Editor's Note: I thought this brief posting in the Nikon Digest deserved to be archived as a guide to calculating f/stop ratios - thanks!]

Date: Mon, 27 Apr 1998
From: Dave Read read@physics.utexas.edu
Subject: Re: Nikkor 85 mm f/1.4

Dave Read wrote:

>There's an easier way to calculate the difference: square the ratio of
>the two numbers.  For instance (1.4/1.8) = 0.778  0.778^2 = 0.60.
>Indeed, that's closer to 2/3 of a stop than 1/3.

Mea culpa, I left off a step in this calculation.  One you have the
squared ratio, if you want f/stops you need to convert the result
by computing the base-2 logarithm of the result.  You can do this
easily on most calculators with this formula:

log  (y)  = ln (y) / ln (x)
   x

In our case, log  (y) = ln (y) / ln (2)
                2

So let's re-work the f/1.4 - 1.8 question from yesterday:
(1.4/1.8)^2 = 0.7778^2 = 0.605
log (0.605) = ln (0.605) / ln (2.0) = -0.503 / 0.693 = -0.725
   2

So f/1.8 is actually 0.725 stops less light than f/1.4.  That's even
worse than the two-thirds-of-a-stop Art Searle mentioned (and I
erroneously confirmed) yesterday.

Thanks to William Allen for prompting me to correct my error.

Cheers (and sorry for the mistake),
Dave   

Date: Mon, 27 Apr 1998
From: Peter.Rongsted@iss-data.dk
Subject: Difference in f-stop

Dave Read, Ph.D. wrote:

<<>>

It may be easier, but it is not correct! (Art's way have the same flaw). Take two f-stops: f4 and f8. Your way will give: (4/8)^2 = 0.25 or (8/4)^2 = 4. The right answer is 2!

Remember the f-stop is a log-scale. The correct way to calculate the difference is:

log(f1) - log(f2)
- -----------------
  log(sqrt(2))

By using this formula you get a difference of 0.725 between f1.4 and f1.8 - closer to 3/4 than 2/3!

HTH,
Peter Rongsted

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