Apparently, I am slow with my NFL posts. It would be nice if I could keep my football blog posts current with recent events in games. Alas, I will try harder in the future.

In this post, I would like to look at the Denver Bronco’s field goal kicker Matt Prater. In the game against the Bears, he had a 59 yard field goal to tie the game. I would embed the video of the kick, but the NFL in their infinite wisdom do not allow their videos to be embedded. So, you have two options. One, you can view the video at NFL.com. Two, you can watch my lego stop-motion recreation of the kick (thanks to my kids for their help).

So, there you have it.

Of course the question: is Prater awesome, or is it easier to kick long field goals in Denver? Does the change in the gravitational field have a big effect? Does the lower air density have a significant effect on the kick? That is what I will look at. Oh, quick note. There have been other long field goals that were not in Denver and I am sure that Prater is an awesome kicker.

## Modeling a Field Goal Kick

Not surprisingly, I have looked at the motion of a football with air resistance before. In that post, I considered a spiral thrown football. It seems that most field goal kicks tumble end-over-end. So, that will change the drag coefficient. Drag coefficient? What is that? Suppose I kick a football. While it is in the air, it will essentially have two forces acting on it. Here is a force diagram.

In that model for the air resistance, the force depends on the size and shape of the ball as well as the density of the air. The gravitational force depends on the gravitational field (*g*) and the mass of the ball. Let me go ahead and state my estimates for these quantities.

- The mass of the ball is 0.42 kg.
- The ball has a small circumference of 0.53 meters and a long circumference of 0.71 meters. I know this isn’t the most helpful dimensions, but it is all that I could find. For a spinning (end over end) ball, I will estimate a cross sectional area of 0.027 m
^{2}. - What about the drag coefficient? There seems to be no agreement here on an accepted value. Perhaps a value of 0.35 might be ok for a tumbling ball. Yes, that is mostly a total guess. However, this should be ok. I am just going to be looking at the change in range due to altitude change.

The rest of the parameters I will change to see how they effect the range.

## Starting Kick

Let me first model a plain field goal kick. Since the forces are not constant (air drag will change with the velocity), I will use a numerical calculation to plot the trajectory. Here is the path of a 30 m/s kick at a 30° angle.

This looks reasonable enough for me. But what angle and speed would you need to kick the ball to make a 59 yard field goal? Here is a plot of range vs. launch angle for a football kicked at 35 m/s.

For this speed, you would get the best range when kicking the ball at about 40° above the horizontal and it would just barely clear a goal post 67 meters away. That would be over a 70 yard field goal. Since you don’t see too many 70 yard field goals, I assume that it is difficult to kick the ball at this optimal angle when people are trying to block you. Or maybe my drag coefficient is off (though I did hear the announcers say that Prater was hitting some 70 yarders during warm up).

Either way, this model seems good enough for me to play with changing the density of air and the gravitational field.

## Variations in Gravity

The gravitational field (*g*) is not constant. Oh, let me go ahead and state that *g* should probably NOT be called “the acceleration due to gravity”. If a book is sitting at rest on a table, its weight (gravitational force) is the mass times *g* even though it is not accelerating. Yes, it is true that if you drop the book, it would have an acceleration of *g*. It is just a good idea to get into the habit of calling it the gravitational field and using units of Newtons/kg. Trust me on this one. It will help later.

Ok, rant off. Why would *g* change? Well, let’s look gravity. Near the surface of the Earth, we often use *mg* as the gravitational force. However, a better model says that the gravitational force decreases as the distance between objects involved increases. Here is a useful model for the magnitude of the gravitational force between two objects.

Here, *G* is the universal gravitational constant. It has a value of 6.67 x 10^{-11} N*m^{2}/kg^{2}. The two *m*’s are the masses of the two objects interacting and *r* is the distance between those objects (for spherical objects, you can use the distance between their centers). If we use this for objects on the Earth, one of the *m*’s is the mass of the Earth (5.97 x 10^{24} kg) and *r* would be about the radius of the Earth (6.38 x 10^{6} m). If you move farther from the center of the Earth, *r* increases.

If you want, you can write the gravitational force on the surface of the Earth like this:

What about Denver (in Colorado – just to be clear)? Denver is about 1600 meters above sea level. What would this do to the gravitational field? If I put in (R_{E} + 1600 m) for the *r* in the gravitational field, I get gravitational field that is just 99.9% the value of *g* at the surface.

The point is, the gravitational field and the gravitational force on the ball don’t change that much. Plus, there is something else to consider. The Earth isn’t a perfect sphere with perfectly spherically distributed density. It is “lumpy”. The gravitational field does not really change uniformly. If you are on a mountain, the higher density of that mountain will have an impact on the local gravitational field. Also, the Earth is rotating. If I want to take into account the rotational effects on the “effective” gravitational field (because we are not in a non-accelerating reference frame), that would be another contributor to variations in *g*.

Here is a gravitational variation map showing these differences.

This map was produced by the European Space Agency’s (ESA) GOCE satellite.

I guess I am trying to say that I don’t think I need to include changes in the gravitational field for this calculation.

## Changes in the Density of Air

Why is the air less dense in Denver than say New Orleans? Basically, it is because the air pressure is less. In Denver, there is less air above it pushing down. The result is lower density. But how much lower?

How do you model the density of the atmosphere anyway? It turns out that this isn’t so straight forward. I blame the weather. However, I have looked at this density stuff before. First, when I calculated the speed of a falling skydiver (in this case the jumper in question was to jump from 120,000 feet – the Red Bull Stratos project).

The model suggested for the density of air looks like this:

#image via https://en.wikipedia.org/wiki/Density_of_air#Altitude

With a mathematical model of:

You can refer to the wikipedia post for an explanation of the constants in that expression. Just for fun, here is a plot of the density from that model for the first 2500 meters.

This is basically linear in this range. Of course, there would be fluctuations due to weather (as I said before).

## Variations in Range Due to Density

Now I can make a plot. Here is a plot of the range of a football kick as a function of height above sea level (ignoring changes in gravitational field). I will keep the kick speed constant (35 m/s) with a launch angle of 40° (even though that might not be optimal).

From this, at an altitude of 1600 m, the range would be 61 meters (66.7 yards). The same kick at sea level would have a range of around 57 meters (62 yards). So, a gain of about 4-5 yards. I would suspect that this is a large enough effect for a kicker to notice. However, I also suspect that there are other more influential effects. Effects like wind, temperature and potentially blocking players.

Oh, and a quick hat tip to Andy (@rundquist and SuperFly Physics) for the suggestion of the lego video. Actually, he suggested I recreate the video with my kids. The lego thing was my idea.

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