In a lever of the second kind, a power of 3 acts at a distance of 12; what weight can be balanced at a distance of 4 from the fulcrum? Here, by No. 6, 3 x 12 = 9 weight. 4 In a lever of the third kind the weight is 60, and its distance 12, and the power acts at a distance of 60 × 12 9 9 from the fulcrum; therefore, by No. 5, 80 the power required. If there be a lever of the first kind, having three weights, 7, 8, and 9, at the respective distances of 6, 15, and 29, from the fulcrum on one side, and a power of 17 at the distance of 9 on the other side of the fulcrum, then a power is to be applied at the distance of 12 from the fulcrum, in the last-mentioned side; what must that power be to keep the lever in balance? = Here (6 × 7) + (15 × 8) + (29 × 9) = 423 the effect of the three weights on the one side of the fulcrum, and 17 x 9 153 the effect of the power on the other side. Now, it is clear that the effect of the weight is far greater than the effect of the power; and the difference, 423—153 = 273, requires to be balanced by a power applied at the distance of 12, which will evidently be found by dividing 270 by 12, which gives 22.5, the weight required. 14. The Roman steel-yard is a lever of the first kind, so contrived that only one movable weight is employed. TABLE showing the Effects of a Force of Traction of 100 pounds, at different Velocities, on Canals, Railroads, and Turnpike Roads.* VELOCITY OF Miles Feet per Hour. 21 per Second. 8 9 10 lbs. lbs. lbs. lbs. 3.66 55,500 39,400 14,400 10,800 7.33 13,875 9,850 14,400 10,800 1,800 13.20 14.66 13.5 19.9 Useful On a level Turn pike Road. Tot. Mass lbs. 1,800 1,800 1,800 Useful effect. lbs. 1,350 1,350 1,350 1,350 1,350 1,350 1,350 1,800 1,350 1,800 | 1,350 1,800 1,350 1,800 1,350 * The force of traction on a canal varies as the square of the velocity; but the mechanical power necessary to move the boat, is usually reckoned to increase as the cube of the velocity. On a railroad or turnpike, the force of traction is constant, but the mechanical power necessary to move the carriage increases as the velocity. TABLE of the Tractive Power of the Locomotive Engine, when the adhesion is from one-fifth to one-fifteenth that of the insistent weight of the Driving Wheels, созвать FOLLOWING 13 ΤΟ 1244.4 1120 1493.3 1344 1792 1568 1991-1 1792 2240 2016 2489 2240 2737.7 2464 2986.6 2688 3235.5 2912 3484.4 3136 3733.3 3360 3982.2 4231.1 f # 8 2240 1866.6 1600 1400 2688 2440 1920 1680 3136 2613.3 2240 1960 3584 2986.6 2560 2240 4032 3360 2880 2520 4480 3733.3 3200 2800 4928 4106.6 3520 3080 5376 4480 3840 3360 5824 4853.3 4160 3640 6272 5226.6 4480 3920 6720 5600 4800 4200 5973.3 7168 5120 4480 7610 6346.7 5440 4760 8064 6720 5760 5040 8512 7093.3 6080 5320 8960 7466.7 6400 5600 4977.7 9408 7840 6720 5880 5226.6 9856 8213.3 7040 6160 5475.5 8586.7 10304 6440 5724-4 10725 7300 8960 7680 5973.3 6720 7000 8000 9333.3 11200 6222.2 4480 4729 3584 3808 4032 4256 4480 4704 4928 5152 5376 5600 2851 3054.5 2800 3280 2986'6 2757 3483.7 3173.3 2929.3 3687.5 3360.0 3101.6 3891 3546.7 3274.0 4094-8 3733.3 3446.3 3920 4298.3 3618.6 4106.7 4502.2 3791.0 4293.3 4705.7 3963.3 4909.5 4480 4135.6 4666.7 4307.6 5091 14 800 960 1120 1280 1440 1600 1760 1920 2080 2240 2400 2560 2720 2880 3040 3200 3360 3520 3680 3840 4000 1/ 746.6 896 1045.3 1194.6 1344 1493.3 1642.6 1792 1941.3 2090.6 2240 2389.3 2538.6 2688 2837.3 2986.7 3136 3285.3 3334.6 3484 3633.3 TO CONSTRUCT AN ECCENTRIC WHEEL. From the centre of the shaft O take O P equal to half the length of the stroke which you intend the wheel to work; and from P as a centre, with any radius greater than P D, describe a circle, and this circle will represent the required wheel. For every circle, drawn from the centre P, will work the same length of stroke, whatever may be its radius; as, whatever you increase the distance of the circumference of the circle from the centre of motion on the one side, you will have a corresponding increase on the opposite side equal to it. Thus, suppose an eccentric wheel to work a stroke of 18 inches is required, the diameter of the shaft being 6 inches; and if 2 inches be the thickness of metal necessary for keying it on to the shaft, then set off, from 0 to P, 9 inches; and 95 14 inches, the radius of the wheel required. Fig. 50. Formulæ. Let S represent the space the end A is moved through by the eccentric wheel, and s the space the slide moves. Then, A B × s = B C × S; and this equation, solved for A B, B C, S, and s, gives the follow ing: A B BCXS S ABX S BC D C B A B (4.) Mode of Setting the Eccentrics on the Main Shaft of Driving Wheels of Locomotives. E F (1.) (2.) FM S We suppose the use of an additional eccentric for working the valves half stroke. Fig. 51. BM ABX S B C BCXS A B (3.) Crank pin. Fig. 50 represents the true position of eccentrics on right-hand side of engine when the rock arm is used. Cylinders and eccentric rod supposed to be horizontal, the crank being on its forward centre, g. |
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