文档介绍：LESSON FIVE FOUNDATIONS OF MACHINE DESIGN (PART 2)
Action Of a Pair Of Mating Involute Teeth
Let A and B be the base circles of a pair of mating involute gears. Line CD is mon tangent to the base circles, while AB is the line of centers. Assume that CD is part of a string being unwound from A and wound upon B, while A and B rotate together in such a manner that string CD remains taut at all times. When A and B start to rotate, point C on the string will leave circle A and move towards circle B, thus describing an involute with respect to circle A.
At the same time, however, one can imagine that, with respect to circle B, point C traces an involute back to its origin B. The involutes form the actual tooth outline in the ideal case. It can be proven that
the basic requirement for proper
gear action , namely , that no
changes in speed ratio occur
during the passage of any tooth,
is fulfilled when the normal to
the mating tooth curves at the
point of contact always passes
through the pitch point.
Let us investigate whether this condition is satisfied by the involute. It is clear that since the string is taut at all times, the path of the point of contact between the two involutes is a straight line. This line intersects the line of centers (AB) at P. Also, the involute is by definition normal to its generating line (., the string) at all times, since the involute is a circular arc with everincreasing radius, and a radius is always perpendicular to its circular arc. Therefore, if we can prove that point P is the pitch point, we have satisfied all the above mentioned requirements.
Triangles ACP and BPD are similar since their corresponding angles are equal. Like the friction drive at the beginning of this chapter, A and B have the same circumferential velocity at points C and D. We may therefore state that
BD/AC = nA / nB
but
BD/AC = BP/AP
and nA/nB is the speed ratio. Therefore, AP and BP must be pitch radii and point P must be the pitch point