1. Advanced Gear Analysis
Epicyclic Gearing
Tooth Strength Analysis

2. Epicyclic Gearset
An epicyclic gear set has some gear or gears whose center revolves about some point.
Here is a gearset with a stationary ring gear and three planet gears on a rotating carrier.
The input is at the Sun, and the output is at the planet carrier.
The action is epicyclic, because the centers of the planet gears revolve about the sun gear while the planet gears turn.
Finding the gear ratio is somewhat complicated because the planet gears revolve while they rotate.
INPUT
CARRIER
Sun
Planet
Ring

3. Epicyclic Gearset Let’s rearrange things to make it simpler:
1) Redraw the planet carrier to show arms rotating about the center.
2) Remove two of the arms to show only one of the planet arms.
INPUT
OUTPUT
INPUT
OUTPUT

4. The Tabular (Superposition) Method
To account for the combined rotation and revolution of the planet gear, we use a two step process.
First, we lock the whole assembly and rotate it one turn Counter-Clockwise. (Even the Ring, which we know is fixed)
We enter this motion into a table using the convention:
CCW = Positive
CW = Negative
INPUT
OUTPUT
Planet
Arm
Sun
Ring

5. The Tabular (Superposition) Method
Next, we hold the arm fixed, and rotate whichever gear is fixed during operation one turn clockwise.
Here, we will turn the Ring clockwise one turn (-1), holding the arm fixed.
The Planet will turn NRing / NPlanet turns clockwise.
Since the Planet drives the Sun, the Sun will turn (NRing / NPlanet) x (-NPlanet / NSun) = - NRing / NSun turns (counter-clockwise).
The arm doesn’t move.
We enter these motions into the second row of the table.
1 Turn
Ring
Planet
Sun
Arm

6. The Tabular (Superposition) Method
Finally, we sum the motions in the first and second rows of the table.
Now, we can write the relationship:
If the Sun has 53 teeth and the Ring 122 teeth, the output to input speed ratio is +1 / 3.3 , with the arm moving the same direction as the Sun.

7. Planetary Gearset n = speed; N = # Teeth
INPUT
OUTPUT
Ring
Planet
Sun
The configuration shown here, with the input at the Sun and the output at the Ring, is not epicyclic.
It is simply a Sun driving an internal Ring gear through a set of three idlers.
The gear ratio is:
n = speed; N = # Teeth
Where the minus sign comes from the change in direction between the two external gears.
If the Sun has 53 teeth and the Ring 122 teeth, the ratio is -1 / 2.3 .

8. Another Epicyclic Example
Here are three different representations of the same gearset:
Given:
1. Ring = 100 Teeth (Input) 2. Gear = 40 Teeth 3. Gear = 20 Teeth 4. Ring = 78 Teeth (Fixed) 5. Arm = Output
Always draw this view to get direction of rotation correct.
See the Course Materials folder for the solution.

9. Epicyclic Tips
If you encounter a gear assembly with two inputs, use superposition. Calculate the output due to each input with the other input held fixed, and then sum the results.
Typically, when an input arm is held fixed, the other output to input relationship will not be epicyclic, but be a simple product of tooth ratios.
OUT
IN 1
IN 2
Use the ± sign with tooth ratios to carry the direction information.

10. Gear Loading
Once you determine the rotational speeds of the gears in a train, the torque and therefore the tooth loading can be determined by assuming a constant power flow through the train.
Power = Torque x RPM, so Torque = Power / RPM
If there are n multiples of a component (such as the 3 idlers in the planetary gearset example) , each component will see 1/n times the torque based on the RPM of a single component.
From the torque, T, compute the tangential force on the teeth as Wt = T/r = 2T/D , where D is the pitch diameter.

11. The Tabular Method - Example 2
First, we lock the assembly and rotate everything one turn CCW     
Next, we hold the arm fixed, and turn the fixed component (Ring 4) one turn CW, to fill in row two.

12. The Tabular Method - Example 2
First, we lock the assembly and rotate everything one turn CCW     
Next, we hold the arm fixed, and turn the fixed component (Ring 4) one turn CW, to fill in row two.

13. The Tabular Method - Example 2    
We turn Ring 4 one turn CW (-1).
Ring 4 drives Gear 3, which turns + N4/N3 x (-1) = - N4/N3 rotations.
Gear 2 is on the same shaft as Gear 3, so it also turns - N4/N3 rotations.
Gear 2 drives Ring 1, which turns + N2/N1 x n2 = + N2/N1 x (- N4/N3) = - N2N4/N1N3 rotations.
The arm was fixed, so it does not turn.
We enter these motions into row two of the table:

14. The Tabular Method - Example 2
Then we add the two rows to get the total motion     
And we can write the relationship:

15. The Tabular Method - Example 2
For our example
Ring 1: N1 = 100 Teeth Gear 2: N2 = 40 Teeth Gear 3: N3= 20 Teeth Ring 4: N4 = 78 Teeth
We compute:     
The output arm rotates almost twice as fast as the input ring, and in the opposite direction.
Output direction is dependent on the numbers of teeth on the gears!