. j Catacombs.—Subterraneous places for burying the dead. Cathedral.—The principal church of a province or diocese, wherein: the throne of the archbishop or bishop is placed. Cavetto.—A concave moulding comprising the quadrant of a circle. Cemetery.—An edifice or area where the dead are interred. Cenotaph. —A monument erected to the memory of a person buried in another place. Centring.—The temporary woodwork, or framing, whereon any vaulted work is constructed. Cesspool.—A. well under a drain or pavement to receive the waste water and sediment. Cham fer—The bevelled edge of anything originally right angled. Chancel.—That part of a Gothic church in which the altar is placed. Chantry.—-A little chapel in ancient churches, with an endowment for one or more priests to say mass for the relief of souls out of purgatory. Chapel.—A building for religious worship, erected separately from a church, and served by a chaplain. Chaplet.—A moulding carved into beads, olives, etc. Cincture.—The ring, listel, or fillet, at the top and bottom of a column, · which divides the shaft of the column from its capital and base. Circus —A straight, long, narrow building used by the Romans for the ex- hibition of public spectacles and chariot races. At the present day, a building enclosing an arena for the exhibition of feats of horsemanship. Clere-story.—The upper part of the nave of a church above the roofs of the aisles. Cloister.—The square space attached to a regular monastery or large church, having a peristyle or ambulatory around it, covered with a range of buildings, . Coffer-dam.—A case of piling, water-tight, fixed in the bed of a river, for the purpose of excluding the water while any work, such as a wharf, wall, or the pier of a bridge, is carried up. Collar-beam.—A horizontal beam framed between two principal rafters above the tie-beam. Colonnade.— A range of columns. Columbarium.—A pigeon-house. Column.—A vertical cylindrical support under the entablature of an order. Common-rafters.—Yhe same as jack-rafters, which вес.

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630 / APPENDIX. Conduit.—A long, narrow, walled passage underground, for secret com- munication between different apartments. А canal or pipe for the conveyance of water. Conservatory.—A building for preserving curious and rare exotic plants. Consoles.— The same as ancones, which see. Contour.—The external lines which bound and terminate a figure. Convent.—A building for the reception of a society of religious persons. Coping.—Stones laid on the top of a wall to defend it from the weather. Corbels.—Stones or timbers fixed in a wall to sustain the timbers of a floor or roof. i | Cornice.—Any moulded projection which crowns or finishes the part to which it is affixed. Corona.—That part of a cornice which is between the crown-moulding and the bed-mouldings. Cornucopia.—The horn of plenty. Corridor. —An open gallery or communication to the different n: ма of a house. Cove.—A concave moulding. Cripple-rafters.—The short rafters which are spiked to the hip-rafter of a roof. Crockéts.—In Gothic architecture, the ornaments placed along the angles of pediments, pinnacles, etc. Crosettes.—The same as azcones, which see. Crypt. —The under or hidden part of a building. Culvert.— An arched channel of masonry or brickwork, built beneath the bed of a canal for the purpose of conducting water under it. Any arched channel for water underground. Cupola.—A. small building оп the top of a dome. Curtail-step.—A step with a spiral end, usually the first of the flight. Сизрз.--Тһе pendants of a pointed arch. " Сута.--Ап ogee. There are two kinds; the cyma-recta, having the upper part concave and the lower convex, and the cyma-reversa, with the upper part convex and the lower concave. Dado.—The die, or part between the base and cornice of a pedestal. Dairy.—An apartment or building for the preservation of milk, and the manufacture of it into butter, cheese, etc. Dead-shoar.—A piece of timber or stone stood — in brickwork, to Support a superincumbent weight until the brickwork which is to carry it has set or become hard. Decastyle.—A building having ten columns in front. Dentils.—(From the Latin, dentes, teeth.) Small аат blocks used in the bed-mouldings of some of the orders. Diastyle,—An intercolumniation of three, ог, as some say, four diameters. Die.—That part of a pedestal included between the base and the cornice ; it is also called а dado. Dodecastyle.—A. building having twelve columns in front. Donjon.—A. massive tower within ancient castles, to which the garrison might retreat in case of necessity.

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GLOSSARY. 631 Dcoks.—A Scotch name given to wooden brick”. Dormer.—A window placed on the roof of a house, the frame being pleni vertically on the rafters. Dormitory.—A. sleeping-room, Dovecote.—A building for keeping tame pigeons. А columbarium. Echinus.—The Grecian ovolo. Elevation.—A geometrical projection drawn on a plane at right angles to the horizon. Entablature.—That part of ап order which is supported by the columns; consisting of the architrave, frieze, and cornice. Eustyle.— An intercolumniation of two and a quarter diameters. Exchange.—A. building in which merchants and brokers meet to transact business. Extrados.—The exterior curve of an arch. Fagade. —The principal front of any building. Face-mould.—The pattern for marking the plank out of which hand-railing is to be cut for stairs, etc. Facia, or Fascia.—A flat member, like a band or broad fillet. Falling-mould.—The mould applied to the convex, vertical surface of the rail-piece, in order to form the back and under surface of the rail, and finish the squaring. Festoon.—An ornament representing a wreath of flowers and leaves. Fillet.—A narrow flat band, listel, or annulet, used for the separation of one moulding from another, and to give breadth and firmness to the edges of mouldings. Flutes. —U pright channels on the shafts of columns. Flyers. —Steps in a flight ot stairs that are parallel to each other. Forum.—In ancient architecture a public market ; also, a place where the common courts were held and law pleadings carried on. Foundry—A building in which various metals are cast into moulds or shapes. Frieze. —That part of an entablature included between the architrave and the cornice. ә Gable.—The vertical, triangular piece of wall at the end of a roof, from the level of the eaves to the summit. Gain.—A recess made to receive a tenon or tusk. Gallery.—A common passage to several rooms іп an upper story. А long room for the reception of pictures. А platform raised on columns, pilasters, or piers. 3 Girder.— The principal beam іп a floor, for supporting the binding and other joists, whereby the bearing or length is lessened. Glyph.—A vertical, sunken channel. From their number, those in the Doric order are called ¢riglyphs. Granary.—A building for storing grain, especially that intended to be kept for a considerable time. !

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632 APPENDIX. Groin.—The line formed by the intersection of two arches, which cross each other at any angle. Guttz.—The small cylindrical pendent ornaments, otherwise called drops, used in the Doric order under the triglyphs, and also pendent from the mutuli of the cornice. Gymunasium.—Originally, a place ж” out and covered with sand for the exercise of athletic games ; afterward, spacious buildings devoted to the mental as well as corporeal instruction of youth. Hall.—'The first large apartment on entering a house. The public room of a corporate body. А manor-house. Ham.—A house or dwelling-place. A street or village: hence Notting- ham, Buckingham, etc. Hamit, the diminutive of am, isa small street or village. | Helix.—The small volute, or twist, under the abacus in the Corinthian capital. Hem.—The projecting spiral fillet of the Ionic capital. Hexastyle.—A. building having six columns in front. Hip-vafter.—A piece of timber placed at the angle made by two adjacent inclined roofs. Homestall.—^A. mansion-house, or seat in the country. Hotel, or Hostel.—A large inn or pisse of public entertainment. А large house or palace. Hot-house.—A glass building used in gardening. Hovel,.—An open shed. Hut.—A small cottage or hovel, generally constructed of earthy materials, as strong loamy clay, etc. | Impost. —The capital of a pier or pilaster which supports an arch. Intagiio. —Sculpture in which the subject is hollowed out, so that the im- pression from it presents the appearance of a bas-relief. Jntercolumniation. —The distance between two columns. Antrados.—The interior and lower curve of an arch. Jack-rafters.—Rafters that fill in between the principal rafters of a roof ; called also common-rafters. . Jail.—A place of legal confinement. //атбз.--Тһе vertical sides of an aperture. Joggle-piece.—A post to receive struts. J'oists.—The timbers to which the boards of a floor or the laths of a ceiling are nailed. Keep—The same as donjon, which see. Key-stone.—The highest central stone of an arch. Кт. А building for the accumulation and retention of heat, in order to dry or burn certain materials deposited within it. | King-post.— The centre-post іп a trussed roof. Knee. А convex bend in the back of a hand-rail. See Ramp.

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GLOSSARY. 633 Lactarium.—The same as dairy, which see. . Lantern.—A cupola having windows in the sides for lighting an apartment beneath. Larmter.—The same as corona, which see. 4 Тайісе.--А reticulated window for the admission of air, rather than light, as in dairies and cellars. Lever-boards.—Blind-slats; a set of boards so fastened that they may be turned at any angle to admit more or less light, or to lap upon each other so as to exclude all air or light through apertures. Lintel —A piece of timber or stone placed horizontally over a door, win- dow, or other opening. Liste], —The same as fillet, which see. Lobby.—An enclosed space, or passage, communicating with the principal room or rooms of a house. Р Lodge.—A small house near and subordinate to the mansion. А cottage placed at the gate of the road leading to a mansion. Loop.— А small narrow window. Loophole is a term applied to the vertical series of doors in a warehouse, through which goods are delivered by means of a crane. Luffer-boarding—The same as Jever-boards; which see. Luther n.—The same as dormer, which see. Mausoleum.—A sepulchral building—so called from a very celebrated one erected to the memory of Mausolus, king of Caria, by his wife Artemisia. Metopa.—The square space in the frieze between the triglyphs of the Doric order. Mezzanine.—A story of small height introduced between two of greater height. Minaret.—A slender, lofty turret having projecting balconies, common in Mohammedan countries. Minster.—A church to which an ecclesiastical fraternity has been or is attached. Moat.—An excavated reservoir of water, surrounding a house, castle, or town. : Modillion.—A projection under the corona of the richer orders, resembling a bracket. р Module.—The semi-diameter of a column, used by the architect as а meas-" ure by which to proportion the parts of an order. Monastery.—A building or buildings appropriated to the reception of monks. Monopteron.—A circular colonnade supporting a dome without an enclos- ing wall. 4 Mosaic.—A mode of representing objects by the inlaying of small cubes of glass, stone, marble, shells, etc. Mosque.—A Mohammedan temple or place of worship. Mullions.—The upright posts or bars which divide the lights in a Gothic window. Muniment-house.—A strong, fire-proof apartment for the keeping and pres- ervation of evidences, charters, seals, etc., called muniments.

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634° APPENDIX. Museunt.—A A dies of natural, scientific, and literary curiosities or of works of art. Mutule.—A projecting ornament of the Doric cornice supposed to repre- sent the ends of rafters. JVave.— The main body of a Gothic church. Newel.—A post at the starting or landing of a flight of stairs. Niche.—A cavity or hollow place іп а wall for the reception of a m vase, etc. JVegs.—W ooden bricks. AVosing.— The rounded and projecting edge of a step in stairs. Nunnery.—A building or buildings appropriated for the reception of nuns.' Obelisk.—A lofty pillar of a rectangular form. Octastyle.—A. building with eight columns in front. Odeum.—Among the Greeks, a species of theatre wherein the poets and musicians rehearsed their compositions previous to the public production of them. Ogee.—See сула. | Orangery.—A gallery or building іп a garden ог parterre fronting the south. Oriel-window.—A large bay or recessed window in a hall, chapel, or other apartment. Ovolo.—A convex projecting moulding whose profile is the quadrant of a circle. Pagoda.—A temple or place of worship in India. Райзаае.--А fence of pales or stakes driven into the ground. Parapet.—A small wall of any material for protection on the sides of bridges, quays, or high buildings. Pavilon.—A turret or small building generally insulated’ and comprised under a single roof. Pedestal,—A square foundation used to elevate and sustain a column, statue, etc. | 1 Pediment.—The triangular crowning part of a portico or aperture which terminates vertically the sloping parts of the roof; this, in Gothic architecture, is called a gae. Penttentiary.—A prison for the confinement of criminals whose crimes are not of a very heinous nature. Piazza,—A square, open space surrounded by buildings. This term is often improperly used to denote a portico. Ріеу.--А rectangular pillar without any regular base or capital. The up- right, narrow portions of walls between doors and windows are known by this term. Pilaster. —A square pillar, sometimes insulated, but more commonly en- gaged in a wall, and projecting only a part of its thickness. Piles.—Large timbers driven into the ground to make a secure foundation in marshy places, or in the bed of a river. "

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GLOSSARY. 635 Рйа».--А column of irregular form, always disengaged, and always deviat- ing from the proportions of the orders; whence the distinction between a pillar and a column. | _ Pinnacle.—A small spire used to ornament Gothic buildings. Planceer.—The same as soffit, which see. Plinih.—The lower square member of the base of a column, pedestal, or wall. Porch.—An exterior appendage to a building, forming a covered approach to one of its principal doorways. Portal, —The arch over а door or gate; the framework of the gate; the lesser gate, when there are two of different dimensions at one entrance. Portcullis.—A strong timber gate to old castles, made to slide up and down vertically. ; Portico.—À. colonnade supporting a shelter over a walk, or ambulatory. Priory.—A building similar in its constitution to a monastery or abbey, the head whereof was called a prior or prioress. Prism.—A solid bounded on the sides by parallelograms, and on the ends by polygonal figures in parallel planes. Prostyle.—A building with columns in front only. Purlines.—Those pieces of timber which lie under and at il angles to the rafters to prevent them from sinking. Pycnostyle.-—An intercolumniation of one and a half diameters. Pyramid. А solid body standing on a square, triangle, or polygonal basis and terminating in a point at the top. Quarry.—A place whence stones and slates are procured. Quay.—(Pronounced ey.) А bank formed towards the sea or on the side of a river for free passage, or for the purpose of unloading merchandise. Quoin.—An external angle. See Rustic quoins. Rabbet, or Rebate—A groove or channel in the edge of a board. Ramp.—A concave bend in the back of a hand-rail. ' ‘Rampant arch.—One having abutments of different heights. . Regula. —The band below the tznia in the Doric order. Riser.—In sfairs, the vertical board forming the front of a step. Rostrum.—An elevated platform from which a speaker addresses an audi- ence. >- Rotunda.—A circular building. Rubble-wall—A wall built of unhewn stone. Rudenture.—The same as cabe, which see. Rustic quoins.—The stones placed on the external angle of a building, p jecting beyond the face of the wall, and having their edges bevelled. Rustic-work.—A mode of building masonry wherein the faces of the stones are left rough, the sides only being wrought smooth where the union of the stones takes place. Salon, or Saloon.—A lofty and spacious apartment comprehending the height of two stories with two tiers of windows.

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636 APPENDIX. Sarcophagus.—A tomb or coffin made of one stone. Scantling.-—The measure to which a piece of timber is to be or has been cut. · Scar fing.—The joining of two pieces of timber by bolting or nailing trans- versely together, so that the two appear but one. Scotia. —The hollow moulding іп the base of a column, between the fillets of the tori. Scroll. —A carved curvilinear ornament, somewhat resembling in profile the turnings of a ram's horn. Sepulchre.—A grave, tomb, or place of interment. Sewer.—A drain or conduit for carrying off soil or water from any place. Shaft.—The cylindrical part between the base and the capital of a column. Shoar.—A piece of timber placed in an oblique direction to support a building or wall. Sidl,—The horizontal piece of timber at the bottom of framing ; the timber or stone at the bottom of doors and windows. Soffit.—The underside of an architrave, corona, etc. The underside of the heads of doors, windows, etc. Sumimer.—The lintel of a door or window; a beam tenoned into a girder to support the ends of joists on both sides of it. Systyle.—An intercolumniation of two diameters. Tenia.—The fillet which separates the Doric frieze from the architrave. Talus.—The slope or inclination of a wall, among workmen called ба tering. Terrace-—An area raised before a building, above the level of the ground, to serve as a walk. Tesselated: pavement.—A curious pavement of mosaic work, composed of small square stones. Tetrastyle. —A building having four columns in front. Thatch.—A covering of straw or reeds used on the roofs of cottages, barns, etc. Theatre—A building appropriated to the representation of dramatic spectacles. i Tile.—A thin piece or plate of baked clay or other materia! used for the external covering of a roof. 70т6.--А grave, or place for the interment of a human body, including also any commemorative monument raised over such a place. Torus.—A moulding of semi-circular profile used in the bases of col- umns. Tower—A lofty building of several stories, round or polygonal. Transept.—The transverse portion of a cruciform church. Transom.—The b2am across a double-lighted window; if the window have no transom, it is called а clere-story window. Thread.—That part of a step which is included between the face of its riser and that of the riser above. Trellis.—A reticulated framing made of thin bars of wood for screens, win- dows, etc.

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GLOSSARY. 637 ` Triglyph.—The vertical tablets in the Doric frieze, chamfered on the two vertical edges, and having two channels in the middle. Tripod.—A table or seat with three legs. Trochilus.—The same as scotia, which see. Truss.—An arrangement of timbers for increasing the resistance to cross- Strains, consisting of a tie, two struts, and a suspending-piece. Turret.—A small tower, often crowning the angle of a wall, etc. Tusk.—A short projection under a tenon to increase its strength. Tympanum.—The naked face of a pediment, included between the level and the raking mouldings. Underpinning.—The wall under the ground-sills of a building. University —An assemblage of colleges under the supervision of a senate, etc. Vault.—A concave arched ceiling resting upon two opposite parallel walls. Venetian-door.—A door having side-lights. Venetian-window.—A window having three separate apertures. Veranda.—An awning. An open = under the extended roof of a building. Vestibule.—An apartment which serves as a medium of communication to another room or series of roorns. Vestry.—An apartment in a church, or attached to it, for the preservation of the sacred vestments and utensils. Villa.—A country-house for the residence of an opulent person. Vinery..—A house for the cultivation of vines. Volute.——A spiral scroll, which forms the principal feature of the Ionic and the Composite capitals. Voussoirs.— Axch-stones. Wainscoting.—W ooden lining of walls, generally in panels. Water-table.—The stone covering to the projecting foundation or other walís of a building. Well.—The space occupied by a flight of stairs. Тһе space left beyond the ends of the steps is called the wed/-hole. Wicket.—A small door made in a gate. Winders.—]n stairs, steps not parallel to each other. Zophorus.—The same as frieze, which see. Zystos.—A mong the ancients, a portico of unusual length, commonly appro- priated to gymnastic exercises.

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638 APPENDIX. TABLE OF SQUARES, CUBES, AND ROOTS. (From Hutton's Mathematics.) CubeR oot, No. Square. | Cube. | Sq. Root. [CubeRoot| No. | Square.| Cube. Sq. Root. i 1 1| 1-000000 1-000000| 68| 4621| 314432| 832462113 4-081653 2 4 8 | 14142136 159599911 69] 4761 328509] §-3066239| 4 10155 3 9 27 | 1-7320508) 1-442950| 70] 4900 343000) 83666003; 4-121285 4 16 64 | 9-0000000| 1:537401 71| 5041| .357911| 84961498! 4:140818 5 95 125 | 22360680 1:709976| 72] 5184 373248| 8:4852814 4-160168 6 36 216 | 9-4494897| 1:817121|| 73] 5329 389017| 8:5410037| 4:179339 T 49 313 | 9-6457513| 1919931! 74 5476 405224| 86023253, 4198336 8 64 512 | 28234271) 2-000000] 75| 5625 4921875) 8-6609540| 4 217163 9 81 729 | 3-0000000' 2080084 76] 5776 433976| 87177979; 4 035824 i 10 100 1000 | 3:1622777| 9:154485| ТТ 5929 456533] 87749644| 4254321 11 121 1331 | 33165243 9-923380 781 6084|: 474552) 88317699! 4:272659 12 144 1728 | 34641016 2239429] 791 6211 a 8:8381944! T a. 13 9 2197 6055213 2-351335) 80) 6400 2000| 8-9442719| 4:3039 ч 196 2744 ж. eee d Bl| 6561 531441| 9-0000000| 4-326719 15 | 995 3375 | 38729833 2-466212| 89 6724 551358 9-0553851| 4344481 16 956 4036 | 4-0000000| 9:519849'| 831 6839 571787| 9-1104336! 4:36207 17 239 4913 | 41931056 2:571932| 84] 7056 592704! 9-1651514' 4:379519 18 324 5332 | 4:9496407| 9-620741 85] 7925 614125) 9:2195445| 4:396830 в POUR EM E D е. s Fs [3 Ei ! oi 2 ә . Ф e 21 n 9261 deme MEM 83 7744 681472) 9:3328315| 4417960 23 484 10643 | 446904153: 2-802033, 89| 7921 704969| 9-4339311| 4-464745 23 529 12167 4:7953315| 2-843367. 90} 8100 7290001 9-4363330| 4181405 24 | 57 13324 | 4-8983795| 9-981499 91} 8281 ded 9539320 CET $ 5 5625 а i 9. б 5 30 : JEJE p ^p € HU tans 27 799 19633 5:1961594 3-000000 94| 8836 830584 9:6953597 EU. 23 784 21952 5 2915026) 3:036599. 95| 9025 TEE dee 232008 S| bel ED EEB E D. 31 | 961 29791 | 55677614| 3-141331! 9З 9604 941192} 9-8994949| 4:610436 32 | 1094 32768 | 56568542 3.174809. 9) . 9801 970299] 9-9498744| 4:626965 33 | 1089 | 35937 | 57445626 3:2)7531| 100] 10000) 1000000! 10-0000000) 4 611589 31 | 1156 39301 | 58229519 3.232619 101] 10201, 1030301| 10-0498756! 4:657009 35 | 1225 42575 | 59160798 3-271066) 102} 104041] 1061208! 10-0995049! 4672329 36 | 1296 46656 | 6-0000000| 3-391927 103] 106091] 1092727| 10-1483916! 4697518 37 | 1369 50653 | 6-0397695! 3-332222 104| 10816] 1124861| 10-1980330! 4:709659 ' 38| 1444 54372 | 6:1644140| 3:361975! 105| 11025| 1157695! 10-2469503! 4:717694 89 | 1591 59319 | 62449980 3-321211| 106| 11236) 1191016 10-2956301! 4:73269 40 | 1600 61000 | 6:3245553, 3419952. 107| 11449| 1225043 10-3440804] 4717459 41 | 1681 63921 6-1031219. 3.448217! 108! 11664| 1959719! 10:3923013: 4762203 ^2 | 1764 74088 | 6-4307407| 3-476027|| 109| 11831} 1295029| 104403255! 477635 43 | 1819 79507 | 6-5574335| 3:503398 110| 12100| 1331000| 10-4880835 4:791420 44 | 1936 85181 | 6:6332496| 3-530318|| 111| 12321| 1367631| 10:535053: 4-805895 45 | 2025 91125 | 67082033, 3:556893.| 112| 195441 1404928] 10:5330052! 4-820984 46 | 2116 97336 | 67823300! 3582М8! 113, 19769| 1442897] 10-6301459! 483458 47 | 2209 AS 6-8556546| 3-608325|| 114| 19996 obe т nd 48 | 2304 110532 | 69232032! 3634241|| 115| 13225. 1520875| 10 7238053, 4:862 49 | 2401 117649 2-0000020 3-659306 116 13456] 1560896 10-7703235| 4-876999 50 | 2500 125000 | 7-0710678| 3-631031) 117| 13689| 1601613) 10:8165533| 4:890973 51| 9601 132651 | 71414984 3-708430)! 118! 13924] 1643032) 10-8627805| 4:904868 52 | 9704 140608 | 72111026! 3:732511| 119| 14161| 1685159, 10-9087121, 4:913685 53 | 9309 148377 | 7-2301099| 37756285| 190) 14400| 1798000| 10-9544513| 4:932424 54| 9916 157461 | 7-3181692 3:779763 121| 14641] 1771561| 11:0000000| 4-946087 55 | 3025 166375 | 7-4161985| 3.302952 | 122| 14884 1815818| 11-0453610| 4059676 56 | . 3136 175616 | 7-4833148| 3-825882! 192. 15129! 1860867| 11:0905355| 4-973190 «NETT; b з ЭБ! 905379 КЕН 3.892990 m T 2000376| 119219722, 5-013298 60 | 3600 216000 | 77459667, 3-914868; 197) 16199 9048383! 11-2694277| 5:026526 61| 3721 226981 | 7-8102197, 3-936497, 198| 16334| 2097159| 11-3137085| 5:039684 62| 3344 233328 | 7-8740079| 3-957891|| 129| 16641! 2146689| 11:3578167| 5:052774 |9 3969 9220017! 7-9372539, 3:979057| 130| 16900) 9197000 ТШ, PORUM 64 96 262144 | 8-0000000, 4-000000/ 13) 17161 9948091 11544559211) 507879 65 D 274625 8:0622577, 4-020726|| 132 17421 9999965) 11-1891253| 5091643 66 | 4356 297496 | 81210331! 4-041240) 133] 17689 22596371 11:5325626 5104169 67 | 4482 300763 | 8-1853523 4-061518]! 131 2406104! 11-5758369| 5:117230 Le | - 17956;

Page 679

No. | Square. Cube, Sq. Root. iCubeK oot, 18225| 9460375) 116189500, 5:129928| 202| 40804 18496] 9515426) 110619033, 5:142563| 203| 41209 8365497 18769) 2571353, 11°7046999| 57155137] 204| 41616) 8189661 19044| 9695072) 117473101, 5°167649|| 205| 42025| 8615125 19321] 2635619| 117898261] 5-189101) 206) 42436) 8711816 19600| 274.4900] 118321596] 5192191) 207] 42819) 8869743 19881) 9803221 118743422, 5204828) 208 43264 8998012. 11-4222051 20164] 9563283 11-9163753| 5217103) 209| 43581 9129329, 14-4568323) 20449] 9924207) 11:9592607| 5-229321) 210} 44100) 9261000! 14:4913707| 5 20736} 2995934! 19-0000000) 5-211483, 211: 4499) 9393931 14:5258390 21025| 3048625| 12:0415946| 57253533] 212} 441944 9528193 14-5602198 21316| 3112136 120830460, 5-265637) 213, 45369| 9663597) 14:5915195| 5:972093 21609| 3176523) 12-1943557 55277632) 914 45796] 9800344: 14:6287338! 5-931421 21901 3241792) 12-1655251 5-289572! 215| 46295 9933375! 14:6623783| 5960726 22201| 3307949| 12-2065558| 5-301459 216| 46656, 10077696) 14-6969395; 6:000000 22500 3375000) 12-2174487, 5-313223, 217 47089 10218313! 147309199 292391 3119951 129332057) 5:322074| 218} 47594) 10360232! 14:7618931 23104; 3211808! 12:3238280, 5:336303, 219) 4796) 10503459) 14-7986486 23409) 3531577 12-3693169 5:318481| 220} 48400; 10618000| 14:8323970 23716; 3652264 121188995 5:360108 92) 43341, 10793861) 14:8660687) ©1025! 3723375) 192:4198994 5:3716535) 222) 4999! 10941048, 14-8926644 24336! 3796416| 12:4329360 5333213 223, 49729| 11039567 14:9331815; 21619| 3869393) 12:5299641 5:324691) 224| 50176 11239424; 14-9666295; 60/3178 21961| 3914312, 12:5538051 5-406120 225| 50625 11330625. 15-0000090] 6:082202 25281) 4019679) 12:6095202 5417501; 226) 51076 11513176, 15:0332961| 6:091199 29600) 4095000 126191106 5:428335| 227| 5152) 11697083, 15-0665192 29921. 4173281 12:6335775 5:440129 | 923] 51981] 11852352 150996639! 20214] 4251593 12°7279221 5451362 229, 52441| 12008939| 15-1327460 26569, 4339747 1277671453 5-462556 230} 52900] 12167000; 15-1657509 26396) 4410944 12-8062485 5473701 93) 53361 123262391) 15:1936349 97225 4499125! 12:8152326 5:43180/| 232| 53824) 12487168, 152315162 97556| 4574996) 12:8310987| 5-495365, 933) 51289 12619337, 152618375 2783) 4657463, 129228180 5:506878, 234| 54755) 12812904 15-2370535 23224 4741632, 12:9614814 5:517818 295 55225 12977875) 15:3297097 23501| 4826809 13:0000000 5328719 226) 55696) 13144256) 15-3622915 28300) 4913000, 13-0381048 5:539658, 237; 56169 13312053, 15-3343013 29241) 5000211! 130766968 5590199) 233) 56644 13481272) 15-4272186 29581! 5083148 13:1143770 5:561993, 239) 57191] 13651919) 15-4596248 29923; 5171717, 19:1529464 5:572055 240) 57600) 13324000) 15-4919334 30276 5268924 13:1909060 5 532770 241) 58081, 13947521 155211741] 30£25| 5359375, 13°2287565| 5:593145 242) 58564) 14172433! 15-5563192; 30976, 5451776! 13:2664992) 5:601072, 913 59049 11313907, 15:5334573 31329| 5515233, 13:3041347| 5:614672/ 944) 59536 14526784) 15-6204994 31681) 5639752, 19:3416541) 5:625226 245| 60025] 14706125, 15:6521753 32041| 5735339 13:3790332) 5635741) 246, 60516] 14836936) 15:6313371 32109, 5832000 134164079 5646216) 947 61009] 15059223 157162333, 32761) 5929741 To iiae 5°656693)| 948 61504 15252902 15-748)157 33121, 6023568 131907376, 5.667051, 249| 62001| 15133219| 157797333 33133; 6123137, 13:5277493. 5-677411 250) 6250) 155230J0| 15:8113333 33326] 6229504! 13:5046600| 5-637731, 951] 63001, 15313251| 15-8429795 31225. 6331625 136014705, 5-693019, 252) 63504| 16003008) 15-8745079 34596) 6431356| 13°6331817) 5-708267) 353, 64009, 16194277) 15-9959737 34969| 65239203 13:0717243 5718179 954 61516| 16337961| 15:9373775 35314] 6644672) 1377113099; 5-723651 955, 65025) 16531375) 15 9687194 35721] 6751269) 13-7477271 5733794 956) 65536) 16777216) 16 0090000 361001 6859000, 13°7810483| 5-748397| 957 65049] 16974593). 16:0312195 36131] 6967371, 13:8202750 5-758965] 253; 66564 17173512, 160623731 36304, 7077838, 13:85641065 57763998, 259| 67031] 17373979 16:034763 37219] 7189027 13-8924440) 5-778996, 260 67600] 17576000) 16:1215155, 6:382504 37036] 73013314, 139283333 5°783960 261) 63121] 17779531, 16°1554944) 6:390676| 38025 7414875| 13:9642400, 5°798390)| 262! 63644) 17981723 38416. 7529536; 14-0000000; 5-808736! 963) 69169] 18191447) 162172747 33809} 17615373) 14:0326683| 5:818648|| 264) 69696] 18399744) 162180763, 6415069 39204] 7769392! 14-0712473, 5-823177| 965 70225) 18609625 16:2738206 6123152 39501} 7880599) 14-1067360| 5°833272)| 965 ДЕТ 18821096; 16:3035961| 6:43122* 40000} 8000000) 141421356, 5:818035 967) 71289, 19031163, 16:3101316| 6432277 40101} 8120601| 14-1774469| 5:8577661 969%) 71894) 19248832) 16°3707055| 6:447306

Page 680

640 269) 970 271 272 E No. | Square. 72361 72900 73441 73384 14529 75076 19625 76176 76729. 77934 71811 18400 18951 19524 80089 80656 81225; 81796 82369 82944 83521 84100 84681 85264 85549 112225 Cube. 19165109 19633000 19902511 20193618 20316417 20570824 20796875 2102457 21953933 21484952 91717639 21952000 22188041 22425768 22665187 22906324! 93149195 93393656 23639903 23387872 24137569 24389900 24642171 24897083 95153757 25412184 25672375 r 25934336 25198073 26163592 26730899 27000009 27270901 27542608 27818127 28094464 28372625 23652616 28934443 29218112 29503629 29791000 30080231 39371328 30664297 30959144 31255375 31551496 31855013 32157432 32461759 32763000 33076161 33336218 33598267 34012224 34323125 34645976 34965783. 35287552 35611239 35937000 36261691 36594368 36926037 37229704 37595375 16:4012195! Sq. Root. 16:4316767 16:4620776 16:4924225 16:5227116 16°5529454 16:5331240 16:6132477 16:6433170 16:6733320 :16:7092931) 16-7332005 16:7630546 16-7923556 16-8226033 168522995 168819430 169115345 16-9410743 16-9705697| 17-0000000 17-0293364 170597991 17:0830075 17:1172498 17-1464292 17-1755640 172046505, 179336879 179696765 179916165 173205081 17:3493516 17-3181472 17-4068952 17-4355959 17-4642492 17-4928557 17-5214155 17-5499288 17'5783958 17-6068169 17:6351921 17-6635217 17-6918060 177900451) 17-7482393] 177763388 17-8044933 17-8325515 17-8605711 178885133 17-9164729 17-9443594 17-9722008 я ao 18:0277564 180551701 18-:0831413 18:1107703 18:1333571 18:1659091 18:1931054 18-2208672 18:2482326 18-2756669 APPENDIX. с ubeRoot. — 6:455315 6:487154 6:495065 6-509357 6-510832 6:518681 6:526519 6:531335 6:512133 6-549912 6:557672 6:565414 6:573139 6:580344 6:538532 6:596902. 6:603354 6:611489 6:619106 6:626705 6:631937 6.641852 6.649400 6:656930 6:664444 6:671940 6:679120 6:685833 6-694329 67701759 | 6:463304| 6:471274| БЕТІН 6:709173 6716570 6-723951 6-731316 6-733664 7145997 6°753313 6-760614 6:767899 6775169 6-782423 6-789661 6-796834 6-804092 6811235 6818462 6-825624 6839771 6-839904 6:847021 6:854124 6:861212 6:875344 6-882339 6:8894I9|| : 6896435 6:903136 6910423 6-917336 6-924356 6-931301 6-933232 18:3030052, 6:945150|. 402 00| 6:868235 | No. | Square. Cube. 112896} - 113569 114244 114921 115600 116281 117649 118336 119025 119716 120409 121104 121801 122500 123201 2} 123904 124609 125316 126025 126736 127449 128164 128881 129600 130321 131769 132496 133225 133956 134639 135194 136161 136900 137641 138384 139129 139876 140625 141376 142129 142884 143641 144400 145161 145924 146639 147456 148925 148996 149769) 150544 151321 152100 159881 153664 154449 155236 156095 155816 157609 158104 159201 160000 160801! 161604! 1169641. 131044! 37933056 38272753 33614172 38928219 39301000 39651821 40001688 40353607 40707584 41063625 41421736, 41781923 42144192 42508549 42875900 43243551 43614208 49982977 44361864 44738875 45118016 45199293 45882712 46268279 46655000 47015381. 47437928 47832147 48223544 48627125 49027896 49130863 49835032 50243109 50653000 51064811 51478818 51895117 52313624 52734375, 53157376 53582633 54010152 54439939 54872000 55306341 55749968 26181837 56623104; 57066625 57512456 57960603 58111072 53363369 59319000 59776471 60236233 60693457 61162984 61629875 62099136 62570773 63041792 63521199 64000000 64481201 64964808 18-3303098| 18:3575598 18:3347763 18:4119526 :18-4390839 18 4661853 18:4932490 18:5202592 18:5472370 18:5741756 18-6547581 18-6815417 18-7082869 18-7319940 18-7616632 18-7382949 18-8148877 18:841443 18:867963 18:8914436 18-9208379 18-9472953 18-9736660 19-0000000 19-0262976 19-0525599 19-0787840 19-1049732 19-1311265 191579441 19-1833261 192093727 19:353341 19-26 13603 19-2373015 19-3132079 19:3390796 19-3649167 19-3307194 194164878 19-4499221| 7 19:4679223 19:4935337 19:5199213 19:5448203 19-5703353 19:5959179 19:6214169 19-6468327 19-6723156 19:6977156 197230829 197484177 1977737199 19-7989899 19:8242276 19:8194332 19-8746069 19:8997487 19-9249598 19-9499373 19-9749844 20-0090000 20 0249341 20 499377 Sq Root. \СиьеК со.

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168 100} 168921 169744 170559 171396 172225 | 173056 173389 174724 | 175561 176100 177241 178084 178929 ! 179776 180625 181476 | 182329 183184 181011 184900 185761 185624 187489 183356 189225 190096; 190969 191814 192721 193600} 194481 ‚ 195364 ‚ 196219 197136 Н 198025; | 193916} 199809; 900704! ; 201601; . 202500. 2 204304 ‚ 905209 | 906116 >; 207025 207936 | 908319 | 209764 21068] 211600 | 212521 , 213444 ‚ 214369 | 215296 ‚ 216225 217156 7 918089 468. 219024; 469| 219961 : 203401. TABLE OF SQUARES, CUBES, AND ROOTS. 65150827 85939964 €613)125 66923116 67419143 67917312; 68417929 63921000 694267 31 699341 23 70444997 70357941 71173375 71991296 72511713 73034632 73560059 74038000 74618461 75151448 75636967 76925091 76165625 77303776 77854483 18402752 78953589 79507000 80062991 80621568 81182737 81746504 82312875 82881856 83453453. 81027672 84601519 85181000, 85765121 86350388 86938307. 87528334 88191195. 88716536 89314623 89915392 90518319 91125000 91733851 92345403 92959677 93576064 91196375. 94818816 95443993 96071912 96702519, 97336000 97972181 98511198 99252347, 99897344 1005 11625, 101194696, 101817563 102503232 10 — 20-0748599 20-0997512 20-1246118 20-1494417 20-1742410 20-1990099 29-2237484 20-2484567 20-2731349 20-2977831 20-3224014 20-3469899 20-3715188 20-3960781 29-4205779 20-4450183 20-4694995 20-4939015 20-5182845 20-5426386 20-5669638 20-5912603 20-6155281 20-6397674 90:663783 20-0831609 207123152 207364414 20-7605395 907816097 20-8086520 20:8226667| 20-8566526 20-8806130 20-9015450 20-9284495 20-9523268 20-9761770 21-0000000 210237960 21-0475652 21-0713075) 21-0950231 211187121! 211493745 21-1660105 21-1896291 212132034 212367606! 21-2502916 219831967 91:3072758 91-3307299 219541505 213775583, 21-4009346) 21-4212353 21-41476106! 214709106 21-4941853 215174318 215406592 215633597 21-5370331 21-6101828 7:336437 7:392549 7:398636 7:404721 7:410795 7:416859 7422914 1428959 717431994 7:441019 7:447034 7:453040 7:459085 7:465092 7:470999 7:476966 7:482924 1488872 1494311 7590741 7506661 . 17512571 1518473 7:524365 1:530248 7:535192 7:511987 75179142 7:553639 7559526 7565355 T571174 7576985 7582786 7:583579 7:934363 7600133 7-605905 2611663 17617412, 7623152, 7-693384: 1621607, 7 610321 7-616027 7051725 7657414. 17663094. 7663766 7:674430! 7680086 | 1:685133 77691372; 177627002 17702625. | | | 7708239, 7713345 7119443 7125032. ШЕР 77130614 77136183 17111753 1141211 77192861 | | н | | 1798102. 21:6333077| 1-763336; 535 Sq. Root. |CubeRoot. No. | Square. | 4701 292900 471 472 221841 222784 413, 223729 474! 224676 475| 476 534 225625 226576 227529 293481 929441 930400 231361 232324 233289 234256 235225 236196 237169 238141 239121 210100 241031 212064 213049 244036 245095 246016 247009 243004 249001 250000 251001 252004 253009 254016 255025, 256036 257049 253061 259081 260100 261191 262144 963169! 964196 265295 260256 267289 963324 269351, 270400 271441 972434 273529 974576) 275625 276676 271199, 273734 279341 280900 281961 291089 285156 286225, 2 и i 3161709, 21-0561078, 7°769462) 536| 287296) Cube. 103323000 104487111 105151048 105823817 106496424 107171875 107850176 108531333 109215352 109902239 110592000 111281641 111930168 112678537 113379304 114081125 114791256 115501303 116214272. 116930169 117649000 118370771 119035138 119823157], 120553784 121287375 122093336 122763413 123595932 124251499 125090090 125751501 126506008 127263527 128024064 198787693 129554216 130323313 131096512 131872229 132651000 133432331 134217728 135005697 | 135796744 136590375 137333096 133185413 138991832 139798359 140608000 141420761 142236618 143053667 143377921 144703125 145531576). 146363183 147197952 148035339 148877000 149721291 150558763 151419437 152273301 153130375 21-6794334 Sq. Root. 217025344 21-7255610 2177485632 91777154111 217944947] 218174242 91:840324: 21-8632111 21-8860686 21:908J023 219317122 19514934 919772610 220000000 22:0227155 22 0451077). 7- 22:0680765 92:02307220 92:1133414 22: 1359135 99:1535193 221819739! 22:2035033, 2272231 103 22°2405955: 92:2710525, 922334963 22:3152135 22:5333)79 22:36U007J8 92:3399293 22:4053565 92:4276615 294499443 92:4722051 29:4914438 22:5166005 22:5333553! 22:5010283 92:5331795 226053991 926274170 22:6195033 22:6115531 226936114) 22 7156334) 2471316340; .8: 927525134 227815715 228035055 22:8254244. 22-8173193 223691933 22:8J10463: 22-9316 1991 229554926 22:9732506) 23-00000 23:04 17289, 230424372 234651222! 93:0867928 23-1084400 23-1300670 153990656, 23:1516738. 8-19309 1.93700 7.94299 1-917574 1-952381 1-989570 8-00521V; 8:014; 801559: 8-072262 8:077374 B-0824 8-087579 8-092677. 809775 8-107913 8:112930 8:113011

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| 329476 993521] у 991600 292631! | 293764 294919! 295936 | 297025 298116 299209 i; 300304 301401 302500 323601 9! 3204704 35809 306916 303025 309136 310249 | 311364 | 312431 313600 31472] 2 315341 | 316969 313096 319225 320356 321489 322624 323761 324900 326011 327184 323329 | 930095. 331776; 332929 334033 335241 336400 | 337561 | 333724 | 339339 i 311056 | 342295 | 313395 314569 345744! 346921 343100, 343281. 359464! 351649! 352836, 354025, 355216, 356109 357604 358801! 360000 361201 362404, 363609; 154851153 155720872 156590819) 157464000 158340421 159920088 160103007 160939184 16 1873625 162771336 163667323 164566592 165169149 166375000 167284151 163196608 169112377 170031464 170953875 171879616 172308693 173741112 174670579 175516000) 176553181 177504323 173453547 179406144 130362125 181321496 182281263 183250432 184220009 185193000 186169411 187149248 183132517 189119294 192109375 191102976 192100033 193100552 194104539 195112000 196122911 197137368 193155237 199176704 200201625 201230055 202262003 203297472 201330102 905379000 2061%5971 207474683 208527857 209581581 210644875 911708736 212776173 213347192 214921799 216000000 21708180] 918167208 919256997 Eun ы Sq. Root. 23-1732605 23:1948270 23:2163735 23:2379001 93:9591067. 232803935: 23-3023604 23°3233076! 8 93:3159351 23°3665429 APPENDIX. CubeRovt | No. | Square. 8-123145 8133187) 8-133223 8-143253 8-148976 8:153294 8:138205 163310 8-168309 8-173302 23-3893031 1| 8:178239 234093998 234307490 234520788 23-4733392 234946802 23-5159520 223:5372016 23:5534330 93:5796529 23-6003474 23 6220236 23-6431808 23-6543191 23-6354336 23-7065392 237276210, 23-74863412 237697283 23:7307515. 528112618! 23:83275.6 23-5537209! 23-8746723 23 8450062 29:9165215 8 23-9374 184. 93:953971 93-9791576 24:0000020 21:0208213 21-01163)6 94:0694188 24-0331891 21-103J416 241216762 241453929 241660919 241867732 242974359 94:930329 219187113 242593222 21:2399156 24-3104916 213310501 24-3515913 24-3791159: 8:183269 8:188944 8:193213 8:198175| 8:203132 8:203032 8:213027 8-217966 8222893 8227825 8-232746] 8237661 8242571 8:247474| 89528711 | 8257263 8:262149/ | 8:267029) 8271904) 8276112 3) 8-281635 8236493. 8:231344 | 8-2J6190 8-3)103) 8:3)5865| 8-310694], 8:315517! 8 320335 8325147) 8:329 8231755 8: 339551! 8314341] 8349126) 8-353905; 8353678): | 83524171! 8:353200!: 8°372967|' 8377719! 8252465: 8-337206! 8391942) 8: 336673}, 8-401398]. 8406118: 24-3326218| 8-410333) 214131112; 24-4335824| 24-4510385 24.4744765 244943274 24:5153)13 21:5356333 24:5560533 8:415542/' 8: 420246]! 8:4 24945). 8:499633! 8:431327! 8-439010!! | G 8: 443683) 8 zur wil 669, 670. 2; 339424 3.4816 356025 307236 368149 369664! 370881! 872100 373321 374554 375769 376996 873222 379156 330689 331924 333161 331400 385641 386834 383129 339376 1 390625 391876 393129 394334 395641 396900 333161 400685 401956 403225 404196 405769 407044 40832] 40960 410881 412164 418419 414736 416025 ; 417316 418509 ‚ 419904 421201 | 422500 423891 | 425104 ! 426409| 2 427716) 429025. | 43.336 431619| | 432964. | 4319811 | 435600 | 436921 2 438244 439569 : 440896 ` 442225 443556 444889 | 446221 447561 448900; Cube. 5р Rect. 220348864) 245764115 221445125) 24-5967478) 999545016! 293648543! 2247557 12! 225866529 22698 1000 298099121 229220928 220246397 231475544 232608375 933744895 934885113 236029032 237176659 233328000 239433061 240641818 911804367 242970624 244140625 215314376 246491833 247673152 248853189 250047000 251239591 252435968 253636127 254810104 256047375 257259456 258174853 259691072 260917119 262144000 263374721 261609233 265817707 267089934 268336125 269586136 270310023 272097792, 273359449 274625000! 275894451 277157808 73145077 219726964 981011375 282300416 953593393. 234890312 936191179. 987496000: 233304781, 290117528. 291434247 292751944: 294079625. 295108296. 296740963. 998077632, 299418309 300763000 24:6170673 246373700 24:6576560 24:6779954 24 6981781 247184142 247386338; 247588358 2417790234 247991935 248193473 248394347 248596058 248797106 218997992 249198716 ‘249399278 21:959:673J 24:0799920 25:0000000 25:019992) 25:0399681 25:0599282 25 0993003] 25:0993008 25:1197134 25:1396102 25:1591913 25:1793566 25:1992€63 25:2190404 25:9333539 25:2585619 25 9731493 25.2932213 953179778 95:3377189 953574147) 95:3771551 25:3968502 25:41653.1| 954321947 95455841 22:4751784) 254950976; 25: 5147016 255342907! 95:5523647! 955734937 25°5929678 25°6 124969) 25°6320112, 25°65 15107} Q3: 6709923. 95694652 257099202 95:7293607! 257487854 251681975 257375939! 23°8069758: 25:8263431| 05:815696)! 258650343 25-8842582, CuleHRoot. 8:54075 8:54531 8:519880 8:551437 8:553990 8:608753 8613248 8:617739 8622225 8:681546 8:688963 8:693376 5:697731 8: 715373 8-719760 8-721141 8-728518 8-732392 87737260 8741625 87145985 87750340

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No, | Square. 671 “67у 673 674 675 -676 677 678 679 630 631 632; 683 450241 451534 452929 454276 455625 456976 458329 459684 461041 462400 463761 465124 466439 684| 467956| 685| 469225 686| 470596 687| 471969 688| 473344 659| 474721 690) 416100 691! 477431 699). 478361 .693| 480219 694! 481636 695| 483525 696! 481416 697| 455309 698| 487204 699| 488601 700! 490000 701| 491401 702! 492304 103! 491209 704! 495616 705! 497005 706| 498436 107| 499549 108; 501264 109| 502681 710| 504100 7111 505521 112| 506914 713) 508369 714| 509756 715| 511225 716! 512656 717 514059) 718! 515524 719! 516961 720| 518400 72]! 519341 722) 521284 123. 5237129. 7241 591116 725| 525645 726. 527076 797) 528529 798. 529984 729! 531411! 730; 532900 731! 534361 732| 535594 232) 5у79м) 134| 532726 735! 540225 736) 541690) 737| 543169 TABLE OF SQUARES, CUBES, AND ROOTS. Cube, 302111711 303464448 304821217 306182024 307546375 308915776 310283733 311665752 312046839 314432000 315821241 517214568 318611987 320013504 321419125 322828350 324242703 325660612 27082769 323509000 329939371 331373338 332819557 334255384 335702375 337153536 338608373 340663392 341532099 343000000 344472101 945945405 347428997 318913661 35040-695 351895316 353393243 354894912 356400829 357911000 35942543! 360944 128 362467097 303994344 365525375 367061696 368601813 37014623: 371694959 373248000 Mag ined 370357048 377933067 379503124 331078125 382657176 384240532 88589835: 337190489 385017000 396617391 309223165 393832837 345446909 337065375 395688256 400315555! — 2 енен Sq. Root. — 259036677, 25:9229628 95:9492435 25:9615100 25-9807621 26:0000000 20192227 26:0381331 26.0576284 260763056 96:0959767! 96:1151297) 261342687! 261533937 26-1725047 26:1916017 26-2106818 262297541 262483095 262673511 26-2868789; 26-3053999 26:3218939 26-3433737 26-3628527 263318119 26:4007576 26-4196896 26:4386081 26-4575131 25-4764046 26-4952826 26:5141472 265399983 26-5518361 26:5706505 26:5394716 26-6082694 26-6270539 26:6458252, 26°6645833 26-68339311 2677020598 767907784 27:739482) 96:7581763 267763557 967955290 268141754 26:8323157 26°85 14432! 26°870U577 268886593 26-9072481 96:9258240 269143872 259629375 “6:9814751 *11:0000000 27-0185122 27-0370117' 27-0554985! QT-0 [39723 970994344 27.1108834 27 1293199 27 1477439 uiis ді | Na, | Square. 87546911] 8759033 8763331 8767719) 8112053. 8776333 8782708 8-785030! 8:789347| 8793659 8-797968 8802272 8-806572. 8810868 8:815160 8819447 8823721 8-823010 == 738 739 740 741 742 743 744! 746 747 743) 749 750 751 752 753 754 195 156 727 758 759 760 761 162 763 764 765 166 767 768 8-836556 8-840823 8:845085 8:849344! 58:853598 8-851849 8862095 8.866337 8:870576 8-874810 58879040 8883266 8-887483 8-891706 8805920 8-900130 8-904337 8-008539 8-912737 8-916931 8-921121 8-925308 8-029490 8-933669 8-037843 8-912014 8-946181 8-950344 8-954503 098658 8-962809 8-966957 8971101 8-975241 8-979377 8-983509 8987657! 8:991762| 5-995883] 9400000) 796 9-004113] 9-008222 798 9019329! 9016131) 800 0-020525 9-091621| 802 9025715! 803 9-032802! 804 745) 3 514614 516121 547600 549031 550564 559009 559504) 5651001 562500 564001 565504 501009 25608516 570025 571536 573449 14564 570081 511006 572121 530644 532169 533696 585225 586756 553259 589824 591361 592900 594441 5,0984! 597529 599076 600625 602176 603729 005284 606841 00340 605961 611524 613089 611656 616225 j| 617796 619369 620944 622521 624100 625631 627264 623349 630430 632025, 634610 685209 636304 635401, 640006 641601 643204 644509 646416 403533419 405224000 406369021 408518488 410172407 411530784 413193025 415160936, 416332723 413508992) 490189749; 421872000 423561751 425259008 425957777 423661064! 430368375) 432081216 433795093 435519512 431245479 433970000 440711051 442450728 444194947 415945714 417697125 449455056 451914663 452984837. 454756609 456533000 452314011 400099648 461835917 465624374 465484375 4072885 16 469097433 470910959 474129139 414992000 416979541 470211768 43045637 4818уо204 4957506:5 455551050 497443405 45930337% 491169069 4990890,0 494915671 496799052 495617251 50056615341: 2459315 504353336 5062615725 50810959% 510682395 512000000 513922401 515915605 517781627 519718164 401947272, 27-1661554! 9:09688 27:1845514! 9:040965 Cube. | Sq. Кост. |CubeRoot. 272029110 272213152 27-2330769 979530253! 272763634 27:294688 | 21:3130066. 21:3313007 21:3195337 217-3015614 27-3361273 274043732 27-4226 134 27°4403455 2/1:45906J4 2741120633, 2174751542 z1:513033J 24:5317998 21:5199546 21:5680975 2753062234 276043475 77 6224546 2400402519» 71: 6580334 21 0707050 z1691;645| 9: 477195179 27735084): 277155739 21 1663865 21 1043850 27: 8023775 213208555 153352 15j rae the 278747197! 273996514 97:9105715 27:9254901 279455772 219642029 279821372 9258:06,0000U 28 0175915 280855915 750535205 25:07 13371 23:0891435 23:1069350 23° 1247222 23:1424946 2:3: 16002597 z3:1780056 98:1957444 259134720 282311884 28:243599> 234609531 28725427 12 94:39019494 28:3195.045 У8:5372510 (8:3545992) 643 904504: 9-049114 9053183 9-16962- 9:1:3532 9:171944 9-18у40. 9193537 y1Y7TZ9u u:z29211 9%06.,79% 9-298021

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644 No. | Square. 643025 j| 649636 651249 659864 654181 656100 657791 659314 660969 662596 “664295 5| 665856 607489 669124 670761 672400 674041 675684 677329 | 678976 680695 632276 633329 685584 687241 6839 4 690561 2| 69:924 693889 695556 | 697225 ` 693896 760569 709244 | 703991 705600 707281 708264 710649 | 712326 714695 715716 717409 719104 720801 729500 794901 795904 | 727609 729316 731025 732736 731449 736161 737381 739600 741391 743044 3 714769 746496 5 748225 719956 | 751689 868! 753124 869) 755161 87С! 756900 871. 758641 1 usu lee ea a ne “йылы. 521060125 523606616 595557942: 597514114) 529175129) 531441000 532411731 535387328 537367797 539353144 511343375 513338496. 545833513 547313432) 549353259 551368000 553387661 555419948 557441767 559476994 561515625 563559976 565609983 967663552) 569722789 571787000 573856191| 75930368! 578009537 580093704 532182875 534277056 586376253 538180472. 590589719, 592704000. 594823321 596947688. 599077107 601211584 603351125 605495736 607645123 609300192 611960019: 614125000! 616295051 618470208 690650477 622835364 695026375 697922016 629429793 631623712 633839779 636056000 638277331 640503928 642735647 644972544 647914625 649161896 651714363 653972032, 656934909 | Cube. | Sq. Root. |Cubekoot. 28 3725219 283901391 98-4077454 28 4953408 234429253 :28:4604989 284780617 23:4956137 23:5131549 28:5306852 23-5482048 23-5657137 28-5832119 23-6006993 28:6181760| - 9-359902! 28-6356421 28-6530976 28-6705424 23-6379766 23-7054002 287228132 287402157 287576077 287749891 28-7923601 28-8097906 28-82707061 98:844410%! 98:8617394 98:8790589! 238963666) 28-9136646 28-9309523 28-9432297 23-9651967 28-9827535 29-0000000 29-0172363 29-0344623 29-0516781 99:0688837| 99-0860791 99:1032644 99-1904296 29-1376046 291547595] 29-1719043 29-1890390! 292061637 292232784 29-2403830 29-2574777 99:2745623. 29:2916370 29-3087018 29:3257566! 29-3493015 29-3598365. 99-3768616. 29-3933769' 29-4108892 99-4978779 29-4448637 29-4618397! 29-4788059 APPENDIX. 9-302177. 9:306323 9:310175 9314019 9-317860 9-321697] 9325532 9323363 9:233192 9-352286 9356095 3-363705 9-373887 9332675 9-336460! 9-390249. 9-394021 9:3. 7796 9:401569 9-405339, 9-409105; 9-4 12869) 9-416630 9:420337 9-491149 9:427834 9-431649 9-435338 9-439131 9-442370 9-446607 9-450341 9-454072 9457800 9-46 1525) 9:465947 9-463966, 9-472682! 9-476336! 9-480106 9-183314. 9487518 1 { 9-491220; 9-491919 9498612! 95093,8! 9:505998: 9:509085 9:513370! 9:517051: 9520730: 9:524406: 9-598079. 9-531750! 9:535417 9-539082. 9-519744! 658503000, 29-1957624' 9546403 660776311! 29:5127021, 9:5500591 No.| Square. 765334: 162129 763376 765625 767376 769129 110884 Si 712641 | 774400 146161 771924 779639 ‚ 781456 | 783225 184996 786769 798544 790391 192100 193881 795664 797449 799236 801025 у 802816 804609 806404 808201 810000 811801 813604 815409 817216 819025 820836 822649 824464 826231 828100 829921 831744 833569 835396 837225 839056 840889 849794 844561 846400 848241 850081 851929 852776 855625 857476! 859329 | 861184 86304 1 864900. 866761: 863624. BTA 872356; 14225. 2 876096; 714516934 Cute. 66354948 665338617 667627624 669921875 679221376 674596132!- 676836159 679151439 681472000: 683797341 636128968! 688465387. 690807101 693154125 695506456 697864103 100927072 709595369 794969000 707347971 709732988 719191957 716917375 719323136 721734273 124150792 726579692). 729000000 731432701 733870808 736314327 738763264 741217625 743677416 746149643 748613312 151089429 153571000 756058031 758550528 761048497 763551944 166060875 763515296 711095213 173620632 116151559 778683000 781229961 783777448 786330167 788839024 191453125 191022776 196597983 79917875: 801765039 804357000 806954491 809557208 8121669371 814730504! 817400375 8200253506 877969 892656<52 879844; 825293672 . 90-0499584 Sq. Reat- lc ubeR oot. — 99-2996461 29-5465734 29-5634910 99:503289 29-5972972 99-6141858 29-6310648! 99:647934, 29-6647939 29-6816442 99,6981848 2)-7153159 29-7321375 29-7489496! 29-7657521 29-7825452 29-7993239 :29:8161039 29°8323678 29-8496231 29-8663690 29:3831056 29-8998328 29°9 165506 2 5332591 23-9499583! 29-9666481 29-9833257 30-0000000 30-0166620 30-0333148 30-0665998 30:0832179 30°0998333 30° 1164407 30°1330383 30° 1496269 30°1662063 30°1827765 30° 1993377 30:2158899 302224399 30-2489669 30-9654919 30-2820079 30-2985 148 303150128 30:331501 30-3479818 39-3644529 30-3809151 30-3373633 30-4138127 3J-43)2431 39-4466747 39-4630924 30-4795013 39-4959014] 30-5 192996! 30:5286750 30:5 150487 30-5514136 39-5777697 30-5541171 306104551 30-6267857! 9:5537 19 9-557363, 9:561011 9-564656 9-568298 9-571938 9:575514 9:579208 9-582840 9-286168 9-590094 .9:593717 9:597337 9-600955 9-604570 9-602182 961791 9-61589% 9:619002 9-622603 9-626202 9-699797 9:633591 9:636981 9-640569 9-644154 9-647737 9-651317 9-654894 9-658468 9-662040 9-665610 9-669176 9-672740 9-676302 9-679860 9:68341 9:636970 9-690521 9-694069 9-697615 9-701158 9-701696 9-708237 9-71177 9.715303, 9-718835 9-72936 9-725888 9.750493 97532398 9-764497 9-767992 9771481 9-774974 9-779162 9781947 9785429 9:73590

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TABLE OF SQUARES, CUBES, AND ROOTS. 645 n_a Nao. | Square.! Cube. | Sq. Root. |CubeRoot.'| No. | Square. Cube. | Sq. Root. [CabeRoor] 881721 827936019 39-6431069! 9:792386 | 970} 940900} 919673000! 31:1448230 9-828983, 883600 830584000! 30:6594194! 9-795861| 971) 942341) 915498611] 31-1608729! 9-90233 885481 833237621) 30:6757233, 9°799334 | 972) 944784] 918330048! 31:1769145| 9:905789 837364. 835896833 30:6990185! 9:809304! 973| 946729| 921167317| 31-1929479| 9-909173 30:7083051| 9-806271| 971) 948676| 924010424! 31:9039731| 9:919571 891136 841232334) 30°7245830! 9°809736 | 975| 950625} 926859375) 31-2249900| 9-915962 893095. 843908625. 30°7408523) 9:813199 | 976| 952576, 929714176! 31:2409987| 9* 894916. 846590536 30-7571130; 9:816659 | 977| 954599, 932574833, 31-2569992 896809 849973123. 30-7733651! 9:820117 | 978) 956484 935441352) 31-2729915 898704: 851971392 30:7896086! 9°823572 | 979] 958141: 938313739 31:2882757| 9:929504 900601! 851670349. 30: 8058436, 9:827025 | 980 960400. 941192000) 31:3049517, 9:932884 902500! 857375000 30° 8220700] 9:830476 | 981) 962361; 944076141) 31-3209195 9-936261 904101! 860085351! 30: 8382879) 9833924 | 982! 964324) 946966168) 31:3368792| 9:939636 906301 869801408. 308544972: 9°837369 | 983) 966289| 919862087) 31-3528308 9-913009 908209 865523177. 30: 8706981 9:840813! 984 952763904) 31:2687743| 9-9163 910116| 868250664: 30: 8868904 9:844954 | 985) 970225, 955671625, 31:3817097, 9:949748 912025! 870983875. 30: :9030743. 9:817692 986 979196, 958285256 31:4006369, 9-953114 913936! 873729816. 309192497 9°851128 | 987; 974169! 961504803) 31:4165561| 9-956477 915849, 876467493, 30-9354166| 9:854562| 288; 976144| 964430972) 31-4324673! 9959839 917764 879217912. 30-95157 251! 9-857993 | 989| 978121) 967361669 31:4183704| 9:963198, 919681| 881974079 30 9577951] 9-861492 | 990| 980100, 970299000| 31:4642654| 9:966555 921500. 884736000' 30: 9338668; 9-864848 | 991; 982081) 973249271, 31:4801525| 9-969909 92352]| 887503681. 31-0000000; 9:368272 999, 981064! 976191488 31:4960315| 9-973262 925444| ` 800977198 31-0161948| 9:871694| 993, 986049| 979146657) 31:5119095| 9-97661 927369; 323056347. 310322413! 9875113 | 994| 988036| 982107784, 31-5277655; 997996 929206 825841314. 31-0483494; 99:878530 | 995) 990025, 985071875, 31-5436206| 9-983305 931295 898632125 31464401 9:881945 | 996! 992016) 988047936! 31:5594677, 9-986619 933156! 901428696 21-0305405! 9:885357| 997| 994009; 991026973| 31:5753068| 9-989990 935089| 904231063 31.1126981 9:883767! 993; 996004| 994011992, 31:5911330| 999332 937094! 907039232 31-1196984| 9:892175 | 999| 998001) 997002999 31:6069613| 9-996666. 938961| 909853209 31-1287648! 9:895580 | 1000|1000000/ 1000000000, 31-6227766|10 The following rules are for finding the squares, cubes, and roots of num- bers exceeding 1000. To find the square of any number divisible without a remainder, Rule.—Di- vide the given number by such a number from the foregoing table as will divide it without a remainder ; then the square of the quotient, multiplied by the square of the number found in the table, will give the answer. Example.—What is the square of 2000? 2000, divided by rooo, a number found in the table, gives a quotient of 2, the square of which is 4, and the square of 1000 is 1,000,000, therefore : 4 X 1,000,000 — 4,000,000: the Ans. Another Example.—What is the square of 1230? 1230, being divided by 123, the quotient will be то, the square of which is тоо, and the square of 123 is 15,129, therefore : Р IOO X 15,129 — 1,512,000: the Ans. To find the square of any number not divisible without a remainder. Rule.— Add together the squares of such two adjoining numbers from the table as shall together equal the given number, and multiply thie” sum by 2 ; then this product, less т, will be the answer. Example.—What is the square of 1487? The adjoining numbers, 743 and 744, added together, equal the given number, 1487, and the square of 743 = 552,049, the square of 744 = 553,536, and these added = 1,105,585, therefore : 1,105,585 x 2 = 2,211,170 — I = 2,211,169: the Ans. To find the cube of any number divisible without a remainder. Rule.—Divide the given number by such a number from the foregoing table as will divide

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646 APPENDIX. it without a remainder ; then the cube of the quotient, multiplied by the cube of the number found in the table, will give the answer. Example. —What is the cube of 2700? 2700, being divided by goo, the quo- tient is 3, the cube of which is 27 and the cube of 900 is 729,000,000, there- fore: 27 X 729,000,000 = 19,683,000,000 : the Ans. To find the square or cube root of numbers higher than is found in the table. Rule.—Select, in the column of squares or cubes,as the case may require, that number which is nearest the given number ; then the answer, when decimals are not of importance, will be found directly opposite, in the column of num- bers. Example.—What is the square root of 87,620? In the column of squares, 87,616 is nearest to the given number; therefore, 296, immediately oppositg in the column of numbers, is the answer, nearly. Another example. —What is the cube root of 110,591? In the column of cubes, 110,592 is found to be nearest to the given number ; therefore, 48, the number opposite, is the answer, nearly. To find the cube root more accurately, Rule.—Select from the column of cubes that number which is nearest the given number, and add twice the number so selected to the given number; also, add twice the given number to the number selected from the table. "Then, as the former product is to the latter, so is the root of the number selected to the root of the number given. - £Example.—What is the cube root of 9200? The nearest number in the col- umn of cubes is 9261, the root of which is 21, therefore: 9261 9200 2 2 18522 18400 9200 9261 As 27,722 is to 27,661, so is 21 to 20:953 +, the Ans. Thus, 27,661 x 21 = 580,881, and this divided by 27,722 = 20-953 +. To find the square or cube root of a whole number with decimais. Rule.—Sub- tract the root of the whole number from the root of the next higher number, and multiply the remainder by the given decimal; then the product, added to the root of the given whole number, will give the answer correctly to three places of decimals in the square root, and to seven in the cube root. Exampie.—What is the square root of 11-14? The square root of r1 is 3-3166, and the square root of the next higher number, 12, is 3-4641; the for- mer from the latter, the remainder is o- 1475, and this by 0-14 equals 0.02065. This added to 3-3166, the sum, 3-33725, is the square root of 11:14. To find the roots of decimals by the use of the table, Rule.—Seek for the given decimal in the column of numbers, and opposite in the columns of roots will be found the answer, correct as to the figures, but requiring the decimal point to be shifted. The transposition of the decimal point isto be performed thus: For every place the decimal point is removed in the root, remove it in the number zwo places for the square root and ¢hree places for the cube root.

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THE REDUCTION OF. DECIMALS. 647 Examples.—By the table, the square root of 86-0 is 9-2736, consequently by the rule the square root of 0-86 is 0.92736. Тһе square root of 9- is 3:, hence the square root df 0:09 is о-3. For the square root of 0.0657 we have 0*25632, found opposite No. 657. So, also, the square root of 0-000927 is 0-030446, fourid opposite No. 927. And the square root of 8-75 (whole num- ber with decimals) is 2-9546, found opposite No. 873. Тһе cube root of о.8 is 0-928, found at No. 800; the cube root of 0:08 is 0-4308, found opposite No. 80, and the cube root of 0-008 is 0-2, as 2:0 is the cube root of 8-0. So also the cube root of 0-047 is о· 36088, found opposite No. 47. RULES FOR THE REDUCTION OF DECIMALS. қ To reduce а fraction to tts equivalent decimal. Rule.—Divide the numerator by the denominator, annexing cyphers as required. Example.—Wlhat is the decimal of a foot equivalent to three inches ? 3 inches is ў; of a foot, therefore : 3... 12)3:00 :25 Ans. Another exanple.—What is the equivalent decimal of 4 of an inch? 4 . . . 8)7-000 -875 Ans. To reduce a compound fraction to its equivalent decimal, Rule.—In accordance with the preceding rule, reduce each fraction, commencing at the lowest, to the decimal of the next higher denomination, to which add the numerator of the next higher fraction, and reduce the sum to the decimal of the next higher denomination, and so proceed to the last; and the final product will be the answer. Example.—What is the decimal of a foot equivalent to five inches, § and ү of an inch ? The fractions in this case are, 4 of an eighth, # of an inch, and 45 of a foot, therefore : 3: eighths. inches. Е dui odds I2) 5-437500 -453125 Ans. The process may be condensed, thus : write the numerators of the given

Page 688

648 i APPENDIX. fractions, from the least to the greatest, under each other, and place each de. nominator to the left of its numerator, thus : i 42 шо 2| 1-0 $e |... 8| 3:5000 Я 12| 5-437500 , :453125 Ans. To reduce a decimal to its equivalent in terms of lower denominations. Rule. —Multiply the given decimal by the number of parts іп the next less denomi- nation, and point off from the product as many figures to the right hand as there are in the given decimal; then multiply the figures pointed off by the number of parts in the next lower denomination, and point off as before, and so proceed to the end ; then the several figures ppipted off to the left will be the answer. -Example.—What is the expression in inches of 0-390625 feet ? Feet 0: 390625 I2 inches in a foot. Inches 4-687500 8 eighths іп an inch. Eighths 5.5000 2 sixteenths in an eighth. Sixteenth 1 T Ans., 4 inches, $ and 4... e example-—What is the expression, in fractions of an inch of- 0:6875 inches? Inches 0*6875 4 8 eignths in ап inch. Eighths 5- 5000 2 sixteenths in an eighth. Sixteenth 1:0 Ans., ё and 75.

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TABLE OF CIRCEES. (From Gregory's Mathematics.) FRoM this table may be found by inspection the area or circumference of a circle of any diameter, and the side of a square equal to the area of any given circle from т to 100 inches, feet, yards, miles, etc. If the given diameter is iu inches, the area, circumference, etc., set opposite, will be inches; if in feet, then feet, etc. Diam. Area. Circum, equal sq. || Diam. Area Circum. equal sq. “25 04908 78539 22155 “75 90:76257 33:77212 9:52693 20 19635 127079 44811 | 11: 95:03317 31:55751 974819 “75! 44178 2°35619 66467 28 99-40195 35:34291 9-97005 1% “4 “78539 3:14159 “88622 5 103-86890 36:12331, 10:19160 95 1592718 3:92699 110778 “75 108:43403 36:91371] 1041316 5 176714 4771238 1:32934|| 19: 113-093733 37:69911] 10:63472 "75 2-40528 5:49773 1:55082 %5 117-85831 38:48451, 10:85627 2- 3°14159 6:23318 1-772145 “5 122:71846 39-96990, 11:07733 95 3:9760% 7:06358 1:99101 5 12767628 40: 05530; 11:29939 5 4:90873 1:85393 89:91556! 13: 13832-73298 ba 8 1070; 11:52095 “75 5:93957 8:63937 2:43712 35 137:88646 11-74250 3: 7:06853 9:42471 2°65368) 5 143:13381 49-41 150) 11:96406 425 8:29576 10:21017 2:83023 “75 143:43934 43:19689] 12:18562 $ 9-62112 10-99557 3:10179|| 14- 153-93804 43:93999] 19:40717 45 11:04466 11:78097 3:32335 25 159°48491 44-76769| 12:62373 4: 12:56637 12:56637 3:51490 5 165:12996 45:55309] 12-85029 %5 14:18625 13:35176 3776616! ЛӘ 170:87318 46-33349| 13:07184 55) 15:90431 14:13716; · 3-98802)| 15: 176771458 47:19338] 13:29340 “75 17712054 14:92256 4:20957! 95 182:65416 47:90928) 13:51196 Жү 19:63195 15:70796 4:43113j Э 188:69190 48:69463 13 73651 95 21:6 4753) 16:49336 4:65269 “79 191:82783 49-4803; 13:95307 S 2315829 17:27875 487424) 16: 201:06192 50:26518 | 14:17963 “75 25.96722, 1806415 5-095580 "09 902:39420 51- 05088) 14:10118 6: 28:27433 18:84955 531736 5 913:89464| ' 51:83627. 14:62274 25) 30:67961 19:63495 553891 -75 220:35327 52-62167 14:84130 "5 33:18307 20-42035 5760477 l; 17: 926:98006 53:40707] 15:06535 "75 35:78170 21:20575 5-93203j' “25 933:70204- 54:19247 15-23741 T 33:48156 21:99114 6: 99358! `5 240:52818 54:97737) 15:50397 %5 41:28919 9777654 6:42514|| -75 947:44950 a34 pease, 15:73059 55 41:17861 23:56194 661670: 13: 954:46900 56:54866! 15:95203 “25! 47:17297 94:34734 6: 86325} "25 261:58667 7383106) 16-17364 8: 50:96518 95:13274 7208981! "5 2035-80258, 53:11916) 16:39519 25 53:45616 25:91813 731137 "75: 97611651, 58:90486 16:61675 5 | 5574501 26:70353 7532921 19 | 282-52873) 59:69096] 16:83331 "75 60:13201 27-48893 115418 25 991:03910 60-47565| 17:05936 9: 63:61725 28:27133 797694] -5 | 298061765 61:26105] 1728142 “95 67:20063 29:0597 8:19759 | 75 306:35437| 62-04645| 17:50298 D 10:83218 29:81513 8:41915|, 20- | 314:15926 62:83185] 17:72153 “75 74:66191 30:63052 8:61071 25) 322-06233| 63:61795] 17:91609 10: 78:53981 31:41592 8:86226 5 330-06257, 64-40264] 18:16765 %5 82:51539 3220132! 9-085382 75 338-16299, 65:18301 18:38920 3 | 86:59014 32:986721 9-30533]: 21- | 346-36059 65:97341| 18:61076 Side of side of

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pane Сл 26: сл сл AN ILI i 5. E TS сл е 12 t5 NUM =... arched * Saori © rnd MT ек; © сл 3l: On ee - QC b 2) C O] I Uo 0) л л сл C CQ сл CA Area. 354:65635 363-05030 371554941 380-13271 388:89117 397:60782, 40649963 415:47559 494:55679 432-3613! 443:01365| 452-38924| 461:86320 471:43524 481:10546 490:87335| 500:74041! 510:20515! 520776306. 530-92915! 54118342 551:54586 562-00147 572:55596 583-20722 5393-95736 60480567 61575216 62679682 637:93965' 649-18066 66051985 671-95721 683-49275 695-12646 706:85831 71868840 "30:61664 74264305 75176763 7566-99039 77931132 79173043 804:21771 816:86317 899-57681 8412-38861 85529859 868:30675 88141308 894-61759 907:99097| 991:32113 931:89016 94841736 962:11975 975:90630 989:79803 1003-78794 1017:87601 1032-06227 1046:34670 1060-72930 1075:21008 1089-78903 1104-46616 1119-2414") Circum. 66:75884 67-54494 68:32964 69:11503. 69-90043! 70-68583. 71-47193 9.95663 73:04909, 73:89749, 74-61989! 75-39899 %6.18362! %6.96909! 77:754411 58.52981 79-32521 80-11061 80-89601 81-68140| 82-46680) 83:95990)! 84.03760! 84:82300 85-60839. 86-39379 87-17919 87:96459! 8874999 89-53539: 90-32078 | 9.10613, 9 39158 92-67698 93-46935 94:94777 | 95-03317 95-81857 96-60397 97:38937. 98-17477 98:96016. 99:74556! 100:53096 101:31636. 102-10176: 102-887 15 103:67255 104:45795 105-24335 106-02875 10681415 107:59954 108-33494 109-17034! 10995904 11074114, | 111:52653| 112:31193| 113: pee 113-8827 Lene 115:45353 116:23899 11709439 117:80972 118:59572 APPENDIX. Side of equal sq. || Diam. mmeo — 18: 83239 38° 19:05337. 19:97543. 19-49699. 19:71854. 19-94010. 20-16166. 90:38321. 90:60477. 90:82633 91:04783. 2196941 91:49100 2171255 91:93411. 99-15567. 99:37729. 99:59878 92:82034. 93-04190: 93:96345. 93-48501: 93:70657! 93:99819. 94:14968 94:37194. 94:59979. 94:81135. 95-03591, 95-95746! 95-47509. 95-70058, 95:99213. 96-14359. 96:36595! 96:58680. 26-80836. 97-05999. OF: 25147 97:47303 07-69459 97:91614: 98:13770: 98:55996. 29:58081 98-50937: 2902393 99:94548. 91-46704! 99:68860! 99:91015 30:13171. 30:35327 30:574821 30-79638. 31-01794 | 31-93949. 31:46105 31:68261, 31:90416. 32-12572 39:34798) 35:56383 32-7903. 33-01195l| 54 33°23350 33:455061. са н> кі C `2? D © 5 398 ACK FAR IF pou pou Qu сл 49: 22 та ңы Еа = Siw © ILE OOE OT сл e ңы = S . oY ЖЕ со юзе сл ор СЫС ЦИШЕ. ОЭК 8А ЫЕ * renew ©л Q — сл Aw ol ед ъл фо Area. 1134:11494| 1149:08660| 1164:15642 1179:32442 1194:59060 1209:95495 1225-41748 1210-97818 1255°63704 1272:39411 1288:24933 ; 1304-29273 1390-9543]! 1336-40406, 1352-65198) | 1368-99808 1385-44236, 1401-98480 1418-62543 1435-36423 1452-20120 1469-13635 1486-16967 1503-30117 1520-53084 1537-85869 1556-93471 1572-80890 1590-43128 1608-15189 1625-97054 1643 88744 1661-90251 165901575; 1698-22717 1716-53677 1734-94454 1753°45048 1772-05460 1790-7569 1809 55736 1828-45601 1847-45989 1866-54782! 1885:71099 1905:83933 1924-42184 1943-90954 1963-49540 1983-17944 2002-96 166 2022-81905 20492-82062 20629-89736 2083-07227 2103-31536 2123-71663 9144-18607 9164-75368 9185-41947 9206-18344 9927-04557 9948-00589 9269-06133 2290-22 104 9311-4758 2332-82889) Circum. 119:38052 120-16591 120-95131 121773671 12252211 193:39751 124-09290 194-87830 125-66370 196-44910 1273450 198-01990! 13:80599) 12959069, 130-37609)| 13116149) 131: 94689! 13273228 8 13351768! 131-30: 308) { 13508348) 1335-87333 136- 65928 137- 44467 133-93007| 129:01517| 139-80087 140-58627 14127166 149:15706 149:91946! 143°72786 14451326 145:99866 146:08405, 14686912! 147:65485 148-44095| 149-22565 150-01104 150-79644 15153181 152°36724 153:15264 153:93804 15472313 155:50883! 156-29493 15707963: 157-96503| 153-650412 159-435932 16022122 16100662 16179902 16257741 163-36981 164-1482] 164-93361 165-71901 166-50441 167-28980 168: 07520 168-86060| 169-64600 170-43140 171-21679 Side of | equal 59. 33-67662 33-89317 34:11973 34:34199 34:56985 34-78440 35-00596 35-99759 3544907 37:99153 37-54398 37 -66464 49:31733 42-53889 49:76044 49:98200 43:90356 43:49511 43:64667 43:86823 44:08978 44:31134 44:53290 44-75445 44:97601 45-19757 45:41919 45:64068 45:89294 46:02230 46:30535 46 59691 46:74847 46:97009 47:19158 47-41314 47:63469 47:85625 48:07781 48:29936

Page 691

Dizm. 5475 53: :95 "5 $5 56° P 5 “75 5T: Ue LL. ОЕ сл сл =. а ЧТИ REESE "оэ сл Тозо o "uc Cus 9015 ол с ВР: D o t ow А Ibn 7-3 LANA Cn л усхо ae tno me 64: ME: с 9” enw C o V 69- Arent. 2354-28005) 237582244! 9397-47698! 2419:92269' 2141:06657| 2463-00864 | 9185-04887! 2507-18728 2520-42337: 9551:75363; 2574:191561 2596-72267 | 2619-35196) 2642:07942; 2664 90505: 2637-82836, 2710-85084: 2733-97100) 2757-18933, 2780-50534 9803 92053, 2327-43338) 2851-04442 2874-75362: 2893-56100, 2922-46656, 2946-47029, 2570-57220 2394-77223 3019-0705 1 3.43 4£697 32367-66157 30929-55135) 3117-2153] 3142:03144 3166-92174 3191-90722, 3216-99087 3242-17270 3267-4527 3292 83088 3318-30724 3313-88176 3369-55447 3395-32534 3421-19439 2417-16162 31,3 29709] 3199 3J060 3523-65235 3552-01228 3373-47033 3605 02655 3631-68110 2658-43373 $685:28153 3712 23350 3739-23065 5766-47597 3793-66947 3521-01118 3348-45100) 3375°98902 3.03-69522 3931-35959 3959-19214 3987-12286} TABLE OF “CIRCLES: Circum. 172:00219 17278159 173:57299 174:3533J 175:14379 175:92918 116771458 1:7:49998 178:28528 179:07078 179:85617 180:64157 181:42657 182 21237 182:99777 183:78317 - 181:56856 182:353.6 186:13936 186:92476 18771016 :183:49555! 189:28095, 192-06635 192:85175 191:63715 192:42255 193:20794 193:99334 19177874 19556414 195:34954 197:13493 19799033 19876573, 199-49113; 200:27653. 201:06 142 901:81739! 209 63272 20341812 20420352 29498392 922577131 206:55571 907:31511 203:13051 208-9159] 209-70130; 210 48570 211:97210 212:05750 2]2-84290 213:62430. 214:41369; 215-19309 215:98149| 916 76939. 217:55529. 213 34068 212519508, 219-91 148 92069683! 42143928 9924%5/64! 992053071 223 535471 . 50-05304 Side of equal sq. ees — MÀ ee ы шы 48:59099, 48-71918. 48:96403. 49-18559 49:40715. 49:62870: 49:85026 50:07182. 50:99337 50-51493 50:73619. 51-17960. 51-40116 51:62271, 51 1 52406583, 59:98738 52:50894 59-73050 53:61672. 53:83328! 21:05984. 512813) 5450235 54:724511 54:94606. 55-16762, 55:33918 55:61073 55:83229. 55-05335 56:97510. 58-49626, 56:71852. 56:94007 57-16163 57-33319 2760475. 57-82630) 53-04786 5326942 53:490971 53:71953 58:93109 53:15564, 59:37720 59:59376. 59:89231. 60-43198 60-70624 60-92310 61:14965 61:37121 61:59277 61:81432 62:03533 62:25744 62:47829 62:70055 62:99211 63:14366 Diam. | ТД; “5 83: V Ағ б о оо e . . LI a "сл ко an сл сл 305 ©л о со тж. NS bou OU OG oO С 88: | 4015-15178. 4043 27833 4071:50407 4099-82750 4198-91909 415676886 4185-33681 4214-10293 4242-91722 4271-82969 4300-84034 4329-94916 4359-15615. 4388-46132! 4417-86466, 4447-36618 4476-96538, 4506-66374 4536-45979, 4566-35400 4526 34640 4626-43586 4656:62571 4656-91962. 4117-99771| 4747-78093, 4778-36242. 4809-01904 4839:81983. 4370-79579 4901-66992 4932-74225 4963-91274’ 4995-18140) 5026-51824! 5053-01325 5039-57644! 512123781 5152-99735 5184-85506 5216-81095 5218-86501 5281-01725 5313 96766. 5345-61624 5378-06301 5410-60794, 5443-25105 54735-69231 5508-83180, 5541-76944, 5574-80325 5607-92923 : 9641-17139 5674-50173. 107-93023. 59711-45632, . 5775-08178 5808-80481 5812-62602 5876-54540 5210:56296 5944 67469; 5973-89260 6013-20468 6017-61494 6082: 12337, 224:62337 225:40027 226-19467 92698006 9227-76546 9228-55086 229-32626 230-12166 230-9076 231:69245 939-47785 23326325 23104865 23483105 935:61944 236°40484 237° 19624 231-97564 рери 2338-76104) 939-54613 240-33183 941411793 241-90263 242-68803 943-47343 244-25382 245 01492 24582962) 9416-61592, 2417-40042; 248:13531 218-97 121, 219-75661 250-34201 25]-32741 959-11981 252-8J820 253-683360 25446900 955-5440 256-0380 25682579 2537-61059 958:32599 959:18139 959-96679 960-75219 961:53753 262-32298 263-10333 263-89378 964:67918 265-46457 966-24997 967:03537 267-8207 268-60617 269-39157 27017696 27096236 27174776 979-53316 273:31856 274:10335 74:88935 975:67475 276:46015| 67:57180 70-67659 70:89815 71:11971 72:00593 72-99749 72-44905 73:99994 74:02150 74:44306 74:66461 19:99395 7621551 76:43707

Page 692

611672393 6151:43476 6186-23772 0221-13585 6256-13315 6291-23563 6326°43129 6361-72512 6397-11712 6432-60730 6463:19566 6503-83219 6533-66639 6575-54977 6611:53282 6647:61005 6683:73745 6720:06303 675643678 6792-90871 6829:47831 6866:14709 6902:91354 6939-77317] Circum., APPENDIX. Side of | equal sq. e — M — аа ee 78:90952!| 94-25 211:24555 278:03094 27881684 279:69174 23033714 23117251 28195794 28274333! 23352873.. 28431413 285:09953 235:83493 286:67032 23145572 238:21112 239-02652 289°31192 ‚ 2390-59732 29133271 99216811 2929535] 993-7339: 294:52431 995:30970| 78°43103 18:65263 Diam. 5 "uS 7887419] 95: 79:09575 79:31730 79-53386 %5 S US 7976042 | 96: 79:98198 80:20353 80°425091} - “25! 5 оф) 80°64669)| 97- 80-8530 81-08976,| 81-31132 81:53237 81:75143] 81-9759} 8219754)! 82-04066 82:86221 83:03377 ” 525 7015:80194 705096109 7083-21842, 712557992 7163-02759 7200:57944 7293:22947 7245:97767 7313:82404 7351: 76859 17389-81131 742795221 746619129 1504:52853 | 7542-96396 % 82:41910|| 99: 529 5 "75 85:80533||100: 7581-49755. 762019933 7653-8509 7 7697-68739 7736-61369 7775-63816 7814-76031! 7353-98163| Circum. 226:09510 296-88050 297-66590 298°45 130 299:23670 320-02209 300-80749 301:59239 302 727829) À 30316369 3053-94908 3041-73448 30551988, 306:30525, 307-09068 307-87608 308:66147 309-446371.. 31093292 91101762 311-80307 31953346 31337336 314-15926| ide of equa} sq. 83:52688 83-74844 83:97909 The following rules are for extending the use of the above table. To find the area, circumference, or side of equal square, of а circle having a diameter of more than 100 inches, feet, etc. Rule.— Divide the given diameter by a number that will give a quotient equal to some one of the diameters in the table; then the circumference or side of equal square, opposite that diameter, multiplied by that divisor, or the area opposite that diameter, multiplied by the square of the aforesaid divisor, will give the answer. Example.—What is the circumference of a circle whose diameter is 228 feet? 228, divided by 3, gives 76, a diameter of the table, the circumference of which is 238-761, therefore : 238-761 5 3 716-283 feet. Ans. Another example.—What is the area of a circle having a diameter of 150 inches? 150, divided by ro, gives 15, one of the diameters in the table, the area of which is 176- 71458, therefore: 176: 71458 100 — IO X IO Ans. 17,671-:458co inches. To find the area, circumference, or side of equal square, of a circle having an intey mediate. diameter to those in the table. Rule.—Multiply the given diameter by a number that will give a product equal to some one of the diameters in the table ; then the circumference or side of equal square opposite that diame- ter, divided by that inultiplier, or the area opposite that diameter divided by the square of the aforesaid multiplier, will give the answer.

Page 693

CAPACITY OF WELLS, CISTERNS, ETC. 653 Example. — What is the circumference of a circle whose diameter is 64, or 6-125 inches? 6-125, multiplied by 2, gives 12-25, one of the diameters of the table, whose circumference is 38-484, therefore : 2)38 -484 19:242 inches. Ans. Another example.—What is the area of a circle, the diameter of which is 3-2 feet? 3:2, multiplied by 5, gives 16, and the area of 16 is 201-0619, therefore : `5 X 5 --25)201-0610(8-0424 + feet. Ans. 200 19 Note.—The. diameter of a circle, multiplied by 3:14159, will give its cir- cumference ; the square of the diameter. multiplied by -78539, will give its area; and the diameter, multiplied by -88622, will give the side of a square equal to the area of the circle. TABLE SHOWING THE CAPACITY OF WELLS, CISTERNS, ETC; The gallon of the State of New York, by an act passed April 11, 1851, is required to conform 10 the standard gallon of the United States government. This standard gallon contains 231 cubic inches. “Іп conformity with this standard the following table has been computed. ГА One foot in depth of а cistern of 5 feet diameter жін -соназсе........22-........-“ 52-872 gallons. 34 E = СЕТЕ. MEE. C eee 71:965 " 4 i шн ZI 93:995 и 44 а “«ҡ 2,04. 2... EM 118-963 E 5 "t 7 ры WIS 146-868 és 54 ài EE . о ЖООДО 727 177:710 ы 6 i st 2 RM OE o 211-490 < 64 “ = cM. Е... 218-207 ш 7 E “- ОЕ... 287-861 А 8 Ju EHE LL 5 M — .375:982 B 9 м ия Же... 475:852 à IO т ч Me on RIB. 587.472 e I2 i е c TESTED ХЕ 815:959 “ Note.—To reduce cubic feet to gallons, multiply by 7:48. The weight of a gallon of water is 8:355 lbs. To find the contents of a round cistern, multi- ply the square of the diameter by the height, both in feet, and this product by 5-875.

Page 694

654 APPENDIX. TABLE OF WEIGHTS. MATERIALS USED, IN THE CONSTRUCTION OR LOADING OF BUILDINGS. WEIGHTS PER Cusic Foor. As per Barlow, Gallier, Haswell, Hurst, Rankine, Tredgold, Wood and the Author. ti ы 5 o 2 M 5 o 2 MATERIAL. E! E E MATERIAL. 51/8 м fa 5 f = 4 | 4 WOODS. Mahogany, St. Domingo.. 43 | 65) 55 Maple Wc oa ЖК a E 33 | 49 41 Асаса с... a : 4r | 51 | 46 | Mulberry....... 4........ 3 55 45 Alder 4.........- ШОО: 35 | st | 38 Oak, Adria c. 5 ососооаас Ж ни 62 B. ао PERITUS 49 51 50 Black IBog....2 0.6 бо 66 63 — a И ЖАТ MI 52 | 49 ШЕ (бапаспал sette б b 54 E. REA TECTITI 39 | 53 | 46 АГС sissies c erare y: И 47 _ ee е 35 | 49 | 42 P SNB. ТЛ... 38; 7o| 54 ЕЕ... a 59 | 65 | 62 MUS e 57| 79 | 68 “ . French...... ЖЕК D. eus 83 M Pedes 22020 47 | 54 51 Brazil-wood . ..............| .. |. | 64 БАТ ыа садыг 43| 57| 50 Седаг.. И 2735 |І. зе à 58 B Canadian .5.........." 47 | 57 | «ЕЛ жане .........°....-.--.. Ea 7 44 6 езе т... зо | 38 | 34 |“Реатг-ттее. ...... ....... 49| 44| 42 * ^ Virginia Red E с ..| 40 Pine, Georgia (pitch) cane 38; 58 | 48 (ncmo AES E. 32 | 46 | 39 Mar Forest .......... es 43 Chestnut, Нөгзе............ 29 | 41 | 85 * Memel and Riga...... 29 | 35 | of Swe M onera. ..27| 55 | 41 SY "Redoute M 37 КОШО .................2 15 W OScoteh oc 22222... 2. 27| st| 39 Cypress HS coant ass os Sane 27 | 41 | 34 * WARE c 21| 55] WS E umsh.......... 40 5 Уе бу... ЖЕ! 27) 3 | So Deal, Ghiristiania........... ЖА ||| от cons «sce б фы з 41| 49 | 45 Euch... ......... NOS | Poplar. esea oog ee б 23| 37| 30 * (Norway Spruce) er | 33 Ж ОЕ... оТ 5 1% 44 ЖЕ! ................. АЛ, || Кеамоой......... - 2 . | "23 Ebony. : -J бо | 255400. Коземооа.......... EE vs Ec 45 ISIS? аоооооссо НААНА бе os | AS | сабаға 22... TECUM. Ж s 30 БЕ ............. 33 Iso | 46 | Satnwo6d..... ere 55| 59| 57 Fir (Norway Spruce)... ... er mess | 7 D раисе 24 | 36 | 80 ӘЗЕР Біпе)............. 39 | 44 | 37 Sycamore... 2222022... 36| 40| 88 O IR Ie оссо T Е ve vadis Е a A 41| бї 51 Gum, Blue Бб || Шіш-мес.................. |.| ЖО Do- ^ Wale? о НА ILE А01 ИИТ л 77| 83| 80 Up i) eesse sesso > Se || Walnut, Black ........... - 261 до| 33 ПОШ ОСОКЕ 22 .2.......... 21 | з | 26 "cl BIOS. con... 8 40| 58| 49 Ыс c eres 49 | 58 | 49 | Whitewood........... eee 25 | 29 97 Тен ы oc ES... 41 | 63 Я % Wi... Se ET TITIUS 5 m 50 areh з... coco TEE 31 5 JR OQ TNAM 31 м | 43 METALS. ER ше... аган. ТЕ .. | 23 || Bismuth, Cast......... А se |... ШІМ Lignum-vite ....,.......... 41 | 83 | 62 || Brass, ar o uH 487 | 525 | 506 I|. MIELE Л 41 | st | 46 " — (Gun- -metal) Nc. ж .. 1544 Gl, rn ans "EE 1| Va Бімешее касы ms. d 528 | 534 | 531 Mahogany, Honduras ......| 35 | 40 | 88 Bronze Т КК әх»... 508 | 524| 516

Page 695

WEIGHT OF MATERIALS. 655 TABLE OF WEIGHTS.—(Continued.) MATERIALS USED IN THE CONSTRUCTION OR LOADING OF BUILDINGS. WEIGHTS PER Совіс Foor. As per Barlow, Gallier, Haswell, Hurst, Rankine, Tredgold, Wood and the Author. ы ы = о ы MATERIAL. z А Б MATERIAL, g 2 2 8 hi 5 | Copper, Cast Ти 537 | 549 | 543 Brick-work КТП. 112 |104 ЕР Hammered......... nae . | 556] О. ee ... 1100 s РАТОР reves [rr ... | 544 s in 1 Сайы: анн. ШШ ... 111% Gald ee ae a a a EL ... 11206 “ in Mortar........ 120 |110 E d Standandeq sses.: f=- .. 11108 Caen Stone ........ ...... |... 1.. ТЕО Günsmetal....... mod |... | SOD Cement, Portland... .. ТЕГ... 81 Шоп, Шат.......... INE S. 475 | 487 | 481 oman, Cast......| ... 100 ИАС Е... 434 | 474 | 454 95 X к АМАНӨВМ!е............[ -- ...] 475 . 118 * Wrought ............- 474 | 486 | 486] CHANTE 9 174 |145 BU NOS eme remm] ..| FOO) ОКШО Т S B і25 |199 " English Cast ........| --- 717, " with Gravel 160 E Nied ecl 713 || Coal, Anthracite. 102 | 96 Mercury at 329................ $51 " Bituminous.,. 83 E DNUS a БЕС 849! “ Cannel....... FM 49 se Be oy UU O УИ og || aos 837 * Cumberland.. 85 Nickel, Cast. ............ „| -- . | QS Bil Соке ............. 62| 54 о ТООРОТ ос .. | 458! Concrete, Cement 135 |130 Platina, сы. ал... 974 һсбанвоь................-..| 7 ... |106 Рше ut NE 1345|| Earth, Common............ 95 | 125 |110 “ * "Rolled........] --- 13479 E OUO S ec]. Еб Plumbago ..................| 177 142 “ with Gravel. ......] ... ... 1126 Silver, Parisian Standard.. . Ее Жиеу................ 71. .. |250 * Pura Case... ww ... 655 Feldspar ..2..- e oe core : ... |160 E " * Hammered] ... 658! Flagging, Si 2.1185 ч“ Standard. ......1..1... ... | Gall Fli ... |163 ый... o 486 | 433 | 489 165 |160 СИБ ccm 456 | 4.4 | 462 а 195 |183 “ис, Саше. ........ uw" ww. 429 | 449 | 439 ... |165 173 |163 18: |174 STONES, EARTHS, Erc. 172 |165 ... |164 Ае. 1... 2...0. 165 | 180 | 173 Egyptian Red... ...|. ... |166 Asphalt, Gritted............| e+ ow | 156 “© Guernsey ...........| oo. .. |185 Pac 1): 7 ............ 57 | 103 80 Quincy..... ... 1166 Barytes, Sulphate of. .... .-.| 250 | 304 | 977 Gravel но Восс 120 |105 E НА ВААС. 155 | 187 | 171 шее NER Lac ... 1184 В@їйыё!оле................. 122 | 156 | 139!) Gypsum........... 145 |140 Béton Coignet.............. 124 | 134 | 129 Lie, айына: soem |. Ow Blue Stone, Common.......] ... ... | 160) Limestone......... T 199 |169 ЈС РЬ ТТТ, 85 | 119} 102 Ы» Aubigné... ... |146 . ИШЕК... с... seis - | 138 9 Limerick.. s. |162 ENR (MP ai hardi.. J ... 107 | Marble.............. 178 |120 М5 Salmon.. comcs ees |] RES * Brocatel....... ... |166 t Philadelphia Front... | е. |... | ТАВ “т (Сатгава............. . |170

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656 APPENDIX. ' TABLE OF WEIGHTS.—(Continued.) MATERIALS USED IN THE CONSTRUCTION OR BUILDINGS. WEIGHTS PER Cunic Foot. As per Barlow, Gallier, Haswell, Hurst, Rankine, and the Author. LOADING OF MATERIAL. Marble, Eastchester........ 4 Egypti...» csse -.. renc hini. oosa? CCS di Таас Мен ЖЕНЕ uscocoooooos ASONTY ...... 2 222222. Міса. aa 2 Bislistone........ c esses Отаг... eb E E dry ....5. ИЕ... © — Hair, incl. Lath and Pails, per foot sup. Наш атт л. s бошт Н соосу әсе * Sand 3 and Lime paste 2 4% “ $ * 2 well beat together. Д... ас Мә -................... Petrified МУ @ОЧ............. сос оосо КОСЕ С ЕЯ iplaster, Савб................!|. T Porphyry, ES ЕНГ |ы Portland Stone.......... as Pumice-stone POZOIA. RII Quartz, Crystallized........] ... Rotten-stone Sand, (COATS dE SSSR See ОЕ Common BI Оны... ...... ood coc Amherst, O.. ...]. en Belleville, N. J... ss Berea, О........|... чч Dorchester, N. S. ke X Little Falls, IMG J- bie Marietta, О. 1... “ Middletown, Сг. | ... E | 9 [| | 51? © E MATERIAL, Б E > » Qt 2009 OS QOO Ол >» 2 OTe OOS nS abd Qoi Que oe OOOD к= Со @ Cr OE mt 9 Фу ол Сл В оло іші кі ome m pa 00 @ >» ка 2 = OD CD кы 9 фа CD м ed mi aF 00 650209525200

Page 697

INDEX. Abscissas of Axes, Ellipse....... 484 Abutments, Bridges, Strength of. 227 Abutments, Houses, Strength of.. 53 Acute Angle Defined........ 349, 544 Acute-angled Triangle Defined... 545 Асше or Lancet Arch........... 5I Algebra, Addition. . sc... ее 398 Algebra, Application of...... ase 393 Algebra, Binomial, Multiplica- fonom ЗИ” ‚ 409 Algebra, Binomial, Square of a... 429 Algebra, Binomial, Squaring а... 410 Algebra рәйпбей..........“..... 392 Algebra, Denominator, Least Common. "AI .. ............. 404 419 419 Algebra, Division, the Quotient.. Algebra, Division, Reduction... . Algebra, Division, Reverse of Multiplication. ......... „88 418 Algebra, Factors, Multiplication of Two and Three.. ^s, eO Algebra, Factors, Multiplication of Three: . 22... ж... > oun. 408 Algebra, Factors, Squaring Differ- ence ото. а жетем 412 Algebra, Fractions Added and Subtract -......... 22. 403 Algebra, Fractions, Denominators 407 Algebra, Fractions Subtracted... 405 Algebra, Hypothenuse, Equality об Squaresyen...............4 Algebra, Letters, Customary Uses - 396 Algebra, Logarithms Explained.. 425 Algebra, Logarithms, Examplesin 426 Algebra, Multiplication, Graphical 408 PAGE Algebra, Progression, Arithmeti- 432 Algebra, Progression, Geometrical 435 Algebra, Proportion Essential.... 347 Algebra, Proportionals, Lever JC’ on о... 2. Algebra, Quantities, Addition and Sete tc i 2. e е”. 424 Algebra, Quantities, Division of. 424 Algebra, Quantities, Multiplica- tion of Algebra, Quantities with Манче Exponents...... Algebra, Quantity, Raising to any НО OCT —— c >: 423 Algebra, Radicals, Extraction of.. 425 -.-Фжегеее,.,ееееееееееое-еее Aigebra, Rules аге General...... 394 Algebra, Rules, Useful Construc- Т ee. 391 Algebra, Sign5....... eor ES 397 Algebra, Signs, Arithmetical Pro- ullo MR 396 Algebra, "stans; Changed when Эшта... 25...” 400 Algebra, Signs, Multiplication of Plus and ‘MO Sin... ........5. 415 Algebra, Squares on Right-Angled Д Ө. sss алайын AE Algebra, Subtraction............ 398 Algebra, Sum and Difference, Pro- ЕСІ of.. <<» tI A ка: ӘЛЕ. 413 Algebra, Symbols Chosen at Pleas- ШШЩ... -„.. с .. 398 Algebra, Symbol, Transferring a.. 399 Algebra, Triangle, Squares on апеей. .... sis... ае. M

Page 698

658 INDEX. PAGE PAGR Alhambra, or Red House, Ancient Anh) О еб И ые. eese TOR 51 Palacevof thg........... ТЕ lr AGH, Каїйрайї.................. 51 Ancient Cities, Historical Ас- Arch, Span ofan....... sex ree c BA counts Of ......... oS 6 "Auch, Spring Тпвйтп»............. 52 Ancient Monuments, their Archi- Arch, Stone Bridges............. 230 tects... ... е. 0e T 172m 6 | Arch-stones, Bridges, Jointing... 233 Angle at Circumference of Circle. 358 | Arch, Strength of. ............... 50 Angle Defined... s.. -SEEN 544| Arch of Titus, Composite Order.. 28 Angle to Bracket of Cornice, To Euch, ОЧКО и -EE 50 ODAN e 5 cas... a ы 343 | Arch, Voussoir іп............... 52 Angle, To Measure a, Geometry.. 348 | Architect and Builder, Construc- Angle.rib to Polygonal Dome.... 223| tion Necessary to............. 56 Angle-rib, ERS of HB вола Architect, Derivation of the Word 5 Domest ot 3 AE... 223 Architects of Italy, 14th Century. 12 Amulet or Fillet, Classic Mould- Architecture, Classic Mouldings D AMET o MEM lo 323 NICO. LV Loo EM с 323 Ante Capt Modern Moulding.... 334 | Architecture, Ecclesiastical, Origin — — Antique Columns, Forms of..... WS 0f.....—. .. Ae TR Ee oo I4 Antiquity of Building...... ees. 5] Architecture, Egyptian, Character Arabian and Moorish Styles, An- of. 2599... Ж. EL 33 tigumMes of ИЕ... тї | Architecture, Egyptian, Features Areostyle, Intercolumniations... 20 ШІ... Б ОЛИ 30 Arc of Circle Defined........... 547 | Architecture, English, Ccttage Мас of Circle, Leñgth, Ruletfor.. 499 Буе; se sw о.о... 35 Arc, Radius of, To Find......... 561 | Architecture, English, Early..... II Arc, Versed Sine, To Find (Geom- Architecture, Grecian and Roman 8 Ее... E... 561 | Architecture, Grecian, History of. 6 (00 COGG ижи... Л... Т... 52 | Architecture, Hindoo, Character йе of Arches, Résistance in.. 752) оҺ.....%................... 30 Arcade in Bridges, Strength of Architecture, Order, Three Princi- ЕС... . 2) Pal ponte Ж......... Ж...” 14 ТЕЛЕ... Е... қо | Architecture, Principles of....... 44 . Arch, Acute or Lancet. ........... 51 | Architecture, Roman, Ruins of... II Amen cAwehivolt іп. Бој Ruler. a OE ЛИС. пе. 88 © Arch, Formation in Bridges...... 51 | Arithmetical Progression (Alge- Arch, Hooke’s Theory of an...... BO | lore). eerie eB cae: 432 Arch, Horseshoe or Moorish..... 51 | Astragal, or Bead; Classic Mould- Arch, Imposti. 2.2272. Л... 52 eee 4923 Arch, Inffadgs of. . o2. o9. ...... 52 | Athens, Parthenon, Columns of.. 48 Arch, Keystone, Position of...... 50| Айіс а Small Order, Top of Arch, Lateral Thrust in...... E. с ж? Building@. -c.e zm. «aoe 5

Page 699

Attic Story, Upper Story.. Axes of Ellipse (Ово |. Axiom Defined (Geometry)....... Axis Ieüned.....- T: ео Balusters, ey Winding Baluster in Round Rail, Winding SU M ss ONES Base, Shaft, and Capital Defined. Bathing, Necessary Arrangements Baths of Diocletian, Splendor of.. Bead, or Astragal, Classic Mould- ings.. 5. : Beams, — be NI fae ЕТОМ еы... uoce eese rre re Beams, Breaking Weight on...... Beams, Framed, Rules for Thick- Beams, Headers Defined......... Beams, Horizontal Thrust, Rules nsus Splicing, Beams, Tail, Defined Beams, Weight on, Proportion of. Beams, White Pine, Table о! Wers... sees eae nue vos Beams, Wooden, Use of Limited. Bearings for Girders...... ...... Binomials, Multiplication of (А1- жыла). s es i Binomials, Square of (Algebra)... Binomials, Squaring (Algebra)... Bisect an Angle (Geometry). ..... Bisect a Line (Geometry)......... Blocking out Rail, Winding Stairs Blondel's Method, Rise and Tread in Stairs INDEX. PAGE I5 250 130 74 235 130 130 130 177 I54 I4I 409 429 410 554 549 301 Bottom Rail for Doors, Rule for Width Bow, Mr. On Economics and Construction Bowstring Girder, Should not be Used Cast - Iron, To Find Braces in Roof, Rule for, Same as Baftet АН. ЖЕ Breaking Weight Defined Brick or Stone Buildings Brick Walls, Modern Bridge Abutments, Strength of... Bridge Arches, Formation of..... Bridge Arch-stones, Joints of.... Bridges, Construction of Various. Bridge, London, Age of Piles Rafters, etc., Bridge, Rib-built Bridge, Rib, Construction of... . Bridge, Rib, Framed,Construction and Distance Bridge, Rib, Radials of.......... Bridge, Rib, Table of Least Rise Bridge, Rib, Rule for Area of.... Bridge, Rib, Rule for Depth of... Bridge, Roadway, Width of...... Bridge, Stone, Arch Construction Bridge, Stone, Arch-stones, Table of Pressures on Bridge, Stone, Arch, Centres for, Bad Construction Bridge, Arch, Spring of.......... Bridge, Stone, Strength of Truss- Bridge, Weight, Greatest on...... Bridge, without Tie-Beam Bridging, Cross-, Additional Strength by... 29 00 Bridging, Cross-, Defined.. Bridging, Cross-, Resistance by Adjoining Beams. ............. 659 PAGE 165

Page 700

660 INDEX. PAGE PAGE Building, Antiquity of........... 5 | Cavetto, Roman Moulding....... 329 Building, Elementary Parts of a.. 46 | Ceiling, Cracking, How to Pre- Building, Expression in a........ 35 RU у Eius c» assi d cci. dis I25 Building by the Greeks.......... 35 | Centre of Circles, To Find (Ge- . Building, Modes of, Defined..... ei) ошету)...Ж ш .....- WM t. 556 Building by the Romans......... 26 | Centre of Gravity, Position of.... 71 Building, Style of, Selected 10 Suit Centre of Gravity, Rule for Find- Destination. ecce 35 ing, ЕЕ аа... оо. 74 Butt-Joint on Handrail to Stairs.. 303 | Chimneys, How Arranged....... 42 Butt-Joint, Handrail, Stairs, Posi- Chinese Structure, The Tent the ТТ” m. OP ШІ) Model Ol Ii sei esie eee ase sse I4 Byzantine Style, Lombard....... Io | Chord of Circle Defined......... 547 Campanile, or Leaning Tower, Twelfth Centaky.. . . е... жй 12 Capital, Uppermost Part of а £y jo ——————— - <0 I5 Carriage Beam, Well-Hole in Mid- “Ше, Find Breadth............. 126 Carriage Beam, One Header, Rule for Breadth. Е. .......... қаны Carriage Beam ог Trimmer De- fned... scc: EIR eoo r30 Carriage Beam, Rule for Breadth. 132 Carriage Beam, Two Sets of Tail Beams, Rule for Breadth.. ... 134 Caryatides, Description and Ori- несі... ғыр... 26 Cast-Iron Bowstring Girder, Should not be Used........... 163 Cast-Iron Girder, Load at Middle, Size of Flanges...... ........ 162 Cast-Iron Girder, Load Uniform, SirElofillanges......- .. „д, „е 163 Cast-Iron Girder, Manner of Mak- ЖжЕЖІС....... ee — к 161 Cast-Iron Girder, Proper Form... 161 Cast-Iron, Tensile Strengthrof.... 161 Cast-Iron Untrustworthy......... 161 Catenary Arch, Hooke’s Theory of. 51 Cathedral of Cologne............ II Cathedrals, Domes of........... 53 Cathedrals of Pisa, Erection in MO... К ч. _. ланады: аЬ 12 Cavern, The Original Place of Shelter ...... то әкі —— 13 Cavetto or Cove, Classic Moulding 323 Cavetto, Grecian Moulding...... 327 Chords Giving Equal Rectangles. 363 Circle, Arc, Rule for Length of... 475 Circle, Area, Circumference, etc., нм ples. 47...........- ВА Circle, Area, Rule for, Length of Are Сеа... 17 478 Circle, Area, To Find........... 475 Circle, Circumference, To Find .. 473 Circle Defined .......... «=. fades 546 Circle, Describe within Triangle.. 5 Circle, Diameter and Circumfer- OT mr AR rs 472 Circle, Diameter and Perpendicu- Lue o a NUT 2... 468 Circle Equal Given Circles, To ИИИ АРИИНЕ А 580 Circle, Ordinates, Rule for.. ...... Circle, Radius from Chord and Verseu ше... m з. Сз. ue. 469 Circle, Sector, Area of........... 476 Circle, Segment, Area of......... 477 Circle, Segment from Ordinates.. 470 Circle, Segment, Rule for Area of. 479 Circles, Ablexof. ....... 4 0. 649-652 Circle through Given Points...... 559 Circular Headed Doors.......... 320 Circular Headed Doors, To Form FIM LC... 321 Circular Headed Windows..... 2. 20 Circular Headed Windows, To Form Soffit 321 Circular Stairs, Face Mould for (1). 282 Circular Stairs, Face Mould for (2).. 285 Circular Stairs, Face Mould for (3). :287 Circular Stairs, Face Mould, First Бесна... оа. [gp --—-— 5) «о ө ө о ө ө ®Ф ө ө о е © © 9699 =

Page 701

Circular Stairs, Falling Mould for Rail Circular Stairs, Handrailing for .. Circular Stairs, Plan of Circular Stairs, Plunib Bevel De- fined Circular Stairs, Timbers Put in after Erection........ Е ү Cisterns, Wells, etc., Table of Capacity of ....... u.s m.m City Houses, General Idea of..... City Houses, Arrangements for... Civil Architecture Defined Classic Architecture, Mouldings Classic Moulding, Annulet or Fil- let. Classic Moulding, Cyma-Recta... Classic Moulding, Cyma-Reversa. Classic Moulding, Ogee. ....... Classic Moulding, Ovolo Classic Moulding, Scotia......... Classic Moulding, Torus......... Comer Wall$........ UC VII S S Cohesive Strength of Materials... Collar Beam in Truss........ ca Cologne, Cathedral of. ......... o Columns, Antique, Form of Column, Base, Shaft and Capital. Columns, Egyptian, Dimensions, Column, Gothic Pillar, Form of.. Column, Outline о/.............. Columns, Parthenon at Athens, НТ... Соп or Pillar’. . x. 1€. C Column, Resistance of Column, Shaft, Form of.......... . Column, Shaft, Swell of, Called Entasis Complex, or Ground Vault....... Composite Arch of Titus Composite, Corinthian or Roman Orde. . 7... "T. „ЧК, INDEX. 661 PAGE PAGE Compression, Resistance to..... > Ug 281 | Compression, Resistance to Crush- 278| ing and Вепдїпд.°............ 85 282 253 653 42 Compression, Resistance to, Pres- sures Classified... (tte: 08 83 Compression, Resistance to, Table ОЁ. c cos. ИК мак CES 79 Compression, Resistance to, in Proportion to Depth........... IOI Compression at Right Angles and Parallel to Length........... .. 206 Compression of Stout Posts...... 89 Compression and Tension, Framed Girders Mosa con AL 174. Compression Transversely to Fi- ае ТТ р ЕРТ 86 Cone Dannedd as з. ...... ЖЫ 548 Gonic-Seetiens............ 2 0 584 Conjugate Axis Defined .... .... 548 Conjugate Diameters to Axes of ЛК С” R 487 Construction Essential.......... 56 Construction of Floors, “Roof, etc., Economy Important....... 123 Construction, Framing, Heavy ҚАНЫН. ise te. o anrd р 56 Construction, Joints, Effect of 0 ж. Iun. s. 128 Construction, Object of Defined.. 123 Construction, Simplest Form Best. 123 Construction, Superfluous Mate- 56 Contents, Table of, General...613-624 Corinthian Capital, Fanciful Ori- gin-of.; oem, . ж. oo СЫ 24 Corinthian Order Appropriate in Бі d img ———MÀMÓÀ 24 Corinthian Order, Character of.... 16 Corinthian Order, Description of. 23 Corinthian Order, Elegance of.... 23 Corinthian Order, The Favorite at Коте. . теа. . . кен 27 Corinthian Order, Grecian Origin o е» ———— аа ЖЕН 16. Corinthian Order, Modification of. 27 Cornice, Angle Bracket, To Ob- tein іһе..:...... ес NUNG 343 Cornice, Eaves, To Find Depth of. 335

Page 702

662 Cornice, Mouldings, Depth of.... Cornice, Projection, То Find™ т. Cornice, Projecting Part of En- tablature Cornice, Rake and Level Mould- ings, Bo Йа ЄЙ. 7... ЕЯ Cornice, Shading, Rule for Cornice, Stucco, for Interior, De- Corollary Defined (Geometry).... Corollary of Triangle and Right ПЕЕ ИС... ИИ Cottage Style, English........... - Country-Seat, Style of a Cross-Bridging Additional Strengiti by. hr. ee. Im. TD. Cross-Bridging, Furring Impor- Cross-Bridging, Resistance of Ad- joining Beams Cross- or Herring-Bone Bridging Defined.. .. И... Cross-Furring Defined......... Е Cross-Strains, Resistance to.... Crushing and Bending Pressure.. Crushing, Liability of Rafter to... Crushing Strength of Stout Posts. Cube Root, Examples in ......... Cubes, Squares and Roots, Table Curb or Mansard Roof Curve Ellipse, Equations to Curve Equilibrium of Dome...... Cylinder, Defined Cyliuder, Platform Stairs Cylinder, Platform Stairs, Lower Cyma-Recta, Classic Moulding... Cyma-Reversa, Classic Moulding. Cyma-Recta, Grecian Moulding . : Cyma-Reversa, Grecian Moulding. Deafening, Weight per Foot Decagon, Defined INDEX. PAGE 342 342 15 344 бі! 340 348 355 35 | Deflection, 9v 137 157 205 89 645 53 257 324 324 327 328 Decimals, Reduction of, Examples Decorated Style, 14th Century.... Decoration, Attention to be given (as cT P RET oo 5: "9.7.9069 ее е Deflection, Differs in Different Ma- terial$ “есвеееееаеееасееоеаеввееевееооов Dwell- First- Deflection, Floor-beams, ings, Dimensions Floor-beams, class Stores, Dimensions . Deflection, Floor-beams, Ordinary Stores, Dimensions. ........... Deflection, Lever, Principle of... Deflection, Lever and Beam, Rela- Deflection, Lever, Breadth or Depth; Load-atiEnd.«. 2... Deflection, Lever, Load Uniform. Deflection, Lever, Breadth or Depth, Load Опйогт.......... Deflection, Lever, (от Certain, көкі Шайотп.......... л Deflection, Load Uniform or at Middle, Proportion of. ........ Deflection, Load Uniform, Breadth andDEpthi te... e Ж e s Deflection, Load Uniform or at Middle, Proportion of......... Deflection, in Proportion to ОО — T Deflection, Resistance to, Rule for sodio PED oS. ns... Deflection, Safe Weight for Pre- vention TERTIO ЗЕЕ Deflection, Weight at Middle, Breadthvand Depth. .«......... Deflection, Weight at Middle, fo AIC QN ECT UL ы Deflection, Weight at Middle, Cer- tain, for Я Deflection, Weight Uniform, for Certain c Mc PAGE 647 II 46 27 112 113 112 127 128 127 110 119 120 I2I 121 122 122 116 119 116 117

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INDEX. 663 PAGE PAGE Deflection, Weight Uniform, Cer- Dome, Small, over Stairways, Form ѓаій for. .. Se eee T өр... Ж. Гужке 220 Denominator, Least Common (Al-- Dome, Spherical, To Form...... 221 Ве), ..... e e a RO 404 | Dome, St. Paul's, London. .. .. 54 Denominator, Least Common Dome, Strains on, Tendencics of. 219 (Fractions). ..... МИР. sae. 384 | Domes, Wooden. ............... 54 Dentils, Teeth-like Mouldings in Doors, Circular Head........... 320 COBDIGOM ...... ж... ЖИЕ О. 20| Doors, Circular Head, to Form Diagonal Crossing Parallelogram бой... 7 321 (Geometry). < r. ae „и Тат 351 | Doors, Construction of.......... 217 Diagonal of Square Forming Oc- Doors, Folding and Sliding, Pro- tagOn........ Ж... "EE 357| portions... t... 316 Diagram of Forces, Example.... 166 | Doors, Front, Location of....... 320 Diameter, Circle, Defined....... 547 | Doors, Height, Rule for, Width Diameter, Ellipse, Defined...... Gio | "ізет. ae m 315 Diastyle, Explanation of the Word то | Door Hanging, Manner of....... 217 Diastyle, Intercolumniation...... 20 | Doors, Panel, Bottom and Lock Diocletian, Baths, Splendor of.... 27 Көш Width. *....... Т. 316 Division, Fractions, Rule for..... 389| Doors, Panel, Four Necessary... 317 Division, by Factors (Fractions).. 381 | Doors, Panel, Mouldings, Width. 317 Division, Quotient (Algebra)..... 419| Doors, Panel, Stylesand Muntins, Division, Reduction (Algebra)... 419| ӨМ... 277 7 316 Dodecagon, Defined............ 546 | Doors, Panel, Top Rail, Width... 317 Dodecagon, To Inscribe......... 569 ; Doors, Stop for. How to Form... 317 Dodecagon, Radius of Circles Doors, Single and Double, height (РОУ) е. гато, . e 452| Ж. 2 E 1 1 172 316 Dodecagon, Side and Area (Raye Doors, Trimmings Explained.... 317 БОЖЕ)... os «ail... e e 453 | Doors, Uses and Requirements of 315 Dome, Abutments, Strength of... 53 | Doors, Width of................. 315 Domes of Cathedrals........ .... 53| Doors, Should not be Winding... 317 Dome, Character of..... | xr. 55 | Doors, Width and Height, Propor- Dome, Construction and Form ПОВНЕ ..... TT ue TN 215 of... TM. UE И 216 | Doors, Width, Rule for, Height Dome, Construction and Strength Given... .. X x... M 316 Of: ......... ERE, НИ 53 | Doric Order, Character of........ 16 Dome, Cubic Parabola computed 219 | Doric Order, Grecian Origin of.. 16 Dome or Cupola, the......... Z.. 53|Doric Order, Modified by the Ro- Dome, Curve of Equilibrium, rule | s "DO M т oe СЕ 27 БЕТ... 218 | Doric Order, Used by Greeks only Dome, Halle du Bled, Paris..... 54 Wt WB. а.о. ен. Й E Ж. Dome, Pantheon at Коте....... 53 Doric Order, Peculiarities of..... 17 Dome, Pendentives of. .......... 53 | Doric Order, Rudeness of....... 30 Dome, Polygonal, Shape of An- Doric Order, Specimen Buildings ee RAD... ПЕСИ NI НЫ е... 19 Dome, Ribbed, Form and Con- Doric Temples, Fanciful Origin . StPüction.. o esee see e + ge ETT 217 II sun 17 Dome, Scantling fa Table of Doric Temples. 22-7. кт. 19 Thickness......... +» Жр . 218! Drawing, Articles Required...... 536

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664. INDEX. PAGE | PAGE Drawing-board Better without Elasticity of Materials........... 84 @зїрє............. n ВЕЙ 537 | Elasticity not Diminished by De- Drawing-board Liable to Warp, fication. Жа... oue І12 How Remedied:.....2.. 4 537 | Elasticity, Result of Exceeding Drawing-board, ^ Difficulty іп Limit... ее... ж-е 120 Steetching/Büper. ..... uem 539 | Elevation, a Front View......... 87 Drawing-board, Ordinary Size.... 536 | Elevated Tie-beam Roof Truss Drawing, Diagrams aid Under- Оһуесмотпай!е-................ 214 іар... ы” 956 [lhpse, Aeg tes... 488 Drawing, Inking itis. о. 542 | Ellipse, Axes, Two, To Find, Di- Drawing, Laying Out the........ 54r ameter and Conjugate Given... 593 Drawing, the Барет........ 227 537 | Ellipse Defined -trge ere. e. a k 481 Drawing in Pencil, To Make Lines 542 | Ellipse, Equations to the Curve.. 482 Drawing, Secure Paper to Board. 537 | Ellipse, Major and Minor pus. Drawing, Shade Lining.......... 543 [XS ned eu ow e лен... OR 481 Drawing, Stretching Paper....... 537 | Ellipse, Ordinates, A of ... 49I Durability in a Building......... 37 | Ellipse, Parameter and Axis, рњ. Dwelling, Arrangement of Rooms 538, lation ОЇ........---........... 485 Dwellings, Floor-beams, To Find Ellipse, Practical Suggestions.... 489 Dimensions eoe: . 2.. Ж” 127 | Ellipse, Semi-major, Axis Defined 486 Dwellings, Floor-beams, Safe ‚| Ellipse, Subtangent Defined..... 486 Weight for. E oL... e LE ы 126 | Ellipse, Tangent to Axes, Rela- D welling-houses, Dimensions and КОШТОН Л ТГ Е АСЕ. аа 485 Bryle...... HEC а m 37 | Ellipse, Tangent with Foci, Rela- bebe e о... түсте сс. 487 Eaves Cornice, Designs for....... 335 | Ellipsis, Axes of, To Find (Geom- Eaves Cornice, Rule for Depth... 335 е ОА CO A ы” 585 Ecclesiastical Architecture, Point- Ellipsis, Conjugate Diameters (Ge- ЕГЕ... Ж.Ж... II ол ме aee... coca, Ld 593 Ecclesiastical Style, Origi of.. £4 | Ellipsis Defined oeuse... ле 548, 585 Echinus, Grecian Moulding..... е. Ellipsis, Diameter Defined....... 549 Economy, Construction Floors, Ellipsis, Foci, To Find.......... 586 2 КОФЕ, Bridges................ і23! Ellipsis, by Intersecting Arcs.... 590 Eddystone апа Bell Rock Light Elipsis, by Intersecting Lines... 588 - - TCWIILERTOQO DIE. 48 | Ellipsis, by Ordinates...... .... 588 Egyptian Architecture........... 30| Ellipsis, Point of Contact with Egyptian Architecture, Appropri- Tangent, То ЕЁїпд.............. 593 ate™Buildings for.............. 33| Ellipsis, Proportionate Axes, to Egyptian Architecture, Character Describe Wilk. грр. о... == 33 | Ellipsis, Trammel, (о Find, Axes Egyptian Ккейикеінге, Огіріп іп білеп. Мек omm. 586 COROT. ае. eT. cames des I4 | Elliptical Arch, Joints, Direction Egyptian Architecture. Principal Of. ee oos ccm us 233 PFoalprestol. Tous. ue conet Бш. + 30 | English Architecture, Early...... II Egyptian Columns, Dimensions English Cottage Style Extensive- and Proportions 3 SEES... 38 ly Used am u 2. тте 35 Egyptian Walls, Massiveness of.. 33 | England and France, Fourteenth Egyptian Wonks of Auto i...... 690! Century... le eere s I2 594 .

Page 705

INDEX. PAGE Entablature, above Columns and Blorizental....... 98 99. т. I4 Entasis, Swell of Shaftof Column 48 Equal Angles Defined ....... .. 349 Equal Angles, Example in....... 350 Equal Angles, in Circle.......... 358 Equal Angles (Geometry)........ 553 Equilateral Rectangle, to De- БЕР... ы...” 568 Equilateral Triangle Defined (Ge. crc n ——— 77 545 Equilateral Triangle, to Construct (apametry)......- ue 568 Equilateral Triangle, to Describe (ОМОШ Еър). . шс. ... ees rent 566 Eqilateral Triangle, to -Inscribe (Greaimiplty). .... cometer 569 Equilateral Triangle (Polygons).. 445 Eustyle Defined.... 20 Exponents, QU сира rat use tive (Algebra) =... sa 423 Extrados of an Arch......... ҘГ.....58 Face Mould, Accuracy of, Wind- OOV DOCENT TTET TIERS С 295 Face Mould, Curves Elliptical, Winding Stairs. ......... We 301 Face Mould, Drawing of, Winding SO ee тп 296 Face Mould, Sliding of, Winding C S CMM RESORTS И 299 Face Mould, Айй сани of, Р1а{- form Stairs. ..... лз. 8. И: 275 Face Mould, a Simple, Kell's Method for ...2529.- r 268 Factors, Multiplication (Algebra) 409 Factors, Two, Squaring Difference of«(Algebray: сее... ғ. . ж 412 Fibrous Structure of Materials... 76 Figure Equal, Given Figure. (Ge- ometry)...... esent 575 Figure, Nearly Elliptical, To Make (Geometry). . «i = жш. инн ыз 591 Fillet or Amulet, Classic Mould- Flanges, Area of, Tubular Iron Girder.. E жарыс « аша ЕР ра ын Flanges, Area of Bottom, Tubular Iron Girder.. .... m. AE Flanges, Load at Middle of Cast- iron Girder, Size&. 25... ae Flanges, Load Uniform on Tubu- lar Iron Girder, Sizes.......... Flanges, Proportion of, Tubular Iron Girder ute... F Flexure, Resistance to, Defined.. Floor-arches, How Constructed.. Floor-arches, Tie-rods, Dwellings, Floor-arches, Tie-rods, First-class Stores Өе S.L. eos sues Floor-beams, Distance from Cen- tres, Sizes Fixed. .......9..... Floor-beams, Dwellings, Safe Wiehe MAOT г... . онй. n Floor-beams, Dwellings, Deflec- tion Given, бірев......2 27... Fioor-beams, First-class Stores, Deflection Given, Sizes....... " Floor-beams, Ordinary Stores, De- flection Given, Sizes........... Floor-beams, cem to Rules og. Cc Floor-beams, Reference to Trans- ЖТТ Г Г у Т. Floor-beams, Proportion Weight en. All. on... . motui Floors Constructed, Single or Floors, Framed, Seldom Used... Floors, Framed, Openings in..... Floors, Headers, Defined........ Floofs, Ordinary, Effect of Fire (0:10 ....„ь„. GOD OS Т conc de Floors, Solid Timber, Dwellings and Assembly, Depth....... я РАСЕ 159 162 156 157 84 205 145 153 153 153 129 126 127 128 127 126 125 126 130 124. 143 I24 130 130 143

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666 . РАСЕ Floors, Solid Timber, First-class Stores; Deptb. ....... 7227 144 Floors, Solid Timber, to Male Fire-proof.. Е «248243 Floors, Tail- — рейшен:. — I30 Floors, Trimmers or Carriage- beams Defined.......989....... 130 Floors, Wooden, More Fire-proof than Iron, Some Савев......... Flyers and Winders, Winding 143 саны Л... o s. 251 Foei Defined ara us... coe. . oom 548 Foci of Ellipsis To Find........ 586 Foci of Ellipse, Tangent........ 487 Force Diagram, Load on Each Support... ..2......g%.. ЖШ 179 Force Diagram, Truss, Figs. 59, 68 and 9007... o Due. cm 179 Force Diagram, Truss, Figs. 60, 70 а кз... EM UE 180 Force Diagram, Truss, Figs. 6r тапа 737 eee. oR. 181 Force Diagram Truss, Figs. 63 ша and 75... . eee owe 183 Force Diagram, Truss, Figs. 64, КОКО... QT... 184 Force Diagram, Truss, Figs. 65, EIN... 1... nomeu 185 Force Diagram, Truss, Figs. 66 Ооа Өт. .....жЖ.....%.... 186 Forces, Parallelogram of. ........ 59 Forces, Composition of.......... 66 Forces, Composition, Reverse of Resolution. . 0 с 20. O Forces, Eion of. „ Ga Ө 9. 59 Forces, Resolution of, Oblique ТЕМИНЕ. . 1... е. 59 Foundations, Description of.... 47 . Foundations in Marshes, Timbers 47 Fractions, Addition, Like Denom- C di cc e ee ERE . es Fractions Added and Subtracted (МЕРЫ, eomm Fractions Changed by Division:. Fractions Defined. .... e ere Fractions, Division, Rule for..... Fractions, Division by Factors... 382 403 380 378 389 381 INDEX. PAGE Fractions Divided Graphically... 388 Fractions Graphically Expressed. 378 Fractions, Improper, Defined.... 380 Fractions, Least Common Denom- inator..... (ON. T. ame. Fractions, Multiplication, Rule... Fractions Multiplied Graphically. 386 Fractions, Numerator and Denom- . mator E КЄ eller 378 Fractions, Reduce Mixed Num- bens ошооо Улоо ОТО ОООО ОС. 381 Fractions Subtracted (Algebra)... 405 Fractions, Subtraction Like De- оа ако Се. rc eere rere Fractions, Unlike Denominators у ТЫ ss. ood 383 Framed Beams, Thickness of, me. A — — у ср I30 Framed Girder, Bays Defined.... 167 Framed Girders, Compression and Tension, Dimensions.......... 174 Framed Girders, Construction Лл m 166 Framed Girders, Height апа Шерін... ee Ce E 167 Framed Girders, Kinds of Pres- ЕЛЕК oues cuum cos erret NS 173 Framed Girders, Long, Construc- tion Өс... IIE emer 174 Framed Girders, Panels on Under Chord, Rable of. bes.. 2"..... 167 Framed Girders, Ties and Struts, Effectof. ж а... Е DNE 174 Framed Girders, Triangular Pres- sure, Upper Сћога.:.......... 168 Framed Girders, Triangular Pres- sure, ВоймФЖога............. 171 Framed Openings in Floors...... 130 Framing Beams, Effect of Splic- їп@ эй... РИТ. 2M М: 235 Framing Roof Truss............ 237 Framing Roof Truss, Iron Straps, JT o —— — осо 239 France and England, Fourteenth Centu. odii ee: Conc “бе 12 Friction, ЕПЗ ич: о... 82

Page 707

INDEX. 667 PAGE PAGE Frieze between Architrave and Geometry, Square Equal Rec- Comite. ......2... ee тесті tangle, To Mabe....... ғо ST, Furring Defined. .... -ore e 125 | Geometry, Square Equal Given | Squares, To make.............. 577 Gable, а Pediment in Gothic Аг- Geometry, Square Equal теле) chbübetur& a. csee. e. L. ur 15| Te Make... 1. Pore, nae 582 Gaining a Beam Defined......... 100| German or Romantic Style, Thir- General Contents, Table of... .613-624 teenth and Fourteenth Centuries. гї Geometrical Progression (Alge- Girder, Bearings, Space Allowed |) MAREC ON WEE 5. 435| for....... ECT 141 Geometry, Angles of Triangle, Girder, Bow-String, Cast-Iron, Three, Equa! Right Angle..... 354| Should not be Used.......... 163 Geometry Chords Giving Equal Girder, Bow-String, Substitute for. 163 Rectangles. ..... i52 2 .-.. 363 | Сігаег, Construction with Long Geometry Defitsd. .......... 09. 544|” Berings M. AR a... T жий 140 Geometry, Divide a Given Line.. 555 | Girder, Cast-Iron, Load Uniform, Geometry, Divisions in Line Pro- оао... 2 ИВ . 163 poitionatem. a... „+... 583 | Girder, Cast-Iron, Load at Middle, Geometry, Elementary...... .... 347 Flanges....5.....- ON 162 Geometry, Equal Angles......... 555 | Girder, Cast-Iron, Proper Form Geometry, Equal Angles, Ex- ЕЕ”... n 161 "-— cn Г = Use | Geometry, Figure Equal to Given Шы. le 140 Pigure, Соһестисі............. 575 | Girder, Different Supports for.... 140 Geometry, Figure. Nearly Ellipti- Girder, Dwellings, Sizes for...... 141 cal by Compasses.......... 99 591 | Girder, Framed, Bays Defined.... 167 Geometry, Measure an Angle. ... 348 | Girders, Framed, Compression Geometry Necessary in Handrail- and Tension, Dimensions. ..... 174 ing, Әнес... го гор MD 257 | Girder, Framed, Construction of.. 140 Geometry, Opposite Angles Equal. 354 | Girder, Framed, Construction and Geometry, Parallel Lines........ 555 uo o ————— T 166 Geometry, а Perpendicular, То Girder, Framed, Construction of Eteot i. -.ж...... M 550| Попд........................ 174 Geometry, Perpendicular,let Falla. 551 | Girder, Framed, Kinds of Pres- Geometry, Perpendicular, Erectat _ Surg... ... cu е 2 Aeg өзен 173 End of Line: ......8 99908 551 | Girders, Framed, Height and Geometry, Perpendicular, Let Fall Depth. . muc ont ты. c 167 ' Near End of Line............. 555 | Girders, Framed, Panels on Under Geometry, Plane Defined (Stairs). 257| Chord, Table of............... 167 Geometry, Point of Contact:..... 558 | Girders, Framed, Triangular Pres- Geometry, Points, Three Given, | sure Upper Chotd.... Лей 168 Find Fonnt .....:... oet 559 | Girders, Framed, Triangular Pres- Geometry, Right Line Equal Cir- sure Both Chords............. 171 cuiifereneek . „сеи. оре ae 566 | Girders, Framed, and Tubular Geometry, Right Lines, Propor- НОһ....... Ж. ткш ll 2 .. 140 tion Between... .. Ж... E 584 | Girders, First-Class Stores, Sizes Geometry, Right Lines, Two | -—X— ИЕ & ер: 141 Given, Find ТҺїгд............. 582! Girders, Sizes, To Obtain.... I4I

Page 708

668 PAGE Girders, Strengthening, Manner I40 Girders, Supports, Length of, Rule. 157 Girders, Tubular Iron, Construc- . I54 Girders, Tubular Iron, Area of Flange, Load at Middle....... I54 Girders, Tubular Iron, Area of > Flange, Load at any Point..... I55 Girders, Tubular Iron, Area.of Flange, Load Uniform......... I56 Girders, Tubular Iron, Dwellings, Area of Bottom Flange........ I59 Girders, Tubular Iron, First-Class Stores, Area of Bottom Flange.. Girders, Tubular Iron, Rivets, Al- lowance fon . «оиа: oS 157 Girders, Tubular Iron, Flanges, Proportion of. . Sov. cq 457 Girders, Tubular = ‘Shearing Strain... EV. Lore I57 Girders, Tubular Iron,Web, Thick- ness о... ЖИГ... 158 Girders, Weakening, Manner of.. Girders, Wooden, Objectionable. 154 Girders, Wooden, Supporting, Mamtier о....22......... 227% TET Glossary of Terms............ 627-637 ШӨНСРАтсһев....2............. 5I Gothic Buildings, Roofs of....... 55 Gothic and Norman Roofs, Con- БӨШТРОГӘ 2...52277.....: 178 Gothic Pillar, Form of........... 48 Gothic Style, Characteristics of... 12 Goths, Ruins Caused by......... 12 Granular Structure of Materials.. 76 Gravity, Centre of, Position...... JI Gravity, Centre of, Examples, and RulSifote. 2.........5... ""-— 71 Grecian Architecture, History of. 6 Grecian Art, Elegance of........ 27 Grecian Moulding, Cyma-Recta.. 327 Grecian Moulding, Cyma-Re- Т -...- Me cess: + os 328 Grecian Moulding, Echinus and Саи... ттер... . 327 Grecian Moulding, Scotia........ 326 Grecian Moulding, Torus........ 326 INDEX. PAGE Grecian Orders Modified by the: аас Др... RM 27 Grecian Origin of the Doric Or асе. ... оо ЫС. 3 16 Grecian Origin of Ionic Order... 16 Grecian Style in America........ 13 Grecian Styles, their Different Ordens. MM e.t. T. om 16 Greek Architecture, Doric Order ' Weeder... Se... ж 19 Greek*BüildingJ................ 35 Greek Moulding, Form of....... 325 Greek, Persian, and Caryatides Отаев и. 5... NET 24 Greek Style Originally in Wood.. 14 Gréek Styles Only Known ich Theft oom 9 и Groined ог Complex Vault...... 52 Halle du Bled, Paris, Dome of... 54 Halls of Justice, N. Y. C., Speci- men of Egyptian Architecture.. 8 Handrailing, Circular Stairs...... Handrailing, Platform Stairs. ... Handrailing, Platform Stairs, Face Майа. . я... Кет 264 Handrailing, Platform Stairs, Lange/Gplinder.. "0... .. а 271 Handrailing .Stairs, Geometry N etossary S amem 257 Handrailing Stairs,“ Out of Wind" елеч 7” К.т... Ж 257 Handrailing Stairs, Tools Used.. 257 Handrailing, Winding Stairs..256, 289 Handrailing Winding Stairs, Bal- usters Under Serol:. Ae. T... Handrailing, Winding Stairs, Centres in Square... .......... Handrailing, Winding Stairs, Face for'Seroll e 0. Bee. eee п 2104 Handrailing, Winding Stairs, Fall- ing ЛИ ИЧ. eoe И... 310 Handrailing, Winding Stairs, Gen- eral Considerations........ ... 258 Handrailing, Winding Stairs, Seroll for, ЖЫ oes 308 Handrailing, Winding Stairs, Scrollvat Жайы 20.12... е 2. 309

Page 709

Handrailing, Winding Stairs, Scroll Over Curtail Step....... Handrailing, Winding Stairs, Scroll for Curtail Step......... Headers, Breadth of............. Headers. Defined... .. WWW. . жк Headers, Mortises, Allowance for Weakening by......295: ат Headers, Stores апа Dwellings, Same for Both. <+.. Э. Hecadecagon, Complete Square (Polygone). =... mmm e Hecadecagon, Radius di Circles (Polygons)... ee.. „ВЕ Hecadecagon, Rules (Polygons). . Hecadecagon, Side and Area (Polyggng)........ T. slm Height and Projection, Numbers of ПЫ ОЕ, .2Ҙ5559%..2.... BieptagonjDe&ned............... Herring-bone Bridging Defined... Hexagon Defined. ............-. Hexagon, To-Inscribe........... Hexagons, Radius of Circles..... Hexastyle, Intercolumniation... Hindoo Architecture, Ancient, Character of... .. ж®........Ж.. Hip-Rafter, Backing of Hip-Roofs, Diagram and Expla- History of Architecture.......... Hogged Ridge in Roof Truss.... Homologous Triangles (Geom- Homologous Triangles (Ratio and Proportion) m. .. рее. 099 Hooke's Theory of an Arch...... Hooke's Theory, Bridge Arch, Pressure Qno Е. ИС Hooke's Theory, Catenary Arch. . Horizontal and Inclined Roofing, Horizontal Thrust in Beams Horizontal Thrust, Tendency of.. INDEX. 669 PAGE PAGE Hut, Original Habitation ........ 13 309 | Hydraulic Method, Testing Wood. . 5... OS Amm 80 зто | Hyperbola Defined..... ..548, 585 130 | Hyperbola, Height, To Find, Base I30| and Axis Given...... Жол 2. 585 Hyperbola by Intersecting Lines. 595 131 om e Equality of Squares (Algebra). . . 416 132 H$ponmmuad JB sra fos poe onometry)s. e... 1. . „э 516 458 | Hypothenuse, Side, To Find (Ge- отеу) г. оа... го 579 455 | Hypothenuse, Right Angled Tri- 459| angle (Geometty)... ..... ne . 355 Hypothenuse,. Triangle (тш 457 пав): . .. . Рог .. ИТ! 518 l 16 | Ichnographic Projection, Ground Plast... ER. m um poem 37 177 | Improper Fractions Defined..... 380 546 | India Ink in Drawing........... 540 157 | Inertia, Moment of, Defined..... 145 546 | Inking-in Drawing.............. 542 569 | Inside Shutters for Windows, Re- 447 аш E TE o o oss 319 го | Instruments in Drawing......... 540 Intercolumniation Defined....... 17 3o | Intercolumniation of Orders..... 20 216 | Intrados of Агсһ................ 52 Ionic Order, Character of........ 16 215 | Ionic Order, Grecian Origin of... 16 44| Ionic Order Modified by the Ro- ЗЕ ТУ ыы 27 Ionic Order, Origin о[........... 20 362 | Ionic Order, Suitable for What Buildings. -st о, Е . 20 370 | Ionic Volute, To Describe an.... 20 50| Iron Beams, Breaking Weight at Middle. . — — 148 5I | Iron Beams, Deflection, To Find, $1| «Weight at Middle... Lco 147 Iron Beams, Deflection, To Find, 190] Weight Uniform.. — пс Iron Beams, рибаи То Find, 74 Weight any Pointe. c... 149 72 | Iron Beams, Dimensions, To Find, 88! Weight Опйогшт.............. 149

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670 INDEX. PAGE PAGE Iron Beams, Dwellings, Distance LawsiofsPressmiec bese rre оа 57 пой Cefitres.. a... а. . NE тт | Laws of Pressure, Inclined, Ex- Iron Beams, First-Class Stores, pim... adesse шы 57 Distance from Centres.. ...... 152 | Laws of Pressure, Vertical, Exam- Iron Beams, Rectangular Cross- PEEN e «АС КЛ... ш: 57 бееШӘп...2...........7. ЕЕ 145 | Leaning Tower or Campanile, Iron Beams, Rolled, Sizes....... ms) Twelfth Centum... =... . sores 12 Iron Beams, Safe Weight, Load Length, Breadth, or Thickness, any Вол. г... E: 148| Relation to Pressure.......... 78 Iron Beams, Safe Weight, Load ° ,Lever, Breadth or Depth, To ИТШ C DIT NCC TET з I5I Findet ee ge. .... III Iron Beams, Table IV........... 146 | Lever, Deflection as Relating to Iron Beams, Weight at Middle, Вани, c II9 Deflection Сізеп............... 146 | Lever, Deflection, Load at End.. 120 Iron Fire-Proof Floors, Action of Lever, Deflection, Load Uniform. 121 Fires... 72. ес ... 143 | Lever, Deflection, Breadth ог Iron Straps, Framing, to Prevent Depth, Dod. atsBud..... un I2I Resting TIn . айы ЛТ 239} Lever, Deflection, Breadth ог Irregular Polygon, Trigon (Geom- Depth, Load Uniform......... 122 try)... 52 ЖМ 2............... 546 | Lever, Deflection, Load Required. 122 Isosceles Triangle Defined....545,584]| Italian Architecture, Thirteenth, Fourteenth, and Fifteenth Cen. lüries........ УИС РУЗЕ 12 Italian Use of Roman Styles...... 13 Italy, Tuscan Order the Principal ЕЛЕНЕ ........... 30 Jack-Rafters, Location of ....... 212 Jack-Rafters and Purlins in Roof. 211 Jack-Rafters, Weight per Superfi- con 3o ............. ce 189 Joists and Studs Defined......... 174 Jupiter, Temple of, at Thebes, Ex- (Ol. ....... ec ct om 33 Kell’s Method, Simple Face МОУ Әішіне................. 268 Keystone for Arch, Position of. ... 50 King-Post, Bad Framing, Effect olg.» ———— "TER 237 King-Post, Location of.......... 213 KinggBost in Roofs >..:..!..... 54 Lamina in Girders Defined....... 174 Lancet Auth... „ноо ке... 5I Lateral Thrust іп Агсһ,.......... 52 Lever Formula, Proportionals іп ТЕ” КЕ 421 Lever Load Uniformly Distrib- Tu CNN TTE NER III Lever, Load at One End......... IIO Lever Principle Demonstrated (Ratioa —— 575 Lever, Support, Relative Strength IIO INDE... me Lac ook cac OE as, 48 Line Defined (Geometry)......... 544 Lines, Divisions in, Proportionate (беотенуге.е..--“---....... 583 Lintel Лоо... . . аео... 49 Lintel, Strength о.............. 49 Load, per foot, Horizontal....... 192 Load on Roof Truss, per Superfi- cial. Көе. Ste tme d. «ыты» 189 Load on Tie-Beam, Ceiling, etc.. 190 Lock Rail for Doors, Width...... 316 Logarithms Explained (Algebra).. 425 Logarithms, Examples........... 426 Logarithms, Sine and Tangents (Polygons): . quem eus ees 464 Lombard, Byzantine Style........ IO Lombard Style, Seventh Century. то London Bridge, Piles, Age of.... 229

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INDEX. PAGE Materials, Cohesive Strength of.. Materials, Compression, Resist- ancewos........ ТТК”. 77 Materials, Cross-strain, Resistance tO qe EUN Pon mio = oO 77 Materials, Structure of.........* 76 Materials, Tension, Resistance to. 77 Materials Tested, General De- scfiption..... X«L о. FO 80 Materials, Weights, Table of.. «« 654 Major and Minor Axes of Ellipse Marshes, Foundation for Timbers ТП... 47 Mathematics Essential.......... 347 Maxwell, Prof. I. Clerk, Diagrams ef"EGrces ec... SSIs Memphis, Pyramids of, Estimate Gi Sme in... vem. me Minster, Tower of Strassburg. ... Minutes, Sixty Equal Parts, to Proportion an Order... ....... I5 Mixed Numbers in Fractions, To ШӨ СЄ... и. СӨ; 381 Modern m First Ap- С 222.22 9 Modern Tuscan, Appropriate for Bisildings...45 44-9 ex 30 Moment of Inertia Defined...... I45 Mono-triglyph, Explanation of the Werk, . э. жн. eee. енен IQ Monuments, Ancient, Their Archi- (есик...... жебеге. aa 6 Moorish and Arabian Styles, Ап- tigflities of). o E Un v fr Mortises, Proper Location of.... 100 Mortising, Beam, Effect оп Strengthwof... a. нога Fae 100 Mortising Beam at Top, Injurious БНС Г. ...... D. om 100 Mortising Beam, Effect of....... 231 Mortising, Beam, Position of.... 236 Mortising Headers, Allowance for Weakening .... 2.0 n И I3I Moulding, Classic, Astragal or Bead... . rer t Qum кем 323 Moulding, Classic, Annulet or Fillet... m... pS DRE eo Sep 671 PAGE . 76 | Mouldings, Classic Architecture. 323 Moulding, Classic, Cavetto ог Сөме,...222... затынан 325 Moulding, Classic, Cyma-Recta.. 324 Moulding, Classic, Cyma-Reversa. 324 Moulding, Classic, Ogee........ 324 Moulding, Classic, Ovolo........ 323 Moulding, Classic, Scotia........ 323 Moulding, Classic, Тогив........ 323 Mouldings, Common to all Or- ders... — ———— и 324 Mouldings Defined.............. 328 Mouldings, Diagrams of......... 330 Mouldings, Doors, Rule for Width. 317 Moulding, Grecian, Cyma-Recta. 327 Moulding, Grecian, Cyma-Re- veréa....... (—Á 328 Moulding, Grecian Echinus and Сатейб......... REI 327 Mouldings, Greek, Form of...... 325 Mouldings, Grecian Torus and Scal е AA. oo... o qoem 326 Mouldings, Modern............. 331 Moulding, Modern, Antz Cap... Mouldings, Modern Interior, Dia- > | 2c. om ге. Е. 332 Mouldings, Modern, Plain....... 333 Mouldings, Names, Derivations of. 324 Mouldings, Profile Defined...... 326 Mouldings, Roman, Forms of.... 325 Mouldings, Roman,Comments on. 329 Mouldings, Roman, Ovolo and Caávetto.. . „а г amras iis 6 е. 329 Mouldings,Uses and Positionsof. 324 Multiplication (Algebra). ........ 408 Multiplication, Plus and Minus (Algebra)e. ss... . г cort 415 Multiplication, Three Factors (Al- a) R ee m 408 Multiplication, Fractions........ 387 Newei Cap, Form of, Winding Staivs: ........ 8 mene нн 31e Nicholson’s Method, Plane Through Cylinder (Stairs)..... 259 Nicholson’s Method, Twists in SMS. a cicer POS COS 259 Nonsgon Deá&ned.......... 546 é

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672 INDEX. PAGE PAGE Normal and Subnormal in. Para- - Order of Architecture, Three м... Mls. КЕР. 496! Principal Patte........ . — 14 Norman and Gothic. Construction Orders of Architecture, Persians of КӨШІБЕ,.-5......... a 878] запи iGaryatidees. .o...0....0... 24 Norman Style, Peculiarities of... rr | Ordinates to an Arc (Geometry)... 563 Nosing and Tread, Position in Ordinates, Circle, Rule for....... 471 ап. Тг. ОИ КГС 241 /Ordinatesiof ellipse. . . ....... „ә 491 Ostrogoths, Style of the.......... 9 Oblique Angle Deineds oS 544 | Oval, To Describe a (Geometry).. 591 Oblique Pressure, Resolution of Ovolo, Classic Moulding......... 323 Forbés4it a Winona. 59 | Ovolo, Roman Moulding........ 329 Oblique Triangle, Тт Two . Angles (Trigonometry)......... 523 | Paper, The, in Drawing, Secure to Oblique Triangle, First Class Board. ies ОЖ. . Ж oU. 537 (ПИ @йотету)..... mem 520 | Pantheon at Rome, Dome of, af Oblique: Triangles, First Gas UNAM: umm... лави 53 Formule (Trigonometry)....... 531 | Pantheon and Roman Buildings, Oblique Triangles, Second Class Wallsrofm:. tw... ге NEL Lom 49 (Trigonometry)... Ai... Eo.. 522 | Parabola, Arcs Described from... 503 Oblique Triangles, Second Class, Parabola, Area, Rule for......... 509 Formula (Trigonometry). ...... 532 | Parabola, Axis and Base, to find Oblique Triangles, Third CESS (Сое) т". T. кке 585 (Trigonometry) ien К... 526 | Parabola, Curve, Equations to.... 493 Oblique Triangles, Third Class, ! Parabola Defined................ 492 Formule (Trigonometry)....... 534 | Parabola Defined (Geometry)..548, 585 Oblique Triangles, Fourth Class | Parabola, Diameters. .. ... . o 497 (Гаропотеиф)............... 528 | Parabola Described from Ordi- Oblique Triangles, Fourth Class, matesi. acess... Ву ОНЕ 504 Formula (Trigonometry)... .... 534| Parabola Described from Diame- Oblique Triangles, Two Sides ter SM. ө ——— НИЕ. РРА 507 (rigonometry) ..... 0X Er 521 | Parabola Described from Points.. 502 Oblique Triangles, Sines and Parabola of Dome Computed.... 219 Sides (Trigonometry).......... 519 | Parabola, General Rules... ..... 499 Obtuse Angle Defined....... 349, 544 | Parabola by Intersecting Lines... 594 Obtuse Angled Triangle Defined. 545 | Parabola Mechanically Described. 500 Octagon, Buttressed, Find Side Parabola, Normal and Subnor- (ББӘПТӨйтТ).........-........:.. 571| Мм CE .. 249... 496 Ociasea Defined. .............. 546 | Parabola, Ordinate Defined... .. 496 Octagon, Diagonal of. Square ParabolajfSub tangent. e. .22.... 496 JOC 4d В erue 357 | Parabola, Rangent .............. 493 Octagon, Inscribe a (Geometry).. 57o0|Parabola, Vertical Tangent De- Octagon, Rules (Polygons)....... дт | Жа ЖҰ Сы. comet 495 Octagon, Radius of Circles (Poly- Parabolic Arch, Direction of рот... m oou о |- о. ctt TOU TE ee age 234 Octastyle, Intercolumniation..... 20 | Parallel Lines Defined. .......*.. 544 Ogee Mouldings, Classic........ 324 | Paralle] Lines (Geometry)....... 555 Opposite Angles. Equal (Geome- Parallelogram, Construct a....... 576 EE): cvs. m De oos. s. 354) Parallelogram Defined........... 545

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Parallelogram Equal to Triangles, Parallelogram of Forces, Strains Parallelograms Proportioned to Bases (Geometry). ........... Parallelogram in Quadrangle (Ceometty). „ж. ЛЛ... T Parallelogram, Same Base (Geom- Parameter, Axes (Ellipse)........ Parthenon at Athens, Columns of. Partitions, Bracing and Trussing. Partitions, How Constructed... .. Partition, Door in Middle, Con- struction Partition, Doors at End, Construc- Owm. Жо н TEN Partition, Great Strength, Con- ИОН. еее Partitions, Materials, Quality of.. Partitions, Plastered, Proper Sup- pou (OT >....... 22-2 Partitions, Pressure on, Rules.... Partitions, Principal, of what Com- Partitions, Trussing in, Effects of. Pedestal, a Separate Substruc- {ШӨ NEM ее 3777772 Pediment, Triangular End of Бала... О. л. Репсії апа Rulers, Drawing..... Pentagon Defined. . ЖЕ. 7. 2” Pentagon, Circumscribed Circles (Polygonm). .. 2/27 oo Om Perpendicular Height of Roof, To Perpendicular, Erect a Perpendicular, Erect a, at End of 1/0988... AU. ACUTE APPEL Perpendicular, Let Palla. .... 8% Perpendicular, Let Fall a, at End САА АБА жа «а Perpendicular Style, Fifteenth Ciütüry. . os. ee В, eee INDEX. 673 PAGE PAGE Perpendicular in Triangle (Pall í 576 рөпе).......... 2 Өш 440 Persians, Origin and Description ES оГ... Де. „Ж. „және 24 Persians and Caryatides, Orders 3601 Used by:Greélis ШЕГШ "EE Piers, x in City Front 364| of Housc.. 44 . | Piers, йө aieiaa ad 352| Sizes... AERE е 228 548 | Piles, London Bridge, Age of.... 229 485| Pine, White, Beams, Table of 48| Weeigbit&forzu 22.2...“ m 176 | Pisa, Cathedral of, Eleventh Cen- : ЛЕТ, om 12 Pisa, Cathedral of, Erection in 175 То1б.. . БЕЙ бкл « ИНИ 12 Pise Wall ofsratce: ...... nta “49 176 | Pitch Board, To Make, for Stairs. 247 Pitch Board, Winding Stairs..... 252 176 | Plane Ей Ее... cae 257 Plane Defined (Geometry)........ 544 175 | Plank,Weight обоп Roof, per foot. 189 175| Plastering, Defective, To what Due.. -. ЛТ TIERS 175 Рав йа, уш %. A 178 177 | Plastering, Weight per foot...... 177 Platform Stairs, Baluster, Posi- 175 #Ө@ӨШЖЮГ..... ТИШ... ЖШ A» Каш. 250 175 | Platform Stairs Beneficial........ 240 Platform Stairs, Cylinder of. ..... 248 I4 | Platform Stairs, Cylinder, Lower Ере... Эти 249 I5 | Platform Stairs, Face Mould, Ap- 540 plication of Plank............ ЖҰЗ 546| Platform Stairs, Face Mould, Handrailing in7...... eei 264 463 | Platform Stairs, Face Mould, Sim- pleMethod.. . ... wes: 2... 267 579| Platform Stairs, Face Mould, 550] Moulded Rails: УЛУ. - 274 Platform Stairs, Face Mould, Ap- Bet) plitation of = eoe Ж 275 551 | Platform Stairs, Face Mould With- | out Canting Р1апК............ 272 . 553 | Platform Stairs, Handrail to. . 269 Platform Stairs, Handrailing кеде ge Султбсг.............. же = 271

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674 INDEX. PAGE PAGE Platform Stairs, Railing’ Where Pressure, Triangular, Framed Rake Meets Level............. 272 Вен ow e o a o o modems Бут Platform Stairs, Twist-Rail, Cut- Pressure, Upper Chord, Triangu- {їтїр A erts soos = 297 ИОГ... во. аео. 168 Platform Stairs, Wreath of Round Prisms Cut by Oblique Plane..... 259 Қой, —— УЕ 267 | Prisms and Cylinders, Stair-Build- Point of Contact (Geometry)..... 558 о... 257 Point Defined (Geometry)........ 544 | Prisms Defined Ше). .--4257,-250 Pointed Style, Ecclesiastical Arch- Prism; Top, Form of, in Bars Месінте. 4.2... om ШК те; ТИ CAO ssl 259 Polygons, Angles of. ............ 462 | Profile of Mouldings Defined.... 326 Polygons, Circumscribed and In- | Progression, Arithmetical TU scribed Circles, Radius of..... 405 | «Ыг foe ee ae 432 Polygons Defined (Geometry).... 546 Polygons, Equilateral Triangle... 445 Polygons, General Rules........ 461 Polygons, Irregular, m (Ge- ometry) MA oo oS oL MU ж 546 Polygons, Perpendicular in Tri- angle ... 5 9. oe). .. Және: Polygon, Regular, Defined (Geom- Polygons, Regular, To Describe (Geometry)... nea m. Polygons, Regular, To Inscribe in Circle (Geometry)............. 572 Polygons, Sum and Difference, MORNE Sine.. сено im 439 Polygons, Table Explained...... 466 Polygons, Table of Multipliers... 465 Polygons, Triangle, Altitude of.. 442 Polygonal Dome, Shapeof Angle- РЕТ. ee . 2-26 Posts, Area, То Find............ 86 Posts, Diameter, To Find........ 92 Posts, Rectangular, Safe Weight.. 92 Posts, Rectangular, To Find Thick- ls cic c occa p MM M 94 Posts, Rectangular, Breadth Less than Мтсіпевее.............. 96 Posts, Rectangular, To Find Битке ыр“. ne 95 Posts, To Find 5іде.. ds 93 Posts, Slender, Safe Weight — 9I Posts, Stout, Crushing Strength of. 89 Pressures Classified............. 85 Pressure, Oblique, Resolution of Бай... comme La PEEL. 522% 59 Progression,Geometrical (Algebra) 435 Projection and Height, Members of Orders of Architecture. ..... Protractor, Useful in Drawing... 541 Purlins and Jack-Rafters in Roof. 211 Purlins, Location of............. 212 Pyramids of Memphis, Amount of Шошефтк» е-е ooh ИБ ы. 33 Pycnostyle, Explanation of...... 20 Quadrangle Defined......... .-..1545 Quadrangle Equal Triangle...... 353 Quadrant Defined............... 547 Quantities, Addition and Sub- traction (А4реБга)............. 424 Quantities, Division of БЕЛЕ 424 Quantities, Multiplication of (Al- abito) mats: MMC on. „жа. 424 Queen-Post, Location of.... .... 212 Queen-Post іп Roof............. 54 Radials of Rib in Bridge........ 226 Radials of Rib for Wedges. ...... 226 Radicals, Extraction of (Algebra). 425 Radius of Arc, To Find..... — 561 Radius of Circle Defined........ 547 Rafters, Braces, etc., Length, To Find. onc и ы М: «d. - suas 580 Rafters, Least ЕТ Rule for... 62 Rafters, Length of, To Find...... » Rafters, Liability to Crush Other Materials... see. es e. ee ' 205 Rafters, Liability to Being Crushed 205 Rafters, Liability to Flexure.... Rafters, Minimum Thrust of..... . 205 62

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INDEX. қ РАСЕ Rafters in Roof, Effect of Weight Reflected Light, Opposite of оп........ Гб 5 179 Shade. . .... iino sucesso MN Rafters in Roof, Strains Subjected Regular Polygon in Circle, To In- ton лга IDEE V ges vs. 205 scribe (Сеоше{гу).............. Rafters and Tie-Beams, Safe | Regular Polygon Defined (Geom- о РОТИ EO S4 787] сету)... Жл Rafters, Uses in Roof........... 54| Regular Polygons, To Describe Rake in Cornice Matched with (Gegm ety). s uiii cf —" Level Mouldings.. „=. Sx, X 2... 344 | Resistance, Capability of —N Railing, Platform Stairs Rake Resistance to Compression, Ар- Meets dkowvel......... de 272| plication of. Ртеввше......7--. Ratio or Proportion, Equals Mul- Resistance to Compression, obl "WA c——-—— — 7... 367 Crushing and Bending........ Ratio or Proportion, Equality of Productse oL. жор, 22% Ratio or Proportion, Equality of ШАМОВ. .. ue. . „аараан аве Ratio or Proportion Equation, Ford Qd. ceu LL VU Ratio or Proportion, Examples... Ratio or Proportion, Four Propor- tionals, to Find Ratio or Proportion, Homologous JUD ML CREER EET IDEE Ratio or Proportion, Lever Prin- Оез... . г e a: Ratio or Proportion, Lever Prin- ciple Demonstrated... . ... ese. Ratio or Proportion, Multiply an Bu GTO aire «Sad EE Ratio or Proportion, Multiply and Divide One Number.......... Ratio or Proportion, Rule of Тітеек..-. кос. еккен Ratio or Proportion, Steelyard as СЕН Ж... e Ratio or Proportioa, Terms of Quantities Ratio or Proportion, Transfer a Factor .. 29. Әль dile Rectangle Defined Rectangle, Equilateral, scribe To De- 370 367 367 366 377 370 372 375 371 367 369 545 568 Beamec...k.cc o OG P ONE I45 Reduction Cubic Fect to Gallons Қам. ms an 653 Reduction Decimals, Examples: . 647 Resistance to Compression, Mate- Resistance to Compression, Pres- sure: Слабае... „о әби Resistance to Compression іп Proportion to Depth........... Resistance to Compression, Stout Posts иі саи: ..... 299 9. л Resistance to Compression, Table Resistance to Deflection, Rule.... Resistance Depending on Com- pactness and Cohesion......... Resistance Depending on Loca- tion, Soil, etc Resistance to Flexure Defined.. . Resistance Inversely in Propor- tion to Length Resistance to Oblique Force..... Resistance, Power of, How Ob- tained Resistance, Proportion to Area... Resistance, Strains, To What Due. Resistance to Tension Greatest in Direction of Length......... Resistance to Tension, Proportion in Materials Resistance to Tension, Table of Materials Resistance to Tension, Materials. Resistance to Tension, Results from Transverse Strains 675 PAGE 572 546 573 IOI 113 145 IO2 206 78 86 78 8r 8r 82 71 82

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676 PAGE Resistance to Transverse Strains, Table ӨР... 83 Resistance to Transverse Strains, · Description of Table. .... X. 84 Resistance Variable in One Ma- |o cM ЖЕ. E oc 79 Reticillated Walls. ж... 49 Rhomboid Defined........ ес... 9u6 Rhombus Defined.........:/.... 545 Ribbed Bridge, Area of Rule.... 225 Ribbed Bridge, Built....... ЖЕ... 224 Ribbed Bridge, Least Rise, Table 0f. 1.25. Т”. T AA. ы 224 Right Angle Defined......... 348, 544 Right Angle in Semicircle (Ge- ЛАР, соз ТЕН РО е 255 Right Angle, To Trisect а....... 554 Right Angled Triangle Defined .. 545 Right 'Angled Triangle, Squares on (Algera) We з... m, 417 Right Angled Triangles (Trigo- nomefty).-. ЖА. 510 Right Angled Triangles, Formula for (Trigonometry)............. 530 Right Lines (Geometry).......... 584 Right Line Equal Circumference. 566 Right Lines, Mean Proportionals ИННАА 584 Right Lines, Two Given, Find Third . ТЕГІ ГГ... | Right ог Straight Line nus. А. Right Prism Defined (Stairs)...... Risers, Number of, Rule to Ob- Сати CSE LL 246 Rise and Tread (Stairs)..:....... 241 Rise and Tread, Connection of (SEES T ee. 248 Rise and Tread, Blondel's Method of Binding (Stairs). ... Ж...... 242 Rise and Tread, Table of, аум Shops and Dwellings (Stairs)... Rise and Tread, To Obtain (Winds jog Stain’). : же. Аы... Rolled Iron Beams, Ө ышы» [75е INDEX. PAGE Roman Architecture, Ruins of.. IL Roman Architecture, Excess of Endo. fT. wa... estes. .. 46 Roman Building............. — О Roman Composite and Corinthian ) Duden... Eom oo cum LoT 28 Roman Decoration.............. 25 Roman Empire, Overthrow of.... 13 Romans, Ionic Order Modified by 27 Roman Moulding, Cavetto...... 329 Roman Mouldings, Comments on. 329 Roman Moulding, Ovolo........ 329 Roman Mouldings, Forms of..... 325 Roman Pantheon, etc., Walls of.. 49 Roman Styles of Architecture.... 26 Roman Styles Spread by the Ital- | p o——— Áo I3 Romantic or German Style, Thir- 4 . teenth and Fourteenth Centu- Meee 20. eee, II Rome, Ancient Buildings of...... 12. Rome and Greece, Architecture Sii. . DONNE CES... I LO. ME 8 EREC Mose eorr ret 54 Roofs, Ancient Norman апа Gothic, Construction .... а. 178 Roof Beams, Weight per Super- ficial eo) acter one. LUN 189 Roof, Brace in, Rule Same as for RIMET e o ie « Жие ee 208 Roofs, Construction of........ ev. 485 Roof Covering, Mode of......... 188 .| Roof Covering, Weights, Table of. 191 Roof, Curb or Mansard.......... 54 Roofs, Diagrams and Description "EL NN M o 398 uM. ЕТУІ. 212 Roof, Gothic Buildings....... TT З Roofs, Gothic and Norman Puild- ings, Соп5{гисйїоп............ 178 Roofs, Hip, Diagram and Expia- ЕЕ m SS +. 215 Roof, Aime жәнее. ж....... 54 Roof, Horizontal Pressure, То Re- move frontene.. Jaa ITa 74 Roof, Tack-Rafters and Purlins... 211 Roof, Kings Post im... -s sa. oor 54 Roof, Load per Foot Horizontal, Ruler ufo m. commos 192

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INDEX. 677 PAGE PAGE Roof, Load, Total per Foot Hori- Roof Truss, Load on........... . I89 zontal, Rule....... Ns 197 | Roof Trusses, Strains, Effect of, on Roofs, Modern, Trusts? Neces- Diffrent ^ т" m 179 5@йу..... Т... RD . 178 | Roof Truss, Weights, Table of, per Roofs, Norman and Gothic Build- | Superficial Poote ak ee 189 TABS. м.а. л... NER 178| Roof Truss, Weight per Superfi- Roof, Pent, To Pindi. 2220. $4| cial Foot. . 252 P 190 Roof, Perpendicular Height, To Roof, Trugsing.in. e eae. ce E 54 та... ОИЕ эч үй 579 | Roof Trussing, Designs for...... 178 Roof Plank, Weight per ее Roof Trussing, Framing for...... 259 fea! Foor. 27: .. Tror T 189 | Roof Trussing, Hogged Ridge.... 238 Roof, РЇаппїй@#а+.......... щл... 188 | Roof Trussing, King-Post, Effect Roof, Pressure оп............... 55| of Bad Framing on............ 237 Roof, Oween-Pogtin.. 2... ККЕ 54 | Roofs, United States........ ... 55 Roof Rafters im r © 7”. 27 54| Roof, Vertical Pressure of Wind Roof, Sagging, To Prevent....... 54 on; Bileet oft. ...... E .. 194 Roof, Slope Should Уагу Ассога- Roof, Snow, Weight per Horizon- їй@ она ешш t... es тот | гав w.. mr... ELS 193 Roof Supports. Distance between. 180 | Roof Weight on Rafter, Effect of.. 179 Roof, Suspension Rods, Safe Roof, Wind, Horizontal and Verti- BLA o M. 210| cal PRESORESL ..... d 193 Косев и іп.....;........ 54 | Roofing, Weight of Horizontal and Roof, Tie-Beam, Tensile Strain lU | GEM o 7 190 ШӨ. ЖИИ ..............:5 204; Roofing, Weight per Superficial Roof Timbers, Mortising ........ о ТТР... 190 Roof Timbers, Scarfing of........ 55 | Roots, Cubes, and Squares, Table Roof Timbers, Splicing of........ 55| CO OU ... 638-645 Roof Timbers, Strains by Parallel- Round Post, Area of............ до ogram of (огсез............... 198 | Rubble Walls............... eese 48 Roof Timbers, Strain Shown Ge- Rulers and Pencil in Drawing.... 540 ometrically.... esse ...... I99, 202! Rupture Compared with Flexure. 84 Roof Truss, Arched Ceiling...... 214 | Rupture, Crushing, Safe Weight.. 89 Roof Truss, Elevated Tie-Beam Rupture, Sliding, Safe Weight.... 57 ОъЪјес(іопаЫе................. 214 | Rupture, Transverse, Safe Weight. 86 Roof Truss, Elevating Tie-Beam, Rusting Iron Framing Straps, To Effect of. . sare e es Т" 187 Prevent.. i eS. с... .. 289 Roof Truss, Force Diagram, Figs. ‘ $9, 08, абу” LL 770 | Safe Load füt-Material......-.. 81 Roof Truss, Force Diagram, Figs. Safe Weight, Allowance for...... 84 60, 70, andy rema. г ы Же 180 | Safe Weight at Any Point, Rule.. ro6 Roof Truss, Force Diagram, Figs. Safe Weight, Beam at Middle.... 103 ОЧ and. 7579... . wed 181 | Safe Weight, Bending. ....... cee gI Roof Truss, Force Diagram, Figs. Safe Weight, Beam, Breadth of, 65, Гена 99 ..... 7. qM 183 To Find.. 47” : IO4 Roof Truss, Force Diagram, Figs. Safe Weight, p Deis To Ow, 77, and 78.00... „НОВ ТО oPind.... oso... 80.0. но ЖЕ 104 Roof Truss, Force Diagram, Figs. Safe Weight, Breadth or бесіне То 56, 80, add 81.2... TER 186| Find, Load at Middle.......... 106

Page 718

6;% WAGE Safe Weight, Болса ос Depth, Тә Riek Leal КИ А assan wr Safe Weight, Cranking Raptare.. Safe Weight Dotard, aee Safe Weight, жікіл, To Pre NOR AA ANANAS Or Grr Gre Gr ere Gr Ucar er are Safe Weighs, Ploar Beams, Dwell О 5 65s CSIR AA. Safe Weight, Floor Beams, Sates, Safe Weight, Load Exeulv Distr т СС Ee 7! Safe Weight, Lead рес pergol Feet ef Bho aa aae Safe Weight, Marin Greater than poo We HE eee Аё Safe Weight, Rafter aad Тіс. Бола, Бұаығыры..................... МУ Sy TO Safe Weight, Rectangular Dosts., 92 Safe Weight, Sender Ромз,..... Qt Safe Weight, Siding Raptate.... $7 Safe Weight, Staa at Middle of Бамы......................... Safe Weight, Suspeusioa Reds ta желі... Зе... Safe Weight, Tensile Sugin...... Safe Weight Teasile Strain, To ДОР 22. чи 92 Safe Weight Transverse Көрісе, 26 Seate ef Equal Pants, for aa TTL TE E T TEES Seale, Use of im Drawing. .....« Malene Triangle Deued, sss Seantling, Thickness for Domes . < Seeria, Classic Морі. о... 323 Scotia, Grecian Moulding. oo... Sectional Area, Explanation of, . . Sector, Area ia Cice. 0...0... 3476 д. Secter Defined. <<. X47 Segment of Cirale, Area, To Find. 479 | Segment of Circle De&ned....... 547 Segwmemt of Circle, To Describe (Geometry). eee eo $60, 862 Segment of Circle from Ordinates, 470 Segment of Cylinder Defined. .... S49 Semimajor Axes of Ellipse De- DOC CHEM 456 Shade Lining, Drawiug. .... e $43 Shadow on Capital of Column... 609 INDEX, WAGE Shadow of Cirewlar Abacus on EE сак wx Sy | Saadiow of Ком on Wall... баң Shadow Qu QUU. aa a A A A 6n Saadow in Ккерізсе............. 008 Shadow of Horisontal Beam... 604 Shadow, Inclination of the Line ef. коб xa, Shactow e£ Moulded Window Line ИВ... eee .5.4%%%4%.% “ж Shadow of Хом et Зер. Г%%%%% ое Shadow e£ Pedestal ow (ер... GOO Shadow of Projection en Crhindite CR LC стт с PRR OR ER IS без Shadow in DEM. сч, лб бочу Shadow in Recess, Back Vertical, Gag Shadows, Rettected Light Opposite Shadew and Shade, Distinction Betwenn aes SOF Shadeve e£ Shelf, , aas ЖЖ | Shadow of Shelf Acute Angled. 000 Shadow ef Shelf, Front Rege re en боз Shadow ef Shelf Inchued, oo... 6% Shadew of Shelf Tactined in Verti- Gal Эмма .................... ©з Shadow of Shelf ow асе Wall, боз | Shadow of Square Abacus on Coal. МАМ, т ххх %5%%4%%4%%.% бот Shadow on Straight Projections aud Mouldingg.... ees SEF | Shadow on Shelf o£ Uneven Width 199 Shadows, Usefulness ө4,......... 596 Shaft, Base aud Capital e£ Colum DARA. 5h. RR BE | Shaft e£ Colum. eee 47 Shaft, Upright Part o£ Column... 1% Shearing Strain, Tubulat Iron LU ANE TTE TETTE weds s OF Shutters, laside, for Windows, Requirements, as esses. 519 Sines and Tangeats, Logarithurs (Фу оз). aee ees 4644 State, Weight per Superfcial Feat. 189 Snow en Roof, Weight per Hort төлі Foot „<<<. 193 Snow on Roof, Weight per Super. ӛсімі КААР. oos MITES A. 190

Page 719

INDEX, 67 thi FAC бейи of Circular Headed Win- Stairs, Mandrailing, Toha Used.. 247 ал” ER o 421 fairs, Light and Уетіміт,.,.,,. 240 HANIA Defined (Geometry), o^ ooo,» 644) Mairs, lox Men and ЗУ отеп, Kise Solid "Timber Floors, FirstClass and Trotha, 7 peoia rur 243 Mores, Герін,,,,,,,,,,,, coor 144 Salts, Mosing and tread, Visi- BAIA Timber Floors, Dwellings бе ere o 241 and Амет! у, Гері, ,,.,.,.,.., 142 Mairs, Ри Moat, Тө Make,,,, 247 Solid Timber Floors, Fíte-Ftoof, Stairs, Pian et, Габтеа,,,.,,,,,,; %7 VO MEME. LL 222, ОАА 143 Maire, Лапе and Cylinder, LITT тит КОРР 52. бе Меобюа,,,,,,,,,,,,,,., 456) Spherical Dome, наре, Тө Vind, 224 Stairs, Patam, Valuster, Vosi- Splicing frente e гг гг * v FF 2% бет a Pe ТӘЙ “yy Splicing, Depth of Indems, Тө Stairs, Platform, VYenefieial,,,,.. 240 Иіні,,,,,,.,..,...,,,..,......?... 26 Фе, Patam, Curing "Twist Splicing, Еее ДАРЕТ CER 23» | RS eet tóstgtstoe t0] ttt 277 Splicing Мед Тітһеге,,,;,,,,,. ££ Stairs, Sathorn, Cylinder %,,,,. 24 Splicing or f«arfing Tie beat... 234 Stairs, Patam, Vare Maid Spring e db... лог, ОТ £2 Without Canting Рімее.,,,,.. 274 Sanare Пебтей,,.,,,..,.,,........ 645 шіт, Viator, Facs Mooid, Ар» Square er Cube Коса, Example, 644 | plication to ИіәтЕ,,,.,,,,,,,,,, 272 Squares, Cubes and Rons, "T ate |! Stairs, Flattorm, Vare Мой, Ар, "LM POPPPPPPPPM LITT plication fo Pank., RRR ERE E 275 Sunare Една Rectangle, To Make — Өзін, Plattorm, Face Movid for (беотелгу). . , „, , eet tecto $41 | ee ee 2% Square Equal Given Henares, To) | Staite, Paton, Vatt Мота, Make (Geotmetiyy, o ooo oo oou, GN Monlded Бай,,,,,.,,.,,,,..,,. 234 Square Vaqual "Triangle, То Make Salve, Viator, Vace Мой, Vitti- (O*ometty), , Lo oo o oo ote tton ga | ple Wothsd.,,.,..o o oec 257 Btability, Principles ОЕ, 2. рава 45 | Stairs, Fatto, Handrail 6,,,,;, 2f Stairs, Circular, V aee Mould, Virst Stairs, Platform, Handrail, Large бесгіов,..................... 2105.5... 0222212722222 с УА faits, Circular, Fare Mould tor Stairs, Viator, Wandishi Where eeteestecceceeeeecceeccba, 206, 252| Кабе Мода Іжбі.,,,.,,.,,,, 272 біліне, Circular, Valling Mould mairs, Plactoreen, Lower Edge A fer Жөй,,,...........,...... MB Суйтбек,..,............... -е Sains, Circular, Wand railing for. 27%! Hairs, Равот W reath fos Келет Stairs, Circular, Plath Ao cover cece 276 O^ „оа етой, арий 267 Mairs, Circular, Viumb Иеуеі De- Stairs, Position and Require. , | ee теегеггегегегғегеггегегғ 2 ou ЮЖ ОЛЛО гар Жаа 24/7 Stairs, Circular, Timbers put іп Mairs, Prisme and Cylinders... 267 Aher Биесйт,,,,,,,.,..,.... 252 Stirs, Rise and fieadh o ,,,,.., tht биліге for Dwellings, Еніс.,.,,,,, 244 Mairs, Rise and Treat, Mondes airs, Face Mould, Ке? Method о, ММ e AG for бітріс,,,,,,,,,,.,.,...... WA Wate, Kise and "Tread, Table for Stairs, Wandeailing, Geometry Shops and Поета. 244 Кесезау,.,,......,».»,,,..;.. 222! ales, Rises, Number of, Ием Stairs, Hard railing, ** Ot Aa Фжбә,,.......“ь»-аһеота. 24^ Wind” Вебтеб,.,.....,...,.,.. 257 fairs, Shops, Ries fot... oor 244

Page 720

12 900 T ETTTEETRRETERTT Y. 400 Stairs, Winding, Flyers and ` ha 00 OLET LETTERS SER. 251 Stairs, Winding, а” String, Grade of...... itte PRETE L- 253 Stairs, Winding, Handrailing.256, 289 Stairs, Winding, Handrailing, 15а1- usters UnderSeroll............ -310 Stairs, Winding, Handrailing, Centres for Square........... . 308 Stairs, Winding, Handrailing, Face Mould for Scroll.... .... 3II Stairs, Winding, Handrailing, Fall- ing Mould for Raking Scroll... 310 680 INDEX. PAGE | PAGE Stairs, Space for Timber and Plas- Stairs, Winding, Handrailing, Gen- Мы өтті. 70. ЖИНДИ 247 eral Considerations............ 258 Stairs, Stone, Public Building.... 240| Stairs, Winding, Handrailing, Stairs, String of, To Make....... 247 Serolis for. СИИИ... Ж 308 Stairs, Tread, То Find, Rise Given Stairs, Winding, Handrailing, —o 9)... 1:34 78 242, 246| Scroll Over Curtail Step....... 309 Stairs, Tread and Riser Connec- Stairs, Winding,, Handrailing, p MEE | MN 248| Scroll for Сипай Step......... 310 Stairs, Width, Rule (ог.......... 241 | Stairs, Winding, Scroll at Newel. 309 Stairs, Winding, Balusters in Stairs," Winding, Illustrations by Round аты... RU En | Plage, ПИРЕ. Е... 261 Stairs, Winding, Bevels in Splayed Stairs, Winding, Moulds for ток...) 319] Оламе ойо... 7. 1. 255 Stairs, Winding, Blocking Out Stairs, Winding, Newel Cap, Form Bell... 2... mU. 301 СИИ... не. „08. биму, хет. 312 Stairs, Winding, Butt-Joint on Stairs, Winding, Objectionable.. 240 Handra. E... Mitos. 303 | Stairs, Winding, Pitch Board, To Stairs, Winding, Butt-Joint, Cor- “ЛОР: La Aa S ПЕШКЕ 252 rect Fimo: m M... 2 307 | Stairs, Winding, Rise and Tread, Stairs, Winding, Diagrams Ex- о ОТ... а. 251 решена. 22...” UL 263 | Stairs, Winding, Sliding of Face Stairs, Winding, Face Mould, Ac- Моша.,..:. ла. EE : 299 cutacy:«of. "П... ..... 295 | Stairs, Winding, String, To Ob- Stairs, Winding, Face Моша, НІ MER. дэрет. гоа 252 Аррасабоп................... 297 | Stairs, Winding, Timbers, Posi- Stairs, Winding, Face Mould, NODOR. а ИЕ ЖЕ 252 Care in Drawifig-............. 295 | Stairs and Windows, How. Ar- Stairs, Winding, Face Mould, ССИ... ern. 42 иткес Elliptical.............. 301 | Stiles of Windows, Allowance for. 319 Stairs, Winding, Face Mould St. Mark, Tenth or Eleventh Cen- E recens er Mes AGO Жаа | Машу ж. JE. Lo гә... ЖЕ... и Stairs, Winding, Face: Mould, Stone Bridge Building, Truss Round КӨйІ................. 203 W ODE... ЕСЕР... 292 Stone Bridge, Building Arch.... 2 Stone Bridge, Centres for, Con- Бїт И... ЖИП, аш. i ves 229 Stone Bridge, Pressure on Arch Stones ИИИ... ПА 230 Stop for Doors... — екан. 317 Stores, Floor Beams, Safe Segi. r20 - Stores, Ordinary, Floor-Beams, Sizes; Тоне: . 0...2. 127 Stores, First-Class, Floor-Beams, Sizes, To Find..... w 128 St. Paul's, London, Dome of.... 54 St. Peter's, Rome, Fourteenth and Fifteenth Centuries..... ЖЕТ... 12 Straight or Right Line Defined... 544

Page 721

Strains, Cross, Resistance to..... Strains on Domes, Tendency of.. Strains Exceed Weights Strains, Graphic Representation.. Strain Greatest at Middle of Beam. Strains by Parallelogram of Бо... О ИТТИ 5 Strains, Practical Method cf De- Шар ТҰҒЫР ы Зе Strains of Rafter іп Roof. ........ Strains, Resistance, To What Due. Strain on Roof Timbers Shown Geometrically... ИШ. TS Strains on Roof Timbers Geomet- rically Applied Strains on Roof Timbers, Parallel- орап dé Forée$.... :::2. 5... Strain, Shearing, Tubular Iron Strain Unequal, Cause of..... оос Straps, Iron, Roof Truss Strassburg, Cathedral of........ 5 Strassburg, Towers of the Min- Strength and Stiffness of Mate- Dc cR TERT 711 oo Structure of Materials. .......... Struts Defined вше апа Ties......... eses me Struts and Ties, Difference Be- St. Sophia, Sixth Gentury.......: Stucco Cornice for Interior. ..... Studs and Joists Defined Styles, Grecian, Only Known by Subtangent, РагаЬо1а............ Subtangent of Ellipse Defined... Subtraction and Addition (Alge- Superficies Defined (Geometry)... Supports, Girders, Length, Rule.. Supports, Position of. ......... 5 Supports, Inclination of, Unequal. INDEX. 681 PAGE PAGE 77| Suspension Rods, Location іп 219 Roof......5... ыы КОШ. LG бі | Suspension Rods in Roof, Safe 165 Weight... 2--- 210 105 | Symbols Chosen at Pleasure (Al- PTA)... o o e dels er "CERE 395 165 | Symbols, Transferring (Algebra). 399 Systyle, Explanation of.......... 20 62 205 | Table of Circles 5 5 ....649-652 78 | Table of Сот(еше...........! 613—621 Tablé оҒ Capacity of Wells, Cis- 109! WEGBHSNEC... .............22 653 Table of Squares, Cubes, апа 202 ооо СВИ НИР. 638—645 Table of Woods, Description of.. 80 198 ; Tail-Beams Defined, ... ......... 130 Tanged Curve, To Describe (Ge- | 157 | onfetry). =... 83 | Tangent to Axes, Ellipse........ 485 239 | Tangent Әейлей............... 547 12 | Tangent with Foci, Ellipse....... 487 Tangent to Ellipse. To Draw..... 592 II; Tangent at Given Point in Cir- | ое... 2 Ж” 557 78. Tangent at Given Point, Without 76 | (Co - «1857 173 ' Tangent of Paravela ae ........ 493 68; Tangents and Sines, Logarithms : (Ре Оп)... .....-.-.- о. 464 69 ; Temples Built in the Doric Style. 10 12 | Temple, Doric, Origin of the..... 17 340 | Temple of Jupiter at Thebes..... 33 174! Tenons and Splices, Knowledge е Importanti. г.г... ттл 88 I6 | Tensile Strain, Area of Piece, To ВИП... РИ 99 14! Tensile Strain, Compressed Ма- | eral. Me УР”. 100 496 | Tensile Strain, Condition of Sus- Goo | pended PEE... Pesos 98 486.| Tensile Strain, Safe Weight... .. 96 Tensile Strain, Safe Weight, To Pa" "Oompüte...T e. eee 97 544 | Tensile Strain, Sectional Area, To | 157 Өег. 22-2 А ЖЕ 97 65 | Tensile Strain, Suspended, Мже- бо! terial Extended............... 100

Page 722

Ties and Struts, Principles of.... 68 Ties, Timbers in a State of Ten- Titus, Composite Arch of. ....... Trimmey, Breadth, To Find, Two Sets Tail-Beams. . . ue ..a- -. 682 INDEX. PAGE PAGE Tensile Strain on Tie-Beam in Roof Top Rail, Doors, Width, Rule.... 317 TRUES в» „5. жэ «сес: 8 204 | Torus, Classic Moulding........ 323 Tensile Strain, Weight of Suspend- Torus, Grecian Moulding........ 326 ей Рірдес...... — — 00 95 | Tower of Babel, History of. ... .. 5 Tensile Strength of Cast Iron.... 16r| Towers of the Minster, Strassburg. її Tension and Compression, Framed Transverse Axis Defined......... 548 Ginder ДИНИНЕН 174! Transverse Strains, Compressed Tension, Resistance to.......... 77 and Extended, Material....... IOO Tension, Resistance to, Table of Transverse Strains, Defined. ..... 99 Майегізів....................: 82 | Transverse Strains, Explanation Tension, Resistance to, Results of Тарс Шы. enr 22-2 IOI ОБРИ, ҮҮ pp ca 82 | Transverse Strains, Greater Tension, Resistance to, Proportion Strength of One Ріесе.......... IOI in Таене и: ... see . 5 8r| Transverse Strains, Neutral Line Tent, Habitation of the Shepherd. 13| Рейпей....................... 100 Testing Machine, Description in Transverse Strains, Proportion to LV ORSUCHEE "Тагы... uana - eet 80 | Breadth an.. жж we IOI Testing Materials, Hydraulic Transverse Strains, | Hatfield's, ЕОС 225.22. ма; ака 8о Reference to..80, I2I, 133, 138, Testing Materials, Dates of....... 80 143, 144, 145, 146, 148 Testing Materials, Manner of.... 80 | Transverse Strains, Resistance to, Tetragep Defined -............. BAG) Жамь оғ... азат ЖЕ 82 Tetragon, Radius of Circles Transverse Strains, Description of (Ро1уропв).................... ШІ) Мый T EST 84 Tetrastyle, Intercolumniation... 20 | Transverse Strains, Strength Di- Thebes, Thickness of Walls at... 33| minished by Division...... p sss 77, Thrust, Horizontal.... ......... 63 | Trapezoid Defined.............. 546 Thrust, Horizontal, Examples.... 64 | Trapezium Defined.............. 546 Thrust, Horizontal, Tendency of.. . 88 | Tread, То Find, Rise Given Tie-Beam in Ceiling, Load on... 190] (${айт5)..................... 242, 246 Tie-Beam and Rafter, Safe Weight. 87 | Tread and Nosing, Position of ТЕР бап іп Roof. ........ -oi 5E ae(Stais) .. 21 за”, meson 241 Tie-Beam in Roof, Tensile Strain. 204 | Tread and Rise, To Find, Winding Tie-Rods, Diameter, To Find.... 164| байв......................... 251 Tie-Rods, Floor Arches, Dwell- Tread and Rise, To Find, Blon- 11605.......................... 153 del's Method.. 2... quw e : Tie-Rcds, Floor Arches, First- Tread and Rise, ТаЬе {ог ашыр Class 5іогев.................. 153 | and Би сінеее............. 245 Tie-Rods, Wrought Iron......... I64 | Tread and Riser, Connection of Ties Бейпей.................... 1781 (fthius)...- 5... exon. - кон 248 Ties and Struts, To Distinguish.. 69 Triangle, Altitude of (Polygons). 442 Ties and Struts, Framed Girders.. 174 Triangles, Base, Formula for (Trig- ОПО) css: „ааа ьа 516 Triangle, Construct, a (Geometry). 587 Triangle, Construct Equal-Sided (Geometny) «d жағаны ve ... 575 Triangle Defined....... 545 Triangle, Examples (Geometry)... 350

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INDEX. 683 : PAGE PAGE Triangles, Equal Altitude........ 361 | Trigonometry, Right Angled Tri- Triangle Equal^Quadtanglet...... 359] "angles....... t" ome 510 Triangles, Equation of (Trigo- Trigonometry, Right Angled Tri- nometry)......... . ERRORS 515| angles, Third Side, To Find... 511 Triangles, Homologous (Geom- Trigonometry, Right Angled Tri- zo 1 "-- тал Е 362 angle, Forthula:.......22...2.. 530 Triangles, Hypothenuse, Formula Trigonometry, Tables... ... ж. 513 ВОГ. оа... ОСОНИИ 516 | Trigonometry, Triangles, Base, Triangles, pa To Find БонпШаяог.. 2)... TN 516 (Tfigonometfy)... ......2 n. € 518 | Trigonometry, Triangles, Equa- Triangles, Puy Peer d To Find tioñs ditt m M. LLL 515 (Trigonometty)........ —— .* 517 | Trigonometry, Triangles, Hypoth- Triangle or Set-Square in Draw- enüse»Potmila........-. .. . 516 imng........ 29. T 539 | Trigonometry, Triangles, Нуре Triangle ог Set-Square, Use of... 541| епиѕе, To Find.. „0... ae Triangles, Terms Defined (Trigo- Trigonometry, A MN ретіне кол). m..." ГИРИ 512) dicüler, "ТӨ Pind om... 7I «CCS Triangles, Three m^ Equal Trigonometry, Triangles, Terms Көтелтріе:.. ж. „ни: 254! DefÓfjyeUe ..... г... ж Ж 512 Triangles, Value of Sides (Trigo- Trigonometry, Triangles, Value of nomet УЙ е... -........ 5їб | Збери И... "EE 516 Trigon, Irregular Polygons (Ge- Trimmer ог Carriage Beam, Ge. 25 OR. IDEE 5461 Breadth, To Find............. 132 Trigon, Radius of Circle (Poly- Trimmer or Carriage Beams De- ОМГ II... 443| СТ... 130 Trigon, Rule (Polygons).......... 441 | Trimmer, One Header, Breadth, Trigonometry, Oblique Triangles, ' To Find, Dwellings and Stores. 133 ЕТте-АТИ ве... Mo- ОЕ 523| Trimmer, Well-Hole in Middle, Trigonometry, Oblique Triangles, ° Breadth, To Ғіпа.......-7... 156 dub 5S. ....: 4g... Im 521 | Trisect a Right Angle........... 554 Trigonometry, Oblique Triangles, Truss, Diagram ef............... 200 ШИС lass: . an eese 360. co 520 | Truss, Force Diagrams, Figs. 50, Trigonometry, Oblique Triangles, osand 00... 796.00. m8 T. 179 Second СЇ Әбу. ecaa 0. TN 522 Figs 60, ;0 End ]г............. 150 Trigonometry, Oblique Triangles, Figs: бт, 75 «60 798. ... 0...1 ISI Third Class®..%@ 5.7.04 олие 526 Figs. 63, 74 eid 75......... 7" 183 Trigonometry, Oblique Triangles, Figs. Өң, 77 a8d*98L...... м 184 Fourth Classe... T HR 528 Figs. 65; 78 dnd 795... а... 185 Trigonometry, Oblique Triangles, Figs. 66, 80 апа 815.3......... 186 Sines апа Sides. ...... 2”. 519 | Truss, Roof, Framing for........ 237 Trigonometry, Oblique Triangles, Truss, Roof, Iron Straps......... 239 Formula, First-.Class«........— 531| Truss, Weight, per Horizontal Trigonometry, Oblique Triangles, Foot, To Fides ...< Sent 192 Formula, Second Class........ 532 | Truss Work, Stone Bridge Build- Trigonometry, Oblique Triangles, ПРЕ cee fm E. 252 Formula, Third Class......... 534 | Trussing and Framing, Gravity Trigonometry, Oblique Triangles, and Resistance... Lc ome cvs 76 Formula, Fourth Class......-.. 534! Trussing Partitions, Effect of.... 175

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684 INDEX. PAGE PAGE Trussing Roofs, Effect of........ 178| Walls, Construction and Forma- T-Square, How to Make......... Бес | Hong. ..... ie... oe... ee 48 Tubular Iron Girder, Area of Bot- Walls, Eddystone and Bell Rock tom Flange, Dwellings. 37. МЫТ ө | Lighthouse&..7...........«* 48 Tubular Iron Girder, Area of Bot- Walls, Egyptian, Massiveness of.. 33 tom Flange, First-Class Stores. 160 | Walls, Modern Brick............ 49 Tubular Iron Girder, Arc of Walls of Pantheon and Roman Flange, Load at Middle........ кен Нет ЖО. .............. 49 Tubular Поп Girder, Area of Walls of Pantheon at Rome...... 53 Flange, Load Any Point....... 155 | Walls, Pisé, of France........... 49 Tubular Iron Girder, Area of Walls, Betienlated .......... ... 49 Flange, Load Uniform......... 135 | Walls, Rubble....... 9... 48 Tubular Tron Girder, Flanges, Walls, 5{теп@ хо... . ола. 7 48 Progortiónvof.. ....... Ж... 09 157 | Walls, Various Kinds............ 49 Tubular Iron Girder, Construction Walls, У©/оодеп................. 49 0f...... o o ...1. . T2 154 | Weakening Girder, Manner of... 140 Tubular Iron Girder, Rivets, Al- Web of Tubular Iron Girder, lowame@Mi@n, ......65.. 008. „8 157 Thicknesewf. ..$.............. 158 Tubular Iron Girder, Shearing Weight of Materials for Building ` Strain... MIEL эз... c ДИ 157 Паре О»... cae. cde ee. Ж 654-656 Tubular Iron Girders, Web of... 158 | Wells, Ciste:ns, etc., Table of Tuscan, Modern, Appropriate for ы E E 653 Buildings .. EE ........®:.: 30| White Pine, Weights of Beams Tuscan Order, Introduction of the. 30 2.015 15. 0 TERT T TT A „н 177 Tuscan Order, Principal Style in | Wind, Greatest Pressure, рег Su. ПРЕ... ЖТТ... P perfcial "Ғөше..... m. 90 Twelfth Century, Buildings in the. Wind on Roof, Effect of Vertical Twist Rail, Platform Stairs....... " ordo cop c MEM 202: 194 Twists, Stairs, Nicholson's Method Wind on Roof, Horizontal and E... E e sess. 250| Vertical Pressure......0.....: 193 — Winders in Stairs, How to Place Undeeagon Defined.............. "o mL T o 9 42 United States, Roofs in.......... 55 | Winders and Flyers, Stairs Менын 251 Windows, Arrangement of. ....... 44 Vault, Simple, Groined or Com- Windows, Circular Headed...... 320 plex... ОТТІ... 52| Windows,. Circular Headed, To Ventilation, Proper Arrangement Eo 1 OMNEM 321 Тос сәт. E. жкезбек..2-.% 45 | Windows, Dimensions, То Find. 318 Versed Sine of Arc, To Find..... 561 *Window-Frame, ОӨЖ, „ита, ж. 318 Vertical Pressure of Wind оп Windows, Front of Building, Ef- ROSE... Ж --...2..... 194| ЖЕсі сой қыт... 320 Vertical Tangent of Parabola De- Windows, Heights, Table of, fined әк. OP. p 495 | WY (GUNG TOI ass. а acres a. 320 Volutes, To Describe the........ 20| Windows, Height from Floor.... 320 Voussoir of an Агсһ............. зе нде, Inside Shutters, Re- | quirement. Paret .. 6.0... 2. swam iig Wall, ЕБ симас ау ..... 48 | Windows, Position and Light Walls, GOL... oe e es ша... " EE o —-— ws . 317

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INDEX. 685 PAGE PAGE Windows and Stairs, How Ar- Winding Stairs, Handrailing, pal- manped. oo. ais cose. RES. 42| usters Under ЗӨВ... 310 Windows, Stiles, Allowance for.. 319 | Winding Stairs, Handrailing, Cen- Windows, Width Uniform, Height tres in Square... . . . Ж eee ОЕ 308 Қызы ле... 319 | Winding Stairs, Handrailing, Face Winding Stairs, Balusters in Mould for,Scroll.- 2 T 31I Bold: Ы... . .......:.. 313 | Winding Stairs, Handrailing, ҒаП- Winding Stairs, Bevels in Splayed ing Mould. о eee 310 ТОС. та oe ms ccm 314| Winding Stairs, ней ы, Winding: Stairs, Blocking Ош General Considerations........ 258 ШШЕ... „о.о ELDER - 301 | Winding Stairs, Handrailing, Winding Stairs, Butt Joint, Posi- Scroll г... р. 308 ПОНОС... ГЕ 303 | Winding Stairs, Handuaailing, Winding Stairs, Butt Joint....... 307| Scroll Over Curtail Step....... 309 Winding Stairs, {Diagram of, Ex- Winding Stairs, — Handrailing, | plained......... XML. s 263| Scroll for Curtail Step......... 310 Winding Stairs, Face Mould for Winding Stairs, — Handrailing, Жаа а 5 290, 293 Scrolls at ХЕФе!.......... С. Winding Stairs, Face Mould, Ас- Winding Stairs, авоии by curacy dí... o. 22... adr; Planes.......-... Ructe MM 261 Winding Stairs, Face Mould, Ap- Winding Stairs, Moulds for Quar- plication Of... os. as esmine.. : 297| tem@ircle..7....... я 255 Winding Stairs, Face Mould, Winding Stairs, Newel Cap, Form Curves Blliptical............-. 301 Ө........... ООС 312 Winding Stairs, Face Mould, Winding Stairs Objectionable.... 240 Drawing..... ССТТ. 296 | Winding Stairs, Pitch Board, То Winding Stairs, Face Mould, OBSin............. 22921 — 252 Балта а... о... 303 | Winding Stairs, Rise and Tread, Winding Stairs, Face Mould, Slid- To Obtain.. . 251 e UNTER 70701 « 299 | Winding Sues, —3 ‘To Онин. 252 Winding Stairs, Face Моша i Winding Stairs, ье. Posi- АДНЫ... Зь. 05.0522. II eor) iden oL....... ЕТ 252 Winding Stairs, Flyers and Wind- Wood, Destruction by Fire...... 37 efs... за aet. 0... ee 251 | Wooden Beams, Use Limited...... 154 Winding Stairs, Front String, Woods, Hydraulic Method of Grade Uf, i ТИРИ ағ. 253 а oC EC 80 Winding Stairs, — Handrailing Wreath for Round Rail, Platform ТОРИ Р occu Ж” Staits ...... E. И dE ае .-» 664 THE END.

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