the depth, d the diameter of the given end or base, and y a quantity the square root of which is the sum of the required base and half the given base; then 1800 5.8751 306.378 solidity in cylindrical feet, and 306.37874 = 118.75 feet. Ans. EXAMPLE. = A measure is to be built in the form of a frustum of a cone, that will hold exactly 1 wine gallon, and the diameter of one of its ends is to be 4 inches, and that of the other 6 inches; what inust be its depth? 1 ÷ (6 × 4 + 1}) × .0034 ≈ 11.61 inches. Ans. Or, EXAMPLE. 231 6 × 4 + 62 + 42 .7854 3 11.61 inches. Ans. A measure in the form of a frustum of a cone holds 1 wine gallon; the diameter of one of its ends is 6 inches, and its depth is 11.61 inches; what is the diameter of the other end? 231 7834 = 294.1176116176 (621)=√49 = 7 — § 4 inches. Ans. CASK GAUGING. CASK-GAUGING, in a general sense, is a practical art, rather than a scientific achievement or problem, and makes no pretensions to strict accuracy with regard to the conclusions arrived at. The aim is, by means of a few satisfactory measurements taken of the outside, and an estimate of the probable mean thickness of the material of which the cask is composed (of which there must always remain some doubt), or by means of a few measurements taken of the inside, to determine, 1st, the capacity of the cask, and, 2d, the ullage, or capacity of the occupied or unoccupied space in a cask but partly full. And the Rule (RULE 6, page 60), which reduces the supposed cask, or cask of supposed curvature, to a cylinder, is as practically correct for the capacity of ordinary casks, as any rule, or set of rules, that can be offered for general purposes. Casks have no fixed form of their own, to which they severally and collectively correspond, nor are they in any considerable degree in conformity with any regular geometrical figure. Some casks a few those having their staves much curved throughout their entire length, are nearest in keeping with the middle frustum of a spheroid; others, slightly less curved than the preceding, correspond in a considerable degree to the middle are frustum of a parabolic spindle; others, again-those having very little longitudinal curvature of stave to their semi-lengths nearly in keeping with the equal frustums of a paraboloid; and others a very few those whose staves are straight from the bung diameter to the heads, or equal to that form, are in accordance with the equal frustums of a cone. The gauging rod, which is intended to be correct for casks of the most common form, gives for all casks, as may be seen in one of the following EXAMPLES, a solidity slightly greater (about 25 per cent.) than would be obtained by supposing the cask in conformity with the third figure above alluded to. The RULE for finding the contents of a cask, by four dimensions, hereafter to be given, is intended as a general Rule for all casks, and, when the diameter midway between the bung and head can be accurately ascertained, will lead to a very close approach to the truth. From the length of a cask, taken from outside to outside of the heads, with callipers, it is usual to deduct from 1 to 2 inches, to correspond with the thickness of the heads, according to the size of the cask, and the remainder is taken as the length of the interior. 4 To the diameter of each head, taken externally, from 1 inch to 1inch should be added for common-sized barrels, 1 inch for 40 gallon casks, and from inch to inch for larger casks, to correspond with the interior diameters of the heads. ΤΟ If the staves are of uniform thickness, any sectional diameter of a cask may be nearly or quite ascertained, by dividing the circumference at that place by 3.1416, and subtracting twice the thickness of the stave from the quotient. For obtaining the diagonal of a cask by mathematical process, the interior length, &c. &c. see Rules, below. In the following formulas D denotes the bung diameter, d the head diameter, and 7 the length of the cask. The solidity of any cask is equal to its length multiplied by square of its mean diameter multiplied by "854. To calculate the contents of a cask from jour dimensions. tlie RULE. To the square of the bung diameter add the square of the head diameter, and the square of double the diameter midway between the bung and head, and multiply the sum by the length of the cask, for its cylindrical contents; the product multiplied by .0034 expresses the contents in wine gallons. EXAMPLE. The length of the cask is 40 inches, its bung diameter 28 inches, head diameter 20 inches, and the diameter midway be tween the bung and head is 25.6 inches; how many gallons' capacity has the cask? 202+282 22 202 +282 + 25.6 × 2 =3805.44 × 40.003486.26 gals. Ans. (D2 + d2 +-2 m2 ) × 7X .7854 cubic contents. 2 By RULE 6, p. 68, this cask will hold 28-20=8X .65=5.2+20=25.2 × 25.2 X 40 × .0034 86.36 gallons. When the cask is in the form of the middle frustum of a spheroid And a cask of this form, having the same head diameter, bung diameter, and length as the preceding, will hold When the cask is in the form of the middle frustum of a parabolic 2 1 spindle. } D2 + } d 2 — √ 2 5 (Dd)?· square of mean diameter. And a cask of this form, having the same head diameter, bung diameter, and length as the preceding, will hold And a cask of this form, having the same head diameter, bung diameter, and length as the preceding, will hold When the cask is in the form of the equal frustums of a cone. § D2 + § ď 2 & (Dd)2= square of mean diameter. And a cask of this form, having the same head diameter, bung diameter, and length as the preceding, will hold 28 × 20 + 21} ×40×.0034 = 79.06 gals To find the contents of a cask the same as would be given by the gauging rod. The gauging rod is constructed upon the principle that the cube of the diagonal of a cask, in inches, multiplied by 644, equals the contents of the cask, in Imperial gallons. The contents in wine gallons of either of the aforementioned casks, therefore, by the gauging rod, would be 31.241.0027-821 gals. The decimal coëfficient to take the place of .0027, for finding the contents of a cask in the form of the middle frustum of a spheroid = .002926; and for finding the contents of a cask in the form of the equal frustums of a cone = .002593. And between these extremes lies the decimal for other casks, or casks of intervening figures. To find the diagonal of a cask, when the interior is inaccessible. RULE. - From the bung diameter subtract half the difference of the bung and head diameters, and to the square of the remainder add the square of half the length of the cask, and the square root of the sum will be the diagonal. EXAMPLE. What is the diagonal of a cask whose bung diameter is 28 inches, head diameter 20 inches, and length 40 inches? 28 — 20=8÷÷ 2 = 4, and 28-4=24, then √(242 +202) = 31.241 inches. Ans. To find the length of a cask, the head diameter, bung diameter and diagonal being given. And the interior length of a cask, whose interior head diameter, bung diameter and diagonal, are as the preceding, will be ✓ (31.2412 — 24o) — 20 × 2 — 40 inches. To find the solidity of a sphere. D2X DX .7854 cubic contents, D being the diameter. (312 + To find the solidity of a spherical frustum. b2 + d2 ×h×.7854 cubic contents, b and d being the bases, and h the height. NOTE. For Rules in detail pertaining to the foregoing figures, and for cther figures, see MENSURATION OF SOLIDS. ULLAGE. The ullage or wantage of a cask is the quantity the cask lacks of being full. To find the ullage of a standing cask, when the cask is half full or more. RULE. — To the square of the head diameter, add the square of the diameter at the surface of the liquor, and the square of twice the diameter midway between the surface of the liquor and the upper head, and divide the sum by 6; the quotient, multiplied by the distance from the surface of the liquor to the upper head, multiplied by .0034, will give the ullage in wine gallons. EXAMPLE. The diameters are as follows-at the upper head, 20 inches; at the surface of the liquor, 22 inches; and at a point midway between these, 21 inches; and the distance from the upper head to the surface of the liquor is 5 inches; required the ullage. ÷ (202+22+ 21.25 X 2) 6448.37 X 5 X .00347.62 gallons. Ans. When the cask is standing, and less than half full, to find the ullagé. RULE. — Make use of the bung diameter in place of the head diameter, and proceed in all respects as directed in the last Rüle, and add the quantity found to half the capacity of the cask; the sum will be the ullage. EXAMPLE. The bung diameter is 28 inches; the diameter at the surface of the liquor, below the bung, is 26 inches; the diameter midway between the bung and the surface of the liquor is 27.3 inches; and the distance from the surface of the liquor to the bung diameter is 5 inches; required the quantity the cask lacks of being half full; and also the ullage of the cask, its capacity being 86.26 gallons. (282 +262 +27.3 × 2) ÷ 6: 740.2 X 5 X .0034 = 12.58 gallons less than full. Ans. And, 86.26 ? 43.13 + 12.58 ≈ 55.73 gallons ullage. Ans. When the cask is upon its bilge, and half full or more, to find the ullage. RULE. Divide the distance from the bung to the surface of the liquor (the height of the empty segment) by the whole bung diameter, and take the quotient as the height of the segment of a circle whose diameter is 1, and find the area of the segment; multiply the area by the capacity of the cask, in gallons, and that product by 1.25; the last product will be the ullage, in gallons, as |