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- Sep 20, 2002
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Was someone asking about this a while back? I'll post this anyway, just incase it helps. Maybe I'll get some good arguments out of it.
As a result of analyses of reciprocity correction determined by Howard Bond
through painstaking experiment, I found that the time to be added to ndicated exposure to correct for reciprocity misbehavior is expressed by the following equation.
Log(tc) = log(tc,1) + 1.62 log(tm)
where tc,1 is the correction at 1 second indicated time.
The factor 1.62 is accurate for all the films tested which were 400TX(0.169),
TMY(0.061), TMX(0.069), HP5+(0.101) and 100Delta(0.046). The numbers in
parentheses are the values of tc,1. This equation is easily plotted on Log-Log
graph paper, or calculated on most pocket calculators. On such a plot, all the
lines are parallel.
Howard found that the factors supplied by manufacturers were not accurate,
possibly because they were not updated with changes in emulsions. Due to the fact that the factor 1.62 works for these diffeent films of different manufacturers, it is my opinion that it will work for any current emulsion to acceptable accuracy. That is to say that I expect it to be within the spread among readings of indicated exposure made by a number of proficient photographers of the same scene. If this is the case, all one needs to know is the reciprocity correction to one indicated exposure to find the correction for any other indicated exposure. The graphical solution is very easy. Plot the correction time at the indicated time on log-log paper and draw a line through that point with a slope of 1.62 vertical inches for each horizontal inch. As you see, the graphical solution does not need the correction at 1 second, and if you plan to do any experimenting, it will be better to use a greater indicated time given the very small corrections required at 1 second for most current films.
Remember, there are no negative coordinates on log-log paper. Negative logs are logs of reciprocals, as log (1/10) = -log(10). If your line goes off the bottom of the graph before it gets to 1 second, just add another piece of paper to give you more cycles. Confused? I am pgainer@rtol.net if you want help without using up forum space.
As a result of analyses of reciprocity correction determined by Howard Bond
through painstaking experiment, I found that the time to be added to ndicated exposure to correct for reciprocity misbehavior is expressed by the following equation.
Log(tc) = log(tc,1) + 1.62 log(tm)
where tc,1 is the correction at 1 second indicated time.
The factor 1.62 is accurate for all the films tested which were 400TX(0.169),
TMY(0.061), TMX(0.069), HP5+(0.101) and 100Delta(0.046). The numbers in
parentheses are the values of tc,1. This equation is easily plotted on Log-Log
graph paper, or calculated on most pocket calculators. On such a plot, all the
lines are parallel.
Howard found that the factors supplied by manufacturers were not accurate,
possibly because they were not updated with changes in emulsions. Due to the fact that the factor 1.62 works for these diffeent films of different manufacturers, it is my opinion that it will work for any current emulsion to acceptable accuracy. That is to say that I expect it to be within the spread among readings of indicated exposure made by a number of proficient photographers of the same scene. If this is the case, all one needs to know is the reciprocity correction to one indicated exposure to find the correction for any other indicated exposure. The graphical solution is very easy. Plot the correction time at the indicated time on log-log paper and draw a line through that point with a slope of 1.62 vertical inches for each horizontal inch. As you see, the graphical solution does not need the correction at 1 second, and if you plan to do any experimenting, it will be better to use a greater indicated time given the very small corrections required at 1 second for most current films.
Remember, there are no negative coordinates on log-log paper. Negative logs are logs of reciprocals, as log (1/10) = -log(10). If your line goes off the bottom of the graph before it gets to 1 second, just add another piece of paper to give you more cycles. Confused? I am pgainer@rtol.net if you want help without using up forum space.
