Re: Light fall-off furmula

Many thanks Bernard and Guillermo.

Guillermo,
I will need a bit of clarification on two points.

First, you state at some point that "M" is the distance away from the center
of the image. In the case of an image 8x9, with an F=5, then an M=4 would
yield a fall-off of 1.43 . Does that mean that at 4 from center the light
is 43 percent less than at the center "C" (roughly 1 stop) or that it is 43
percent (roughly half) of one stop less than at C?

Second, you state that M is the distance away from C. But later on, you
state that
" BTW, "M" would be the diameter of the circle you want.". Is M the radius
or diameter?

One last question, if I may. Does your formula cover all the sources of
light fall-off (e.g. increased distance from pinhole to film, transformation
of circle into oval and lost luminosity due to thickness of brass shim?0 or
does it focus only on distance from lens to film?

Thanks for this very useful information.

Best regards,

Guy

P.S. I did get your 55mm pinhole sieve yesterday in the mail. Can't wait
to see what kind of results I'll be getting compared with a standard
pinhole!
----- Original Message -----
From: "Guillermo" <penate@rogers.com>
To: <pinhole-discussion@spitbite.org>
Sent: Tuesday, July 26, 2005 3:22 AM
Subject: Re: [pinhole-discussion] Light fall-off furmula

> ** Warning ** Warning **
> ** Middle High School Math bellow **
> ** Scroll down at your own risk **
>
> ----- Original Message -----
> From: "Guy Glorieux" <guy.glorieux@sympatico.ca>
>>
>> I'm looking for a formula to calculate the light fall-off from center to
>> edges.
>
> Similar questions have been raised by others in the past, this answer, for
> instance, is related to your question:
> http://spitbite.org/pinhole-discussion/2003/0306/0899.html
>
>> I'll be working on a giant pinhole this week-end (8 feet x 9 feet) and I
>> need to anticipate the amount of dodging required at the center of the
> print
>> in order to get a fairly even exposure across the paper negative.
>>
>> As a rule of thumb, I've been working on the basis of Eric Renner's
>> book -
>> where he states that the maximum diameter is equal to 3.5 times the focal
>> length. However, I note that Larry Fratkin's calculator works on the
> basis
>> of just under twice the focal length.
>
> As I have said before, the "factor", 3.5 for Renner and 2 for Larry (as
> you
> mention) is not something cast in stone, that factor depends of the
> maximum
> fall off you are willing to accept and still consider that area of the
> total
> projected image is still part of the "usable" image.
>
>> What I'm looking for is a formula that will calculculate the amount of
>> fall-off from the center of the image.
>
> Before I try to give you an formula I have to say that Bernard is in the
> right track, cos^4 give us the fall off, unfortunately not in terms of
> stops. Here is my attempt to answer your request:
>
> If you call "M" the distance from a point off-center to the film's center
> (called "C") and if you call "F" the distance from the center of the film
> ("C") to the pinhole (this would be the focal length of your pinhole
> camera), then:
>
> The fall off at a point "M" units of length away from "C" (center of film)
> would be given by:
>
> FallOff = ( LOG [ (M / F)^2 + 1 ]^2 ) / 0.3
>
> LOG stand for Logarithm base 10 (or also known as common logarithm).
>
> For instance, assuming your 8x9 feet camera would have a focal lenght
> distance (distance "F") equal to 5 feet, the fall off at a point 4 feet
> away
> from the center would be:
>
> F = 5'
> M = 4'
>
> FallOff = ( LOG [ (M / F)^2 + 1 ]^2 ) / 0.3
> FallOff = ( LOG [ (4 / 5)^2 + 1 ]^2 ) / 0.3
> FallOff = ( LOG [ (0.8)^2 + 1 ]^2 ) / 0.3
> FallOff = ( LOG [ 0.64 + 1 ]^2 ) / 0.3
> FallOff = ( LOG [ 1.64 ]^2 ) / 0.3
> FallOff = ( LOG [2.6896] ) / 0.3
> FallOff = 0.42968 / 0.3
> FallOff = 1.43
>
> any point in the film that is 4 feet away from the center of the film
> would
> have a fall off of 1.43 stops.
>
>> Alternatively, what would be the
>> diameter of the circle with minus-1 stop exposure, minus-2 stop exposure,
>> minus-3 stop exposure times.
>
> In other words, how far from the center do we have to go to have 1, 2 and
> 3
> stops of fall off. The answer would be given by a formula derived from
> the
> one above. BTW, this time we need to find the distance "M" for any given
> FallOff and focal length "F":
>
> M = F * SQRT ( { SQRT [ 10^( 0.3 * FallOff ) ] } - 1 )
>
> SQRT stands for square root of.
>
> For instance: What would be the distance "M" at which the fall off is
> 1.43,
> for a camera with focal length equal to 5' (we know the result should be
> 4', see above):
>
> M = F * SQRT ( { SQRT [ 10^( 0.3 * FallOff ) ] } - 1 )
> M = 5 * SQRT ( { SQRT [ 10^( 0.3 * 1.43 ) ] } - 1 )
> M = 5 * SQRT ( { SQRT [ 10^( 0.429 ) ] } - 1 )
> M = 5 * SQRT ( { SQRT [ 2.6853 ] } - 1 )
> M = 5 * SQRT ( 1.6387 - 1 )
> M = 5 * SQRT ( 0.6387 )
> M = 5 * 0.799
> M = 3.995 quite close to the expected answer of 4'
>
> BTW, "M" would be the diameter of the circle you want.
>
> I'll calculate "M" for your camera and a Fall Off of 3 stops:
>
> M = F * SQRT ( { SQRT [ 10^( 0.3 * FallOff ) ] } - 1 )
> M = 5 * SQRT ( { SQRT [ 10^( 0.3 * 3 ) ] } - 1 )
> M = 5 * SQRT ( { SQRT [ 10^( 0.9 ) ] } - 1 )
> M = 5 * SQRT ( { SQRT [ 7.94328 ] } - 1 )
> M = 5 * SQRT ( 2.81838 - 1 )
> M = 5 * SQRT ( 1.81838 )
> M = 5 * 1.34847
> M = 6.7423
>
> I'd let you do the ones for 1 and 2 stops of fall off.
>
> If you don't have a calculator with LOGARITHM functions, you could
> download
> this one http://www.calculator.org/download.html , that is the one I use.
>
>> Guillermo, how does one calculate the light fall-off with a pinhole
>> sieve?
>
> Same as above.
>
> Let me know if any questions or corrections to the above.
>
> Guillermo
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Received on Wed Jul 27 05:42:53 2005