What is there to say about the 1,500-meter run? I guess I could say the distance is 1,500 meters — well, about 1,500 meters. The cool thing about the long-distance events is the runners need not stay in their lanes. They tend to run in a group. And this raises the question: When you are in a lane with other runners, does it help to be behind someone, drafting as bike racers often do?

First, let me start with speed. How fast do these runners go? The current record, set by Hicham El Guerroug of Morocco in 1998, is 3:26.00. The next nine fastest runners are within four seconds of that. So figure three and a half minutes. If I use a time of 210 seconds and a distance of 1,500 meters, we get an average speed of:

That is almost 16 mph. Wow. That is faster than most people ride a bike. I guess there is a reason I am not an Olympic athlete. OK, but what about the power needed to run like this? You could look at three things a person needs to use energy on during the run:

Breathing and other bodily functions. Think of this as “overhead”.

Moving the legs. This is probably where most of the energy goes. The runner has to repeatedly increase and decrease the speed of the legs to get this whole running thing to work. (QWOP anyone? )

Air resistance. As the runner moves, he or she has to push through the air. This isn’t free.

I guess I should say something about difference between power and energy. Please don’t think I am using these terms interchangeably. In short, power is the rate that energy is used.

Depending on the situation, I can use the change in energy or the work done on an object. In the case of the 1,500-meter run, it might make the most sense to look at power instead of energy. So, how much power would the runner have to use simply to account for air drag? Here is a force diagram for a person running at a constant speed.

Illustration: Simon Lutrin

Since the runner is moving at a constant speed, the magnitude of the frictional force must be the same as the magnitude of the air resistance. For the air resistance, I will use the common model that has the magnitude:

Here, ρ is the density of the air. A is the cross-sectional area of the object moving and C is a drag coefficient that depends on the shape. So a sphere would have a different drag coefficient than a similarly sized plate. I can get a value for the density of air — 1.2 kg/m3 is a pretty good estimate on the Earth’s surface. What about the other values?

I don’t need to find A and C separately. Instead, I can find the product of A*C. How can I get the area of a runner? How can I get the drag coefficient? Essentially, I can’t. When you can’t get what you want, you do something else. In this case I can look at something close — a skydiver.

A person falling from a plane isn’t so different than a person running, right? Oh sure, they would be going much faster but they are at least close to the same size and shape as a runner. The key point is I can find the product of A*C for a skydiver if I know the terminal speed. At terminal speed, the skydiver falls at a constant velocity. This means the gravitational force has the same magnitude as the air resistance force. In case it wasn’t clear, the air resistance force on a skydiver is in the opposite direction as the velocity (and thus gravity).

Assuming a skydiver falls at about 120 mph (about 54 m/s), then I can find AC.

If I use a mass of 70 kg and a value of 9.8 Newtons per kg for g, then AC would have a value of 0.39 m2. The point is I can now calculate the air resistance force on my 1,500m runner. Oh, I think you could argue that the AC value for a runner could be higher or lower than the skydiver. With that in mind, I will just use the same value. So, if a runner is going 7.1 m/s, then:

Now we have something to work with. What about the work needed to push against this air resistance? What about the power? Suppose the runner moves a distance s. During this time, the air resistance would do work on the runner (negative work since the force is in the opposite direction the runner is moving). The runner would need to do work just because of this air resistance. The amount would be:

Remember, this is just the part of the total work that deals with the air resistance. For power, I just need to divide this work by how long it took to move this distance s. Of course, the distance over the change in time is the speed.

I know the magnitude of the air resistance and I know the speed. This gives a power of 84 Watts. That is enough to power a lightbulb, but the total power is going to be on the order of 500 Watts. This would indicate that the air resistance is significant.

Now, what about the drafting? I know, I know. You’re thinking, “Finally.” Let’s say that one runner is running behind another. What does this do? Well, the first runner pushes the air some. How much? This is tough to say. Suppose that by running behind another person could reduce the effective air speed by maybe 1 m/s. In this case, the air force would also decrease from 11.8 Newtons to 8.7 Newtons. The power would to compensate for the air drag would be 62 watts. Not too bad. Reducing output by 20 watts could help with a sprint at the end. It’s not much of a savings, but this is the Olympics we are talking about here, and every bit matters in an event where less than four seconds separates the 10th fastest time ever from the fastest time ever.