[Leica] F-stops

[R.P.A.C.Savenije@AVV.RWS.minvenw.nl]

Some maths!!!

f stops are calculated using the surface area of the lens.
which is related to the diameter (pi r square  as you might remember)

This you can recognize looking at the f stops at the lens:

1 -- 1.4 -- 2 -- 2.8 -- 4 -- etc.

which equals

2^0 -- 2^0.5 -- 2^1 -- 2^1.5 -- 2^2 -- etc. (2^0 means 2 raised to the power
of 0)

so ...

2^0.5 which also can be noted as square root 2 actualy is 1.1415... 
Al the figures engraved on your lens are truncated. You can see this when
looking at the more smaller apertures. It starts with each second figure
multiplied by two but at the end it changes: f1.4 f2.8 f5.6 f11! f22 f45!
f11 is 2^3.5= 11,3.
Look at this: f5.6 is 2^2.5. However 2^2.5 equals 5,65685 which is rounded
off as 5.7
So f5.6 is actually closer to f5.7

The question is, what f stop is 1.5?
or ... 2 raised to which power equals 1.5?
This is caluclated by 2 log 1.5 (sorry for the math) which equals 0.58496
We can find this number somewhere between 0.5 and 1 thus between f14. and f2.
... right.

0.58496 is 0.08496 higher than 0.5 (fstop 1.4)
The next stop is 0.5 further (2^0.5 -- 2^1, one fstop, remember?!)
and 0.08496/0.5 = 0.16992.

This means f1.5 is 0.16992 stop further than 1.4
which equals 1/5.88512 or more or less 1/6

so ...

f 1.5 is +/- 1/6 smaller compared to f 1.4
(possible truncation of 1.5 and inaccuracies of measuring the
actual apperature not taken into account)


Yet another answer!!!


Richard.