Modern Machine-Shop Practice Part 235

If the amount of valve travel is given, however, all the other elements may readily be found by the following construction:

[Ill.u.s.tration: Fig. 3311.]

Suppose that in Fig. 3311 a D valve is to be designed to cut off the steam when the piston has travelled from position B' to R', or at three-quarters of its stroke. Then to find the position the crank pin will be in when the cut off occurs, we draw a circle, B D, representing the path of the crank on the same scale that the length of the piston stroke is represented. The straight line from B to D will, therefore, represent the piston stroke without drawing the piston or cylinder at all (this being done in the figure to make the explanation clear). When the crank is on its dead centre, B, the piston, will be at B', and the valve in the position shown (supposing it to have no lead). As soon as the crank and valves begin to move, the steam will enter steam port _a_, and to find where the crank will be when the piston is at three-quarters stroke, and is, therefore, in position R', we mark a point at R three-quarters of the distance from B to D. Then, taking no account of the length of the connecting rod, we draw a vertical line Y from R to the circle, and this line gives at H the position the crank will be in when the piston is at R. We have so far, therefore, that while the piston travels from B' to R', the crank will travel from B to H. Now, it will be found that if we set a pair of compa.s.ses from B to F, which is half-way from B to H, and then rest the compa.s.ses at D, and mark an arc V, then a line from V to the centre of the crank will give us the proper position of the eccentric. As the centre of the crank pin and also the centre of the eccentric both travel in a circle, we may, therefore, take a circle having a diameter equal to twice the throw of the eccentric, (or, what is the same thing, equal to the full travel of the valve), and let it represent the paths of both the eccentric centre and the crank pin centre, the latter being drawn to a scale that is found by dividing the length of the piston stroke by the travel of the valve; thus, if the travel is 3 inches and the stroke 30 inches, the diameter of a 3 inch circle will represent the valve travel full size, and the piston stroke one-tenth full size, because 30 3 = 10. It has been shown on page 376 that the length of the connecting rod affects the motion of the piston by distorting it, and it is necessary to take this into account in constructing the actual diagram, which may be done as follows:

The valve travel and point of cut off being given, to find the required amount of lap, there being no lead, draw a circle equal in diameter to the travel of the valve, and draw the line of centres B D, Fig. 3312; mark on the line of centres a point R, representing the position the piston is to be in at the time the cut off is to take place.

Set a pair of compa.s.ses to represent the length of the connecting rod on the same scale as the circle B D represents the path of the crank; thus, if the connecting rod is three times the length of the stroke, the compa.s.ses would be set to three times the diameter of the circle B D.

[Ill.u.s.tration: Fig. 3312.]

A straight line from B to D and pa.s.sing through the centre C of the crank will represent the line of centres of the engine, which must be prolonged to the right sufficiently to rest the compa.s.ses on it and draw the arc Y, which will give at H the position of the crank when the piston is at R, and the cut off is to occur.

We have thus found that the amount of circular path the crank will move through from the dead centre to the point of cut off is from B to H, and as the eccentric is fast upon the same shaft, it will, in the same time, of course, move through the same part of a circle.

One half of its motion will be to open and one half to close the port, so that we may by means of the arcs at F get the point F, which is midway between B and H, and with the compa.s.ses set from B to F, mark from D the two arcs V and V' whose distance apart will obviously be the same as from B to H.

Then from V to V' draw the line P, and from this line to the centre C of the crank shaft is the amount of steam lap necessary for the valve, while from this line (P) to D is the width of the steam port.

The proof of the diagram is as follows:

When the crank is on the dead centre, the centre of the eccentric is at V, its throw line being represented by the line from V to C, and the valve is about to open the port as shown in the figure.

While the eccentric is moving from V to D, the valve will move in the direction of the arrow and will fully open the port, while the crank pin will move from B to F.

Then, while the crank moves from F to H, the eccentric will have moved the valve to the position it occupies in the figure, having closed the port and effected the cut off.

We have here found the amount of lap and the position of the eccentric necessary for a given point of cut off when the latter is given in terms of the piston stroke. If, however, the point of cut off had been given in terms of the crank pin position, we might find the required amount of lap at once, by simply drawing a line from the centre B, the point to H where the crank pin is to be when the cut off occurs.

From this line we could then draw the dotted circle G, and just meeting the line P, which would give the eccentric position.

To find the piston position, the arc Y would require to be drawn by the same means as before.

[Ill.u.s.tration: Fig. 3313.]

[Ill.u.s.tration: Fig. 3314.]

If the valve is to have lead, the diagram may be constructed as in Fig.

3313, in which the circle has a diameter equal to the travel of the valve and the cut off is to occur when the piston is at R and the crank at H.

When the valve is at the end of its travel and has fully opened the port, the eccentric will be at D, hence from D we mark an arc G distant from D to an amount equal to the width of the steam port, drop the vertical _m_ from G, and at its lower end V' is the position of the eccentric centre at the point of cut off. Then draw a line P, distant from _m_ equal to the lead, which will give at V the position of the eccentric when the crank is on the dead centre, and the valve is open to the amount of the lead. The lap is obviously the distance from the centre C of the crank shaft to the arc G.

We have here found all the points necessary except the point at which the valve will open the port for the lead, and this we may find by setting a pair of compa.s.ses to the radius B H (or to radius V V', as both these radii are equal), and from V as a centre, mark at A an arc, which will give the crank pin position at the time the port first opens for the lead, or in other words it will give the position. The proof of the construction is, that if we set the compa.s.ses to the distance between the crank pin position on the dead centre and the point of cut off (or from B to H), we may apply the compa.s.ses to the points V, V', which represent the eccentric position when the port is opened to the amount of the lead, and when the cut off occurs.

If the point of cut off only is to be found, we mark from C, Fig. 3314, an arc G representing the amount of valve lap and arc S representing the lead. A vertical P gives the eccentric position V when the crank is on the dead centre at B, and a vertical _m_ from G gives at V' the eccentric position at the point of cut off. Then with the compa.s.ses set to the points V V', we may mark from B an arc, locating at H the position of the crank at the point of cut off, and from this with compa.s.ses set to represent the length of the connecting rod on the same scale as the circle represents the path of the crank, we may, from a point on the line of centres, mark an arc Y giving at R the piston position at the point of cut off.

When, therefore, the lap is given, we mark it from the center C of the crank shaft, and find the other elements from it, whereas, when the lap is to be found, we mark the width of the port from the end D of the valve travel, and find the other elements from that.

A proof of all the constructions is given in Fig. 3314, in which the letters of reference correspond to those in the previous figures, and the positions of the parts are marked in degrees of angle.

To find the piston position at the point of cut off, measured in inches, of the piston stroke it must be borne in mind that as the circle B D represents the full travel of the valve, the diagram gives all the positions of the eccentric and valve full size, but that as it represents the crank path on a reduced scale, therefore we must multiply the measurement on the diagram by that scale.

Suppose, for example, that the piston stroke is 10 inches, and the valve travel 2-1/2 inches, and the circle being 2-1/2 inches in diameter, is, when considered with relation to the eccentric motion, full size, but when considered with relation to the piston or crank motion, it is only 1/4 the size, hence to find the piston position at the time of cut off, we must multiply the distance from B to R by 4.

[Ill.u.s.tration: Fig. 3315.]

LINK MOTION FOR STATIONARY ENGINES.

The ordinary mechanism employed to enable a stationary engine to be reversed or run in either direction is the Stephenson link motion. Other forms of link motion have been devised, but the Stephenson form has become almost universal.

[Ill.u.s.tration: Fig. 3316.]

Fig. 3315 represents this link motion or reversing gear with the parts in position for the full gear of the forward motion, and Fig. 3316 represents it in full gear for the backward motion.

The meaning of the term full gear is that the parts are in the position in which the steam follows the piston throughout the longest or greatest part of the stroke. When in full gear the link motion operates the valve almost precisely the same as if the eccentric rod was attached direct to the valve spindle and no link motion was used.

Besides enabling the engine to run in both directions, however, the link motion provides a means of reducing the amount of valve travel and thus causes the live steam to be cut off earlier in the piston stroke, thus using the steam more expansively. This is done by moving the reversing lever more upright, the earliest point of cut off being obtained when it is upright and the latch is in the notch marked O on the sector in Fig.

3315. If with the engine standing still we move the link motion from full gear forward to full gear backward and watch the valve, we shall find that the valve lead increases as the reversing lever approaches the upright position, or mid gear as it is termed, and that after pa.s.sing that point it gradually diminishes again, the valve being so set that the lead is the same for full gear forward as it is for full gear backward.

The reversing lever is used to move the link into the required position and to hold it there (the end of the latch fitting into the notches in the sector being the detaining or locking device); as the link is suspended by its saddle pin S and the link hanger, therefore its motion is to swing or partly rotate on the pin S, and at the same time ending in the arc of a circle whose centre of motion is in the pin at the upper end of the link hanger which is pivoted to the lower arm of the lifting shaft (which is sometimes termed the tumbling shaft). It will clearly be seen that with the position the parts occupy in Fig. 3315, and the crank motion being in the direction of the arrow, the forward eccentric will move the top of the link to the right and therefore the valve will move to the right, while the backward eccentric will move the bottom end of the link to the left.

In full gear, however, the bottom eccentric rod has but a very slight effect indeed on the motion of the valve because both the link hanger and the link block will permit the link to swing on centre of the link block pin as a pivot. If now we turn to Fig. 3316 for the full gear backward, we shall see that these conditions are reversed and the backward eccentric becomes the effective one, being in line with the valve spindle. By s.h.i.+fting the link from one gear to the other, therefore, we have merely changed the direction in which the link will move the valve, and, therefore, the direction in which the engine would run.

In Fig. 3315 for the full gear the parts are shown in position, with the piston at the crank end of the cylinder, and the crank pin on the dead centre, and the eccentrics must be set as shown in the cut, the eccentric rods being open and not crossed. When, however, the crank is on the other dead centre and the piston at the head end of the cylinder, the rods will cross each other, and it is necessary to remember that the rods should be open when the piston is at the crank end of the cylinder.

If, however, the running gear contains a rock shaft, or rocker (as is the case in American locomotives), then these conditions are reversed, and the eccentric rods will cross when the piston is at the crank end of the cylinder.

In setting the slide valve of an engine having a link motion, there are two distinct operations. First, to put the crank on the respective dead centres, which will be fully described on page 394 and need not be repeated; and second, to set the eccentrics in their proper positions on the shaft, and correct, if necessary, the lengths of the eccentric rods.

The crank being on the dead centre, with the piston at crank end of the cylinder, the eccentric should be moved around on the shaft by hand until there is the desired amount of lead at the crank end port, and temporarily fastened there, a set screw usually being provided (in the eccentric) for this purpose. The lead is best measured with a wedge, W, Fig. 3315. The crank is then put on its other dead centre, and the lead for the head end port is measured. If the lead is to be made equal for the two ports (as is usually the case in horizontal engines) and it is found to come so, the valve setting for the forward gear is complete. If the lead is not equal, the forward eccentric rod or else the valve spindle must be altered so as to make the lead equal. In some engines adjusting screws are provided for the purpose of regulating the length of either the eccentric rod or else of the slide spindle; it does not matter which is altered. The link motion is then put in full gear for the backward motion, and, with the crank on the dead centre (it does not matter which dead centre), the eccentric is moved by hand upon the crank shaft until there is the required amount of valve lead. The eccentric is then fastened on the shaft and the crank put on the other dead centre, and the lead tried for the other port, and made equal by lengthening or shortening the backward eccentric rod. It is to be noted that altering the length of the eccentric rod or of the valve spindle makes it necessary to reset the eccentric, as it affects the amount of lead at both ports; hence, if any alteration of rod length is made, the whole process here described must be repeated after each alteration of rod length.

FLY BALL OR THROTTLING GOVERNORS.

An isochronal governor is one in which the two opposing forces are equal throughout the whole range of governor action, or, in other words, equal, let the vertical height of the plane in which the b.a.l.l.s revolve or swing be what it may.

A dancing governor is one that acts spasmodically. Such an action may occur from undue friction in the parts of the governor or of its throttle valve.

The friction offers a greater resistance to starting the parts in motion than it does to keep them in motion after being started; hence, the parts are apt to remain at rest too long, and to move too far after being put in motion.

Rule to find the number of revolutions a governor should make. Divide the constant number 375.36 by twice the square root of the height of the cone in inches. The quotient is the proper number of revolutions per minute.

_Example._--A governor with arms 30-1/2 inches long, measuring from the centre of suspension to the centre of the ball, revolves, in the mean position of the arms, at an angle of about thirty degrees with a vertical spindle forming a cone of about 26-1/2 inches high. At what number of revolutions per minute should this governor be driven? Here the height of the cone being 26.5 inches, the square root of which is 5.14 and twice the square root 10.28, we divide 375.36 by 10.28, which give us 36.5 as the proper number of revolutions per minute at which the governor should be driven.

The construction of the Pickering governor is as follows:

[Ill.u.s.tration: Figs. 3317, 3318.]

In Fig. 3317 it is shown in its simplest form, and in Fig. 3318 with the driving pulley and speeder (or engine speed regulating device) attached.