Table 158.-FRACTIONAL PARTS OF A POUND AVOIRDUPOIS AND THEIR SIZE AND WEIGHT OF BRICKS AND TILES. London stock bricks, size in inches 8 × 4× 2 weight each 6 lbs. Table 159.-FRACTIONAL PARTS OF A HUNDREDWEIGHT AND THEIR DECIMAL EQUIVALENTS. DECIMAL APPROXIMATES, ETC. Area of a circle = diameter 2 x 7854. Area of a circle x 6366 area of inscribed square. Area of an ellipse = the product of the two axes × 7854. Circumference of a circle = diameter x 3'1416. The circumference of a circle is nearly equal to 22 times the diameter divided by 7. Circumference of a circle x 2821 = side of a square of equal area. Diameter of a circle = circumference 3'1416. Diameter of a circle = square root of the quotient of the area divided by *7854. The diameter of a circle is nearly equal to 7 times the circumference divided by 22. The difference of the diameters of any two circles, multiplied by 31416, will give the difference of their circumference. Cubic inches x '028848 = pints. Diameter of a sphere x 6667 length of equal cylinder. Side of a square × 11284 = diameter of a circle of equal area. Side of a square multiplied by 3'545 circumference of a circle of equal area. Side of an inscribed square x 1'4142 = diameter of the circumscribing circle. Side of an inscribed square × 4'4430 = circumference of the circumscribing circle. = Circular inches multiplied by 7854 square inches. Feet multiplied by 15 feet. links. = Square links multiplied by 4356 square feet. Miles multiplied by 87 knots. = Statute acres multiplied by 4840 Grains multiplied by 0001429 square yards. lbs. avoirdupois. Pounds avoirdupois multiplied by 7000 = grains. tons. bushels. lbs. avoirdupois. lbs. avoirdupois. Area of egg-shaped sewer = one-half the square of the depth. Gravity.—To find the velocity in feet per second acquired by a falling body.-Rule: Multiply the time in seconds by 32°2. To find the height of the fall in feet.-Rule: Multiply the square of the time in seconds by 16.1. To find the time in falling in seconds.-Rule: Divide the velocity in feet per second by 32°2. To find the velocity in feet per second for a given height.-Rule: Multiply the height of the fall in feet by 644, and take the square root of the product. Work accumulated in a Moving Body.-To find the force acquired by a weight in falling freely from a given height.-Rule: Multiply the weight in lbs. by the square of the velocity in feet per second, and divide by 64 4. The result is the accumulated work in foot pounds. Or another rule for the same is: Multiply the weight in lbs. by the height in feet of free fall. The product is the accumulated work in foot pounds, or the force that would raise a similar weight to a similar height. The following examples of accumulated work show the application of these rules : To find the distance in feet a ball will traverse before coming to a state of rest, say, on a bowling green, at a velocity of 50 feet per second; weight frictional resistance to its motion being th the 502 velocity x 20 lbs. weight of ball, 20 lbs., and the weight of the ball; then = 338 feet. 2 lbs. frictional resistance × 64'4 To find the distance in feet a train will move on a level rail, whose frictional resistance is 8 lbs. per ton, and supposing that there is no other resistance; the weight of the train being, say, 100 tons, and its velocity when the steam is shut off, 50 feet per second; 502 velocity × 100 tons weight of train × 2,240 lbs. then 100 tons weight x 8 lbs. per ton frictional resistance × 64°4 108695 feet before coming to rest. Punching and Shearing Iron, &c., Plates.-Punching.-The resistance of a wrought-iron plate to punching is about the same as its resistance to tearing. Taking the maximum resistance at 25 tons per square inch, and the resistance to the punch being the area of the metal separated, or the circumference of the hole multiplied by the thickness of the plate, the force in tons required to punch a plate of wrought-iron is = circumference of the hole x its depth x 25. And a simple rule to find the force required to punch a plate is :-Multiply the diameter of the hole in 16ths of an inch by the thickness of plate in 16ths, and divide the product. by 10; which result multiply by 31 for wrought iron plates; by 45 for steel plates; and by 25 for copper plates. The final product will be the required force in tons. The compressive strength of a hardened steel punch is 100 tons per square inch, or four times greater than the maximum tensile strength of wrought-iron plates. The smallest size of hole that can be punched, is that of which the diameter is equal to the thickness of the plate. Shearing. The resistance of a wrought-iron plate to shearing is 20 per |