EXAMPLE. Required the content of an oblate, and of an oblong, fpheroid; the axes being 50 and 30. First, 50 x 30 x·78539816 = 1178.09724 the area of the ellipfe. Then 1178 09724 × × 3023561.9448 oblong fpheroid. the And 1178.09724 X X 50 = 39269.908 = the oblate one. femi-fphere, as in prob. 24 fect. 1.—And in like manner fand r being fuppofed equal to each other in the values of the fruftums and fegments of a fpheroid, in the preceding corollaries, will give the values of the like parts of a sphere. Corol. 5. All fpheres and fpheroids are to each other as the fixed axes drawn into the fquares of the revolving axes. Corol. 6. Any fpheroids, and fpheres, of the fame revolving axe, as alfo their like or correfponding parts cut off by planes perpendicular to the faid common axe, are to one another as their other or fixed axes. This follows from the foregoing corollaries. Corol. 7. But if their fixed axes be equal, and their revolving axes unequal, the fpheroids and spheres, with their like parts terminated by planes perpendicular to the common fixed axe, will be to each other as the fquares of their revolving axes. Corol. S. An oblate fpheroid is to an oblong fpheroid, generated from the fame ellipfe, as the longer axe of the ellipfe is to the fhorter. For, if r be the tranfverfe axe, and c the conjugate; the oblate fpheroid will be T2 c, and the oblongncT; and thefe quantities are in the ratio of r to c. Corol. 9. And if about the two axes of an ellipfe, be generated two fpheres and two fpheroids, the four folids will be continual proportionals, and the common ratio will be that of the two axes of the ellipfe; that is, as the greater fphere, or the fphere upon the greater axe, is to the oblate fpheroid, fo is the oblate spheroid to the oblong fpheroid, fo is the oblong fpheroid to the lefs fphere, and fo is the tranfverfe axe to the conjugate. For these four bodies will be as T3, T2c, TC2, c3, where each term is to the confequent one, as r to c. PROBLEM XII. To find the Content of the Fruftum of a Spheroid; its Ends being Perpendicular to one of the Axes, and one of them paffing through the Center. RULE I. To the area of the lefs end, add twice that of the greater; multiply the fum by the altitude of the fruftum, and of the product will be the content.— By corollary to the laft problem. That is, (2D2+ d2) × an = the fruftum whofe ends are perpendicular to the fixed axe. Where D is the diameter of the greater end, d that of the less, a the altitude, and n785398. And (2Tc +tc) Xan the fruftum whofe ends are parallel to the fixed axe. Where T and c are the tranfverfe and conjugate axes of the greater end, and t and c thofe of the less end. Note. It is evident that the double of the fruftum will give the content of the zone, or spheroidal cask. EXAMPLE I. There is a cafk in the form of the middle fruftum, or zone, of an oblong fpheroid; the bung diameter is 30, the head diameter 18, and the length of the cafk 40 inches, what is the content in ale and wine-gallons? 18, and a = 40. Here D 30, d = Therefore (2D+ ď2) × ÷an = (2 × 302 + 183) × 2618 X 402124 x 10'472 = 22242 528 content in inches. the Then, fince the gallon ale meafure contains 282 cubic inches, and the wine gallon 231, we have 22242 528282 78.874 the ale gallons. And 22242 52823196.288 the wine gallons. EXAMPLE II., If a veffel, in the form of the middle fruftum of an oblate fpheroid, have the diameter of each end 40, in the middle 50, and its length 18 inches; what is its content in ale and wine gallons? Here (2D2+ d) × ‡an = (2 × 502 + 402) × •2618 × 18 = 118800 x 2618 31101.84 cubic inches. Then 3110184282110.29 ale gallons. And 31101 84÷231 = 134·64 wine gallons. EXAMPLE III. In the fruftum of an oblong spheroid, the greater end is the generating ellipfe, whofe axes are 50 and 30, the axes of its lefs end 40 and 24, and its 3༠, height 9 inches; required the content in Winchester bufhels. Here (2TC +tc) Xan = (2 × 50 X 30 +40 X 24) × 2618 × 9 = 9330*552 cubic inches. But 268.8 inches a gal. or 2150.4 a corn bufhel. And therefore 9330*552÷2150°44'339 bufhels. In the fruftum of an oblate fpheroid, the greater end is the generating ellipfe, whofe axes are 50 and 30, and the height is 20 inches; required the folidity. Here, by the nature of the ellipfe, as 50: 30:: ✔(25 + 20).X (25 20)=√45 × 5 = 15:9 the femi-conjugate axe of the lefs end. And, And, by prop. I fect. 2, as 30: 50 :: 9:15 the femi-tranfverfe axe of the lefs end. X 2618 X 20 Then (2TC +tc)× na = (2 × 50 × 30 + 30 × 18) 118 X 30 X 20 X 2618 = 18535'44 cubic inches. 70800 X 2618 From 3 times the fquare of the femi-axe perpendicular to the ends of the fruftum, fubtract the fquare of the height of the fruftum; then multiply the difference by of the height, and the product by_3*14159 &c; and call the last product P. Then I 1. If the ends be parallel to the fixed axe, As the revolving axe is to the fixed axe, fo will P be to the content of the fruftum. 2. When the ends are perpendicular to the fixed axe, As the fquare of the fixed axe is to the fquare of the revolving axe, fo is to the content of the fruftum. That is, 3ff-bb x phrr will be the fruftum whofe ends are perpendicular to the fixed axe. And 3rr - bb 3r x pbf the fruftum whofe ends are parallel to the fixed axe: ƒ being the fixed, and r the revolving femi-axe, b the height, and p = 3.14159. EXAMPLE I. If the axes of an oblong fpheroid be 50 and 30, required the content of a fruftum whofe ends are perpendicular to the fixed axe, and one of them paffing through the center, the height of the fruftum being 20 inches. Here X 20 X 152 1475 X 3.14159 × 12 = 1112123799 the content required. ड EXAMPLE II. If from the fame fpheroid be cut a fruftum whose height is 9, the ends being parallel to the fixed axe, and one of them paffing through the center, what will be its content? 3r Here3rr-bb xpbf= X9X25×3.14159 3×15 = 594 × 5 × 3∙14159 = 933053018 the content required. EXAMPLE III. If from an oblate fpheroid, whofe axes are 50 and 30, a fruftum, whofe height is 9, and its ends perpendicular to the fixed axe, be cut; what will be its folidity? In the fame fpheroid, it is required to find the content of a fruftum whofe height is 20, the ends being parallel to the fixed axe. 314159 202 3 × 253xpbf= 3 × 25 X 20 X 15 X 1475 × 4 × 3·14159 = 18535*39665 the content required, PRO |