Since Earth is a sphere, one has only a limited visibility radius. How far is that, actually?

This Q&A was inspired by this question, about whether or not Legolas can see the 24km distant Riders of Rohan.

  • $\begingroup$ I have up-voted both the question and the answer by M.Herzkamp, but I also think that answer is somewhat more complicated than it needs to be. Accordingly I've added my own answer. $\endgroup$ Commented Aug 18, 2014 at 15:23
  • $\begingroup$ Both answers need an additional factor of about two. You don't have to be able to see the orcs' toes, just the tops of their heads. $\endgroup$ Commented Aug 18, 2014 at 16:05
  • $\begingroup$ The factor of two is only right, if the Orcs are of the same height as you (or elevation). As I remember, Legolas stood on a hill, while the Orcs rode over the plains of Rohan. $\endgroup$
    – M.Herzkamp
    Commented Aug 19, 2014 at 6:34
  • $\begingroup$ @M.Herzkamp - that's why I said about two. $\endgroup$ Commented Aug 19, 2014 at 14:08
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    $\begingroup$ Maybe Middle Earth has a planar geometry. :-) $\endgroup$
    – peterh
    Commented Aug 20, 2014 at 11:45

4 Answers 4


I have up-voted the answer by M.Herzkamp, but I also think he makes it somewhat more complicated than it needs to be. The distance from the center of the earth to your eye is $r+h$, where $r$ is the radius of the earth and $h$ is the height of your eye above the ground. The distance from the center of the earth to a point on the horizon is $r$. The distance from your eye to the point on the horizon let us call $d$. The three sides of a right triangle are then the legs, $r$ and $d$, and the hypotenuse $r+h$. Applying the Pythagorean theorem, we have $$ r^2 + d^2 = (r+h)^2. $$ It follows that $$ d^2 = (r+h)^2 - r^2 $$ so $$ d=\sqrt{(r+h)^2-r^2}. $$ This admits simplifiction: $$ d=\sqrt{(r+h)^2-r^2} = \sqrt{(r^2+2rh+h^2) - r^2} = \sqrt{2rh + h^2}. $$ When $h$ is tiny compared to $r$, we can say $$ d \approx \sqrt{2rh\,{}}. $$

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    $\begingroup$ I might be worth clarifying that any objects beyond the horizon can of course be visible if they are high or tall enough - an object at height $H$ is visible to an oberver at height $h$ at distance $d \approx \sqrt{2rh} + \sqrt{2rH}$. $\endgroup$
    – JohannesD
    Commented Aug 18, 2014 at 16:08
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    $\begingroup$ @JohannesD I guess an object on the precise polar opposite of you, could be as high as you want, and you still couldn't see it. $\endgroup$
    – Cruncher
    Commented Aug 18, 2014 at 16:34
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    $\begingroup$ @Red_Shadow you also need to allow your vantage point to be arbitarily high. At any finite height the planet casts a shadow covering some angle. $\endgroup$
    – jxnh
    Commented Aug 18, 2014 at 19:11
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    $\begingroup$ @Red_Shadow Actually to see anything (except something perfectly opposite you) both you and the object must be arbitrarily tall/high. If your height is bounded, then you can only see things that are not "below" the infinite conic surface defined by your position and the horizon circle. $\endgroup$
    – JohannesD
    Commented Aug 18, 2014 at 19:20
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    $\begingroup$ A handy formula implementing the above result is $$d\,[{\rm km}]=3.56\cdot\sqrt{h\,[{\rm m}]}\ .$$ $\endgroup$ Commented Aug 19, 2014 at 18:03

Let us suppose, an observer of height $h$ stands on a perfectly spherical planet of radius $r$:


Edit: here is an easier way, making use of the right angle between the line of sight and the radial ray. You can just use the definition of the cosine:

$$ \cos(\theta_T) = \frac{r}{r+h} \qquad \Rightarrow \qquad s = r\cdot\theta_T = r\cdot\cos^{-1}\!\!\left(\frac{r}{r+h}\right) $$

which is equivalent to the solution obtained by the complicated method. /Edit

The distance $s$ to the farthest point he can then see is determined by the tangent to the semi circle through his head. If you describe the semi circle in a cartesian coordinate system by $$ y^2+x^2 = r^2, $$ the observer's head is at $y=r+h,\ x=0$.

To obtain the slope of the tangent, we plug the tangent equation $y=mx+r+h$ into the circle equation and solve for $x$: $$ x_{1/2} = -(r+h)\frac{m}{1+m^2} \pm \sqrt{\frac{(r+h)^2m^2}{(1+m^2)^2}+\frac{r^2-b^2}{1+m^2}} $$ Those are two intersection points, and in order to have a tangent, they must be equal. That is the case, if the term under the square root is zero. The resulting equation can be solved for $m$: $$ m_{\pm} = \pm \sqrt{\frac{(r+h)^2}{r^2}-1} $$ Let's take the negative solution for the tangent on the right (it does not matter), and calculate the tangent point: $$ x_T = -(r+h)\frac{m_-}{1+m^2_-} = \frac{r}{r+h}\sqrt{(r+h)^2-r^2} $$ The viewing distance angle is $\theta_T = \text{asin}(x_T)$. To get the viewing distance, we observe that $$ \frac{s}{2\pi r} = \frac{\theta_T}{\text{full angle}} = \frac{\theta_T}{2\pi}\text{, with angle in radian} $$ $$ \Rightarrow s(h) = r\cdot\text{asin}\left(\sqrt{1-\frac{r^2}{(r+h)^2}}\right) $$ If you plot this for $h$ small compared to $r$, it resembles a square root function, and indeed, $$ \lim_{h\rightarrow0^+}\frac{s(h)}{\sqrt{h}} = \sqrt{2r} $$ which means that for small heights, the viewing distance can be described as $$ s(h) \approx \sqrt{2rh} $$ On Earth ($r\approx6371\text{km}$), a normal person ($h\approx1.8\text{m}$) can see the surface about 4.8km away. Not much further. If you climb a hill or tree ($h\approx 50\text{m}$), your range increases to 25km!

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    $\begingroup$ I'd have called it the Pythagorean theorem rather than the "circle equation". ${}\qquad{}$ $\endgroup$ Commented Aug 18, 2014 at 15:15
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    $\begingroup$ Now we only need to know the height of an average orc ;-) $\endgroup$ Commented Aug 18, 2014 at 15:25
  • $\begingroup$ This approach, involving solving an equation for $x$, is unnecessarily complicated. I've shown a simpler way in my answer. $\endgroup$ Commented Aug 18, 2014 at 16:00
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    $\begingroup$ Note that this calculates the distance to the horizont where the completely smooth Earth begins to be in the way of itself. In order words, the distance where you can see the whole orc, including soles of his shoes. If you can live with seeing less of the orc, the distance is greater. $\endgroup$ Commented Aug 19, 2014 at 14:24
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    $\begingroup$ Holy cow this is freaking complicated. You should've thrown general relativity in there too! $\endgroup$
    – user541686
    Commented Aug 19, 2014 at 22:43

Atmospheric refraction cannot be neglected. As mentioned here the effect of this can be taken into account approximately by pretending as if the Earth's radius is larger by a factor of 7/6. This makes the distance $d$ to the horizon when the height $h$ is much less than the Earth's radius $R$ equal to

$$d = \sqrt{\frac{7}{3} R h}$$

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    $\begingroup$ Physicist invasion! Regardless of how true this is, it's a comment; not an answer. (Unless you provide an answer, and factor the effect due to refraction too) $\endgroup$
    – OJFord
    Commented Aug 18, 2014 at 19:19
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    $\begingroup$ I've updated the post. $\endgroup$ Commented Aug 18, 2014 at 19:36

If you go sailing, you'll have about 5 nautical miles of visibility (1nm = 1852m). I personally find the nomogram a delightful invention:

enter image description here

Simply draw a straight line between the height of the observer and the height of the object on the horizon (in this case = 0), then read off the geographical range.

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    $\begingroup$ Unfortunately the scales are very nearly too large for a 1.8m observer watching a 1.8m object. On the other hand, your chart appears to show about 5 nautical miles, so about 9.26km, which roughly matches M.Herzkamp's answer of 9.6km. $\endgroup$ Commented Aug 19, 2014 at 22:24
  • $\begingroup$ It's most odd that they don't expand the left and center scales (there's no reason for the left scale to be half the height of the right one, a simple manipulation of the determinant produces a bigger scale on the left and center axes, though perhaps it may have been done so that the most common calculations for intended purpose would result in nearly horizontal lines (trading off one source of error for another). $\endgroup$
    – Glen_b
    Commented Mar 12, 2017 at 0:38
  • $\begingroup$ @Glen_b That's a good point. The actual positioning of each line, and scale of the measurements would be easier to draw with a computer. I see that the left and right lines have the same scale. $\endgroup$
    – ahorn
    Commented Mar 12, 2017 at 6:38

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