Walking to the Horizon

I am subscribed to David Horvitz’s new project entitled IDEA SUBSCRIPTION in which he posts almost-daily simple instructions. Yesterday’s instructions read as follows:

CALCULATE HOW FAR THE HORIZON IS FROM YOU. THIS CAN BE DONE BY THE FOLLOWING: DETERMINE THE HEIGHT OF YOUR EYES FROM THE GROUND (IN FEET). MULTIPLY THIS BY 1.5. FIND THE SQUARE ROOT OF THIS NUMBER. THE FINAL NUMBER IS THE DISTANCE IN MILES TO THE HORIZON SPECIFICALLY FROM YOU. NOW WALK THIS EXACT DISTANCE. AS YOU DO THIS THINK ABOUT HOW YOU ARE WALKING THE ENTIRE RANGE OF WHAT IS VISIBLE FROM WHERE YOU STARTED. AT THE END TAKE ONE LAST STEP AND ENTER OVER THE HORIZON, INTO DISAPPEARANCE. NOW WALK BACK.

I do not profess to have spent much time researching this in the past, but I had never heard of this approximation before. The approximation is so concise that I was curious as to its error. The approximation is obviously incorrect for very tall heights since it is unbounded: $\lim_{h \rightarrow \infty} \sqrt{1.5 h} = \infty,$ vi&., in actuality an enormously tall person (whose eyes were almost an infinite distance away from the surface of the Earth) would only be able to see a quarter of the Earth’s circumference in front!

I therefore spent the last 5 minutes formalizing a bound on the error of this approximation. The results, which follow, were quite surprising.

First I had to find a more exact approximation for the distance one would have to walk to get to the horizon (making the assumption that the surface of the Earth is smooth and of uniform altitude). This turned out to be a relatively simple geometric exercise. Let’s say the point on the earth at which the person is standing is $p$ and the person’s eyes are $h$ feet above the ground. The radius of the Earth at point $p$ is $r$, therefore making his/her eyes $r+h$ feet away from the center of the Earth. Now imagine a ray emanating from the person’s eyes; that ray will be tangent to the surface of the Earth at the horizon. Let’s call that horizon point $q$. The distance from $q$ to the center of the Earth will be $r$. Since the ray from the eyes to $q$ is tangent to the Earth, it meets the radius of the Earth at a 90 degree angle. We can therefore draw a right triangle between the eyes, $q$, and the center of the Earth, with one side of length $r$ and the hypotenuse of length $r + h$. The angle between the hypotenuse and the side, $\alpha$, is therefore

$\alpha = \cos^{-1}\left(\frac{r}{r+h}\right).$ The ratio of $\alpha$ to 360 degrees should tell us the percentage of the Earth’s circumference that is visible: $d = \frac{\alpha}{2\pi} 2 \pi r = r \cos^{-1}\left(\frac{r}{r+h}\right).$ $d$ will therefore be a good approximation of the distance, in feet, to the horizon.

$d$ can be made even more accurate, however, by taking into account the fact that the radius of the Earth, $r$, is not a constant; in reality, the Earth is not perfectly spherical. The radius at the Equator is ~6378137 meters, whereas the radius at the poles is ~6356752.3 meters. In order to take this into account, we need to define the radius as a function of the latitude of the point at which the person is standing

$r : [-\pi, \pi] \rightarrow [6356752.3\mbox{m}, 6378137\mbox{m}].$ I approximated this thus: $r(lat) \mapsto \sqrt{\frac{(a^2\cos(lat))^2 + (b^2\sin(lat))^2}{(a\cos(lat))^2 + (b\sin(lat))^2}},$ where $a$ is the polar radius and $b$ is the equatorial radius (converted to feet, of course).

Substituting the new $r$ function into the original equation for $d$ gives the approximation up with which I ended. The error in the original (supposedly faulty) approximation is therefore

$\delta(h, lat) \mapsto \left|r(lat) \cos^{-1}\left(\frac{r(lat)}{r(lat)+h}\right) - f\sqrt{1.5h}\right|,$ where $f$ is the number of feet in a mile. It is important to note that this error asymptotically approaches positive infinity with respect to height; therefore, the taller the human the more error. We shall therefore obviously consider Bao Xishun (鲍喜顺), the tallest man alive. For simplicity, let’s say his eyes are 7.75 feet off of the ground. $\mathop{\arg\,\max}_{lat \in [-\pi, \pi]} \delta(7.75, lat) = 0,$ therefore, the most error will occur at the equator. The maximum error is $\delta(7.75, 0) \approx 21.96\ \mbox{feet}.$ This is very surprising to me: The relatively simple approximation has a worst case error of only ~22 feet!

The error is almost zero at latitudes of ± ~60 degrees (59.33623821 to be exact), so if you want the the simple approximation to work best you’ll need to go to Stockholm.

The entrance to Östermalms Saluhall would be a delicious place to start (just remember to walk either due East or due West).