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# Power Meter Puzzler

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# Power Meter Puzzler

02-04-12, 09:28 PM
#1
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Power Meter Puzzler

How can you have two rides of equal duration (50 minutes), both indoors on a trainer using the exact same equipment, where you go farther (i.e. faster) on one, yet produce fewer Kilojoules (i.e. less work)? Here are the particulars:

Jan 24- 12.8 miles, Avg power 127, Normalized power 165.... 413 Kj
Feb 2- 13.3 miles, Avg power 136, Normalized power 149.... 408 Kj

Jan 24 ride was a series of 1-minute intervals
Feb 2 ride was a set of 5 minute intervals.

Perhaps I do not understand Kilojoules... can someone please clarify?
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02-05-12, 07:01 AM
#2
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Possible explanation #1: power and effort do not increase in a linear fashion, so doing those 1 minute power intervals averages out a little higher.

Possible explanation #2: those numbers are well within the margin of error.
02-05-12, 09:59 AM
#3
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Originally Posted by Nikephoros
Possible explanation #1: power and effort do not increase in a linear fashion, so doing those 1 minute power intervals averages out a little higher.
This. Most trainers are designed to simulate outdoor riding so resistance increases with the square of speed. Putting out 30% extra power will only give you a 10% increase in speed.
02-05-12, 10:43 AM
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Or in another way of saying... your burnt energy vs. work performed is not linear depending on level of effort. (Guess that's why you train for efficiency on certain discipline).
02-05-12, 01:54 PM
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Originally Posted by billydonn
How can you have two rides of equal duration (50 minutes), both indoors on a trainer using the exact same equipment, where you go farther (i.e. faster) on one, yet produce fewer Kilojoules (i.e. less work)? Here are the particulars:

Jan 24- 12.8 miles, Avg power 127, Normalized power 165.... 413 Kj
Feb 2- 13.3 miles, Avg power 136, Normalized power 149.... 408 Kj
Work (J) equals power (W) times time (s). It doesn't matter how the power was applied; total work is the integral of instantaneous power over the duration of exercise (i.e., linear). For the two cases, 127 W over 49:56 (2996 seconds) is 380.5 kJ. 136 W gives 408 kJ.

So the question is where did your 413 kJ number come from?
02-05-12, 04:25 PM
#6
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Originally Posted by asgelle
Work (J) equals power (W) times time (s). It doesn't matter how the power was applied; total work is the integral of instantaneous power over the duration of exercise (i.e., linear). For the two cases, 127 W over 49:56 (2996 seconds) is 380.5 kJ. 136 W gives 408 kJ.

So the question is where did your 413 kJ number come from?
It comes from here... it is from Cycleops Power Agent software. it was attached to my post but here it is again:
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Last edited by billydonn; 02-05-12 at 04:30 PM.
02-05-12, 05:12 PM
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Originally Posted by billydonn
It comes from here... it is from Cycleops Power Agent software. it was attached to my post but here it is again:
You can write or publish it as many ways as you want; it doesn't add any more value. What I was asking was how was the data measured, recorded, stored, and manipulated. Errors in any of those could cause errors in the displayed average power and work. In your case, it's clear there's an error in the calculated work. Without examining the data train, there's no way of knowing for sure where that's coming from. You might want to start by checking total work against the product of average power and time to see how many are off.
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