[Leica] OT : Stop fractions - how long and tedious can you get?......very

["Garrelfs, R (Rick)" <Rick.Garrelfs@utc.rabobank.com>]

> 
> Marc - I'm curious...How does one calculate these "stop fractions?"
> 
> B. D.
> 

I have just thought about it a bit, and I think it is a little easier than
we might think. As someone already stated, it is a logaritm thing, and we
don't need to calculate any actual lens diameters. Here is what I think may
work:

we know that subsequent f-stop numbers are calculated by multiplying the
previous number by the square root (sqrt) of 2. That has to do with the
pi*r*r that has been mentioned before, and that is how we get the f-stop
number sequence:
1.0; 1.0*(sqrt 2)= 1.414; 1.414*(sqrt 2)= 2; 2*(sqrt 2)= 2.818.....etc.

or, for instance: 2.8xxxx = 1.0 * (sqrt 2) * (sqrt 2) * (sqrt 2) = 1.0 *
(sqrt 2)^3
(^3 meaning: to the 3rd order).

a difference of 3 f-stops equals: the original f-stop * (sqrt 2)^3
a difference of n f-stops equals: the original f-stop * (sqrt 2)^n

example: 16 = (sqrt 2)^3 * 5.6, or: a difference of 3 stops

so, if one f-stop is Y, the other X:  Y = (sqrt 2)^N * X

The thing is: we want to know N for every for every Y and X we can think of.
So, fiddling around a bit, we can get (and here comes the logaritm  (Ln is
natural logaritm)):

Y = (sqrt 2)^N * X
Y/X = (sqrt 2)^N

Ln(Y/X) = Ln{(sqrt 2)^N} 
Ln(Y) - Ln(X) = N * Ln(sqrt 2)


N = {Ln(Y) - Ln(X)} / Ln(sqrt 2)

Tataa. Flash out the pocket calculator, and:

What is the difference between a 1.4 summilux and a 1.5 Noktor?

Y = 1.5
X = 1.4

Ln(1.5) - Ln(1.4) = 0.06899;  Ln (sqrt 2) = 0,34657

difference is:   0.06899/0,34657= 0,19906, or about 0,2 f-stop.


Other example: what is the difference in f-stops in a zoom between its 4.5
and 6.3 setting?

{Ln(6.3) - Ln(4.5)}/0,34657 = 0,97085 , almost a full stop. Sounds about
right, doesn't it?

Good luck using it. If you have actually read this post up to here: I admire
you patience, hope it was worth it.


Rick Garrelfs

































==================================================
The information contained in this message may be
confidential and is intended to be exclusively for the
addressee. Should you receive this message unrightfully,
please do not use the contents herein and notify the
sender immediately by return e-mail.


==================================================