You write:
>Scott J. Fine" <[EMAIL PROTECTED]> wrote:
>
>"I am under the impression that f-stops are measured in a scale that is
>either based on 1 or 1.4, as in:
>
>1, 1.4, 2, 2.8, 4.0, 5.6, 8, 11, 16, 22, 32..."
>
>It's actually based on the square root of 2 (1.414) and has to do with how
>the area of the aperture is calculated. The f stop is the ratio of the
>aperture diameter to the focal length. An f stop of 1.0 (theoretical) with a
>50mm lens at infinity would mean an aperture diameter of 50mm. If you wanted
>to halve the area of the aperture (to reduce the amount of light reaching
>the film by half), you wouldn't halve the diameter because the area of a
>circle is:
> pi x r**2. Halving the radius would result in an aperture one fourth the
>area (or 2 stops). Therefore, to halve the aperture area you divide the
>diameter (and therefore the radius) by the square root of 2. The new f stop
>becomes f/1.414 (f/1.4) as the diameter is 1/1.414 of the focal length.
>Divide by the square root of 2 again and you get an f stop of f/2 as the
>diameter is now 1/2 the focal length. Keep dividing by the square root of 2
>and you get the rest of the scale.
Please tell me if this is correct - I subtracted one stop from the next
largest, divided that number by 2 and added the result to the larger
aperture. I cam up with the following 1/2 stop scale.
1.414 (1.4)
1.706 (1.7)
1.999 (2)
2.413 (2.5)
2.827 (2.8)
3.412 (3.5)
3.997 (4)
4.824 (4.5)
5.652 (5.6)
6.822 (6.7)
7.992 (8)
9.646 (9.5)
11.301 (11)
13.640 (13.5)
15.980 (16)
19.288 (19)
22.596 (22)
30.273 (?)
31.951 (32)
And if I take the difference between apertures, multiply by .333 and add
that result to the larger aperture, I get 1/3 stops? As in f/9 being 1/3
of a stop smaller than f/8? ((11.301 - 7.992)(.333)) + 7.992 = 9.093 [f/9]
And thank you all for your answers, even the ones with links to information
that I have no hope of understanding :)
s/
http://www.eyecafe.net
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