# What Is the Power to Run Through Football Players — from “How To”

I’m continuing to work on explanations for the Randal Munroe book How To — Absurd Scientific Advice for Common Real-World Problems (Penguin Press). Basically, it’s a book that just has a ton of fun answering weird questions with some cool cartoons.

There is a chapter about scoring in an NFL football game. One idea is to just use a horse and literally plow through the opposing team. It’s just like a scene from Lord of the Rings with a horse running through a sea of orcs. Honestly, I can’t remember exactly which movie this was.

So, here’s the stuff from How To. First he uses the orcs as though they were a drag force with the following equation (for the magnitude of the drag force).

In this expression:

• ρ is the density of the medium. If you were moving through air, it would be the density of air. In this case, it’s the crowd density of orcs.
• A is the cross sectional area of the object. For a horse, this would be the front surface area.
• C is a drag coefficient that depends on the aerodynamic shape of the thing.
• v is the relative velocity of the object with respect to the medium.
• Oh, the 1/2 is the number 0.5. Ha.

Using that, he gets the following equation for the power needed to move through the orcs.

Finally, he uses some estimated values to get a power of 130 hp (which is more than the power from 1 horse).

OK, let’s go over these things.

Drag Force

I think a horse moving through orcs is a great way to visualize the drag force — it’s the same drag force your car experiences as it moves through the air. In both cases, there are collisions between things (orcs or molecules) and these collisions exert a net force on the object.

Here is a diagram to help us model this collision force. Instead of having the object move, I have it at rest with the orc-air balls moving towards it.

These orc-balls are spaced out with a distance s between them and moving with a velocity v-1 towards the object. When they hit it, they just stop. Yes, it’s very possible that they could shot off to the side or even at some angle — but for now, they just stop.

If the orc-ball has a mass of m, then when this orc-ball collides with the object there will be a change in momentum (where momentum is p = mv). According to the Momentum Principle, the net force on this orc-ball is equal to the time rate of change of the momentum.

Let’s use this for an orc-ball moving in the x-direction (so I will write the x-component of this equation).

If you are worried about the signs — remember that the orc-ball was initially moving in the negative x-direction. But anyway, what about that time? Suppose the orc-ball collides and then another orc-ball collides (the next one)? If I use the time between collisions for my Δt, then I will get the overall average impact force.

Since the orc-balls are moving with a velocity v-1 and are spaced out a distance of s, the following would be true.

Solving for Δt and substituting into the Momentum Principle equation, I get:

Now think of orc-balls in some 3D space. It’s a volume swept out by the object with a cross sectional area of A. I can define the orc-ball density as mass divided by volume. For this swept out area, the volume would be A*s.

Solving this for m/s and then substituting into the drag force:

OK, let’s switch back to the frame with the orc-balls at rest. That means the object is moving with some velocity — I’ll just call it v. Also, we can account for the different type of orc-ball deflections with a constant C — this is the drag coefficient. For a small drag coefficient, the orc-balls would sort of just slightly deflect to the side and not actually come to a stop. You could have very large C values for orc-balls that even bounce back. But that’s your drag force. Oh, what about that factor of 1/2? I think it just comes from a different definition of C. It’s all cool though.

Now, what about the power? Let’s suppose you have a horse running with some constant velocity v. In order to move at a constant velocity, the net force must be zero. That means that some forward pushing frictional force (from the horse’s feet interacting with the ground) must equal the backwards pushing orc-force. Notice that I’ve switched all the way to orc-collisions. No more balls.

During this time, the horse (yes, my horse might look like a cat — I’m cool with that) moves a distance of Δx. OK, let’s start with the definition of power — it’s the rate of work.

You can’t really calculate the work done by friction (because that would mean the ground is giving energy to the horse — that would be super silly). Actually, the horse is decreasing in internal energy. However, the important thing is that the change in energy of the horse would be “like” the work done by friction and the frictional force is equal to the drag force.

Since the forces are in the same direction as the displacement (Δx), then the work would be:

This would give a power of:

But check it out. Δx/Δt is just the velocity. Since there is a v² term in the drag force, this gives a v³ term in the power. Putting it all together:

Got it.

Now let’s check the numbers from the book. Here’s what Randall uses:

Clearly, I have a problem with the mixing of units. Imperial and metric units like wearing plaid and stripes. Well, who am I kidding — you can wear those together if you are feeling it. But still, imperial units are just bleh. But still, I can handle this. Let me make some comments:

• 1 orc per square meter — I’m cool with this.
• 200 pounds per orc. That seems fine. I think he is going to use the 200 lbs as a mass and not a force (but pounds is a unit of force).
• This 2.5 feet is throwing me off. I get that it’s the width of the horse, but I’m not sure how this works with all the units. Oh! I think I get it. Instead of orc volume density — he just uses orc area density. To compensate, he uses just the cross sectional length instead of the cross sectional area.
• 25 mph for the speed — got it.

Quick unit check. Suppose I use metric units. This is how they would combine.

That actually works. Think of energy as kinetic energy. The kinetic energy has units of kg*m²/s². Divide this by seconds can you get power in units watts. So, it checks. But notice that you have to convert lbs to kg — which is weird and technically wrong.

OK, let’s get to work. I’m going to do this in python because I like it that way. Here is my code.

Looks like it all checks out.

There is one small problem — and I will leave this for a homework assignment. What coefficient of friction would you need to have for the interaction between the horse’s feet (hooves) and the ground so that it could push forward through all these orcs?

Physics faculty, science blogger of all things geek. Technical Consultant for CBS MacGyver and MythBusters. Former WIRED blogger.

## Read everything from Rhett Allain — and more.

Upgrade to Medium membership to directly support independent writers and get unlimited access to everything on Medium.

Become a member

## More from Rhett Allain

Physics faculty, science blogger of all things geek. Technical Consultant for CBS MacGyver and MythBusters. Former WIRED blogger.

## The easiest explanation to the Monty Hall problem

Get the Medium app