Simulating a car in an intersection

Question

I'm somewhat confused. I want to simulate in real-time an intersection where cars have to turn left, right or go straight. What I have are 2 way points: One at the beginning of the intersection on the incoming street and the other at the end of the intersection on the outgoing street. As I know the next way point on the outgoing street, I know which direction the car should be pointing.

How would I slow the car down to the optimal speed, calculate its steering angle and correct it in a time interval so that the car drives an optimal curve?

A resource I have found, that seems quite good for this is the following paper.

I just don't really understand the first part of the paper where the Circular track is calculated. At which point is the steering angle applied?

Answer 1

Consider a steady corner shown below:

The coordinates of any point along the curve are $x = r - r \cos \varphi$ and $y = r \sin \varphi$. The angle $\varphi$ is computed from the distance traveled $s = r \varphi$ where $r$ is the cornering radius.

If the final point $B$, and orientation $\theta$ are given, then the radius is

$$ r = \frac{x_B-x_A}{1-\cos\theta} $$

and

$$ y_B-y_A = r \sin \theta $$

The velocity and acceleration vectors at $P$ are:

$$ \vec{v} = v(t) (\sin \frac{s(t)}{r}, \cos \frac{s(t)}{r}) $$ $$ \vec{a} =( \dot{v} \sin \frac{s(t)}{r}+\frac{{v(t)}^2}{r} \cos \frac{s(t)}{r}, \dot{v} \cos \frac{s(t)}{r}-\frac{{v(t)}^2}{r} \sin \frac{s(t)}{r} ) $$

Now since most people don't corner with more than $a_N =\frac{{v(t)}^2}{r}= 0.2g$ then the target cornering speed is

$$ v(t) = \sqrt{ a_N \; r } $$

If you are accelerating $\dot v > 0$ or decelerating $\dot v <0$ to reach the target speed, the make sure you do not exceed the desired cornering acceleration $a_N$ by checking

$$ \left(\frac{v^2}{r}\right)^2 + \left(\dot{v}\right)^2 \le a_N^2 $$