## Epicyclic (Planetary) Gear Trains

Epicyclic gear trains differ from ordinary ones in that at least one of their axes is not fixed to the frame, but instead rotates about another axis. This allows several speed reductions to be achieved with a single set of gears, and yields very high power densities.

<img src="./figs/imagenes_tema_08/Tren_Epicicloidal_gif.gif" alt="Tren_Epicicloidal_gif" width="200px">

In an epicyclic gear train we distinguish 3 fundamental rotational speeds:

*   $\omega_{input}$ ($\omega_O$): angular velocity of the first central axis (e.g. the Sun gear).
*   $\omega_{output}$ ($\omega_Z$): angular velocity of the last central axis (e.g. the Ring gear).
*   $\omega_{carrier}$ ($\omega_b$): angular velocity of the planet-carrier support (Carrier/Arm).

<img src="./figs/imagenes_tema_08/Tren_Epicicloidal_Esquema.png" alt="Components: Sun, Planets, Carrier, Ring gear" width="600px">

### Willis' Formula and the Apparent Ratio ($\mu_a$)

To solve the kinematics, we use the **kinematic inversion** method (Willis' formula). It consists of mentally subtracting the carrier's velocity from the whole system so as to bring it to rest ($\omega_b - \omega_b = 0$). The train then momentarily behaves as an ordinary gear train.

The velocity ratio in this fictitious state is called the **Apparent Gear Ratio ($\mu_a$)**:

$\mu_a = \frac{\omega_{output} - \omega_{carrier}}{\omega_{input} - \omega_{carrier}} = (\pm 1) \cdot \frac{\Pi Z_{driving}}{\Pi Z_{driven}}$

> **Note:** $\mu_a$ (also called $\mu_{fixed\_train}$) is the ratio the train would have if the carrier were held fixed. Its sign depends on whether the meshes are external (-) or internal (+).

### Classification and Behavior Depending on the Fixed Element

Although the system has 2 degrees of freedom, in many industrial applications one of them is constrained (one shaft is held fixed) to use it as a reducer or multiplier. As we already know, if it is the carrier that is fixed, the result is an ordinary gear train:

<video src="_static/gears/Tren_Bloqueado.mp4" height="300" controls autoplay muted loop></video>

However, by fixing the first or the last gear instead, we obtain the following cases.

#### A. Simple Epicyclic Train (First shaft fixed, $\omega_{input} = 0$)

The first gear is fixed (for example, the Sun gear). The input is the carrier and the output is the last gear (e.g. the Ring gear).

$\frac{\omega_{output}}{\omega_{carrier}} = 1 - \mu_a$

**Behavior depending on $\mu_a$:**
*   $\mu_a < 0$: speed multiplication.
*   $0 < \mu_a < 1$: speed reduction (very high if $\mu_a \approx 1$).
*   $\mu_a > 1$: reversal of rotation direction.
*   $\mu_a > 2$: multiplier with reversal.

<video src="_static/gears/Tren_Simple.mp4" height="300" controls autoplay muted loop></video>

Examples of a simple epicyclic train:

<img src="./figs/imagenes_tema_08/ejemplo_simple.png" alt="Simple train example" width="300px">

#### B. Rocker Train / "tren de balancín" (Last shaft fixed, $\omega_{output} = 0$)

The last gear (normally the Ring gear) is prevented from rotating. The input is usually the Sun gear and the output is the Carrier.

$\frac{\omega_{input}}{\omega_{carrier}} = 1 - \frac{1}{\mu_a}$

**Behavior depending on $\mu_a$:**
*   $\mu_a < 0$: multiplication (very high if $\mu_a \approx 0$).
*   $\mu_a > 1$: no reversal of rotation. The closer $\mu_a$ is to 1, the greater the reduction.

<video src="_static/gears/Tren_Balancin.mp4" height="300" controls autoplay muted loop></video>

Examples of rocker trains:

<img src="./figs/imagenes_tema_08/ejemplo_balancin.png" alt="Rocker train example" width="600px">

### Differential Gear Trains

These are pure epicyclic trains where **no gear is fixed**. The system keeps its 2 degrees of freedom.

**Willis' formula for differentials:**
Solving the general equation for $\omega_{carrier}$:

$\omega_{carrier} = \frac{1}{1-\mu_a} \cdot \omega_{output} - \frac{\mu_a}{1-\mu_a} \cdot \omega_{input}$

**Main functions:**
*   **Combining motion (2 inputs $\rightarrow$ 1 output):** Adds or subtracts the velocities of two shafts onto a third one.
*   **Distributing motion (1 input $\rightarrow$ 2 outputs):** The case of the automotive differential.

### The Automotive Differential

It allows the driving wheels to rotate at different speeds while cornering. In a symmetric differential ($\mu_a = -1$), the carrier's speed (connected to the engine) is the average:

$\omega_{carrier} = \frac{\omega_{left\_wheel} + \omega_{right\_wheel}}{2}$

Differentials are fitted between the engine and the driving wheels of automobiles so that each wheel can turn a different number of revolutions while still receiving power. For example, when the car turns to the right, the left wheel travels a greater distance than the right one, so it must turn more revolutions than the right wheel, or else one of the two wheels will skid.

<img src="./figs/imagenes_tema_08/Trenes_Diferenciales.png" alt="Differential diagram" width="600px">

The differential gear train can take several forms, but the most common one is the one shown in the figure on the left. Schematically, it can be represented as shown in the center figure. Gear $e$ corresponds to the input shaft coming from the gearbox. Gear 2 is attached to the housing which, in turn, carries the axles of the planet gears 1. The two gears 1 are redundant and are fitted for symmetry, although kinematically a single one would suffice.

In the usual case where the car is travelling in a straight line, the planet gears behave as if their axis were locked and could not rotate about itself; they simply drag the $i$ and $d$ shafts and make them rotate at the same speed. When the car goes around a bend, gears 1 acquire a certain rotational speed of their own, which allows the left and right wheels to rotate at different speeds.

To solve the differential's motion mathematically, we use the method described in the previous section, consisting of bringing the moving axes to rest — in this case, those of gears 1. To do this, we stop gear 2 and obtain the train shown in the figure on the right, from which we deduce:

$ \frac{\omega_d - \omega_2}{\omega_i - \omega_2} = - \frac{Z_i}{Z_d} $

But since the numbers of teeth $Z_i$ and $Z_d$ are equal, this gives:

$ \omega_d - \omega_2 = -(\omega_i - \omega_2) $

Since the bevel gears 2 and $e$ mesh directly, we can write:

$ \frac{\omega_2}{\omega_e} = \frac{Z_e}{Z_2} $

Solving for $\omega_2$ and substituting into the first equation, we obtain:

$ \omega_i + \omega_d = 2 \omega_e \frac{Z_e}{Z_2} $

This expression relates the angular velocity of the input shaft to the two output shafts. From it we deduce that, for a given input speed, the sum of the output speeds is fixed, but not each output speed individually.

It can also be verified that if the engine is stopped ($\omega_e = 0$) and, for example, the left wheel is made to rotate in one direction, the right wheel will rotate at the same angular speed but in the opposite direction.