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/
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