If the pinion rotates at 120 rpm. Determine;

i) The minimum number of teeth on each wheel to avoid interference.

ii) The number of pairs of teeth in contact.

Subject: Kinematics of Machinery

Topic: Gears and Gear Trains

Difficulty: High

Given $\phi=20^{\circ}$; $N_{P}$=120 rpm; VR=3; Addendum=1.1 m, m=3mm, $\alpha_{w}$=1.1

(i) $T=\frac{2a_{w}}{\sqrt{1+\frac{1}{G}(\frac{1}{G}+2)\sin^{2}\phi-1}}=\frac{2\times1.1}{\sqrt{1+\frac{1}{3}(\frac{1}{3}+2)\sin^{2}20^{\circ}-1}}$=49.44

Taking the higher whole number divisible by the velocity ratio,

i.e T = 51, and t=51/3 =17

(ii) Number of pairs of teeth in contact,

n= Arc of contact / circular pitch = (Path of contact / $\cos\phi$) $\times \frac{1}{\pi m}$

$n=\frac{\sqrt{(R_{a})^{2}-R^{2}\cos^{2}\phi}-R\sin\phi+\sqrt{(r_{a})^{2}-r^{2}\cos^{2}\phi}-r\sin\phi}{\cos\phi\times\pi m}$

$R=\frac{mT}{2}=\frac{3\times 51}{2}=76.5 mm$

$R_{a}=R+ 1.1 m=76.5 + 1.1 \times 3 = 79.8 mm$

$r=\frac{mT}{2}=\frac{3\times 17}{2}=25.5 mm$

$r_{a}=25.5 + 1.1 \times 3 = 28.8 mm$

$n=\frac{\sqrt{(79.8)^{2}(76.5\cos 20^{\circ})^{2}}-76.5\sin20^{\circ}+\sqrt{(28.8)^{2}-(25.5\cos 20^{\circ})^{2}}-25.5\sin20^{\circ}}{\cos20^{\circ}\times\pi 3}$

$n=\frac{34.646-26.165+15.977-8.720}{\cos20^{\circ}\times\pi 3}=1.78$

Thus, 1 pair of teeth will always remain in contact whereas for 78% of time 2 pairs of teeth will be in contact.