EXAMPLES. 1. Required the solidity of the parabolic frustrum BCd, the diameter AB of the greater end being 58, that of the less cnd dc 30, and the height no 18. n B Here (582 +30) x 18 x.3927=(3364+900) x 18%.3927 =4264 X 18X.3927=76752x.3927=30140.5104=solid. ity required. 2. What is the solidity of the frustrum of a parabolic conoid, the diameter of the greater end being 60, that of the less end 48, and the distance of the ends 18? Ans. 41733.0144. PROBLEM XXI. To find the solidity of an hyperboloid. RULE. * To the square of the radius of the base add the square of the middle diameter between the base and the vertex; and this sum multiplied by the altitude, and the product again by .5236, will give the solidity. B Bd ment is AB—ab. But 1–5:-a (d) :: B: A= Bbd and B-b:d::b:a= by the nature of the paraboloid, and these values of A and a being substituted for them will db2-db make AB-) ab= 2B4-25 =fdx (B+b) which is the same as the rule. Q. E. I) * Demon. Let t=transverse, and c=conjugate diame. ter of the generating hyperbola, p=3.1416, y, y, the or EXAMPLES. 1. In the hyperboloid ACB, the altitude Cr is 10, the radius Ar of the base 12, and the middle diameter nm 15.8745; what is the solidity? At+A? في dinates, or semi-diameters of the ends of any frustrum of the hyperboloid, x=its altitude, and =distance of the less ordinate y from the vertex of the whole solid. (t+1+x)(1+x) Then since yo= xc, we shall have At + AP + 2ax + the fluxion of the solid = pro; =pc*: * ť x2 At+A? + Ax+ šta + far and its fluent = pc x x ť Yo , for and for f Att A2+2AX +tx + x2 becomes (vo + y2 - ) x 2px 12 3t solidity of the frustrum. But to convert this into the rules given in the text, let D, 8, d, be the greatest, middle, and least diameters, x=ab. scissa whose ordinate is d, and a=altitude. Then we shall nave these three equations : #p=0xt+xxx tp’=coxt+x+jax æ+ja From the sum of the two latter of which subtract tho double of the former, and there will result t X D-282+d aʼca 2D482 + 2d2 = taʼcé : and hence Which being 3t2 3 19+48+do substituted for it in the theorem above will give 6 ap for the content of the frustrum ; which is the same As the following rule given in the text. And if d the least diameter be supposed to become in. Here 15.87452 + 122 X 10 X.5236=251.99975 +144 * 10x.5236 = 395.99975 X 10 X.5236 =3959.9975 X 5236 =2073.454691=solidity required. 2. In an hyperboloid the altitude is 50, the radius of the base 52, and the middle diameter 68; what is the solidity? Ans. 191847 PROBLEM XXII. To find the solidity of the frustrum of an hyperbolic conoid RULE.* Add together the squares of the greatest and least semi. diameters, and the square of the whole diameter in the middle, then this sum being multiplied by the altitude, and the product again by .5236, will give the solidity. D' +48 finitely little, or nothing, the rule will become 6 X ap=p?+482 xax.5236. Q. E. D. * The demonstration of this rule is contained in that of the last problem. Or, if D=middle diameter, m=that at $ of the length, 8=generating area of the hyperbola, L= length of the spindle, and p=3.1416. 4m_302 38 Then will (d'+ X -D) * ipl = solidity of 4m-3D the spindle. And if the generating hyperbola be equilate L +D ral, then will (38 * -L?) * pl=solidity of the spin dle. L LD EXAMPLES. 1. In the hyperbolic frustrum ADCB, the length rs is 20, the diameter AB of the greater end 32, that DC of the less end 24, and the middle diameter nm 28.1708; required the solidity C Here (16+122 +28.17082) x 20 x.52359= (256+144 +793.5939) X 20 X.52359=1193.5939 X 20 X.52359= 23871.878X.52359=12499.07660202=solidity required. 2. Required the solidity of the frustrum of an hyperbolic conoid, the height being 12, the greatest diameter 10, the least diameter 6, and the middle diameter 81. Ans. 667.59. And, if l-length of the frustrum, s=generating area, and the other letters as before; then will (2D? + d + 4m2_3D_ď? 3s 1 4m—3—*ī+d-D) ia pl=solidity of the middle frustrum of an hyperbolic spindle. But if the generating hyperbola be equilateral, the frus 3s 7+D 1 trum will be =(7d?_I+ Х -) D-d Note. The content of any spindle formed by the revolution of a conic section about its axis may be found by the following rule: Add together the squares of the greatest and least diameters, and the square of double the diameter in the middle between the two, and this sum multiplied by the length and the product again by .1309 will give the solidity. And the rule will never deviate much from the truth when the figure revolves about any other line which is not the axis. 6 pl. 3. What is the content of the middle frustrum of an hyperbolic spindle, the length being 20, the middle or greatest diameter 16, the diameter at each end 12, and the diameter at of the length 141 ? Ans. 3248.938. 4. Required the content of the segment of any spindle, its length being 10, the greatest diameter 8, and the middle diameter 6. Ans. 272.272 Miscellaneous Questions in Solids. 1. If the diameter of the earth be 7930 miles, and that of the moon 2160 miles, required the ratio of their sur. faces, and also of their solidities, supposing both of them to be globular, as they are very nearly. The surfaces of all similar solids are to each other as the squares of their like dimensions ; such as diameters, circumferences, like linear sides, f-c. c. And their solidities, as the cubes of those dimensions. Hence the surface of the moon : surface of the earth :: 21602 4665600 1 21602 : 79302 and* 79302 62884900 13.47 or, As 1 : 131 nearly. Also the solidity of the moon : solidity of the earth :: 21603 1 21603 : 79303 and 79303 nearly, or, As 1 : 494 49.5 Ans. 2. Three persons having bought a sugar-loaf, want to divide it equally among them by sections parallel to the base: it is required to find the altitude of each person's snare, supposing the loaf to be a cone whose height is 20 inches. * The ratio of one quantity to another may be expressed by divid. ing the antecedent by the consequent. |