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If we have an object at certain distance D in front of a camera. The object has a horizontal diameter d, giving us an angular diameter theta (2*arctan(d/2D)) of 50 degrees (converted from radians). And a camera of 100 FoV, horizontally.
Given those figures, does that mean that the horizontal image size of the object will be exactly 50 % of the image width?
This is an interesting question, as intuitively we would think that the image size of the object will be 50% of the image width. However, upon some playing around with triangles, I found this to not be the case.
For example, take a look at the diagram below. Here we have two overlapping triangles, one with angular diameter of 50 degrees (representing the object's angular diameter), and one with angular diameter 100 degrees (representing the camera's FoV). As we can see, the object has a width of 1 unit as seen from the camera, however the image width is around 3 units - therefore in this case the horizontal image size of the object will be around 33% of the image width.
If you were to play around with different values for D and d, you would find that the angular diameter ratios would vary. There would be a way to calculate this ratio directly with some trigonometry, however I do not know of how much use that would be to you.
Hope this helps in some way :)
