tpk wrote:gregsmith01748 wrote:If your generously assume that rowing burns 1000 calories an hour ...
One thing I've never been able to get a straight answer on is, do calorie per hour values (e.g. here
http://www.mayoclinic.org/healthy-livin ... t-20050999) typically include or exclude basal metabolic rate. Assume a BMR of B (something in the order of 100 cals/hr maybe?). And assume a quoted R calories per hour for "calories per hour burned by rowing" (in your example, 1000). Then is the typical measurement quoted such that the *total* cals burned in the hour of rowing is B+R, or is it simply R? I've always assumed the latter, but as I say I've never been able to get a straight answer.
tpk
The following quote from "The Physics of Rowing" {
http://www.atm.ox.ac.uk/rowing/physics/ ... #section11} does not give the breakdown, but it does discuss the extra Calories/hour shown on the C2 monitors.
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
Note that the number of total extra C/hr is 300 and the whole thing is multiplied by 4 to account for an estimated 25% efficiency of the human body as a fuel burning machine. If 100 C/hr are being used just to keep the "engine" alive, then that 300 C/hr presumably includes 25 C/hr for BMR. The rest is for external work done that is not measured by the monitor, mainly the work done moving the body up and down the slide.
Bob S.