Plotting Mechanical Motions

Let it be required to find how much motion an eccentric will give to its rod, the distance from the centre of its bore to the centre of the circumference, which is called the throw, being the distance from A to B in Figure 257. Now as the eccentric is moved around by the shaft, it is evident that the axis of its motion will be the axis A of the shaft. Then from A as a centre, and

with radius from A to C, we draw the dotted circle D, and from E to F will be the amount of motion of the rod in the direction of the arrow.

This becomes obvious if we suppose a lead pencil to be placed against the eccentric at E, and suppose the eccentric to make half a revolution, whereupon the pencil will be pushed out to F. If now we measure the distance from E to F, we shall find it is just twice that from A to B. We may find the amount of motion, however, in another way, as by striking the dotted half circle G, showing the path of motion of B, the diameter of this path of motion being the amount of lateral motion given to the rod.

In Figure 258 is a two arm lever fast upon the same axis or shaft, and it is required to find how much a given amount of motion of the long arm will move the short one. Suppose the distance the long arm moves is to A. Then draw the line B from A to the axis of the shaft, and the line C the centre line of the long arm. From the axis of the shaft as a centre, draw the circle D, passing through the eye or centre E of the short arm. Take the radius from F to G, and from E as a centre mark it on D as at H, and H is where E will be when the long arm moves to A. We have here simply decreased the motion in the same proportion as one arm is shorter than the other. The principle involved is to take the motion of both arms at an equal distance from their axis of motion, which is the axis of the shaft S.

In Figure 259 we have a case in which the end of a lever acts directly upon a shoe. Now let it be required to find how much a given motion of the lever will cause the shoe to slide along the line x; the point H is here found precisely as before, and from it as a centre, the dotted circle equal in diameter to the small circle at E is drawn from the perimeter of the dotted circle, a dotted line is carried up and another is carried up from the face of the shoe. The distance K between these dotted lines is the amount of motion of the shoe.

In Figure 260 we have the same conditions as in Figure 259, but the short arm has a roller acting against a larger roller R. The point H is found as before. The amount of motion of R is the distance of K from J; hence we may transfer this distance from the centre of R, producing the point P, from which the new position may be marked by a dotted circle as shown.

In Figure 261 a link is introduced in place of the roller, and it is required to find the amount of motion of rod R. The point H is found as before, and then the length from centre to centre of link L is found, and with this radius and from H as a centre the arc P is drawn, and where P intersects the centre line J of R is the new position for the eye or centre Q of R.

In Figure 262 we have a case of a similar lever actuating a plunger in a vertical line, it being required to find how much a given amount of motion of the long arm will actuate the plunger. Suppose the long arm to move to A, then draw the lines B C and the circle D. Take the radius or distance F, G, and from E mark on D the arc H. Mark the centre line J of the rod. Now take the length from E to I of the link, and from H as a centre mark arc K, and at the intersection of K with J is where the eye I will be when the long arm has moved to A.

In Figure 263 are two levers upon their axles or shafts S and S'; arm A is connected by a link to arm B, and arm C is connected direct to a rod R. It is required to find the position of centre G of the rod eye when D is in position E, and when it is also in position F. Now the points E and F are, of course, on an arc struck from the axis S, and it is obvious that in whatever position the centre H may be it will be somewhere on the arc I, I, which is struck from the centre S'. Now suppose that D moves to E, and if we take the radius D, H, and from E mark it upon the arc I as at V, then H will obviously be the new position of H. To find the new position of G we first strike the arc J, J, because in every position of G it will be somewhere on the arc J, J. To find where that will be when H is at V, take the radius H, G, and from V as a centre mark it on J, J, as at K, which is the position of G when D is at E and H is at V. For the positions when D is at F we repeat the process, taking the radius D, H, and from F marking P, and with the radius H, G, and from P as a centre marking Q; then P is the new position for H, and Q is that for G.

In Figure 264 a lever arm A and cam C are in one piece on a shaft. S is a shoe sliding on the line x, and held against the cam face by the rod R; it is required to find the position of the face of the shoe against the cam when the end of the arm is at D.

Draw line E from D to the axis of the shaft and line F. From the shaft axis as a centre draw circle W; draw line J parallel to x. Take the radius G H, and from K as a centre mark point P on W; draw line Q from the shaft axis through P, and mark point T. From the shaft axis as a centre draw from T an arc, cutting J at V, and V is the point where the face of the shoe and the face of the cam will touch when the arm stands at D.

Let it be required to find the amount of motion imparted in a straight line to a rod attached to an eccentric strap, and the following construction may be used. In Figure 265 let A represent the centre of the shaft, and, therefore, the axis about which the eccentric revolves. Let B represent the centre of the eccentric, and let it be required to find in what position on the line of motion x, the centre C of the rod eye will be when the centre B of the eccentric has moved to E. Now since A is the axis, the centre B of the eccentric must rotate about it as denoted by the circle D, and all that is necessary to find the position of C for any position of eccentric is to mark the position of B on circle D, as at E, and from that position, as from E, as a centre, and with the length of the rod as a radius, mark the new position of C on the line x of its motion. With the centre of the eccentric at B, the line Q, representing the faces of the straps, will stand at a right angle to the line of motion, and the length of the rod is from B to C; when the eccentric centre moves to E, the centre line of the rod will be moved to position P, the line Q will have assumed position R, and point C will have moved from its position in the drawing to G on line x. If the eccentric centre be supposed to move on to F, the point C will move to H, the radii B C, E G, and F H all being equal in length. Now when the eccentric centre is at E it will have moved one-quarter of a revolution, and yet the point C will only have moved to G, which is not central between C and H, as is denoted by the dotted half circle I.

On the other hand, while the eccentric centre is moving from E to F, which is but one-quarter of a revolution, the rod end will move from G to H. This occurs because the rod not only moves endwise, but the end connected to the eccentric strap moves towards and away from the line x. This is shown in the figure, the rod centre line being marked in full line from B to x. And when B has moved to E, the rod centre line is marked by dotted line E, so that it has moved away from the line of motion B x. In Figure 266 the eccentric centre is shown to stand at an angle of 45 degrees from line q, which is at a right angle to the line of motion x x, and the position of the rod end is shown at C, J and H representing the extremes of motion, and G the centre of the motion.

If now we suppose the eccentric centre to stand at T, which is also an angle of 45 degrees to q, then the rod end will stand at K, which is further away from G than C is; hence we find that on account of the movement of the rod out of the straight end motion, the motion of the rod end becomes irregular in proportion to that of the eccentric, whose action in moving the eye C of the rod in a straight line is increased (by the rod) while it is moving through the half rotation denoted by V in figure, and diminished during the other half rotation.

In many cases, as, for example, on the river steamboats in the Western and Southern States, cams are employed instead of eccentrics, and the principles involved in drawing or marking out such cams are given in the following remarks, which contain the substance of a paper read by Lewis Johnson before the American Society of Mechanical Engineers. In Figure 267 is a side view of a pair of cams; one, C, being a full stroke cam for operating the valve that admits steam to the engine cylinder; and the other, D, being a cam to cut off the steam supply at the required point in the engine stroke. The positions of these cams with relation to the position of the crank-pin need not be commented upon here, more than to remark that obviously the cam C must operate to open the steam inlet valve in advance of cam D, which operates to close it and cause the steam to act expansively in the cylinder, and that the angle of the throw line of the cut-off valve D to the other cam or to the crank-pin varies according as it is required to cut off the steam either earlier or later in the stroke.

The cam yoke is composed of two halves, Y and Y', bolted together by bolts B, which have a collar at one end and two nuts at the other end, the inner nuts N N enabling the letting together of the two halves

of the yoke to take up the wear. It is obvious that as the shaft revolves and carries the cam with it, it will, by reason of its shape, move the yoke back and forth; thus, in the position of the parts shown in Figure 267, the direction of rotation being denoted by the arrow, cam C will, as it rotates, move the yoke to the left, and this motion will occur from the time corner a of the cam meets the face of Y' until corner b has passed the centre line d. Now since that part of the circumference lying between points a and b of the cam is an arc of a circle, of which the axis of the shaft is the centre, the yoke will remain at rest until such time as b has passed line d and corner a meets the jaw Y of the yoke; hence the period of rest is determined by the amount of circumference that is made concentric to the shaft; or, in other words, is determined by the distance between a and b.

The object of using a cam instead of an eccentric is to enable the opening of the valves abruptly at the beginning of the piston stroke, maintaining a uniform steam-port opening during nearly the entire length of stroke, and as abruptly closing the valves at the termination of the stroke.

Figure 268 is a top view of the mechanism in Figure 267; and Figure 269 shows an end view of the yoke. At B, in Figure 268, is shown a guide through which the yoke-stem passes so as to be guided to move in a straight line, there being a guide of this kind on each side of the yoke.

The two cams are bolted to a collar that is secured to the crank-shaft, and are made in halves, as shown in the figures and also in Figures 270 and 271, which represent cams removed from the other mechanism. To enable a certain amount of adjustment of the cams upon the collar, the bolts which hold them to the collar fit closely in the holes in the collar, but the cams are provided with oblong bolt holes as shown, so that the position of either cam, either with relation to the other cam or with relation to the crank-pin, can be adjusted to the extent permitted by the length of the oblong holes.

The crank is assumed in the figures to be on its dead centre nearest to the engine cylinder, and to revolve in the direction of the arrows. The cams are so arranged that their plain unflanged surfaces bolt against the collar.

The method of drawing or marking out a full stroke cam, such as C in Figure 267, is illustrated in Figure 272, in which the dimensions are assumed to be as follows:

Diameter of crank shaft, 7-1/2 inches; travel of cam, 3 inches; width of yoke, 18 inches.

The circumference of the cam is composed of four curved lines, P, P', K 1, and K 2. The position of the centre of the crank shaft in this irregularly curved body is at X. The arcs K 1 and K 2 differ in radius, but are drawn from the same point, X, and hence are concentric with the crank shaft.

The arcs P, P', are of like radius, but are drawn from the opposite points S, S', shown at the intersection of the arcs P, P', with the arc K 1. Thus arcs P, P', are eccentric to the crank shaft.

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.

The engravings from and including Figure 274 illustrate 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 assumed to be 4 feet. Figure 274 illustrates 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, passing 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.

With X as a centre, draw the circle E, having a diameter of 18 inches, equal to the space 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.

Fig. 276.

Fig. 277.

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.

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 compass 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 compasses 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.

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.

Fig. 278.

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.

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.

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.

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.

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 compasses 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 compass 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.

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 compasses 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.

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.

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.

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.

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.

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.

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