the sum of the greatest and least depths of the liquor; then 5535.36 × 38.4 = 212557.824; and 212557.824 ÷ 294.12 = 722.690, the content in wine gallons. EXAM. 3. Here 63.8o = 4070.44, the square of the diameter 28.8 + 25.2 of the base; and 2 = 54 2 = 27, half the sum of the greatest and least depths of the malt; then 4070.44 x 27 = 109901.88; and 109901.88 ÷ 2738 40.138, the content in malt bushels. EXAM. 4. Here 65o = 4225, the square of the diameter of 25.8 the base; and = 12.9, half the greatest depth 2 of the liquor; then 4225 × 12.9 = 54502.5; and 54502.5 359.05 = 151.796, the content in ale gallons. PROBLEM XXIII. To find the content of a cylindrical ungula or hoof, when its base is less than a semicircle. EXAM. 2. By Rule 1, Problem XV., Part IV., we have 96 × 6900 36 × = 3456 × % = 3 =2300, two thirds of the product of the chord and versed sine; and 365 = 46656 96 × 2 192 =243, the cube of the height divided by twice the chord; then 2300 + 243 = 2543 square inches, the area of the base of the hoof. 100 Also, 3650 36 = 14, the difference between half the diameter of the vessel and the versed sine of the hoof's base; and 2543 x 14 35602, the area of the base multiplied by the said difference. 221184 Again 485 x = 110592 x } = 3 73728, two thirds of the cube of half the chord; and 73728 35602 = 38126; then 38126 x = 38126 x 3 = 108 36 114378, the content in cubic inches; hence 114378 231 = 495.142, the content in wine gallons. PROBLEM XXIV. To find the content of a cylindrical ungula, when its base is a semicircle. To find the content of a cylindrical ungula, when its base is greater than a semicircle. EXAM. 2. By Rule 2, Prob. XV., Part IV., we have 64 ÷ 80.8, the tabular height; then by Note 2, of the same Problem, 1 and the Area Seg. answering to .2 is .111823, which being taken from .785398, .8 2; leaves .673575, the Area Seg. corresponding to .8; hence 673575 x 80% = .673575 x 6400 = 4310.88, the area of the ungula's base, in square inches. Also, 64 40 = 24, the difference between the versed sine of the ungula's base and half the diameter of the vessel; and 4310.88 x 24 = 103461.12, the area of the base multiplied by the said difference. Again, 325 x = 32768 x = % 65536 3 21845.33, two thirds of the cube of half the chord; and 103461.12 + 21845.333 = 125306.453; then 92 11528193.676 125306.453 X = 64 64 = 180128.026, the content of the ungula, in cubic inches; hence 180128.026 ÷ 231 = 779.775, the content in wine gallons. PROBLEM XXVI. To find the content of a pyramidical ungula, or the quantity of liquor contained in a vessel in the form of the frustum of a square or rectangular pyramid. when it is placed in such a position that the liquor just covers the bottom, and rises up three of the sides in an oblique direction, forming a figure that is called a cuneus or wedge. EXAM. 2. Here 30 x 2 + 33 = 60 + 33 = 93, twice the length of the base added to the length of the upper 93 x 27 x 22 2511 x 22 edge of the liquor; then 55242 = 6 = 6 = 9207, the content of the ungula in Here 58 x 2 + 38 = 116 + 38 = 154, twice the length of the base added to the length of the upper edge of the liquor; then 154 × 45 × 42 42 6 6930 X = 6930 × 7 = 48510, the content of the 6 ungula in cubic inches; and the content in ale gallons. 48510 = 172.021, 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. EXAM. 2. Here √82.6 × 58.7 = √ 4848.62 = 69.62, the the mean proportional between the diameters; and 4848.62 x 69.62 69.62 337560.9244, the product of the two diameters multiplied by the mean proportional between them. = Also, 82.65 = 563559.976, the cube of the 563559.976 337560.9244 bottom diameter; and 225999.0516 23.9 82.6 58.7 = 9456.027; then 9456.027 × 47.3 x .001133447270.0771 × .001133 = 506.75699735 wine gallons, the answer required. 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. EXAM. 2. Here 67.8 x 42.3 = √2867.94 = 53.553, the mean proportional between the diameters; and 2867.94 x 53.553 = 153586.79082, the product of the two diameters multiplied by the mean proportional between them. Also 42.35 = 75686.967, the cube of the 153586.79082-75686.967 67.8 42.3 bottom diameter; and 77899.82382 25.5 =3054.895; then 3054.895 × 54.6 x .0009283166797.267 x .0009283154.8379029561 ale gallons, the content required. 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. EXAM. 2. To find the content of the frustum. By Rule 2, Prob. VIII., we have 40 x 30 1200, the product of the diameters; and 40 + 30} = |