FJCamper wrote:Put the plywood edge on roller bearings and the textbook answer of equal effort flat or on edge is moot.
Lay the plywood flat and it will take the same force to pull over the roller bearings as it does on edge. What's your point?
Hydraulic engineers have two basic ways of exerting disk brake clamping forces, small pads and high pressure and large pads and lower pressures, both modified by piston size.
No, there's only one way of applying force and that's hydraulic pressure times piston area (F = PSI x A), the area units cancel: (lbs-f = (lbs-f / in^2) x in^2). Yes, a small pad will have higher PSI against the rotor than a large pad, but they will apply the same total force (given equal line pressure, piston size). It's that total force (Fn), not pad PSI that you use when calculating the force from friction (Ff <= u x Fn).
Given the same hydraulic pressure, and equal piston sizes, a larger pad might not exert more clamping pressure, but the modifier is pad material ... just like tires were the modifier in lateral G forces.
Well of course if you use pads with a greater coefficient of friction they will have greater braking force given the same clamping force. But the coefficient of friction has nothing to do with pad area, it's a dimensionless quantity.
You'll notice I never mentioned anything about pad material, I was assuming they were the same. Since you never mentioned anything about differences in pad material in your post, I'm not sure why you think I should have taken that into consideration in my reply.
Cranking up the pressure on a smaller pad only leads to faster pad failure once the pad material thermal limits are reached.
Yes. More pressure equates to more heat. And because the small pad has a higher PSI against the rotor than a large pad, it generates more heat per square inch even with the same pressure. Make the pad area really small and you've got yourself a chop saw.