Originally Posted by
cyccommute
This is also not a high school physics problem.
Yes it is. In it's simplest form (which is being argued here) it's a very straightforward equilibrium problem. Hell, it's not even three-dimensional.
Originally Posted by
cyccommute
Even if you managed to stop the wheel but were in front of the contact patch, the tire won't slide. The bearings that we so carefully adjust on the wheels allow for free movement and the momentum of our mass will carry us over the bars before the tire slides.
This (bolded text) is untrue. Think about it, you've locked the wheel up with the brake so the bearings aren't turning.
Originally Posted by
cyccommute
And I have explained to you your error in thinking that the "pitch over" point is when the rear wheel just leaves the ground. A "pitch over" can also be called an endo, a header (from the days of the ordinary), a face plant or a few other colorful things. They all describe crashes in which the rider leaves the bike by going over the handlebars. To do that your rear wheel has to be quite a bit further in the air then the instant your rear wheel has zero force on the ground. That's a skid if the rear brakes have locked the rear wheel and isn't nearly as disastrous as a faceplant.
Again, (bolded) is inaccurate, although only a tangential point. Locking up the rear wheel with the rear brake is NOT equivalent to unweighting the rear wheel through use of the front brake. The rear brake will lock up well before the rear wheel is carrying zero weight. When it does braking suffers because sliding friction is weaker than static friction but it'll still slow you down, which couldn't happen if there was no weight on it.
In fact it's an excellent demonstration of why the front brake is inherently more powerful.