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Message: <blockquote data-quote="pompadour" data-source="post: 2756049" data-attributes="member: 104665">Could someone please explain - thanks. Sometimes the apparent field of view, AFOV is calculated as the real field of view, RFOV x magnification. For the actual ranges of RFOV and magnificn. found in bins, commonly 4° to 9°, and 6x and more, respectively, how would this give a relevant result?From all I can find the stated x power in a bin is the linear magnificn. in the plane at right angles to the direction of view, i.e. it's the magnificn. of the real width (and of the real height) of the subject area in the image. It's not the magnificn. of its real angle. Is that correct?The ISO formula goes by that. In final, directly usable form it's a = 2 x arctan (m x tan 1/2 r) where a and r are the apparent and real angles and m the magnificn.The simpler a = r x m result is always, within the actual ranges mentioned above, significantly more.At first glance using this seemd to me to be a mistake made by some possibly because it works for small angles such as those within the actual RFOV range in bins: i.e. the width of the field at 8° is 2x the width of the field at 4°, etc. (tan varying close enough to linearly with the angle).But I find that is how Zeiss and apparently Swarovski calcualte the AFOVs stated in their websites (and elsewhere?) Seems the latter actually adjust this down to exclude distortion at the edge for each model separately! (I asked them about their figures through the query submit facilty in the website and they kindly explained.) Baffled.</blockquote>

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