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The rear wheels of a car always face in the direction the car is moving. The front wheels are able to turn left or right and thus can point in the direction the car is moving towards. What I don't understand is how a car can turn with all four wheels rotating (not skidding). That is, how is it possible that the front two tires can face in one direction, the rear two tires in another direction, with the four tires all connected by rigid rods and with all four tires rotating without skidding?

I'm trying to visualize this assuming the car is moving very slowly, but even then the situation just seems impossible to me. Is it in fact that the rear tires are skidding in just very small micro-steps so that we don't actually observe it happening?

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I was going to give a wikipedia link to differential (en.wikipedia.org/wiki/Differential_%28mechanical_device%29) but I think the article there unnaturally lacks obvious visual explanation. –  Yrogirg Aug 17 '12 at 13:24
maybe I misunderstood the question. Do you concerned that while turning outer wheels accomplish greater path than the inner ones? –  Yrogirg Aug 17 '12 at 13:35
No, I was not concerned with the fact that the outer wheels rotate at a different speed than the inner wheels via a differential. What I'm asking is the following: Let the rear tires be facing 0 degrees and the front tires be turned to 5 degrees with the car moving at some reasonably slow velocity such that all of the tires are rotating. How is it that the rear tires can (after some time) be facing 5 degrees without any skidding? How is this possible since the rear tires cannot turn? –  mcFreid Aug 17 '12 at 13:42
How is it that the front tires can (after some time) be facing 5 degrees without any skidding? Both front and rear tires have to turn and skid around the vertical axis to actually turn. Is that the skidding around the vertical (not horizontal) axis you were concerned about? –  Yrogirg Aug 17 '12 at 13:48
I might think about it like from a horse pulling a sleigh (which is not a bad model for a front wheel drive car). The sleigh skids, no matter what direction it's going -- it has no wheels, so everything is a skid. When we put some wheels on the back of the sleigh, it is able to move without skidding in (and only in) the direction of the wheels' rotation. If the sleigh is moving in a direction other than straight ahead, the sleigh still skids, except that now it's actually the tires that skid since they're in contact with the road instead of the sleigh. –  J.T. Grimes Aug 17 '12 at 20:08

4 Answers 4

up vote 15 down vote accepted

The key here is that you think there is no skidding. In fact, there is skidding, although for normal automobiles this is barely noticeable. For normal cars, the rear wheels simply skid a lot less than would the front wheels when a turn would be fully forced.

You can see this also in trucks, where it becomes necessary to have dual or triple-axle steering when doing tight turns while manoeuvring.

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You can hear the sound when they slowly turn tight. –  huseyin tugrul buyukisik Aug 17 '12 at 13:31
that's true, good one –  Rody Oldenhuis Aug 17 '12 at 13:34
Although this isn't completely an answer, it is a correct start to the answer. I remember spending countless hours trying to formalize the problem in basic calculus. If solved fully and correctly, it may not so much describe how a car turns, but instead give an optimal algorithm for a car cornering system, as a naive solution would wear down the tires surprisingly fast. –  AlanSE Aug 17 '12 at 13:41
mcFreid, youtube.com/… (time 1.29) She throws a hoop. It rotates and skids for a certain time, then stops skidding and rotating comes back. I couldn't find a better video, maybe youtube.com/… too –  Yrogirg Aug 17 '12 at 15:31
Thank you! After one second of watching, it clicked. –  mcFreid Aug 17 '12 at 15:38

Simplify it. Think bicycle, not car.

The two axle lines intersect at a point C. Each wheel travels in a circle about that point. There's no skidding involved.

enter image description here

EDIT: As a result of comments, I thought it might be helpful to show what I think the road looks like from the viewpoint of the tire. This is an exaggerated view of the contact patch of the tire against the road. From the tire's point of view, the roadway material is traveling in a circle about center C. So a piece of rubber comes down straight, makes contact with the road, travels in an arc, and then breaks contact with the road and continues in a straight line. It can do this because it's made of flexible rubber.

At no time does it slide against the road - i.e. skid, except for the tiny amount due to the material at the outside edge of the patch actually having to travel farther than the material at the inside edge of the patch.

enter image description here

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Actually, we can simply my question even further. Let there be one wheel on a road which is fixed at the origin upon which it pivots about the vertical axis. Clearly upon spinning the rod, the wheel does rotate (assuming a small enough wheel and a long enough rod). However, I just can't see how the wheel actually changes direction without skidding. I'm trying to visualize the process in small steps, and it seems to me that the wheel must skid. From Rody's answer above this appears to be the case. Now I am trying to understand how both skidding and rotation can occur at the same time. –  mcFreid Aug 17 '12 at 14:24
@mcFreid: As long as we're simplifying, idealize the wheel to be zero width (a knife edge of super-hard material). It rolls forward and backward without slipping. The path on which it rolls does not have to be perfectly straight, does it? From instant to instant, it can turn a little bit without slipping, because it contacts the road at an infinitessimal point. –  Mike Dunlavey Aug 17 '12 at 14:32
@mcFreid: Actually, since a tire contacts the road in a patch, not a point, the side of the patch away from C does have to travel a larger distance. So it does skid a little bit that depends on how wide the tire is. –  Mike Dunlavey Aug 17 '12 at 14:40
At zero width, I still can't seem to visualize it turning without slipping. I can accept the fact without understanding it, but I'd much rather understand it :/ It just seems to me that a rotation has to be in a straight line. If it's not in a straight line, there has to be some kind of movement that isn't a rotation... –  mcFreid Aug 17 '12 at 14:43
@RodyOldenhuis: Then it can turn in a circle about the center of the rear wheel. Then what I said earlier about there being some skidding at the rear wheel due to the width of the contact patch applies. –  Mike Dunlavey Aug 17 '12 at 16:20

Mind you also that the front wheels, which are turning, do not turn to one direction. Both front wheels will be aligned separately, to ensure that the curvature of the trajectory they follow leads to no skidding (see Ackermann steering geometry). The rear wheels, as stated above, are prevented from skidding by the rear differential.

In four wheel driven cars, you typically find three differentials: one for rear wheels (left+right), one for front (left+right), one for front+rear overall.

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This is not exactly what I was asking (see comment in original post). Thanks for pointing this out anyway as I did find the link interesting! –  mcFreid Aug 17 '12 at 13:45
@mcFreid: Indeed this is one of the Ackermann geometry properties: the rear wheels are always just going "straight", as the center of curvature is aligned with the rear wheel axes. I think the illustration in Wikipedia shows this quite neatly. With this geometry, it is only the front wheel steering angles that need to be considered. –  Schedler Aug 17 '12 at 13:50
I think the point that the OP is uncomfortable with is that the rear wheels (and indeed the front wheels on Ackermann steering geometry too!) are slipping (minimally...) due only to their having nonzero width. –  Steven Lu May 6 at 16:26

The way skidding is minimized is by using a differential steering system. Here's an article about the topic and an awesome 1937 video which gives a good explanation about the concept.

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