The following TABLE exhibits the coefficients of weight, in pounds, of one foot in length, of various thicknesses, of different kinds of pipe, of any diameter whatever. To obtain the weight of pipes by means of the above TABLE — RULE. - Multiply the diameter of the pipe, taken from the interior surface of the metal on the one side to the exterior surface on the opposite, (interior diameter + thickness,) in inches, by the number in the table under the respective metal's name, and opposite the thickness corresponding to that of the pipe-the product will be the weight, in pounds, of ONE foot in length of the pipe, and that product multiplied by the length of the pipe, in feet, will give the weight for any length required. EXAMPLE. Required the weight of a copper pipe whose length is 5 feet, interior diameter and thickness 3 inches, and thickness inch. 31 = 25 3.125 X 1.516 X 5 23.687 lbs. Ans. of an EXAMPLE. Required the weight of a cast iron pipe, 10 feet in length, whose interior diameter is 3 feet, and whose thickness is 1 inch. 37 X 9.829 X 10 =3636.73 lbs. Ans. 36 +1 WEIGHT OF CAST IRON AND LEAD BALLS. To find the weight of a sphere or globe of any material · RULE. Multiply the cube of the diameter, in inches, or feet, by the weight of a spherical inch or foot of the material. The weight of a spherical inch of Therefore · EXAMPLE. — Required the weight of a leaden ball whose diameter is of an inch. 1 1 × 1 × 1 = 64 .015625 .215 .00336 lb. Ans. 4 EXAMPLE. is 8 inches. Required the weight of a cast iron ball whose diameter 83 × .1365 — 69.888 lbs. Ans. EXAMPLE. — How many leaden balls, having a diameter 4 of an inch cach, are there in a pound? EXAMPLE. What must be the diameter of a cast iron ball, to weigh 69.888 lbs? 69.888.1365 3 512 8 inches. Ans. EXAMPLE. — What must be the diameter of a leaden ball to equal in weight that of a cast iron ball, whose diameter is 8 inches? [Lead is to cast iron as .215 to .1365, as 1.575 to 1.] WEIGHT OF HOLLOW BALLS OR SHELLS. The weight of a hollow ball is the weight of a solid ball of the same diameter, less the weight of a solid ball whose diameter is that of the interior diameter of the shell. EXAMPLE. Required the weight of a cast iron shell whose exterior diameter is 6 inches, and interior diameter 44 inches. 61 25 X 25 X 25 244.14 X .1365 33.33 10.48 22.85 lbs. Ans. Or, If we multiply the difference of the cubes, in inches, of the two diameters - the exterior and interior-by the weight of a spherical inch, we shall obtain the same result. EXAMPLE.Required the weight of a cast iron shell whose ex terior diameter is 10 inches and interior diameter 8 inches. 10383 X .1365 = 66.612 lbs. Ans. Woods of most descriptions vary little from 80 per cent. volatile matter, and 20 per cent. charcoal. TABLE Exhibiting the Weights, Evaporative Powers, &c., of Fuels, from Report of Professor Walter R. Johnson. MENSURATION OF LUMBER. To find the contents of a board. RULE. Multiply the length in feet by the width in inches, and divide the product by 12; the quotient will be the contents in square feet. EXAMPLE. A board is 16 feet long and 10 inches wide; how many square feet does it contain? To find the contents of a plank, joist, or stick of square timber. RULE. — Multiply the product of the depth and width in inches by the length in feet, and divide the last product by 12; the quotient is the contents in feet, board measure. EXAMPLE. A joist is 16 feet long, 5 inches wide, and 21⁄2 inches thick; how many feet does it contain, board measure? To find the solidity of a plank, joist, or stick of square timber. RULE.Multiply the product of the depth and width in inches by the length in feet, and divide the last product by 144; the quotient will be the contents in cubic feet. EXAMPLE. A stick of timber is 10 by 6 inches, and 14 feet in length; what is its solidity? 10 X 6 60 X 14840 ÷ 144 = 5 feet. Ans. NOTE.--If a board, plank, or joist is narrower at one end than the other, add the two ends together and divide the sum by 2; the quotient will be the mean width. And if a stick of squared timber, whose solidity is required, is narrower at one end than the other (A+a+√ Ãa)÷ 3 = mean area. a being the areas of the ends. To measure round timber. A and RULE (IN GENERAL PRACTICE.) - Multiply the length, in feet, by the square of the girt, in inches, taken about the distance from the larger end, and divide the product by 144; the quotient is considered the contents in cubic feet. For a strictly correct rule for measuring round timber, see MENSURATION OF SOLIDS · Frustum of a Cone. EXAMPLE. A stick of round timber is 40 feet in length, and girts 88 inches; what is its solidity? 88÷4=22× 22 = 484 × 40≈ 19360 ÷ 144 — 134.44 cub. ft. Ans. The following TABLE is intended to facilitate the measuring of Round Timber, and is predicated upon the foregoing RULE. To find the solidity of a log by help of the preceding TABLE. RULE. Multiply the tabular area opposite the corresponding 4 girt, by the length of the log in feet, and the product will be the solidity in feet. EXAMPLE. The girt of a log is 22 inches, and the length of the log is 40 feet; required the solidity of the log. 3.362 × 40 = 134.48 cubic feet. Ans. NOTE. Though custom has established, in a very general way, the preceding method as that whereby to measure round timber, and holds, in most instances, the solidity to be that which the method will give, there seems, if the object sought be the real solidity of the stick, neither accuracy, justice, nor certainty, in the practice. Thus, in the preceding example, the stick was supposed to be 40 feet in length, and 88 inches in circumference at the distance from the larger end, and was found, by the method, to contain 134.44 cubic feet: now 88 — 3.1416 = 28 inches, 28 inches, the diameter at f the distance from the greater base, and retaining this diameter and the length, we may |