Mathematical model of a mechanical differential

This page describes and explains 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 gears $Z_4$ 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:

$$ \left\lbrace \begin{array}{r c l} \omega_4 &=& \omega_2+\omega^c_4 \\ \omega_5 &=& \omega_2+\omega^c_5 \\ \omega^c_4 &=& -\omega^c_5 \end{array} \right. $$

$$ \left\lbrace \begin{array}{r c l} \omega_4 &=& \omega_2-\omega^c_5 \\ \omega_5 &=& \omega_2+\omega^c_5 \end{array} \right. $$

$$ \left\lbrace \begin{array}{r c l} \omega^c_5 &=& \omega_2-\omega_4 \\ \omega_5 &=& \omega_2+\omega^c_5 \end{array} \right. $$

Solving the system provides the final equation of the model:

$$ \omega_2 = \frac{1}{2}(\omega_4+\omega_5) $$

Torque

The torque 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: