777fly: Work out the math(s)
F = uR
F = braking force
u = coefficient of static friction
R = weight (the vertical force due to gravity and other stuff*)
F = ma
a = acceleration (deceleration) resulting from F acting on mass m
R = mg (*see above)
so:
ma = uR = umg
m cancells out:
a = ug
The maximum deceleration available, a (at max antiskid brake pressure) depends on the coefficient of static friction, u. Obviously, this a linear simplification of real life.
Capt Pit Bull:
Hmm. But u is only constant for static friction. Once things are sliding everything gets way more complicated. Materials heat up / liquify / vaporise and this is where the big tyres make a diference.
An approximation of static friction is what you have as long as you don't skid. Its an approximation because under braking forces, the tires begin to flex and there is some scuffing between them and the pavement. This results in heating and a resulting change in u (sometimes an increase as the tires warm up and get a bit softer).
I'll stand behind my statement that, for the simplified case, braking 'effectiveness' doesn't depend on aircraft mass. Mass does affect factors such as the tire contact patch area, tire heating and change in u, weight shifting due to braking (a much larger effect in automobiles) and a number of other factors. Some of these have highly nonlinear relationships to weight (downward force on the contact patch) and depend on which side of the peak on some graph an engineer has placed the braking characteristics.
Look at the case of a truck: More tires (and wider). Heavier (greater R per tire), and yet they tend to have longer stopping distances (at maximum no-skid braking) than a passenger car. That is because the engineers (due to economics) have designed closer to, or on the other side of, the point of maximum braking effectiveness. Which is why you see more 'road alligators' (tire carcasses) from trucks than cars on the highway.
