A worm and wormgear set is to be designed to produce a velocity ratio of 40. It has been proposed that the diametral pitch of the wormgear be 8, based on the torque that must be transmitted. (This will be discussed in Chapter 10.) Using the relationships presented in this section

Question

A worm and wormgear set is to be designed to produce a velocity ratio of 40. It has been proposed that the diametral pitch of the wormgear be 8, based on the torque that must be transmitted. (This will be discussed in Chapter 10.) Using the relationships presented in this sectionspecify the following:
Worm diameter, [latex]D_W[/latex]
Number of threads in the worm, [latex]N_W[/latex]
Number of teeth in the gear, [latex]N_G[/latex]
Actual center distance, C
Face width of the gear, [latex]F_G[/latex]
Face length of the worm, [latex]F_W[/latex]
Minimum thickness of the rim of the gear

Accepted Answer

Many design decisions need to be made, and multiple solutions could satisfy the requirements. Presented here is one solution, along with comparisons with the various guidelines discussed in this section. This type of analysis precedes the stress analysis and the determination of the power-transmitting capacity of the worm and wormgear drive which is discussed in Chapter 10. Trial Design: Let’s specify a double-threaded worm: [latex]N_W[/latex] = 2. Then there must be 80 teeth in the wormgear to achieve a velocity ratio of 40. That is,

[latex]VR = N_G/N_W = 80/2 = 40[/latex]
With the known diametral pitch, [latex]P_d = 8[/latex], the pitch diameter of the wormgear is
[latex]D_G = N_G/P_d = 80/8 = 10.000[/latex] in
An initial estimate for the magnitude of the center distance is approximately [latex]C[/latex] = 6.50 in. We know that it will be greater than 5.00 in, the radius of the wormgear. Using Equation (8–27) ([latex]1.6<{\frac{C^{0.875}}{D_W}}<3.0[/latex]), the recommended minimum size of the worm is

[latex]D_W = C^{0.875}/3.0 = 1.71[/latex] in
Similarly, the maximum diameter should be
[latex]D_W = C^{0.875}/1.6 = 3.21[/latex] in
A small worm diameter is desirable. Let’s specify [latex]D_W = 2.25[/latex] in. The actual center distance is
[latex]C = (D_W + D_G)/2 = 6.125[/latex] in
Worm Outside Diameter
[latex]D_{oW} = D_W + 2a = 2.25 + 2(1/P_d) = 2.25 + 2(1/8) = 2.50[/latex] in

Whole Depth
[latex]h_t = 2.157/P_d = 2.157/8 = 0.270[/latex] in
Face Width for Gear
Let’s use Equation (8–28) ([latex]F_G=(D^2_{oW}-D^2_W)^{1/2}[/latex]):
[latex]F_G = (D_{oW}^2 - D_W^2 )^{1/2} = (2.50^2 - 2.25^2)^{1/2} = 1.090[/latex] in
Let’s specify [latex]F_G= 1.25[/latex] in.

Addendum
[latex]a = 1/P_d = 1/8 = 0.125[/latex] in
Throat Diameter of Wormgear
[latex]D_t = D_G + 2a = 10.000 + 2(0.125) = 10.250[/latex] in
Recommended Minimum Face Length of Worm

[latex]F_W = 2[(D_t/2)^2 - (D_G/2 - a)^2]^{1/2} = 3.16[/latex] in
Let’s specify [latex]F_W = 3.25[/latex] in.

Minimum Thickness of the Rim of the Gear
The rim thickness should be greater than the whole depth:
[latex]h_t > 0.270[/latex] in

Question 8.5: A worm and wormgear set is to be designed to produce a veloc...

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