## February 01, 2012

I hate running.  But, I've learned that there are few ways to exercise which are easier and cheaper than running.  I mean, all you really need to run are a decent pair of shoes and a place to run, right?  (Barefoot runners need even less.)  Most of the time when I run (and, trust me, I don't run, I jog slowly) I am on a treadmill.  Knowing physics can be useful when you're on a treadmill.  I can pace myself and try to speed up or slow down as much as I think I need to when I want to.  But knowing physics can also be a curse when on the treadmill, since I can calculate how much longer I'll be running when I don't really want to be there in the first place.

To try to keep my mind off the running, I'm often staring at all the displays on the treadmill. One of the displays keeps track of how many calories (really kilocalorie, which I abbreviate as kCal) I have burned in my workout. The treadmill displays how many calories I can burn in an hour at whatever speed I'm going at.

One day last week when I was on the treadmill I started to think about how that rate at which I burn calories (really my power) should vary with speed.  It seems obvious that the faster I run, the more energy per hour will be required to maintain that higher speed.  But how does it vary with speed?  I made a naive guess that it would be proportional to my average kinetic energy:

$$KE = {1 \over 2} m v^2$$

$$P = {KE \over t}$$

So if my speed doubles, I would expect the rate at which I convert the stored internal energy into the energy used move at that speed to increase by a factor of four. When I asked on twitter and facebook last week what people thought about how power varies with speed on a treadmill, I only got one response from someone who said "I respectfully disagree with you." Time for some data.  I turned on the treadmill and started recording the rate of calories burned at each speed from 0.1 miles per hour to just over 7 miles per hour.  Here is that graph:

So, as you can see, the graph has two distinct regions, which are both linear.  The discontinuity occurs at around 3.7 mph, roughly where it becomes too difficult to maintain a walk, and I have to switch to a jog to stay on the machine.

I'm a little surprised by this graph. In case you're wondering, I did not enter in my weight or height or age into the treadmill.  I really don't even care if the power values are correct right now.  I'm was puzzled by why the trend is so linear.  Let's go back to the definition of power, I said power was related to the kinetic energy.  It would have been more correct to say power is related to the work I am doing:

$$P = {W \over t}$$

Since I assumed that all of my work went into kinetic energy, I don't feel too bad about the earlier statement.  Work is $\vec{F} \cdot \vec{d}$ or just $Fd$ if the force and displacement are in the same direction, so power is:

$$P = {Fd \over t}$$

But $d \over t$ is my speed!  Oh, so I should write my power as:

$$P = Fv$$

where $v$ is my speed.  Oh, this is linear.  I guess the treadmill designer got it right.  Bummer, I was really hoping to burn more calories by just increasing the speed a little bit.