324x,7854-254,4696 inches, area of the base. 20 length in feet. 144)5089,3920(35,343 solid feet, Ans. ART. 10. To find how many solid feet a round stick of timber, equally thick from end to end, will contain when hewn square. RULE. Multiply twice the square of its semi-diameter in inches by the length in feet, then divide the product by 144, and the quotient will be the answer. EXAMPLE. If the diameter of a round stick of timber be 22 inches and its length 20 feet, how many solid feet will it contain when hewn square? 11x11x2x20÷144-33,6+ feet, the solidity when hewn square. ART. 11. To find how many feet of square edged boards of a given thickness, can be sawn from a log of a given diameter. RULE. Find the solid content of the log, when made square, by the last article-Then say, As the thickness of the board including the saw calf is to the solid feet :: so is 12 (inches) to the number of feet of boards. : EXAMPLE. How many feet of square edged boards, 13 inch thick, including the saw calf, can be sawn from a log 20 feet Long and 24 inches diameter ? 12×12×2×20÷144-40 feet, solid content. As 13: 40; 12: 384 feet, the Ans. ART. 12. The length, breadth and depth of any square box being given, to find how many bushels it will contain. RULE. Multiply the length by the breadth, and that product by the depth, divide the last product by 2150,425 the solid inches in a statute bushel, and the quotient will be the answer. EXAMPLE. There is a square box, the length of its bottom is 50 inches, breadth of ditto 40 inches, and its depth is 60 inches; how many bushels of corn will it hold? 50x40x60 2150,425-55,84+ or 55 bushels, three pecks. Ans. ART. 13. The dimensions of the walls of a brick building being given, to find how many bricks are necessary to build it. RULE. From the whole circumference of the wall measured round on the outside, subtract four times its thickness, then multiply the remainder by the height, and that product by the thickness of the wall, gives the solid content of the whole wall; which multiplied by the number of bricks contained in a solid foot, gives the answer. EXAMPLE. How many bricks 8 inches long, 4 inches wide, and 24 inches thick, will it take to build a house 44 feet long, 40 feet wide, and 20 feet high, and the walls to be one foot thick ? 8×4×2,5=80 solid inches in a brick, then 1728÷80 =21,6 bricks in a solid foot. 44+40+44+40=168 feet, whole length of wall. -4 four times the thickness. 164 remains. Multiply by 20 height. 3280 solid feet in the whoe walk Multiply by 21,6 bricks in a solid foot. Product, 70848 bricks. Ans. ART. 14. To find the tonnage of a ship. RULE. Multiply the length of the keel by the breadth of the beam, and that product by the depth of the hold, and divide the last product by 95, and the quotient is the ton nage. EXAMPLE. Suppose a ship 72 feet by the keel, and 24 feet by the beam, and 12 feet deep; what is the tonnage ? 72×24×12÷÷95=218,2+tons. Ans. Multiply the length of the keel by the breadth of the beam, and that product by half the breadth of the beam, and divide by 95. EXAMPLE. A ship 84 feet by the keel, 28 feet by the beam; what 'is the tonnage ? 84x28×14-95-350,29 tons. Ans. ART. 15. From the proof of any cable, to find the strength of another. RULE. The strength of cables, and consequently the weights of their anchors, are as the cube of their peripheries. Therefore; As the cube of the periphery of any cable, Is to the weight of its anchor; So is the cube of the periphery of any other cable, EXAMPLES. 1. If a cable 6 inches about, require an anchor of 24 cwt. of what weight must an anchor be for a 12 inch cable ? As 6×6×6: 24cwt.:: 12×12×12: 18cwt. Ans. 2. If a 12 inch cable require an anchor of 18 cwt. what must the circumference of a cabie be, for an anchor of 24 cwt. ? cwt. cwt. in. As 18: 12x12×12 :: 2,25 : 216 216 6 Ans. ART. 16. Having the dimensions of two similar built ships of a different capacity, with the burthen of one of them, to find the burthen of the other RULE. The burthens of similar built ships are to each other, as the cubes of their like dimensions. EXAMPLE. If a ship of 300 tons burthen be 75 feet long in the keel, I demand the burthen of another ship, whose keel is 100 feet long? T.cwt.grs.lb. As 75×75X75: 300 :: 100X100X100: 711 2 0 24+ Is DUODECIMALS, OR CROSS MULTIPLICATION, a rule made use of by workmen and artificers in casting up the contents of their work. RULE. 1. Under the multiplicand write the corresponding denominations of the multiplier. 2. Multiply each term into the multiplicand, beginning at the lowest, by the highest denomination in the multiplier, and write the result of each under its respective term; observing to carry an unit for every 12, from each lower denomination to its next superior. 3. In the same manner multiply all the multiplicand by the inches, or second denomination, in the multiplier, and set the result of each term one place removed to the right hand of those in the multiplicand. 4. Do the same with the seconds in the multiplier, set ting the result of each term two places to the right hand of those in the multiplicand, &c. |