So I was building this city atop a mountain plateau with 1900 to 2300 meters high. That mountain rests on a small island, in the middle of the sea, near the continent.

More information about this city in here: Water supply on an mountain fortress

What I would like to ask is... how far from the continent can I put this island for the city to still be visible from land?

  • 2
    $\begingroup$ It depends on the details about your atmosphere as well as the size of your planet - can you add any details on these two topics? $\endgroup$
    – dot_Sp0T
    Nov 4 '16 at 23:17
  • $\begingroup$ Sorry. Same size and atmosphere as earth. $\endgroup$ Nov 4 '16 at 23:40
  • $\begingroup$ Let's imagine a clear, sunny day. $\endgroup$ Nov 4 '16 at 23:50
  • $\begingroup$ Depends on how high you are. From my own experience, on a really clear day it's possible to see Mt. Shasta (4,321 m) from the top of Mt. Rose (3,287 m), near Reno, Nevada, a distance of 298.2 km. Of course you can't make out detail at that distance. $\endgroup$
    – jamesqf
    Nov 5 '16 at 6:10

To expand a bit on @kingledion 's answer:

The angular resolution of the naked eye is about $\frac{\pi}{10800}$, or one arc minute.

You mentioned a civilisation on the level of ancient Rome, which had walls of around $10\left(m\right)$ high.

From basic trigonometry we know that $\tan{\frac{\pi}{10800}}=\frac{10\left(m\right)}{d\left(m\right)}$, so $d\left(m\right)=\frac{10\left(m\right)}{\tan{\frac{\pi}{10800}}}=34377.5\left(m\right)$, or more generally:

$$d\left(m\right)=\frac{h\left(m\right)}{\tan{\frac{\pi}{10800}}}=3437.75 h\left(m\right)$$

This means that on a clear day, the Maximum distance at which the average human eye can resolve an object of a $5\left(m\right)$ radius is around $34\left(km\right)$. However, at this range, the walls would probably just look like a thin line of different colour from the rock.

The formula scales linearly, so if the walls could be made out properly with 4 "pixels", the distance would be about $8.6\left(km\right)$

This isn't all though, after all, you mention that the city is on a $~2000\left(m\right)$ high plateau, this changes the formula. The angle at which we are looking is now $tan^{-1}{\left(s+\frac{h}{2}\right)}=\alpha=\tan^{-1}{\frac{2005\left(m\right)}{d\left(m\right)}}$

We need to multiply the distance with the cosine of this value. So:


Solving this numerically with Mathematica:

Solve[Cos[ArcTan[(s + h/2)/d]] h/Tan[Pi/10800] == d]

d -> 1/2 Sqrt[-h^2 - 4 h s - 4 s^2 + 4 h^2 Cot[\[Pi]/10800]^2]


With[{s = 2000, h = 10}, 
  Solve[Cos[ArcTan[(s + h/2)/d]] h/Tan[Pi/10800] == d]] // N

d -> 34318.9

Plot[{d, Cos[ArcTan[2005/d]] 10/Tan[Pi/10800]}, {d, 0, 35000}]

enter image description here

The distance for a variable height of the plateau (s) is givien by:


enter image description here

  • $\begingroup$ Ok, I'm sorry, but I don't follow. "Solving this numerically with Mathematica"... what's the answer? $\endgroup$ Nov 5 '16 at 18:46
  • $\begingroup$ 35 Km? Am I interpreting it correctly? $\endgroup$ Nov 5 '16 at 18:48
  • 2
    $\begingroup$ @PedroGabriel 34.3 km, call me a nerd, but I like calculating these things as precisely as possible. $\endgroup$
    – Feyre
    Nov 5 '16 at 20:42
  • $\begingroup$ OK, thank you. Hum... I know you posted the equations, but I'm having a little difficulty getting the hang of them, and I don't have Mathematica... could you please calculate for the range 1.900-2.300 m, instead of just 2.000 m? $\endgroup$ Nov 5 '16 at 23:44
  • $\begingroup$ @PedroGabriel I've added a generic formula, and a list of tables. $\endgroup$
    – Feyre
    Nov 6 '16 at 10:35

Distance to the horizon is calculated by $d \approx 3.57\sqrt{h}$ where $d$ is distance in km and $h$ is height of eye in meters. The same formula works in reverse, for an observer at sea level and an object at height $h$.

For 2000 m, $d \approx 3.57\sqrt{2000} = 160 km$.

Of course, you can't actually SEE anything at that point, everything in the city, including the city itself, is far too small without some powerful optics. And any sort of atmospheric disturbance, even high humidity will distort the picture and make things hard to see even with powerful optics. But on a perfectly bright, perfectly clear day, with a spectacular telescope, you could see the city 160 km.

Anecdote: Mt. Cameroon is very tall and very much right at the edge of the sea. I saw a sunrise over Mount Cameroon from about 50 miles off. This is what it would look like, to see your city from so far.

Sunset over a mountain

(I know this picture is sunset, but close enough)

Edit, regarding the human eye. According to wikipedia, the angular resolution is about 1 arcminute. This corresponds to 0.3m at 1km distance, for 30m at 100km distance.

Depending on what you mean by 'see', you could see that a well lit city was there at night from the full 160km (thanks @notstoreboughtdirt), and you could resolve a city size splotch of color during the day from that far away too. To actually tell that there are buildings and not just a blotch of color, you might have to resolve the ~5m distance between buildings which you could do at about 15km, based on the arc-minute resolution.

  • 1
    $\begingroup$ Thank you for your answer. My question concerns visibility with a naked eye. $\endgroup$ Nov 4 '16 at 23:52
  • $\begingroup$ With electric lights a city might be visible at night. $\endgroup$
    – user25818
    Nov 4 '16 at 23:53
  • $\begingroup$ The limiting factor for daytime visibility for something as large as a city is going to be atmospheric haze: 160 km visibility corresponds to exceptionally clear seeing conditions. $\endgroup$
    – Mark
    Nov 5 '16 at 0:35
  • 1
    $\begingroup$ Note the sun in the picture, which is 93,000,000 miles from the photographer. Part of distance visibility is (or can be) the amount of light being outputted by the observed object. If the sun were only 1 mile across but were still able to put out the same level of light/radiation, it would probably still be visible from that distance. As a further example stars are visible, and they don't have a very large apparent visible diameter either. $\endgroup$ Nov 6 '16 at 17:08

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