This is how I calculated the width of the area used by TOP sight with "triangles", the picture should clarify my thoughts:
Let's imagine that you are a camera directly above a soldier, who is a single point in a 2D coordinate system. Soldiers horizontal FOV forms a triangle, which one angle is 91 degrees as the horizontal FOV in the game (phase 1 in the picture). If you draw a single line in the middle of the triangle, you get a right-angled triangle, which one angle is 45,5 degrees. From that you can calculate, that when triangle's bottom leg is 1x, then the other leg is ~1,02x (phase 2). So in practice, an object 1 meter away should be 1,02 meters wide to fill half of the horizontal width of the screen in the game.
TOP sight has 2,5 magnification, so the same object should fill same amount of screen when it's 2,5 times longer distance from the soldier. So now we can imagine a right-angled triangle with a bottom leg of 2,5x and the other leg same ~1,02x. So soldier's "half FOV" is 22,15 degrees (phase 3), and whole FOV is 44,3 when the same ~1,02x object fills a half of the screen.
Now we only have to calculate what is the width of the area seen from the TOP sight. When the FOV in that tank optic is 15 degrees, we again can make a right-angled triangle by drawing a line from the middle of it. So we get a right-angled triangle with one angle of 7,5 degrees. When the bottom leg is 2,5x (TOP sight magnification), the other leg is ~0,33x (phase 4).
Now what does that mean at the monitor? Well if ~1,02x is a half of monitor's width (960 pixels), then ~0,33 is 310,5 pixels, and that's only one half. So by that the aiming reticle should be 621 pixels wide (phase 5).
Things that worry me in these calculations are "did I calculate the problem in a right way" and "how those reticles are drawn on the screen". I've modded another game succesfully with these calculations, but the aiming reticles may be drawn in a different way in RO2.
And aaz, shouldn't you calculate the area covered by a reticle with the formula from the area of a circle? Pi times the square of the radius; π
r^2