Question 14.14: A 0.2-module, 20° pinion of 42 teeth drives a gear of 90 tee...

Given:  m=0.2 mm , a=20^{\circ}, z_{1}=42, z_{2}=90 .

d_{1}=m z_{1}=0.2 \times 42=8.4 mm , d_{2}=m z_{2}=0.2 \times 90=18 mm .

r_{b 1}=r_{1} \cos \alpha=4.2 \cos 20^{\circ}=3.95 mm , r_{b 2}=r_{2} \cos \alpha=9 \cos 20^{\circ}=8.46 mm .

r_{a 1}=r_{1}+m=4.2+0.2=4.4 mm , r_{a 2}=r_{2}+m=9+0.2=9.2 mm .

\text { Length of path of contact, } L_{p}=\left(r_{a 1}^{2}-r_{b 1}^{2}\right)^{0.5}+\left(r_{a 2}^{2}-r_{b 2}^{2}\right)^{0.5}-\left(r_{1}+r_{2}\right) \sin \alpha .

=\left[(4.4)^{2}-(3.95)^{2}\right]^{0.5}+\left[(9.2)^{2}-(8.46)^{2}\right]^{0.5}-(4.2+9) \sin 20^{\circ} .

=1.9384+3.6150-4.5147=1.0387 mm .

Base pitch,    p_{b}=\frac{2 \pi r_{b 1}}{z_{1}}=2 \pi \times \frac{3.95}{42}=0.5909 mm .

Contact ratio    =\frac{L_{p}}{p_{b}}=\frac{1.0387}{0.5909}=1.758 \simeq 2 .