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Re: [OM] 50mm/55mm/500mm Lens questions -Reply

Subject: Re: [OM] 50mm/55mm/500mm Lens questions -Reply
From: "Ian A. Nichols" <I.A.Nichols@xxxxxxxxxxxxx>
Date: Fri, 18 Aug 2000 16:27:04 +0100 (BST)

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

-- 
________________________________________________________________________
*             |                                                        |
|  /  | |/-\  |                      Ian A. Nichols                    |
| |   | |   | |                                                        |
|  \-/| |  /  |                  i.a.nichols@xxxxxxxxxx                |
|             *                    iann@xxxxxxxxxxxxx                  |
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