Worm gear drive numerical

Worm drive relations

For a worm with $z_1$ starts (threads), axial pitch $p_x = \pi m$, and diameter $d_w = qm$:

• Lead $L = z_1p_x$.
• Lead angle $\tan\lambda = \dfrac{L}{\pi d_w} = \dfrac{z_1}{q}$.
• Wheel pitch diameter $d_g = mz_2$; the wheel helix angle equals $\lambda$ (for 90 degree shafts, with the same hand).
• Centre distance $C = \dfrac{d_w + d_g}{2}$; velocity ratio $= z_2/z_1$.

Calculation

Axial pitch: $p_x = \pi \times 6 = 18.850$ mm.

Lead: $L = 3 \times 18.850 = 56.55$ mm.

$$\lambda = \tan^{-1}\frac{z_1}{q} = \tan^{-1}\frac{3}{10} = 16.70^\circ$$

Worm diameter: $d_w = qm = 10 \times 6 = 60$ mm. Wheel diameter: $d_g = 6 \times 45 = 270$ mm.

$$C = \frac{60 + 270}{2} = 165.0\ \text{mm}$$ $$\text{Velocity ratio} = \frac{z_2}{z_1} = \frac{45}{3} = 15$$

Efficiency

$$\eta = \frac{\cos\phi_n - \mu\tan\lambda}{\cos\phi_n + \mu\cot\lambda} = \frac{\cos 20^\circ - 0.05 \times 0.3000}{\cos 20^\circ + 0.05/0.3000} = \frac{0.9247}{1.1064}$$ $$\eta = 83.6\ \%$$

Advantages and disadvantages

Answer: $L = 56.55$ mm, $\lambda = 16.70^\circ$, $d_w = 60$ mm, $d_g = 270$ mm, $C = 165.0$ mm, VR $= 15$, $\eta \approx 83.6\ \%$.