To find the Content of a Spheroidal Cafk, not full, ftanding upon its End, the Axe being Perpendicular to the Horizon. That is, of the Fruftum of an Oblong Spheroid, the Ends being Perpendicular to the Axe, but neither of them paffing through the Center. Multiply the difference of the fquares of the diameters of the ends, by 4 times the fquare of the difference between the height of the liquor and half the length of the cafk, and divide the product by the fquare of the length of the cafk; fubtract the quotient from 3 times the fquare of the bung diameter, and multiply the remainder by the aforefaid difference between the height of the liquor and the length of the cafk; then the product multiplied by 261799 will give the quantity by which the cafk is more or lefs than half full; and which, therefore, being added to, or taken from, half the content of the cafk, will give the content of the part filled. That is, [(28a + b2) 11±d( 3ba — 4dd (b®? — b2) ] × ↓ n the content of the part filled, called the ullage; uling the upper or under figns, according as the cafk is more or less than half full; where b and b the bung and head diameters, the length of the cafk, n = 785398, and d = 11~w, w being the wet part of 1, or the height of the liquor in the cafk.* 2 For, if n be the diameter at the furface of the liquor, by the property of the ellipfe bb : 4l2 :: b2 — D2: d; and hence D2 = b2 4dd x (bb). But, by the last problem, (262 + b2) × !In = half the con tent If a spheroidal cafk, whofe head and bung diameters are 18 and 30 inches, and length 40 inches, be filled to the height of 30 inches; how many ale and wine gallons are in it? X 48 X 12) X 10 (2700 144) X 10 25560. And (262 +b2) × 17 = (1800 + 18 × 18) × 20 = 42480. Confequently (42480 +25560) × n = 68040 X 2617993878 = 17812.8303458 is the content in inches; and being divided by 282 and 231, gives 63 166065 ale, and 77.111819 wine gallons. EXAMPLE II. If the height of the liquor in the fame cask be only 10 inches, required the ullage. Hered w 20-10 10 the fame as before, and therefore the part which in the laft example was added, muft here be fubtracted; fo that (42480-25560) × n = 16920 X •2617993878 = 4429 6456415 is the content in inches 15.707963 ale, and 19.175955 wine gallons. PRO tent of the cask, and [362 – 4dd (62 -(b2 — b2)] × } dn the fruf tum by which the part filled is more or less than half the cask. Confequently [(2b2 +b2) ¿1 ± (362 the part filled. 4dd × (b2 = b2) d)] PROBLEM XIV. To find the Solidity of the Segment of a Spheroid whofe Bafe is Perpendicular to one of the Axes. RULE I. From 3 times the femi-axe perpendicular to the bafe of the fegment, take the height of the fegment, multiply the remainder by of the fquare of the height; then multiply the product by 3.14159, and call the last product o Divide the axe which is parallel to the base by the other axe, and call the quotient q. Then for the fegment whofe bafe is perpendicular to the fixed axe, multiply Q by the fquare of q; and for the other fegment, multiply That is, Q99 = 3f-b × prrbb by q. the fegment whose base is perpendicular to the fixed axe. And og = 3r-b × pfbb the other fegment, whose base is parallel to the fixed axe. Where is the height of the fegment, and the other fymbols as in the 12th problem. EXAMPLE. I. If from an oblong fpheroid, whofe axes are 50 and 30, be cut a fegment whofe bafe is perpendicular to the fixed axe, its height being 5; required the content of it. Here 3f-bx pr rbb = 3 x 25—5×152×52×3•14159 3ff = 70 × 3 × 3.14159 required. 3×252 659°73445 the content If from the fame fpheroid be cut a fegment whose height is 6, what will its content be, fuppofing its bafe to be parallel to the fixed axe? Here 37-b 3r = 3 × 15 × 25×6×3.14159 = 39 × 20 × 3.14159 = 2450°44226 the content required. EXAMPLE III. Required the folidity of the fegment of an oblate fpheroid, whofe axes are 50 and 30; the height of the fegment being 6, and its base perpendicular to the fixed axe. 6 × 3.14159 = 1300 x 3.14159 4084 07044 the content required. EXAMPLE IV. To find the content of a fegment of the fame fpheroid, its bafe being parallel to the fixed axe, and its height 5. xpfbb=3X25-5×15×5×5×3*14159 3×25 = 70 × 5 × 3.14159 = 1099°55742 = the content required. RULE II. For the Segment whofe Bafe is Parallel to the Fixed Axe; having given its Height and the Diameters of its End. To the fquare of the height add 3 times the fquare of the lefs or greater femi-axe of the bafe, according according as it is the fegment of an oblong or oblate fpheroid; divide the fum by the fame femiaxe, and multiply the quotient by the other femiaxe, the product by the height, and this product by *52359, will give the content of the fegment. That is, ment. 3A2+b2 xpвb the content of the feg Where is its height, a the lefs or greater femidiameter of the bafe, according as the fegment is that of an oblong or oblate fpheroid; B is the other femi-diameter, and p = 3.14159. EXAMPLE I. It is required to find the content of the fegment of an oblong fpheroid, whofe bafe is parallel to the fixed axe; its height being 6, and the axes of its elliptic bafe 40 and 24. Here 3AA+bb 6A XpBb = 3 × 122+62 6×12 X 20×6×3.14159 = 13 × 60 × 3*14159 = 2450°44226 the content required. EXAMPLE II. Required the content of the fegment of an oblate fpheroid, whofe bafe is parallel to the fixed axe; its height being 5, and the diameters of its bafe 18 and 30. |