It is always cool to repeat classic experiments. How many of us know the radius of the Earth because we determined it on our own? I haven't done this, at least not yet. For all my life I have just trusted the values given to us. You know, if the ancient Greeks could calculate the radius of the Earth, why can't I?

If you are not familiar with Lake Pontchartrain, it is the huge lake just north of New Orleans. There are a few important features of this lake (which were useful to me):

There is the Lake Pontchartrain Causeway – essentially a 24 mile bridge that goes straight across the lake.

It isn't uncommon for the lake to be quite still without waves.

There is a very nice beach on the north shore. That is only important in that it put me in the right place.

So there I was, in the lake having fun. Here is what I noticed. These are two pictures. One taken just above the water and one at about eye-level above the water.

The first thing I noticed that was that the causeway at some point goes lower than the water. The location of this point where it goes lower depends on the height of the camera (or eye). Why? Curvature of the Earth. Also, you can see the bascule (draw bridge) sticks out even more and is farther away. I guess it should be pointed out that at this viewing location, the causeway is slanting away. The more to the left, the further away the bridge is. Here is a map.

The red arrow shows the location of the camera and the other yellow pin is the bascule. You can see the causeway as the north-south line.

But here is the real question. If I know where the causeway is, can I use this to calculate the radius of the Earth? That would be cool. But where to start? How about a diagram.

What is this mess? Well, this shows a side view of me and the bridge. Really. This shows the point at which the bridge goes below the horizon. Some notes:

h1 is the height of the camera above the surface (the water).

h2 is the height of the bridge above the water.

x 1 and x 2 are the distances from the objects to the visible horizon.

and x are the distances from the objects to the visible horizon. R is the radius of the Earth.

Let me make one assumption – that the arc length in the diagram (like s 1 ) is essentially the same length as the straight-line distance (x 1 ). Clearly, this is not absolutely true – but hopefully it will be true enough.

So, from the diagram, I get two giant right triangles. Using the pythagorean theorem, I get:

I hope you can see that the hypotenuse was the h 1 + R term. After that, I just simplified. Now, I can solve for R:

A quick check. This expression has the correct units. Also, if h 1 is larger than x 1 , I would get a negative number for the radius. That is ok, because if the height is greater than the distance to the horizon we wouldn't really even be doing this problem.

Next I can do the exact same thing for the triangle on the other side. This gives:

Remember that I don't actually know x 1 or x 2 . However, I do know that when they are added together they should give the distance to the bridge. Let me call this distance d so that:

I will now use this expression to remove x 1 from the first equation.

Using this and the other expression for R, I can solve for x 2 :

Sorry you had to see that, but it had to be done. This way, when I get some crazy answer at the end someone can figure out where I went wrong. At least the units seem to work. Oh, I didn't finish. But I did get it into a form so that I could use the quadratic equation. I am not even going to write out the solutions for x 2 .

Assume that I did. I could then use those to find R from the previous equation. I guess that one of the values of x 2 will give me an unrealistic answer.

Estimated Values —————-

So what do I need? First, I need d – the distance from where I was to where the bridge seems to go below the water. Although this is easy to pick out in real life, it isn't so easy in the video. Instead I will look at the bascule since I know where that is. It seems that the top of the bascule disappears when the camera is about 10 cm above the water. So there is h 1 . The distance from my location can be determined with Google Earth. With that, I get d = 11,400 meters.

What about the height of the bascule? The official Causeway site lists the clearance of the bascule at 45 feet. I assume this is for the case with the bridge down. So, maybe the top rail of the bascule at that point is perhaps 50 - 55 feet above the water (not counting the towers and stuff which are difficult to see). Let me just call this (h 2 ) 16 meters.

Ok, now that I have h 1 , h 2 and d I can calculate the possible values for x 2 .

The Results ———–

I know. This is what you have been waiting for. You really need to know how big the Earth is so you can plan your voyage from Europe to India. Ok. I calculated two different values for x 2 which gave two different values for R.

Not too bad. The accepted value for the radius of the Earth is around 6.38 x 106 meters. I think I could get a much better answer if I took more careful measurements.

Homework Questions:

Or things I could do with more time.

Using the accepted radius of the Earth, at what point on the photo 150 cm above the water should the causeway go below the water? You can assume the bridge to be 15 feet above the water.

Use shadows from Google Earth images to find the ratio of the normal roadway height to the height of the bascule. See image below.

Estimate values for the uncertainty in each starting quantity. Use this to obtain an uncertainty in the radius of the Earth.

Here is your extra picture. I cut an image from the flat part of the bridge and pasted it next to the highest part to help you.

See Also: