I have been repeatedly told when dealing with some data for Navigation purposes that I can assume it follows a "bicycle model". I am not very familiar with vehicle dynamics so I consulted online. I saw many different models varying on the level of sophistication but most of them seems to consist in planar Dynamics similar to the following:

\begin{equation} \begin{bmatrix} x_{k+1} \\ y_{k+1} \\ \theta_{k+1} \end{bmatrix} = \begin{bmatrix} x_{k} \\ y_{k} \\ \theta_{k} \end{bmatrix} + \begin{bmatrix} v\cdot\Delta t\cdot\cos{\theta_k} \\ v\cdot\Delta t\cdot\sin{\theta_k} \\ f(\gamma,v) \end{bmatrix} \end{equation}

where $\gamma$, and $v$ are the steering angle and the velocity, both inputs from the vehicle sensors, and $\theta$ is the angle respect to the origin of a fixed frame. $f(\gamma,v)$ is some function relating both angles.

My doubt arises from the fact that the data I have been granted contains only the following fields: $x$,$y$,$\theta$. My assumption is that, if we know it must follow a bicycle model it should hold a correlation like the following:

\begin{equation} \begin{bmatrix} x_{k+1} \\ y_{k+1} \\ \theta_{k+1} \end{bmatrix} = \begin{bmatrix} x_{k} \\ y_{k} \\ \theta_{k} \end{bmatrix} + \begin{bmatrix} d_k\cdot\cos{\theta_k} \\ d_k\cdot\sin{\theta_k} \\ \Delta\theta_k \end{bmatrix} \end{equation}

Where $d_k$ and $\Delta\theta_k$ are known at every time step and are the distance traveled and the change in the angle respectively.

If I generate data this way, starting from an initial point $(x_0,y_0)$ and building up, it doesn't seem to correlate well with the actual data I have been granted. My question is if the deduction exposed has any conceptual mistake, so I can clear out if the problem is that the data given is very noisy and deviates after a while or that what I think is a bicycle model is not

  • $\begingroup$ if you are given x,y,theta then you don't need to solve the system you are showing here. Its not really clear what you are trying to do. $\endgroup$ – agentp Mar 27 '17 at 20:54
  • $\begingroup$ Hi @agentp. I understand it may sound confusing. The data given has errors, so what I want to do is to compare the data values from the ones you would retrieve expanding the time series form the beginning using the "perfect" dynamics without noise. However since I don't know the Dynamics beforehand, only know they are from a "bicycle model", I am trying to find what this generally means in this context $\endgroup$ – A. Frenzy Mar 28 '17 at 0:19
  • 1
    $\begingroup$ "bicycle model" simply means you are not concerning yourself with the details of steering geometry, accounting for the slight differences between inside and outside wheel angle and so on. $\endgroup$ – agentp Mar 28 '17 at 0:29
  • $\begingroup$ Thank you @agentp. Therefore an agent that moves using the above Dynamics would be following a bicycle model? As a contextualization, the real agent is supposed to be a pedestrian, whose movement has been simplified to the "bicycle model" $\endgroup$ – A. Frenzy Mar 28 '17 at 1:56

Your problem may be simplified to this: You have an initial position and velocity (including angular displacement) for discrete time points. And, you wish to find the position and direction at any given time instant.

The bicycle model you have discussed here is basically a first order approximation solution. It means you are ignoring the change in velocity between one time-step and the other. This is still acceptable if the time-step is small. If it is large, the approximation may no longer be accurate. If the time-step is too... small, there may be truncation errors on account of the precision of your computer (you may not possibly worry about this error).

The conclusion can be that either the resolution of your data is low or else, as you have pointed out, the data may be too noisy.

| improve this answer | |
  • $\begingroup$ My doubt is precisely if the model in question is called or not the "bicycle model". The fact that it is an approximation of planar dynamics is clear to me, but thanks anyway. $\endgroup$ – A. Frenzy Mar 30 '17 at 23:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.