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,

Question

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.

Accepted Answer

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

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

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

Question 8.4: A wormgear has 52 teeth and a diametral pitch of 6. It mates...

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