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Message: <blockquote data-quote="tdodd" data-source="post: 1534237" data-attributes="member: 55450">My understanding of FOV comes from binocular optics, where you might see the FOV described as 345 feet at 1000 yards for an 8X pair of bins - examples here - <a href="http://www.sportoptics.com/Swarovski-Pocket-Binoculars.aspx" target="_blank">http://www.sportoptics.com/Swarovski-Pocket-Binoculars.aspx</a>.AOV is pretty obviously an angular description. While the two are directly related, they are two very different ways of describing what is effectively the same thing.  If I've remembered my trig right, from 30+ years ago,FOV = 2 x tan(AOV/2) x subject distanceso it is possible to convert directly back and forth between the two methods of expressing the view, should you be so inclined.To my mind, it is far easier to visualise FOV than AOV. e.g. if the FOV is 345' at 1000 yds then it will be 34.5' at 100 yds and 3.45' at 10yds. Bins of 10X mag have a FOV of 285' per 1000yds, which in very roughly rounded terms is close to 1yd of width for every 10yds of distance, or 1m of width for every 10m of distance. That information easily allows me to judge whether those optics will be suitable for viewing birds, or anything else, of any given size at any given distance. If, on the other hand, you told me my optics had an AOV of 1 degree, 2 degrees, and so on, then I'd have no clue how useful those optics would be to me in practice, without looking up FOV equivalency.Of course, it follows that if you could memorise the FOV for a given focal length - say 100mm - when coupled to a specific size of sensor, then you could easily derive any other FOV for other focal lengths by simple multiplication/division. e.g. a 400mm lens would have an FOV 1/4 as wide, a 600mm lens would be 1/6th as wide, a 50mm lens would be 2X as wide etc..</blockquote>

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