Factors affecting motorcycle grip when turning

Hi everyone.

I've been having this discussion with a pal regarding what affects the grip of a motorcycle's tires when turning and I was hoping someone in here could enlighten us :)

So, we have a motorcycle that is making a turn at a constant speed and leaned in an angle.
How do you calculate if the bike is able to make the turn or if the tires will skid and the bike fall? Clearly, making everything else constant, there will be a velocity cutoff that the bike will not be able to make it.
From wikipedia I got this equation that gives the lean angle with respect to the velocity and curve radius:
∂ = arctan( v^2 / (gr) )

Because the tires are not spheres, the bike can not lean indefinitely. Assuming the lean is still inside the tire limits, what affects the grip?

To me, this problem is solved as follows:
We have a centripetal force that is causing the bike to turn.
We have the weight of the bike equal to the normal force of the ground.
therefore, let
Fn be Normal force,
Ff be Friction Force,
μ be the static friction of the tire/asphalt
m the mass of the bike
r the radius of the turn
ac centripetal accelaration
v velocity

we have:
ac = v^2/r
(fricion force is the friction coeficient times the normal force) (=) Ff = μ Fn
(on the vertical axis, the resulting force is 0) (=) Fn - mg = 0 (=) Fn = mg
(on the horizontal axis the bike is turning) (=) Ff = m ac (=) μ Fn = m v^2/r

if we solve this for v, we get v=√(μ g r)

So, this means that
a) the mass of the bike is irrelevant
b) the lean angle is irrelevant. The Friction does not increase with the lean angle

However, I've seen in some places people say that the friction actually increases with the lean. How is this possible?
(e.g. http://genjac.com/BoomerBiker/Two%20...%20Physics.htm
 A myth exists that leaning a motorcycle reduces the cornering grip (friction) of the tires. Notice that the mass of the vehicle does not change—if it weighed 600 lbs. on a straightaway, it still weighs 600 lbs. in a curve. However, gravity works in corners to help us out. Friction is actually increased when g-forces come into play. Camber thrust and total grip is also affected by centrifugal force as the motorcycle leans when going around a corner. Cornering causes centrifugal force to press the tires downward into the asphalt, compressing both the front and rear suspension springs, reducing ground clearance. (...) Camber thrust literally compresses the motorcycle tire as the rubber tries to roll under the wheel rim. Cars can't do this trick since the inside tires lose as much grip as the outside tires gain. Traditional car suspensions lose camber thrust completely in turns. The faster a motorbike rider goes through the same turn, the more grip he has to work with (until the bike drags). At a 45 degree lean angle, a motorcycle has nearly 50% more "weight" pressing the tires into the pavement, and thus benefits from nearly 50% more grip than it does when it is vertical, thanks to centrifugal force and camber thrust.
How is this possible?

Thank you!
 The quoted passage appears to use the idea that the coefficient of static friction is the ratio of frictional force to total force. As such, it is flatly wrong. If that sort of notion were supportable then a coefficient of static friction above 1.0 would ensure that a tire could never detach from the pavement under any cornering speed or lean angle.
 Wouldn't the motorcycle's centrifugal force push sideways: parallel to the ground? I don't see how it would add to the weight and thus the grip in the tire. Unless the road was banked.

Factors affecting motorcycle grip when turning

That is my idea.
The leaning is irrelevant to the friction. The weight of the bike does not change because it is leaning, therefore the Friction force must remain the same.
I might be missing something though..