Pete's experiment shows that heat in the tyre does not appear to be a factor. If it were, the count would run down with each run, and it clearly does not. Which leaves us with surface friction/tyre slip as the cause of the discrepancy.
quote:You ask:In order for R=H, the length of the contact region needs to be shorter than what I have shown in Figure 1. How does this happen?
It is shorter because the tyre does not slip and its longitudinal compression is constant everywhere along the length of the contact patch. For the motorbike formula to be true, the friction with the road is not able to provide the necessary compression of the tyre near the end of the contact patch so the tyre slips and has a lower average value of compression, this means that the contact region is in fact longer than the no slip case, which has a 0.6% smaller value for your angle alpha. There have to be changes to the tyre geometry outside the contact patch which enables this to happen without changing H. As I see it the whole non contacting part of the tyre changes slightly so that it is no longer an arc of a circle.
This is what I am finding difficult to calculate and then draw.
I see where this is going. As the tread surface is flattened, its deformation causes the tread to creep backward (opposite the direction of rotation). The already-compressed bit at the center will not slip, due to available traction, so the leading edge is forced to do so. This action slows the rotation of the wheel.
Roll a ball of dough across a table, and you'll notice a fat little ridge develops at the leading edge where the dough meets the table. It develops because you are forcibly rolling it. With a bicycle tire, this action would be much less pronounced, but the data supports it.
I'm not surprised you're having a hard time calculating and drawing it.
The available traction being the limiting factor in slowing the wheel's rotation, different surfaces would produce different calibrations.
This is getting good.