Now to 34.4 inches correspond, on the rod, 907 ale gallons, or 111 wine gallons, the content required. Note.—The contents exbibited by the rod, answer to the most common form of casks, and fall in between the 2d and 3d varieties following. OF CASKS AS DIVIDED INTO VARIETIES. It is usual to divide casks into four cases or varieties, wbich are judged of from the greater or less apparent curvature of their sides ; namely, 1. The middle frustrum of a spheroid, And if the content of any of these be computed in inches, by their proper rules, and this be divided by 282, or 231, or 2150.4, the quotient will be the content in ale gallons or wine gallons, or malt bushels, respectively. Because 282 cubic inches make 1 ale gallon. 1 wine gallon. 1 malt bushel. And the particular rule will be for each as in the followiug problems : PROBLEM XII. of the head diameter add doublo the square of the bung diameter, and multiply the sum by the length of the cask. Then let the product be multiplied by .00094, or divided by 1077, for ale gallons ; and multiplied by .00113, or divided by 882 for wino gallons. EXAMPLES. 1. Required the content of a spheroidal cask, whose length is 40, and bung and head diameters 32 and 24 inches. By the Gauging Rule. against the same 38.0 sum 97.3 ale gallons. And having set 40 on C, to the wine gauge 29.7 on D. against the same 46.5 119.1 wine gallons. Ex. 2. Required the content of the spheroidal cask, whose length is 20, and diameters 12 and 16 inches. 12.136 ale gallops, Ans. 14.869 wine gallons. { PROBLEM XIII. To find the content of a cask of the second form. To the square of the bead diameter, add double the square of the bung diameter, and from the sum take for 1% of the square of the difference of the diameters; then multiply the remainder by the length, and the product again by .00094 for ale gallons, or by .00113 for wine gallons. EXAMPLES. 1. The length being 40, and diameters 24 and 32, required the content. By the Gauging Rule. Having set 40 on C, to 32.82 on D, against 8 on D stands 2.4 on C; the of which is 0.96. This taken from the 97.3 in the last form, leaves 96.3 ale gallon's. And having set 40 on C, to 29.7 on D, against 8 on D, stands 2.9 on C; the 1% of which is 1.16. This taken from the 119.1 in the last form, leaves 117.9 wine gallons. Ex. 2. Required the content when the length is 20, and the diameters 12 and 16. 12.018 ale gallons. Ans. 14.724 wine gallons. { PROBLEM XIV. To find the content of a cask of the third form. To the square of the bung diameter add the square of the head diameter; multiply the sum by the length, and the product again by .0014 for ale gallons, or by .0017 for wine gallons. EXAMPLES 1. Required the content of a cask of the third form, when the length is 40, and the diameters 24 and 32. 64000 108.8 wine. By the Gauging Rule. Set 40 on C, to 26.8 on D; then against 24 on D, stands 32.0 on C. sum 89.3 ale gallons. And having set 40 on C, to 24.25 on D; then against 24 on D, stands 39.1 on C. 32 on D, stands 69.8 on C. sum 108.9 wine gallons. {13: Ex. 2. Required the content when the length is 20, and the diameters 12 and 16. 11.2 ale gallons. Ans. 13.6 wine gallons. PROBLEM XV. To find the content of a cask of the fourth form. Add the square of the difference of the diameters to three times the square of their sum; then multiply the sum by the length, and the product again by .00023; for ale gallons, or by .000283 for wine gallons. EXAMPLES. 1. Required the content, when the length is 40, and the diameters 24 and 32 inches. 9408 378880° ale 87.90016 gall. 107.34933 wine. BY THE SLIDING RULE. 8 on D, stands 0.6 on C 29.1 sum 87.9 ale gallons. |