FOR THE DRY ULLAGE. Bung Diam. Dry Inches Quotient 35) 15.00 (428, which is less than 500 Defect Divide by 4) 0720 (018 one fourth the defect Multiply by 150 content of the cask 20500 410 61.500 gallons, DRY ULLAGE. PROOF. 88.35 gallons wet 61.50 gallons dry Sum 149.85 gallons, WHOLE CONTENT. PROBLEM IV. To Ullage a Cask Standing, by the Pen. RULE. Divide the number expressing the wet inches, by the length of the Cask, and if the quotient be less than •500, subtract from the quotient one-tenth part the decimal number by which it falls short of 500. Then multiply the remainder by the content of the Cask, and the product will be the wet ullage. If the quotient arising from the wet inches be greater than 500, then one-tenth part its excess above 500 must be added to the quotient; and in this case the sum multiplied by the content of the Cask will be the wet ullage. EXAMPLE. A Cask, of which the length is 50 inches, capable of holding 120 gallons, being partly filled, and standing on its head, wets 20 inches of the dipping rod. content of liquor in the Cask is required. SOLUTION. Length Wet Inches Quotient The If the dry ullage had been required, the operation would have been as follows: PROOF. 45 gallons wet Sum 120 gallons, WHOLE CONTENT. PROBLEM V. To Inch a Standing Cask of the First Variety. RULE. From the area of the bung diameter subtract the area of the head diameter, and divide the remainder by three times the square of half the length of the Cask, reserving the quotient. Then multiply the square of the proposed distance from the bung by the reserved quotient, and subtract the product from the area of the bung diameter. Lastly multiply this remainder by the distance from the bung, and the product will be the content of liquor in the Cask above the middle of the bung. EXAMPLE. The bung and head diameters of a Cask of the First Variety are respectively 30 inches, and 20 inches, the length of the Cask being 40 inches. The method of obtaining the content, in wine gallons, on every inch of the depth standing, is required. 20 x 20 x 3 = 1200) 1.7000 (001416 quotient. 2 Multiply by the distance from the bung CONTENT of two inches from the bung 6.108672 W.G. For the third inch from the bung. From the area of the bung diameter 3.060000 Subtract 001416 x 3 x3 = 0.012744 Remainder 3.047256 3 Multiply by the distance from the bung CONTENT of three inches from the bung 9-141768 W.G. 4 Multiply by the distance from the bung CONTENT of four inches from the bung 12-149376 W.G. As soon as the content of the fourth inch has been found, the content of each of the remaining inches may be determined very easily by the method of differences; for if each content be subtracted from the content before it, the remainders will form a set of First Differences: and the differences of the first differences will be the Second Differences: in like manner, the differences of the second differences will be the Third Differences. But all the third differences will be equal; wherefore, by continually adding this difference to the first of the second differences, we shall have all the second differences: and by continually subtracting any second difference from the corresponding first difference, we shall obtain the next first difference. Lastly, by adding the first difference to the content before it, we get the next content, and in this manner we proceed through the upper half of the Cask. The content at every inch of this half being known, it follows that, by supposing the Cask reversed, we may obtain the content on every inch up to the bung by subtraction; and then by adding to |
