----- Original Message -----
From: "Joao Ribeiro" <jribe...@greco.com.br>
> Now, what I intend to do is to place 2 sheets of film not at the end of
> the camera, but at the walls that are perpendicular to the pinhole
> plane. I imagine that the light absorbed by these walls will also form
> images, certainly distorted (I believe someone in this list have already
> done that). How far from the pinhole plane the film should be to be
> completely covered by light? My imaginations tells me that if it is too
> close I'll have a "V" shape image.
> Is this formula you sent me able to give me that info? I'm asking because
> it takes into account the film diagonal and I believe this cone is
> independent of the film diagonal, I'm not sure.
Joao,
It was, I believe, a bit hard to formulate the question without
help of drawings, the case is the same for when stating an answer. Let me
see if a can make sense of what I am thinking, when I put it in writing:
As you know, COS^4 law dictates the fall off at the edges of a FLAT film
plane NORMAL (90 degrees) to the lens axis (pinhole, ZP or glass). There
are 3 factor that make up the factor for "off lens axis" imaging, namely:
1) The round pinhole looks like an oval whose size decreases by a factor of
COSine of the angle
2) The maze of light fall obliquely on the film plane,
therefore covers and area 1/COSine larger and finally
3) the distance pinhole film increases by a factor of 1/COSine, which
translates into a lose
of light of 1/COS^2 (due to inverse square law).
All the above adds up to the well known COS^4 LAW factor and for your
particular application, is valid for calculations of the fall off on the
"base" of the camera. For the fall
off on the walls of the camera we have to make some analysis:
1) the round pinhole still looks like an oval as we "climb" the wall
starting from the "base", that give us a factor of COSine of the angle in
light lose, as above.
2) The maze of light stills falls obliquely, but this time the film plane
being vertical rather than horizontal causes the lose of light to be
proportional to 1/SINE of the angle, rather than 1/COSine as above.
3) and finally, the distance pinhole to (vertical) film plane increase
changes at a rate of
K / (2 * SIN of angle) , where "K" is the amount of times the focal length
fit in the "base"
of the camera, in other words, in the width of the film plane.
All the above adds up to a factor of:
Light fall off FACTOR = [ 4 * SIN^3 (angle) * COS (angle)] / K^2
(sorry I didn't simplify the above a bit more, but what I remember of my
grade 8 trigonometry class does not help, probably it does not allow more
simplification, but if someone out there thinks otherwise, I'd love to know
how)
You could use the above FACTOR result to find out the number of stops that
angle will cause as light fall off on the camera walls.
Number of fall of stops = [LOG ( 1/ FACTOR )] / 0.3
BTW, it's been said that for flat film planes the maximum image circle is
3.5 times the focal length of the camera. In such a camera, the fall off at
the extreme edges would be 4 stops with respect to the center of the film
plane. I mention this in case you want to find the angle that causes a
similar fall off on the edges of the walls.
Hope it helps Joao.
(Answer is open to corrections)
Guillermo
