Maths anyone?

robkit said:

%MHR = 37 + (0.64 * %VO2 Max)

but can any coach help me with the following:

- convert VO2 numbers to energy expenditure in kcal of kjoule

- convert any of the above to power output, assuming say 21% efficiency.

Thus I can understand such statements as "the very best time triallists can maintain over 420 watts for an hour indicating a VO2max of over 5l/min".

Thanks!

The precise energy yield per volume of O2 consumed varies slightly (~7% maximum) depending on the fuel mixture that is oxidized. For your purposes, however, you can simply assume that for every 1 L/min of O2 taken up, 5 kcal or 20 kJ of energy will be produced (actually, released) via aerobic metabolism. Assuming an efficiency of 21%, external power output would be (20,000 J x 0.21)/(1 L/min x 60 s/min) = 70 W for every 1 L/min of O2 taken up. With that rather low efficiency, a sustained power output of 420 W therefore would require a VO2max of at least 6 L/min.

acoggan said:

The precise energy yield per volume of O2 consumed varies slightly (~7% maximum) depending on the fuel mixture that is oxidized. For your purposes, however, you can simply assume that for every 1 L/min of O2 taken up, 5 kcal or 20 kJ of energy will be produced (actually, released) via aerobic metabolism. Assuming an efficiency of 21%, external power output would be (20,000 J x 0.21)/(1 L/min x 60 s/min) = 70 W for every 1 L/min of O2 taken up. With that rather low efficiency, a sustained power output of 420 W therefore would require a VO2max of at least 6 L/min.

Interesting.

I'm concerned I have a low VO2max because my short term power seems to be what's limiting me in mass start events.

On a climb that starts at around 6000', I can consistently do 295W for 20 min. I weigh 68 kg. How low could my VO2max then be?

acoggan said:

The precise energy yield per volume of O2 consumed varies slightly (~7% maximum) depending on the fuel mixture that is oxidized. For your purposes, however, you can simply assume that for every 1 L/min of O2 taken up, 5 kcal or 20 kJ of energy will be produced (actually, released) via aerobic metabolism. Assuming an efficiency of 21%, external power output would be (20,000 J x 0.21)/(1 L/min x 60 s/min) = 70 W for every 1 L/min of O2 taken up. With that rather low efficiency, a sustained power output of 420 W therefore would require a VO2max of at least 6 L/min.

thanks this is very useful. (well ok it's useful in an academic context assuming i will never have the time to fully realise my potential!) it really drives home the importance of high VO2max as a measure of potential before starting to think about threshold levels and body mass.

i suggested an efficiency figure of 21% because the paper by coyle re development of lance armstrong suggested he started out at 21% efficiency at age 21, only hitting 23% at age 28 after years of intensive training. therefore i'm guessing 21% would be on the realistic/optimistic side for most amateurs.