Again, 33 mean proportional Multiply by 31 least diameter 33 99 Product 1023 to be subtracted. And, 35 greatest diameter Multiply by 35 greatest diameter 175 105 Product 1225 square of the greatest diameter 1023 product to be subtracted 202 remainder Multiply by 35 greatest diameter 1010 606 7070 product. Multiply by 13 one-third the altitude 21210 35314) 91910 product Quotient 22977.5 Wherefore it will be, W.G. circ. divisor 294) 22977.5 (78 15 wine gallons. A. G. circ. divisor 359) 22977·5 (64.00 ale gallons. M. B. circ. divisor 2738) 22977.5 ( 8.39 malt bushels. For the Less Hoof. The mean proportional between the greatest and least diameters, as already found, is, 33 mean proportional Multiply by 35 greatest diameter 165 99 Product 1155 31 × 31 = 961 sq. of the least diameter And, W.G. circ. divisor 294) 19545-5 (66-48 wine gallons. A. G. circ. divisor 359) 19545·5 (54.44 ale gallons. M. B. circ. divisor 2738) 19545·5 (7·13 malt bushels PROBLEM XVIII. To determine the Content of the Ungula of a Square Pyramid. RULE. FOR THE GREATER HOOF. Multiply the greatest breadth by the least, and to the product add twice the square of the greatest: then multiply the sum by one-sixth part the altitude or depth, and divide the product by the number of cubic inches in the proposed integer. FOR THE LESS HOOF. Multiply the greatest breadth by the least, and to the product add twice the square of the least: then multiply the sum by one-sixth part the altitude or depth, and divide the product by the number of cubic inches in the proposed integer. EXAMPLE. Let ABD be the greater hoof, and ACD, the less hoof of the frustum of a square pyramid; the greatest width, AB, being 43 inches; the least, CD, 25 inches; and the perpendicular depth, 24 inches: the content of each hoof is required in wine gallons, ale gallons, and malt bushels. B SOLUTION. For the Greater Hoof. 43 greatest breadth Multiply by 25 least breadth 215 86 Product 1075 43 x 43 x 2 = 3698 twice the square of AB. Sum 4773 Multiply by 4 one-sixth the depth W.G. divisor 231) 19092 (82.65 wine gallons. A. G. divisor 282) 19092 (67-70 ale gallons. M. B. divisor 2150) 19092 (8.88 malt bushels. For the Less Hoof. 43 greatest breadth Multiply by 25 least breadth 215 86 Product 1075 25 × 25 × 2 = 1250 twice the square of CD. Sum 2325 Multiply by 4 one-sixth the altitude W.G. divisor 231) 9300 (40-26 wine gallons. A. G. divisor 282) 9300 (32.97 ale gallons. M. B. divisor 2150) 9300 (4.32 malt bushels. PROBLEM XIX. To determine the Content of the Segment of a Sphere. RULE. By the Pen. To the square of the altitude, add three times the square of half the diameter of the base, and multiply the sum by the altitude; then divide the product by the spherical divisor for the proposed integer, and the quotient will be the content. |