E

(ab) (ac) (ad) (ae) (bc) (bd) (be)

VERY link of a mechanism has a motion relative to every other link, and, as shown in April number, in Plane Motion, Part I, every one of these motions is a rotation about a center either permanent or instantaneous. In order, therefore, to completely investigate the motions of the various moving parts of a mechanism, to determine the linear velocities of various points or the angular velocities of the links, it is necessary to locate some or all of the instantaneous centers. We shall in this article attempt to develop a general method for finding any desired center or centers.

We first inquire as to the number of centers in a mechanism with a given number of links. First, suppose a mechanism has three links a, b, and c. Each link has a motion relative to each of the other two; thus, a has a motion relative to b which gives rise to the center (a b), and a motion relative to c about the center (a c).

In this and any future article, the center belonging to two links will always be denoted by enclosing in parenthesis the letters denoting the links, thus: (ab), (cd),

etc.

For this system of three links there are evidently three centers, viz.. (a b), (a c), and (bc). Let us, for convenience, arrange them in the form of a triangle, thus:

(ab) (ac) (bc)

For a mechanism with four links, a, b, c, and d, there are six centers, which may be arranged in the following form:

(ab) (ac) (ad) (bc) (bd) (cd)

The formation of the table of centers is obvious in the first line the letter a is paired with each of the remaining letters b, c, and d; in the second line the letter b is paired with each of the letters c and d which follow it, and so on.

If we add a fifth link e to the mechanism, we must add to the table of centers another vertical row, containing the centers of link e with respect to each of the other links. The table becomes :

centers of any system of three links lie in a straight line. Thus, if we consider the three links a, b, and d, we obtain, by taking them in pairs, the three centers (ab), (ad), and (bd), and according to the law given in the previous article, these centers must lie in a straight line.

The process of finding the unknown centers may be best illustrated by an example. In Fig. 1 is shown a mechanism composed of two spur gears, two straight links, and a frame or bed to which the gears are connected. The mechanism has thus five links and 5 X 4 10 centers, which we will now 2 attempt to locate. For the sake of clearness,

the mechanism is shown in a skeleton form in Fig. 2, and the links are lettered as shown. The fixed frame a is represented by the single line connecting the centers of the gears, and the gears are represented simply by the circles b and c. We first form the following table containing all of the centers, and indicate those centers that are at once known by underscoring them with dashes. (ab) (ac) (ad) (ae) (be) (bd) (be)

(ed) (ce) (de)

We observe in the first place that the joint connecting any two links is the center for the relative motion of those links; thus,

m

centers (bc), (cd), and (bd), which we know must lie in the same straight line. Therefore, we draw through the known centers (be) and (ed) a straight line. Somewhere on this line lies the unknown center (bd). In the same way we combine with the links b and d the third link e, forming a system of three centers (be), (de), and (b d). The unknown center (bd) must lie in the straight line connecting the known centers (be) and (de); therefore, since (bd) lies in both lines (bc)-(ed) and (be)-(de), it must lie at their intersection. We may express this fact by the following arrangement :

(bd) { (bc)-(cd)

(be)-(de)

(ae)

the motion of the gear b relative to the frame a is a rotation about the point (ab), which is the center of the journal; the motion of the link d relative to the link e must be a rotation about the joint (de) that connects them. We find, therefore, the five centers (ab), (a c), (cd), (de), and (be) by simple inspection. Furthermore, the instantaneous center of the relative motion of the gears b and c is their point of contact (be); that is, if we consider the gear b as stationary, every point of the gear c is rotating about the point of contact as a sort of fulcrum.

We have now to find the unknown centers (ad), (bd), (ae), and (ce) by drawing straight lines. Consider first the center (bd), which involves the two links b and d. By adding another link c, we get a system of three links b, c, and d, with the three

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(bd)

points in a rotating body are proportional to their distances from the center of rotation. Suppose we know the velocity of the end (be) of the link e; then, we can find the velocity of the other end (de) by the proportion vel. (de) distance (de)-(a e)

vel. (be) distance (be)-(a e)'

By means of the various centers we are enabled to find the velocity and direction of motions of any kind of a moving link. This application of the instantaneous center will be fully treated in a subsequent article.

The mechanism shown in Fig. 1 has various applications in practice. If the gears have equal diameters, and the links d and e are equal, the point (de) will move in a straight line, provided the points (be) and (cd) are so located that the lines (ac)-(cd) and (ab)-(be) lie on the same side of (ab)(ac) and make equal angles with it. The path of the point (de) in Fig. 2 is the curve mn. If some of the links of a mechanism have a motion of translation relative to the fixed link or to some other link, the corresponding centers will lie at an infinite distance. (See Plane Motion, Part I.) This fact, however, in no way complicates the problem of

So.

(ab)

a

(ad)

(bd)

L(be

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pendicular

to

the direction of the sliding. The centers (ab), (ad), and (bc) are known, they being the

joints connect

ing adjacent links.

To find the unknown instantaneous center (bd), we have the combination,

(bd) { (be)-(ed) (ab)-(ad) That is, (bd)

lies at the intersection of the

lines (ab)-(ad) and (bc)-(c d). The first line is readily drawn, as the centers (ab) and (ad) are known; the second line joins the known center (bc) to the center (cd), which lies at an infinite distance. We cannot, of course, draw a line to an infinitely distant point, and in this case we are not required to do We have merely to draw a line through (bc) in the direction of the infinite point (cd); as we have seen, the direction is perpendicular to the direction of the motion of the link c, relative to. d. Therefore, a line drawn through (bc) perpendicular to the direction in which c is sliding, cuts the line (ab)-(ad) in a point which is the desired center (bd).

The center (a c) is located from the combination :

What is the simplest way of making a perspective drawing from the enclosed sketch? 1 believe there is a method by which a perspective drawing can be made on a very small board, the vanishing points being found indirectly.

The answer to this question-sent in by a subscriber is too long to be published in the Answers to Inquiries columns, but, as it will probably interest a large number of our readers, we insert it here.

Make first a faint, freehand sketch of the monument, showing approximately (see Fig. 2), the size and position of the perspective drawing wanted. Draw a vertical center line ab through the middle of the sketch. Draw a horizontal base line cd perpendic

ular to the center line ab through the lowest point e of the sketch. Draw fg parallel to cd, locating it at that position in the sketch where all lines which are in reality level are horizontal in the sketch. This is the horizon line and must be drawn as high above cd as the eyes of the person who is looking at the object are assumed to be. All vanishing points of level lines are on this horizon line. The intersection a is the point of sight. Using a straightedge, draw the base line eh over the sketched base line, meeting the horizon line in g. This point g is the vanishing point of all edges parallel to eh. Draw a straight line over the sketched base line e i, meeting the horizon, if convenient. This meeting point is generally, as in the present case, too far away to come within the limits of a small board. Instead of this vanishing point, a vertical line jf may be used. This line is found in the following manner: Draw the vertical line ee through e, inter

secting fg in e'. Bisect ee' in k and draw a parallel to ei through k, meeting ƒg in ƒ. In case ƒ does not fall within the limits of the paper, a smaller part of ee' may be taken from e' down (preferably a quarter). Draw a vertical line fj, intersecting ei continued, in j. Next, the standpoint n

must be found: Draw a parallel to eh through k, meeting ƒg in l. Make gm equal to lf (unless ek was made one quarter of e'e, then make gm equal to twice lƒ) and make m n equal to m g. The point n is the standpoint, and na is the distance at which the perspective view is taken, in order that it shall appear about like the first rough sketch. Draw op parallel to fg, a short distance above the top of the sketch and intersecting the center line a b in a', and e e' in e''. The line op represents the plane of the picture seen on edge from above, and a plan A of the object must be drawn behind this plane, or (looking at the drawing) above the line op. The required position of the plan A is found as follows: Through e' draw e''h' parallel to a line through n and g, and make e''h' equal to the given width of the plinth-in this case 3' 4"-to a scale of, say, inch = 1 foot. Draw e'' at right angles to ''h', and make a̸'i' equal to the front of the plinth, 5' 3'', to the same scale. The angle h'e'i' represents the front part of the plan of the plinth. Draw the center line v q parallel to e'' and bisecting e''h'. The remainder of the plan A is added as wanted. Make a'n' equal to an, and draw lines from all exposed points in the plan to n', intersecting op in i', 61, etc. Draw vertical lines from these points of intersection into the perspective В where wanted. To find the perspective heights of points on these vertical lines, proceed as follows: Make a'b' equal to ab, and draw b't perpendicular to b'a'. Extend ee" above op tor intersecting the lines of the plan, as shown. Through these intersections draw parallels to op intersecting b't. Number the points in the plan A which are to be shown in the perspective B, and place the same numbers on the corresponding intersections on b't, in C. As point 1 is on the plane of the picture, its height above ed is

the same in the perspective B as in the projection drawing C, which is a sectional elevation; therefore, e-1 in B is made equal to the given height of the plinth, and, by drawing 1-g, in B, the correct perspective of the top of the plinth is obtained on this side. The perspective of the rear corner, 2, B, is found by drawing 2-n' from 2 in A, intersecting op in 2, and then drawing a vertical line down from 2, on op to h. The front corner 4 in A, of the bottom of the socket or base just above the plinth, is found by drawing a parallel to op from the intersection 4 of the line 4-5 with e'r to b't, then setting off the height of this point from 4 on b't to 42. Draw 4-n' intersecting op in 43. Set off b'-43 on ee' from e up to 4, B. Draw g-4, in B; then draw n'-4, A, intersecting op, and from this intersection draw a vertical line to g-4, B, continued, intersecting it in 41, which is the per

spective of 4, A. To find the perspective of the line 4-6, A, continue 6-4, A, to intersecter, and transfer this point to o p as before by drawing a parallel to o p intersecting b't, setting off its height thereon from b't and drawing a line to "'. The point thus found on op will be nearer to b' than 43. Bisect its distance from b', and make j-42 equal to the half so found; then draw 41-42, B. All the other points and lines are found in a similar manner; thus, j 1, is one-half of e-1. In case the desired position of the plan A in reference to the front line or plane of the picture op, and the distance a'n' of the standpoint n' from op are given, then it is not necessary to begin by making a freehand sketch, and the perspective may be found by first drawing the plan A and locating the vanishing points g etc. by drawing a parallel to "h', A, through n.