I was watching a video regarding Flat Earthers giving curvatures of Earth that sounded way too big, so I decided to calculate how much the surface of Earth would drop with respect to a line perpendicular to the ground per a certain distance from this point.

enter image description here

I'll use variables as defined in my awfully drawn diagram.

Let $s = 1km$.

Given that the radius of Earth is 6371km, we can find

$\theta = \frac sr = \frac {1km}{6371 km} = \frac {1}{6371} rad = 1.56*10^{-4} rad$

I think h could found by

$h = 6371\sin(\frac{\pi}{2}) - 6371\sin(\frac{\pi}{2}-1.56*10^{-4}) = 7.84*10^{-5} km$

which converted to meters would result in a curvature of $0.07\frac{m}{km}$.

EDIT Note that the formula is not linear, so in order to get the "fall" of Earth for 2 kilometres, multiplying the result by 2 is not enough.

The formula is $h = 6371-6371*\sin(\frac{\pi}{2} - s/6371)$ where $s$ is the distance to the object in kilometers.

Did I make any mistake? This value seems quite small and I'd like to make sure. Thanks in advanced.

  • $\begingroup$ Googling gives an estimate of 8 inches per mile = 0.13 meters per kilometer (see davidsenesac.com/Information/line_of_sight.html). So that's the right order of magnitude, though there's about a factor of two difference. $\endgroup$ Aug 18, 2016 at 15:31
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    $\begingroup$ Without having looked at this problem at all, I think it may have been the same calculation I did when also watching a flat-earth'er "doc", and as I recall it it wasn't actually the math that was completely off, but the assumptions: It was regarding that "you shouldn't be able to see the statue of liberty from a distance of 60 miles if the Earth had curvature" or something like that, and that would be true if not for the fact that you of course can't see the statue of liberty from 60 miles away at all! $\endgroup$ Aug 18, 2016 at 15:31
  • $\begingroup$ @Semiclassical i really can't see where the error is though, my logic seems valid $\endgroup$ Aug 18, 2016 at 15:35
  • $\begingroup$ @Lovsovs none of their assumptions are true, of course $\endgroup$ Aug 18, 2016 at 15:36
  • $\begingroup$ @Semiclassical maybe he is assuming the spectator has some height, I don't wish to take that into account. $\endgroup$ Aug 18, 2016 at 15:47

1 Answer 1


Your answer is correct.

See, e.g. the approximate formula given in the Wikipedia entry for horizon, which lists $d \approx 3.57 \sqrt{h}$. We see that a horizon of 1 kilometer is approximately corresponding to a height of

$$ (1 / 3.57)^2 = 0.0785 \text{ meters} $$

Incidentally, I would've used $\cos$ instead of $\sin$ when writing the formula. Then you get

$$ h = r * (1 - \cos \theta) \approx r \cdot \frac12 \cdot (\frac1r)^2 $$

where $r$ is the radius of the earth, and $\theta = 1/r$ is the angle in radians. The approximation uses Taylor expansion for $\cos\theta \approx 1 - \frac12 \theta^2 + \ldots$. So you get immediately

$$ h \approx 1/r $$

when $h$ and $r$ are given in the same units.

Let me give finally a remark concerning the discrepancy with this comment. The TL;DR is basically that the quantity $h$ is not linear as a function of the distance. Since the approximation is done by approximating the circle by a parabola, we actually have that the approximate height scales like the square of the distance to the horizon. For a quadratic relationship

$$ 0.0785 \text{ meters } * \frac{1.6092^2}{1^2} \approx 0.20 \text{ meters} \approx 8 \text{ inches } $$

we see that the 8 centimeter per kilometer estimate is actually compatible with the 8 inches per mile estimate.

  • $\begingroup$ Thank you very much, that is another way of solving the problem I guess. I used $\sin$ because I was perceiving the model as if the starting position was in the north pole. I could have used $\cos$ too, yeah. $\endgroup$ Aug 18, 2016 at 15:54
  • $\begingroup$ BTW, I've just checked with a procedure using Pythagoras, and setting the values of the calculation to what I specified, the result are identical, so yes, I am right. $\endgroup$ Aug 18, 2016 at 15:57
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    $\begingroup$ It looks like the difference between the result found above and that found via Google is a consequence of that horizon formula being nonlinear: An altitude of 0.0785 meters $\approx$ 3 inches gives a horizon distance of about 1km, whereas an altitude of 8 inches gives a horizon distance of 1 mile. (Morally, this suggests that units like "inches per mile" are probably not that useful.) $\endgroup$ Aug 18, 2016 at 16:09

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