the space contained between the bottom and a horizontal plane touching its crown, from which the depth of the vessel is taken. The content of this portion is most easily found by measuring the quantity of water required to fill it, till the bottom is covered. A similar method is adopted for a falling crown. 2. Find the content of a still, from the dimensions below, the uppermost portion being considered a frustum of a sphere, the cross diameters at its two ends being given, and also those at the middle of other four portions, the quantity of water required to cover its rising crown being 35 gallons: Depth of portions in inches. 8 aaa Cross diameters. 27 55.2 59.8 63.8 64 62 271 55 60.2 64.1 64.3 62.2 Mean diameters. 43.5 91.77 104.74 105.4 115.07 10.5 45.5 whole depth. Ans. 495.48. 161. When the sides of the vessel are sloping and straight, though the vessel be circular or oval, if two corresponding diameters at the top and bottom are measured, those at any intermediate depth are easily found. Thus, if e denotes the excess of the top diameter above that at the bottom, and if h is the depth of the vessel, and h' the depth of any other place, reckoning from the bottom, and e' the excess of the diameter there above the bottom diameter; then e h e' being thus found, if it is added to the bottom diameter, the result is the diameter at the given depth. If h' = 10 inches, then e' is the difference of diameters for every 10 inches; and the diameters for every 10 inches of depth are therefore easily found. The cross diameters are computed in the same manner; and then the content at every inch of depth can be found and registered in a table. CASK GAUGING. 162. Casks are usually divided into four varieties:-The first variety is the middle frustum of a spheroid; the second, the middle frustum of a parabolic spindle; the third, two equal frustums of a paraboloid united at their bases; and the fourth, two equal conic frustums united at their bases. The rules for calculating the contents of the middle frustums of circular, elliptic, and hyperbolic spindles, are too difficult for the purposes of practical gauging, and they are therefore omitted in treatises on this subject. When the cask is much curved, it is considered to belong to the first variety; when less curved, to the second; when still less, to the third; and when it is straight from the bung to the head, to the fourth variety. First Variety. 163. PROBLEM IX.-To find the content of a cask of the first or spheroidal variety. To twice the square of the bung diameter add the square of the head diameter; multiply the sum by the length of the cask; and divide the product by 1059-108, and the quotient is the content in imperial gallons.' Or, C=(2 B2+H2) L ÷ 1059.108, where H, B, are the head and bung diameters, and L the length of the cask. This is just the rule given in article 407 (Part I.) in the first case, only instead of multiplying by 2618 or of ⚫7854, and then dividing by 277-274 for imperial gallons; the divisor 1059-108 is taken, which is 3 times the divisor 353-036 for circular areas. The square root of this number, or 1059-10832-54, is used with the sliding rule. By the Sliding Rule. Set the length on C to 32.54 on D, then opposite to the bung and head diameters on D are two numbers on C, and the sum of the latter, and twice the former, is the content in imperial gallons.' EXAMPLE. What is the content of a cask, whose bung and head diameters are 32 and 24, and length 40 inches. C=(2 B2+H2) L÷ 1059-108=(2 × 322 +242) × 40 1059·108(2048 +576) × 40 ÷ 1059·108 99.1 imperial gallons. Or set 32.5 on D to 40 on C, then 32 on D gives 387 on C and 38.7 × 2=77·4 Set 32.5 on D to 40 on C, then 24 on D gives 21.7 Content in imperial gallons = 99-1 EXERCISES. 1. Find the content in imperial gallons of a cask, whose bung and head diameters are 30 and 18, and length 40 inches. Ans. 80.2. 2. What is the content of a cask, whose bung and head diameters are 24 and 20, and length 30 inches? Ans. 43.96. Second Variety. 164. PROBLEM X.-To find the content of a cask of the second variety. To twice the square of the bung diameter, add the square of the head diameter, and from the sum subtract of the square of the difference of these diameters; multiply the remainder by the length, and the product, divided by 1059.108, will give the content in imperial gallons.' C = { 2 B2 + H2 — } (B — H)2 } L÷1059.108. The rule in this case is the same as that in article 431 (Part I.), which is easily reduced to this form. By the Sliding Rule. 'Find the content by the rule for the first variety, then set 32.5 on D to the length on C, and against the difference of the diameters on D is a number on C, of which subtracted from the preceding content will give the required content.' EXAMPLE.-Let the dimensions of a cask of the second variety be the same as those given in the example for the first variety to find its content. =(2 x 322 + 242 C={2B2+H2-2 (B-H)2 } L÷ 1059.108 x 82) x 40 1059-108 2 ÷ — (·2624 — 25·6) 40 × 1059·10898-1 imp. gallons. Set 32.5 on D to 40 on C, and opposite to 8 on D is 2.5 on C, and × 2·5 = 1, which, subtracted from the content for the first variety 99-1, gives 98.1 for the content. EXERCISES. 1. Find the content of a cask, whose bung and end diameters are 48 and 36, and length 60 inches. Ans. 331.2. 2. What is the content of a cask, whose bung and head diameters are 36 and 20, and its length 40 inches? Ans. 109.12. Third Variety. 165. PROBLEM XI.-To find the content of a cask of the third variety. 'Add the square of the bung diameter to that of the head diameter, multiply the sum by the length, and divide the product by 706-0724 for its content.' C=(B2+H2) L÷ 706-0724. The formula is the same as that in article 402 (Part I.), only, instead of multiplying by × ·7854, and dividing by 277-274, the equivalent divisor 706-0724 is used. The square root of this number, or 706·072426-6, is used with the sliding rule. By the Sliding Rule. 'Set the length on C to 26-6 on D, then opposite to the bung and head diameters on D are two numbers on C, whose sum will be the content." EXAMPLE.-What is the content of a cask of the third variety, of the same dimensions as that in the example for the first variety? C=(B2+H2) L÷706-0724=(322+242)×40÷706-0724 706-0724-90-6. (1024+576) x 40 Set 26-6 on D to 40 on C, and at 32 on D is 57.9 on C 24 32.7 Content in imperial gallons 90.6 = EXERCISES. 1. Find the content of a cask, whose bung and head diameters are 30 and 24, and its length 36. Ans. 75.25. 2. What is the content of a cask, whose bung and head diameters are 29 and 15, and its length 24 inches? Ans. 36.2. Fourth Variety. 166. PROBLEM XII.-To find the content of a cask of the fourth variety. 'Add together the product of the bung and head diameters, and their squares; multiply the sum by the length, and divide the product by 1059-1086 for the content.' C = (B2+BH+H2) L÷ 1059.1086. For this is the formula of article 344 (Part I.), except that, instead of the factor 2618 or 3 × ·7854, and the divisor 277-274, the equivalent divisor 1059-1086 is taken. The square root of this number, or 325, is used with the sliding rule. By the Sliding Rule. Find a mean proportional M to the head and bung diameters H and B (57), then set 32-5 on D to the length on C, and opposite to the bung and head diameters, and their mean proportional on D, are three numbers on C, whose sum is the content." EXAMPLE. Find the content of a cask of the fourth variety, of the same dimensions as that in the example for the first variety. C= (B2+ BH + H2) L÷ 1059.1086 Set 24 on C to 24 on D, and then 32 on C is at 27.7 on D; which is M. Set 32.5 on D to 40 on C, and at 32 on D is 38·7 on C 24 21.7 27.7 29. Content in imperial gallons, 89.4 ... ... ... ... |