> This involves setting up a double integral and solving it (calculus). I
>consulted with a colleague, a theoretical physicist, this morning and
>we did a few calculations for a simplified case. We ended up with an
>expression involving the arctangent of the dimensions of the lightsource
>divided by the distance to the center of the source multiplied by the
>reciprocical of the distance to the center of the source and a few
>other factors. When the arctangent is expanded, one gets a series of
>terms. The first is an inverse square term and the rest are inverse
>fourth, sixth and eighth power etc. terms.
Argghhhhhh! I'm sorry, I didn't mean to cause this to happen.
> When the distance to the source is equal
>or nearly equal to the source dimensions, one gets a fairly complex
>mathematical relationship.
Ok, say we have a 4 foot square light bank with 12 40 watt fluorescents, I
assume we are talking about the bank being about 4 feet away for the
complex relationship to take place. In most of the discussions, the
distances talked about are in inches, that is from 1 inch to a foot or so
from the print. I gather from your study of this, that there will be very
little fall off in a 4 foot sq bank in any distance from 1 inch to 2 feet? I
would find it odd that anyone with a 4 foot bank would place it 4 feet away
unless they're trying to expose a 8 foot sq print.
Ah the wonderful world of alt photo!
Dick S.
Bostick & Sullivan
1541 Center Dr.
Santa Fe, New Mexico
87505