Calculating the amount of teeth on ring gear in planetary gear system

Question

I need to make a planetary gear system such as the following:

The sun and planet gears will be stationary (so that they can only rotate around their centers, not around the sun gear). My goal is to make it so that whenever the sun gear rotates $90^{\circ}$, each planet gear will rotate $-180^{\circ}$ and the ring gear will rotate $-90^{\circ}$. I am fairly certain from my own simulations that a $2:1$ ratio for the sun-to-planet gear system will produce the $90^{\circ}:-180^{\circ}$ rotation I need.

Therefore I have two questions:

1. Will a $2:1$ ratio make the sun and planet gears rotate in a $90^{\circ}:-180^{\circ}$ ratio?
2. How many teeth are needed on the ring gear (or what is the gear ratio) to make the gears rotate in a $90^{\circ}:-180^{\circ}:-90^{\circ}$ fashion?

Answer 1

For the first question: yes you are correct, provided indeed the centres of the planet gears stay fixed.

As for the other question...

The problem is that the needs you are specifying for planetary gearing is in conflict with each other, i.e. the problem is over-constrained. You can define a speed ratio between two types of gears (e.g. sun and planet) but not for three gear types; the third gear will already be determined in order to satisfy the gear ratio between two gear types while also satisfying geometric constraints: that is, the annulus (ring) gear must have a number of teeth equal to that of the sun gear plus twice that of a planet gear; gear pitch radius is proportional to number of teeth as teeth size on meshing gears must be the same.

There are initially six unknowns here: angular velocities of the sun, planet and annulus gears; and the number of teeth of the sun, planet and annulus gears:

$$\omega_s, \omega_p, \omega_a, N_s, N_p, N_a$$

Six unknowns require six independent equations for a unique solution.

Kinematics, or the relations of speeds of gears, must be valid, and so this enforces 1 of the 6 equations:

$$\omega_s N_s + \omega_p N_p = \frac{1}{2} \left( \omega_s N_s + \omega_a N_a\right)$$

Geometry must be satisfied; i.e. the sun and planet gears must be able to fit inside the annulus gear: equation 2 of 6:

$$N_a = N_s + 2 N_p$$

Now, you wish to constrain the motion so that the centres of the sun and planet gear do not move; equation 3 of 6, noting pitch radius is proportional to tooth number:

$$\omega_s N_s = -\omega_p N_p$$

Equations 4 and 5 come from specifying speed ratios:

$$\omega_p = -2 \omega_s$$

$$\omega_a = -\omega_s$$

At this point, it's tempting to think: oh the problem's under-constrained, we still need one more equation. However, whenever you substitute equations into each other, two variables will cancel out in the process: one of the angular velocities and one of the tooth numbers. This is okay: it is the ratios of speeds between gears that matter, not the absolute speeds; same for teeth numbers.

With two unknowns dropping out, this leaves us with 5 equations and 4 unknowns: the problem is over-constrained. This means that, apart from pure coincidence, contradictions will arise whenever you try to solve the equations. An example would be like trying to solve the following:

$$x + y = 3$$ $$x + 2y = 5$$ $$2x + y = 5$$

Two of the equations give a solution, the third creates a contradiction.

How do you resolve this over-constraining? Remove one of the five equations! You can't remove the kinematic or geometric equations since they are fundamental laws for planetary gears. Instead, remove the constraint where the centres of the planet gears must be fixed, or enforce a speed ratio between only two gear types. Then you get four equations, four unknowns.

Answer 2

Have a look at the excellent Woodgears site. Here he explains:

Working out the tooth counts for planetary gears is actually not that complicated, so I initially neglected to mention how to do it. But having received the question a number of times, I'll elaborate.

For convenience, let's denote R, S, and P as the number of teeth on the gears.

• R Number of teeth in ring gear
• S Number of teeth in sun (middle) gear
• P Number of teeth in planet gears

The first constraint for a planetary gear to work out is that all teeth have the same pitch, or tooth spacing. This ensures that the teeth mesh.

The second constraint is: $$ R = 2 × P + S $$

That is to say, the number of teeth in the ring gear is equal to the number of teeth in the middle sun gear plus twice the number of teeth in the planet gears.

In the gear at left, this would be 30 = 2 × 9 + 12

In your drawing you have constrained R and S so you get

$$ P = \frac {R - S}{2} = \frac {24 - 12}{2} = 6 $$

The article also covers "Working out planetary gear turns ratios" and from this we learn that the ratio of angle turned by the planets relative to the ring is given by

$$ T_P = \frac {R T_R}{R+S} = \frac {24T_R}{24+12} = \frac {2}{3}T_R $$

where Tp is the angle of rotation of the planets and Tr is the angle of rotation of the ring. So the planets will turn 2/3 that of the ring.

You can't get the ratios you wish for.

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Another way of looking at this is that if the planets rotate by a number of teeth on the sun gear they will "throw" the ring by double that number of teeth. (It may be easier to imagine this on a linear rack and pinion with the bottom rack (sun) fixed and the upper rack (ring) moveable.

Now if we rotate the planets one complete turn around your sun they will have traveled 12 teeth. The ring will have been thrown by 24 teeth but its diameter is 4/3 times that of the planet's centre so its rotation will be 24 * 3 / 4 = 18 teeth or 1.5 times that of the planets.