I always wondered about that.This means that a cyclist can only go at about 45.5 km/h around a [flat] corner without losing grip
That calculation only works when:
The radius of a turn on a Olympic velodrome is around 20 meters,
Now Short Track Ice Speedskaters with lots more traction on flat ice can corner way faster!
Track - the other off-road
http://www.lavelodrome.org
If you go from the straightaway into a constant radius turn you go instantly from no centripetal acceleration to V^2/R, and it would be instant changes from no "compression" to max compression and then suddenly back. Instead they're a cycloid (or close to it) so the centripetal acceleration comes on slow, hits a max at the middle of the turn, and decreases slowly as you exit.
Another thing that's fun is conservation of angular momentum-- it helps in the paceline transitions in the corners (in addition to the elevation change effects).
Track - the other off-road
http://www.lavelodrome.org
It is a good start to the basics. The coefficient of friction is a variable on a bicycle. The author's first equation assumes that turning a bicycle is flat. Turning a bicycle on flat ground requires a lean which changes force vectors and friction. He/She doesn't come back to this until the end of the article where a correction is added.
I think there should have been material on the basics of superelevation.
the lion from within must guard his palace, because everybody's going to try to take a sip from his chalice
Also found this if anyone is interested
http://en.wikipedia.org/wiki/Track_transition_curve
Taras
To turn fast, you must lean. To go (X) amount of speed, you must lean (C) amount of degrees. (C) amount of degrees is already farther than a bike can lean (A). So to get to the (C) degree angle needed, the track must me leaned as well (B). Therefore (A)+(B) = (C) and we can turn while going (X) amount of speed.
the lion from within must guard his palace, because everybody's going to try to take a sip from his chalice