f-stop calculations and reciprocity failure numbers

Time for some geek-speak...

I had read the discussion of f-stop increments and rounding pinhole
calculated f-stops to the 'standard' ones in a chart, in part because
pinhole exposures tend toward (and I can't remember over or under) being one
side of exposed.

This intrigued and bothered me. It made me curious what was the origin of
the f-stop increments and also wonder why one's pinhole f-stop number
couldn't be exactly what it's supposed to be.

If focal length and aperture define f-stop, rounding to a standard f-number
equates to rounding off one's f.l. or aperture diameter...they're all
interrelated.

I haven't completely satisfied myself with regard to answering all my
questions (like some "why's?"), but this is what I have thought up.

if f 1.0 is the theoretical 100% light getting in, (although I did see a
Canon rangefinder supposedly with an f 0.95 lens...must have had some kind
of amplification), the f-stop increments appear to come from an exponential
relationship based on powers of the square root of two. I haven't derived
where that comes from, but I figured out how to calculate the 'standard'
table numbers and how many factors of exposure multiplication result from
using higher f-stops.

The camera I am converting to pinhole has a continuously adjustable f-stop
range (which of course doesn't get used in this application), despite
'standard' f-numbers. The aperture can be placed in-between the 'standard'
stops.

Since reciprocity failure has to be factored in and detemining exposure is
pretty variable, all this boiling down to a time multiplier, I question the
necessity of choosing an f-number that 'fits' the chart.

How many conformists do we have here, anyway (anybody with me on this?).

Anyway, the f-numbers appear to appear as follows. This is getting too wordy
already and I can't explain it in other-than-math language so just look at
the results...I see it but don't know what to call it

exponent 0 1 2 3 4 5 6 7
8 9
f-stop 1 1.414 2 2.828 4 5.656 8 11.313 16
22.63

These look familiar? I think f11 looks better on a lens than f11.3.

Anyway, I am trying to figure out how to put my variation of a calculator on
the web...maybe one of you cats (oops, that would be the jazz discussion
group)music that knows Java or Javascript can help me

What it figures out is, what number to multiply exposure time by when
using a f-stop off the chart, with respect to another f-stop, say f-16, or
any other.

I have tried to simplify my logarithm algebra but some subconscious error
keeps creeping in when I try to simplify it and I'm sick of trying to polish
any further.

basically, take natural logarithm (call it "ln", apologies to followers of
European notation) of the larger f-stop number and divide it by ln(2^0.5).

Now do the same with the smaller f-number and subtract the result from the
first resulting number, and multiply by two.

This is the factor by which exposure time has to be multiplied when shifting
from one fstop to another.

If you use the 'accurate' f-numbers as shown in my chart (which may get
butchered by email programs everywhere), this is exactly accurate. If you
use 'convenient' numbers like 4,5.6,8,11,16,22, 32 there is error.

I have it in a MS Works spreadsheet right now, and could put it on Excel if
desired. If anyone wants it , email me.

Oh, almost forgot my question -

Does anyone have reciprocity failure multipliers for common 35 mm films?
Kodak wouldn't address my question unless I gave them a specific product,
and I didn't know what to use yet. Fujii is mailing me datasheets, but who
knows how long that will take.

Thanks

Murray

Murray
Murray
Received on Fri Jul 20 19:56:40 2001