What is the difference between F- stop and Aperture Value
Posted 29/01/2013 - 17:54
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I too am puzzled. Surely someone on here knows the answer.
Posted 29/01/2013 - 18:26
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Take as an example an F-stop of f13. The corresponding "aperture value" will read 7.4.
Why? Av is formally defined as 2 x log(base2) of the aperture. So 2 x ln(13)/ln(2) is ~7.40.
Why log(base2)? Because each whole number increase represents a doubling, i.e. one stop.
Why multiply by 2? Because the amount of light let in by a given f-stop is proportional to the aperture's area rather than its diameter, so the aperture squared is the key variable.
A fuller explanation is here:
http://en.wikipedia.org/wiki/APEX_system
Why? Av is formally defined as 2 x log(base2) of the aperture. So 2 x ln(13)/ln(2) is ~7.40.
Why log(base2)? Because each whole number increase represents a doubling, i.e. one stop.
Why multiply by 2? Because the amount of light let in by a given f-stop is proportional to the aperture's area rather than its diameter, so the aperture squared is the key variable.
A fuller explanation is here:
http://en.wikipedia.org/wiki/APEX_system
Posted 29/01/2013 - 18:50
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Many thanks. Now I understand - well, almost.
Posted 03/02/2013 - 12:40
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Quote:
Take as an example an F-stop of f13. The corresponding "aperture value" will read 7.4.
Why? Av is formally defined as 2 x log(base2) of the aperture. So 2 x ln(13)/ln(2) is ~7.40.
Why log(base2)? Because each whole number increase represents a doubling, i.e. one stop.
Thanks for that. it took me a while, but I think I understand it now.Take as an example an F-stop of f13. The corresponding "aperture value" will read 7.4.
Why? Av is formally defined as 2 x log(base2) of the aperture. So 2 x ln(13)/ln(2) is ~7.40.
Why log(base2)? Because each whole number increase represents a doubling, i.e. one stop.
Regards, Horst
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14 years
Melbourne,
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Why is this so?
Regards, Horst