Power Data Estimates for the climbing stages

[indurain666]

Because I'm cranky and just finished a stage race at altitude I will suggest another evidentiary symptom of a cleaner race...attacks are shorter in duration and the recovery intervals are longer. That's if what I saw on screen is as believable as the mountains of data being analyzed.

It is a long shot to talk about a fully clean peloton but the facts are: these guys train better, race less, use lighter/faster equipment and are still much slower than 90s generation riders.

"Something" is making them slower...

 

[Oreknan]

and in the classics the times of mur de huy and others are improving or not?

 

[simoni]

What are we seeing?
Contador climbs the Telegraph all out at 6.38 watts/kg (30 min. effort), 1st climb of the day.

A few years back Pantani was climbing AdH after a few hard passes at about 6.9 watts/kg, while L.A. was between 6.6 and 6.7 watts/kg.

To do 6.38 watts/kg at 92% VO2 max you probably need a VO2 max of roughly 87 ml/mn.kg, not a VO2 max of > 92 ml.

It is quite likely that a few of the TdF top contenders have a clean VO2 max approaching 90 ml/mn.kg.

I was shocked to see throngs climbing AdH in less than 40 min. at the end of a hard stage, but not so much seeing a few top guys doing it in about 42 min after a shorter stage.

I'm a competitive runner competing at national (and occasional international) level. My VO2max has been measures at 79ml/kg/min. I'm certainly not doped!! (If I was it'd be the least effectively doping programme in sporting history!) I've no trouble in believing that highly talented, better trained, full time professional athletes could have values in the high 80s.

 

[halamala]

and in the classics the times of mur de huy and others are improving or not?

Dr. Michele Ferrari:

Fleche Wallonne 2011

Published: 21 Apr 2011

The dominance of Philippe Gilbert on the Muur de Huy was pretty clear: at the finish of a tactically perfect race, the Belgian champion rode the last Km (112m of climbing) in 2’40”, with a VAM of 2520m/h, corresponding to about 570w, assuming a body weight of 68 Kg.
An excellent performance, but quite far from the one supposed by Eugenio Capodacqua (800 average watts): after all, the good Capodacqua is one prone to exaggerating...

This time on the last Km is the fastest ever: the previous "record" was 2'44" (Rebellin in 2007).

http://www.53x12.com/do/show?page=indepth.view&id=116

 

[Alex Simmons/RST]

Dr. Michele Ferrari:http://www.53x12.com/do/show?page=indepth.view&id=116

For 112m VAM over 1km, assuming bike + kit of 8kg, I get a power at zero wind of 585W (8.6W/kg) for 160 seconds.

Headwind ranging from -2.5 to +2.5 m/s gives a power range from:
560W - 625W (8.5 - 9.2W/kg).

 

[acoggan]

Stop trolling. SoS never claimed Boardman's power at 6.4 W/kg was incorrect

How could I have ever missed this?

Anyway, the post I was referring to was this one:

http://forum.cyclingnews.com/showpost.php?p=603359&postcount=337

in which the statement is made "Maybe the SRM data are also wrong?"

, but rather that the value you plucked out of your rear orifice of 9W/kg was incorrect (where were the "quality data" to back this up?).

For the 9,999th time: 9 W/kg is simply the value at which you arrive if you use the same approach to estimating maximal sustainable cycling power output as Mike Joyner did to arrive at maximal sustainable running speed (i.e., marathon pace). The "quality data" to back it up can be found throughout the scientific literature; the only assumption required is that extreme values for VO2max, LT, and efficiency would occur in one individual. The point is not to claim that such a power is to be expected, but simply to illustrate that, based on what we presently know about the human body, such a power output is not IMpossible.

 

[Tyler'sTwin]

The science of sports guys are a wee bit hypocritical since they think Bolt's results aren't implausible because someone concluded that the human physiological limit is lower than 9.58. This is pretty much the same logic as Coggan's 9 W/kg limit which they dismiss. They'll defend their "above 6.2 W/kg is impossible" assertion with arguments like it has only been done in the rotten armstrong era and the 90's, but destroying the fastest dopers of all time in another sport with an equally rotten history is perfectly fine. Another argument is that you cannot have the VO2 max and efficiency combo to sustain a wattage at 6.2 W/kg or higher for 30-40 minutes. If your max is high enough, your efficiency will be too low and vice versa. They base this on studies of riders with a sub 6 W/kg FTP. Of course someone with an FTP below 6.2 W/kg isn't going to have the physiology to sustain 6.2 W/kg. It's like studying 9.9-10s 100m sprinters and concluding that 9.5s is "physiologically impossible". They also have an unwarranted cleaner view of distance running.

 

[Le breton]

The science of sports guys are a wee bit hypocritical since they think Bolt's results aren't implausible because someone concluded that the human physiological limit is lower than 9.58. This is pretty much the same logic as Coggan's 9 W/kg limit which they dismiss. ..........

I don't know if the two are equivalent.
We are talking about 2 completely different regimes.
I don't for one minute believe in 9.58, not any more than in Florence Griffiths 10.54 ( It is well known that on the occasion of her 10.49 the anemometer went on strike).

The 9 Watts/kg dependon 3 assumptions, 2 of which are believable and the third one not.

The highest ever pre EPO VO2 max was 94 ml/min.kg (Mieto, cross-country skier).

95% of VO2 max over say 30 min does not shock me.

Which means 89 ml/min.kg available over 30 min.

Knowing that 1 liter oxygen (burning carbs) ~ 350 watts

89 ml/min.kg corresponds to 31.1 watts

Therefore you need to assume an efficiency of ~29% to get 9 watts/kg.

As far as I know the efficiency of isolated muscle cells is only 32%, so that getting 29% for 30 minute while supporting a whiole organism ( and its "overhead") seems to be quite a feat.

The weak link in the 9 W/kg is the 29% efficiency (when "everybody" among endurance type cyclists seems to be around 22-24%).

Welle, sun is coming out and meteox.com shows rain in 2 hrs, I better get on my bike, sorry for not proofreading, you'll understand.

 

[neineinei]

I don't know if the two are equivalent.
We are talking about 2 completely different regimes.
I don't for one minute believe in 9.58, not any more than in Florence Griffiths 10.54 ( It is well known that on the occasion of her 10.49 the anemometer went on strike).

The 9 Watts/kg dependon 3 assumptions, 2 of which are believable and the third one not.

The highest ever pre EPO VO2 max was 94 ml/min.kg (Mieto, cross-country skier).

95% of VO2 max over say 30 min does not shock me.

Which means 89 ml/min.kg available over 30 min.

Knowing that 1 liter oxygen (burning carbs) ~ 350 watts

89 ml/min.kg corresponds to 31.1 watts

Therefore you need to assume an efficiency of ~29% to get 9 watts/kg.

As far as I know the efficiency of isolated muscle cells is only 32%, so that getting 29% for 30 minute while supporting a whiole organism ( and its "overhead") seems to be quite a feat.

The weak link in the 9 W/kg is the 29% efficiency (when "everybody" among endurance type cyclists seems to be around 22-24%).

Welle, sun is coming out and meteox.com shows rain in 2 hrs, I better get on my bike, sorry for not proofreading, you'll understand.

And Mieto is part ent.

 

[Le breton]

And Mieto is part ent.

I don't understand "part ent".
Can you be clearer?

 

[indurain666]

I don't understand "part ent".
Can you be clearer?

Lord of the Rings...he was trying to be funny in an effort to deviate his brain from the fact that he could not really understand your argument

 

[Le breton]

......... he could not really understand your argument

Thanks for the explanation, but don't be mean to him, I was in a hurry and left out a number of intermediate steps as for example 1liter O2 <==> ~350 watts.

Why?
Because 1 liter O2 when used to burn glycogen provides roughly 21 kJoules (you read numbers between 20.9 and 21.3 in the literature).

But we are talking about 1 liter O2/minute
so 21000 joules / minute = 21000/60 joules/sec = 350 joules/sec = 350 watts.

If burning fat (or protein -muscle) you get less energy per liter O2.

Anyway my ride went OK, but was a bit disappointing :
- the air pressure was real low (960 mb sea-level equivalent, ie 910 mb real air-pressure),
- 22°C,
- wind less than 5 km/h,
- little traffic, hit the 3 red lights OK
- plus I had my aero wheels on my KG381 Look and
- I forgot my helmet and did not get back to get it as I had little time available.

So, with all these positive elements I expected to be faster (than usual this year) by about 1 km/h as everything seemed to be in my favour (although 30°C would have been nice).

Instead I was only marginally faster, so it looks like in my old age 24hrs is not quite enough to recuperate fully from a hard 100km ride (yesterday)

 

[Alex Simmons/RST]

Thanks for the explanation, but don't be mean to him, I was in a hurry and left out a number of intermediate steps as for example 1liter O2 <==> ~350 watts.

I know you showed the steps but really you can only equate a volume of O2 to kJ of work done, not power (since power is a rate of doing work).

You need to specify a volume of O2 per unit time in order to equate it to power.

 

[Le breton]

I know you showed the steps but really you can only equate a volume of O2 to kJ of work done, not power (since power is a rate of doing work).

You need to specify a volume of O2 per unit time in order to equate it to power.

Come on Alex, don't be silly. You know exactly what I mean.

When I 1st wrote before going biking

Knowing that 1 liter oxygen (burning carbs) ~ 350 watts



I, of course meant, as I explained after my ride

But we are talking about 1 liter O2/minute
so 21000 joules / minute = 21000/60 joules/sec = 350 joules/sec = 350 watts.

for people who could have missed it and been confused between liters O2 and liters O2 per min., but if there is someone who couldn't possibly have missed it, it would be you.

 

[Alex Simmons/RST]

for people who could have missed it and been confused between liters O2 and liters O2 per min., but if there is someone who couldn't possibly have missed it, it would be you.

were you expecting someone who would miss it to pick it up?

Anyway, just to help others, here's the calculation method to convert rate of O2 utilisation to power, which anyone can pop into a sheet to make it easy:

 

[acoggan]

were you expecting someone who would miss it to pick it up?

Anyway, just to help others, here's the calculation method to convert rate of O2 utilisation to power, which anyone can pop into a sheet to make it easy:

You could make that spreadsheet even more sophisticated by including a LOOKUP table to determine the energy yield per liter of O2 based on the RER.

Then again, it wouldn't really be any more accurate, since 1) presumably most people (here) are interested in performing such calculations for conditions where carbohydrates are the dominant fuel, and 2) the conversion of O2 uptake to energy production is based on the conditions that are assumed to exist in cells, i.e., assumed delta G values, which means that the precise energy yield is never really known (which is why it is termed indirect calorimetry in the first place).

Sorry, just navel-gazing...carry on!

 

[Alex Simmons/RST]

You could make that spreadsheet even more sophisticated by including a LOOKUP table to determine the energy yield per liter of O2 based on the RER.

Then again, it wouldn't really be any more accurate, since 1) presumably most people (here) are interested in performing such calculations for conditions where carbohydrates are the dominant fuel, and 2) the conversion of O2 uptake to energy production is based on the conditions that are assumed to exist in cells, i.e., assumed delta G values, which means that the precise energy yield is never really known (which is why it is termed indirect calorimetry in the first place).

Sorry, just navel-gazing...carry on!

Yeah - it's just a quick n dirty with calcs shown to save me sending people a spreadsheet. I originally had power listed as FTP - where we are operating on pretty much 100% CHO.

 

[halamala]

Henao, Leipheimer climbed 5.6 W/kg

Tour of Utah 2011, Stage 5, Final Climb Little Cottonwood Canyon, last 10.5 Km

Sergio Henao

Elevation / Höhenmeter [m] : 800 m
Distance / Streckenlänge [Km] : 10.5 Km
Time in seconds / Fahrzeit in Sekunden [sec] : 1890 = 31 min 30 sec = 31:30
Weight rider / Gewicht Fahrer [kg] : 61 kg
Weight bicycle, clothes etc. / Gewicht Fahrrad [kg] : 8 kg

Grade / mittlere Seigung : 7.6 %
Average speed / mittlere Geschwindigkeit : 20.0 Km/h
Total weight / Gesamtgewicht : 69.0 kg

Power : 347.5 Watt
Power / kg : 5.6 Watt / kg

Levi Leipheimer

Elevation / Höhenmeter [m] : 800 m
Distance / Streckenlänge [Km] : 10.5 Km
Time in seconds / Fahrzeit in Sekunden [sec] : 1890 = 31 min 30 sec = 31:30
Weight rider / Gewicht Fahrer [kg] : 62 kg
Weight bicycle, clothes etc. / Gewicht Fahrrad [kg] : 8 kg

Grade / mittlere Seigung : 7.6 %
Average speed / mittlere Geschwindigkeit : 20.0 Km/h
Total weight / Gesamtgewicht : 70.0 kg

Power : 351.9 Watt
Power / kg : 5.6 Watt / kg

Source: [ http://www.rst.mp-all.de/bergauf.htm ]

Henao's power output probably higher = possible headwind. Leipheimer's power output likely lower = "wheelsucking".

 

[Le breton]

Tour of Utah 2011, Stage 5, Final Climb Little Cottonwood Canyon, last 10.5 Km

Sergio Henao

Elevation / Höhenmeter [m] : 800 m
Distance / Streckenlänge [Km] : 10.5 Km
Time in seconds / Fahrzeit in Sekunden [sec] : 1890 = 31 min 30 sec = 31:30
Weight rider / Gewicht Fahrer [kg] : 61 kg
Weight bicycle, clothes etc. / Gewicht Fahrrad [kg] : 8 kg

Grade / mittlere Seigung : 7.6 %
Average speed / mittlere Geschwindigkeit : 20.0 Km/h
Total weight / Gesamtgewicht : 69.0 kg

Power : 347.5 Watt
Power / kg : 5.6 Watt / kg

.

Hello Halamala,

What was the starting altitude for that 800m climb?
I assume it was quite high (climbybike 1640?, 1840?)

In any case, sufficiently high that a correction for altitude would be in order, don't you think?

 

[Alex Simmons/RST]

Hello Halamala,

What was the starting altitude for that 800m climb?
I assume it was quite high (climbybike 1640?, 1840?)

In any case, sufficiently high that a correction for altitude would be in order, don't you think?

Using an altitude of 1600m above sea level would drop the power required on such a climb by ~6W (compared to starting at sea level) or 0.1 W/kg.

I'd like to know what assumptions are being used for Crr & CdA.

I get ~ 5.5W/kg with no wind, Crr 0.005 & CdA 0.33m^2 and starting at 1,600m

 

[halamala]

Vuelta a Espana 2011, Stage 4, Final Climb Sierra Nevada

Distance : 23.0 Km
Grade : 5.75 %
Eleevation : 1322 m

Daniel Moreno : 57:20 , 24.07 Kph , VAM 1383 m/h , 5.37 Watt / kg
Chris Anker Sorensen : 57:23 , 24.05 Kph , VAM 1382 m/h , 5.37 Watt / kg
Peloton : 57:31 , 24.00 Kph , VAM 1379 m/h , 5.35 Watt / kg

Note: Dr. Michele Ferrari's formula.

 

[Le breton]

Using an altitude of 1600m above sea level would drop the power required on such a climb by ~6W (compared to starting at sea level) or 0.1 W/kg.

...

I was not refering to the reduced drag, but to the reduced efficiency of the human engine and the correction needed to compare to sea-level efforts ( about 10% at 2200 meters).

 

[Sophistic]

Vuelta a Espana 2011, Stage 4, Final Climb Sierra Nevada

Distance : 23.0 Km
Grade : 5.75 %
Eleevation : 1322 m

Daniel Moreno : 57:20 , 24.07 Kph , VAM 1383 m/h , 5.37 Watt / kg
Chris Anker Sorensen : 57:23 , 24.05 Kph , VAM 1382 m/h , 5.37 Watt / kg
Peloton : 57:31 , 24.00 Kph , VAM 1379 m/h , 5.35 Watt / kg

Note: Dr. Michele Ferrari's formula.

Dr.Ferraris formula is crap, at 5,7% a 70kg guy would have a lower watt/kg number than a 55kg guy at the same speed.

 

[euanli]

Care to explain that.

 

[Le breton]

Care to explain that.

a 55 kg guy will not be allowed to use a 5.5 kg bike, he will have to use a 7 kg bike, just like his heavier opponent.
So, roughly speaking, considering than gravity accounts for 80% of the power expenditure on such a slope(5.7% at ~2000m altitude) the lighter guy will be at a disavantage having to produce 80% of 1.5/62 the amount of watts/kg of his opponent.
0.8 times 0.024 = 2%

so, if the heavier one produces 5.8 watts/kg, the lighter one will need to produce 1.02 times 5.8 = 5.9 watts/kg.

Of course, if the lighter guy sucks the wheel of the heavier one at 25 km/h he might save as much as 0.3 watts/kg and need to produce a mere 5.6 watts/kg to keep up with him.