PROBLEM I To Find the Weight of a Body from its Magnitude. As one cubic foot, or 1728 cubic inches, So is its tabular specific gravity, To the weight of the body. EXAMPLES. EXAM. 1. Required the weight of a block of marble, whose length is 63 feet, and breadth and thickness each 12 feet being the dimensions of one of the stones in the walls of Balbeck? Ans. 683,7 ton, which is nearly equal to the burden of an East-India ship. EXAM. 2. What is the weight of 1 pint, ale measure, of gunpowder ? Ans. 19 oz nearly. EXAM. 3. What is the weight of a block of dry oak, which measures 10 feet in length, 3 feet broad, and 21 feet deep; Ans. 43351 lb. PROBLEM III. To find the Specific Gravity of a Body. CASE 1. When the body is heavier than water, weigh it both in water and out of water, and take the difference, which will be the weight lost in water. Then say, As the weight lost in water, Is to the whole weight, So is the specific gravity of water, EXAMPLE. A piece of stone weighed 10lb, but in water only 61lb, required its specific gravity? Ans. 2609. CASE 2. When the body is lighter than water, so that it will not quite sink, affix to it a piece of another body, heavier. than water, so that the mass compounded of the two may sink together. Weigh the denser body and the compound mass separately, both in water and out of it; then find how much each loses in water, by subtracting its weight in water from its weight in air; and subtract the less of these remainders from the greater. Then say, As ! As the last remainder, Is to the weight of the light body in air, To the specific gravity of the body. EXAMPLE. Suppose a piece of elm weighs 15lb in air; and that a piece of copper which weighs 181b in air, and 16lb in water, is affixed to it, and that the compound weighs 6lb in water; required the specific gravity of the elm? Ans. 600. PROBLEM IV. To find the Quantities of Two Ingredients, in a given Compound. TAKE the three differences of every pair of the three specific gravities, namely, the specific gravities of the compound and each ingredient; and multiply the difference of every two specific gravities by the third. Then say, as the greatest product, is to the whole. weight of the compound, so is each of the other products, to the two weights of the ingredients. EXAMPLE. A composition of 1121b being made of tin and copper, whose specific gravity if found to be 8784; required the quantity of each ingredient, the specific gravity of tin being 7320, and of copper 9000? Ans. there is 100lb of copper in the composition. and consequently 12lb of tin OF THE WEIGHT AND DIMENSIONS OF BALLS AND SHELLS. THE weight and dimensions of Balls and Shells might be found from the problems last given, concerning specific gravity. But they may be found still easier by means of the experimented weight of a ball of a given size, from the known proportion of similar figures, namely, as the cubes of their diameters. PROBLEM I. To find the Weight of an Iron Ball, from its Diameter. An iron ball of 4 inches diameter weighs 91b, and the weights being as the cubes of the diameters, it will be. as 64 VOL. If 32 (which A (which is the cube of 4) is to 9 its weight, so is the cube of the diameter, of any other ball, to its weight. Or, take of the cube of the diameter, for the weight. Or, take of the cube of the diameter, and of that again, and add the two together, for the weight. EXAMPLES. EXAM. 1. The diameter of an iron shot being 6.7 inches, required its weight? Ans. 42-294lb. EXAM. 2. What is the weight of an iron ball, whose diameter is 5.54 inches? Ans. 24lb nearly. PROBLEM II. To find the Weight of a Leaden Ball. 14 A leaden ball of one inch diameter weighs of a lb; therefore as the cube of 1 is to, or as 14 is to 3, so is the cube of the diameter of a leaden ball, to its weight. Or, take of the cube of the diameter, for the weight, nearly. EXAMPLES. 14 EXAM. 1. Required the weight of a leaden ball of 6.6 inches diameter ? Ans. 61.606lb. EXAM. 2. What is the weight of a leaden ball of 5:30 inches diameter ? Ans. 32lb nearly. PROBLEM III. To find the Diameter of an Iron Ball. MULTIPLY the weight by 74, and the cube root of the product will be the diameter. EXAMPLES. EXAM. 1. Required the diameter of a 421b iron ball? Ans. 6.685 inches. Ans. 5.54 inches. EXAM. 2. What is the diameter of a 24lb iron ball? To find the Diameter of a Leaden Ball. MULTIPLY the weight by 14, and divide the product by 3; then the cube root of the quotient will be the diameter. EXAMPLES. EXAMPLES. EXAM. 1. Required the diameter of a 64lb leaden ball? Ans. 6.684 inches. EXAM. 2. What is the diameter of an 8lb leaden ball? 64 PROBLEM V. Ans. 3.343 inches. To find the Weight of an Iron Shell. TAKE of the difference of the cubes of the external and internal diameter, for the weight of the shell. That is, from the cube of the external diameter, take the cube of the internal diameter, multiply the remainder by 9, and divide the product by 64. ÉXAMPLES. EXAM. 1. The outside diameter of an iron shell being 12-8, and the inside diameter 9.1 inches; required it weight? Ans. 188.941lb. EXAM. 2. What is the weight of an iron shell, whose external and internal diameters are 9.8 and 7 inches ? PROBLEM VI. Ans. 24 lb. To find how much Powder will fill a Shell. DIVIDE the cube of the internal diameter, in inches, by 57.3, for the pounds of powder. EXAMPLES. EXAM. 1. How much powder will fill the shell whose internal diameter is 9.1 inches? Ans. 13,2lb nearly. EXAM. 2. How much powder will fill a shell whose internal diameter is 7 inches? Ans. 6lb. PROBLEM VII. To find how much Powder will fill a Rectangular Box. FIND the content of the box in inches, by multiplying the length, breadth, and depth all together. Then divide by 30 for the pounds of powder. EXAMPLES. EXAM. 1. Required the quantity of powder that will fill a box, the length being 15 inches, the breadth 12, and the depth 10 inches? Ans. 60lb. EXAM. 2. How much powder will fill a cubical box whose side is 12 inches? PROBLEM VIH. Ans. 573lb. To find how much Powder will fill a Cylinder. MULTIPLY the square of the diameter by the length, then divide by 38.2 for the pounds of powder. EXAMPLES. EXAM. 1. How much powder will the cylinder hold, whose diameter is 10 inches, and length 20 inches? Ans. 5211b nearly. PROBLEM IX. To find the Size of a Shell to contain a Given Weight of Powder, MULTIPLY the pounds of powder by 57.3, and the cube root of the product will be the diameter in inches, EXAMPLES. EXAM. 1. What is the diameter of a shell that will hold 131 of powder? Ans 9.1 inches. EXAM. 2. What is the diameter of a shell to contain 61b of powder ? Ans. 7 inches. PROBLEM X. To find the Size of a Cubical Box to contain a given Weight of Powder. MULTIPLY the weight in pounds by 30, and the cube root of the product will be the side of the box in inches. EXAMPLES. EXAM. 1. Required the side of a cubical box, to hold 50lb of gunpowder? Ans. 11·44 inches. EXAM. 2. Required the side of a cubical box, to hold 400lb of gunpowder? Ans. 22.89 inches, PROBLEM XI. To find what Length of a Cylinder will be filled by a given MULTIPLY the weight in pounds by 38.2, and divide the product by the square of the diameter in inches, for the length. EXAMPLES. |