Per the answers here we know that four-wheel steering is a mainstream (even if uncommon) feature in commercial autos. Generally it appears that at high speeds and for small turning rates it will turn the rear wheels in the same direction as the front wheels, causing the car to "snake" or "crab" sideways without introducing a yaw moment. (I can't find a canonical term for this type of steering. So I'll stick with "snake" for now.)

It suspect that snaking is preferable to turning for emergency lane-change maneuvers for two reasons:

  1. It is less demanding of traction: It loads the outside two wheels equally and doesn't introduce a yaw moment. In contrast, a turning lane-change first loads the front outside, then the rear outside, and then does the same sequence on the other side as it tries to straighten out in its new lane. As speed increases this is the perfect recipe for inducing a spin (or, with a high center of gravity, a rollover) when one approaches the limit of traction. I would expect that a vehicle can execute an emergency lane-change by snaking at significantly higher speeds and rates than it can by turning.
  2. I think a snaking lane-change can be performed in a shorter distance, because both outside tires are loaded simultaneously, and the translational moment is always less than that translation plus a rotational moment.

Are either or both of these suspicions correct? And if so, can anyone provide numbers to indicate how much faster and/or shorter snaking can maneuver than turning?

(Note that, if correct, these features mean that a car that can execute a snaking lane-change would be able to avoid obstacles that it could not in the same situation using a turning lane-change.)

  • I think you should ask this on engineering.stackexhange.com. Most people here are skilled, experienced mechanics at best. They can give an educated guess, but that's about it. For dynamic road behaviour, you need to think at least in first and second order processes. This kind of complex math is mostly only understood by engineers and such. That being said, the automotive sector is one of the most innovative ones. If it's not applied often, there's probably a good reason, after costs.
    – Bart
    Commented Jan 24, 2017 at 8:00
  • 4
    @Bart I think you're over thinking the problem. The question isn't asking for a numerical solution to a set of differential equations. It's really just setting up a reasonably straightforward suspension under load problem. Turns out that's a common problem in motor sports, making this a good place to ask.
    – Bob Cross
    Commented Jan 24, 2017 at 12:25

1 Answer 1


So let's think through this really logically. We'll start by defining how much the back wheels can turn by comparing with the statistic on the Wikipedia article you referenced: they said that at low speeds, the turning radius can be reduced by as much as 25%. So let's start by defining how far the back wheels are turning at low speeds.

If you view a circle in your mind:

enter image description here

Let's say the radius (arbitrarily) is 10 units. That means that the distance from the center (where the two lines intersect) to the edge of the circle is 10 units.

Now let's subtract 50 percent from the radius:

enter image description here

Because circumference is directly and exponentially related to radius:

c = 2πr

...it's a decent assumption that in order to cut the radius of the circle in two, you would have to double the steering power. In other words, the back wheels would have to turn exactly the same as the front wheels.

Therefore, to knock of 25% of the steering angle, you would have to turn the back wheels 50% of the angle of the front wheels. Now let's transfer that over into the snaking. Let's presume that the maximum steering aid from the back wheels would be 50% of the max turning of the front wheels.

We all know that at high speeds, you never reach anywhere close to 50% of your turning tolerance (at least, I don't remember the last time I had to do hand-over-hand on the autoroute... :) ). Therefore, for the sake of argument, we can say that the back wheels have the same tolerance as the front wheels.

Suppose your front wheels can bring you into a new lane in 1 second (for argument's sake) at your reaction speed turning (in case of emergency). A large section of that 1 second will be "wasted" in a normal vehicle because the back tires are being brought into following. However, depending how long your vehicle is, they will take more or less time following into the lane which you are in.

It would also depend exactly how your particular driver steers. Thus I conclude: mission impossible. It would indeed be faster to make a lane change on a highway with snaking, but how much would have to be determined by too many ungiven factors (factors which would be very difficult to give, incidentally. :) )

Bottom line:

Great question.

Point one is absolutely right. Friction is reduced, as you said, because the back tires are actually leading and not just following: in a sense, being scraped along sideways.

Point two is right, but for an exact calculation, it would totally depend on the vehicle's length and the driver's driving habits, consequently being undeterminable with the variables given.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .