I have been implementing a basic motorcycle physics model and have been predicting the turning circle of both wheels based on the current angle of the front wheel.
However, this fails to predict the turning circle once the motorcycle begins to lean. How can I calculate the radius of the turning circle of each wheel, given I know the angle of the motorcycle lean?
The explanation by @user121330 uses arguments in the non-inertial frame of the rider. That'll work but I feel it is an overly complicated way to do it. Here is how you work it out in a simpler, inertial, frame of reference where you don't need to invoke any "fictitious" forces like the centrifugal force that they are talking about.
The motorcycle+rider going around the corner needs to satisfy two conditions:
Sum of forces = ma Sum of torques (about an axis parallel to the direction of motion) = 0
The former condition is needed to turn the corner. Assuming a flat road the only horizontal force is a static friction pointing to the center of the curve. $a = v^2/R$ to go around the corner.
The second condition is needed for the bike to not fall over. There are three forces acting on the bike+rider: gravity, static friction (parallel to the ground, towards the centre of the curve), normal force (up). Take torques about the centre of mass. So gravity causes no torque since there is a zero moment arm. Here's the free body diagram:
So we have
$ma = m\frac{v^2}{R} = f_s$
$n - F_G = n - mg = 0$
$f_s r \sin(\theta) - n r \sin(90^\circ - \theta) = 0$
or, with a slight trig. identity
$f_s r \sin(\theta) - n r \cos(\theta) = 0$
which we can solve to get
$\frac{n}{f_s} = \tan{\theta}$
Since you can solve for $f_s$ if you know the radius of the curve and the speed, and you can solve for $n$ if you know the mass of the bike+rider, this lets you find the angle of lean, $\theta$. Basically this translates to the condition that the vector sum of the friction and the normal force must pass through the centre of mass.
This is using the approximation that the tires are "thin" and so we can think of the bike + rider as just a thin rod. @user12330 says that the answer would depend on the shape of the tire. This is correct. The thickness/shape of the tires would displace the actual point of contact between the road and the tire relative to what my calculation assumes. But I would think this is usually a pretty small correction. Even making this correction you can still use the fact that the vector sum of the friction and normal need to point through the centre of mass. But now the angle of this vector sum is not the same as the angle of lean of the bike, which makes for some yucky geometry.
This doesn't quite get you where you want, which is the turning radii of both wheels. You are going to need a much more detailed analysis then where you consider the forces on the individual wheels, and the sum of torques about a vertical axis with the condition that the angular velocity of the bike rotating about the vertical axis matches the angular velocity of the bike traveling around the corner. That'll also be pretty yucky! Good luck!
Fundamentally, the shape of the tire will impact the way that a bike turns and how much turning results from leaning and how much comes from turning the front wheel. Fortunately, you can ignore all that. The rider experiences a [left]ward force when they turn [right] and a simple free-body diagram will tell you how large that force is. Assume uniform circular motion and you can solve for the radius of the turn.
