Nature’s straightedge (with a penguin!)

(Not the punk subculture. The horizon.) A couple years ago I was looking out at the ocean and wondered how one might estimate the size of the earth given the distance to the horizon. I couldn’t do the math at the time, but the question came back to me and I decided to try again. Here’s a picture I drew to illustrate how the horizon “works”; you can see how the earth blocks itself from view. (I’m ignoring any atmospheric effects. Does light bend around the earth or something? I don’t know. Let’s pretend it doesn’t.)

When I try to solve a problem in physics or chemistry (this is neither, but I’m just saying), I like to think about it in two or three distinct phases: first you make the physical, geometric, or otherwise “real-world” observations. They should be independent from one another, and if you have too few then you won’t be able to get an answer. Next you translate these equations into math, and then it’s just a matter of algebra. For the first step, you can forget about the math and just look at the big picture, and for the third part, you can forget about the big picture and just look at the math.

The three observations are:

1. The line of sight is perpendicular to a line from the center of the earth to the point (x, y) at the surface indicated in the diagram (at the horizon).
2. The point mentioned above falls on a circle of radius R.
3. The distance to the horizon is the length of an arc with radius R and subtended angle θ.

These then become:

1. $-x/y=(y-R-h)/(x).$
2. $x^2 + y^2 = R^2,$ or $x=R\sin\theta$ and $y=R\cos\theta.$
3. $H=R\theta.$

Now we do math. First we take Eq. (1) and change it up to $x^2=yR+yh-y^2,$ then we move over the $y^2$ term and use the first of Eqs. (2) to get $R^2=y(R+h).$ Then use Eq. (3) and the third of Eqs. (2) to get $R/(R+h)=\cos (H/R).$

Here’s where I probably got stuck the first time I tried this a couple years ago, before I had heard of Taylor series. What we need is the fact that $\cos\theta = 1 - \theta^2/2! + \theta^4/4! - \theta^6/6! + \cdots.$ In this problem we can disregard all but the first two terms. After all, H/R is extremely small (it is the ratio of the distance to the horizon to the radius of the earth), so the higher-order terms are negligible. We want to keep the second term (even though it is extremely small itself) so that we still keep θ-dependence; if you were to try the approximation $\cos\theta\approx 1$ then you would not find a useful answer.

We are then left with the equation $R/(R+h)=1-(H/R)^2/2,$ which is a quadratic equation in R whose two solutions can be expressed as $R=(H^2/4H)(1\pm\sqrt{1+8(h/H)^2}).$ The root is greater than one, so we must choose the “plus” of the plus-or-minus in order to get a meaningful answer. (The earth is not inside-out.)

I would test this out, but I fear that data online is probably based on similar math, so it would be pretty circular. Maybe if I ever go sailing or something, I’ll watch a clock and note the speed. (I suppose a mathematically minded pirate would have calculated the size of the earth in knot-hours.)