Maximum angle for highway lane change I am preparing to fight a traffic ticket from a speed camera. There is a lot more to that story, but the info I need right now involves the angle at which a car can change lanes, in terms of vehicle angle relative to the direction of traffic flow. I have a 2009 VW Golf Kombi (this is the Jetta Sportwagen in the US). In my case, the relevant speed is 90 km/h, but I'm also interested in learning how to calculate this for any speed.
For example, I believe my vehicle has a steering ratio of approximately 16:1. Thus, if I turned the wheel 16 degrees, I would only deviate from my original lane by 1 degree. That would get me to the other lane eventually, but it is likely to be quite leisurely. On the other hand, if I turned too fast I would likely lose traction.
I know the actual maximum is dependent on many factors (tires, weather, etc). So my question is what type of angles can I expect to achieve in a car similar to mine? And, on a perfect day, with the best tires, what is the highest angle I could achieve without crashing?
 A: The fastest possible way to do a lane change is to fully steer one way on the traction limit and then steer on the opposite way again on the traction limit.
The traction limit is $\mu g = \frac{v^2}{r}$ where $g=9.81\;{\rm m/s^2}$ is gravity, $\mu=0.8\ldots0.9$ is the coefficient of friction (half it in the rain), $v$ is the speed in meters per second and $r$ the radius of turn measured from the center of the car along the rear wheels.
The means the maximum wheel steering angle $\theta$ (to maintain control) at speed is $$\tan \theta = \frac{\mu g \ell}{v^2}$$ where $\ell$ is the wheel base of the car (in meters).
To move the car sideways one lane width $w$ you need to maintain the steering angle $\theta$ one way to trace an arc of radius $r=\frac{v^2}{\mu g}$ for an angle $$\psi=\arccos\left(1- \frac{w}{2 r}\right)$$ The time it takes for this part is $$t=\frac{\psi r}{v} =  \frac{v}{\mu g} \arccos\left( 1-\frac{\mu g w}{2 v^2} \right) $$
For the full lane change double the time. The total distance traveled parallel to the lanes for the full lane change is equal to $$d=\sqrt{ w (4 r-w)} =w \sqrt{ \frac{4 v^2}{\mu g w}-1} $$
