On Fri, 18 Aug 2000, David Irisarri Vila wrote:
> Hi Scott,
Scott? who's this Scott person? my name is Ian, and those are my
initials too!.
> I´d very pleased if you could explain me how do you
> calculate this.
> I know that f2 goes to 1,414 and then to 0,9999=1
> Now lets calculate intermediate stops!!!
> f2 goes to f1,80466 (The famosus Zuiko f1,8) and then to f1,60933
> and then f1,414 and f1,27614 and f1.13807 and finally f1.00000!!!
> So I think f 1,27614 is one 1/3 stop faster than Zuiko f1,4
> isn´t it?
Let's define scale of sequentially numbered f-stops like this
Stop number f/
0 1
1 1.4
2 2
3 2.8
4 4
.... and so on. The f-numbers are a geometric progression with a common
ratio of sqrt(2) (sqrt being short for "the square root of", as used in
most spreadsheets).
Therefore
f=sqrt(2)^s
(f is f-number, s is stop number, "^" means raised-to-the-power-of)
with a bit of mathematics, which I'll leave as an exercise for the
reader, it can be shown that
s=log10(f^2)/log10(2)
actually, the base of the logarithms doesn't matter, since converting
logarithms between bases requires only multiplication by a constant.
putting f=1.2 into this formula gives 0.53, i.e. 0.47 stops brighter
than f/1.4
--
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