PROBLEM XIV. To find the solidity of an elliptic spindle. RULE. From 3 times the square of the middle or greatest diameter take 4 times the square of the diameter at of the length, or equally distant between the middle and one end; and from 4 times this last diameter take 3 times the middle diameter; and of the quotient arising from dividing the former difference by the latter, will give the central distance. Find the axes of the ellipse; and also the area of the generating segment. Divide 3 times this area by the length of the spindle; from the quotient subtract the middle diameter; and multiply the remainder by 4 times the central distance, before found. Subtract this product from the square of the middle diameter; multiply the remainder by the length of the spindle, and the product thus obtained by .5236, for the solidity. EXAMPLES. 1. What is the solidity of the elliptic spindle ACBDA; the length AB being 40, the middle diameter CD 12, and the diameter EF, at of the length, 9.49546 ? Here (122 x3- 9.495462 x 49.49546x4-12×3 x = (432 360.0546 37.9818436) × = 1 (71.3454 1.98184) x 36 x 9, the central × 1 distance OG. 12 Also, 9+ x2=9+6×2 = 15 x 2 = 30, the conjugate diameter CH. 6= By Problem I, Case 4, Part VII, we have 30 24 GH, the greater absciss; and 24 × 6 = √144 = 12; then, as 12: 20 (AG) :: 30 (CH); 50, the transverse diameter KL. Again, by Problem 22, Part II, we have 6 30 =2, the tabular height; and the Corresponding Area Seg.is .111823; then .111823 × 50 × 30 = 5.59115 × 30 = 167.7345, the area of the generating segment ACB. Now, (16 167.7345x3 40 X 36 (12.5800875 (144 12) × 36 = .5800875 × 36 = 20.88315; then (122 20.88315) × 40 × .5236 = 20.88315) × 40 × .5236 123.11685 X 40 X 52364928.674 x .5236 = 2578.5593064, the solidity required. 2. The length of an elliptic spindle is 80, the middle diameter 24, and the diameter at of the length 18.99094; what is the solidity? Ans. 20628.4744512. PROBLEM XV. To find the solidity of a parabolic conoid. RULE. Multiply the square of the diameter of the base by the altitude, and the product thence arising by .3927, (one half of .7854;) and the last product will be the solidity. Note 1. This Rule will give the solidity of any segment of a paraboloid, whose base is a circle. 2. A paraboloid is equal to half of its circumscribing cylinder. EXAMPLES. 1. What is the solidity of the paraboloid ADB; the height Dm being 42, and the diameter of the base AB 24? m Here 24 × 42 × .3927 = 576 × 42 × .3927 24192 .3927 = 9500.1984, the solidity required. x 2. What is the solidity of a paraboloid, the height of which is 30, and the diameter of its base 50? Ans. 29452.5. 3. Required the solidity of the segment of a paraboloid, the height of which is 9, and the diameter of its Ans. 508.9392. base 12. PROBLEM XVI. To find the solidity of the frustum of a paraboloid, when its ends are perpendicular to the axis of the solid. RULE. Multiply the sum of the squares of the diameters of the two ends by the height of the frustum, and the product thus obtained by .3927; and the last product will be the solidity. Note. If the lower frustums of two equal paraboloids be joined together at their greater ends, they will form a figure which, by Gaugers, is called a cask of the third variety; hence this Rule may be applied in Gauging. EXAMPLES. 1. What is the content of the parabolic frustum ABCD; the diameter AB of the greater end being 30, that of the less end DC 24, and the height mn 9? Here (302 +242) × 9 × .3927 = (900 + 576) × 9 X .3927 1476 x 9 x .3927 = 13284 .3927 = 5216.6268, the content of the frustum. 2. What is the solidity of the frustum of a paraboloid; the diameter of the greater end being 80, that of the less end 40, and the altitude 45? Ans. 141372. 3. The length of a cask, in the form of two equal frustums of a parabolic conoid, is 30, the bung diameter 24, and the head diameter 18 inches; what is its content in ale and wine gallons? Ans. 37.5989 ale, and 45.9 wine gallons. PROBLEM XVII. To find the solidity of a parabolic spindle. Multiply the square of the middle diameter by the length of the spindle, and the product thence arising by .41888, and it will give the solidity. 15 Note. A parabolic spindle is equal to of its circumscribing cylinder; hence we obtain the multiplier .41888, which is of .7854. EXAMPLES. 1. The length AB of the parabolic spindle ACBD is 81, and the middle diameter CD 27; what is its solidity ? G g C B D Here 272 x 81 x .41888 = 729 x 81 x .41888 = 59049 .41888=24734.44512, the solidity required. 2. What is the solidity of a parabolic spindle, whose length is 20, and middle diameter 8? PROBLEM XVIII. Ans. 536.1664. To find the solidity of the middle frustum of a parabolic spindle. RULE. Add 8 times the square of the middle diameter, 3 times the square of the end diameter, and 4 times the product of those diameters into one sum; then this sum being multiplied by the length, and the product thence arising by .05236 (one-fifteenth of .7854,) will give the solidity. Note. A cask in the form of the middle frustum of a parabolic spindle, is caled a cask of the second variety; and is the most common of any of the varieties; hence the above Rule is very useful in cask-gauging. EXAMPLES. is the solidity of the middle frustum ABCD, of a parabolic spindle; the diameter of the end AB being 12, the middle diameter EF 16, and the length |