# Geometry on the High Seas and Elsewhere

Last week I went on a short cruise with my wife. We’ve done this kind of vacation several times now, and at I kind of hate to admit that I’m starting to really like it. Nothing about being stuck in a floating hotel with thousands of loud people really seemed that appealing on paper. I’m really more of a fall/winter person. But there’s something refreshing about blasting Led Zeppelin’s Immigrant Song through my headphones while leaning on the rail of a balcony and taking in the waves in the North Atlantic. (We went to Canada once). Also, you need to supply your own audio for that link to have full emotional effect.

During this last sailing, staring out at the emptiness of the open ocean, I began to wonder how far away the horizon was. Most ships aim to keep you at least 12 nautical miles (nearly 14 miles, or 22.2 km) from shore for as long as possible so the casino and duty-free store can be (legally) open for business. Depending on where you’re going, you might be substantially farther from land on any given sea day. What governs how far you can see is mainly your height above sea level, though any number of more complicated factors are going to influence the visibility. Let’s assume the water is quite calm –basically a spherical surface and neglect any atmospheric effects. This will be just a “best case scenario” estimate. Everything that follows is based on some very straight-forward trigonometry –simple enough to work out on a cocktail napkin.

Imagine you’re standing at some height $h$ above the water level. Your horizon is the point on the surface at which light would have to travel tangent to the surface at the horizon in order to make it to your eye by traveling in a straight path. Beyond this point, light would have to go through the Earth to make it to you without somehow deflecting, so you wouldn’t see these sources. You can even make a pretty diagram labeling everything clearly:

Here $R$ is the radius of the Earth, and we’ll take the (curved) arc length $s$ to be the distance of the horizon. In practice, you won’t often find yourself high enough that $s$ differs appreciably from the straight line connecting you with the horizon. This is not the “cruise ship approximation” –more of the “any situation that doesn’t put you in outer space” approximation, but more on this below. Since the path connecting you to the horizon is tangent to the surface at the horizon, it will form a right angle with the line going straight to Earth’s center (of length $R$). So now we have a right triangle and can do some good, old-fashioned trigonometry. By the very definition of a radian, we have the result that if $\theta$ is measured in radians,

$s = \theta R.$

From the right triangle we can say

$\displaystyle \cos\theta = \frac{R}{R+h} = 1-\frac{h}{R+h}.$

We could proceed exactly, but since the radius of Earth is a whopping 6400 km (approximately), we can safely take $h \ll R$ and simplify things a bit.

$\displaystyle \cos\theta \approx 1 - \frac{h}{R}.$

This doesn’t look like a big help, but if we appeal to the small-angle approximation (leading-order Taylor expansion) for cosine $\cos\theta \approx 1 - \frac{\theta^{2}}{2}$ we can identify

$\displaystyle \theta \approx \sqrt{2h/R},$

which gives $s \approx \sqrt{2hR}$. The Royal Caribbean Quantum Class ships have a North Star viewing platform which reportedly swings you up to 300 ft $\approx$ 90 m above sea level. Putting this in, your horizon should be at a distance of about

$\displaystyle s \approx \sqrt{2(90\mbox{ m})(6370\times 10^{3}\mbox{ m})} \approx 34\mbox{ km},$

That’s about 21 miles, which seems pretty far. But is it worth it? On a sea day, you’re likely at least this far from land, so you aren’t going to see much aside from some other scattered ships. How does it compare to the view on the top deck (or the best view from a ship without the beloved North Star)? Looking through photos online (scroll about halfway down), a really crude estimate places the actual top deck height at about half this height, or about 150 ft. If $h\rightarrow h/2$, the horizon distance is reduced by a factor of $\frac{1}{\sqrt{2}}$, or

$\displaystyle h\rightarrow \frac{34 \mbox{ km}}{\sqrt{2}} \approx 24\mbox{ km}.$

For comparison, if you’re of average height of 1.7 m and just standing at sea level, you can see as far as

$\displaystyle s \approx \sqrt{2(1.7\mbox{ m})(6370\times 10^{3}\mbox{ m})} \approx 4.7\mbox{ km},$

or almost three miles. One might object to the approximations, but we can compare this to the exact result (not using the small-angle approximation)

$\displaystyle s_{exact} = R\cos^{-1}\frac{R}{R+h}.$

Plotting both of these on the same axes, a visual assessment suggests it turns out to be a pretty good approximation for heights $h<0.1R$.

Looking more carefully at the difference between the two predictions, the small-angle approximation overshoots the estimate by about 5% when $h=0.1R$, so this is a sensible upper bound. A height of 10% of Earth’s radius is about 640 km, which is sixteen times the height from which Felix Baumgartner jumped to break the sound barrier in 2012 –so it’s a good approximation even for his height of almost 40 km where the horizon should have appear about 714 km away (444 miles). Also, if you knew (or could estimate) the height of some reference object, you could run this analysis sideways and actually measure the radius of the Earth.

As soon as I scribbled all of this down, I thought of another consistency test which might ring a bell for anyone familiar with Grandfather Mountain in North Carolina. The “Mile High Swinging Bridge” is one of the big western North Carolina attractions. One photograph in the Visitor Center (or maybe it’s in the Nature Museum) that really struck me as a kid: a photo –by Hugh Morton, himself– that captured the skyline of Charlotte, about 82 miles away “as the crow flies.” For years I tried squinting until I saw Charlotte, but it never worked. Granted, I was always there a warm summer afternoon and didn’t have a fancy telephoto lens (photo was reportedly from a cold December morning). If one stood a mile above sea level, the actual horizon should be about

$\displaystyle s \approx \sqrt{2(1\mbox{ mi})(3960\mbox{ mi})} \approx 89\mbox{ mi},$

which just happens to safely include downtown Charlotte. The city actually has an elevation of about 750 ft, and portions of those buildings would be visible even further than 89 miles away (we’re computing the $s$ based on an object at sea-level one can just barely see), but these numbers all being roughly similar is encouraging. Something I wasn’t aware of until trying to dig up a link to that Charlotte photo is that people have reportedly been able to capture the tops of buildings in Winston-Salem from the same location. Winston-Salem is closer to 90 miles away (and a little higher in elevation than Charlotte), pushing the limits of what should be visible. Again, portions of those skyscrapers would be visible even farther away, but working out how far an object of height $H$ can be seen from an elevation of $h$ is a more-involved calculation (also carefully discussed here). In a quick-and-dirty version, assuming Earth’s radius is much larger than either your elevation or the height of what you see past the horizon, you just extend that ray of light that grazes the horizon back to its point of origin on some tall object.

Now the total distance $s = s_{1} + s_{2}$ is given by the sum of two parts. The analysis above carries over for the second part, and the largest distance from which you can just barely make out the top of a building of height $H$ is

$s \approx \sqrt{2hR} + \sqrt{2HR}.$

The tallest building in Winston-Salem is about $490\mbox{ ft} \approx 0.087 \mbox{ miles}$, so standing a mile above sea level, you could expect to see the top at a distance of

$s \approx 89\mbox{ mi} + \sqrt{2(3960\mbox{ mi})(.087\mbox{ mi})} \approx 115\mbox{ mi}$.

That’s enough buffer room to make it seem plausible that it was, in fact, (parts of) the Winston-Salem buildings in the distance.