ute the weight of an engine or tender to two or more axles, and prevent excessive load upon one axle by reason of inequalities of track or bed. Locomotives having two driving axles have these two equalized together; those with three or more commonly have the forward driving axle equalized with the leading truck. Equalizers are always used for four-wheeled engine trucks, and frequently with tender trucks. The British name is compensating beam. They are designated "transverse" when they connect the equalizing systems on the two sides of the locomotive. They are also designated by their position as equalizer, top of box; equalizer between drivers, etc.

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Fig. 255. Equalizing Weight on Drivers.

Q. If there are three pairs of drivers with two pairs of equalizers, and a weight of 96,000 pounds is applied vertically at the center of each lever, how much comes on each wheel?

A. On each center wheel 24,000 pounds, on each of the others 12,000.

Q. How must the load be applied so as to give each wheel the same load?

A. At points two-thirds away from the center wheel pair; this will put 16,000 pounds on each wheel of the first and third pairs and twice 8,000 = 16,000 on each one of the center pair. (Fig. 255.)

Q. In equalizing an engine, should the spring hanger length be changed?

A. No; if the spring is weak, liners should be put under the gib in the hanger; when it has full set the liners should be taken out.

Q. Why should the change be made in the spring and not in the hanger?

A. Because the hanger never changes, while the spring constantly does so.

Q. What is a traction increaser?

A. An arrangement for transferring a portion of the weight from the leading or traveling truck to the drivers, to increase the tractive power in starting. It consists of a cylinder supplied with compressed air and containing a piston the rod of which, by operating a set of levers, shifts the fulcrum of the equalizing beam that connects the driving and trailing truck springs on a 4-4-2 type of engines. Locomotives of the 2-6-2 type have the piston rod of one traction-increaser cylinder applied to the equalizer connecting the forward truck with the front transverse equalizer, and that of the other applied to the trailing trucks as above stated. The device is operated from the cab; in some designs the air valve is so connected with the reverse lever that it is closed and the normal distribution of weight on the drivers restored as soon as the reverse lever is moved back to a certain point; say when cutting off at about 60 per cent.

Q. What is the average weight of an American steam passenger locomotive?

A. About 60 tons without tender.

Q. How much of this is on drivers?
A. About 38 tons.

Q. Of freight engines?

A. About 68 tons on drivers.

CHAPTER LXII

COUNTERBALANCE

Q. What is the object of counterbalancing?

A. To equalize the centrifugal force of those reciprocating parts which also have a rotating motion at one end.

Q. What is centrifugal force?

A. The tendency of a revolving body to fly off at a tangent if the restraining force that keeps it at a certain distance from the center, was removed.

Q. What is the rule for figuring the centrifugal (or tangential) force of a revolving body?

A. Multiply the weight of the revolving body (in pounds) by the square of the number of turns per minute, the radius in feet of the center of weight from the center of revolution, and 0.00034; that will give the centrifugal or tangential force in pounds. Thus supposing a weight of 600 lbs., making 200 turns a minute, concentrated at a point 5 ft. from the center of revolution, we have 600 X5 X 40,000 X 0.00034 = 40,800 lbs. (Practically the same result will be obtained by dividing by 2,934, as by multiplying by 0.00034.)

Q. Illustrate by a diagram the force tending to accelerate or to retard the piston during a full stroke.

A. Fig. 256 shows in A B the length of the stroke on any convenient scale, and in A C, B D, the centrifugal force. The distance of a diagonal line from C to D from A B represents the accelerating or the retarding force at various points; being zero at mid-stroke, E.

Q. How would you show the effect of the momentum and inertia of the reciprocating parts, in modifying the effect of centrifugal force?

A. As in Fig. 257, in which the centrifugal force is ac

celerated at one end of the stroke and retarded at another; the straight line C D being replaced by a curve F H, cutting the stroke line at G, before mid-stroke E.

Q. In a two-cylinder locomotive at high speed, what is the tendency of the heavy end of each connecting rod as it rotates?

A. To raise the entire engine on that side when the rod goes up, and to hammer the track as it goes down; one side lifting the engine and the other hammering the track, at the same time; thus also causing a "wee-wawing" or swinging of the entire engine from side to side of the track.

Q. How is this counteracted?

A. It cannot be entirely counteracted on a two-cylinder engine; but the moving weight of the connecting rod may be partly counterbalanced so as to lessen the hammer blow on the track, while increasing the tendency to jerk the train back and forth.

Q. Where are the counterbalance weights placed? A. In the driving wheels, opposite each crank pin. Q. How much counterbalance weight should be thus placed opposite each crank pin?

A. Such a weight as, multiplied by the distance of its center of gravity from the axle center, will equal the weight at the crank pin multiplied by half the piston stroke.

Q. Can the lack of balance in the reciprocating parts be counteracted by giving either lead or compression?

A. No; nothing but weight will remedy it even in part; and the only way by which weight may be made to do it effectually is to have for each crank pin another one connected to rods and parts of equal weight, going in exactly the opposite direction; so that for every pound that goes up there will be another pound coming down at the same time and speed; and for every pound going forward there shall be another coming back at the same time and speed. Q. Suppose that you have a segment-shaped counterweight; how can its center of gravity be found?

A. By cutting out a wood or cardboard templet of even

thickness, of the same size and shape as the weight, and suspending it from several points in its surface, near its rim, by a bradawl thrust through it at right angles to its face; dropping plumb-lines from this awl in the several positions, and marking where they cross the templet face. Where two of these lines intersect will be the center of gravity of the templet, and should be that of the piece of regular thickness which it was made to match.

Q. Where there are two segment-shaped counterweights separated by a spoke, will their common center of gravity be at the same distance from the axle center as that of either one of them?

A. No; it will be nearer the axle.