Concept2 Forum

A Watt is a Joule per second and a kilocalorie (food calorie) is 4,184 joules.

If I work out for 30 minutes averaging 175 watts over 60 minutes. My calorie expenditure doesn't match the wattage.

Example:

175 watts = 175J/sec X 30' X 60 sec/min = 315,000 joules of energy expended
315,000J divided by 4,184 joule/Kcal = 75.5Kcal
My machine display shows I burned ~ 400 Kcals

I'm at a loss at the difference. I weigh 165 # so that value is essentially insignificant. Can you explain where i went wrong?

Billius_53 wrote:A Watt is a Joule per second and a kilocalorie (food calorie) is 4,184 joules.

If I work out for 30 minutes averaging 175 watts over 60 minutes. My calorie expenditure doesn't match the wattage.

Example:

175 watts = 175J/sec X 30' X 60 sec/min = 315,000 joules of energy expended
315,000J divided by 4,184 joule/Kcal = 75.5Kcal
My machine display shows I burned ~ 400 Kcals

I'm at a loss at the difference. I weigh 165 # so that value is essentially insignificant. Can you explain where i went wrong?

The monitor throws in a free bonus of 300 kcal/hr to account for the work done accelerating your body back and forth on the rail. There is quite a bit of energy used for that and it is not measured by the monitor, which can only detect the work actually done to make the wheel spin. Obviously the amount of Calories generated would vary with the rower's weight - also with the stroke rate and the length of travel of the center of mass of the body (+ seat). C2 uses sort sort of standard rower, but I don't remember the details. I think that there might be conversion tables available for "non-standard rowers," but I don't know where. I don't use Calories so have not looked into it other than when I first stumbled on the problem. Much like you, I couldn't figure out at first why there was no agreement between the units.

Bob S.

I just took a closer look at your numbers and note that there is still a discrepancy. The "bonus" for 30 minutes should be 150 kcal not the 324.5 that your figures would indicate. I don't know any answer to that. The bonus would have to be 649 kcal/hr to account for that. I am sure that it is not. Note: I don't disagree with your figures. I just gave the calculations a cursory mental check, but they look OK to me.

By the way, what did you mean by the "over 60 minutes?"

Bob S.

Billius_53 wrote: I'm at a loss at the difference. I weigh 165 # so that value is essentially insignificant. Can you explain where i went wrong?

You're applying science to the subject of metabolic (k)calories and you're assuming that human bodies are 100% efficient.

The biggest nonsense with (k)calorie counting (apart from the fact the that units are scaled by 1000) is that the calorific testing of food is done in a pure oxygen stream so that every last joule of energy is extracted from the sample being tested. If we humans ran that efficiently we'd need to eat a single pea to get our whole days energy. The food "scientists" have also stopped using calorimeters and now simply determine fat vs protein vs carbs then apply a mulitplier for each component to get the (k)calorie count (not much good science there). The whole game then becomes "common law" and joins the other non-science stuff like 220-age which is a gross generalisation.

Concept2 came up with their own formula for PM2/PM3/PM4 (k)calories which doesn't seem to match any empirical value. HR wrist watches seem to be slightly better at doing the (k)calorie estimation so if you want to follow those mythical numbers the I'd recommend using the values from your watch.

Here's an interesting NYT article on the subject: http://www.nytimes.com/2010/01/12/healt ... .html?_r=0

Bob S. wrote:I just took a closer look at your numbers and note that there is still a discrepancy. The "bonus" for 30 minutes should be 150 kcal not the 324.5 that your figures would indicate. I don't know any answer to that. The bonus would have to be 649 kcal/hr to account for that. I am sure that it is not.

Aha! I just remembered the efficiency factor. I believe that 25% was the figure used. The assumption is that the body uses up a 100 kcals of fuel to produce 25 kcal of equivalent work. That would mean that 302 kcals of fuel were used up to produce the 315 kj of work. With the 150 kcal bonus for the unmeasured work done on body movement it comes out to 452 kcal. Still not the 400 that you report, but at least in the ball park.

I find all of this a good reason for ignoring the Cal/hr numbers shown on the monitor. In all fairness to those that use them, I think that they are probably a reasonable guideline for folks who are keeping track of those things, especially if the proper modifications for size and rate are taken into account.

Citroen wrote: The biggest nonsense with (k)calorie counting (apart from the fact the that units are scaled by 1000) is that the calorific testing of food is done in a pure oxygen stream so that every last joule of energy is extracted from the sample being tested.

At least the original research on it took into account that food is not used efficiency by the body. Obviously there are still a lot of calories left in excrement. It has even been widely used for fuel - buffalo chips of the American plains, cattle dung in India, and yak dung in Nepal. I wondered about the Calorie values that have been reported for the various food and found that the guy that seems to have done the first research on this did a careful, thorough study, which involved calorimetric measurement of both intake and outgo.

That still begs the question of variability of the efficiency of food metabolism from one person to another. It appears to show up during sleep - not much work is being done, but some folks stay quite warm with little cover, obviously burning up fuel, while others conserve fuel and have to bundle up to stay warm.

Bob S.

Bob:

Over a 60 minute period. Just an expression meaning average calories burned during a 60 minute time span.

BTW

Thanx for your comments

Bill

Back to the origin of this thread, here is the full scoop:

Quote:
11. Power v. Indicated Calories

The Concept Model C also has a 'Calories' display as a (very) rough guide to how many calories an average individual has burned up in a piece. This is not the same as the mechanical work done.
Mechanical work W (a type of Energy) is defined as the average Power x time:

(11.1) W = P t

If Power P is measured in Watts and time t in seconds, then the Work W is obtained in Joules. So, rowing a steady 200W for 30 minutes, you would generate an amount of mechanical work

(11.2) W = 200 x 1800 = 360 000 J = 360 kJ

In physics, a 'calorie' is defined as the amount of heat energy required to raise the temperature of 1 gramme of water by 1 degree centigrade, giving 1 calorie = 4.2 Joules. Dieticians, on the other hand, use the term 'calories' differently - their 'calories' are 1000 times bigger ('kilo-calories', kC), so dividing 360 kJ by 4.2 gives the mechanical work done in terms of 'dietary calories': 85.6 kC

However, for the above workout you would actually get a displayed value approaching 500 kC, i.e. a factor 5 - 6 times larger. This is because the computer attempts to calculate the number of calories you burn up (effectively chemical energy contained in fats and carbohydrates) in order to generate the mechanical work. It uses the formula

(11.3) E = ( 4 W + 0.35 t ) / 4.2 [kC]

where E is the displayed number of calories [kC], W is the mechanical work in kJ, calculated according to Eq. (11.1), t is the time in seconds. This assumes that the body actually requires 4 units of chemical energy to generate 1 unit of mechanical energy (i.e. 25% efficiency) plus a background consumption of 0.35 kJ/sec (=300 kC/hour).

Comment Jon Williams of Concept2 (12 Aug 04)
The 300 kC/hour has always been our best approximation for keeping alive and awake and going through the rowing motion at a reasonable stroke rate on an erg with the flywheel removed. This was arrived at from internal experiments and observations, data from Fritz Hagerman and studies done at Ball State.
For the above workout this would give

(11.4) E = ( 4.0 x 360 + 0.35 x 1800 ) / 4.2 = 493 [kC]


The 'Calorie' output on a Concept ergometer is an approximate guide to calories [kC] burned rather than mechanical work performed.
Unquote

The quote is from this site:

Physics of rowing (ergometers):

http://www.atm.ox.ac.uk/rowing/physics/ergometer.html

I was pleased to see that my vague memory was correct. They do use an efficiency factor of 4 and a bonus of 300 kcal/hr.

Bob S.

Bob:

That was a very cogent explanation for me. It made perfect sense. The assumptions seem valid as well. I haven't read the article yet on Physics of Ergonometers, but I'm sure it will elucidate the answer even more.

Thanx and Cheers!

Bill
Sydney, NS
Canada

Billius_53 wrote:Bob:

That was a very cogent explanation for me. It made perfect sense. The assumptions seem valid as well. I haven't read the article yet on Physics of Ergonometers, but I'm sure it will elucidate the answer even more.

I can't claim credit for the explanation. That was a direct quote from the article. It was the entire section of the article that dealt with the Calorie count, so there is no need to look up the original article. I included the URL to show where it came from. However the article as a whole has a lot of other interesting items and might be worth checking out. Since I live at 4,000' of elevation, I was interested in what it had to say about altitude effects.

Bob S.

Useful thread.

Recently, I started to track my daily mechanical work (Watts * Time), and find this single number especially useful as I tweak drag or pace or other workout variables. For me, this works out to about 180 kJoules (42.9 kCal) - mechanical energy (which I thought far too low till reading this thread: chemical vs. mechanical energy)

(I assume doing more daily mechanical work leads to better health/weight loss - all other factors like diet, carbs being controlled separately.)

Curious if others do the same, what do you track?

Here's how I do it:

Typical workout is 60 Watts (3:00, yes slow!) for 50 minutes (3000 sec), so the mechanical energy is

Mechanical Work = Watts * Time (in seconds), or
= 60 watts x 3000 sec
= 180,000 joules or
= 180 kJ or
= 42.9 kCal (divide by 4.2 joules per calorie)

Wow, you guys are incredible, I read your posts and feel like I just flunked junior high science! Just one comment from my own limited rowing, the number being accurate is not such a big deal in my case, previous numbers (calories or whatever) are just targets to reach or to compare to.
Wayne