Re: Light fall-off furmula

----- Original Message -----
From: "Guy Glorieux" <guy.glorieux@sympatico.ca>

> 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?

You are correct in that IF the light were 43% less that at the center that
would mean roughly 1 stop (being 50% less light exactly equivalent to 1
stop), but when the formula gives you a fall-off of 1.43, it means that
there are 1.43 stops at any point 4 feet away from the center "C". If
knowing the fall-off in stops, you wanted to know the percentage of light at
a point "M" units away from "C", like the one in question, you would find it
this way:

Light% = 100 / (2^n)

Where n=FallOff in stops

In the case above, that is FallOff = 1.43, we would have

Light% = 100 / (2^1.43) = 100 / 2.6944 = 37.1%

Light at a point 4 feet away from "C" is 37% of the light at the center "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?

Up until I say: " BTW, "M" would be the diameter of the circle you want",
"M" represents the distance that separates a point (any) from the point "C"
or center of the film. All the existing points "M" units away from "C"
would create a circle around "C", therefore "M" is the radius and not the
diameter, that was my mistake........ I did it on purpose tho, I just wanted
to know if you were paying attention, Guy, congratulations!! you passed the
test :-)

> 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,

Yes (it causes Cos^2 fall-off)

> transformation of circle into oval

In 2 ways. First, from points off center the pinhole appears to be an
oval, no longer a circle (it causes Cos^1 fall off). Second, because light
falls obliquely on the film, it covers a larger area spreading light a bit
thinner (it causes Cos^1 fall-off).

The net effect of the above is the known Cos^4 fall off
(Cos^2 x Cos^1 x Cos^1 = Cos^4)

> and lost luminosity due to thickness of brass shim?

No, It does not cover this time of mechanical vignetting. I really wouldn't
worry about this vignetting unless the pinhole is made using a very thick
material compared to the diameter of the pinhole. For a camera 5 feet in
focal length you need roughly a 1.5mm pinhole (optimum size), even if you
were to use the aluminum of soda can, the ratio thickness of the aluminum
versus the diameter of the hole wouldn't be such to cause too much
vignetting.

> or does it focus only on distance from lens to film?

No, see above.

> Thanks for this very useful information.

It is my pleasure, Guy.

PS: I got your mail too.

Guillermo
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Received on Wed Jul 27 07:02:02 2005