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post #7 of (permalink) Old 03-30-2012, 07:21 PM
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Originally Posted by Doc7 View Post
Is this how the equation in the original post of this thread is meant to be used? It makes sense to use the "effective angle" for PAR (half the optic angle, so 20 for a 40, and 40 for a 60), but making sure that Hoppy did not already account for this in his equation?
I used the exact formula in the original post by Hoppy, which I assume indicates the use of the "half" angle of a lens. I may have misinterpreted the formula, but from the examples that I saw, I thought the angle was suppose to be half of the optics.
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