3661.3258 465 1·5022 466 1.5042 393 1-3711-429 1-4349 | 472 15157 473 15176 438 1'4514 474 1.5196 439 14533 475 1.5215 440 14551 476 1.5235 4771-5254 320 1.2539 357 1-3112 321 1-2554 358 1.3128 ·322 1·2569 359 1.3144 323 1.2584 360 1.3160 324 1.2599 361 1.3176 397 1.3780 325 1.2614 362 1.3192 :398 1.3797 326 1.2629 363 1.3209 399 1.3815 435 1.4459 327 1.2644 364 13225 400 1.3832 .436 1 4477 328 1.2659 365 1.3241 401 13850 .437 1.4496 329 1.2674 402 1.3867 330 1.2689 403 1.3885 404 13902 -4051-3920 406 13937 -442 14588 478 1-5274 407 13955 443 14607479 1-5293 408 1.3972 444 1'4626480 1.5313 409 1.3990 445 1'4644 481 1.5332 4101 4008 446 1'4663 482 1-5352 411 14025 447 14682 483 1-5371 412 1·4043 448 14700 || 484 1.5391 413 14061 449 14719 414 14079 450 14738 415 14097 451 1'4757 416 14115 || 452 1.4775 488 1.5470 417 14132 453 14794 489 1-5489 418 14150 454 14813 490 1.5509 419 1-4168 455 14832 || 491 1 5529 4201·4186 456 14851 4921-5549 421 1-4204 457 14870 4931-5569 331 1.2704 368 13291 332 1.2720 369 1.3307 333 12735 370 1.3323 334 12750 371 13340 335 12766 372 13356 336 12781 37313373 337 12786|| 374 13390 338 12812 375 13406 339 12827 376 13423 || 340 12843 377 13440 341 12858 378 13456 342 12874 379 13473 343 12890 38013490 344 12905 381 13507 345 12921 38213524 346 12937 38313541 485 1-5411 4861-5430 487 1.5450 496 1-5628 347 12952 384 13558 3431'2968 385 1 3574 34912981 386 13591 422 14222458 14889 494 1.5585 350 13000 387 13608 423 1-4240 459 1'4908 495 1.5608 351 13016 388 13625424 1.4258 460 1.4927 352 13032 389 13643 425 1-4276 461 1-4946 353 1-3047 390 13660 426 1.4295 462 1.4965 354 13063 3911'3677 ·427 1·4313 463 1-4984 3551-3079 392 13694 || 428 1·4331 464 1.5003 500 1.5708 3561.3095 497 1.5648 498 1-5668 499 1-5688 To find the Length of an Arc of a Circle by the foregoing Table. Rule.-Divide the height by the base, and the quotient will be the height of an arc, of which the base is unity. Seek in the table for a number corresponding to the quotient, and take the length of that height from the next right-hand column. Multiply the number, thus found, by the base of the arc, and the product will be the length of the arc or curve required. Example. The profiles of the intradoses of the arches of a bridge are each a semi-ellipse; the span of the middle arch is 150 feet, and the height 38 feet: required the length of the curve. 38÷150953, and 253, as per table=1.1628. Hence, 11628 × 150=174-4200, the length required. TABLE Of the Proportions of the Lengths of Semi-elliptic Arcs Height Length Length Height Length Height Length Height 635 1.7850820 100 10416265 1-2306450 1.4931 Length of Arc. 2.0971 2.1060 2.1148 2.1237 2.1326 2.1416 2.1505 2:1595 2.1685 3.1775 2.1866 2 1956 2.2047 2'2139 2:2230 2.2322 2'2414 2'2506 2'2597 2'2689 2'2780 2'2872 2 2964 2'3056 2'3148 2'3241 2 3335 2'3429 2'3524 2.3619 2'3714 2'3810 2'3906 2'4002 2'4093 To find the Length of the Curve of a Right Semi-ellipse. Rule. The rule for circular arcs in the preceding table is equally applicable here. The two last tables are not entirely confined to works which may be carried into practice, but are useful in estimating, to a very minute degree of accuracy, the quantity of work which is to be executed from drawings to a scale. As the tables, however, do not afford the means of finding the lengths of the curves of elliptic arcs, which are less than half of the entire figure, the following geometrical method is given to supply the defect. To find the Length of an Elliptic Curve, which is less than half the figure. Let the curve, of which the length is required to be found, be a b c. Produce the versed sine, b d, to meet the centre of the curve in e. Draw the right line, c e, and from the centre, e, with the distance, e b, describe an arc, b h. Bisect c h in i, and from the centre, e, with the radius, e i, describe the arc, i k, meeting e b produced to k; then, ik is half the arc a b c. NOTE. When the quotient is not given in the column of heights, divide the difference between the two nearest heights by 5; multiply the quotient by the excess of the height given, and the height in the table first above it, and add this sum to the tabular area of the least height. = = 00515 00102 × (118115) 00306, which, added to 10567,105976, the length for 118. · Haswell. SECTION III. OF SOLIDS BOUNDED BY PLANE SURFACES. The mensuration of solids is divided into two parts. I. The mensuration of the surfaces of solids. II. The mensuration of their solidities. The measure of any solid body is the whole capacity or contents of that body, when considered under the triple dimensions of length, breadth, and thickness. A cube, whose side is one inch, one foot, or one yard, &c., is called the measuring unit; and the contents or solidity of any figure is computed by the number of those cubes contained in that figure. DEFINITIONS. 1. A cube is a right prism, bounded by six equal square faces, of which any two, opposite to each other, are parallel. 2. A parallelopiped is a prism bounded by six quadrilateral planes, every opposite two of which are equal and parallel. 3. A prism is a solid, whose ends are parallel, similar, and equal, and the sides connecting these are parallelograms. 4. A pyramid is a solid, whose base is any plane figure, and whose sides are triangles, having all their vertices meeting together in a point above the base, called the vertex of the pyramid. 5. A frustrum or trunk of a pyramid is a portion of the solid that remains after any part has been cut off parallel to the base. 6. A wedge is a solid of five sides, two of which are rhomboidal, and meet in an edge, a rectangular base, and two triangular ends. 7. A prismoid is a solid, whose ends or bases are parallel, but not similar, and whose sides are quadrilateral. OF CUBES AND PARALLELOPIPEDS. PROBLEM I. To find the Lateral Surface of a Prism. Rule.-Multiply the perimeter of the base into the altitude, and the product will be the convex, or lateral surface. When the entire surface of the prism is required, add to the convex surface the area of the bases. Example.-Required the lateral surface of a prism whose base is a regular hexagon, and whose sides are each 2 feet 3 inches, the height being 11 feet? 2 ft. 3 in.=27 in. and 27×6=perimeter of the base. U ft. =132 inches height. Then, 132 × 162=21384 square inches. 21384 144-148 50 sq. ft. Ans. e PROBLEM II. To find the Solidity of a Cube or Right Prism. Rule.-Multiply the area of the base by the perpendicular height, and the product will be the solid contents. NOTE. The capacity of a vessel, in gallons or bushels, of any given dimensions, may be readily ascertained by calculating its contents in inches, and then dividing the contents by the number of cubic inches in one gallon or bushel. Examples.-1. Required the number of ale gallons there are in a cistern which is 6 feet 8 inches deep, and whose base is 5 feet 4 inches square? 6 ft. 8 in. 80 in. Then, 642=4096, and 2. What is the solidity of a prism of granite, 9 feet 2 inches long, and 16 by 12 inches side dimension, and what will be its weight, reckoning 169 lbs. to the cubic foot? 110 in. = length. 9 ft. 2 in. area of base. 192×110=21120 solidity in in. 21120÷1728 12 22 cubic ft. Ans 12.22 × 169=2065 lbs. Ans. OF PYRAMIDS. PROBLEM III. To find the Lateral Surface of a Regular Pyramid. Rule.-Multiply the perimeter of the base oy the slant height, and half the product will be the surface. If the whole surface be required, add to this the area of the base. Example.-What is the lateral surface of a regular triangular pyramid, a b c, whose slant height, d a, is 20 feet, and the sides of whose base are each 8 feet? To find the Lateral Surface of the Frustrum of a Regular Pyramid. Rule.-Multiply the perimeters of the two ends by the slant height of the frustrum, and half the product will be the surface required. To this add the surface of the two ends when the entire surface is required. |