Calculations for fstop enlarging timing
I'm trying to build an enlarger timer that can do quarter-stop exposure calculations. Working on the web (and glancing at Tim Rudman's f-stop calculation table), I've concluded that the relevant formula is:
[t2] = 2 ^ (-1* ( ([scale] -1) * log2 ([f]^2) + [scale]*log2(1/[t1]) ) )
[t1] = initial time
[f] = f-stop
[scale] = the amount by which one wants to scale (e.g. 1.5 would offer a 1/2 stop increase)
[t2] = the final time required (keeping the f-stop constant)
This is based on the idea that EV = log2( [f]^2 ) + log2 ( 1/[t] )
I've checked my values for Dr Rudman's table, and I'm correct per his table for 2 seconds basic exposure and above, when INCREASING f-stops.
BUT, 3 questions:
1. If I start with a basic exposure of [t1]=1s, log2(1/[t1])=0 and therefore no matter how large my [scale] value, my [t2] remains 1. (i.e. My formula only works for values over 1s initial exposure)
2. My formula does not handle decreasing exposure. If I have a [scale] of 0.5, and [t1]=2s, it suggests [t2]=1.4s, whereas of course 1s is correct: half the exposure.
3. My formula includes the f-stop, and is (heavily) effected by the f-stop. In fact, at an f-stop of 2, my times DECREASE as I attempt to INCREASE the exposure - clearly nonsense. Dr Rudman's table seems, therefore, based on the f-stop being f1, and not changing. I could simply exclude it for the purposes of my calculation, but I would like to understand why it is included in the calculation, and the logic of excluding it before I do so.
Many thanks for any help,
I don't think it is as complicated as your formula is. Look at http://unblinkingeye.com/Articles/Te.../testexpo.html. The basic relationship is t2=2^f*t1. f= exposure change in f-stops. Use a negative number for a decrease in f-stops.