then 51842738 1.893, the content in malt bushels. PROBLEM XVIII. To find the content of a vessel, in the form of the frustum of a parabolic conoid. EXAM. 2. Here 100 x 100 = 10000, the square of the diameter of the greater end; 50 x 50 = 2500, the square of the diameter of the less end; and (10000 + 2500) × 56 = 12500 × 56700000, the sum of the squares of the diameters multiplied by the depth; then 700000 718.1 =974.794, the content in ale gallons; and 700000588.24 = 1189.990, the content in wine gallons. EXAM. 3. Here 28 × 28784, the square of the bung diameter; 21 x 21 441, the square of the head diameter; and (784 + 441) × 36 = 1225 x 36 = 44100, the sum of the squares of the two diameters multiplied by the length; then 44100 718.1 = 61.412, the content in ale gallons; and 44100 ÷ 588.24 = 74.969, the content in wine gallons. PROBLEM XIX. To find the content of a circular vessel when its sides are a little curved. EXAM. 2. Here 84 × 847056, the square of the bottom. diameter; 81.5 × 81.5 = 6642.25, the square of the top diameter; and 94 × 94 × 4 = 8836 × 4 = 35344, four times the square of the middle diameter; then (7056 + 6042,25 + 35344) × 32 = 49042.25 × 52 =2550197; and 2550197 ÷ 2154.3 = 1183.770, the content in ale gallons; and 2550197 ÷ 176479 =1445.100, the content in malt bushels. EXAM. 3. 35 x 4 = of the bung Here 30 x 30 x 2 = 900 x 2 = 1800, twice the square of the head diameter; and 35 1225 x 4 = 4900, four times the square diameter; then (1800 + 4900) × 38 = =254600; and 254600 ÷ 2154.3 = 118.181, the content in ale gallons; and 254600 ÷ 1764.72 = 144.272, the content in wine gallons. 6700 × 39 PROBLEM XX. To find the content of a circular vessel, when its sides are much curved, by teking five diameters at equa! distances from each other. EXAM. 2.. Here 5246o = 2704 + 2116 = 4820, the sum of the squares of the top and bottom diameters. Also, 632 x 2 = 3969 × 2 = 7938, twice the square of the middle diameter. And, 58 +60° x 4 = 3364 + S600 × 4 = 6964 × 427856, four times the sum of the squares of the diameters, taken at one-fourth, and at threefourths of the depth. Then, (48207938 +27856) x 60 = 40614 × 60 = 2436840; and 2436840 ÷ 4308.6 = 565.575, the content in ale gallons. EXAM. 3. Here 762 +86o = 5776 + 7396 = 13172, the sum of the squares of the top and bottom diameters. Also, 1042 x 2 = 10816 x 2 = 21632, twice the square of the middle diameter. And, 95o 100o x 4 = 9025 + 10000 x 4 19025 x 4 = 76100, four times the sum of the squares of the diameters taken at one-fourth, and at three-fourths of the depth. Then, (13172 + 21632 +76100) × 82110904 X 82 = 9094128; and 909412832856 = 276.787, the content of the vessel in malt bushels. PROBLEM XXI. To find the content of a circular vessel, when its sides much curved, by taking a competent num are very ber of equidistant, parallel sections. Then, 24.0883 + 32.7264 = 56.8147, the sum of the two extreme areas; (35.3940 + 42.2729 + 38.6483) × 4 = 114.3152 x 4 = 457.2608, four times the sum of all the even areas; and (40.641943.1697) × 2 = 83.8116 x 2 = 167.6232, twice the sum of all the odd areas; also 114.6 ÷ 6 = 19.1, the common distance of the sections; hence, (56.8147 + 457.2608 + 167.6232) 19.1 681.6987 × 19.1 13020.44517 X 3 = Then, 29.4066 + 39.9519 = 69.3585, the sum of the two extreme areas; (40.7668 + 51.6060 + 47.1812) X 4 = 139.5540 x 4 = 558.2160, four times the sum of all the even areas; and (49.6149 + 52.7008) × 2 = 102.3157 x 2 = 204.6314, twice the sum of all the odd areas; hence (69.3585 + 19.1 832.2059 x 19.1 558.2160 204.6314) X = 3 3 = 5298.37756, the content in wine EXAM. 3. By Rule I. Here 108.4 x .7854 = 11750.56 x .7854 = 9228.889824, the area of the first section; 123.6a × .7854 = 15276.96 x 7854 = 11998.524884, the area of the second section; 130.8 x .7854 = 17108.64 .7854 13437.125856, the area of the third section; 136.62 x .7854 = 18659.56 × 7854 14655.218424, the area of the fourth section; 136.22 14569.515576, the X .7854 18550.44 x .7854 area of the fifth section; 131.82 x .7854 = 17371.24 x.7854 13643.371896, the area of the sixth section; 124.22 x .7854 = 15425.64 × .7854 = 12115.297656, the area of the seventh section; 114.8% X .7854 = 13179.04 x .7854 10350.818016, the area of the eighth section; and 96.82 x .7854 = 9370.24 x 7854 7359.386496, the area of the ninth or last section. Then 9228.889824 + 7359.386496=16588.276320, the sum of the areas of the two end sections; (11998 .52438414655.218424 + 13643.371896 + 10350 -818016) x 4 = 50647.932720 x 4 = 202591 .730880, four times the sum of all the even sections; and (13487.125856 +14569.515576 +12115.297656) X240121.939088 x 2 = 80243.878176, twice the sum of all the odd sections; hence (16588.276320 18.3 +202591.730880 + 80243.878176) × 3 299423.885376 x 6.1 1826485.7007936, the content in cubic inches; consequently, 1826485.7007936 282 6476.8996482, the content in ale gallons; 1826485.7007936 231 = 7906.8645056, the content in wine gallons; and 1826485.7007936 .42 849.36231, the content in malt bushels. 2150 To find the content of the hoof of a cylinder, or the ·quantity of liquor contained in a cylindrical vessel, placed in an inclining position. |