The math would be: horizontal field of view=37° (normal 105mm)/100 (1° spot)<horizontal field of view=5°(tele 800mm)/6 2/3 (15° spot)? Correct?
Let me answer that myself: No, it is not correct, but it does begin to describe the question.
Oh, Im terrible at math, have always been.
We know that a hand held 1° spot meter (normal) horizontal field of view is 37°
When we divide that by 100=0.37°
We know that at some point (distance) the diameter of this spot becomes smaller than the 15° TTL spot as viewed through a telelens of X mm.
... I donīt know how to do it?
What I can do is: I can draw it, illustrate it by applying the horizontal fields of views of a 400mm 6X7 through to 1000° 6X7 (10° - 4°) and then measure when the spot is larger than the 1° spot.
Or I can skip all that and make a guess based on the 645=344mm ... ad 1/3=500mm 6X7
Or ask some one with a hand held 1° spot meter experience: When shooting with a super-tele does it make sense to use such a meter?
Last edited by jt_cph_dk; 01-31-2017 at 02:18 AM.