Can I see curvature of the Earth on a picture 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:
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&lt;&gt;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&lt;&gt;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!