Converting Power to Kcal (calories used)
Karl2010
Posts: 511
After reading the other post called "Low Level Rides" i have a question about how to work out the calories used during a session.
I have a G*rmin 705 and i know this massively over estimates the calories consumed during a session. (one of the mods on the G*rmin forums even says so)
One of the ways of adjusting the displayed Calories Consumed is to reduce your weight in the Rider Profile setting menu on the G*rmin
One of my motivations for buying a power meter is so i can better calculate how many Kcal i have burnt on a ride / session.
I know my Garmin will give a Kcal consumed figure regardless of weather or not im using the HR strap or not.
So i guess my question is what is the best way to work out Kcal consumed.?
And is it more accurate with a power meter?
Is there some other software that can download the G*rmin data and use a differant algorithm to calculate Kcal consumed based on, Power, HR, Rider & Bike Weight + Speed & Cadance?
I just dont trust the G*rmin to give accurate Kcal readings.
Thanks chaps. And all the best for 2011.
I have a G*rmin 705 and i know this massively over estimates the calories consumed during a session. (one of the mods on the G*rmin forums even says so)
One of the ways of adjusting the displayed Calories Consumed is to reduce your weight in the Rider Profile setting menu on the G*rmin
One of my motivations for buying a power meter is so i can better calculate how many Kcal i have burnt on a ride / session.
I know my Garmin will give a Kcal consumed figure regardless of weather or not im using the HR strap or not.
So i guess my question is what is the best way to work out Kcal consumed.?
And is it more accurate with a power meter?
Is there some other software that can download the G*rmin data and use a differant algorithm to calculate Kcal consumed based on, Power, HR, Rider & Bike Weight + Speed & Cadance?
I just dont trust the G*rmin to give accurate Kcal readings.
Thanks chaps. And all the best for 2011.
0
Comments

A power meter will give you a read out of how many Kjs you've used. But, that's only for the effort going through the power meter. Its very unlikely that you're 100% efficient.
I'm not sure if there's a reasonable percentage to work off for an estimation? Presumably efficiency varies with effort too?Racing for Fluid Fin Race Team in 2012  www.fluidfin.co.uk0 
incog24:
We need a scientist...!!! I might go to the University and get some advice..0 
It has been shown in various studies that the human body is somewhere between 19% and 26% efficient, so if you have the number of Kj expended, you can use this as a proxy for calories used.
Just a note about Garmins, the calories used varys massively if you don't have a HRM strap on. I would say mine underestimates calories used, or I am not very powerful at all.
Soon see when I get my PM.0 
You could roughly calibrate your garmin by using one of the calculators on the web
this onne for instance
http://www.noping.net/english/
Not perfect but it will give you a rough idea that is better than Garmins guestimate.0 
incog24 wrote:A power meter will give you a read out of how many Kjs you've used. But, that's only for the effort going through the power meter. Its very unlikely that you're 100% efficient.
I'm not sure if there's a reasonable percentage to work off for an estimation? Presumably efficiency varies with effort too?
Just take power meter reading of kJ of mechanical work and multiply by 1.10 +/ 0.05 and that'll be about the right range for Calories (kcal) metabolised. It might be a bit more or less than that range if your efficiency is at the high/low end of typical.
That ratio takes into account the typical gross efficiency of reasonably trained cyclists and the conversion of kJ to Cal.0 
Can you give me a Calculation Bezza?
Example.. Over 60 Mins
200w/hr = 720 kj
100 (%) Minus 23 (%) = 77 (%)  (so 77% loss through inefiiency)
720 kj devided by 100 (%) = 7.2 kj
7.2 Multiply by 77 (%) = 554.4 kj
554.4 kj ADD 720 kj = 1274.4 kj
1274.4 converted to Kcal = Aprox 318 kcal (calories)

If this is wrong plase feel free to correct me.
Also i used 77% as the ineficiency (energy used but not recorded through power meter)0 
Karl2010 wrote:Can you give me a Calculation Bezza?
Example.. Over 60 Mins
200w/hr = 720 kj
100 (%) Minus 23 (%) = 77 (%)  (so 77% loss through inefiiency)
720 kj devided by 100 (%) = 7.2 kj
7.2 Multiply by 77 (%) = 554.4 kj
554.4 kj ADD 720 kj = 1274.4 kj
1274.4 converted to Kcal = Aprox 318 kcal (calories)

If this is wrong plase feel free to correct me.
Also i used 77% as the ineficiency (energy used but not recorded through power meter)
200 watts at crank for 1 hour (3600 seconds) = 200 x 3600 = 720,000 joules = 720 kJ of mechanical work done.
1 kcal (Cal) = 4.18kJ
So 720kJ = 720 / 4.18 Cal = 172 Cal
If we are 23% efficient, then
172 Cal of mechanical work / 23% = 749 Cal of energy metabolised.
guess what?
749 / 720 = 1.04
(Note that 23% is towards the upper end of typical efficiency range)0 
Alex:
Im confused how does 172 cal devided by 23% become 749 cal?0 
Karl2010 wrote:Alex:
Im confused how does 172 cal devided by 23% become 749 cal?
The amount of work done at the crank is 172Kcal, multiply the inefficiencies and you get 749 Kcal.
In short the body has burnt 749Kcal to produce only 172Kcal of work at the crank, alot of calories have been expended by heat and other body movements not associated with actually pedalling.0 
I understand that the body has burnt 749Kcal to produce only 172Kcal of work at the crank.
I couldnt work out how 172 Cal of mechanical work / 23% became 749 Cal of energy metabolised.
But i got it now its:
172 Divided 23 = 7.478
7.478 Multiply by 100 = 747.8 Cal0 
Karl2010 wrote:I understand that the body has burnt 749Kcal to produce only 172Kcal of work at the crank.
I couldnt work out how 172 Cal of mechanical work / 23% became 749 Cal of energy metabolised.
But i got it now its:
172 Divided 23 = 7.478
7.478 Multiply by 100 = 747.8 Cal
23% = 0.230 
Alex_Simmons/RST wrote:Karl2010 wrote:I understand that the body has burnt 749Kcal to produce only 172Kcal of work at the crank.
I couldnt work out how 172 Cal of mechanical work / 23% became 749 Cal of energy metabolised.
But i got it now its:
172 Divided 23 = 7.478
7.478 Multiply by 100 = 747.8 Cal
23% = 0.23
But if two trains leave London and Bristol at 6 pm, which one will get cancelled first due to the snow?Le Blaireau (1)0 
DaveyL wrote:But if two trains leave London and Bristol at 6 pm, which one will get cancelled first due to the snow?0

Alex_Simmons/RST wrote:DaveyL wrote:But if two trains leave London and Bristol at 6 pm, which one will get cancelled first due to the snow?
Damn you
I'm up in Scotland for the holiday break. The snow has finally thawed, to reveal a set of totally trashed roads. Brilliant.Le Blaireau (1)0