Originally posted by Andy
Lets say the tire presssure is equal and say the footprint of the bigger tire is larger.
"Why say it if it ain't so", he replies in a totally smart ass tone.
Just kidding. I think we're heading in a similar direction with this in that, if fewer square inches were engaged with the road, and those same fewer inches were still carrying the same weight, the pounds per square inch loading would be higher, resulting in more friction per square inch, netting out to approximately the same gross friction as the larger footprint at the lower PSI loading. Is that the theory?
When I said you can play with air pressures, that is basically what I was getting at. It has the effect outlined above. Although, experience playing with air pressures suggests that the relationship between surface area, a fixed load, and the resulting friction is not a wash. Within reason, decreasing air pressure increases the footprint and traction, increasing air pressure reduces the footprint and traction. Another factor involved shows in heat in the tire and the resulting effect on the friction but I don't think you were going into that detail... were you?
What I was getting at originally was that due to the arc design of the tire and a minimal change in o'all width (approx. 1/4" per side), either tire will flex as much as the air pressure in it will allow, resulting in the same square inches of tire engaging the pavement to support the equal load. A minor allowance could be made for differences in carcass stiffness but it would be miniscul. The "real world" difference is in lifting the bike up on that wider edge at extreme lean angles but there is virtually no difference in the amount of rubber meeting the road. That's my story and I'm sticking to it.
Your turn... show me the error in my ways.