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 from the fulcrum; therefore, by No. 5,

80 the power required.

9

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?

=

the

Here (6 × 7) + (15 × 8) + (29 × 9) = 423 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.*

3.66 55,500 39,400 14,400 10,800
4.40 38,542 27,361 | 14,400 | 10,800

1,800

1,350

1,800

1,350

5.13 28,316 20,100 14,400 10,800
5.86 21,680 15,390 14,400 10,800 1,800
7.33 13,875 9,850 14,400 10,800 1,800

1,800

1,350

1,350

1,350

8.80 9,635 6,840 14,400 10,800 1,800 10.26 7,080 5,026 14,400 10,800 1,800

1,350

1,350

8

11.73 5,420

3,848 14,400 10,800

1,800 1,350

4,282 3,040 14,400 10,800
3,468|| 2,462 14,400 10,800
|
1,900
1,350 14,400 10,800

* 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.

TRACTION IN LBS. WHEN THE ADHESION IS IN THE FOLlowing Ratios.

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.

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

Mode of Setting the Eccentrics on the Main Shaft of Driving Wheels of Locomotives.

We suppose the use of an additional eccentric for working the valves half stroke.

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.