# Estimating VO2 Max from FTP/Kg ratio

Discussion in 'Power Training' started by bbrauer, Jul 9, 2009.

1. ### bbrauer New Member

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I believe this has come up before, and someone once posted a means of estimating VO2 max from what I remember as your max 5 min power. I was having a discussion with someone who made the argument that, for example, Lance's ave power of approximately 465 watts for 30 minutes during the 04 L'Alpe d'Huez TT, based on his weight, would give him an estimated VO2 Max over 100, which would be beyond human physical limits.

I think his actual VO2 Max is supposedly 82, which he also argued that, though excellent, was not a world beater and so therefore his achievements are not natural.

I'm not interested in another "did Lance dope?" argument, but that reverse estimate of VO2 from power/weight made me think a little. Is it possible to achieve a ballbark figure of VO2 Max from FTP/weight?

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2. ### daveryanwyoming Well-Known Member

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It would be an estimate based on an estimate and probably not very accurate, but sure you could ballpark VO2 Max in ml/kg/min from FTP in watts or watts/kg.

It would look like

VO2 Max as a function of 5 minute power and 5 minute power as a function of FTP which of course varies quite a bit through training.

But as a really rough swag:

VO2 Max in ml/kg/min = 12* 5 minute MMP (w/kg) + 3.5

(I've seen estimation methods that vary the multiplier from ~ 11 to 12, but since this is a ballpark swag and the next term is even more variable 12 is just easier)

And 5 minute MMP (w/kg) ~ FTP(w/kg) *1.25

(that 1.25 is a huge swag as 5 minute MMP could vary anywhere from 105% to maybe 140% of FTP or more based on genetics, training, AWC, etc.)

So for the ballpark VO2 Max estimate:

VO2 Max in ml/kg/min = 12*(FTP*1.25) + 3.5 where FTP is defined in w/kg

So based on those assumptions, VO2 Max for 4 w/kg FTP is somewhere near 63...

-Dave

3. ### bbrauer New Member

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Just what I needed. Thanks. BTW, where are the sources for the different methods you reference? Fascinating stuff.

The 1.25 figure gives a VO2 Max over 90 based on an assumed Armstrong body weight of 160 lbs. I plugged in the 1.05 figure, which gave me 80 and change. Honestly, I would go with the lower multiplier with the assumption that Lance's threshold is a very high percentage of his VO2 Max.

And threshold power was estimated at 441.75 based on the .95 multiplier of the original 465 figure, which is from my vague memory of calculating his power up that TT using that online calculator which doesn't exist anymore (was it Kruezotter or something?) So it's an estimated calculation based on multiple guesses.

4. ### daveryanwyoming Well-Known Member

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That formula and variations on it have been discussed here and on the Google wattage lists a few times. Here's one link, but there was a longer thread with more discussion (including discussion about the error of adding the +3.5 ml/kg/min baseline 02 term twice in the formula linked).

Using the corrected form of the formula linked the multiplier is 1.8*6.12 or rouglhly 11 instead of the 12 I used above.

-Dave

5. ### SolarEnergy New Member

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I think it is a fair assumption.

The great Lance is also known for having an unusually high economy too, that plays a role.

One may have a Vo2Max little lower, if a higher percentage of o2 consumed contributes to forward propulsion, then that plays a role. This is why at least to me, power at Vo2Max is a variable that speaks louder.

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7. ### bbrauer New Member

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That's pretty handy. Do you have the Excel file on your blog anywhere for saving and plugging in different values?

8. ### Alex Simmons Member

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I can't post the excel easily.

That's the reason the formula are shown in the last column, so it's easy to set up yourself.

9. ### bbrauer New Member

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I was thinking. Instead of multiplying by a factor of anywhere from 1.05-1.40, why couldn't I take the two numbers and plug them into a Monod-Scherrer calc and get a five minute estimate? If I have an estimated power from the 04 time trial of 465 watts which was, say, 30 minutes in duration, and I multiply by .95 to get a FTP estimate of 441.75, I could plug in those numbers and get a 5 minute power of 489. Then I just use that number for the calc. That gives me an 84.99 VO2 Max figure.

I forget. Would two numbers at the long end of the duration scale in the CP curve underestimate power for shorter durations because there isn't a real sample that incorporates AWC?

10. ### Alex Simmons Member

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You should think again. There are so many things that could introduce error with your proposed estimation idea I'm not sure where to start.

If you want a 5-min estimate, go and do a 5-min effort.
Only takes 5-min.

Here are a few items about traps people fall into when estimating FTP.
Alex's Cycle Blog: The Sins of Sins (Testing FTP #2)

11. ### bbrauer New Member

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Yeah, that's all well and good, but unfortunately I can't just call Lance on his cell in France in the middle of the Tour and ask him to go in his time machine back five years to perform a 5 min max test for me.

All of these numbers are based on equations that act as analogues for the real thing. I wouldn't presume to use the 12*(watts/kg) + 3.5 = VO2 Max as a substitute for a real test.

If Lance has a reported VO2 Max of 84, the exercise was to determine if his best performance in the Tour is generally in line with this figure. So using Dave's suggestion of using that Level 5 range of 1.05-1.40*FTP, to estimate a 5 min MMP, and settling on the 1.10 number based on Lance's presumed high threshold as a % of VO2, I get 83.88.

If I used the estimate of 5 min CP from Monod and plug that into the equation, I get a figure of 84.37.

So in terms of roughly correlating with his reported VO2 Max of 84, I'd say both numbers are more or less in line with that number. Or more specifically, his actual performance is in the range of what is possible based on his physiology, which was the original point of my exercise.

12. ### daveryanwyoming Well-Known Member

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Yeah, but to use the Monod model you need two independent points each representing a maximal effort for the duration (so no 10 minute climbs partway through a 5 hour stage for instance). You really have one data point, not two. You're also implicitly taking a guess at his Monod curve to try to create two points from one. All the x minutes * y% methods of estimating FTP (e.g. 20 min * 95%) are really just guesses at the shape of someone's Monod curve. So you really shouldn't guess at the shape of someone's Monod curve and use it as an input to a Monod curve. You aren't really buying any additional accuracy in your guesses.

I think you're still closest with Alex's method or something like the rough swag I gave above.

-Dave

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