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# 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:

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

The relation on torques is given by:

$$\frac{\Gamma_2}{2} = \Gamma_4=\Gamma_5$$