Is it legitimate to convert an athlete's continuous  oxygen consumption in mlOfO2/(kg-min) to watts/kg?  

No, in my opinion it is not. I think you are overlooking a couple of things:

1.  O2 is not the only source that can form ATP

The amount of energy used during exercise is actually the amount of ATP-usage.

This can be translated to the amount of O2 that is necessary to form all the ATP, if one assumes that the P/O ratio is constant, if the P/O ratio is 2.835 the amount of ATP formed with 1 ml O2 (equals 22.4 mmol) is 22.4 /(2*P/O) = 3.95. The value of 3.95 is often called the cPCr. This means that the total amount of energy used during exercise can bet translated by:

This VO2tot is not equal to the VO2 because the glycolyses can form ATP, and creatine phosphate can work as a buffer to form ATP. Even if you assume that you are in a steady state (the VO2 and other processes are at equilibrium) the VO2 is not exactly linear with the VO2tot. An example, (see in the blue areas in  figure below, ) produced with the equations of Mader A. & Heck 1986. The amount mitochondria’s and the amount of glycogenic enzymes vary from person to person. So the assumption that VO2 is linear with the VO2tot is only realistic if:

1.  here is a steady state situation

2.  at a (very) low VO2tot where the influence of the glycolyses is neglectable

2.  the power output is not the only energy you use.

Even in rest you are breathing and using O2 this vary from 6 to 20 ml* min *kg^-1 ,so this means that there is a lot of O2 usage that doesn’t have anything to do with the power output. The resting O2 (VO2rest) must therefore be taking into account. In the equation

The VO2rest is the resting VO2 and the rcVO2 is determine the metabolic efficiency.

3.  the metabolic efficiency vary from person to person.

By using the assumption that the P/O ratio is constant (and therefore not too much extra heat is produced), the metabolic efficiency is also influenced, by co-contraction, pedal rate.

The calculations in short:

In the example from you:

70 kg

280 Watts

48 ml* min *kg^-1 = 48/3.5= 12.85 METS

With the ACSM Leg Ergometry Equation (VO2 =10.8 * watts) / kgbw + 7) the VO2 is only 1 ml* min *kg^-1  of so the VO2rest would be around the 6 ml*min*kg^-1 or the efficiency score of 10.8 is a bit lower.

If we assume that the VO2rest is 6 the efficiency of someone with a higher amount of glycogenic enzymes (VLamax) and the same VO2max and percentage of muscles used during the exercise would have to have a less efficient movement.     

In the example figures below the VLamax is changed from .3 mmol *min*kg^-1 to .6 mmol *min*kg^-1 this gives a VO2tot 50 and 54 ml*min*kg^-1 at a VO2 of 48. So the rcVO2 is than (50-6)/(280/70)=11 and (54-6)/(280/70)=12 the difference between these 2 values is very difficult to estimate without mathematical modeling (to estimate this from VCO2 is also possible but on blood lactate levels above +_2 mmol/l sometimes takes 10 to 15 minutes to reach a steady state).

Martijn Carol

VO2max=70;VLamax=.3;Volrel=.4;

VO2max=70;VLamax=.6;Volrel=.4;

Ref: Mader A., & Heck H. (1986). A theory of the metabolic origin of “Anaerobic Threshold”. Int J Sports Med, 7(6), 45-64.

-----Original Message-----
From: sportscience@yahoogroups.com [mailto:sportscience@yahoogroups.com]
Sent: 24 October 2009 14:49
To: sportscience@yahoogroups.com
Subject: Digest Number 1078

There is 1 message in this issue.

Topics in this digest:

1. Converting oxygen consumption to watts/kg and human efficiency?   

    From: TJACMC@...

Message

________________________________________________________________________

1. Converting oxygen consumption to watts/kg and human efficiency?

    Posted by: "TJACMC@..." TJACMC@... ted_andresen

    Date: Fri Oct 23, 2009 10:41 am ((PDT))

Is it legitimate to convert an athlete's continuous  oxygen consumption in

mlOfO2/(kg-min) to watts/kg?  

If 1 mlOfO2/(kg-min)x(4.8 Kcal/1000 mlOfO2)x(4186  Joules/Kcal)x(1

min/60)*(watt*sec)/Joule equals 0.33 watts/kg.  

Therefore, to convert oxygen consumption to watts/kg,  just divide the

oxygen consumption by 3.

Also, although oxygen consumption (mlsOfO2/(kg-min)) and  METS are referred

to as energy, aren't they actually power, i.e., more  specifically mass

specific power in terms of watts per kg?

It would be very useful to convert O2 consumption to  power.  Just an

example, if a 70-kg  cyclist consumes 45 mlsOfO2/(kg-min), he or she is consuming

15 watts/kg of  metabolic power.  If the athlete is  generating 280 watts

of power, they generating 4 watts/kg, so they have a  metabolic-to-mechanical

efficiency of approximately 26%?

Am I overlooking something?  I have never seen this in the  literature.

Ted Andresen

St.  Petersburg, Florida

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