From 2 on arc H, we mark with the compasses 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 compasses 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.

Fig. 286.

Fig. 287.

Fig. 288.

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 passing 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 compasses 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.

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 passed 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 passes, 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 pass 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.

Fig. 291.

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, passing 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.

Fig. 293.

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.