Before I bore alot of people with techno babble, let me invite folks to see some scans that I have done over the last year or so:
http://hiddenworld.net/pinhole Thanks to the several people who responded to my reciprocity problem. I have not had good results, especially in low light conditions, trying to guesstimate exposure, to my satisfaction. At the risk of sounding like a techno geek, here is what I am trying to do: What I am developing is a small exposure calculator using a palm pilot and a shareware program called "mathpad". What mathpad allows you to do is type out equations, and solve for unknown variables. Since the Finney has an adjustable bellows, and multiple pinholes from which to choose, I thought this might be helpful to me. Currently I have been using this to calculate either my fstop, aperature (A),and Focal Length(F) using my Finney: A=Sqrt(13.5*F)/100 // Gives optimal aperature for pinhole fstop=F/A // Gives fstop, given F and A Using this, I can enter in either A, F, or fstop, and get the other values quickly. I was thinking of adding functionality to this by calculating in filter factor correction ( i like to use filters) as well as reciprocity factors. This would be somewhat simplified by the fact that I use a single film (TMax 100) from which I have some pretty good exmple compensation times already. I would like to enter in a meter reading, and have it give methe number of seconds (or minutes) for the exposure with all filter factors and reciprocity factors adjusted in. Factoring filters is easy, but I am hung up on reciprocity. >From the responses I have received, it appears that I need to do some sort of nonlinear regression curve. I did some quick research on the net for information on that, I quickly became bewildered with alot of information. I might be biting off more than I can chew! I am not sure if this can definately be constructed with the program I am using or not. Obviously an outside program might not be to helpful, as I am trying to get exposure information "on the fly" in the field. Thanks again for the responses, and my long winded posts.. Jeff
