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 Wheeled Physics.htm How is this possible? Thank you!