3 The content of that part of a compound figure forming the frustum of a pyramid, may be found by Problem VI.; the content of that part forming a prismoid, by Problem IX.; and the content of that part forming a wedge, by the last Problem. 4. The content of the compound figure ABFE, may also be found by subtracting the content of the dry wedge EFH from the content of the frustum or prismoid A BFH. (See Problem XXIX.) PROBLEM XXVII. To find the content of a conical ungula, or the quantity of liquor contained in a vessel in the form of the frustum of a cone, when it stands upon its greater base, and in such a position that the liquor just covers the whole of its bottom. RULE. By the Pen. Multiply the product of the top and bottom diameters by the mean proportional between them; subtract the last product from the cube of the bottom diameter; divide the remainder by the difference of the diameters; multiply the quotient by the perpendicular height of the ungula, and the product thence arising by .0009283, and .001133; and the respective products will be the content in ale and wine gallons. Note. When the liquor does not rise to the top of the vessel, the diameter at the upper extremity of the liquor, measured parallel to the top of the vessel, must be taken for the top diameter, in order to determine the content of the ungula. EXAMPLES. 1. What is the content of the ungula A B C, in ale gallons; the diameter A B being 60 inches, the diameter CD 40 inches, and the perpendicular height Cm 35 inches? A D B Here √60x40=2400=48.98979, the mean proportional between the diameters; and 48.98979 × 2400= 117575.496, the product of the two diameters multiplied by the mean proportional between them. Also, 603=216000, the cube of the bottom diameter; and 216000-117575.496 98424.504 20 60-40 = 4921.2252; then 4921.2252 x 35 x .0009283 = 172242.882 x .0009283= 159.8930673606, ale gallons, the answer required. 2 Required the content of a conical ungula, in wine gallons; the bottom diameter being 82.6 inches, the top diameter 58.7 inches, and the perpendicular height 47.3 inches. Ans. 506.756 wine gallons. PROBLEM XXVIII. To find the content of a conical ungula, or the quantity of liquor contained in a vessel in the form of the frustum of a cone, when it stands upon its less base, and in such a position that the liquor just covers the whole of its bottom. RULE. By the Pen. Multiply the product of the two diameters by the mean proportional between them; from the product thus obtained subtract the cube of the bottom diameter; divide the remainder by the difference of the diameters; multiply the quotient by the perpendicular height of the ungula, and the product thence arising by .0009283, and .001133; and the respective products will be the content in ale and wine gallons. EXAMPLES. 1. What is the content of the ungula ABC, in wine gallons; the diameter A B being 30 inches, the diameter CD 50 inches, and the perpendicular height A m 40 inches? Here ✓50 × 30=√1500=38.729, the mean proportional between the diameters; and 38.729 × 1500=58093.5, the product of the two diameters multiplied by the mean proportional between them Also, 30327000, the cube of the bottom diameter; and 58093.5-27000 31093.5 =1554.675; then 1554.675× 50-30 20 40x.001133 62187x.001133=70.457871, wine gallons, the content required. 2. What is the content of a conical ungula, in ale gallons; the bottom diameter being 42.3 inches, the top diameter 67.8 inches, and the perpendicular height 54.6 Ans. 154.838 ale gallons. inches? PROBLEM XXIX. To find the quantity of liquor contained in a conical vessel placed in an inclining position, so that the liquor intersecting the opposite sides of the vessel, in an oblique direction, disposes itself into a compound figure consisting of the frustum of a cone, and a conical ungula. RULE. By the Pen. Find the content of that part of the liquor forming the frustum of a cone, by Problem VIII.; and the content of the ungula by either of the last Problems, as the case requires; and the sum of the contents thus obtained will be the content of the whole compound figure. Note. For the method of finding the content of a compound figure consisting of the frustum of a square pyramid and a wedge, or of a prismoid and a wedge, see Problem XXVI. EXAMFLES. 1. What is the content of the compound figure ABCDE, in ale gallons; the diameter A B being 50 inches, the diameter C D 40 inches, the diameter EF 30 inches, and each of the perpendiculars mn and En 20 inches? To find the Content of the Frustum ABCD. By Rule I., Problem VIII., we have 50 x 40 x 3=2000 x3=6000, three times the product of the top and bottom diameters; and 50-40° 10° 6000+100 x 20 difference; then 100, the square of their 6100 x 20 =113.261,the content in ale gallons. To find the Content of the Ungula C D E. By Problem XXVII., we have 40 x 30=1200= 34.641, the mean proportional between the diameters; and 1200 x 34.641=41569.2, the product of the two diameters multiplied by the mean proportional between them. 22430.8 Also, 403=64000, the cube of the bottom diameter; and 64000-41569.2 =2243.08; then 2243.08 × 20 40-30 x.0009283=44861.6 x .0009283=41.64502328, the content in ale gallons. 10 Lastly, 113.261+41.64502328=154.90602328 ale gallons, the content of the whole compound figure ABCDE. Note. The content of the compound figure ABCDE may also be |