head diameters being 40 and 20 inches, and its length 50 inches. Ans. 30.555 mean diam.; 158.7 wine; 130.1 ale; 132.2 imp. gallons. The content of any cask may be found very nearly by the general rule for the middle frustum of the spindles. See T56. DIMENSIONS OF CASKS. To find the dimensions of a cask: Take the measure of the head diameters close to the outside, and for small casks add three-tenths of an inch, and for casks of thirty or forty gallons, add four-tenths, and for larger casks add five or six tenths of an inch, to half the sum of the head diameters for the mean head diameter on the inside. Then measure the bung diameter with a rule or rod, and observe whether the stave opposite the bung is thicker or thinner than the rest, by moving the rod to and fro; and if it differ from the others, make the necessary allowance, Then measure the length of the cask between the outside of the heads with a rule, (or with callipers made for the purpose,) and allow for the thickness of both heads,-for casks of twenty, thirty, or forty gallons, one inch; and for larger casks allow 1.5; and for casks of more than 110 gallons allow two inches. Having taken the dimensions of a cask, as directed above, To find its capacity in ale and wine gallons, if it be of a medium curvature:-Add the square of the head diameter to twice the square of the bung diameter; and from the sum subtract four-tenths of the square of the difference of the diameters; multiply the remainder by one-third of the length, and the product by .0034 for wine, by .0028 for ale, and by .002833 for imperial gallons. ULLAGE OF CASKS. When a cask is not full, the whole capacity is divided into two portions, one part being occupied with liquor and the other empty; either of which is called the ullage To find the ullage of the filled part of a lying cask :— Divide the number of wet inches by the bung diameter, and if the quotient is less than .5, subtract from it one-fourth of what it wants of .5; but when the quotient exceeds .5, add one-fourth of that excess to it; then, if the remainder in the former case, or the sum in the latter, be multiplied by the content of the whole cask, the product will be the ullage of the part filled. To find the ullage of a standing cask : Divide the number of wet inches by the length of the cask; then if the quotient is less than .5, subtract from it one-tenth part of what it wants of .5; but if the quotient is greater than .5, add to it one-tenth of the excess above .5; then multiply the remainder in the former case, or the sum in the latter, by the content of the cask, and the product will be the ullage of the part filled. EXAMPLES. 1. How many gallons does a lying cask contain, whose whole capacity is 40 wine gallons, the bung diameter being 20 inches, and the number of wet inches under the bung 8 ? Ans. 15 gallons. 2. The content of a lying cask is 98 gallons, the bung diameter 32, and wet inches 10; required the ullage of the part filled. Ans. 26.03 gallons. 3. The content of a lying cask is 90, its bung diameter 36, and wet inches 27; find the ullage of the part filled. Ans. 73.1. 4. The content of a standing cask is 105 wine gallons, the length 45 inches, and the wet inches 25; what is the ullage of the part filled, in wine, ale, and imperial gallons? Ans. 58.8; 48.166; and 49 gallons. To find the length of a cask, its content in gallons and its head and bung diameters, being given : Place the mean diameter of the cask on D, under the number of gallons on C; then over the respective gauge points on D will be found the length of the cask on C. EXAMPLES. 1. What is the length of a cask which will hold 40 wine gallons, the mean diameter being 21 inches? Ans. 32.5 inches. 2. The bung and head diameters of a cask of the fourth variety are 30 and 24 inches, and its content 60 ale gallons; its length is required. Ans. 29.5 inches. 59. TONNAGE OF SHIPS. By a law of congress, the tonnage of ships is to be computed in the following manner : 66 If the vessel be double-decked, take the length thereof from the fore part of the main stern to the after part of the sternpost above the upper deck; the breadth thereof to be taken at the broadest part above the main wall, half of which breadth shall be accounted the depth of such vessel; then deduct from the length three-fifths of the breadth; then multiply the remainder by the breadth, and this product by the depth; divide the last product by 95, and the quotient will be the true tonnage of such vessel." If the vessel be single-decked, take the length and breadth as above directed, in a double-decked vessel; and deduct from the length three-fifths of the breadth; and, taking the depth from the under side of the deck-plank to the ceiling in the hold, multiply and divide as above directed, and the quotient will be the true content or tonnage of such vessel." EXAMPLE.-What is the tonnage of a double-decked vessel, whose length is 80 feet, and breadth 24 feet? Ans. 198.871 tons. Ship carpenters usually compute the tonnage of a vessel by the following rule :— Multiply the length of the keel by the breadth of the main beam, and the product by the depth of the hold, and divide the last product by 95, and the quotient is the number of tons.—In double-decked vessels, half the breadth of the main beam is taken for the depth of the hold. 60. COAL-PIT. A coal-pit, well built, is very nearly equal to 3 of a cylinder, whose diameter is equal to that of the base of the pit, and length equal to the perpendicular height of the pit. Therefore, To find the solidity:-Multiply the square of the diameter of the base by the height of the pit, and the product by .32413. To gauge a coal-pit by the sliding rule:— ་ Place the height of the pit in feet over 1.756 on D, then over the diameter of the base found on D, will be found its solidity in feet on C. Or:-Place the height in feet over 19.87 on D, (calling the number of feet in height so many cords,) and over the diameter in feet found on D, will be found its contents in cords on C. EXAMPLES. 1. How many cubic feet are there in a coal-pit, whose diameter is 12 feet, and height 8 feet ? Ans. 373.4 feet. 2. How many cords of wood in a coal-pit, whose height is 10, and the diameter of its base 15 feet? Ans. 5.68 cords. 3. How many cords of wood in a coal-pit 12 feet in height, and 16 in diameter at the base? Ans. 7.78 cords. Grindstones, in the form of cylinders, are sold and bought by the stone, 24 inches in diameter and 4 inches in thickness being called one stone. To find how many stones of the above dimensions are contained in any stone: Multiply the square of the diameter in inches by its thickness in inches, and divide the product by 2304. To find the number of stones in any grindstone by the sliding rule: Divide the thickness of the stone by 4, and place the quotient over 24 on D; then over the diameter of the stone in inches found on D, will be found the number of stones which it contains on C. EXAMPLES. 1. How many stones 24 inches in diameter and 4 in thickness are contained in a stone 36 inches in diameter and 8 inches thick? Ans. 4.5 stones. Ans. 6.95. 2. How many stones are contained in a grindstone 10 inches in thickness and 40 in diameter ? |