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 is labelled with name and is composed of 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 , , and . The term is the angular velocity of gear expressed in the carrier’s referential.

According to the previous direction of rotation, the transmissions between gears , , and 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 is redundant with the gear . The relation between gear and 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 , 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 . In the carrier’s referential, the angular velocity of gear gear is equal to . The angular velocity of gear 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 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 , and 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 :

$! \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 , the output torque can be rewrite as:

$! P_{out}=P_4+P_5=\Gamma_4.\omega_4+\Gamma_5.\omega_5 $

As , the previous equation can be simplified:

$! P_{out}=\Gamma_4.(\omega_4+\omega_5) $

As , 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 $