Why does speed affect turn radius when turning a motorcycle given a fixed lean angle?

Question

I have been looking at a formula which is supposed to calculate the lean angle and turn radius which is

$$\theta=\arctan\left(\frac{v^2}{gr}\right)$$

I do not understand how velocity effects turning radius at a fixed lean angle. I would think that turn radius is proportional to lean angle regardless of velocity (within limits of no traction loss or tyre warp or falling over)

My reasoning is that I understand motorcycles turn due to lean. At lean the outer tyre contact patch has a shorter wheel radius than the inner tyre radius. The bike turns in an arc resolve the differences in outer tyre radius and inner tyre radius. (think of a cone rolling along a flat surface, the cone will roll in a turning direction). And that tyres are parabolic in shape which means greater leans increase the difference between inner and outer contact patch which makes smaller (sharper) turning circle.

If there is no loss of traction and no warping in the tyre (like the cone example). Would the speed of the rolling cone make any difference in the turning circle of the cone? Would not the cone make the same turning circle radius regardless of speed? and the motorcycle tyre do the same?

I can imagine however that at higher speeds the force on the inner part of the tyre is increased (due to centripetal force of the turn), forcing the contact patch to have less of an angle which will increase the circular radius (have wider turns). But how is the formula above describing this, should it not take into account tyre pressure and and mass to calculate the warp in the tyre?

Is the formula above simply describing something else; like the minimum speed you need to hold a lean at a given lean angle without falling over?

Answer 1

It helps to understand where the $\theta$ formula comes from. Consider the free body diagram:

The bike (represented by the thick red line) is rotating about the axis of rotation (AoR, dashed black line). The distance of the CoG to the pivot point $P$ is $R$. The distance of the CoG to the AoR is $r$. The lean is $\theta$, with respect to the vertical.

There are now $2$ torques, acting in opposition about the point $P$. To prevent the bike from flipping over either way, both torques have to cancel each other, as per N2L: $$mgR\sin\theta=F_c R\cos\theta$$ The $R$ cancel, so: $$\Rightarrow mg\sin\theta=\frac{mv^2}{r}\cos\theta$$

So that:

$$\boxed{\tan\theta=\frac{v^2}{gr}}\tag{1}$$

I can imagine however that at higher speeds the force on the inner part of the tyre is increased (due to centripetal force of the turn), forcing the contact patch to have less of an angle which will increase the circular radius (have wider turns). But how is the formula above describing this, should it not take into account tyre pressure and and mass to calculate the warp in the tyre?

The formula makes one important assumption, primarily that there's enough friction between the pivot point $P$ and the road, to maintain the rotation about the AoR. Precisely how this is achieved is irrelevant to the lean and the formula doesn't need to concern itself with that.

The lean simply arises from the balance of torques shown in the free body diagram.

Answer 2

The static friction between the tires and road causes the centripetal acceleration. For minimum radius of turn at a given speed, you want the maximum allowed friction, μN. At any speed you want the resultant force from the road to point through the center of mass. This gives the condition for the angle of lean from the vertical, tan(θ) = f/N =((mv^2)/R)/(mg). For maximum friction, tan(θ) = μN/N = μ. Given that no road surface is perfectly uniform, one should not try turning at this minimum radius .