Prob. 2. The linear dimension, or solidity, of a body being given, to find its weight. Multiply the cubic inches contained in the body, by the tabular weight corresponding to the name of the same kind, and the product will give the weight in pounds avo.rdupois. Er. What is the weight of a piece of oak, of a rectangular form, whose leogth is 50in. the breadth 18in. and the depth 12in. ? Now 56 x 18 x 12-12096 inches. OF THB FIVE REGULAR SOLIDS. Definitions. 1. A regular solid, is a body that either may be incribed or circumscribed by a sphere, in such a manner as to be contained under equai and similar planes ; alike posited, and equally distant from the centre of the sphere. 2. Tlie Tetraedron, is contained under four equilateral triangles. 3. The Heraedron, is contained under six equal squares. 4. The Octaedron, is contained under eight equilateral triangles. 5. The Dodecaedron, is contained under twelve equilateral and equiangular pentagons. 6. The Icosaedron, is contained under twenty equilateral triangles. Prob. 1. To find the superficies, and solidity, of any of the five regular bodies. To find the superficies. Multiply the area (taken from the following table) by the square of the linear edge of the solid, for the superficies. To find the solidity. Multiply the tabular solidity by the cube of the linear edge, for the solid content. Ex. If the linear edge or side of a tetraedron be 3, required its superficial and solid content. Thus 1.73205 x 9=15.58845 superficies. Answer { superficies = 24 Er. 2. What is the surface and solidity of the hexaedron, whose side is 2 ? } i 8 Er. 3. Required the superficies and solidity of the octaedron, whose linear side is 2. Answer { superficies = 13-8564 } 3.7712 Er. 4. What is the superficies and solidity of the dodecaedron, whose linear side is 2 ? superficies = 8258292 } Er. 5. What is the superficies and solidity of an icosaedron, whose linear side is 2? Answer { superficies = 34:64) } = 17.45352 OF MEASURING IRREGULAR SURFACES AND SOLIDS. Def. An irregular surface, or solid, is such a surface or solid which have their bounds by lines or surfaces in any manner whatever, of no particular kind of form or shape, but merely accidental, according as they are to be found or given. Prob. 1. To measure any irregular surface whatever, by means of equidistant ordinates. Method 1. To the half sum of the two outside ordinates, add the sum of all the other remaining ordinates ; multiply the whole sum by the distance between any two ordinates, and the product will be the superficial content. Er. 1. Let fig. 1 be the curve proposed, whose equidistant ordinates, AB, CD, EF, GH, IK, LM, and NO, are respectively 5ft., 5ft. Oin., 6ft., 7ft., 9ft., and 8ft., and the distance of AC, CE, EG, or CI, is 3 feet, required the area of the curve. AB=5 2)13 6 6 half the sum of the outside ordinates IK= 90 44 0 132 superficies. Er. 2. Let ABCD, fig. 2, be a circle, whose diameter AC, or BD, is 10 feet, it is required to find the area by means of equidistant ordinates, marked 3ft. 4ft. 4:5ft. 4.9ft. and 5ft. being at the distance of 1 foot from each other. 0 2,5 2-5 half the sum of the outside ordinates. 18:9 area of one quarter. 4 75.6 feet, area of the whole. If the diameter, which is 10 feet, be multiplied by 7854, the pro. duct, 78.54, will be the area. From hence it appears, that this mode of operation, by means of equidistant ordinates, is exceedingly near the truth in measuring irregular planes ; for it will produce the area of a circle, which is one of the most oblique curves possible, as the ends raise quite perpendicular to the axis, from only 10 equidistant spaces within the 1.26th part of the truth ; and would be still nearer when applied to measuring any plane surface, where it is bounded partly by concave and partly by convex curves: because, if wholly bound by a convex curve, or curves, the area will be some. thing less than the truth; but if bounded by a concave curve or curves, the arta will be something greater than the truth ; and if the extremi Tt ties of the ordinates are joined by straight lines, the area so found will be exactly true: but the following is a method of approximation still nearer to the truth, whether the curve be concave or convex to the axis. Method 2. Divide the given curve, by ordinates, into any even number of equal parts; then add into one sum four times the sum of all the even ordinates ; twice the sum of all the odd ordinates, except the first and last, and also the first and last ordinates; and if onethird of that sum is multiplied by the common distance between any two ordinates, the product will be the answer. Ex. 1. Let fig. 1, be a curve of any kind, whose equidistant ordinates, AB, CD, EF, GH, IK, LM, and NO, are respectively 5ft. 5ft. hin. Oft. 7ft. Oft. 10ft. and 8ft. and the distance between the ordinates is 3ft., required the area of the curve. CD, GH, and LM, will be the even ordinates ; that is, the second, fourth, and sixth ; EF, and IK, the odd ordinates, that is, the third and fifth ; AB, and NO, the fust and last. f. in. f. EE= 6 IK= 9 15 2 30 Now by coni aring this area, viz. 133 feet, with the area found in Method 1, Example viz. 132 feet, there appears to be a differeuce of 1 foot; but this last method is the most correct. Er. 2. In fig. 3, let ACEGILN be a concave curve, whose equidistant ordinates, A, BC, DE, FG, HI, KL, and MN, are respectively - do 3. 6, 10, 15, 21, and the common distance 2; required the a. 90 the area, very near the truth. Ex. 3. Let AHI, fig. 4, be a parabola, whose ordinates, A, BC, DE, FG, and HI, are respectively 0, 7, 12, 15, and 16, and their common distance 6; reouired the area of the curve. 7 12 15 88 four times the sum of the even ordinates, 16 sum of the end 423 Tt2 |