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Chapter 8

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.

Step-by-Step

Verified Solution

Circular Pitch
p = π/P_d = π/6 = 0.5236 in
Axial Pitch
P_x = p = 0.5236 in
Lead
L = N_WP_x = (3)(0.5236) = 1.5708 in
Lead Angle

λ = tan^{-1}(L/πD_W) = tan^{-1}(1.5708/π2.000)
λ = 14.04°
Gear Pitch Diameter

D_G = N_G/P_d = 52/6 = 8.667 in
Center Distance
C = (D_W + D_G)/2 = (2.000 + 8.667)/2 = 5.333 in
Velocity Ratio
VR = N_G/N_W = 52/3 = 17.333
Gear rpm
n_G = n_W/VR = 1750/17.333 = 101 rpm