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95
Parallel Motion.

CHAPTER IV.




PARALLEL MOTION.


The forms of parallel motion which have been used are so various, that we shall not attempt to enumerate the whole. In what follows, however, we shall treat of those varieties that have been most commonly adopted in land or marine engines; and in doing so, it will be our object both to explain their principles of action, and give practical rules for calculating the lengths of the various parts in every assignable case.

The forms of parallel motion that we shall discuss are four in number; one adapted to land engines, and represented in fig. 3; the others used for marine engines, and represented in figs. 4, 5, 6. It will be found that all these different arrangements act on the same fundamental principle; and therefore, before going farther, we shall proceed to explain it. Let G C and H F be two levers moving round centres at C and H, and connected together by the rod or link G F. We shall suppose that G C represents the beam of an engine standing in a horizontal position at half stroke. The rod H F represents the radius bar of the parallel motion, and at half stroke it must be horizontal, or parallel to G C. In the present case we shall suppose it to be equal to G C in length. If the beam be made to turn on its centre, so as to cause G to move towards G′, the point E in the middle of the link G F will move upwards nearly in a perpendicular line R′ S′, bisecting the versine D K. This will be evident from the following considerations:—As the point G revolves in the circle G G′, the point F revolves with nearly equal rapidity in the equal circle F F′. Both points, it is evident, will, when the motion commences, approach nearer the line R′ S′; and since the circles F F′ and G G′ are of equal radius, and the arcs described are nearly equal, the point F will move towards the line R′ S′ at the same rate at the point G. Both points will cross that line at the same time, and both will respectively reach F′ and G′ simultaneously. F′ L will evidently be equal to G′ K. Now, if the points F and G are always on opposite sides of the line R′ S′, and at the same distance from it, it is clear that the centre of the link E must always be in it; and, accordingly, this would be mathematically true, if G F, by changing its inclination to the perpendicular, did not alter the perpendicular distance between the points F and G. This, however, it does; and as the points G and F ascend the link, G F gets more nearly perpendicular; thus increasing the perpendicular distance between G and F; in consequence of which, at the point where F crosses the perpendicular R′ S′, it has not described quite so large an arc as the point G; or, when F is crossing R′ S′, the point G is still at a small distance from it. This renders the motion of the point E not perfectly coincident with the perpendicular line; but the shortening of the perpendicular distance between G and F, by the vibration of the link G F, when the length of the link is considerable, and the stroke of the beam not greater, in proportion to its length, than usually obtains in steam engines, is so exceedingly small as to be unworthy of the slightest consideration in practice. At either end of the stroke the link G F will have the same inclination as at half stroke, and the point E will therefore be exactly in the perpendicular line R′ S′, at both these points. Its motion will be represented by the curved line R′ O S', fig 7, cutting the perpendicular at R′, O, and S′, and deviating from it by an almost imperceptible quantity at X and W. From the fact, however, of a deviation being caused by the vibration of the link G F, we may infer the propriety of making that vibration as small as possible: and one very obvious precaution to be taken on this ground is, so to place the levers H F and G C at half stroke that the line R′ S′ shall bisect both their versines F L and D K. By this means the link G F vibrates equally on both sides of the perpendicular, and, consequently, never deviates from it so much as if it were to vibrate in one direction during the motion of G to G′. It is not always convenient to have the radius bar H F of the same length as the beam G C: it is sometimes longer, and sometimes shorter. In fig. 2 it is represented one half the length of G C. In this case, H F being shorter in proportion to the arc F″ F F′, which it describes, the versine F L is of course greater than in the former case. The lever should still be so placed, however, that the line R′ S′ bisects its versine L F. Owing to the inequality of the versines L F and G K, the middle point of the length G F can no longer move in the line R′ S′; but there is a point E, nearer the end of the link G, that does move nearly in this line. This point is so situated, that G E : E F :: F H : G C. For, within the limits usually assigned to the arcs described by engine-beams,[1] the versines are nearly in the inverse ratio of the radii; that is to say, vers. L F : vers. D K :: G C : H F. Now the vibrations of the points F and G on each side of the line R′ S′ are equal to half the versed sines of the arcs described by these points respectively; that is, to F f and G g. Hence, F f : G g :: G C : H F, and by similar triangles F f : G g :: F E : G E; hence, F E : G E :: G C : H F.

As this proportion holds true at any part of the stroke, it proves that the point E, which divides the link in this proportion, moves in the perpendicular line R S′. When, therefore, the lengths of the levers are given, it is easy to find the point E; but in practice it generally happens (as will be shown subsequently) that the point E is fixed: and we have to find H F. This will be done from this proportion, F E : E G :: G C : H F = \tfrac{GC \cdot EG}{FE} … (A). This proportion is founded on the assumption that the versed sines of the beam and radius bar are inversely as the radii. This, though nearly true, is not exactly so. The versed sines of the smaller radii increase in a greater proportion than the inverse ratio of the radii. The consequence of this is, that the point E in the link G F will be moved too much toward H, and will deviate from the perpendicular in that direction at each end of the stroke. In small engines, and when H F is nearly equal to G C, this deviation will be so small as scarcely to be worthy of notice; but in large engines, and when there is a considerable difference between the lengths of the radius rod and beam, it will be of some consequence. From what has already been said, it will be observed that the deviation is all on one side of the perpendicular; so that the curve described by the point E will be of the shape shown in fig. 8. By making a small addition to the length of the radius bar, we can make the deviation take place equally on each side of the perpendicular. We can add so much to its length as to make the point E move towards H too slowly at half stroke; or, in other words, deviate towards C till the beam has got about quarter stroke from the horizontal position, when the point E will cross the perpendicular, and deviate to an equal amount towards H. The curve described by E will then be of the shape shown in fig. 9, where the deviations at m, n, o, p, are all equal, and only half the deviation at r and s in the last figure.[2] To facilitate the making of this correction in the length of the radius bar, we give the following table of the amounts to be added to its length for the various proportions that may obtain between it and the beam C G. It is to be observed too, that when the radius bar H F is longer than the beam G C, these amounts are to be subtracted from its length.


TABLE (A.)
This column gives F HC G when C G is the greater, and C GF H when F H is the greater. Correction to be added to or subtracted from the calculated length of the radius bar, in decimal parts of its calculated length.
 1.0   0
 .9 0.0034
 .8 0.0075
 .7 0.0163
 .6 0.0270
 .5 0.0452
 .4 0.0817
  1. The beam of an engine is usually made at least three times the length of the stroke; so that the sine of the angle which it makes with the horizontal line at the extreme of its stroke, is equal to ½ of the radius. The angle whose sine is ½ of the radius, is 19° 28′.
  2. Let G C = r G' G'' = s, and  \mathrm{arc}\ G'' G G' = \phi ; then G K = \mathrm{vers.}\ \phi;
    but by Euclid (III. 35), GK(2 G C-G K) = (\tfrac{1}{2} G' G^2)^2
    \therefore \mathrm{vers.}\ \phi (2r - \mathrm{vers.}\ \phi) = \tfrac{1}{4}s^2 \therefore 2r\ \mathrm{vers.}\ \phi = \tfrac{1}{4}s^2 + \mathrm{vers.}^2\ \phi
     \therefore r=\tfrac{\tfrac{1}{4}s^2 + \mathrm{vers.}^2\ \phi}{2 \mathrm{vers.}\ \phi}= \tfrac{\tfrac{1}{4}s^2 + \tfrac{1}{2} \mathrm{vers.}^2\ \phi}{\mathrm{vers.}\ \phi}; or, r=\tfrac{\tfrac{1}{2} s^2}{\mathrm{vers.}\ \phi}+\tfrac{1}{2} \mathrm{vers.}\ \phi
    \tfrac{1}{2}s^2 being a constant quantity, this equation shows that the radii are inversely as the versed