Earth Horizon Calculations

A friend recently questioned how we could calculate what would be visible above the horizon, given a known distance and height of the obstacles. Figuring this should be fairly simple, I’ve presented my calculations here for discussion – based on two towers of height h1 and h2, and a distance d apart.

If I make changes I’ll note the edit. Please feel free to question the maths if you think I’ve made a mistake – it’s late and I’m tired!


During the analysis I’ll be using the following constants/assumptions:

Light travels in straight lines from point to point, no atmospheric effects are accounted for. They may or may not be present, may or may not help the viewer see more or less of the h2 height. This may be refined later if I have time.

Mean earth radius, (R) straight out of Wikipedia from the “equatorial radius” –  6,378,137m.

Pi (π): defined as 3.141592654. No I’m not going further, that should be enough.

No rounding mid-calculation, only at the final answer – even tiny decimals are expected to have a large impact on the final answer due to the ratio of the scales involved (i.e. 1m height vs ~6 million metres earth radius) – a fraction of a degree off and you’ll be several tens of metres out.


And I’ll be using the following mathematical equations:

2πR = Circumference of Circle

Here’s my basic model of the earth – green sphere, grey R radius lines to any point on the surface. H being a special one, the radial line which intersects the horizon as determined by the tangent from h1 (i.e. the straight line view of someone stood at the top of h1. As you can see, the person can see anything above h3 at the second location, so anything h2 tall has h2h3 of it’s height visible.


t1 is the angle between the radial line at h2 and the radial line at the horizon. t2 is the angle between the radial lines at h1 and h2. d is the distance, over the surface of the sphere as the crow flies between the point h1 and h2 (specifically NOT the distance “through” the planet which would be a bit shorter).


Given a situation where we know two places are d miles apart on the earth’s surface,and knowing R from Wikipedia, we can see that if:

The full circumference (360 degrees) is C = 2πR, the distance between two points, d, will be given by


Which we can re-arrange to give:

t2=360x(d/2πR))                  (Equation 1)

Time to introduce a second bit of geometry – tangents.


So we can see that, taking one half of this pair of tangents, that we know h (we’ll use h1) and R. From this and Pythagoras, we can say:

Cos(t4) = R/(R+h1), so

t4 = cos^-1 (R/(R+h1))                             (Equation 2)

Jumping back to the first image I put up, you can see that t4 in this image is the same as t2 – t1 in that image, and that t1 = t2 – t4.

This tells us that the visible horizon distance, dh1, from h1 can be identified by the angle t4 using the original formula:

dh1=2πR(t4/360)                             (Equation 3)

We can identify the angle between the h1 horizon point and h2, we called this t1, as t4t2.

We can re-insert this angle into the tangent formula, in order to try to find h3. Because we know R and t1 we can use

cos (t1) = R/(R+h3) and so

h3 = (R/cos (t1)) – R                              (Equation 4)

This means that any part of the h2 tower at a height of h3 and above can be seen from the top of the tower at h1.

So, let’s do an example with some actual numbers and see how far it gets us:

Take this image as an example…

Two mountains, Mt Rainier and Brunswick Mountain, 193.77 miles apart as the crow flies – that’s 311,846m in metric. Rainier is a not-insignificant 4,394m high, while Brunswick is a slightly more tame 1,788m.


The suggestion was that surely at nearly 200 miles away, the mountain would be behind the horizon and so not visible. Well, let’s calculate it to see.

Remembering equation 1, we can say:


And so

t2 = 360x(311846/(2π*6,378,137))

t2 = 2.801360 degrees.

Remembering equation 2;

t4 = cos^-1 (R/(R+h1))

t4 = cos^-1 (6378137/(6378137+1788))

t4 = 1.356512 degrees.

And now using equation 3 we can identify how far away the horizon is from h1


dh1 = 2π * 6378137*(1.356512/360)

dh1 = 151006m – about half of the way to Mt Rainier. But that doesn’t mean we can’t see Mt Rainier, because it’s huge. Really huge. So we might be missing the botom of it, but we can use equation 4 to see at what height up Mt Rainier we can start to see again.

h3 = (R/cos (t1)) – R

h3 = (6378137/cos (t2-t4))- 6378137

h3 = (6378137/cos (2.801360 -1.356512))- 6378137

h3 = 2028.5m

So we can clearly see we can see the top 2365.5m of the mountain. Given the base of the mountain isn’t at sea level, it’s nearer 500m (see Ashford, WA town altitude) it’s visible protuberance is nearer 3894m, you’re seeing the top 60% of the mountain’s height.

 Let’s look at something a little closer to home and smaller, see how the numbers compare. Blackpool tower – at 158m. Legend has it that it can be seen from the Wirral – Hoylake beach for example. Precisely 46,958m away. The beachgoer with a scope might try to spot the tower from the beach. Let’s say the beach is at sea level and the beachgoer has the scope set at 1.5m height.

t2 = 360x(46985/(2π*6,378,137))

t2 = 0.422073 degrees

t4 is going to be tiny, due to the low observer height…

t4 = cos^-1 (6378137/(6378137+1.5))

t4 = 0.039295 degrees – phew, that’s tiny. And it translates to a distance over the ground of…

dh1 = 2π * 6378137*(0.039295/360)

dh1 = 4374.3m

Can we see any of the tower at Blackpool?

h3 = (6378137/cos (t2-t4))- 6378137

h3 = (6378137/cos(0.422073-0.039295))-6378137

h3 = 142m

So technically you should be able to see the top 16m of the tower, or this:


But we know we can see more than this. So what’s happening here? Well this is where refraction of light causes some complexities. Light doesn’t travel in straight paths – we’re not living in a vacuum. We know that when you pass light through a material with a different density, that light bends at the interface between the two materials – remember the old high school physics experimemt where you put a pencil in a glass of water and see it’s bent mid-way?


Ptolemy knew this in 150AD, though he didn’t understand why or how.

This happens in air too, and not just at distinct boundaries or layers – a smooth density change from sea level to some height will create a gentle curvature of the light path as it passes through materials of different densities. This tends to curve light in such a way that it more closely (not perfectly) follows the curve of the earth. This is known as atmospheric refraction, a heavily researched area – tough to model at larger scales due to the fact that air density isn’t consistent across even a few miles. Consider how hard it would be to get perfectly still air with a perfect gradient, over more than a few feet in a room. A decent description of this is given in these educational materials – I’ll use them as a basis of further calculations soon, but for now – have a read…

In order to demonstrate this principle, coolphysicsvideos on YouTube has created a video in which they produce a tank of fluid with varying density, similar to the way the earths atmosphere varies in density – they’ve used more readily available materials to make the gradient (a sugar water solution with varying sugar content with height).

Check out their video


This means, ultimately, that you can see below the geometric horizon, a little way. The further you are away, the more the light bends with distance, countering the curvature of the earth to some extent, but not completely.

So, has anyone got a real photo from Hoylake of Blackpool tower so we can see if the visible matches the calculated, and if refraction can account for the difference?

Ramblings of a mad man.