RULE. To the sum of the areas of the two ends add four times the area of the middle section parallel to them; multiply this sum by the perpendicular depth, and } of the product will be the content in cubic inches; which divide by 282 for ale, 231 for wine gallons, and 2150 for malt bushels. Note. The length of the middle section is equal to half the sum of the lengths of the two ends, and its breadth is equal to half the sum of their breadths, EXAMPLES. 1. The length and breadth of the bottom of a vessel in the form of a prismoid, measures 72 and 64, the length and breadth of the top 96 and 82, and the perpendicular depth 65 inches; what is its content in ale and wine gallons, and malt bushels ? Here 72 x 64+ 96 × 82 = 4608+ 7872 = 12480, the areas of the two ends. 72 +96 168 Also, = = 84, the length of the middle 2 2 middle section; then 84 x 73 x 4 = 6132 x 4 = 24528, four times the area of the middle section; whence (12480 65 37008 × 65 + 24528) × the content in cubic inches; then 1421.7021 282 Note. This and some of the following Problems may be performed by the Sliding Rule; but as the operations are too tedious for practice, they are omitted. 2. Each side of the bottom of a cistern, in the form of the frustum of a square pyramid, measures 86, each side of the top 78, and the perpendicular depth 74 inches; what is its content in ale and wine gallons, and malt bushels? Ans. 1765.8534 ale gallons, 2155.7171 wine gallons, and 231.6142 malt bushels. 3. The length and breadth of the bottom of a vessel, in the form of a prismoid, are 36 and 32, the length and breadth of the top 48 and 40, and the depth 60 inches; what is its content in ale and wine gallons, and malt buhsels? Ans. 323.4042 ale gallons, 394.8051 wine gallons, and 42.4186 malt bushels. PROBLEM V. To find the content of a vessel in the form of the frustum of a cone, in ale and wine gallons, and malt bushels. RULE. To three times the product of the top and bottom diameters, add the square of their difference; multiply the sum by the depth, and divide the product by 1077 for ale, 882 for wine gallons, and 8214 for malt bushels. Note 1. When the frustum of a cone is cut by a diagonal piane passing through the opposite extremities of the ends; or when a vessel of that shape is placed in an inclining position, so that the liquor just touches the opposite extremities of the top and bottom, the two parts or sections thus formed are called elliptic hoofs; and their contents may be found by the following Rule: Multiply the product of the diameters of the ends of the frustum by a mean proportional between them; and cube the diameter of the hoof's base. From the greater of the numbers thus found subtract the less; and divide the remainder by the difference of the diameters; then the quotient being multiplied by the perpendicular height of the hoof, and the product by 2618, will give the content in cubic inches; which divide by 282, and 231, respectively, and you will obtain the content in ale and wine gallons. 2. If the greater end of the frustum be considered as the base of the hoof, the product of the diameters by a mean proportional between them, will be less than the cube of the diameter of the base; but the contrary will be the case, when the less end of the frustum represents the base of the hoof. 3. If a vessel be placed in such a position that the liquor just covers the bottom, and part of one side, measure the diameter at the bottom, the diameter at the upper extremity of the liquor, and also the liquor's perpendicular depth, or height of the hoof; then proceed with these dimensions as directed in Note 1, and you will obtain the content. EXAMPLES. 1. The bottom diameter of a vessel, in the form of the frustum of a cone, is 46, the top diameter 62, and the depth 60 inches; what is its content in ale and wine gallons, and malt bushels? Here (62 × 46 × 3 + 32 462) × 60 = (2852 × 3 +162) × 60 = (8556 +256) × 60 = 8812 × 60 = 528720; then 528720 1077 599.4557 wine gallons; and 490.9192 ale gallons; 528720 882 = bushels. 2. The bottom diameter of a guile tun is 98, the top diameter 115, and the depth 75 inches; how much ale will it contain ? Ans. 2374.5821 gallons. 3. The bottom diameter of a wine-vat is 64, the top diameter 78, and the depth 72 inches; what quantity of wine will it contain? Ans. 1238.5306 gallons. PROBLEM VI. To find the content of a circular vessel, whose sides are curved. CASE I. When the sides are not much curved, as in the following figure. RULE. To the sum of the squares of the top and bottom diameters, add four times the square of the diameter taken exactly in the middle between them; and this sum being multiplied by the depth, and the product thence arising by .1309, will give the content in cubic inches; which divide by 282 for ale, 231 for wine gallons, and 2150 for malt bushels. Note 1. As vessels of this kind are seldom perfectly round, it is best to measure two diameters at right angles to each other; and take half their sum for a mean diameter. 2. The Rule given in this Problem is the same, in substance, as that given in Problem 10, Part IV. Sect. I. See the Scholium in that Problem. EXAMPLES. 1. The diameter AB, of the following vessel, measures 50, the diameter CD 56, the diameter EF 54, and the depth mn 46 inches; what is its content in ale and wine gallons, and malt bushels ? E C m B n Here 502 +542 + 562 × 4 = 2500 + 2916 +12544 =17960; and 17960 × 46 × .1309826160 ×.1309 108144.344, the content in cubic inches; = 108144.344 = 383.4905 ale gallons; bushels. 108144.344 2150 then 108144.344 231 50.2996 malt 2. The bottom diameter of a mash-tun is 75, the top diameter 81, the diameter taken in the middle between them 84, and the depth 56 inches; what is its content in malt bushels ? Ans. 137.7774 malt bushels. CASE II. When the sides are very much curved, as in the following figure. RULE. Add the square of the top diameter to the the bottom diameter; and call the sum A. of the To the square of the diameter taken at depth, add the square of that taken at of the depth; and call the sum B. 12 Square the diameter taken in the middle, or at of the depth, and call it C; and put D = the depth, and .7854 E = = .06545; then (A + 4B +2C) ×D× E will be the content of the vessel, in cubic inches; which divide by the proper divisors, and the respective quotients will be the content in ale and wine gallons, and malt bushels. Note. This Rule I have deduced from the method of equidistant ordinates, described in Problem 23, Part II.; and it may be used for all vessels of an ordinary size; but if a vessel be very deep, and great accuracy be required, find the areas of as many equidistant, parallel sections as you think proper' by multiplying the square of the mean diameter of each section by .7951; and proceed with these areas as if they were equidistant ordinates, and the result will be the content required. See Prob. Í., Case 2, Part VI. EXAMPLES. 1. The diameter AB, of the following vessel, measures 49, CD 62, EF 68, GH 65, KL 56, and the depth mn 48 inches; what is its content in ale and wine gallons, and malt bushels? |