Gill
you have WAY toooooo much time on your hands......
but a great explaination..I'll save this one
thanks
andy

-----Original Message-----
From: pinhole-discussion-admin@p at ???????
[mailto:pinhole-discussion-admin@p at ???????]On Behalf Of Guillermo
Sent: Wednesday, October 31, 2001 10:19 AM
To: pinhole-discussion@p at ???????
Subject: Re: [pinhole-discussion] f stops



----- Original Message -----
From: "ragowaring" <ragowar...@btinternet.com>

> > They follow a geometric progression that make the f/stops increase by a
> > factor of square root of 2
>
> Wow!, is it really that simple. Of course, it all makes sense now and it
> will be very helpful.  Thankyou for your replies.  If it is not too much
> trouble a complete explanation would be very welcome Guillermo.

Although John Yeo already gave you an explanation, risking being redundant,
I will give you another one:

In general terms, f/stop is just a ratio that tell us how many times the
diameter of the aperture the focal length of your lens is.  i.e.. a 50mm
lens
with an aperture diaphragm opening of 25mm in diameter would have an f/stop
of f/2   (50/25 = 2).  Another example: a 90mm focal length pinhole camera
with a 0.35mm pinhole would have an f/stop = f/114  ( 90 / 0.35 = 114 )

f/stop = focal length / aperture diameter

Both Focal length and Diameter must be given in the same units of measure.

The amount of light that any f/stop let through is double the one the
immediate closed down full f/stop let through.  For instance, f/8 let
through double the amount of light than f/11 let through and that means the
area enclosed by the round aperture opening at f/8 is twice the one for
f/11.  The area enclosed by a circle is proportional to the square of its
diameter  (Area = 0.7854 * D^2), therefore, to double the area inside a
circle (allowing double the light, meaning opening up 1 f/stop), we need to
increase the diameter of the aperture by just square root of 2 = 1.414

We now know the diameter of the aperture from one stop to the next, increase
by a factor of 1.414, we also know   f/stop = focal length / diameter,
therefore the f/stop numbers will increase also by a factor of 1.414

Starting with f/1 (which BTW is neither the theoretical nor practical
maximum aperture), to find out the next full stop, just multiply the
preceding one by 1.414 and approximate the result as required, like this:

1 = f/1
1 x 1.414 = f/1.4
1.4 x 1.414 = f/2
2 x 1.414 = f/2.8
2.8 x 1.414 = f/4
4 x 1.414 = f/5.6
5.6 x 1.414 = f/8
and so on.

As you can notice, the numerical value of the f/stop doubles every other
full f/stop, so after you find the first 2 (f/1 and f/1.4) you no longer
need to multiply but 1.414  just double the f/stop 2 stops behind.  ie. the
next stop after f/8 would be equal to double f/5.6 = f/11  (5.6 x 2 = 11),
the next stop after f/11 would be double f/8 = f/16 and so on.

As an added information, I'd like to mention how to find out intermediate
f/stops.
When we divide  the focal length by the pinhole diameter, most likely than
not we get an f/stop number that is not a "full" f/stop.   Some paragraphs
above I mentioned the following example: "a 90mm focal length pinhole camera
with a 0.35mm pinhole would have an f/stop = f/114  ( 90 / 0.35 = 114 )", it
is clear that we need to approximate that f/114 to either a full stop, 1/2
stop, 1/3 stop or whatever you like.  I usually approximate to the next 1/3
or 1/2 stop,  the question becomes: how to know which of these f/114 is
closer to?   Get your "slide ruler" or your scientific calculator (I use
CALC98 http://www.calculator.org/) 'cause we will get some logarithms.

To find where f/114 falls with respect to the closest larger full f/stop
(larger means smaller numerical value, remember), we just divide f/114 by
that larger full f/stop (f/90 in our case), then get the common logarithm
(LOG in most calculators, as suppose to Ln) of the answer and finally divide
that by 0.15

Let see:

114 / 90 =  1.266666

LOG (1.266666) =  0.10266

0.10266 / 0.15  = 0.684

The 0.684 means f/114 is 0.684 stops smaller than f/90 , knowing this helps
me to know that f/114 is just  a bit larger (numerically) than f/90 2/3
(2/3 = 0.666)  and also f/114 is just smaller than f/90 3/4 stops  (3/4 =
0.750).  Knowing all this may be an overkill for most of us, but it doesn't
hurt knowing it, anyway.

I better stop here.

Guillermo










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