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#1 |
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Registered User
Join Date: Feb 2007
Location: East Coast
Posts: 127
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It has occured to me that a relationship between long term power (let's call it aerobic power) and short term power (the sum of aerobic power plus stored energy power) might simply be given by this equation:
Pt = Pa + E/t Where Pt is the average power output of a rider over any given period of time, Pa is the rider's aerobic power limit, E is the rider's stored energy (watt-min in the tank) and t is time. By pluging this equation with data from hard efforts over two different periods of time, and then solving the two simultaneous equations, it's then possible to determine Pt and E. Here's an example for one minute and five minute maximum efforts: P1 = 566 watts average over one minute P5 = 344 watts average over five minutes 566 = Pa + E/1 344 = Pa +E/5 Solving these equations we determine that Pa for this rider is 288.5 watts, and E is 277.5 watt-minutes. Now, with these two numbers, it's possible to calculate the rider's potential power output over any given period of time. For example, FTP is power over 60 minutes, so for the rider above, FTP would be: P60 = 288.5 + 277.5/60 = 293 watts I've tested this relationship on my own data and the calculated power over any given time comes very close to what I've actually produced. I would be very curious to know the results others get using their data and this formula. If the relationship described by this formula turns out to be valid, then a practical application would be another, shorter and easier test for FTP. Just go hard for one minute, then when rested, go hard for five minutes. Actually any two time intervals could be used, but 1 minute and five minutes are easiest for me and I'd think that intervals of less than about 30 seconds would be to greatly influenced by technique and other physical limitations (you might have the energy in the tank but not the muscles, tendons and bones to take it out more rapidly). What do you think? |
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#2 | |
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Registered User
Join Date: Feb 2007
Location: Germany
Posts: 18
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Quote:
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#3 |
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Registered User
Join Date: Jul 2003
Posts: 342
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I think using such short durations will skew results and that there arguably is little difference between 1m and 5m power when compared to 60m power.
My maths stink, so I didn't try to solve the equations, but how do the following #'s fit your idea? 1m= 580w 5m= 385w Then run it with 10m= 335w 15m= 325w My guess is that, like Monod, the accuracy of your formula depends on the durations used and the power profile of the individual. If I use 1, 5, 10m in Monod I get 60m= 312w, if I use 5, 10, 15m I get 302w. Not a huge difference, and well within the vagaries of PM measurement and daily performance fluctuations, but a difference nonetheless. |
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#4 | |
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Registered User
Join Date: Jul 2003
Posts: 2,588
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Quote:
His formula is the Monod formula. |
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#5 |
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Registered User
Join Date: Jul 2003
Posts: 342
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doh!
Well, I did say my maths stink... |
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