# Track Cycling: Velodrome Racing and Training Area - Anyone for physics? (article on banking...)

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joshpants
05-14-12, 10:25 AM
http://www.theaftermatter.com/2012/05/physics-of-cycling-why-does-velodromes.html

Enjoy. :beer:

-josh

carleton
05-14-12, 10:58 AM
This means that a cyclist can only go at about 45.5 km/h around a [flat] corner without losing grip

Speedskater
05-14-12, 01:08 PM
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!

bitingduck
05-14-12, 04:33 PM
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!

Speedskates have long sharp edges that dig into the ice in one axis and slide nicely across it on the other.

The article leaves out a lot of interesting physics about why the turns generally aren't semicircles, as well as some interesting angular momentum effects.

chas58
05-15-12, 11:07 AM
The article leaves out a lot of interesting physics about why the turns generally aren't semicircles, as well as some interesting angular momentum effects.

So, what are the other interesting bits of physics, and why are velodrome turns not a constant radius?

Dolamite02
05-16-12, 10:43 AM

Note too, that this is only on a wood surface, not concrete.

bitingduck
05-16-12, 11:21 AM
So, what are the other interesting bits of physics, and why are velodrome turns not a constant radius?

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).

San Rensho
06-11-12, 07:32 PM

But you are going to strike a pedal way before 45 kph.

KrautFed
07-04-12, 09:19 AM
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.

taras0000
07-04-12, 09:28 AM
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).

Is it a cycloid? I always thought it was some form of Euler curve. I've never thought if a cycloid in being used for a banking. They both make sense to me in some respects, but not being a rocket surgeon, I wouldn't know which is best.

taras0000
07-04-12, 11:45 AM
Also found this if anyone is interested

http://en.wikipedia.org/wiki/Track_transition_curve

David Broon
07-04-12, 07:06 PM
I'm a simplistic kind of guy. The kind who rides bikes in a circle. Fast.

Somebody want to explain to us ordinary mortals what's going on?

KrautFed
07-05-12, 04:44 PM
I'm a simplistic kind of guy. The kind who rides bikes in a circle. Fast.

Somebody want to explain to us ordinary mortals what's going on?

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.