Given the position of the vehicle (𝑥,𝑦) at different time points, the speed of the vehicle (m/s), the direction the vehicle is facing (heading — in degrees), the track width of the vehicle, and the wheelbase of the vehicle, how can I calculate the steering angle of the vehicle given the vehicle is rear-wheel steering?
This is very similar to my previous question.
Here is how to solve the geometry problem of getting the wheel steering angles from the known radius of curvature $r$ of a path.
I used $w$ for the track, and $b$ for the wheelbase, for a rear-wheel steering vehicle.
The wheel angles from straight equal in measure the angles shown above as $\theta_1$ and $\theta_2$ (due to similar triangles).
$$\begin{aligned} \tan \theta_1 & = \frac{b}{r-w/2} \\ \tan \theta_2 & = \frac{b}{r+w/2} \\ \end{aligned} $$
Now given a path of coordinates $(x,y)$ of the point where I drew the blue velocity vector, you need to fit a cubic spline through the points in order to get a list velocities and acceleration. Each point needs to be a function of an independent quantity such as time, so you can get a table of values for $$\begin{array}{c|c|c} \text{time} & \text{position} & \text{velocity} & \text{acceleration} \\ \hline t & (x,y) & (\tfrac{\rm d}{{\rm d}t} x, \tfrac{\rm d}{{\rm d}t} y) & ( \tfrac{{\rm d}^2}{{\rm d}t^2} x, \tfrac{{\rm d}^2}{{\rm d}t^2} y) \\ \ldots \\ t_i & (x_i,\,y_i) & (\dot{x}_i\,\dot{y}_i) & (\ddot{x}_i, \, \ddot{y}_i) \end{array} $$
Then for each point the tangent vector is given by the direction of the velocity vector
$$ \boldsymbol{\hat{e}} = \frac{ ( \dot{x}_i , \dot{y}_i )}{\sqrt{ \dot{x}_i^2 + \dot{y}_i^2 } } $$
The radius of curvature $r$ by
$$ \frac{1}{r} = \frac{ \ddot{x}_i \dot{y}_i - \ddot{y}_i \dot{x}_i }{ \left( \dot{x}_i^2 + \dot{y}_i^2 \right)^{3/2} } $$
