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?

  • 1
    $\begingroup$ You are travelling at 90km/h and you want to know the angle for a lane change. Estimate how long it takes you to change lanes, then estimate the distance between the middle of two adjacent lanes. Then use that to get the speed perpendicular to the direction of traffic. Some fancy trigonometry will then easily allow you to figure out the angle your car was at $\endgroup$
    – Jim
    May 21, 2015 at 16:18
  • $\begingroup$ I'm more concerned with what the tires can handle. How fast could one change lanes without skidding? It would be pretty hard to experiment on this safely (or with much accuracy in the measurements). $\endgroup$
    – Eric Marsh
    May 21, 2015 at 17:06
  • $\begingroup$ There's not enough information available for us to answer that question. But anecdotal evidence would suggest they can handle quite a bit. I've seen people change lanes in under half a second. $\endgroup$
    – Jim
    May 21, 2015 at 17:30
  • $\begingroup$ I have to imagine in the vehicle dynamics world there are plenty of people who would know a rough answer off the top of their heads. I guarantee they have a computer model of my vehicle at VW that can simulate the suspension, tires, and weight to determine a very precise value. I'm interested in something in between. What can we do with basic physics based on properties such as the friction of the tires and the mass of the vehicle? $\endgroup$
    – Eric Marsh
    May 21, 2015 at 17:39
  • $\begingroup$ Indeed, if you tell us what the end goal is, what you are ultimately trying to show, we might be able to tailor answers to provide the physics for that end $\endgroup$
    – Jim
    May 21, 2015 at 17:44

1 Answer 1


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 $$d=\sqrt{ w (4 r-w)} =w \sqrt{ \frac{4 v^2}{\mu g w}-1} $$

  • $\begingroup$ Is the coefficient of friction much different for different types of tires? Or maybe it's negligible for our purposes? If it is relevant, would you agree that the front tires are more important than the back tires (I have a different type in front than in rear)? $\endgroup$
    – Eric Marsh
    May 21, 2015 at 21:25
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    $\begingroup$ All season tires will have lower coefficient of friction compared to performance (summer) tires. A minimum of 0.85 can be assumed for most tires. $\endgroup$ May 22, 2015 at 0:40
  • $\begingroup$ A simpler calculation is to estimate it takes for example $t=3$ seconds to do the maneuver and the lanes centers are $w=3$ meters so the average angle is $$\psi =\frac{w/t}{v}$$ where $v$ is the speed in meters per second and the angle is in radians. The maximum angle can be estimated to be twice the average angle. $\endgroup$ Dec 20, 2015 at 14:46
  • $\begingroup$ Your first sentence sounds plausible: should it be obvious, or is there some nontrivial optimization problem that you haven't shown in full? $\endgroup$ Jul 11, 2016 at 14:48
  • $\begingroup$ Also known in the racing world as the "bang-bang" method. You can prove it mathematically, or you can read up in controls, path planning, robotics and cam mechanisms that optimal is always at the constraint limits. $\endgroup$ Jul 11, 2016 at 19:10

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