Path of contact and contact ratio numerical

Derivation

Let $R_a$, $r_a$ be the addendum circle radii of the wheel and pinion, $R$, $r$ the pitch circle radii, and $R_b = R\cos\phi$, $r_b = r\cos\phi$ the base circle radii. The line of action $N_1N_2$ is tangent to the pinion base circle at $N_1$ and to the wheel base circle at $N_2$, and passes through the pitch point $P$. Contact starts where the wheel addendum circle cuts the line of action (point $K$) and ends where the pinion addendum circle cuts it (point $L$).

   N1 ..........K....P....L......... N2
        approach | recess

In the right-angled triangle $O_2N_2K$ (wheel centre $O_2$): $KN_2 = \sqrt{R_a^2 - R_b^2}$. In triangle $O_2N_2P$: $PN_2 = R\sin\phi$. So

$$\text{Path of approach } KP = KN_2 - PN_2 = \sqrt{R_a^2 - R_b^2} - R\sin\phi$$

Similarly, from the pinion triangles $O_1N_1L$ and $O_1N_1P$:

$$\text{Path of recess } PL = LN_1 - PN_1 = \sqrt{r_a^2 - r_b^2} - r\sin\phi$$ $$\text{Path of contact } KL = KP + PL = \sqrt{R_a^2 - R_b^2} + \sqrt{r_a^2 - r_b^2} - (R + r)\sin\phi$$

Arc of contact $=\dfrac{\text{path of contact}}{\cos\phi}$; contact ratio $=\dfrac{\text{arc of contact}}{p_c}$.

Data

$r = 40$ mm, $R = 80$ mm, $r_a = 44$ mm, $R_a = 84$ mm, $\phi = 20^\circ$.

$$r_b = 40\cos 20^\circ = 37.588\ \text{mm}, \quad R_b = 80\cos 20^\circ = 75.175\ \text{mm}$$

(a) Path of approach and recess

$$KP = \sqrt{84^2 - 75.175^2} - 80\sin 20^\circ = 37.479 - 27.362 = 10.117\ \text{mm}$$ $$PL = \sqrt{44^2 - 37.588^2} - 40\sin 20^\circ = 22.873 - 13.681 = 9.192\ \text{mm}$$

(b) Path of contact

$$KL = 10.117 + 9.192 = 19.31\ \text{mm}$$

(c) Arc of contact

$$\frac{KL}{\cos 20^\circ} = \frac{19.31}{0.9397} = 20.55\ \text{mm}$$

(d) Contact ratio

Circular pitch $p_c = \pi m = 12.566$ mm.

$$\text{Contact ratio} = \frac{20.55}{12.566} = 1.64$$

A contact ratio of 1.64 means that two pairs of teeth are in contact for about 64% of the time and one pair for the rest. Since it is greater than 1, the drive is continuous.

Answer: approach $= 10.117$ mm, recess $= 9.192$ mm, path of contact $= 19.31$ mm, arc of contact $= 20.55$ mm, contact ratio $= 1.64$.