Expert: Steve Johnson - 4/25/2006

Question
Steve, your knowledge is incredible!  Most impressive.  Your answer was very helpful!

In my textbook, "Exercise Physiology: Energy, Nutrition, and Human Performance" by McArdle, Katch, and Katch (3rd edition), on page 808-809, it says for somebody my size (163 pounds), I would burn 10 kcal/min at 11:30 pace, 14.3 kcal/min at 9 min/mile pace (so 1 mile would be 128.7 kcal), 15.4 kcal/min at 8 min/mile pace, 16.8 kcal/min at 7 min/mile pace, etc.  

I hope that helps!

Thanks again!

Gratefully yours,
David


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Followup To
Question -
Hi Steve. I have a practical question. Yesterday, I ran in a race. I was running at a 9:00 min/mile pace northward fighting a 15 mph wind from the NE. I am curious to know how much extra energy I expended running into this wind? If all at once, the wind totally died down while expending the same amount of effort on my part (with the wind and w/o the wind), how much faster could I have run? I'm just curious! I studied Math through Calculus and had Physics many years ago, but I'm not sure how to solve this problem!

Thanks Steve!
David

Answer -
Hi David,

This is an interesting question. There will be many approximations and hedges in my response.

The wind resistance is called Drag, same as in discussions about aerodynamic streamlining of cars and planes. Drag is given the symbol D and the formula is
D = (1/2)* rho*(V^2)*A*Cd.
Rho is density of the fluid, air in this case, which is approximated in tables as 0.0024 slugs/ft^3.
V is the velocity, in ft/sec, relative to the fluid.
V = (1 mi / 9 minutes) + (15 mph)* cos 45 degrees.
V = ((1 mile * 5280 ft/1 mile) / (9 min * 60 sec/1 min)) + 15 mph * ((88 ft/sec)/60 mph)*.707
v = 9.8 ft/sec + 15.5 ft/sec = 25.3 ft/sec
A is your frontal area. Let's assume you're 6 feet tall and 1 foot wide as an average, for an area of about 6 ft^2.
Cd is the Drag Coefficient. Tables I found showed the 2006 Chevrolet Corvette Z06 with an 0.34 coefficient, a smooth brick with 2.1, and a bicyclist and his bicycle at 0.9. Let me use 1.0 for you.
So the Drag on you was (1/2)*0.0024*(25.3^2)*6*1.0 = 4.6 slugs*ft/sec^2 or 4.6 pounds.

If there had been no wind, your drag would have been .7 pounds. So the wind caused you to continually exert about 4 pounds of force. So how much work did the wind cause you to do?
Work, W = F*d = 4 lb * 5280 ft = about 21,000 ft lbs for every mile you ran.
Converting to calories, that's 21,000 ft lbs * 1 calorie/3.087 ft lbs = 6800 calories. Runnersworld.com says that if you weigh 160 lbs, you burn 100 calories per mile. (Other sources I saw said about 120.) Clearly something wrong here.

This is where I needed help. I was expecting perhaps a 30% increase in the calories required - this is almost 70 times as much. I consulted with Kevin Johnson, another of the experts, and he told me that the "calories" used in discussions in diet and fitness writing are kilocalories to a physical scientist. So we now have the 6800 calories you used to overcome the wind versus 100,000 total calories burned in a normal 1 mile run. A 7% increase. (Don't expect that we can calculate the different ways the 100 Kcalories are spent. Work fighting the wind is real work, but most calories are spent in ways that are harder to calculate. Much is spent in doing things like working your lungs. You generate a lot of heat. I think physiologists measured the 100 Kcalories by instrumenting a runner, measuring the oxygen he takes in.)

You would like to know how much faster you could run in calm wind and reach the finish line equally close to exhaustion. That would mean a calculation of what speed increase those 6800 calories/mile (that didn't have to go to fighting the wind) would power. I don't have the information to support an answer. I'm surprised that Runnersworld.com in their 'calories expended' guide gives it only according to weight of the runner. It seems that higher speed would yield a higher calorie consumption per mile. But their information doesn't indicate that. Neither does other sites I looked at. If I went to a track meet and asked someone who had just finished running in a 1 mile race if they were ready to do a marathon minus 1 mile run I wonder what the reaction would be. Anyway, this part is more for a physiologist to answer.

I hope I provided some insight.
Steve

Answer
Hello again David,

OK, that's some interesting data you have there. I noticed something interesting. You converted the data to yield the kcals burned per mile for the 9 minute mile rate. So I did the same for the other speeds and found that as speed increases, kcal consumption per mile decreases! Run faster and it's easier per mile?

That puzzled me until I realized that besides running during the elapsed time for a mile, they also continued to do all the routine bodily things that the people watching were also doing - digesting, filtering, etc. So 11.5 minute runners did 3.5 minutes more of that routine stuff during their mile than the 8 minute runner.

If you have kcal numbers for someone at rest, perhaps that could be subtracted out, leaving kcals going specifically to support running. And then maybe we could see how much you could speed up when the wind quits. If you have kcal/min for someone at rest, send it and we'll see where it can take us.

Steve