Assembly condition of planetary gear trains

Condition for equal spacing

Let the sun have $z_s$ teeth and the ring $z_r$ teeth. Fix the ring and place the first planet in mesh with both. Now rotate the arm to bring the first planet to the next planet position, a fraction $1/N$ of a revolution away (angle $360^\circ/N$). The sun turns through

$$\theta_s = \frac{360^\circ}{N}\left(1 + \frac{z_r}{z_s}\right)$$

(the arm turns $1/N$ revolution, and with the ring fixed the sun turns $1 + z_r/z_s$ times as fast as the arm).

A second planet can be placed in the vacated position only if the sun has turned through a whole number of tooth pitches, that is, $\theta_s = k \times \dfrac{360^\circ}{z_s}$:

$$\frac{z_s + z_r}{N} = k \quad \text{(integer)}$$

So the sum of the teeth on the sun and ring must be divisible by the number of planets. For the geometry (equal module) we also need $z_r = z_s + 2z_p$.

Check: $z_s = 24$, $z_r = 72$

Geometry: $z_p = (72 - 24)/2 = 24$ teeth, which agrees with the given planets.

$$z_s + z_r = 24 + 72 = 96$$

The planets must also not touch each other. The planet centre radius is $m(24+24)/2 = 24m$, and the planet outside diameter is $m(24 + 2) = 26m$. For $N = 6$, the distance between neighbouring planet centres is $2 \times 24m\sin 30^\circ = 24m$. This only equals the pitch diameter, so the tooth tips (26m) would overlap: 6 planets cannot be fitted. For $N = 4$ the distance is $2 \times 24m\sin 45^\circ = 33.9m > 26m$, and for $N = 3$ it is $41.6m$, so both are fine.

Result: 3 and 4 planets can be assembled; 5 fails the tooth condition; 6 meets the tooth condition but the planets would interfere.

Applications

1. Automatic transmissions of vehicles (speed change by locking elements).
2. Differentials of automobiles (bevel epicyclic train).
3. Reduction gearboxes of wind turbines, hoists and cranes.
4. Turbine speed reducers; watch mechanisms and epicyclic screwdrivers/hand drills.