Mathy explanation of the Anamorphic Pinhole
I decided to draw a slightly better diagram of how the anamorphic camera works, which led me to work out how the exposure works a little more formally. I'm going to post the more mathy bits over on the lensless-imaging forum at www.f295.org/Pinholeforum/forum/Blah.pl once I have an understandable explanation.
Here's the result with no explanation:
N = √2^(1.6226 + n) × r ⁄ d
Here's a minimal explanation:
N is the f-number, r the camera radius (as above), d the pinhole diameter, and n the exposure tolerance, 1 indicating no more than 1 stop of overexposure.
The exposure of the film has a maximum at 60° (which is covered by most of the actual anamorphic cameras I could find measurements for), which is 1.6226 stops more than given by the theoretical f-number of r ⁄ d.
So, to find an f-number for purposes of exposure that will give you no more than n stops of over-exposure. Add 1.6226+n, then increase your base f-number r ⁄ d by this many stops.
d = 0.24mm
r = 14mm
N = √2^(2.662) × 14 ⁄ 0.24 = 2^1.3113 × 58⅓ = 144.76
Theory says: f/144.76
The rule of thumb is to use the distance to the centre of the film to calculate the f-number.
Rule of thumb says: f/141.72
The difference is 0.06 of a stop, which is vanishingly small. I doubt that the shutter on any of my cameras is that accurate, let alone all the variations and noise involved in a pinhole camera!
Even in theory, the rule of thumb is all you need for cameras less than about 1.5 times as tall as they are wide. Taller than that, and you start getting a possibly significant (>0.5 stop) difference between theory and rule-of-thumb, as well as a large variance in exposure over the height of the film, which could yield situations like the bottom of the film requiring reciprocity correction, but not the top (which would obviously be troublesome).