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