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I've been piloting a drone yesterday and made a video hovering at around 100m, the visibility was very good. I've shown the video to a friend of mine and we were arguing whether it's the curvature of the Earth visible at horizon, or is it just camera lens disturbing the image.

I'm rather sure it's curvature of the Earth, but still don't know how to prove it. I already checked that on the lower part of the picture the buildings-roofs are lined up straight, no disturbance. Yet the horizon is slightly bent (checked by putting a piece of paper on the screen). Which I think already proves the point: should lens disturb the picture, the disturbance would be visible both on the upper and lower part of the picture symmetricly.

I've read How high must one be for the curvature of the earth to be visible to the eye?, but it doesn't help me just yet. As far as I understand, the above link help calculating the distance to the horizon depending on the altitude. But why is the horizon bent?

I would appreciate anyone helping me straighten my thinking :).

The picture: enter image description here

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  • $\begingroup$ The curve you're seeing is in all probability a distortion effect. You'll need a much better lens than the cheap ones in drones or mobiles. (Those are even designed to have a wide angle.) $\endgroup$ – Jan Doggen Oct 21 at 14:51
  • $\begingroup$ Curvature of earth shouldn't be visible from this height. You're almost certainly seeing distortion from your lens. Lens distortion will vary across the picture - though usually it will be symmetrical, so a straight ~horizontal line at the bottom should have similar (and opposite) distortion to one at the top. $\endgroup$ – Semidiurnal Simon Oct 21 at 17:28
  • $\begingroup$ Actually, the visibility may be very good for your locality (England, perhaps?), but in general it's not very good. Hereabouts (western US) it's not unusual to have visibility around 100 miles/160 km from the top of a mountain, or a light plane. And you can't see noticable curvature at those altitudes. $\endgroup$ – jamesqf Oct 22 at 3:09
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Make your own numbers... I will take one sample, just change your figures.

I have one 4K resolution dron camera. So I can register 3840 x 2160 pixels in one shot.

Considering the Earth as an spherical body, the distance from you to the horizon depends on Earth Radius and your height by:

Distance = (Radius + height) * Sinus {arc-cosinus [Radius / (Radius + height)]}

So depends on your height:

| Height (m) | Dist. To Hor. (Km) |
|:----------:|:------------------:|
|    1       |         3,6        |
|    10      |        11,3        |
|    100     |        35,7        |
|    1000    |        112,9       |
|    10000   |        357,1       |

My camera have a 24mm focal (on 35mm relative), so in the photograph, I can see 84º degrees. So the distance on between the edges of my horizon is:

| Height (m) | Dist. To Hor. (Km) | Dist. Edge<>Edge (km) |
|:----------:|:------------------:|:---------------------:|
| 1          |         3,6        |          4,8          |
| 10         |        11,3        |          15,1         |
| 100        |        35,7        |          47,8         |
| 1000       |        112,9       |         151,1         |
| 10000      |        357,1       |         477,9         |

Once that you have this data, you just need to calculate the arrow of the arc expected:

Arrow Circ. Arc = Radius * Cosinus [ Arc-Sinus (Dist/2/Radius)]

So, with this data and the initial data of the camera:

| Height (m) | Dist. To Hor. (Km) | Dist. Edge<>Edge (km) | Arrow (km) | Curvature (pixel) |
|:----------:|:------------------:|:---------------------:|------------|-------------------|
| 1          |         3,6        |          4,8          |    0,000   |         0         |
| 10         |        11,3        |          15,1         |    0,004   |         1         |
| 100        |        35,7        |          47,8         |    0,045   |         4         |
| 1000       |        112,9       |         151,1         |    0,448   |         11        |
| 10000      |        357,1       |         477,9         |    4,482   |         36        |

So finally... If you have a perfect visibility condition, your camera is completely leveled, there is no fish-eye distortion... On a 4K camera at 10 km height, the Earth curvature it will be 2% -> 36 pixels on 3840 pixels wide.

If you are interested on it, that is not the way. Human eye can not perceive clearly the Earth Sphericity at a commercial flight altitude. So forget your dron, yo should go for a balloon.

Hope it helps!

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  • $\begingroup$ Thank you for the great answer. Unfortunately I cannot upvote yet, as I need 15 reputation to do so :(. $\endgroup$ – mmm Oct 22 at 8:51
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A drone can't fly high enough to show the curvature of the Earth. When the Americans were testing German V2 rockets at White Sands in the late forties and early fifties, some reached altitudes of well over 100 miles, but even from that height the curvature of the horizon was barely perceptible. Whatever illusion of a curved horizon you may perceive in your drone photo is only what you would see from a high building.

If you can't get to the heights reached by V2 rockets, the best way to see the curvature of the Earth is to watch a tall ship approaching from over the horizon. On a clear day, the tips of the masts or funnels will be visible when most of the ship is hidden by the curvature of the Earth. Likewise when the sun goes down beyond the marine horizon and your town is in twilight, a high flying aircraft overhead will be illuminated by the sun that you can't see because the curvature of the Earth is in the way.

From a high building the horizon will appear to curve because it surrounds you in a 360 degree circle, and circles by definition are curved.

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