[Ill.u.s.tration: Fig. 273.]

To draw the cam place one point of the dividers at X, which is the centre of the crank shaft, and draw the circle E equal to width of yoke, 18 inches. Through this centre X, draw the two right lines A and B. On the line B, at the intersection of the curved line E, draw the two vertical lines A 1, A 1. With a radius of 10-1/2 inches, and with one point of the dividers at X, draw the arc K 1. With a radius of 7-1/2 inches, and one point of the dividers at X, draw the arc K 2. With a radius of 18 inches, and one point of the dividers at the intersection of the arc E, with the vertical line A 1 at S, draw the arc P opposite to S, and let it merge or lose itself in the curved line K 2. Draw the other curved line P' from the other point S, and we have a full stroke cam of the dimensions required, and which is represented in Figure 273, removed from the lines used in constructing it.

[Ill.u.s.tration: Fig. 274.]

The engravings from and including Figure 274 ill.u.s.trate the lines embracing cut-off cams of varying limits of cut-off, but all of like travel and dimensions, which are the same as those given for the full stroke cam in Figure 272.

In drawing cut-off cams, the stroke of the engine plays a part in determining their conformation, and in the examples shown this is a.s.sumed to be 4 feet. Figure 274 ill.u.s.trates the manner of finding essential points in drawing or marking out cut-off cams. With X as a centre, and a radius of 2 feet, draw the circle E 1, showing the path of the crank-pin in making a revolution. This circle has a diameter of 4 feet, equal to the stroke of the engine. Draw the horizontal line B, pa.s.sing through the centre of circle E 1. Within the limits of circle E 1, subdivide line B into eight equal parts, as at 1, 2, 3, 4, etc. Draw the vertical lines, 1, 2, 3, 4, etc., until they each intersect the circle E 1.

[Ill.u.s.tration: Fig. 275.]

With X as a centre, draw the circle E, having a diameter of 18 inches, equal to the s.p.a.ce in the yoke embracing the cam.

From the centre X draw the series of radial lines through the points of intersection of the vertical lines 1, 2, 3, 4, etc., from the circle E 1, and terminating at X. We will now proceed to utilize the scale afforded by Figure 274, in laying off the cut-off cam shown in Figure 276, of half stroke limit.

[Ill.u.s.tration: Fig. 276.]

With X as a centre, draw the circle E, Figure 275, having a diameter of 18 inches. Bisect this circle with the straight lines A and B, which bear the same relation to their enclosing circle that the lines A, B, do to the circle E in Figure 274.

[Ill.u.s.tration: Fig. 277.]

It will be observed, in Figure 274, that the vertical line A is (at the top half) also No. 4, representing 4/8, or half of the stroke. With a radius of 18 inches, and one point of the dividers placed at V, which is at the intersection of the circle E with the horizontal line B in Figure 275, draw the arc P. With the same radius and with one compa.s.s point rested at V', draw the arc P'; then two arcs, P and P', intersecting at the point S.

With the same radius and one point of the compa.s.ses at S, draw the arc H H. The arcs K 1 and K 2 are drawn from the centre X, with a radius of 10-1/2 for K 1 and 7-1/2 inches for K 2, and only serve in a half stroke cam to intersect the curved lines already drawn, as shown in Figure 275. In practice, the sharp corner at S would be objectionable, owing to rapid wear at this point; and hence a modification of the dimensions for this half stroke cam would be required to obtain a larger wearing surface at the point S, but the cam of this limit (1/2 stroke) is correctly drawn by the process described with reference to Figure 275, the outline of the cam so constructed being shown in Figure 276.

[Ill.u.s.tration: Fig. 278.]

In Figure 278 is shown a cam designed to cut off the steam at five-eighths of the piston stroke, the construction lines being given in Figure 277, for which draw circle E and straight lines A and B, as in the preceding example. By reference to Figure 274 it will be observed that the diagonal line drawn through circle E at 5 is drawn from the straight line marked 5, which intersects circle E 1, and as this straight line 5 represents five-eighths of the stroke laid off on line B, it determines the limit of cut-off on the five-eighths cam in Figure 277.

[Ill.u.s.tration: Fig. 279.]

Turning then to Figure 274, take on circle E the radius from radial line 4 to radial line 5, and mark it in Figure 277 from the vertical line producing V'.

Now, with a radius of 18 inches, and one point of the dividers fixed at point V, forming the intersection of the circle E with the horizontal line B, draw the arc P. With the same radius, and one point of the dividers fixed at point V', draw the opposite arc P'. With a radius of 10-1/2 inches from the centre X, draw the arc K 1, intersecting lines P P', at S S. With a radius of 7-1/2 inches, draw the curved line K 2, opposite to curved line K 1. Now, with a radius of 18 inches, and one point of the dividers fixed alternately at S S, draw the arcs H, H, from their intersection with the circle E, until they merge into the curved line K 2. These curved lines embrace a cut-off cam of five-eighths limit, shown complete in Figure 278.

[Ill.u.s.tration: Fig. 280.]

From the instructions already given it should be easy to understand that the three-fourths and seven-eighths cams, shown in Figures 279, 280, 281 and 282, are drawn by taking the points of their cut-off from the same scale shown in Figure 274, at the diagonal points 6 and 7, intersecting circle E in that figure; and cut-off cams of intermediate limit of cut-off can be drawn by further subdividing the stroke line B, in Figure 274, into the required limits.

[Ill.u.s.tration: Fig. 281.]

Cut-off cams of any limit are necessarily imperfect in their operations as to uniformity of cut-off from opposite ends of the slides, not from any defect in the rule for laying them off, but from the well-known fact of the crank pin travelling a greater distance, while driven by the piston from the centre of the cylinder, through its curved path from the cylinder, over its centre, and back to the centre of the cylinder, than in accomplishing the remaining distance of its path in making a complete revolution; and, although the subdivisions of eighths of the stroke line B, in Figure 274, does not truly represent a like division of the piston stroke, owing to deviation, caused by inclination of the connecting rod in traversing from the centres to half stroke, still it will be found that laying off a cut-off cam by this rule is more nearly correct than if the divisions on stroke line B were made to correspond exactly with a subdivision of piston stroke into eighths.

[Ill.u.s.tration: Fig. 282.]

The cut-off in cams laid off by the rules herein described is greater in travelling from one side of the slides than in travelling from the opposite end, one cut-off being more than the actual cut-off of piston stroke, and the other less; and in practical use, owing to play or lost motion in the connections from cam to valve, the actual cut-off is less than the theoretical; hence cut-off cams are usually laid off to compensate for lost motion; that is, laid off with more limit; for instance, a five-eighths cam would be laid off to cut-off at eleven-sixteenths instead of five-eighths.

[Ill.u.s.tration: Fig. 283.]

Figure 283 represents the motion a crank, C, imparts to a connecting rod, represented by the thick line R, whose end, B, is supposed to be guided to move in a straight line. The circle H represents the path of the crank-pin, and dots 1, 2, 3, etc., are 24 different crank-pin positions equidistant on the circle of crank-pin revolution. Suppose the crank-pin to have moved to position 1, and with the compa.s.ses set to the length of the rod R, we set one point on the centre of position 1, and mark on the line of motion _m_ the line _a_, which will be the position rod end B will have moved to. Suppose next that the crank-pin has moved into position 2, and with the compa.s.s point on the centre of 2 we mark line 2, showing that while the crank-pin moved from 1 to 2, the rod end moved from _a_ to _b_; by continuing this process we are enabled to discern the motion for the whole of the stroke. The backward stroke will be the same, for corresponding crank-pin positions, for both strokes; thus, when the rod end is at 7 the crank-pin may be at 7 or at 17. This fact enables us to find the positions for the positions later than 6, on the other side of the circle, as at 17, 16, 15, etc., which keeps the engraving clear.

[Ill.u.s.tration: Fig. 284.]

In Figure 284 a pinion, P, drives a gear-wheel, D, on which there is a pin driving the sliding die A in the link L, which is pivoted at C, and connected at its upper end to a rod, R, which is connected to a bolt, B, fast to a slide, S. It is required to find the motion of S, it moving in a straight line, dotted circle H' representing the path of the pin in the sliding die A, arc H representing the line of motion of the upper end of link L, and lines N, O, its centre line at the extreme ends of its vibrating motion. In Figure 285 the letters of reference refer to the same parts as those in Figure 284. We divide the circle H' of pin motion into 24 equidistant parts marked by dots, and through these we draw lines radiating from centre, C, and cutting arc H, obtaining on the arc H the various positions for end Z of rod R, these positions being marked respectively 1, 2, 3, 4, etc., up to 24. With a pair of compa.s.ses set to the length of rod R from 1 on H, as a centre, we mark on the line of motion of the slide, line _a_, which shows where the other end of rod R will be (or in other words, it shows the position of bolt B in Figure 284), when the centre of A, Figure 284, is in position 1, Figure 285.

[Ill.u.s.tration: Fig. 285.]

From 2 on arc H, we mark with the compa.s.ses line _b_ on line M, showing that while the pin moved from 1 to 2, the rod R would move slide S, Figure 284, from _a_ to _b_, in Figure 285. From 3 we mark _c_, and so on, all these marks being above the horizontal line M, representing the line of motion, and being for the forward stroke. For the backward stroke we draw the dotted line from position 17 up to arc H, and with the compa.s.ses at 17 mark a line beneath the line M of motion, pursuing the same course for all the other pin motions, as 18, 19, etc., until the pin arrives again at position 24, and the link at O, and has made a full revolution, and we shall have the motion of the forward stroke above and that of the backward one below the line of motion of the slide, and may compare the two.

[Ill.u.s.tration: Fig. 286.]

[Ill.u.s.tration: Fig. 287.]

[Ill.u.s.tration: Fig. 288.]

[Ill.u.s.tration: Fig. 289.]

Figures 286 and 287 represent the Whitworth quick return motion that is employed in many machines. F represents a frame supporting a fixed journal, B, on which revolves a gear-wheel, G, operated by a pinion, P.

At A is an arm having journal bearing in B at C. This arm is driven by a pin, D, fast in the gear, G; hence as the gear revolves, pin D moves A around on C as a centre of motion. A is provided with a slot carrying a pin, X, on which is pivoted the rod, R. The motion of end N of the rod R being in a straight line, M, it is required to find the positions of N during twenty-four periods in one revolution of G. In Figure 288 let H'

represent the path of motion of the driving pin D, about the centre of B, and H the path of motion of X about the centre C; these two centres corresponding to the centres of B and C respectively, in Figure 287. Let the line M correspond to the line of motion M in Figure 286. Now since it is the pin D, Figure 287, that drives, and since its speed of revolution is uniform, we divide its circle of motion H' into twenty-four equal divisions, and by drawing lines radiating from centre C, and pa.s.sing through the lines of division on H' we get on circle H twenty-four positions for the pin X in Figure 286. Then setting the compa.s.ses to the length of the rod (R, Figure 286), we mark from position 1 on circle H as a centre line, _a_; from position 2 on H we mark line _b_, and so on for the whole twenty-four positions on circle H, obtaining from _a_ to _n_ for the forward, and from _n_ to _y_ for the motion during the backward stroke. Suppose now that the mechanism remaining precisely the same as before, the line M of motion be in a line with the centres C, B, instead of at a right angle to it, as it is in Figure 286, and the motion under this new condition will be as in Figure 289; the process for finding the amount of motion along M from the motion around H being precisely as before.

[Ill.u.s.tration: Fig. 290.]

In Figure 290 is shown a cutter-head for a wood moulding machine, and it is required to find what shape the cutting edge of the cutter must be to form a moulding such as is shown in the end view of the moulding in the figure. Now the line A A being at a right angle to the line of motion of the moulding as it is pa.s.sed beneath the revolving cutter, or, what is the same thing, at a right angle to the face of the table on which the moulding is moved, it is obvious that the highest point C of the moulding will be cut to shape by the point C of the cutter; and that since the line of motion of the end of the cutter is the arc D, the lowest part of the cutter action upon the moulding will be at point E.

It will also be obvious that as the cutter edge pa.s.ses, at each point, its length across the line A A, it forms the moulding to shape, while all the cutting action that occurs on either side of that line is serving simply to remove material. All that we have to consider, therefore, is the action on line A A.

It may be observed also that the highest point C of the cutter edge must not be less than 1/4 inch from the corner of the cutter head, which gives room for the nut N (that holds the cutter to the head) to pa.s.s over the top of the moulding in a 2-1/2 inch head. In proportion as the heads are made larger, however, less clearance is necessary for the nut, as is shown in Figure 291, the cutter edge extending to C, and therefore nearly up to the corner of the head. Its path of motion at C is shown by dotted arc B, which it will be observed amply clears the nut N. In practice, however, point C is not in any size of cutter-head placed nearer than 1/4 inch from corner X of the cutter-head.

To find the length of the cutter edge necessary to produce a given depth of moulding, we may draw a circle _i_, Figure 292, equal in diameter to the size of the cutter head to be used, and line A A. The highest point of cutting edge being at _e_, and the lowest at g, then circles _d_ and _f_ represent the line of motion of these two points; and if we mark the cutter in, the necessary length of cutting edge on the cutter is obviously from _a_ to _b_.

[Ill.u.s.tration: Fig. 291.]

[Ill.u.s.tration: Fig. 292.]

Now the necessary depth of cutter edge being found for any given moulding, or part of a moulding, the curves for the edge may be found as follows: Suppose the moulding is to be half round, as in the end view in Figure 290. The width of the cutter must of course equal the width of the moulding, and the length or depth of cutting edge required may be found from the construction shown in Figure 292; hence all that remains is to find the curve for the cutting edge. In Figure 293, let A A represent the centre of the cutter width, its sides being F F', and its end B B. From centre C draw circle D, the upper half of which will serve to represent the moulding. Mark on A the length or depth the cutting edge requires to be, ascertaining the same from the construction shown in Figure 292, and mark it as from C to K'. Then draw line E E, pa.s.sing through point K. Draw line G, standing at the same angle to A A as the face _h b_, Figure 292, of the cutter does to the line A A, and draw line H H, parallel to G. From any point on G, as at I, with radius J, draw a quarter of a circle, as K. Mark off this quarter circle into equal points of division, as by 1, 2, 3, etc., and from these points of division draw lines, as _a_, _b_, _c_, etc.; and from these lines draw horizontal lines _d_, _e_, _f_, etc. Now divide the lower half of circle D into twice as many equal divisions as quarter circle K is divided into, and from these points of division draw perpendiculars _g_, _h_, _i_, etc. And where these perpendiculars cross the horizontal lines, as _d_, will be points through which the curve may be drawn, three of such points being marked by dots at _p_, _q_, _r_. If the student will, after having drawn the curve by this construction, draw it by the construction that was explained in connection with Figure 79, he will find the two methods give so nearly identical curves, that the latter and more simple method may be used without sensible error.

[Ill.u.s.tration: Fig. 293.]

[Ill.u.s.tration: Fig. 294.]

When the curves of the moulding are not arcs of circles they may be marked as follows:

Take the drawing of the moulding and divide each member or step of it by equidistant lines, as _a_, _b_, _c_, _d_, _e_, _f_, _g_, in Figure 294; above the moulding draw lines representing the cutter, and having found the depth of cutting edge for each member by the construction shown in Figure 292, finding a separate line, _a b_, for each member of the moulding, transfer the depths so found to the face of the cutter; divide the depth of each member of the cutter into as many equal divisions as the corresponding member of the moulding is divided into, as by lines _h_, _i_, _j_, _k_, _l_, _m_, _n_. Then draw vertical lines, as _o_, _p_, _q_, _r_, etc.; and where these lines meet the respective lines _h_, _i_, _j_, etc., are points in the curve, such points being marked on the cutter by dots.

CHAPTER XIII.