> I re-checked everything. It all looks correct. The radius
> isn't much more than 60'. Speed is close to 40 mph. There
> is slight banking to the track, but not much.
> I'll have to think about it some more. Is there some part
> of this equation that is being left out?
Let's say Q (read theta) is the bank angle, v is speed, M is total
mass, r is radius, g is gravity and mu is friction.
Resolving forces relative to the track surface, I get
M (v^2/r cos Q - g sin Q) needed down the slope, and
M (g cos Q + v^2/r sin Q) download on the tyres
Check at Q = 0 => sin Q = 0, cos Q = 1 leaving
Mv^2/r across and Mg into the ground. Sound good.
So adding tyre friction, to corner at the limit we have
mu M (g cos Q + v^2/r sin Q) = M (v^2/r cos Q - g sin Q)
Since I can now see a trig identity coming on, and I don't like
those ;-), cheating and using Excel gives me a bank angle a little
under 19 degrees for mu = 0.9 and the data you've already given.
Less, obviously, if the tyres are stickier than 0.9.
See if that sounds plausible!
Jonny