Fig. 257.

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.

Fig. 258.

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.

Fig. 259.

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.

Fig. 260.

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.

Fig. 261.

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.

Fig. 262.

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.

Fig. 263.

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.

Fig. 264.

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.