bitingduck

I've had a lot of fun races there, but I definitely think it's a harder track to ride than VSC. You won't slide down in the corners, but it will bite you a lot of other ways.

tobukog

Another thing to remember is that the minimum safe speed is dependent on how high you are on the track. The higher you are, the faster you have to go to maintain grip.

ruudlaff

You sure about that? If it's a uniform gradient and smooth then surely it would be the same?

tobukog

The amount of compression you get in the tighter turn at the bottom of the track is significantly different than at the top.

rensho3

Huh? I usually try not to be blunt, but have you ever ridden a high angle track? The angle is the same between the entrance and exit of the turny end. 15 mph will make you slip off the track in the Pole lane between turns 1 and 2 (and between 3 and 4, for that matter) the same as it does it at the top of the track.

bitingduck

a=(V^2)/R, and the higher you are the further out you are in the radius, so the centripetal force is smaller. The curves are also not semicircles, so the acceleration varies throughout the turn.

tobukog

Uh yes. If you ride enough madisons you figure this out pretty quickly.

tobukog

Let me revise that. I've always been able to ride at a walking pace anywhere on shallower 333s, 400s, and 500s. It usually only comes into play on steeper 250s and smaller tracks.

ruudlaff

I'm not saying anyone is wrong, was just surprised to hear it

carleton

Minimum speeds are depenent on the angle, though. The angles are consistent from bottom to top of the track. So, if you can ride at 11mph at the bottom of the track, you can ride at 11mph at the top of the track.

TDinBristol

Not exactly. It's also a matter of centrifugal force. The smaller arc at the bottom of the track will create greater force than the longer arc at the top, and your in-leaning angle will increase, slightly. But I suspect the differences in force are so small it's not especially significant.

bitingduck

Not quite.

You stay on the track due to the friction between your tire and the track, while gravity is trying to pull you down off the track. The force holding you up is the friction coefficient between your tire and the track times the force you're applying to the track, which has two components-- a downward component from gravity and an outward component from centripetal/centrifugal force. The downward component is always the same for a given angle (given round tires). The outward component is your mass times the acceleration, which is (V^2)/R. Higher up on the track, the radius of the turn (the R) is larger because you're farther from the center of the circle (assuming a circle for convenience of description), so to get the same frictional force to keep you from sliding down you have to have a higher speed.

If you draw the free body diagram and do the math (just a few sines and cosines) you'll see it's correct - it's a standard freshman physics problem but it's usually phrased in terms of a car on a banked track and at what speed will they slide off the top. It's similar to skidding a car on the flat when you take a tight turn too fast (you just have to set the banking to zero degrees).

If you solve the whole problem you'll also see that the mass cancels out - the minimum speed is independent of your mass, which is why a big rider and small rider can both creep along together on relief in a madison or in a match sprint, and the one with less skill will slide first.

bitingduck

It's not insignificant on a 7 m wide track.

[edit- I just looked up some numbers and alpenrose is ~16 m inner turn radius; other 250s will be in the 15-17 m range, too. A 5 m difference (because the bank is 45 degrees) is a significant change in the acceleration from the inside to the outside.]

bitingduck

And one more practical point when it comes to riding the VSC - when you're riding up at the balustrade you not only have to be going faster to avoid sliding, but you also have to put a little extra pressure on the pedals to keep your speed up as you go into the turn at the top of the track because you're going uphill. If you're leading a paceline, the polite thing to do is a little extra pressure on the way into the turn and float slightly on the way out (because it's downhill) to keep your speed constant.

carleton

Ah. Good point!

So, how many MPH/KPH difference would you estimate there would be between the minimum speed at the measurement line for a given rider and the minimum speed at the boards, in the middle of turns 1/2?

bitingduck

Assuming a 15 m turn inner radius and a 5 m wide track*, about 15% faster at the outside than the inside to keep the same frictional force. so if the minimum is 15 mph at the black line, it will be around 17 at the top.

*that's 5 meters inner radius to outside radius, which you get from a 7 m wide riding surface at 45 degree banking.

carleton

ruudlaff

I've just remembered why I questioned this...I was told it back during accreditation!

If I remember correctly this was used as a tool to encourage us newbies to not fear going high..."If you can ride the black, you can ride the top" type statement.

Poppit

Our track has a 23m radius and 42deg banking

bitingduck

so that will be about a 10% difference from inside to outside.

rensho3

I agree, and can go slow on those kind of tracks. One quirk I have noticed is that when I am on a 333, or even Encino (a shallow 250) while I can ride slow, my brain starts to go into panic mode because it is used to VSC. I once got pinned on the rail at San Diego because of that; after taking a couple of panic breathes, my opponent gave me just enough to accelerate out of it. It's amazing how you learn something new after each race, even when you've been doing for more years than the fathers of some of the juniors have been alive!

rensho3

carleton

@bitingduck, does weight factor into this? I assume that lighter riders can cruise slower than heavier because the heavier riders can break the coefficient of friction easier which keeps the slide from initiating, right?

bitingduck

Nope - the magnitudes of the forces involved all have mass in them, so it ends up cancelling out. I'm too lazy to type it all out, but the related problem of a car skidding off the high side of a banked track is written out here: More circular motion

I can creep along just fine behind riders 2/3 my weight when riding relief and not slide.

taras0000

Post #63, last paragraph. F=MA works in both directions simultaneously, canceling weight out of the equation.