# Model of a mecanical differential

This post describes and explain the mathematical model of a car differential. We will consider the following device: ## Framework

First, let’s define the framework, axis and positive direction of rotation: Let’s name each gear. The gear number $i$ is labelled with name $Z_i$ and is composed of $Z_i$ teeth: ## Right angle transmission

Let’s have a look at the right angle transmission: This is a classical right angle gear, the transmission is given by: $\omega_1.Z_1 = \omega_2.Z_2$

and $\frac {\Gamma_1}{\Gamma_2} = \frac {Z_1} {Z_2}$

We will no longer consider this transmission in the following.

## Carrier’s frame

We will now work in a new referential attached to the carrier and focus on gears $Z_3$, $Z_4$, $Z_5$ and $Z_6$. The term $\omega^c_i$ is the angular velocity of gear $Z_i$ expressed in the carrier’s referential. According to the previous direction of rotation, the transmissions between gears $Z_3$, $Z_4$, $Z_5$ and $Z_6$ are given by: $\begin{matrix}\omega^c_4.Z_4=\omega_3^c.Z_3 \\\omega^c_3.Z_3=\omega^c_5.Z_5\end{matrix}$

Note that the gear $Z_6$ is redundant with the gear $Z_3$. The relation between gear $Z_4$ and $Z_5$ is given by: $\omega^c_4.Z_4=-\omega^c_5.Z_5$ $\frac{\omega^c_4}{\omega^c_5}=-\frac{Z_5}{Z_4}=-1$

As $Z_4=Z_5$, the relation between angular velocities is : $\omega^c_4=-\omega^c_5$

## Global frame

Let’s now come back in the global referential. The carrier angular velocity is given by $\omega_2$. In the carrier’s referential, the angular velocity of gear gear $Z_4$ is equal to $\omega_4$. The angular velocity of gear $Z_4$ in the global referential is thus given by: $\omega_4=\omega_2+\omega^c_4$

With the same reasoning, we can also express the angular velocity of gear $Z_5$ in the global referential: $\omega_5=\omega_2+\omega^c_5$

Gathering the previous equations gives the following system: $\left \{ \begin{array}{r c l}\omega_4 &=& \omega_2+\omega^c_4 \\\omega_5 &=& \omega_2+\omega^c_5 \\\omega^c_4 &=& -\omega^c_5\end{array} \right .$ $\left \{ \begin{array}{r c l}\omega_4 &=& \omega_2-\omega^c_5 \\\omega_5 &=& \omega_2+\omega^c_5 \end{array} \right .$ $\left \{ \begin{array}{r c l}\omega^c_5 &=& \omega_2-\omega_4 \\\omega_5 &=& \omega_2+\omega^c_5 \end{array} \right .$

Solving the system provides the final equation of the model: $\omega_2 = \frac{1}{2}(\omega_4+\omega_5)$

## Torque

The relation between gears $Z_3$, $Z_4$ and $Z_5$ is given by: $\begin{matrix}\frac{\Gamma_4}{\Gamma_3}=\frac{Z_4}{Z_3} \\\frac{\Gamma_5}{\Gamma_3}=\frac{Z_5}{Z_3}\end{matrix}$

As $Z_5=Z_4$ : $\frac{\Gamma_4}{\Gamma_3}=\frac{\Gamma_5}{\Gamma_3}$

And finally : $\Gamma_4=\Gamma_5$

## Power

The input power must be equal to the output power: $P_{in}=P_{out}$

As $P=\Gamma.\omega$, the output torque can be rewrite as: $P_{out}=P_4+P_5=\Gamma_4.\omega_4+\Gamma_5.\omega_5$

As $\Gamma_4=\Gamma_5$, the previous equation can be simplified: $P_{out}=\Gamma_4.(\omega_4+\omega_5)$

As $\omega_4+\omega_5=2.\omega_2$, the previous equation can be rewriten: $P_{out}=\Gamma_4.(2.\omega_2)=\Gamma_2\omega_2=P_{in}$

Finally: $2.\Gamma_4=\Gamma_2$

## Conclusion

The relation on velocities is given by: $\omega_2 = \frac{1}{2}(\omega_4+\omega_5)$

The relation on torques is given by: $\frac{\Gamma_2}{2} = \Gamma_4=\Gamma_5$