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Calculating Calories Burned through Exercise

Date: 05/11/2004 at 13:00:18
From: Michael
Subject: Calorie Physics

I wanted to know how many Calories are required to perform a "Bench 
Press" and various other exercises.  Therefore, I figured out the 
formula and wanted someone to check it for accuracy.

Using Newton's Second Law, I found the difficult part is converting 
the values to Calories.  The values seem within reason but somehow I 
feel something is missing...

Problem:  How many Calories are required to complete a Bench Press 
using 3 sets of 5 repetitions at 205 lbs.?

Known values:  Weight = 205 lbs. or 92.986kg, Time (to complete one 
rep) = 2 sec., Distance = 24 inches or 0.610m (0.305 meters per 
second).

If Force (required to complete one rep) = 1kg * m/s2 = 1 Newton, then
92.986kq * 0.610m/2/2 = 14.18 Newtons.

However if I use the Fgrav = kg * 9.8N I get 911N which translates 
to W = F * d * Cos(deg) which is = 911N * (0.610/2) * Cos(0) = 277.9 J 
which converts to (3 sets of 5) = 995.8 Calories.  

This result seems high in my opinion...am I close or way off?



Date: 05/11/2004 at 15:56:53
From: Doctor Edwin
Subject: Re: Calorie Physics

Hi, Michael.

There are a couple of minor problems with the work you've done so far, 
but you're close.  First, the amount of time it takes you to move the 
weight has no effect on how much energy it uses.  I know, that's not 
how human muscles work, but that's how the physics of the situation 
works.  For example, to hold a 10kg weight for two hours takes a lot 
of calories.  But a two-by-four can do the same job without expending 
any energy at all.  So from a pure physics perspective, you can ignore 
the time it takes to move the weight.

The m/sec/sec factor actually is part of the measure of gravity, and 
is not related to the distance you moved the weight, or to the time 
you were moving it.  The formula you want is

energy =       force                 *  distance

       =     mass * gravity          *  distance

       = 92.986kg * 9.8m/sec/sec     * .61m

       = 555.870308 kg m^2 /sec^2
       
       = 555.870308 Joules

Your conversion to calories was correct.  I use a free units converter 
at www.convertit.com. 555.8703 J gives me 132.7673 calories.  Not bad 
for one rep!

However, dietary calories are actually kilocalories (whose idea was
THAT, anyway?).  So each rep is burning 0.132 calories.  Remember, 
that's the theoretical minimum energy any device could expend moving
that weight that distance.  The real situation has us spending more
energy.  Here are some other factors that need to be considered:

 - Googling around the web, I find that human muscle tissue is about 
24% energy efficient, so you can multiply the energy so far by a 
factor of 4.

 - Lifting at 80% of your RM (rested maximum? Real Man? Romanian 
meatloaf?) requires 12 times as much energy as lifting at 20%, partly 
because you are using more stabilizing muscles.  So you are only 1/3 
as efficient at high load.  Let's assume that the 24% represents the 
efficiency at some average load, and I know I don't routinely work at 
50% of my capacity, so let's assume that the average load is actually 
on the low side.  So if you're lifting 80% of your RM, then you can 
multiply the theoretical minimum expenditure by another 3.  (Sorry, I 
lost the URL for this piece of information.)  So we're up to about
.132 * 15 * 4 * 3 = 23 dietary calories to do 15 reps.

 - Even though the theoretical energy expenditure for moving the 
weight doesn't depend on the time, the real expenditure certainly 
does.  I have no way to account for this, but it's going to be a major 
factor.  Just holding the bar for two seconds will require you to 
expend calories.

 - Actually, the theoretical minimum energy for doing the bench press 
is zero, since the weight ends up where it started.  You COULD design 
a machine that stored the energy when it lowered the bar (maybe by 
compressing a gas cylinder) and then used that same energy to move it 
back up for the next rep.  However, again, bodies don't do that.  It 
requires almost as much energy to lower the bar back down as it does 
to pick it up.

I hope this helps!

- Doctor Edwin, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Physics/Chemistry

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