# Simulating a car in an intersection

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?

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It is not clear to me that this is a physics question. Could you say a little more about exactly what you are trying to model here and why? –  dmckee Sep 13 '12 at 13:49
I do not agree with the paper in that people are not proportional controllers in adjusting their speed. I think a more reasonable assumption is a) constant (fractional) power acceleration and b) constant value deceleration. –  ja72 Sep 13 '12 at 15:30
I think you need a diagram or a sketch in order to define the problem as you have it. –  ja72 Sep 13 '12 at 16:40

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

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

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Most people corner under 0.2g? That explains a lot. –  Colin K Oct 13 '12 at 20:50
@ColinK: If you have a smartphone and can see the accelerometer values, take a ride with your friends and make some observations of your own. –  ja72 Oct 14 '12 at 0:16
I've gotten up to 0.5 on streets. But I think 0.2 is probably right for normal people. I max out at 0.96 steady state, >1 peak, on street tires in competition. So my standards are not normal :) –  Colin K Oct 14 '12 at 3:24