As far as the physics, I have no training other than high school physics over 50 years ago..., but what little physics I think I understand would indicate that the lower the rear wheel is, the greater the braking force possible.

The reason is that the center of gravity is raised and moved forward based on any rotation caused by the wheel lifting while the direction of force (momentum) remains relatively constant. As with any other lever, the closer the center of mass is to the fulcrum (front axle) the less force is required to lift the mass. While technically the weight is always the same distance from the fulcrum in this case, it seems intuitive that it is effectively closer during braking if relating it to the direction of the force creating the rotation (momentum going forward).

I have not performed any experiments, or read the results of anyone else's... However, my hypothesis is that if you have the front wheel secured, and apply horizontal force directly forward (on the saddle), it would require significant force to lift the rear wheel... also, if you apply vertical force upward on the saddle, it should require less force to start rotation. Starting the bicycle at a 45 degree angle, I believe that the two forces would be closer to the same... So in short, the forces needed to rotate a bicycle using forward momentum should decrease as the distance of the rear wheel from the ground increases (until completely vertical). Since less momentum is required, I believe it makes sense that maximum braking possible is less.

What the impact of angle of force versus rotation is at various points in the rotation, I leave to real physicists.

I do also accept that changing the forward momentum to the upward rotation of lifting the rear wheel does take some energy, so this must be considered in the whole equation, and not just static forces as in my hypothesis... I just have no means to hypothesize the impact of that momentum change.

Thank you for taking the time to read my uninformed post.