Originally Posted by
cyccommute
Yes, the rider center of gravity is shifted forward but that has no effect on the rate of deceleration. If the rider does nothing, the deceleration will remain the same regardless of the height of the rear wheel. Moving the CG forward increase the risk of going over the handlebars because it is easier to lift the rear wheel but it doesn't cause the deceleration to decrease.
It doesn't change the rate of deceleration, it changes the rate of deceleration
at which the rear wheel stops rising. A cent
Originally Posted by
cyccommute
Think about what happens when you start down a hill. Depending on how steep the hill is, a bicycle can easily have a vertical differential of 2" to 6". Does this mean that the bike's ability to slow down is less than on level ground? Of course not. If what Wilfred Laurier says is true, a bike headed down an incline would reach a point where it couldn't be stopped.
Firstly, there
is a point where a bicycle can't be stopped going downhill. A hill steep enough that you're CG is over the front wheel. Secondly, those two situations are not equivalent. The force of friction between two surfaces is parallel to the contact between those surfaces. The normal force is called the normal force precisely because it is "normal" (i.e. perpendicular) to those surfaces. On flat ground, the force of friction acts horizontally. On a slope, it acts parallel to the slope (and the normal force is angled off-vertical). So the angle between the traction force and the force of gravity is different, which changes the moment forces.
Originally Posted by
cyccommute
The CG of a bicycle rider is located around the mid-torso of the rider and moves forward but not up when the rear wheel lifts.
Wut?
Originally Posted by
cyccommute
Moving the CG forward increases the normal force...i.e. gravity...on the contact patch.
Yes, but if you're braking hard enough to lift the rear wheel (even if the wheel hasn't lifted yet) then the normal force at the front wheel is already at it's maximum because it's supporting 100% of your weight.
Originally Posted by
cyccommute
With the CG behind the contact patch, the rear wheel can still be pulled down to the ground if the brakes are released which means that there is still a part of the normal force pulling the rear wheel down.
That's not what "normal force" means, or does. Gravity is pulling the rear wheel down, but there is no normal force on it when it isn't touching the ground.
Originally Posted by
cyccommute
If the CG is directly over the contact patch, the normal force is at it greatest on the contact patch.
Yes, but it doesn't reach it's maximum there, it reaches it's maximum as soon as the rear wheel is completely unweighted.
Originally Posted by
cyccommute
But none of this has any effect on the rate of deceleration.
Not the actual rate of deceleration, that's determined by the limits of traction and the torque that the brake puts on the wheel. It
does affect the maximum rate of deceleration possible without tipping further forward.
Originally Posted by
cyccommute
You and Wilfred Laurier assume that the bicycle is going to naturally stop when it gets to the point of a nose wheelie
That's the exact opposite of what I'm assuming. It's also the exact opposite of what I said.
Originally Posted by
cyccommute
It won't. The rider has to feather the brakes and delicately adjust the CG to avoid falling forward. If the braking force doesn't change and if the rider is over the deceleration needed to put them over the handlebars, there is nothing that is going to stop them from falling.
That's what I said.
Seriously, just let me put my dinner in the oven and find my tablet and I'll draw this out for you. It's clear that you have some misconceptions about the geometry of the problem, at least, and I don't think you have a grasp on the dynamics. We can calculate everything, at every point in time, easily.
*Edit: Bah. I suck at digital drawing. Lemme find some paper.