Calculating an angle to rotate some tangent around an axis

ok you have the exitAperture, which I'll assume is your starting angle/vector. And you have your target vector. To find the angle between these two vectors you can use the Vector2.Angle function (https://docs.unity3d.com/ScriptReference/Vector2.Angle.html)

angleBetweenApetureAndTarget = Vector2.Angle(exitApertureVector, targetVector)

Is this what you were looking for?

Not quite. I'm using that now which gives me angle Theta in the drawing. And that would work if I were pivoting at the exitAperture. But I'm pivoting at some other point r (radius) units away, leaving me the problem illustrated above.

Ah- I was assu$$anonymous$$g center was (0,0,0) So, it sounds like you want to know the angle between:

(exitApertureVector - center) (vector from center to exit aperture)

and

(targetVector - center) (vector from center to target)

Note, when center IS (0,0,0), we have the same results I assumed earlier.

OR, are you are looking for the angle between: the vector from aperture to target: targetVector - exitApertureVector and the vector from center to aperture: exitApertureVector - center (from your drawing it looks like this one is what you want, but I don't see why. I don't think this is the right angle to rotate the circle to make the aperture point at the target.)

I’m kinda lost, no clue what’s happening in the image above, but from the sound of it you are trying to find the $$anonymous$$imum angle a object (weapon) should rotate so that a point, previously out of it’s FOV, is visible. Am I correct?

Lordlycastle you are exactly correct.

Glurth, your first scenario is close (exitA - center to target - center), except I want the $$anonymous$$imum angle I need to rotate, as lordly points out.