Reciprocity Tables
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Reciprocity Tables
Hi folks
I have assembled a small list of exposure compensation figures for reciprocity failure fo the major Fuji and Ilford films. I am going to put them together in a nice .pdf file so that they can be printed and laminated.
But before I do, could I ask if there are any other films that we need to add to this list ?
Ilford seem to use the same chart for all of the following films :
FP4, HP5, Delta 100
Polaroid 54
Polaroid 55
Any more ?
I have assembled a small list of exposure compensation figures for reciprocity failure fo the major Fuji and Ilford films. I am going to put them together in a nice .pdf file so that they can be printed and laminated.
But before I do, could I ask if there are any other films that we need to add to this list ?
Ilford seem to use the same chart for all of the following films :
FP4, HP5, Delta 100
Polaroid 54
Polaroid 55
Any more ?
Last edited by Joanna Carter on Wed Sep 06, 2006 2:00 pm Etc/GMT-1+01:00, edited 2 times in total.
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I must say something here!
They say that reciprocity exists when the blackness ( D ) are a function of the exposure
( H ) which is independent of the light ( E ) are strong or low or the exposure time ( t ) very short or very long! Mainly that one value H(=E*t) gives D value!
Mr Schwartschild shown that this wasn’t a case at all! An a exposure of 1000 Lux under one second was infact more effective and gives higher density than 1 Lux exposed under 1000 seconds! To achieve the same D density the effective exposure time must increase which works after certain formula. This means sometimes a couple of F stops!
This of course pure sience and contains nothing else like compensating for bellows which may effect the quality of your negative.
There is two type of failure!
A
When the film exposes to a high intensity of light through short time t and than the Ag Hal crystals can’t receive that propitiated amount of energy. That means a lot of energy got lost!
B
Under long exposure takes time for those photons which should activate and disturb the AgHal crystal structures to build up a latent image and the very limited amount of reorganized crystals in this case could not build up a complete latent image!
This can also be called fading!
So the best alternative is to operate with your f stop and not with the longer exposure times!
Remember that you should only stop down which is necessarily to get the image sharp never more at all time! Don’t use the smallest f stop as your solution for final resolution!
All lenses have its best working aria and it’s often f22 in modern large format lenses! Those means the lens giving its best performance and image circle on that or very close f stops.
Now it had been stated that Ilford using the same table to all films! It might be incorrect as the hp5 is the old type of classic film and I don’t believe that has the qualities or works as any modern delta film.
I might check this one out if I where you!
I don’t know much about those films as I’m practically a Kodak-Agfa–Forte man and in principle I never shut colors as I have this ethical problem to letting somebody else involving in my work and of course the archival status of any existing color print!
Also very important that the content of tables like this is not exchangeable! All film has its own character and emulsions differ!
They say that reciprocity exists when the blackness ( D ) are a function of the exposure
( H ) which is independent of the light ( E ) are strong or low or the exposure time ( t ) very short or very long! Mainly that one value H(=E*t) gives D value!
Mr Schwartschild shown that this wasn’t a case at all! An a exposure of 1000 Lux under one second was infact more effective and gives higher density than 1 Lux exposed under 1000 seconds! To achieve the same D density the effective exposure time must increase which works after certain formula. This means sometimes a couple of F stops!
This of course pure sience and contains nothing else like compensating for bellows which may effect the quality of your negative.
There is two type of failure!
A
When the film exposes to a high intensity of light through short time t and than the Ag Hal crystals can’t receive that propitiated amount of energy. That means a lot of energy got lost!
B
Under long exposure takes time for those photons which should activate and disturb the AgHal crystal structures to build up a latent image and the very limited amount of reorganized crystals in this case could not build up a complete latent image!
This can also be called fading!
So the best alternative is to operate with your f stop and not with the longer exposure times!
Remember that you should only stop down which is necessarily to get the image sharp never more at all time! Don’t use the smallest f stop as your solution for final resolution!
All lenses have its best working aria and it’s often f22 in modern large format lenses! Those means the lens giving its best performance and image circle on that or very close f stops.
Now it had been stated that Ilford using the same table to all films! It might be incorrect as the hp5 is the old type of classic film and I don’t believe that has the qualities or works as any modern delta film.
I might check this one out if I where you!
I don’t know much about those films as I’m practically a Kodak-Agfa–Forte man and in principle I never shut colors as I have this ethical problem to letting somebody else involving in my work and of course the archival status of any existing color print!
Also very important that the content of tables like this is not exchangeable! All film has its own character and emulsions differ!
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A few years back, Ilford were in the process of revamping the reciprocity curves to give data for each film type rather than a blanket answer. This was before the last "splat" in the company's history so I don't know whether anything was finished (reciprocity-wise.)
Somewhere, I've got the equation of the line for Ilford although it's not too difficult to find if you plot it in excel and curve fit it through either power or exponential regression (I can't remember which at the moment.)
Somewhere, I've got the equation of the line for Ilford although it's not too difficult to find if you plot it in excel and curve fit it through either power or exponential regression (I can't remember which at the moment.)
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Polaroid 54 & 55?
Also, with the exception of the Ilford graph, I assume that it is aperture changes you have listed. It would be nice to also have shuter speed changes.
I was also in the process of compiling myself this type of data, to add to my little collection of laminated charts for use in the field.
Steve
Also, with the exception of the Ilford graph, I assume that it is aperture changes you have listed. It would be nice to also have shuter speed changes.
I was also in the process of compiling myself this type of data, to add to my little collection of laminated charts for use in the field.
Steve
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Steve
I have added the tables for Polaroid 54 & 55.
The problem with specifying shutter speeds would be that +1/3 stop is not easy to achieve on most LF shutters. If your uncorrected reading was 1/4 sec, what would you set for +1/3 stop? +1 stop would be 1/2 sec, so I guess +1/3 would be around 5/12 sec, something I can't find on my shutter
But as reciprocity seems to kick in at around the 1 sec mark for some films, greater for others, all you have to do is to multiply the number of seconds/minutes by 1+n where n is the f-stop correction factor.
So 1 sec @ +1/3 stop becomes 1 x 1.33 = 1.33 sec; 10 secs becomes 13 secs; etc. This kind of thing is best either done on a graph which is subject to reading errors and inaccuracies, or just use a small pocket calculator
I have added the tables for Polaroid 54 & 55.
The problem with specifying shutter speeds would be that +1/3 stop is not easy to achieve on most LF shutters. If your uncorrected reading was 1/4 sec, what would you set for +1/3 stop? +1 stop would be 1/2 sec, so I guess +1/3 would be around 5/12 sec, something I can't find on my shutter
But as reciprocity seems to kick in at around the 1 sec mark for some films, greater for others, all you have to do is to multiply the number of seconds/minutes by 1+n where n is the f-stop correction factor.
So 1 sec @ +1/3 stop becomes 1 x 1.33 = 1.33 sec; 10 secs becomes 13 secs; etc. This kind of thing is best either done on a graph which is subject to reading errors and inaccuracies, or just use a small pocket calculator
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The best treatise I have seen on the subject is:
http://silvergrain.org/Photo-Tech/reciprocity.html
Stick with it, the imortant bit is the stuff at the end (the graphs).
Maybe I will do a test for the delta stuff at some time in the future...
Marc
http://silvergrain.org/Photo-Tech/reciprocity.html
Stick with it, the imortant bit is the stuff at the end (the graphs).
Maybe I will do a test for the delta stuff at some time in the future...
Marc
Real Photographers use METAL cameras.....
...and break their backs in the process...
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...and break their backs in the process...
http://homepage.mac.com/mjjs/Photography/
Joanna,Joanna Carter wrote:...The problem with specifying shutter speeds would be that +1/3 stop is not easy to achieve on most LF shutters. If your uncorrected reading was 1/4 sec, what would you set for +1/3 stop? +1 stop would be 1/2 sec, so I guess +1/3 would be around 5/12 sec, something I can't find on my shutter
Yes, I do agree, and it can be a problem. But it is normally the shutter speed that you want to increse. If you are talking about 1/3 stop, opening the lens a little is usually fine.
However, once you get into longer exposures and larger corrections, the aperture change is not optimal, and exposures measured in seconds can be easily set.
I don't think that this is correct. The problem is that if you give a longer exposure, you are introducing more reciprocity failure. Ansel Adams does in fact refer to this in "The Negative", p. 42. (Sorry, it just happens to be my bedtime reading at the moment @-) ). For small corrections, then the difference is small, but the differtence grows with the original reciprocity error.But as reciprocity seems to kick in at around the 1 sec mark for some films, greater for others, all you have to do is to multiply the number of seconds/minutes by 1+n where n is the f-stop correction factor.
So 1 sec @ +1/3 stop becomes 1 x 1.33 = 1.33 sec; 10 secs becomes 13 secs; etc. This kind of thing is best either done on a graph which is subject to reading errors and inaccuracies, or just use a small pocket calculator
I have been trying to work out if there is a mathematical conversion that can be applied to get the correct time, but I am not sure if I have got it yet. I have created some graphs which I think are a bit closer, but this is just theoretical at the moment. As an example, for Polaroid 54, an indicated 10s exposure requires ~ 2.2 stops, which is 46s. My chart is showing ~110s.
Steve
Last edited by keffs on Wed Sep 06, 2006 4:47 pm Etc/GMT-1+01:00, edited 1 time in total.
The vertical axis is the speed change. Thus as the -ve value increases in magnitude, it means slower speeds and thus more exposure. Ceratinly a slightly unconventional way of displaying the data .Joanna Carter wrote:Just an observation on the Polaroid figures.
It would seem from the charts that the reciprocity is in the negative direction as time increases. If correct, this indicates that you have to reduce the exposure for longer times
Steve
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The final table!
Should this thread ever result in a final(ish) table that I could use, I would gratefully receive it, and with thanks to all involved.
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My experience with long exposure and reciprocity is that you can hardly overexpose. So if you take a ballpark compensation figure, just add some more to it and shoot that.
Also, at first I was always watching the clock/timer with anticipation etc.. But now I know that a minute difference doesn't make that much difference for a 7 minutes exposure, and so on...
For my favourite film (Shanghai) reciprocity seems to match exactly the fp4 formula : T' = T ^ 1.48. I use a derivative of that I can calculate in my head in the field, but it's probably only useful for computer trained people :
Example:
T = 18 seconds
+ I take the closest power of two to T, that is 16
+ 16 is 2 ^ 4 : 4 is this thus the number of half/stop to compensate, or, two stops.
+ So T' = ((18 * 2) * 2) = 72 seconds; and strangely enough, 18 ^ 1.48 = 72.078
Not bad
For the non-computer people, you can figure out which power of two to use by counting on your fingers 2,4,8,16,32,64 for 1,2,3,4,5,6 half stops.
With that method, I get pretty close without too much trouble :
T = 60 -> 64 -> 2^6 -> 6 * 1/2 stops -> 3 stops -> 480 seconds
T = 60 ^ 1.48 = 428 seconds...
Again, not bad, and on the good side or the error margin too (more light in)
Also, at first I was always watching the clock/timer with anticipation etc.. But now I know that a minute difference doesn't make that much difference for a 7 minutes exposure, and so on...
For my favourite film (Shanghai) reciprocity seems to match exactly the fp4 formula : T' = T ^ 1.48. I use a derivative of that I can calculate in my head in the field, but it's probably only useful for computer trained people :
Example:
T = 18 seconds
+ I take the closest power of two to T, that is 16
+ 16 is 2 ^ 4 : 4 is this thus the number of half/stop to compensate, or, two stops.
+ So T' = ((18 * 2) * 2) = 72 seconds; and strangely enough, 18 ^ 1.48 = 72.078
Not bad
For the non-computer people, you can figure out which power of two to use by counting on your fingers 2,4,8,16,32,64 for 1,2,3,4,5,6 half stops.
With that method, I get pretty close without too much trouble :
T = 60 -> 64 -> 2^6 -> 6 * 1/2 stops -> 3 stops -> 480 seconds
T = 60 ^ 1.48 = 428 seconds...
Again, not bad, and on the good side or the error margin too (more light in)