Page 1 of 1
Watts -"Stupid Question"
Posted: October 17th, 2006, 9:07 pm
by flan48
Sorry for what seems, to me

,ignorance or at least naivete', but when obtaining a watts reading on the PM3 (I have a C2 D) what does the reading signify?
In other words is it watts per a certain time, watts per distance, etc?
I'm confused because everything else presented on the monitor is straightforward, be it split time, calories/hour, total distance and so on.
Thanks for your help.
Best regards
Posted: October 17th, 2006, 9:49 pm
by johnlvs2run
Per time.
One watt = 1 joule per second.
Posted: October 17th, 2006, 10:47 pm
by flan48
[quote="John Rupp"]Per time.
One watt = 1 joule per second.[/quote]
John,
Thanks for yur response. So if at the end of my session the PM3 reads 147 watts, that means I burned an average of 147 joules/second or 147 x 60 = 8820 joules/minute. So for a 30 minute session we are talking about
8820 x 30 = 264600 joules. This translates to 63301 calories or 63.301 kcalories (what we usually describe as calories).
Something is not right(?).
Best regards
Posted: October 17th, 2006, 11:24 pm
by haboustak
That looks reasonable to me. It's normal that it won't match what the monitor displays as Calories.
The monitor's value for Calories is intended to be an estimate of energy expended by the rower. The value you calculated is the actual energy, in Calories, imparted by the rower on the flywheel as torque.
You could theoretically divide the effective Calories by the expended Calories and get a number that kind of represents physiological efficiency. But the monitor's calculation of Calories is such a ROUGH estimate that it's probably just a waste of time.
In English, the value of Power in Watts is the amount of energy you're able to apply to the flywheel per unit time.
Mike
Posted: October 18th, 2006, 12:37 am
by jamesg
Barry,
The formula used by the PM3 etc seems to be:
Total Heat generated in kCal = 300 + kWh x 4 x 850, where:
300 kCal/h is our credit for getting on the machine and moving the handle;
4 or rather 0.25 is our efficiency seen as fuel cells (the chemical reactions in us that generate work also generate heat - 3 x as much is the guesstimate)
850 kCal is as far as I remember the heat equivalent of 1 kWh.
So if you work for ½h at 200W = 0.1 kWh you will be credited with
300*½ + 4*0.1*850 kCal = 490 kCal.
After ½h @ 147W, you would see about 400 kCal on the monitor.
Posted: October 18th, 2006, 10:12 am
by flan48
Mike/James,
Thanks for your responses and explanations. While I understand what you have written, and certainly do not expect "wattage" to calculate exactly to the calorie readout, it is way off.
Once again, suppose I work our for 1/2 hour at a wattage of 150.
150 watts=150 joules/second. So 150 joules/sec. x 60 seconds/minute x 30 minutes = 270000 joules. There are 4.18 joules in a calorie.
Therefore, 270000 joules x 1 calorie/4.18 joule = 64593 cal = 64.593 Kcal.
It is impossible to believe that working out at this energy level on the erg for 30 minutes only burns 65 calories! That is why I posed the question originally, i.e. is the wattage readout average for that workout or for something else.
Thank you again.
Posted: October 18th, 2006, 10:28 am
by haboustak
The only problem is that it's not 65 Calories burned, it's 65 Calories imparted to the flywheel/fan. As James pointed out, C2 has a formula to estimate a value for Calories burned given flywheel Power.
The reason it's so much lower is because it's an inefficient way of generating Power. Every time Energy is converted from one form or another there is Energy that is lost. And there are a lot of conversions going on (e.g. chemical > electrical > mechanical > heat&sound). Plus there's energy expended that's not applied directly to the chain (e.g. maintaining balance, moving the slide, noise from the rollers, heating your environs).
So after all your effort converting Energy and generating heat you were able to apply ~65 Calories of Energy to the fan (regardless how much Energy you spent to do it).
Edited to add:
To try and answer your Watts question. The Watts reading on the monitor is the amount of Power (Work done per time) as detected by the fan. It can display instantaneous or average values. I think the "instantaneous" value is Power for the previous stroke, while the average value is for the entire session. Even if the Work done by the rower is approximately the Work performed on the flywheel, the values for flywheel Energy and Calories burned will vary significantly (because of the required conversions and inherent lossy nature of our powerplant).
Mike
Posted: October 18th, 2006, 11:29 am
by johnlvs2run
Thanks for the great explanations.

Posted: October 18th, 2006, 11:45 am
by PaulS
haboustak wrote:The only problem is that it's not 65 Calories burned, it's 65 Calories imparted to the flywheel/fan. As James pointed out, C2 has a formula to estimate a value for Calories burned given flywheel Power.
The reason it's so much lower is because it's an inefficient way of generating Power. Every time Energy is converted from one form or another there is Energy that is lost. And there are a lot of conversions going on (e.g. chemical > electrical > mechanical > heat&sound). Plus there's energy expended that's not applied directly to the chain (e.g. maintaining balance, moving the slide, noise from the rollers, heating your environs).
So after all your effort converting Energy and generating heat you were able to apply ~65 Calories of Energy to the fan (regardless how much Energy you spent to do it).
Edited to add:
To try and answer your Watts question. The Watts reading on the monitor is the amount of Power (Work done per time) as detected by the fan. It can display instantaneous or average values. I think the "instantaneous" value is Power for the previous stroke, while the average value is for the entire session. Even if the Work done by the rower is approximately the Work performed on the flywheel, the values for flywheel Energy and Calories burned will vary significantly (because of the required conversions and inherent lossy nature of our powerplant).
Mike
Small point of correction. The "instantaneous" (Per stroke) figure is in fact an Average, calculated by taking the total Joules absorbed by the Flywheel divided by the total Stroke Time. The Watts can be plotted for each time period between sensor pulses by tools such as ErgMonitor, and this plot looks much like a sine wave, rising during the drive and falling during the recovery.
At a 2:00 Pace, the 202watts being displayed is an average, the actual power production could range from something like 100-300watts through the entire stroke cycle.
Posted: October 18th, 2006, 12:14 pm
by Sitwronge
PaulS wrote:the actual power production could range from something like 100-300watts through the entire stroke cycle.
PAUL S DOES THIS MEAN 100 WATTS ARE PRODUCED FROM THE RECOVERY PART OF THE STROKE!!!!!!!??????
Posted: October 18th, 2006, 2:25 pm
by PaulS
Sitwronge wrote:PaulS wrote:the actual power production could range from something like 100-300watts through the entire stroke cycle.
PAUL S DOES THIS MEAN 100 WATTS ARE PRODUCED FROM THE RECOVERY PART OF THE STROKE!!!!!!!??????
No John Rupp, there is no power input to the Flywheel during the recovery, however it does continue to be dissipated. However the energy dissipated during the recovery does tend to be quite similar to that input during the drive, but over a longer period of time.
Posted: October 18th, 2006, 2:42 pm
by johnlvs2run
Paul, that previous message came from Sitwronge.
You're getting more confused in your old age.

Posted: October 18th, 2006, 2:45 pm
by johnlvs2run
Seems to me that you'd been better off using all that energy on the drive,
instead of making the recovery longer to waste more of it.
Posted: October 18th, 2006, 2:57 pm
by Ducatista
Sitwronge wrote:!!!!!!!??????
There once was a troll on the forum
And though people tried to ignore him,
He soon found a way
To keep ruining their day:
HE LET A SOCK PUPPET TALK FOR HIM!!!!!!!
Posted: October 18th, 2006, 3:18 pm
by Sitwronge
I AM NOT PAUL SMITH!!!!!!!!!!!!