I am currently looking for a way to do odometry using wheel encoders, for a car that has no differential drive. The two rear wheel would rotate at the exact same speed. I have been looking everywhere for an answer, but unable to find any. I am starting to think that this is not a possible thing to do. I would be very grateful if anybody can point me in the correct direction.

  • $\begingroup$ how does the vehicle turn? $\endgroup$
    – jsotola
    Apr 17, 2020 at 16:29
  • $\begingroup$ Do you mean a car with ackermann steering but a locked differential (solid rear axle)? $\endgroup$
    – Ben
    Apr 19, 2020 at 12:29

1 Answer 1


So long you have a sensor to read the angle of the steering wheel, you can use the bicycle kinematic model to compute odometry for an Ackermann drive platform, e.g. a car:

$$ \begin{align} \\ \dot{x} & = v \; cos(\psi + \beta) \\ \\ \dot{y} & = v \; sin(\psi + \beta) \\ \\ \dot{\psi} & = \frac{v}{l_r} sin(\beta) \\ \\ \beta & = tan^{-1} \left ( \frac{l_r}{l_f + l_r} tan(\delta_f) \right ) \end{align} $$

Where $v$ is linear forward velocity, $\dot{x}$ and $\dot{y}$ are the linear velocities alongside the Cartesian axes, $\psi$ and $\dot{\psi}$ are the heading angle and angular velocity, $l_f$ and $l_r$ are the distances from the center of mass to the front an rear axles, and $\delta_f$ is the steering wheel angle.

                          enter image description here

Assuming $v$ and $\delta_f$ are retrieved using encoders or other sensors, the vehicle pose can be approximated by:

$$ \begin{align} \\ x_t & = x_{t - 1} + \dot{x} \; \Delta t \\ \\ y_t & = y_{t - 1} + \dot{y} \; \Delta t \\ \\ \psi_t & = \psi_{t - 1} + \dot{\psi} \; \Delta t \\ \\ \end{align} $$


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