9.15.6 Descending II Braking at maximum lean
Description
This article is from the Bicycles FAQ, by Mike Iglesias with numerous contributions by others.
9.15.6 Descending II Braking at maximum lean
For braking in a curve, take the example of a rider cornering with
good traction, leaning at 45 degrees, the equivalent of 1G centrifugal
acceleration. Braking with 1/10g increases the traction demand by one
half percent. The sum of cornering and braking vectors is the square
root of the sum of their squares, SQRT(1^2+0.1^2)=1.005 or an increase
of 0.005. In other words, there is room to brake substantially during
maximum cornering. Because the lean angle changes as the square of
the speed, braking can rapidly reduce the angle and allow even more
braking. For this reason skilled racers nearly always apply both
brakes into the apex of turns.
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