Re: Light fall-off furmula

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** Middle High School Math bellow **
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----- 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 Tue Jul 26 00:22:40 2005