Question 20.1: A pair of worm gears is designated as,1/30/10/8 Calculate (i...

\text { Given } \quad z_{1}=1 \quad z_{2}=30 \text { teeth } \quad q=10 \quad m=8 mm .

Step I Centre distance
From Eq. (20.9),

a=\frac{1}{2} m\left(q+z_{2}\right)                 (20.9).

a=\frac{1}{2} m\left(q+z_{2}\right) \frac{1}{2}(8)(10+30)=160 mm        (i).

Step II Speed reduction

i=\frac{z_{2}}{z_{1}}=30 .           (ii).

Step III Dimensions of worm

d_{1}=a m=10(8)=80 mm            (a).

d_{a 1}=m(q+2)=8(10+2)=96 mm              (b).

\tan \gamma=\frac{z_{1}}{q}=\frac{1}{10} \quad \text { or } \quad \gamma=5.71^{\circ} .

d_{f 1}=m(q+2-44 \cos \gamma) .

=8[10+2-4.4 \cos (5.71)] .

= 60.9747 mm               (c)

p_{x}=\pi m=\pi(8)=25.1327 mm                     (d).

Step IV Dimensions of worm wheel

d_{2}=m z_{2}=8(30)=240 mm .            (a)

d_{a 2}=m\left(z_{2}+4 \cos \gamma-2\right) .

=8[30+4 \cos (5.71)-2] .

= 255.8412 mm          (b)

d_{f}=m\left(z_{2}-2-0.4 \cos \gamma\right) .

=8[30-2-0.4 \cos (5.71)] .

= 220.8159                   (c).