HIght. Area Seg. Hght. Area Seg. Hght. Area Seg Hght. Area Seg. Hght. Area Seg. Ight. Area Seg. 331850 127 057991190 103900 253 156149 315 212011 377 270951439 128 058658191104685 254157019 316 212940 378271920 440 332843 129 059327 192 105472 255 157890 317 213871 379 272890 441 333836 130 059999193 106261 256158762 318 214802 380 273861 442 334829 131 060672194 107051257159636 319 215733 381 274832 443 335822 132 061348·195 ·107842 258 160510320216666 382 275803 444 336816 133 062026 196 108636 259 161386 321217599 383 276775 445 337810 134 062707-197 -109430 260 162263 322218533 384 277748 446 338804 135 063389 198 110226 261 1631 40 323 219468 385 278721 447 |·136 064074-199·111024 262 164019 324 220404 386 279694448 137 064760 200 111823 263 164899 325 221340387 280668 449 138 065449-201·112624 264165780 326 222277 388 139 066140 202 ·113426 265 166663 327 223215 389 140 066833 203·114230 266 ·167546 328 224154390 283592452 339798 340793) 341787 281642 450 342782 343777 344772 284568453 345768 285544 454 346764 286521455 347759 287498 456 348755 457 | 349752 458 350748| 459 351745 288476 291 411 141 067528204 115035 267·168430 329 225093 391 142 068225205·115842268·169315 330 226033 392 ∙143 068924 206·116650 269 170202 331 226974 393 •144 069625207·117460 270 171089 332 227915 394 ∙145 070328208118271 271·171971 333 228858 395 146 071033209·119083 272 172867 334 229801 396 147 071741| 210-119897273 173758 335 230745 397 148 072450 211120712274 174649 336 231689 398 460 | 352742 149 073161 2121 21 529 275 175542 337 232634 399 292390 461 | •353739 150 073874 213·122347 276·176435 338 233580 400 293369 462 354736 151 074589 214·123167 277 177330 339 234526 401 | ·294349 463 355732 152 075306 215123988278 178225 340 235473 402 295330 464|356730| 153 076026 216 124810279·179122 341 | 236421403296311 465 357727 154 076747 217125634 280 180019 342 237369 404 297292 466 358725 155 077469 218·126459 281180918 343 238318405 298273 467 ⚫359723 156 078194 219 127285 282 181817 344 239268406299255 468 360721 157 078921 220 128113283 182718 345 240218 407 300228 469 ⚫361719 158 079649 221128942284·183619346 241169408 301220 470 362717 159 080380 222 129773 285 184521 347 242121 409 302203471|363715 160 081112 223 130605 286 185425 348 243074 410 303187 472 ·364713 161 081846 224 131438 287 186329 349 244026 411 304171 473365712 162 082582 225·132272288 187234 350 244980412·3051 55 474366710 163 083320 226 133108 289 188140 351 245934 413 306140 475 367709 164 084059 227 133945 290 189047352 246889414·307125 476 368708 •165 084801 228 134784 291 189955 353 247845 415 308110477 369707 166 085544 229 135624 292190864 354 248801|416309095478 370706 167 086289 230 136465 293 191775 355249757 417 310081 479 371704 168 087036 231 137307 294192684 356 250715 418 311068 480 372704 169 087785 232 138150295193596 357|251673419312054 481|373703| ∙170 088535 233138995 296 194509 358 952631 | 420 313041482 374702 171 089287 234 139841 297195422 359 253590421 314029 483 375702 172 090041 235 140688 298·196337 360 254550 422 315016 484376702 173 090797 236 141537 299 197252 361 255510423316004 485 377701 174 091554237·142387 300 ⚫198168362256471 424316992486 | 378701 175 092313 238 143238301·199085363257433 425 317981 487 379700 176 093074239 144091 302 200003 364 258395426 318970 488 380700 177 093836 240·144944203 200922 365 259357 427 319959 489 381699] 178 094601 241 145799 304 201841 366 260320 428 320948490 382699 179 095366242·146655 305 202761367 261284 429 321938 491 383699| •180 096134243·147512 306 203683 368 181 096903 244148371 307 204605 369 182 097674 245 ⚫149230 308 205527 370 183 098447 246 150091 309 206451 371 184 099221 247 150953 310 207376 372 185 099997 248151816 311 208301 373 267078 186 100774 249152680 312 209227 374268045 187101 55325) ·153546 313 210154 375 269013437 188 102334 251154412 314 211082 376 269982438 350858·500|·392699 189 103116 252 155280 262248 430 322928 492384699| 435 436 324909494 | 386699| 325900 495 387699| 326892496388699| 327882497-389699| 328874498390699 329866 499 391699| 1226. PROBLEM XIII. To find the area of an ellipsis. Rule. Multiply the longest and shortest diameter together, and their product by 7854, which will give the area required. This rule is founded on Theorem S. Cor. 2. in Conic Sections. (1098, 1100.) Example. Required the area of an ellipse whose two axes are 70 and 50. Here 70 x 50 x 7854=2748'9. 1227. PROBLEM XIV. To find the area of any elliptic segment. Rule. Find the area of a circular segment having the same height and the same vertical axis or diameter; then, as the said vertical axis is to the other axis (parallel to the base of the segment), so is the area of the circular segment first found to the area of the elliptic segment sought. This rule is founded on the theorem alluded to in the previous problem. Or, divide the height of the segment by the vertical axis of the ellipse; and find in the table of circular segments appended to Prob. 12. (1224.) the circular segment which has the above quotient for its versed sine; then multiply together this segment and the two axes of the ellipse for the area. Example. Required the area of an elliptic segment whose height is 20, the vertical axis being 70, and the parallel axis 50. Here 20÷702857142, the quotient or versed sine to which in the table answers the segment 285714. Then 285714 x 70 × 50 = 648 342, the area required. 1228. PROBLEM XV. To find the area of a parabola or its segment. Rule. Multiply the base by the perpendicular height, and take two thirds of the product for the area. This rule is founded on the properties of the curve already described in conic sections, by which it is known that every parabola is of its circumscribing parallelogram. (See 1097.) Example. Required the area of a parabola whose height is 2 and its base 12. MENSURATION OF SOLIDS. 1229. The measure of every solid body is the capacity or content of that body, considered under the threefold dimensions of length, breadth, and thickness, and the measure of a solid is called its solidity, capacity, or content. Solids are measured by units which are cubes, whose sides are inches, feet, yards, &c. Whence the solidity of a body is said to be of so many cubic inches, feet, yards, &c. as will occupy its capacity or space, or another of equal magnitude. 1230. The smallest solid measure in use with the architect is the cubic inch, from which other cubes are taken by cubing the linear proportions, thus, 1728 cubic inches=1 cubic foot; 1231. PROBLEM I. To find the superficies of a prism. Multiply the perimeter of one end of the prism by its height, and the product will be the surface of its sides. To this, if wanted, add the area of the two ends of the prism. Or, compute the areas of the sides and ends separately, and add them together. Example 1. Required the surface of a cube whose sides are 20 feet. Here we have six sides; therefore 20 × 20 × 6=2400 feet, the area required. Example 2. Required the surface of a triangular prism whose length is 20 feet and each side of its end or base 18 inches. Here we have, for the area of the base, 1·52 — ·752 = (2·25 — ·5625 = ) 1·68752 for the perpendicular of triangle of base; and √1.6875=1.299, which multiplied by 1.5=1.948 gives the area of the two bases; then, 3 x 20 x 1 ·5 + 1 ·948 = 91 ·948 is the area required. Example 3. Required the convex surface of a round prism or cylinder whose length is 20 feet and the diameter of whose base is 2 feet. Here, 2 x 3.14166*2832, and 3·1416 × 20=125 664, the convex surface required. 1232. PROBLEM II. To find the surface of a pyramid or cone. Rule. Multiply the perimeter of the base by the length of the slant side, and half the product will be the surface of the sides or the sum of the areas of all the sides, or of the areas of the triangles whereof it consists. To this sum add the area of the end or base. Example 1. Required the surface of the slant sides of a triangular pyramid whose slant height is 20 feet and each side of the base 3 feet. Here, 20 x 3 (the perimeter) × 3÷2=90 feet, the surface required. Example 2. Required the convex surface of a cone or circular pyramid whose slant height is 50 feet and the diameter of its base 8 feet. Here, 85 x 31416 × 50÷2=667.5, the convex surface required. 1233. PROBLEM III. To find the surface of the frustum of a pyramid or cone, being the lower part where the top is cut off by a plane parallel to the base. Rule. Add together the perimeters of the two ends and multiply their sum by the slant height. One half of the product is the surface sought. This is manifest, because the sides of the solid are trapezoids, having the opposite sides parallel. Example 1. Required the surface of the frustum of a square pyramid whose slant height is 10 feet, each side of the base being 3 feet 4 inches and each side of the top 2 feet 2 inches. Here, 3 feet 4 inches x 413 feet 4 inches, and 2 feet 2 inches x 4 =8 feet 8 inches; and 13 feet 4 inches + 8 feet 8 inches = 22. Then 222 x 10-110 feet, the surface required. Example 2. Required the convex surface of the frustum of a cone whose slant height is 12 feet and the circumference of the two ends 6 and 8.4 feet. Here, 6 +8·4=14·4; and 14·4 × 12·5÷2=180÷2=90, the convex surface required. 1234. PROBLEM IV. To find the solid content of any prism or cylinder. Rule. Find the area of the base according to its figure, and multiply it by the length of the prism or cylinder for the solid content. This rule is founded on Prop. 99. (Geometry, 980.). Let the rectangular parallelopipedon be the solid to be measured, the small cube P (fig. 518.) being the measuring unit, its side being 1 inch, 1 foot, &c. Let also the length and breadth of the base ABCD, and also let the height AH, be divided into spaces equal to the side of the base of the cube P; for instance, " here, in the length 3 and in the breadth 2, making 3 times 2 or 6 squares in the base AC each equal to the base of the cube P. It is manifest that the parallelopipedon will contain the cube P as many times as the base AC contains the base of the cube, repeated as often as the height AH contains the height of the cube. Or, in other words, the contents of a parallelopipedon is found by multiplying the area of the base by the altitude of the solid. And because all prisms and cylinders are equal to parallelopipedons of equal bases and altitudes, the rule is general for all such solids whatever the figure of their base. Example 1. Required the solid content of a cube whose side is 24 inches. Here, 24 x 24 × 24=13824 cubic inches. D Fig. 518. P Example 2. Required the solidity of a triangular prism whose length is 10 feet and the three sides of its triangular end are 3, 4, and 5 feet. Here, because (Prop. 32. Geometry, 907.) 32 + 42=5%, it follows that the angle con 3x4 tained by the sides 3 and 4 is a right angle. Therefore x 10 60 cubic feet, the content required. Example 3. Required the content of a cylinder whose length is 20 feet and its diameter 5 feet 6 inches. Here, 5.5 x 5.5 × 7854=23·75835, area of base; and 23-75835 × 20=47′5167, content of cylinder required. 1235. PROBLEM V. To find the content of any pyramid or cone. Rule. Find the area of the base and multiply that area by the perpendicular height. One third of the product is the content. This rule is founded on Prop. 110 (Geometry, 991.) Example 1. Required the solidity of the square pyramid, the sides of whose base are 30, and its perpendicular height 25. Example 2. Required the content of a triangular pyramid whose perpendicular height is 30 and each side of the base 3. and 4.5-3=1.5, one of the three equal remainders. (See Trigonometry, 1052.) but 45 x 15 × 1·5 × 1·5 × 30÷3=3·897117 x 10, or 38.97117, the solidity required. Example 3. Required the content of a pentagonal pyramid whose height is 12 feet and each side of its base 2 feet. Here, 1-7204774 (tabular area, Prob. 6. 1218.) × 4 (square of side) = 6·8819096 area of base; and 6.8819096 × 12=82-5829152. Example 4. Required the content of a cone whose height is 10 feet and the circumference of its base 9 feet. Here, 07958 (Prob. 9. 1222.) × 81=6·44598 area of base, And 3-5 being of 10 feet, 6·44598 × 3·5=22·56093 is the content required. 1236. PROBLEM VI. To find the solidity of the frustum of a cone or pyramid. Add together the areas of the ends and the mean proportional between them. Then taking one third of that sum for a mean area and multiplying it by the perpendicular height or length of the frustum, we shall have its content. depends upon Prop. 110. (Geometry, 991.). This rule It may be otherwise expressed when the ends of the frustum are circles or regular polygons In respect of the last, square one side of each polygon, and also multiply one side by the other; add the three products together, and multiply their sum by the tabular area for the polygon. Take one third of the product for the mean area, which multiply by the length, and we have the product required. When the case of the frustum of a cone is to be treated, the ends being circles, square the diameter or the circumference at each end, and multiply the same two dimensions together. Take the sum of the three products, and multiply it by the proper tabular number, that is, by 7854, when the diameters are used, and 07958 when the circumferences are used, and, taking one third of the product, multiply it by the length for the content required. Example 1. Required the content of the frustum of a pyramid the sides of whose greater ends are 15 inches, and those of the lesser ends 6 inches, and its altitude 24 feet. and 8125 × 24 (altitude)=19.5 feet, content required. Example 2. Required the content of a conic frustum whose altitude is 18 feet, its greatest diameter 8, and its least diameter 4. Here, 64 (area gr. diam.) +16 (less. diam.) + (8 × 4)=112, sum of the products; Example 3. 3 Required the content of a pentagonal frustum whose height is 5 feet, each side of the base 18 inches, and each side of the upper end 6 inches. Here, 152 +1.52 + (1·5 × ·5)=2.5625, sum of the products; 1-7204774 (tab. area) x 2 5625 (sum of products) x 5 3 but, 9.31925, content required. 1237. PROBLEM VII. To find the surface of a sphere or any segment of one. Rule 1. Multiply the circumference of the sphere by its diameter, and the product will be the surface thereof. This and the rules in the following problems depend on Props. 113. and 114. (Geometry, 994, 995.), to which the reader is referred. Rule 2. Rule 3. Square the diameter, and multiply that square by 3·1416 for the surface. Square the circumference, then either multiply that square by the decimal 3183, or divide it by 3.1416 for the surface. Remark. For the surface of a segment or frustum, multiply the whole circumference of the sphere by the height of the part required. Example 1. Required the convex superficies of a sphere whose diameter is 7 and circumference 22. Here, 22 x 7=154, the superficies required. Example 2. Required the superficies of a sphere whose diameter is 24 inches. Example 3. Required the convex superficies of a segment of a sphere whose axis is 42 inches and the height of the segment 9 inches. Here, 13-1416::42: 131 9472, the circumference of the sphere; Example 4. Required the convex surface of a spherical zone whose breadth or height is 2 feet and which forms part of a sphere whose diameter is 12 feet. Here, 13-1416::12·5; 39-27, the circumference of the sphere whereof the zone is a part; and 39.27 × 2=78.54, the area required. 1258. PROBLEM VIII. To find the solidity of a sphere or globe. Rule 1. Multiply the surface by the diameter, and take one sixth of the product for the content. Rule 2. Take the cube of the diameter and multiply it by the decinal 5236 for the content. Example. Required the content of a sphere whose axis is 12. 1239. PROBLEM IX. To find the solidity of a spherical segment. Rule 1. From thrice the diameter of the sphere subtract double the height of the segment, and multiply the remainder by the square of the height. This product multiplied by 3236 will give the content. Rule 2. To thrice the square of the radius add the square of its height, multiply the sum thus found by the height, and the product thereof by 5236 for the content. Example 1. Required the solidity of a segment of a sphere whose height is 9, the diameter of its base being 20. Here, 3 times the square of the radius of the base =300; and the square of its height =81, and 300 +81=381; but 381 x 9-3429, which multiplied by 5236=1795-4244, the solidity required. Example 2. Required the solidity of a spherical segment whose height is 2 feet and the diameter of the sphere 8 feet. Here, 8 x 3-4=20, which multiplied by 4=80; and 80 x 5236-41888, the solidity required. It is manifest that the difference between two segments in which the zone of a sphere is included will give the solidity of the zone. That is, where for instance the zone is included in a segment lying above the diameter, first consider the whole as the segment of a sphere terminated by the vertex and find its solidity; from which subtract the upper part or segment between the upper surface of the zone and the vertex of the sphere, and the difference is the solidity of the zone. The general rule to find the solidity of a frustum or zone of a sphere is to the sum of the squares of the radii of the two ends add one third of the square of their distance, or the breadth of the zone, and this sum multiplied by the said breadth, and that product again by 1.5708, is the solidity. SECT. VII. MECHANICS AND STATICS. 1240. It is our intention in this section to address ourselves to the consideration of mechanics and statics as applicable more immediately to architecture. The former is the science of forces, and the effects they produce when applied to machines in the motion of bodies. The latter is the science of weight, especially when considered in a state of equilibrium. 1241. The centre of motion is a fixed point about which a body moves, the axis being the fixed line about which it moves. 1242. The centre of gravity is a certain point, upon which a body being freely suspended, such body will rest in any position. 1243. So that weight and power, when opposed to each other, signify the body to be inoved, and the body that moves it, or the patient and agent. The power is the agent which moves or endeavours to move the patient or weight, whilst by the word equilibrium is meant an equality of action or force between two or more powers or weights acting against each other, and which by destroying each other's effects cause it to remain at rest. PARALLELOGRAM OF FORCES. 1244. If a body D suspended by a thread is drawn out of its vertical direction by an horizontal thread DE (fig. 519.), such power neither increases nor diminishes the effort |