Question 12.4: Geometric Quantities of a Worm A triple-threaded worm has a ......

For a 90° shaft angle, we have \lambda=\psi .

a. The velocity ratio of the worm gearset is

r_s=\frac{N_w}{N_g}=\frac{3}{48}=\frac{1}{16}

b. Using Equation (12.20),

L=p_w N_w        (12.20)

p_w=\frac{L}{N_w}=\frac{75}{3}=25  mm

From Equation (12.2) with m_n=1 / p_n , we obtain

P p=\pi, \quad P_n p_n=\pi, \quad P_n=\frac{P}{\cos \psi}, \quad P=\frac{N}{d}      (12.2)

p_n=\pi m_n=7 \pi=21.99  mm

Equation (12.1) results in

p_n=p \cos \psi, \quad p_a=p \cot \psi=\frac{p_n}{\sin \psi}        (12.1)

\cos \lambda=\frac{p_n}{p_w}=\frac{21.99}{25}=0.88 \quad \text { or } \quad \lambda=28.4^{\circ}

Application of Equation (12.21) gives

\tan \lambda=\frac{L}{\pi d_w}=\frac{V_g}{V_w}      (12.21)

d_w=\frac{L}{\pi \tan \lambda}=\frac{75}{\pi \tan 28.4^{\circ}}=44.15  m

Through the use of Equation (12.18),

r_s=\frac{\omega_g}{\omega_w}=\frac{N_w}{N_g}=\frac{L}{\pi d_g}        (12.18)

d_g=\frac{L}{\pi r_s}=\frac{75}{\pi / 16}=381.97  mm

We then have

c=\frac{1}{2}\left(d_w+d_g\right)=\frac{1}{2}(44.15+381.97)=213.1  mm