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I Hate to do this but Brad Wiggins

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Jul 19, 2009
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Escarabajo said:
I have you on record now.

I think is a bit too much at this time of the race where recuperation seems to be a bigger factor than anything else. Both look strong now, so everything is possible. I have never would have bet a dollar for Wiggins to be battling out with Contador at this time of the race.

By the way - make sure to quote this one too - my calculations also tell me that Armstrong will beat Contador by 1'09" in the TT!
 
May 13, 2009
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dienekes88 said:
If you look at the difference in power output between the two: 417W (estimated by Ferrari) and 466W (estimated by me using an extrapolation of Ferrari's method)... you can compare their velocity on the flat. The ratio of the power is equal to the ratio of the squares of their velocity.

Science nerd alert:

You have to be careful here to distinguish force and power, as well as different types of frictional (drag) forces.

Let's start with different frictional forces. We have:
1) Newtonian friction, where the frictional force is proportional to the speed (and a friction coefficient). F=eta*v, where F=force, v=velocity and eta=coefficient of friction. The type of friction which typically obeys this law is friction of mechanical parts of the bike or of tire on road surface, etc.
2) Air resistance, where the frictional force is proportional to the velocity squared (creating air turbulence in the wake), the effective surface area A, and some coefficient alpha. So we have F=alpha*A*v^2

For a flat, fast time trial, I think air drag is quite a bit larger than Newtonian friction, so one might neglect the latter and only consider the former.

Now, when talking about power to overcome the frictional force, one has to consider that P=F*v, i.e. that power is equal the product of the frictional force times the velocity. So, for air drag, power becomes proportional to velocity cubed. (See for instance Wikipedia)

Meaning that for otherwise equal conditions, power is proportional to effective area*velocity^3. (If your starting point is power/weight numbers, you'd have to multiply those with the weights of the riders to get numbers for power first.)

Let's assume rider A has 9% more power (not power/weight, just plain power) than rider B, but also 3% more effective area (he's bigger), then the increment in top velocity would be only one third of the difference, i.e. percent increase in velocity would be (9%-3%)/3=2% (This simple formula is only valid for small percent changes of area and power).

I think you made a mistake when you calculated the velocity ratios from the power ratios, and also, you forgot to take into account the difference in sizes (effective surface area) for the two riders. Based on that, I'd think relative to your prediction, Contador will do better (and Armstrong worse).

end science nerd alert
 
Jul 19, 2009
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Cobblestones said:
Science nerd alert:

You have to be careful here to distinguish force and power, as well as different types of frictional (drag) forces.

Let's start with different frictional forces. We have:
1) Newtonian friction, where the frictional force is proportional to the speed (and a friction coefficient). F=eta*v, where F=force, v=velocity and eta=coefficient of friction. The type of friction which typically obeys this law is friction of mechanical parts of the bike or of tire on road surface, etc.
2) Air resistance, where the frictional force is proportional to the velocity squared (creating air turbulence in the wake), the effective surface area A, and some coefficient alpha. So we have F=alpha*A*v^2

For a flat, fast time trial, I think air drag is quite a bit larger than Newtonian friction, so one might neglect the latter and only consider the former.

Now, when talking about power to overcome the frictional force, one has to consider that P=F*v, i.e. that power is equal the product of the frictional force times the velocity. So, for air drag, power becomes proportional to velocity cubed. (See for instance Wikipedia)

Meaning that for otherwise equal conditions, power is proportional to effective area*velocity^3. (If your starting point is power/weight numbers, you'd have to multiply those with the weights of the riders to get numbers for power first.)

Let's assume rider A has 9% more power (not power/weight, just plain power) than rider B, but also 3% more effective area (he's bigger), then the increment in top velocity would be only one third of the difference, i.e. percent increase in velocity would be (9%-3%)/3=2% (This simple formula is only valid for small percent changes of area and power).

I think you made a mistake when you calculated the velocity ratios from the power ratios, and also, you forgot to take into account the difference in sizes (effective surface area) for the two riders. Based on that, I'd think relative to your prediction, Contador will do better (and Armstrong worse).

end science nerd alert

Very good point! I had thought about this, but the pure multiplication that you advocate only works in the case that air velocity and ground velocity are the same, an instance that only works when there is no wind. Given that the ground velocities are quite similar (differing by less than 5%), I wanted to preserve the relationship in the aerodynamic drag component, which I considered to be the most important as it was approximately an order of magnitude greater than the velocity component. P(AC) = Fd(AC) * v(AC)g; P(BW) = Fd(BW) * v(BW)g... with v(AC)g ~ v(BW)g.

The basis of this observation is in these articles by Dr. Ferrari:
http://www.53x12.com/do/show?page=indepth.view&id=89
http://www.53x12.com/do/show?page=indepth.view&id=47

AND

This article in the Journal of Applied Biomechanics by Martin et al (including Coggan, who apparently posts here...!)
http://www.recumbents.com/WISIL/MartinDocs/Validation of a mathematical model for road cycling.pdf

Perhaps I erred in doing this... or perhaps maintaining the difference between ground and air velocity was appropriate. I suppose the physics is simple enough that each person should make his own decision in which model to back.

Using the cubes, as you suggest, the difference is much smaller. Wiggins would take 1'38" out of Contador on the flat section and lose 23" to him on the climb. That leads to a more reasonable margin of 1'15". The more conservative margin backed by the cubes-model may turn out to be closer due to the fact - as Escarabajo points out - that a lot of the time trial performance will depend on the ability to recover from these monster efforts. That's a wrench in the sprockets...
 
Jul 21, 2009
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dienekes88 said:
Well, so far it has averaged 40.57kph. Wiggins @1'46" back has averaged 40.55kph. While not record breaking, this is faster than any of Armstrong's first 5 Tours..

With the hardest stages of the tour still to come, and the traditional go slow on Sunday, it's a bit early to make average speed predictions for this year.

dienekes88 said:
The '07 Tour (at 39.23kph average for the winner) had the slowest winner since 1994? That's actually incorrect. '00 @ 39.22kph was slower than '07. It's mincing stats, but the implications of what you're saying are broader than just speed...

You're not mincing stats here, just making them up. The '00 Tour averaged 39.56 kph. ref cyclingnews

dienekes88 said:
Anyway, is that supposed to suggest that the winner was clean? It doesn't take a historian to remember that Contador won that Tour, and people love to hate Contador. The next year's winner averaged a blistering 40.50kph.

Only suggests, that on average, the Tour is slowing down. Maybe the riders aren't as fit as they used to be and the bike technology has hit the rocks? The winners time is 99% a function of the speed of the peloton, so the fewer riders on the gas the fairer race for everyone. Teams win the Tour not just the rider in the yellow jersey.
 
May 13, 2009
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Ok, wind definitely changes the equation (literally). What one then has to consider is the direction of the wind. The two extremes being head and tail wind. For the drag force one has to consider the velocity relative to the wind squared (if it's not precisely head or tailwind, then you have to consider angles as well). This becomes the drag force which you still have to multiply with ground velocity to get power.

Say for a head wind, the drag force would be proportional to (v_ground+v_wind)^2 which you then had to multiply with v_ground. For a tail wind, you would have to multiply (v_ground-v_wind)^2 with v_ground.

At this point, it becomes harder to establish simple rules as in my previous post. You'd almost have to put in specific numbers for wind speed to get a good estimate for the expected time differences. For speeds of, say, 45 km/h and an assumed wind speed of, say, 15 km/h (head wind), the factor which relates percent gains between power and velocity drops for instance to 2.5, e.g. for a power increase of, say, 10%, the rider gains an increase in velocity of 10%/2.5=4%. For a tailwind of the same speed, the factor goes up to 4, e.g. a rider with 10% more power has an increase in velocity of only 10%/4=2.5%. In average (on a round course) wind favors stronger riders. No wind at all gives the factor 3 in my last post. So any kind of wind in Annecy is bad news for Contador as the smaller (less powerful rider).
 
Jul 19, 2009
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mercuryrising said:
With the hardest stages of the tour still to come, and the traditional go slow on Sunday, it's a bit early to make average speed predictions for this year.

You're not mincing stats here, just making them up. The '00 Tour averaged 39.56 kph. ref cyclingnews

I'm a bit confused. Sorry. As Cobblestones will have you know, I'm not very good at math.

Winning time = 92h 33m 8s
= 92.5522 hours

The splash page for the 2000 Tour says it was 3,630km

3,630km/92.5522h = 39.22kph

Please let me know where I went wrong.

Only suggests, that on average, the Tour is slowing down. Maybe the riders aren't as fit as they used to be and the bike technology has hit the rocks? The winners time is 99% a function of the speed of the peloton, so the fewer riders on the gas the fairer race for everyone. Teams win the Tour not just the rider in the yellow jersey.
 
Jul 19, 2009
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Cobblestones said:
Ok, wind definitely changes the equation (literally). What one then has to consider is the direction of the wind. The two extremes being head and tail wind. For the drag force one has to consider the velocity relative to the wind squared (if it's not precisely head or tailwind, then you have to consider angles as well). This becomes the drag force which you still have to multiply with ground velocity to get power.

Truly, we can complicate this as much as we want. We could even add in things like chain friction... if we want to try to model everything.

This is why I used the ratio of powers and the ratio of speeds (to a certain power). Using the ratio of speeds (squared or cubed), you can cancel out the ground speed as it's an order of magnitude smaller than the drag force, which uses air speed squared... except in a block tail wind. Of course you can argue that too.

Say for a head wind, the drag force would be proportional to (v_ground+v_wind)^2 which you then had to multiply with v_ground. For a tail wind, you would have to multiply (v_ground-v_wind)^2 with v_ground.

At this point, it becomes harder to establish simple rules as in my previous post. You'd almost have to put in specific numbers for wind speed to get a good estimate for the expected time differences. For speeds of, say, 45 km/h and an assumed wind speed of, say, 15 km/h (head wind), the factor which relates percent gains between power and velocity drops for instance to 2.5, e.g. for a power increase of, say, 10%, the rider gains an increase in velocity of 10%/2.5=4%. For a tailwind of the same speed, the factor goes up to 4, e.g. a rider with 10% more power has an increase in velocity of only 10%/4=2.5%. In average (on a round course) wind favors stronger riders. No wind at all gives the factor 3 in my last post. So any kind of wind in Annecy is bad news for Contador as the smaller (less powerful rider).

We can agree on the last bit.

The variations and the changes are part of why I went with what I did. The ratio of their ground speeds remains relatively consistent... so they can drop out due to their being less important than the drag in anything but a monster tailwind where the effective air speed is quite low. Even in a windless TT, the differences in the velocity squared will be so much bigger than the differences in the ground velocity. The point is simplicity...

The variations in wind and the changes in wind direction are the same for both riders. That's why we're only talking about rider speed ratios with a made up average speed for the calculation of time... as opposed to predicting the actual speeds at which they will ride. The benefit and penalty of the wind... will average out into their average speeds.

I guess what I'm trying to say is that you can parse it into specific sections and get really into the calculations and model everything to the point that you have to use a computer simulation with historical wind data and a topopgraphical map... or you can just go with the average speed squared with the ground velocity dropped out for the reasons explained above.

Considering that I have a job and have to train, I'll go with the latter for my own quick calculatiosn. ;) Clearly, you're better at math than I am, so I'll lean on you for the more exact calculations!

So what are your predictions? Discussing the model is fun... but not as much fun as applying it. At the end of the day, the proof is in the pudding, and let's get a guess out of you!
 
Mar 10, 2009
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Wiggins is having a fantastic Tour; he's one of the few highlights in an otherwise measured, predictable Tour.

<side note> I guess measured and predictable is what happens when you take drugs out of the equation.

Anyway, I believe Wiggins is riding clean.


Go Wiggo!
 
dienekes88 said:
So what are your predictions? Discussing the model is fun... but not as much fun as applying it. At the end of the day, the proof is in the pudding, and let's get a guess out of you!

Original message not directed to me but I did the calculations anyway. Here are my predictions:

Input Data:
Wiggins Weight: 71 Kg, No head wind or tail wind
Contador's Weight: 61 Kg, No head wind or tail wind

If both Wiggins and Contador are fully recuperated (100% to what they showed in Verbier, W/kg) then:

1':8" Wiggins over Contador

But as we all know there is horrible mountain stage before the TT, so maybe I expect Contador to be more recuperated than Wiggins so the prediction for this scenario will be:

0:38" Wiggins over Contador

Let's wait and see what happens in the stage to Grand Bornand and make better predictions with those numbers then.:)
 
Jul 19, 2009
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Escarabajo said:
Original message not directed to me but I did the calculations anyway. Here are my predictions:

Input Data:
Wiggins Weight: 71 Kg, No head wind or tail wind
Contador's Weight: 61 Kg, No head wind or tail wind

If both Wiggins and Contador are fully recuperated (100% to what they showed in Verbier, W/kg) then:

1':8" Wiggins over Contador

But as we all know there is horrible mountain stage before the TT, so maybe I expect Contador to be more recuperated than Wiggins so the prediction for this scenario will be:

0:38" Wiggins over Contador

Very conservative estimate. Are you predicting their weight loss through the Tour too!? Hahaha. Good call.

Can't you dream big? Don't you believe in miracles? ;)

You have been pretty convincing with the effect of the tough tough days... and I recognize that 2'08" might be a little too hopeful. However, you guys have to remember that Contador lost 2'18" to Levi and 1'27" to Cadel Evans at the end of the 2007 Tour. He has improved since then, but so has Twiggo.

Modern twiggo v. Leipheimer? We saw that in the opening TT. Even with the weight penalty, Twiggo beat Leipheimer by 11 seconds. Contador's massive weight advantage equalized the difference in power.

Let's wait and see what happens in the stage to Grand Bornand and make better predictions with those numbers then.:)

Good idea, but it'll be tough to control for the fact that this is a TT, i.e. not at the end of a stage.

I wonder if that's a good argument for using the performances at Verbier. Today's so hard that nobody will even come close to their peak outputs.
 
Apr 11, 2009
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So after today, is Wiggins still doped to his eyeballs?

Physics says he can do it on one stage (watts/kg), experience/pedigree says not on consecutive ones. Refills would allow him the latter.

He's clean.
 
Mar 10, 2009
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Parrot23 said:
So after today, is Wiggins still doped to his eyeballs?

Physics says he can do it on one stage (watts/kg), experience/pedigree says not on consecutive ones. Refills would allow him the latter.

He's clean.

He only lost a minute to Kloeden and Armstrong after 4 1cat climbs and the last 2 climbs were 8km at 8-9%. I wouldn't go out on a limb yet, although I agree it 'looks' more trustworthy.
 
Jul 19, 2009
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Parrot23 said:
So after today, is Wiggins still doped to his eyeballs?

Physics says he can do it on one stage (watts/kg), experience/pedigree says not on consecutive ones. Refills would allow him the latter.

He's clean.

Hmmm... inconsistency = clean? I guess Kashechkin and Vinokourov were clean. Hamilton and Landis too. ;)

The suggestion that consistency is doping would implicate a lot of riders, e.g. Evans (up until this Tour), Sastre (again up until this Tour), etc.

I don't think strict performance can be a measure of clean vs. doped the way the LeMond wishes it could be. You have to go out there and measure the blood directly. There are TOO MANY factors that influence a performance on a given day.

Some guys actually get rest and recover their legs on rest days. Knowing how you have to ride and recover on a rest day is something only experience can give you. Vande Velde and Schleck the Elder clearly handled the rest day pretty well. They improved their level and were closer to some of the younger guys who rode away on the slopes of Verbier: Nibali, Contador, Schleck the Younger. Or you could just say they were doped (I'd rather not think so), 'cause how could Vande Velde hold onto a pretty elite group when he lost 2'41" at Verbier? If you're going to use consistency as a marker for doping, you should apply it to everyone... even your favorites.

Back to Twiggo, though... I think I might've acted like an overly excitable Cat. 4. We'll see how he recovers from today's efforts. I'm hoping he just rode within himself to limit his losses, but I didn't see pictures of him cresting the final climb or finishing.

At the moment, I have to say the advantage is going in the way of the more conservative estimates.

C'mon, TWIGGO!
 
dienekes88 said:
...'cause how could Vande Velde hold onto a pretty elite group when he lost 2'41" at Verbier? If you're going to use consistency as a marker for doping, you should apply it to everyone... even your favorites.
The problem is that not everyone dopes the same. Amount, frecuency, type, quality, etc. Others take bigger risks. So we can only judge based on individual efforts.
 
Apr 11, 2009
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The pt. with consistency is that Wiggo is an interloper; the others are not. A relative newbie to GTs riding consistently with the leaders on consecutive mountain days doesn't make a whole lot of sense, whatever the absolute engine/weight. With time/experience in GTs (and/or refills, etc.), Wiggo might develop better ability to "recover" during a GT. Right now he has neither; if he did, there would be a lot more grounds for suspicion--which isn't to say there are none.
 
May 13, 2009
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dienekes88 said:
So what are your predictions? Discussing the model is fun... but not as much fun as applying it. At the end of the day, the proof is in the pudding, and let's get a guess out of you!

Oh well, now I'm on the line. Usually I don't do prediction. Only under very rare circumstances. Now, what I'm saying is not based on any model whatsoever, simply because I have no real data beside viewing stages.

Anyway, I think that Bertie didn't exert himself too much, I'm not so sure about Wiggins. Also, I'm not sure how well Wiggins will recover in the third week. He hasn't done too many GT and his third week in the Giro wasn't great (probably on purpose though). Hard to say. Let's say Wiggins gains 30" over Contador who's a very accomplished TT'ist himself. A. Schleck will lose 1'30" on Bertie but only 30-40" on LA. Kloden is hard to read. It looked like he bonked today, so maybe one should put him around a LA time. On the other hand, he did a great ITT in the first stage. I hope Nibali can recoup some time as well. He's only 1'14" behind LA. I would hope he could make time on him. I'd love to see a bunch of young riders in the top 5. Contador, Schleck*2, Wiggins, Nibali. It looks like the time of Evans, Sastre, Menchov, Leipheimer is coming to an end. They should take LA with them.

Anyway, as my last prediction, I predict that reality will disagree with everything else I wrote above. In that way, I'm guaranteed to be right on at least one account.
 
Jul 19, 2009
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Cobblestones said:
Oh well, now I'm on the line. Usually I don't do prediction. Only under very rare circumstances. Now, what I'm saying is not based on any model whatsoever, simply because I have no real data beside viewing stages.

Hahaha... There's no harm in hazarding a guess. Some people may skewer you for it, but if they do, then they're missing out on the fun!

Anyway, I think that Bertie didn't exert himself too much, I'm not so sure about Wiggins. Also, I'm not sure how well Wiggins will recover in the third week. He hasn't done too many GT and his third week in the Giro wasn't great (probably on purpose though). Hard to say. Let's say Wiggins gains 30" over Contador who's a very accomplished TT'ist himself. A. Schleck will lose 1'30" on Bertie but only 30-40" on LA. Kloden is hard to read. It looked like he bonked today, so maybe one should put him around a LA time. On the other hand, he did a great ITT in the first stage. I hope Nibali can recoup some time as well. He's only 1'14" behind LA. I would hope he could make time on him. I'd love to see a bunch of young riders in the top 5. Contador, Schleck*2, Wiggins, Nibali. It looks like the time of Evans, Sastre, Menchov, Leipheimer is coming to an end. They should take LA with them.

So you're saying:
Wiggins
Contador @ 30"
(hopefully) Nibali?
Kloeden @ 1'20"
Armstrong @ 1'20"
Schleck @ 2'00"

To put it in writing... my somewhat excitable predictions were:
Wiggins
Armstrong @ 53"
Contador @ 2'02"

I got some work to do, but I'll look at Andy Schleck's, Nibali's, and Kloeden's rides at Verbier to predict some TT times later tonight... possibly adjusted for how they looked today (very qualitative assessment = introduce a lot of error).

You're right. Kloeden looked pretty gassed. I don't know if it was a bonk, though. I think he just went so far over the limit to match the acceleration of the Schleck Brothers and was riding at an uncomfortably high pace. That really sapped his energy. I wouldn't be surprised if we measured a blood lactate at the top of the climb in excess of his maximal lactate steady state... 10-12 mmol/L? Possible. Higher than that? Again, possible.

Armstrong looked pretty good. Sitting on Twiggo's wheel (he's tall and has a decent draft) at the lower climbing rate meant that he could recover and surge to catch Kloeden.

Nibali... very impressive, but he's headstrong. He took off on the descent and was picked up later. I don't think that is a good thing for his TT... though he's a pretty talented time triallist. He may risk it all and go over the limit too early. He was good on Verbier and solid today, but he's still young. Wait for it... and he may have a lot to learn... ;) about his limits and his abilities in the third week of a Grand Tour.

Wiggo. I really have to reiterate that I hope he rode within himself. Otherwise my prediction is going to be CRAP! :D

Anyway, as my last prediction, I predict that reality will disagree with everything else I wrote above. In that way, I'm guaranteed to be right on at least one account.

HA. Nice.
 
Jul 19, 2009
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Looking back at my quick calculations from before, I realized that I had used Dr. Ferrari's estimates of Lance's wattage without checking it against my statistical model. I corrected my math (which some of you take issue with anyway...), and it actually increased the time gap between LA and AC. Interesting.

Based on today's form, I made a new set of estimates. I'm still going to say that Twiggo is going to win, but by a slightly smaller amount. Beyond that, I still hope against all hope that he'll beat Contador by 2'00"!

My "trying to maintain hope and excitement for Twiggo" predictions:
Wiggins -
Armstrong @ 35"
Kloeden @ 45"
A. Schleck @ 1'08"
Contador @ 2'00"
Nibali @ 2'15"

Kloeden was very clearly riding for Armstrong on the climb, so I can't make an estimate based on his climb of Verbier... his time is pretty much a random guess.
 
Mar 13, 2009
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Parrot23 said:
So after today, is Wiggins still doped to his eyeballs?

Physics says he can do it on one stage (watts/kg), experience/pedigree says not on consecutive ones. Refills would allow him the latter.

He's clean as a mumbai slum.

there, fixed for you