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 Cast Iron =.1365 lbs. 1 × 1 × 1 = 64: = .015625 X .215.00336 lb. Ans. EXAMPLE.Required the weight of a cast iron ball whose diameter is 8 inches. EXAMPLE. - How many leaden balls, having a diameter of an inch each, are there in a pound? 1.00336 = 100000 = 298. Ans. What must be the diameter of a cast iron ball, to EXAMPLE. weigh 69.888 lbs? 69.888.1365 = 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. Ans. 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 61 inches, and interior diameter 4 inches. 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 exterior 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 ? 16 X 10160 ÷ 12 = 13. Ans. 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 2 inches thick; how many feet does it contain, board measure? 5 × 2.5 × 16 ÷ 12 = 1612. Ans. 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 660 X 14 = 840 ÷ 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+✔Aa)÷3=mean area. A and a being the areas of the ends. To measure round timber. RULE (IN GENERAL PRACTICE. - Multiply the length, in feet, by che square of the girt, in inches, taken about the distance from 1 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 X 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 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 X 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 883.141628 inches, the diameter at the distance from the greater base, and retaining this diameter and the length, we may suppose, with sufficient liberality, and without being far from the general run of such sticks, the diameter at the greater base to be 30 inches, and that of the less to be 24 inches, and By a correct rule the stick contains — 30 X 24 720+12=732.7854 X 40-22996-144-159.7 cubic feet, or 19 per cent. more than given by the method under consideration; and we need hardly add that the nearer the stick approaches to the figure of a cylinder, the wider will be the difference between the truth and the result obtained by the method referred to. Thus, suppose the stick a cylinder, 28 inches in diameter, and 40 feet in length; and we have, by the fallacious rule, as above, 134.44 cubic feet; and By a correct method, we have 282.7854 X 40=24630÷144=171 cubic feet, or over 27 per cent. more than furnished by the erroneous mode of practice. Again: suppose the stick in the form of a cone, 30 inches at the base, and tapering to a point at 150 feet in length; and we have, by a correct rule — 3023=300.7854 X 150=35343144245.44 cubic feet; and by the ordinary method of gauging, or the aforementioned practice, we have — 203.141662.832÷4=15.7082 × 15037011.19÷144=257 cubic feet, or nearly 4 per cent. more than the stick actually contains. In short, without taking into account anything for the thickness of the bark, that may be supposed to be on the stick, the method is correct only when the stick tapers at the rate of 5 inches diameter per each 10 feet in length, or over inch diameter to each foot in length of the stick. If, however, we suppose the stick as before, (30 inches at the greater base, 24 inches at the smaller, and 40 feet in length,) and suppose the bark upon it to be 1 inch thick, we shall have, by the usual method, 134.44 cubic feet, as before. And, exclusive of the bark, by a correct method, we shall have. 30-2X 24-2=616+12=628 X.7854 X 40=19729144 = 137 cubic feet, or only about 2 per cent. more than that furnished us by the usual practice. The following simple rule for measuring round timber is sufficiently correct for most practical purposes: RULE. Multiply the square of one-fifth of the mean girt, (exclusive of bark,) in inches, by twice the length of the stick in feet, and divide the product by 144; the quotient will be the solidity in feet. To find the solidity of the greatest rectangular stick that may be cut from a given log, or from a stick of round timber of given dimen sions. RULE.-Multiply the square of the mean diameter of the log, in inches, by half the length of the log, in feet, and divide the product by 144. EXAMPLE. The diameter (exclusive of bark) of the greater base of a stick of round timber is 30 inches, and that of the less base is 24 inches, and the stick is 40 feet in length; required the solidity of the greatest rectangular stick that may be cut from it. 30 × 24+ (30—24)2=732=square of mean diameter,* and 732 X 20=14640÷144-101 cubic feet. * Except in the case of a cylinder, there is a difference betwixt the solid having circular bases, and the middle diameter of that solid. reduces the solid to a cylinder; the middle diameter is the diameter two bases. Ans. mean diameter of a The mean diameter midway between the |