Marginal Planetary Gear Set

I have to calculate a ratio of a planetary gear set. It's an existing gear set and had been already implemented for years ago(works fine). Somehow it does not fit the general formula of planetary gear set which is

R = 2*P + S (number of teeth)

On my gear set: R = 62 P = 21 S = 18
62 != 21*2 + 18

So how can i calculate the correct ratio?

• are you sure you didn't miscount? Aug 3 '17 at 13:34
• I'm sure. It's written on technical drawing but anyway, I have counted on both technical drawing and CAD file. Aug 3 '17 at 14:17

While R = 2*P + S is typical, the deviation doesn't stop you from calculating the gear ratio. The carrier is over sized from a tight fit, but it seems to be functioning. A 22 gear planet would help, but these are unusual. So let's just calculate:

The overall gear ratio of a simple planetary gearset can be calculated using the following two equations, representing the sun-planet and planet-ring interactions respectively:

$$S\omega_s + P \omega_p - (S + P)\omega_c = 0$$ $$R\omega_r - P \omega_p - (R - P)\omega_c = 0$$

where $\omega_r,\omega_s,\omega_p,\omega_c$ is the angular velocity of the Ring, Sun, Planet and Planet Carrier respectively, and $R,S,P$ are the Number of teeth of the Ring, the Sun gear and each Planet gear respectively. From the above we can deduce that:

$$S\omega_s + R\omega_r = (S + R)\omega_c$$ $$-\frac{R}{S} = \frac{\omega_s - \omega_c}{\omega_r - \omega_c}$$

Assuming $\omega_{r} \neq \omega_c$. If we know which are fixed, we can go further. Calling $n=\frac{R}{S}$

Case 1, Sun fixed: $$\frac{\omega_c}{\omega_r}=\frac{n}{1+n} =\frac{R}{R+S}$$ Case 2, Carrier fixed: $$\frac{\omega_r}{\omega_s}=-\frac{1}{n}=-\frac{S}{R}$$ Case 3, Ring Fixed: $$\frac{\omega_c}{\omega_s}=\frac{1}{1+n} =\frac{S}{R+S}$$

In your example, $n=62/18 = 3.44$, so Case 1 is 31:40, Case 2 is 9:31, and Case 3 is 9:40.