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 auswer. 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? 50×40×60÷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 neces sary 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 2 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. $280 solid feet in the whole wall. 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 that quotient by 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? 72x24×12÷95 RULE II. 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? 84x28x14-95-350,29 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 : 2‡cwt. ;: 12×12×12: 18cwt. Ans. 2. If a 12 inch cable require an anchor of 18 cwt. what must the circumference of a cable be, for an anchor of 24 cwt. ? cwt. cwt. in. As 18: 12×12×12 :: 2,25 : 216 216=6 Ans. ART. 16. Having the dimensions of two similar bui ships of a different capacity, with the burthen of o of them to find the burthen of the other. ART. 14. To find the tonnage of a ship. RULE. nage. EXAMPLE. Suppose a ship 72 feet by the keel, and 24 feet by the beam, and 12 feet deep; what is the tonnage ? 72x24×12÷95-218,2+tons. Ans. EXAMPLE. A ship 84 feet by the keel, 28 feet by the beam; what is the tonnage ? 84×28×14-95-350,29 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: 18cit. Ans. 2. If a 12 inch cable require an anchor of 18 cwt. what must the circumference of a cable be, for an anchor of 21 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 af 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 500 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 75x75x75: 300 :: 100×100×100: 711 2 0 24+ DUODECIMALS, OR CROSS MULTIPLICATION, Is a rule made use of by workmen and artificers in casting up the contents of their work. RULE. 1. Under the multiplicand write thé 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, setting the result of each term two places to the right hand of those in the multiplicand, &c. |