It is well known in racing that driving the car on the ideal "slip angle" of the tire where it is crabbing slightly from the pointed direction produces more cornering speed than a lower slip angle or a higher one.

(More explanation as requested) I'm considering two main effects on the tire when in a turn:

  1. The tread of the tire is twisted from the angle of the wheel it is mounted to. There is more force as speed increases, and generally, more twisting.

  2. The tire slides somewhat at an angle on the road surface rather than rolling.

At low speeds, the angle between the pointed direction of the wheel (90 degrees to the axis of rotation) and the direction of travel is nearly 0. When the speed increases to the point the angle reaches about 10 degrees, the tire generate more grip and the car goes faster around the turn. (Higher angles produce lower grip)

So the grip is higher at 10 degrees of slip than at 0 or 20 degrees.

What is the physical effect that causes this increase in grip?

  • $\begingroup$ Not to say there's anything wrong with your question, but since a lot of people here may not be familiar with the mechanics of car racing (at least I think it's not just me :P), it would make your question better if you add some more details and/or perhaps a picture to explain the situation you're describing. $\endgroup$
    – David Z
    Commented Feb 25, 2011 at 2:53
  • $\begingroup$ I assume he means that a tire can extert more lateral force, if its direction of motion includes some lateral slip. I don't know if this is true, but it is my take on the physics of the question. $\endgroup$ Commented Feb 25, 2011 at 3:10
  • $\begingroup$ @David Sure, see above. $\endgroup$ Commented Feb 25, 2011 at 22:33

5 Answers 5


At small slip angles, the tire deforms linearly with the side-force demanded of it. This effect saturates when the rubber is essentially fully deformed. The tire then starts to slide over the surface of the road, which reduces the side-force somewhat. Now, this force acts perpendicular to the wheel plane, which is not quite perpendicular to the car's trajectory. This effect reduces the force available to turn the car and causes an "induced drag" that slows it down. The combination of these effects gives rise to the concept of an optimum slip angle.

  • $\begingroup$ You describe why a high slip angle reduces grip, but not why a small slip angle increases grip over zero slip angle. $\endgroup$ Commented Feb 25, 2011 at 19:32
  • 1
    $\begingroup$ The tire is not rigid, so you can't have a zero slip angle with nonzero force. It's not really that the slip is producing a force so much as the force is producing the slip. Increasing the force increases the deformation. $\endgroup$
    – Stingray
    Commented Feb 25, 2011 at 20:56

The simple answer

The tyre is deformed sideways due to the sideways force like Stingray describes, but only around the contact with the road, the rest of the tyre is not deformed so.

The tyre constantly rotates. As the contact pad is lifted off the road the force is taken off this pad and thus it undeforms, meanwhile a new part of the tyre get into contact with the road and is therefore deformed instead. A new deformation happening always moves the car a little sideways, and as new deformations have to happen as the tyre rotates this sideways motion depend on the speed of the car.

The more exact event description

My description has been stepwise, but this is of course a fluid motion. Lets consider a small part of the tyre on its way through the contact area with the road. During the contact phase the tyre part gradually deform more and more, this process must be linear, as the car is moving sideways at a certain speed any point of the tyre that is stationary relative to the road must deform at this speed. The lateral force on the tyre part increase with this deformation.

For the first part of the contact phase this actually looks like a pretty neat pattern, the vertical force increase from the start of the contact pad to the middle of it, so the same coefficient of friction can gradually support more lateral force. Unfortunately towards the back of the contact area the vertical force decreases while the lateral force continue to increase, at some point before the end of contact area the coefficient of friction can't keep up and that part of the tyre slips.


Part of the tyre is slipping, and thus causing degradation, disproportionately more so the close you get to the friction limit. But the grip come from the part of the tyre that is not sliding, this part is however in a constant motion of deformation that effectively makes this part of the tyre rotate in a slightly different direction than the rest of the tyre.


The is a complex question and to cover it we need to look in depth at how tires generate tangential force on the road (as opposed to normal force that is perpendicular to the road).

The answer is of course friction $F=\mu N$ but in this case the answer is complicated by the fact that the rubber surface of the tire deforms, and thus rather than simply taking the normal force multiplying by a friction coefficient and arriving at our peak force, we actually have to integrate the shear stress across the contact patch.


The coordinate system is defined relative to the contact patch.

$$\begin{align*} x &\equiv \text{nominal distance from front of contact patch} \\ y &\equiv \text{distance from left edge of contact patch} \\ s &\equiv \text{length of the contact patch} \\ v_t &\equiv \text{nominal tire surface velocity} \\ v_r &\equiv \text{road velocity} \\ \epsilon &\equiv \text{displacement of tire surface from nominal} \\ \sigma &\equiv \text{surface stress at the tire/road interface} \\ E &\equiv \text{tire stiffness} \\ \mu_s &\equiv \text{static coefficient of friction} \\ \mu_d &\equiv \text{dynamic coefficient of friction} \\ P &\equiv \text{contact pressure} \\ \gamma &\equiv \text{slip rate} \equiv \frac{v_r-v_t}{v_t} \end{align*}$$


Constant tire stiffness

Second assumption: To calculate what the shear stress is at the tire surface we're going to presume the the shear stress is proportional to the displacement of the tire surface from where it would be if it were not touching the road surface.

$$\sigma = E \epsilon\tag{1}$$

Uniformity across the tire width

The contact pressure is nearly constant across the width (at least if you ignore treads) so if we pretend everything is uniform in that direction then we can take our 2D problem and make it 1D

$$\frac{\partial}{\partial y}=0$$

Steady state

We're going to ignore dynamic effects like relaxation length and assume everything that's happening to the tire right now has been happening for a while.

$$\frac{\partial}{\partial t}=0$$

Slipping or not slipping

We're going to assume for each little segment of the tire either it moves along with the road, not slipping at all, or that it is sliding and thus has a shear stress equal to the normal pressure multiplied by the dynamic coefficient of friction.

For the no slip condition this results in the tire surface moving at $v_r$ instead of $v_t$

so relative to a spot on the tire that moves within this no slip region:


Note that this implies an ever increasing displacement, and therefore stress. Once this stress exceeds the static friction, this spot will begin slipping and as the non zero velocity difference would ensure that it remained slipping.

$$\frac{dx}{dt}=v_t$$ $$\frac{dt}{dx}\frac{d\epsilon}{dt}=\frac{v_r-v_t}{v_t}=\gamma$$ $$\frac{d\epsilon}{dx}=\gamma\tag{2}$$

$$\sigma \leq \mu_d P\tag{3}$$

For the slipping region:

$$\sigma = \mu_s P\tag{4}$$

Putting it together

Since we know that the contact pressure at the front and rear of the contact patch approach zero, we know that the shear stress must also approach zero.


$(1)$ & $(5)$


If we assume that the tire is in the no slip condition at the beginning then we have $(2)$ & $(6)$:


$(7)$ & $(1)$

$$\sigma(x)=x\, E\,\gamma\tag{8}$$

$(8)$ & $(3)$

$$\mu_s P \geq x\, E\,\gamma$$

$$x \leq \frac{\mu_s P}{E\,\gamma} $$

This implies that there's a maximum x such that before that x the rubber does not stick and after that the tire does slip. It would be useful to define this distance $x_t$

$$x_t \equiv \frac{\mu_s P}{E\,\gamma}\tag{9}$$

$(4)$, $(8)$, & $(9)$

$$\sigma(x)=\left\{ \begin{array}{ll} x\, E\,\gamma & \quad x \leq x_t \\ \mu_d P(x) & \quad x \gt x_t \end{array} \right.$$

Our total friction force is given is the integral of the stress so:

$$F=\int_0^{x_t} x\, E\,\gamma dx + \int_{x_t}^s \mu_d P(x) dx$$

$$F=\frac12 E\,\gamma \, {x_t}^2 + \mu_d\int_{x_t}^s P(x) dx$$

At this point it would be helpful to define a pressure profile for the tire. Unfortunately the shape of this profile depends on a number of factors including tire type, the level of inflation, and current loading.


The simplest non-zero polynomial that has at least two zeros. A parabola is a decent approximation for highly inflated stiff tires.

$$P(x)= \frac{6L}{s^3} (s\,x - x^2)$$

Where L is the total normal load on the tire.

$$F=\frac12 E\,\gamma \, {x_t}^2 + \mu_d \,L \left(1-3\left(\frac{x_t}{s}\right)^2+2\left(\frac{x_t}{s}\right)^3\right)$$

$$x_t = \frac{\mu_s P(x_t)}{E\,\gamma}$$


Substituting and a lot of simplification yields:


Lets take a look at these variables:


Now $\frac{\gamma \, s}2$ is the average displacement that would occur there was no slipping. And $E\,s\,\epsilon_{ave}$ is the force. So $\frac{E\,s^2\,\gamma}2$ is the force that would occur if there was no slipping.

$L\,\mu_s$ is the force that would occur if the entire patch was slipping. Thus we can call this variable a normalized slip rate $\gamma'$

Note the ratio of static to dynamic friction also appears multiple times. We will define this ratio as $\mu'$

$$\frac{F}{L\,\mu_d} = (3\mu'-2){\gamma'}^3+(3-6\mu'){\gamma'}^2+3\,\mu'\,\gamma'$$

This equation can now easily be plotted for a for a given friction ratio

Force vs slip friction ratio 2 to 1

or a more reasonable

Force vs slip friction ratio 1.3 to 1

Note that as the slip goes towards full slip the plot levels off at one to indicate that the force levels off at full slipping force once the tire starts fully slipping. Also note that before the tire is fully slipping part of the tire is not slipping and thus providing the higher coefficient of static friction. Thus, the real reason that tires provide more cornering power while partially slipping. This allows a maximal use of static friction.

Also note that in this analysis a direction was never assumed for $v_r-v_t$, $\epsilon$, $\sigma$, and $F$ other than that they all were in the same direction. Thus with the approximations used, accelerating, braking, and cornering forces are all treated identically. If you'd like to use this for pure cornering then $\gamma=tan(\theta)$ where theta is the slip angle.

As for the pressure distribution assumed lets see if there's always a maximum before it starts slipping. There will always be a maximum if the slope is ever negative as it will be decreasing from a maximum.


$$\frac{dF}{dx_t}=\frac12\mu_s x_t \frac{d P(x_t)}{dx_t}+\left(\frac12\mu_s - \mu_d\right) P(x_t)$$

So this quantity must always be non-positive:

At $x_t=0$


Therefore if $\frac{d^2F}{{dx_t}^2}\gt 0$ at $x_t=0$ there will be a maximum.

$$\frac{d^2F}{{dx_t}^2}= (\mu_s-\mu_d) \frac{d P(x_t)}{dx_t} + \frac12 \mu_s x_t \frac{d^2 P(x_t)}{{dx_t}^2}$$

At $x_t=0$

$$\frac{d^2F}{{dx_t}^2}= (\mu_s-\mu_d) \frac{d P(x_t)}{dx_t}$$

Which must be positive for all continuous pressure distributions. So for any realistic pressure distribution there will be a maximum force that occurs before the tire is fully slipping.


Not familiar with racing but my initial thought is that the vectors are in a more ideal direction. Maybe the kinetic friction plays a small role (as opposed to static friction)?


The effect sounds the same as braking with ABS. Maximal deceleration is achieved when a wheel rolls slightly slower than the speed of the car. The lateral effect in corners works the same; both are friction effects between rubber and road surface. Both are isotropic materials, i.e. the direction of the friction force doesn't matter.

  • $\begingroup$ I believe this to be true. But why is this true? Is it true for a non-pneumatic tire as well, i.e. a property of the tire material? Or are there other forces at work? $\endgroup$ Commented Mar 2, 2011 at 0:43
  • $\begingroup$ While the materials used are isentropic the tire geometry is not. If you take a look at my answer, the tire geometry effects E and causes the lateral and longitudinal effects to differ in magnitude. However, you are correct that the effect has the same root cause. $\endgroup$
    – Rick
    Commented Jul 17, 2015 at 17:28

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