Calorie, or kilocalories as the are accuratly called. What is the formula for calculating the calorie (as a unit of energy) for weight and distance the weight travels. I looked but could not find the info.
To be more clear, I want to calculate, for instance, how many kilocalories it takes to move 100 lbs, 2.5 feet, as in a sqat movement. Anyone, know?
I think this is what you’re looking for:
(LBS/2.205)9.81(FT/3.2808)*(1/4.1868) = kcal
This assumes perfect mechanical efficiency, however, which is not likely.
[quote]pat36 wrote:
Calorie, or kilocalories as the are accuratly called. What is the formula for calculating the calorie (as a unit of energy) for weight and distance the weight travels. I looked but could not find the info.
To be more clear, I want to calculate, for instance, how many kilocalories it takes to move 100 lbs, 2.5 feet, as in a sqat movement. Anyone, know?[/quote]
The “work” or energy it takes to move 100 lbs 2.5 feet is not the same as the kcals that your body must convert into force to apply the that 100 lbs. The work required to move a given object is constant and absolute. In this case it would take a minimum of 100 pounds of force to move that weight, assuming there is no friction.
On the other hand, the caloric expenditure to produce the necessary amount of work will vary. If one is extremely strong and conditioned it will take fewer kcals than if one is weak and untrained. There would probably be a minimum amount required and maybe a maximum, but I don’t believe there is a direct correlation between work produced and kcals combusted from person to person. Also remember that the body weight of the lifter must also be lifted, especially in the squat and deadlift.
TopSirloin
[quote]Chest Rockwell wrote:
I think this is what you’re looking for:
(LBS/2.205)9.81(1/3.2808)*(1/4.1868) = kcal
This assumes perfect mechanical efficiency, however, which is not likely.[/quote]
Where does the body weight of the lifter come into play??? Would you just add it to the total weight lifted?
TS
Substitute the your bodyweight into the LBS place and the distance your center of mass(around your hips) moves vertically for the FT part.
As the poster before me eluded, this is not going be the most accurate measure of how many calories you are expending when lifting.
Thanks Chest, that is exactly what I was looking for.
Ok, Chest. A couple of questions. Where does one substitute distance, or does this just calculate kinetic energy? Oh yea, I figured out 2.205 is for conversion in to kilograms, and 9.81 is the acceleration of grvity, but what do the other pieces of the formula stand for? Thanks you math wiz you.
Just as an aside, the human body is approximately 20% efficient in converting Calories (this is actually equivalent to kilocalories when capitalized) to work. Most is lost to heat. Don’t know if that plays into what you’re trying to find out.
-Dan
[quote]pat36 wrote:
Ok, Chest. A couple of questions. Where does one substitute distance, or does this just calculate kinetic energy? Oh yea, I figured out 2.205 is for conversion in to kilograms, and 9.81 is the acceleration of grvity, but what do the other pieces of the formula stand for? Thanks you math wiz you.[/quote]
I just edited my original post, I had forgotten to put the FT in for distance. Finally a use for my thermodynamics education.
Thanks again for the direct answer.
I am just using it as a base line for camparison between movements. I don’t expect to get a pure accurate result on what I am burning during a particular excercise.