# Does a tire need to slip to generate force?

Recently, I have been doing some research on racing and tire modelling. While I was doing this, I encountered many curves like those shown below. Tire Curves http://www.insideracingtechnology.com/Resources/bhvrdrvbrkslip.gif

While I understand the need of slip angles to generate lateral forces (those perpendicular to the direction of motion), the need to have slip in the direction of motion in order to produce longitudinal force has me confused.

After all, wouldn't a tire with no slipping (i.e. perfect traction), be able to produce forward forces far more efficiently than one that didn't use all of that potential?

So, does anyone have an explanation for this phenomenon?

• Notice that car handling is most influenced by the reaction moments produced by lateral forces. Those make the reaction center of the tire patch be behind the centroid of the patch such that the tire tracks (centers) as it rolls. – ja72 Oct 22 '14 at 13:52
• Can someone please explain to me what exactly a force or slip 'percentage' means ? – Gaurav Dec 16 '14 at 11:39
• driving slip = "distance you would expect to have traveled given the rotation and diameter of the tire"/"distance actually traveled"-1 braking slip = 1-"distance you would expect to have traveled given the rotation and diameter of the tire"/"distance actually traveled" Driving force % is just normalized to the peak frictional force the tire can generate. – Rick Dec 17 '14 at 16:25
• Curiously, the question tom proposes has been voted as a duplicate of this very question (cf. this review history). I personally think this question is a better version (answer + question) than what tom has proposed & am voting to leave this one open & hoping others vote to close the other one as a dupe of this one. – Kyle Kanos Jul 22 '15 at 15:03

The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is.

To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the tire, as shown in the following figure: Now if this tire is rolled along the ground with no tangential force between the tire and the ground, we expect the pattern left on the ground to look like this:

where the spacing between the marks on the ground is the same as the spacing of the marks along the circumference of the tire. Now we know that since there are twenty splotches along the circumference of the tire, the distance separating twenty (well twenty-one, technically) splotches on the ground corresponds to the distance the car travelled in one revolution of the tire. Since the spacing of the splotches on the ground is equal to the spacing along the circumference of the tire, we see that distance the car travels in one revolution of the tire is indeed one circumference of the wheel. Thus there is no slipping when there is not tangential force between the tires and the ground.

Now imagine there is a tangential force between the tire and the ground (as will inevitably happen due to air resistance). What happens to the tires? Well the road is providing a force on the tire. This will cause the tire to deform and the bottom of the tire you will get scrunched up in the direction of the force. This is shown (exaggerated) in the figure below:

As the wheel rolls forward (to the right), the scrunched up parts will leave their mark on the ground. This means that our little green splotches in this case are twice as close together, as shown below: When the tire makes a full revolution, it must leave twenty splotches on the ground. On the other hand, these splotches have only half the separation that they did before. Thus the car only moves forward by half of a tire circumference when the wheels rotate once. Therefore, we conclude that the wheel has slipped.

Since there will always be some scrunching given a non-zero tangent force, you will always get slip for a nonzero tangent force.

Now for high enough slip ratios, the wheel will actually slide across the pavement, but until you get to the this point, static friction is still in play, so the car is accelerating from static friction.

Both graphs have 0% force at 0% slip. That implies that no force is being applied to the tire.

As applied force is increased until a certain point the increase in slip remains nearly linear because the coefficient of friction remains nearly constant. Thus effective force also increases nearly linearly.

After that point the coefficient of friction gets reduced as the applied force increases. Thus traction decreases faster than applied force increases and effective force actually decreases.

No idea if that helps or is even correct, but that is how I am reading the graphs.