Free collection of beautiful vector icons for your web pages.

Mathematical model of a mechanical differential

This page describes and explains the mathematical model of a car differential. We will consider the following device:

3D render of a mechanical differential


First, let's define the framework, axis and positive direction of rotation:

Frames and axis used for modeling the mechanical differential

Let's name each gear. The gear number \(i\) is labelled with name \(Z_i\) and is composed of \(Z_i\) teeth:

Number of teeth for each part of the mechanical differential

Right angle transmission

Let's have a look at the right angle transmission:

Mechanical differential - right angle transmission

This is a classical right angle gear, the transmission is given by:

$$ \omega_1.Z_1 = \omega_2.Z_2 $$


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

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.

Mechanical differential - carrier's frame

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:

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


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 :

$$ \Gamma_4=\Gamma_5 $$


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:

$$ P_{out}=\Gamma_4.(2.\omega_2)=\Gamma_2\omega_2=P_{in} $$


$$ 2.\Gamma_4=\Gamma_2 $$


The relation on velocities is given by:

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

The relation on torques is given by:

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

See also

Last update : 10/21/2020