----- Original Message ----- From: "Gregg Kemp" <gregg.kemp@???????>
> Guillermo, I don't remember ever seeing your correction formula before. Is > that theoretical, or have you tested it? Theoretical, Gregg, as I haven't done any close up work myself. For those (few) people interested in were that correction factor comes from, here it is: ****************** Pinhole is in most ways, not different than glass lens photography. The lens conjugate equation is: 1 / F = 1 / I + 1 / O where I = distance pinhole to Image plane; F = Focal length and O = distance pinhole to Object being photographed. We can simplify that to: F = I x O / I + O When the Object being photographed is at infinity ( O = infinity ): I + O = O therefore the formula F = I x O / I + O becomes: F = I x O / O and that becomes F = I In other words, when the Object is far away (more than 10 times the distance pinhole to film, in practical terms), the Focal length of the camera is equal to the distance pinhole to film plane. Now, when the Object being photographed is close to the pinhole lens (less than 10 times the distance pinhole to film, in practical terms), the Focal length of the camera is given by (as I stated above): F = I x O / I + O For close up work, then, "F" has to be substituted (in any of the formulas for optimum pinhole size) by: I x O / I + O For instance, the formula I use is: D = 0.0073 x SQRT( F ) where D = pinhole diameter in inches; F=pinhole camera focal length and SQRT stands for square root of For close up work that formula becomes: D = 0.0073 x SQRT( I x O / I + O ) As I said before, F = I for infinity, therefore I can write the infinity formula as D = 0.0073 x SQRT( I ) Base on that, I can also rewrite the close-up formula as: D = 0.0073 x SQRT( I ) x SQRT ( O / I + O ) There you have the correction factor: SQRT ( O / I + O ) BTW, there are many formulas for optimum pinhole size but all have within them "SQRT(F)", therefore, the above correction factor should apply fine to all of them. Let's see an example: Our pinhole camera has a distance pinhole to film I = 8" and the Object being photographed is O = 12" away from the pinhole, what size of pinhole is the optimum to use: The optimum pinhole size for infinity is: D = 0.0073 x SQRT ( 8 ) D = 0.020" (aprox) Correction factor is: SQRT ( 12 / 8 + 12 ) Correction factor is: 0.774 Pinhole size for close up work (object 12" from pinhole lens) D = 0.020" * 0.774 D = 0.0155" BTW, all the above is nothing but a sort of "bellows" correction. For people with mathfobia but that have read this msg up to this point, here are some corrections factor based on how many times the camera "pinhole-film" distance the object is away from the pinhole lens: less than 10 times => correction factor = 0.95 less than 9 times => correction factor = 0.94 less than 8 times => correction factor = 0.94 less than 7 times => correction factor = 0.93 less than 6 times => correction factor = 0.92 less than 5 times => correction factor = 0.91 less than 4 times => correction factor = 0.89 less than 3 times => correction factor = 0.86 less than 2 times => correction factor = 0.81 less than 1 times => correction factor = 0.70 less than 0.5 times => correction factor = 0.57 Another couple of formulas than may help the original's question poster are: Magnification M = I / O Therefore the Correction factor can also be written as: Correction factor = SQRT [ 1 / (M+1) ] So for instance, if she wanted to photograph an object 2" tall with a camera having 8" between pinhole and film and get a magnification of 2X, she will need to position the object at a distance: O = I / M = 8 / 2 = 4" O = 4 inches and the pinhole size (diameter) should be: D = 0.0073 SQRT(8) * SQRT[ 1 / (2+1)] D = 0.012" The image size will be 4" and therefore the minimum format size needed is 4x5 (portrait) or perhaps -better- 5x7" to allow for some cropping. Guillermo
