Re: f stops

on 31/10/01 3:18 pm, Guillermo at penate@home.com wrote:

>
> ----- Original Message -----
> From: "ragowaring" <ragowaring@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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Thankyou John Yeo and Guillermo for your exhaustive explanations.
I new some of it but it seems now I will know all of it (or have I said the
wrong thing). Never again shall I want for knowing. Thankyou again.

All the best

Alexis
Received on Wed Oct 31 13:46:43 2001