Model of a mecanical differential

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

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Framework


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

Frame

Let’s name each gear. The gear number i is labelled with name Z_i and is composed of Z_i teeth:

Differential

Right angle transmission

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

Link_1

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.
Link_4

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

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