280 Mr. Arry on the Forms of the Teeth of Wheels. 
of the teeth, and CD the normal of both at that point; therefore 
they will touch at D, and the line of action CD, will pass through 
the fixed point C. If now we give equal velocities to the cir- 
cumferences CH, CK, the same will be found at all times to be 
true. These forms then are proper for the teeth of wheels. 
Suppose then this problem proposed. Given the form of the 
teeth of one wheel, to find the form of those of another, that 
they may work together correctly. The followimg is the obvious 
solution. Divide the line joining the centers of the circles at C, 
into two parts, whose proportion is the mechanical power. De- 
scribe the circles CH, CK. Find the curve which by revolving 
upon CH, will generate the given tooth HD. Make the same 
curve revolve in CK, and with the same describing point let it 
generate KD; KD is the form required. 
The usual construction of the involute of a circle, would 
seem to require that the circles 4H, and BK, should be separated. 
If however DH be the involute formed in the usual way from 
the circle MN, (Fig. 5.) the normal CM will be inclined at a 
E : P J 
constant angle to C4, (since its sine = Lea and the construc- 
tion given before shews that the involute HD may be generated 
by the revolution of a logarithmic spiral upon CH, the describing 
point being the pole of the spiral, and the angle between its 
radius and tangent, the same as the angle made by MC, with 
the tangent of the circle at C. In the same way the revolution 
of this spiral in the second circle will generate another involute ; 
and hence if the teeth of one wheel be involutes, those of the 
other wheel must also be involutes. The generating circles of 
the involutes must have radii proportional to 4C, BC. 
It will be seen immediately, that we may if we please sup- 
pose successive parts of the curve described by different generating 
curves; or we may make one curve revolve on the outside of 
