H'=H÷B===8; and hence C'0024927, and C = C′B2L = ·0024927 × 302 × 36=80·7. The same answer as that to the example in art. 168. EXERCISES. 1. What are the contents of each of four casks of the four varieties, their diameters being 32 and 24, and length 40 inches? Ans. The contents will be the same as for the four casks in the exercise to the preceding problem. 2. Find the contents of each of four casks of the four varieties, their diameters being 31.5 and 24.5, and the length 42 inches. Ans. Content for the first = 102.6; the second = 101.9; the third = 94·9; and the fourth = 94. 3. What is the content of a pipe of wine, whose length is 50 inches, head diameter 22-7, and bung diameter 31.7, the cask being of the first variety? Ans. 119.5.* GENERAL METHOD FOR A CASK OF ANY FORM. 171. PROBLEM XV.-To find the content of a cask of any form, by one method, independently of tables. 'Add together 39 times the square of the bung diameter, 25 times the square of the head diameter, and 26 times the product of the diameters; multiply the sum by the length, and divide the product by 31773-25 for the content in imperial gallons.+ C = (39 B2 + 25 H2 + 26 BH) L÷31773-25. EXAMPLE. Find the content of a cask, whose diameters are 32 and 24, and length 40 inches. C(39 B2 +25 H2 +26 BH) L+ 31773.25 =(39 × 322 +25 × 242 +26 x 32 x 24) x 40+÷31773-25 =93-53 imperial gallons. EXERCISE. Find the content of a cask, whose diameters are 36 and 48, and length 60 inches. Ans. 315-7. *The two last exercises and the preceding table are taken from Symons' useful treatise-The Practical Gager. For the principle of this rule, see Hutton's Mensuration, p. 451. ULLAGE OF CASKS. The ullage of a cask is the content of the part occupied by liquor in it when not full, or of the empty part. Only two cases are usually considered, namely, when the cask is lying, or when it is standing. When the ullage of the part filled is found, that of the empty part can be obtained by subtracting the ullage found from the content of the whole cask. 172. PROBLEM XVI.-To find the ullage of the filled part of a lying cask in imperial gallons. 1 Divide the number of wet inches by the bung diameter, and if the quotient is under 5, deduct from it of what it wants of 5; but when the quotient exceeds 5, add 1 of that excess to it; then if the remainder in the former case, or the sum in the latter, be multiplied by the content of the whole cask, the product will be the ullage of the part filled.' Let WFK the wet inches, R = W÷B, H C'the content of the cask, = then U (RD) C, = K using when R 5, and + when R 5. By the Sliding Rule. Set the bung diameter on the slider B of Segments Lying to 100 on the line S. L.; then opposite to the wet inches on the slider is a segment on the line S. L. Set 100 on A to the content of the whole cask on B, and at the segment just found on A will be the ullage on B.' EXAMPLE. The content of a lying cask is 98 gallons, the bung diameter 32, and wet inches 10; required the ullage of the part filled. R=W÷B=32=3125, D=5—3125·1875, D=0469. Hence, U = (R— 1 D) C = (·3125 —·0469) × 98 =2656 × 98 = 26.03. Or, set 32 on B to 100 on S. L., and at 10 on B is 24.6 on S. L. Set 100 on A to 98 on B, and at 24.6 on A is 24.3 on B. This result differs by 17 gallons from that found by the calculation, as the same rule is used for all varieties. Let L, L = the length of given and experimental cask used in constructing the lines S. S. and S. L. (149), their capacities; and hence C' = 100. the capacity of a portion of the given cask, when lying, to be ullaged, and of a similar portion of experimental cask; W, W' the wet inches for these portions. Then L: WL: W', and log. L-log. W= log. L'log. W'. Hence, since the slider for the line S. L. is a logarithmic line, the distance from L to W on it is equal to that from L' to W'; and when L on the slider is opposite to C' or 100 on S. L., W on the slider will be opposite to the same number on S. L. that W' would be opposite to when L' is opposite to 100 on S. L.; that is, W would be opposite to U', the ullage of a similar portion of the experimental cask, which is therefore obtained by the above rule. Again, C' or 100 : C = U': U; and since C', C, and U', are known, therefore U, their fourth proportional, can be found by means of the lines A, B, according to the rule in article 54. C, C' EXERCISE. The content of a lying cask is 90, its bung diameter 36, and the wet inches 27; find the ullage of the part filled. Ans. 73.1. 173. PROBLEM XVII.-To find the ullage of the filled part of a standing cask in imperial gallons. 'Divide the number of wet inches by the length of the cask, then if the quotient is less than 5, subtract from it part of what it wants of 5; but if it is greater than 5, add to it of its excess above 5; then multiply the remainder in the former case, or the sum in the latter, by the content of the cask, and the product will be the ullage.' Let W GH the wet inches, and let C, U, and D, have the same when R5, and + when using R5. By the Sliding Rule. 'Set the length of the cask on the slider C of Segments Standing to 100 on the line S. S., then opposite to the wet inches on the slider will be a segment on the line S. S. Set 100 on A to the content of the whole cask on B, and opposite to the segment just found on A will be the content on B. This rule is proved in exactly the same manner as that of the preceding problem. Or, set 48 on C to 100 on S. S., on S. S. EXAMPLE. The content of a standing cask is 120 gallons, its length is 48, and the wet inches 40; required the ullage of the part filled. R=W÷L=10 吾 ='83', hence D = ·3′, and D·03′. Hence, U (R+D) C = (·83′ +·03′) × 120 =86' × 120 104. and at 40 on C is 85.4 Set 100 on A to 120 on B, and at 85-4 on A is 101.2 on B. EXERCISE. The content of a cask is 105 gallons, its length is 45 inches, and the wet inches 25; what is the ullage of the part filled? Ans. 58.8. MALT GAUGING. 174. Barley to be malted is steeped in water in a cistern for not less than 40 hours. When sufficiently steeped, it is then removed to a frame called a couch frame, where it remains without alteration for about 26 hours; it is then reckoned a floor of malt, till it is ready for the kiln. During the steeping, the barley swells about of its original bulk, or of its bulk then; after being less than 72 hours out of the cistern, it is considered to have increased of its bulk at that time; and after being out a longer time, it is considered to have increased by of its bulk then; and hence the rule in the following problem. 175. PROBLEM XVIII. Having given the cistern, couch, or floor-gauge of a quantity of malt, to find the net bushels. 'Multiply the cistern or couch bushels by 8, and the floor bushels by, when it has been out of the cistern for less than 72 hours, or by, when it has been out a longer time." EXERCISES. 1. The number of couch bushels of malt is 420, what are the net bushels? Ans. 336. 2. If the number of floor bushels, which has been 30 hours out of the cistern, is 524, what are the net bushels? Ans. 3493. 3. What is the number of net bushels corresponding to 636 bushels that have been out of the floor for more than 72 hours? Ans. 318. During the process of malting, the malt is repeatedly gauged, and the duty is charged on the greatest gauge, after the legal deductions are made, whether that arises from the measurements taken in the couch, frame, or floor. The greatest gauge can be determined by the following problem. 176. PROBLEM XIX.-Having given the greatest cistern or couch gauge or the greatest floor gauge, to determine which is the greatest or duty gauge. 'Multiply the greatest cistern or couch gauge by 1.2, if the floor gauge has been taken before the malt was 72 hours out of the cistern, or by 1-6, if taken after that time; then, if this gauge is greater than the floor gauge, it is to be taken for the duty gauge, otherwise the floor gauge is to be taken.' Since the cistern and couch gauge are each to be multiplied by the same number 1-2, to obtain the floor gauge, |