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For example, I want to measure the distance between two points on a Tennis ball from one picture, assuming it is a perfect sphere and a given radius. I acknowledge there is a post Great arc distance between two points on a unit sphere making use of great-circle distance but I don't quite understand how to apply it here.

sonic12
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    The exact details of the calculation will necessarily depend a lot on the exact form those points are given in. Is it a longitude and latitude on the ball? Is it 3D Cartesian coordinates? Something else? – Arthur Feb 13 '23 at 11:57
  • Since you talk about the picture, I assume you can identify the points by their distances from the sides of the picture. But the problem comes in figuring out how far the focus of the camera is from the ball and from the plane of the picture. You might be able to determine this (at least, the necessary ratios) using the seam pattern, but on a featureless ball, there would be no way to tell. You can just assume the camera is far enough away to pretend the picture is an orthogonal projection of the ball, but that will always entail some error. – Paul Sinclair Feb 14 '23 at 19:15
  • If I only want to find the arc distance on the image, i.e. in terms of pixels, can it be calculated? Assuming the picture is taken from a very far distance. – sonic12 Feb 15 '23 at 11:27
  • Sorry, but I don't quite understand what you mean by that. "arc distance" is on the ball, not on the image, and "pixels" exist on the image, not the ball. The image is a flat plane, there is no arc. To find that distance, just use the Euclidean formula $\sqrt{(x_1 - x_2)^2 + (y_1-y_2)^2}$ where $(x_1, y_1)$ and $(x_2, y_2)$ are the pixel addresses of your two endpoints. Either that, or you want $\max{|x_1 - x_2|, |y_1 - y_2|}$, depending on exactly what you mean by "distance in terms of pixels". – Paul Sinclair Feb 15 '23 at 21:21

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