On 2007-02-11, wrote:
[...]
I'd rather not go to the numbers because they are irrelevant. What is
relevant is a coefficient of traction enough to cause an endo on solid
pavement and that this can be as much as 1g considering the rider CG
(belly button) can be on a 45° line from tire contact with the road.

Half of that load is on each dropout pushing up and rearward at 45°.
At the same time a downward force of the entire front wheel load times
the ratio o f wheel diameter to disk diameter (up to 4x) is pulling
one side of the wheel downward. Therefore, roughly 8x times the
upward force on that one dropout.

This is correct, but I misunderstood it.

If we consider the direction of the net force on the front wheel, it's
[1, 0] from braking, [0, 1] from ground reaction, and [sin(theta),
-cos(theta)] * R/r from the disk. Not 2R/r.

We're assuming maximum braking here, and so the net torque on the wheel
is 0. The fact that the disk is only one side makes no difference to
that calculation.

Jobst's description is considering net force on the axle, which is
pushing the rider back along 45 degrees with both dropouts, but is
resisted at only one.

Correcting my calculations again, I get 34 degrees from vertical for Ben
Micklem's wheel (which is 54 degrees from the dropout exit angle), and
17 degrees from vertical for the wheel with the caliper at 3 o'clock.

I think the numbers do matter, particularly the difference in angle
between the direction of the ejection force and the exit direction
provided by the dropout. If the difference is 45 degrees, it seems to
me the wheel would certainly not eject however large the magnitude of
the force. This is because the component pushing it out of the dropout
will be exceeded by the component burying it into the steel/aluminium of
the dropout.

function phi = ejection_angle(theta, R, r)
% theta is angle of caliper in radians, measured anticlockwise with 0 at 3
% o'clock. R is tyre radius, r is disk radius. phi is angle of ejection
% force from vertical in degrees.

% The caliper force. x +'ve is rearward, y -'ve is downward.
c = [sin(theta), -cos(theta)];

% Its magnitude is scaled by difference in radius between disk and tyre,
c *= R / r;

% Add the braking force and the contact force (friction coefficient is 1.0)
% to give the net force f on the wheel under braking.
f = c + [1, 0] + [0, 1];

% Work out that vector's direction from vertical
phi = atan(f(1) / f(2));
phi *= -360 / (2*pi);
endfunction

% Caliper at 2:30.
theta = (0.5 / 12) * 2*pi;
disp(ejection_angle(theta, 675.14, 160));

% Caliper at 3:00
disp(ejection_angle(0, 675.14, 160));