Tooth, The term also applied in mechanics to the small protruding parts on the circumference of a wheel, by means of which it acts on the leaves of its pinion to turn it.

The shape of gear teeth is critical and must be treated with great care in making machines. The ratios of wheels and pinions may have been calculated precisely, as well as the action to be produced by a given force in a machine, but if the shape of the wheel teeth and the pinion leaves they act upon is not such that the motion of the pinions is uniform, in other words if the effort of the wheels to turn them is not constantly the same, calculation will tell nothing of the actual performance of the machine. Since the effort of the wheels is sometimes greater and sometimes smaller, the performance of the machine can be predicted only in the worst case, and the performance will often be very difficult to measure. This shows how essential it is that the teeth have a suitable shape. However, although machines employing gears have been built for centuries, students of mechanics have completely overlooked these considerations and left this part of machine building to the workmen, whose only rule was to make gear teeth in such a way that the wheels and pinions meshed freely and never stalled. M. [Philippe] de la Hire, of the Académie Royale des Sciences, is the first to speak of the topic. He examines the subject at great length in his Traité des épicycloïdes et de leurs usages dans les Méchaniques [1730], but among the various curves he determines for wheel teeth and various kinds of pinions, only the one he gives for teeth driving a lantern pinion is practical. M. Camus has made up for the shortcomings of M. de la Hire's Traité . This learned academician, in his Mémoire, published in the Mémoires de l'Académie Royale des Sciences , 1733 , determines the curves for the teeth of a wheel and the leaves of a pinion to obtain uniform drive, whether the tooth meets the leaf on line RI (Plate XIX, [key SS,] Figure 102), called the center line ; or (Figure 99) before the center line and drives it beyond; or finally (Figure 98) the tooth meets the pinion leaf before the center line and drives it as far as the line; he may be said to have rendered a very great service to horology. Although skilled clockmakers had fairly sound ideas on the subject, the precise shape of wheel teeth was always a kind of problem for them.

We wish we could reproduce the Mémoire here, for we have found much information in it, but it is a bit too long, and its language is somewhat too abstract for most clockmakers, so we shall attempt to supplement it by demonstrating in another way the shapes of wheel teeth and pinion leaves.

Given a wheel, REV (Figures 98, 100), and a pinion, PIG, I claim that in order for the wheel to drive the pinion uniformly, at any moment during the meshing of the tooth and the leaf the perpendiculars to the faces of tooth and leaf at their point of contact must all meet and lie at the same point, M, on the center line; and point M must be so located on the center line that RM is to MI as the number of teeth on the wheel is to the number of leaves on the pinion.

To demonstrate this, assume line LO drawn perpendicular to the face of the leaf at point G, where the tooth touches it, and lines IO and RL dropped perpendicular to this line from I and R, the centers of the pinion and the wheel. Lines RL and IO will express respectively the lever arm with which the wheel drives the pinion (RL) and the lever arm with which the pinion is driven (IO). This will be obvious if it is observed that lever RL moves perpendicular to line OI, and consequently the length of the infinitely small arcs described in an instant by points L and O will be the same, as happens when a lever acts directly on another on the perpendicular. Therefore since RL expresses the lever with which the wheel drives the pinion, and IO the one with which the pinion is driven, it is clear that at every point of the meshing if the pinion lever and the wheel lever have the same ratio, the action of the wheel on the pinion at all the various points will be uniform. The value in degrees of each of the arcs described in the same time by levers RL and OI is inversely proportional to their length, or as OI is to RL; and the value in degrees of the arcs described by the wheel and the pinion during the same time is again as lever OI is to lever RL. However, since similar levers IO and RL are always in the same ratio at every point of meshing, the value in degrees of the arcs described in the same time by the wheel and the pinion will be so as well. Now the angular velocities of the pinion and the wheel are like arcs. Furthermore, we know from the principles of mechanics that in order to have equilibrium between two forces they must be in inverse proportion to their velocities; therefore, if constant forces acting in opposition, one on the wheel and one on the pinion, are in equilibrium at any point of the meshing, they will be proportional to the velocities of the pinion and the wheel at that point; but since the velocities are always in the same ratio at every point of the meshing, the forces will always be in equilibrium. Therefore, the force with which the wheel will drive the pinion at all these points will always be the same; therefore, the pinion will be driven uniformly.

Once this principle of mechanics has been clearly understood, let us imagine the tooth (Figures 98 and 100) in a random location EG, and that the perpendicular to point G goes through a point, M, on the center line. RL, as we have seen, will be the lever with which the wheel drives the pinion, and OI the lever by which the pinion is driven. Let us assume further that the tooth and the leaf are on the center line and touch at the same point, M. RM will at this point be the lever with which the wheel drives the pinion, and MI the one by which the pinion is driven. But because triangles RLM and MOI are similar, we have RL:OI = RM:MI, so by the preceding principle the wheel will drive the pinion uniformly at points M and G, since the ratio between levers RM and MI at point M is the same as the ratio between levers RL and OI at point G. The same will be demonstrated for all the other points of the meshing, provided that the perpendiculars to the tooth and the leaf meet at point M. Furthermore, the revolutions or velocities of the pinion and the wheel must be inversely proportional to their numbers, and since the wheel must drive the pinion uniformly, their respective velocities at a given point of the meshing must still be in the same ratio. Once these numbers are given, the respective velocities of the pinion and the wheel will be given as well. Now the angular velocity of the pinion at point M is to that of the wheel at the same point as lever MR is to lever MI, therefore MR must be to MI as the number of the wheel is to that of the pinion. Otherwise, the angular velocity of the pinion at this point would not be to that of the wheel as the number of the wheel to that of the pinion. Therefore, point M must intersect line RI in such a way that RM is to MI as the number of the wheel is to that of the pinion. Therefore, if a wheel is to drive its pinion uniformly, at every point of the meshing the perpendiculars to the tooth and the leaf must meet at point M on the center line, at such a location on the line that RM is to MI as the number of the wheel is to that of the pinion.

Q.E.D.

This demonstration, it will be seen, applies to all three cases since the tooth has been considered in any position, before or beyond the center line. It is clear, therefore, that either the tooth and the leaf meet on the center line, or they meet before the line and separate on it, or finally they meet before the center line and separate beyond it. The pinion will be driven uniformly if throughout the meshing the perpendiculars to the contact points between tooth and leaf meet a single point, M, on the center line, so located that RM is to MI as the number of the wheel is to that of the pinion. Furthermore, this demonstration applies to all kinds of gearing in which the wheel is to drive the pinion uniformly, whatever the shape of the wheel teeth and the pinion leaves.

It follows from the preceding demonstration (see Figures 103 and 104) that if the perpendicular to the leaf at any point, G, where the tooth touches it passes through any point, F, between R and M instead of through M, the force required for the wheel to turn the pinion at this point, G, will be greater than when the tooth and the leaf were on the center line and met at M; on the contrary, if the perpendicular passes through a point, T, between M and I, the force required will be less. This is obvious since in the first case the pinion will turn slower, its velocity in relation to the wheel being as RF to FI, as we have said; in the second case it will turn faster, its velocity being to that of the wheel as RT is to TI.

We could have demonstrated this more briefly and in more geometrical terms, but we were guided by clarity and the need to be understood by persons of the craft.

We have seen the conditions that gears must fulfill for the wheel to drive the pinion uniformly; now we shall demonstrate that when the tooth meets the leaf on or beyond the center line, uniformity requires that the face of the leaf be a straight line to the center, and the face of the tooth a portion of an epicycloid generated by a point on a circle whose diameter is the radius of the pinion, rolling on the outside circumference of the wheel.

If circle CQ (Figure 97, No. 2) rolls outside the circumference of another circle, ALE, or internally as at M, a point, C, on the circumference of the first circle will describe a line called an epicycloid . See Epicycloid. If circle COQ has as its diameter the radius of a circle, ALE, then when rolling inside the circumference, as at M, it will describe a straight line, the diameter of circle ALE. See Epicycloid. Having established this, let circles PIG and RVE (Figure 95, No. 2) [1]represent the wheel and the pinion respectively, their diameters HI and HR having the same relationship as the numbers of their teeth . Assume two small circles, COQ [Figures 96, No. 2 and 97, No. 2], having the radius of the pinion as their diameter, and one placed so accurately atop the other that only one is seen. Let their centers be precisely at point O on the center line, and point C at H or D on the same line [Figures 94, No. 2 and 95, No. 2]; then imagine (Figure 94, No. 2) that the wheel and the pinion revolve on their centers from M to X, and the two small circles revolve also, one inside on the circumference of the pinion, the other outside on the circumference of the wheel, in such a way that for every arc described by the pinion and the wheel, they describe completely equal arcs in the opposite direction. In other words, when the wheel and the pinion have described arc MH and MD respectively, the two circles COQ have also described in the opposite direction arc MC equal to arc MH or MD, the pinion outside on the surface of the wheel and the wheel inside on the surface of the pinion. It will follow from this motion of the two circles COQ that their center, O, will not leave center line RI, since whenever the motion of the wheel and the pinion tends to remove them from it along any arc, they will always be brought back by rolling an arc of the same length in the other direction. Now let us assume for a moment that the wheel turns from M to H and drives the pinion by mere friction on its circumference; the effect will still be the same, and the pinion will be driven uniformly since it and the wheel may be considered a pair of rollers, one turning the other by contact. But the small circles, by their movement in the pinion and on the circumference of the wheel respectively, will be in the same situation as the circles COQ M (Figure 96, No. 2) and COQ, which rolled inside and outside the circumference of circle ALE. Thus point C of circle COQ rolling inside the pinion will describe straight line DS, the diameter of the pinion, part of which, like CD, will correspond to arc CM described by the circle at the same time. Similarly, point C of circle COQ, rolling on the circumference of the wheel, will describe an epicycloid, part of which, like CH, will also correspond to arc MH, equal to CM. But since these two circles have the same diameter and always describe equal arcs in the same direction because of the uniform motion of the pinion and the wheel, the point describing C on the circle turning inside the pinion will be located at the same place as the point describing C on the circle turning on the circumference of the wheel. Therefore point C on part DI of straight line DS, and point C on part CH of the epicycloid will be described at the same time. Now at any location of the point describing C, line MC drawn from point M on the center line will be perpendicular to line CD or ID, since these two lines will always form an angle whose apex will be at the circumference of circle COQ and touch its diameter. Likewise, line MC will also be perpendicular to the infinitely small portion of epicycloid CK described at the same time, since MC will be like the radius describing an infinitely small arc, CK. Therefore, if the faces of the leaf and the tooth are generated by a point on a circle whose diameter is equal to the radius of the pinion, and which rolls on its inside circumference and on the outside circumference of the wheel, they will have the same properties as lines CS and CH, and consequently in all situations where they will be, the perpendiculars to their points of contact will all meet and pass through the same point, M. By construction, this point M will divide the center line in the ratio of the numbers of the pinion and the wheel. Therefore, if the face of the leaf is a straight line from the center, and that of the tooth an epicycloid described by a circle whose diameter is the radius of the pinion and rolls on the outside circumference of the wheel, the wheel will drive the pinion uniformly since the perpendiculars to the pinion leaf and the face of the tooth at every point where they meet will converge and always pass through the same point, M, on the center line, dividing the line according to the requisite conditions.

It can easily be seen that this demonstration applies to all kinds of epicycloids; in other words, a wheel will always drive its pinion uniformly if the faces of its leaves are any epicycloids generated by a point on a circle rolling inside the pinion, and those of the tooth other epicycloids generated by the same circle rolling on the circumference of the wheel. Since the action of the wheel in driving the pinion is always uniform, it is clear conversely that the action of the pinion in driving the wheel will be so as well. If at any point of the meshing the action of the pinion were different from any other point, the contrary action of the wheel would be so as well; it would not always act uniformly, which is contrary to the assumption.

In the case where pinion PIG drives wheel REV (Figure 102), it is clear that the leaf would meet the tooth before the center line and would drive it to the line, whence it is easily concluded that a wheel whose tooth meets the leaf before the center line and drives it to the line, is in exactly the same situation. But we have just seen that the pinion drove the wheel uniformly when the faces of the leaves were straight lines from the center, and those of the teeth portions of epicycloids generated by a point on a circle having the radius of the pinion as its diameter and rolling on the outside circumference of the pinion. Likewise, again, when (Figure 99) the tooth drives the leaf before and beyond the center line, its shape must have two lines, one straight, GK, from the center of the wheel, that drives the leaf before the center line, and the other one curved, GE, that drives it beyond; the pinion leaf must also have two lines, one curved, GS, which the tooth drives before the line, and the other straight, DG, from the center of the pinion, with which it drives beyond. The curve of the tooth must be an epicycloid described by a circle whose diameter is the radius of the pinion, rolling on the outer circumference of the wheel; the curve of the pinion must be an epicycloid described by a circle whose diameter is the radius of the wheel, rolling on the outer circumference of the pinion.

We have shown the curves the teeth of the wheel and the leaves of the pinion must be given in the three different situations where the tooth may meet the leaf; it remains only to decide which of these situations is the most advantageous. Clearly, it is the one where the tooth meets the leaf on the center line because 1) the friction of the tooth on the leaf is much less, since there is no butting, as in the other two; and 2) any dirt is pushed outside, rather than being pushed inside, as in the other cases. There is only one case where meshing before and beyond the center line is to be preferred, and that is when the pinion has too few leaves, like 6, 7, etc. up to but not including 10; in such small pinions, assuming the tooth meets the leaf on the center line, it is easily seen that there can be no meshing because the interval between the tips of the teeth is greater than the interval between the leaves at the same point. In order to confirm this by calculation, note that in triangle RIG (Figure 102), knowing the two sides and their angle, it is easy to find the third, which will give the quantity of the mesh as well as the angle IRG; for the meshing to occur on the center line, angle IRG must be smaller by at least two degrees than half the angle described by the tips of two neighboring teeth .

As for the curve to be given the wheel teeth that drive pinions in a different plane, as in a crown wheel , for example, it must be a portion of a cycloid; assuming the face of the pinion leaf is a straight line to the center, the cycloid must be generated by a circle with the radius of the pinion as diameter. The reason for this will be readily grasped if the preceding has been understood at all.

There is much more to say on this subject, which has been greatly neglected although it is much broader than is generally believed, but that would further lengthen this article, which is already rather long. The article Lantern Pinion deals with the shape of teeth that mesh with this type of pinion. See Gears, Wheel, Pinion, Lantern, Leaves, Mesh, Epicycloid, Cycloid, etc.

1. RVE omitted in the figure (translator's note).