I noticed that while riding my bike on the wet streets the other day, any water that was on the tire would bead up on the middle of the tread. At my typical speed, this didn’t last very long, unless there was quite a bit of water on the tire. I figured there was enough cohesion to hold the water on, and that centrifugal force, imaginary tho it may be, pulled the water to the center. What I’m wondering about is what happened on turns. To turn a bike, you have to lean into the turn several degrees. When this happened, the bead of water moved to the opposite side on the top of the tire. Which way did the bead move on the bottom side of the tire?
I think that in order to find out where the bead would move, you’d have to sum the force vectors of the centrifugal force…actually i’m not sure what it’s called but you have to use the force that is perpendicular to the axis of rotation (Right-Hand Rule and all that) and then add to that the tangential force resulting from the turn (i.e., it has to be in the same plane [x, y, z]).
My WAG would be that while sleeping is right, it doesn’t have to be that complicated. When you are riding in a straight line, centrifugal force pulls the water away from the center of the wheel, so it would slide toward the center. When you are turning, centrifugal force also pulls sideways. So, if you were turning left, the water should move a little to the right, on both the top and bottom of the tire. Again, this is only a WAG, and someone who knows the answer will be here in a moment.
Don’t forget gravity in all this; I think I’m right in saying that the bead of water becomes a spray and leaves the wheel somewhere at the front - the bead exists because (in very crude terms) centrifugal force is pushing the water up and gravity is pulling it down.
There should be no bead on the bottom of the tyre.