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## Q. 8.4

A wormgear has 52 teeth and a diametral pitch of 6. It mates with a triple-threaded worm that rotates at 1750 rpm. The pitch diameter of the worm is 2.000 in. Compute the circular pitch, the axial pitch, the lead, the lead angle, the pitch diameter of the wormgear, the center distance, the velocity ratio, and the rotational speed of the wormgear. See Figure 8–28.

## Verified Solution

Circular Pitch
$p = π/P_d = π/6 = 0.5236$ in
Axial Pitch
$P_x = p = 0.5236$ in
$L = N_WP_x = (3)(0.5236) = 1.5708$ in
$λ = tan^{-1}(L/πD_W) = tan^{-1}(1.5708/π2.000)$
$λ = 14.04°$
$D_G = N_G/P_d = 52/6 = 8.667$ in
$C = (D_W + D_G)/2 = (2.000 + 8.667)/2 = 5.333$ in
$VR = N_G/N_W = 52/3 = 17.333$
$n_G = n_W/VR = 1750/17.333 = 101$ rpm