EXAMPLE. What number of revolutions ought a governor to make per minute whose pendulums are 24 inches long? 375 √24 = 76 ÷ 2 = 38 revolutions per minute. The Revolutions per Minute of a Governor given, to find the Length of Pendulums. ROLE. Divide 375 by twice the number of revolutions per minute; and the square of the quotient will be the length required. EXAMPLE. When the velocity of a governor is 38 revolutions per minute, what ought to be the length of pendulums ? The effective power obtained by means of a high pressure engine, is nearly two-thirds of the force of the steam; one-third being expended in friction, &c. ;* hence, multiply the cylinder's area in inches by the force of the steam in pounds, and by the velocity of the piston in feet per minute; deduct of the product, and divide the remainder by 33,000, the quotient will be the force of the engine expressed in horses' power. EXAMPLE.-Required the power of an engine, the cylinder being 8 inches diameter, and stroke 2 feet; the engine making 50 revolutions per minute, and the ⚫ There is always a resistance of steam on the piston of a high pressure engine equal to the pressure of the atmosphere, but this cannot be taken into account unless we also take into account the pressure of the atmosphere upon the boiler. weight upon the safety valve equal 30 lbs. per square inch. 8 inches diameter 50.2656 inches area; and 50 revolutions x 4 feet = 200 feet velocity; then, 50.2656 x 30 lbs. x 200 x 2 201062.4 3 6.09 horses' power nearly. 33000 Or, multiply 49,500 by the number of horses' power required; divide the product by the force of the steam in pounds, multiplied by the velocity of the piston in feet per minute, and the quotient will be the area of the cylinder. EXAMPLE.-Required the diameter of a cylinder for an engine of 12 horses' power, working pressure 35 lbs. per square inch, length of stroke 2 feet 6 inches, and making 45 revolutions per minute. 49500 × 12 594000 = = 75.43 inches area, or 45 × 5 × 35 7875 To find an Equivalent Force of the Steam, when the Engine is working expansively. RULE 1. Divide the length of the stroke in inches by the distance (also in inches) that the piston moves before the steam is shut off, and divide the pressure on the boiler in pounds by the quotient. 2.-Add 1 to the hyperbolic logarithm of the number of times to which the steam is expanded, and multiply the logarithm by the number of pounds to which the steam is expanded, and the product is the uniform force of the steam acting throughout the whole stroke. EXAMPLE.-Let the steam in the boiler of a high pressure engine equal 45 lbs. per inch, the length of stroke 4 feet, and the steam to be shut off after the piston has moved 16 inches; required an equivalent force of the steam in the cylinder. 4 feet = 48 inches, and 48 16 = 3. Then 453 = 15 lbs. And, I + 1.0986123 = 2.0986123 × 15 = 31.4791845 lbs. uniform force of the steam. 14 .2231435 34 1.1786549 5 1.6582280 11122223 .4054651 3 14 .5596157 3 .6931472 4 .8109302 4 24 .9162907 4 24 1.0116008 44 24 1.0986123 5 74 1.9810014 72.0149030 74 2.0476928 1.2527629 5 1.7047481 MISCELLANIES. Approximate Rules for finding the Weight of Round, Square, and Rectangular Beams, Bars, &c. of Cast and Wrought Iron. RULE 1.-Multiply the square of the diameter in inches by the length in feet, and by 2.6 for wrought iron, or 2.48 for cast iron; and the product will be the weight in pounds avoirdupois nearly. 2.-Multiply the area of the cross section in inches by the length in feet, and by 3.32 for wrought iron, or 3.16 for cast iron; and the product will be the weight in pounds avoirdupois nearly. EXAMPLE 1.-Required the weight of a round bar of wrought iron 14 feet long and 2 inches diameter. 2.52 x 14 = 87.50 × 2.6 = 227.5 lbs. EXAMPLE 2. The length of a piece of cast iron is 9 feet, its breadth 7 inches, and thickness 24, required its weight. 2.25 × 7 = 15.75 × 9.5 = 149.625 × 3.16 = 472.815 lbs. The Dimensions of a Cast Iron Ring being given, to find its Weight nearly. RULE.-Multiply the breadth of the ring added to the inner diameter by .0074, and that again by the breadth and by the thickness, and the product will be its weight in cwts. nearly. N EXAMPLE.-Required the weight of a ring whose dimensions are 8 feet 4 inches interior diameter, 5 inches broad, and 4 inches thick. 8 feet 4 inches .777 x 53.885 x inches. 100+ 5 = 105 × .0074 = 4: = 15.52 cwts. nearly. To find the Weight of any Cast Iron Ball whose Diameter is given. RULE.-Multiply the cube of the diameter in inches by .1377, and the product will be the weight in avoirdupois pounds nearly. EXAMPLE.-Required the weight of a ball 7 inches diameter. 73343 x .1377 47.2211 lbs. To find the Diameter of a Cast Iron Ball when the Weight is given. RULE.-Multiply the cube root of the weight, in pounds, by 1.936, and the product will be the diameter in inches nearly. EXAMPLE.-Required the diameter of a ball that will weigh 64 pounds. 364 = 4 x 1.936 = 7,744 inches diameter. |