A theorem about reducing double exposure with different filter## to a single exposure

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Lets assume I made two exposures from a good full scale negative or very long step wedge:

1) First exposure = T1 seconds with filter # F1 2) Second exposure = T2 seconds with filter # F2

My question is: do exist such an exposure time T3 and such a filter #F3 that the effect of the single exposure (T3, F3) is completely equivalent to the effect of the double exposure mentioned above?

No burning and dogging at all. For sake of discussion we can assume the filter value can be changed continuously (not discretely as in the realty: #00, #0, #1/2,...)

I'd like to warn that effect of the double exposure cannot be reduced to an addition of two characteristic curves (of #F1 and #F2), since there is no separate emulsion for each filter value. For example Ilford tells there are only 3 emulsions in the MG paper.

Thank you

-- Andrey Vorobyov (AndreyVorobyov@yahoo.com), October 26, 2001

Answers

Split filtering, as you described [(F1 * T1) + (F2 * T2)], is the functional equivalent of using a filter factor of (F1 + F2)/2 * (T1+T2)/2. Or, to make it clearer, let's use specific values for "T" and "F": assign F1=#1 filter, assign F2=#2 filter. Assume equal exposure filter factors for T1 and T2 (T1=T2), as is the case with most variable filter sets. Therefore, we may assign T1=T2=10 sec. Now, in this case, expose through each filter for 5 seconds. Your net effect would be a contrast grade of 1.5, or the same value you would have achieved using a #1.5 filter for 10 seconds. No point.

Now, if you really needed to split hairs, you could take a #1 and a #1.5 filter to achieve a grade of 1.25. Would it be worth it? Only you can decide that.

-- Ted Kaufman (writercrmp@aol.com), October 28, 2001.


Split filter printing doesn't hold up against imperical data but does work. A 5 sec exposure at grade 2 and a 7 sec exposure at grade 5 has a better black without dulling the highlights than any single filter exposure. Why I don't know but I have been printing this way for some time now and there is a difference. James

-- james (james_mickelson@hotmail.com), October 29, 2001.

Phil Davis had an article some time back about this issue. I remember the article was accompanied by two pictures, one the split filtered and the other though one filter. The pictures looked identical to my eye in the reproduction and Phil Davis claimed there was no difference based on his tests (which included things like step wedges nd reflection density readings). So, how come so many people like split filtering? Nary a clue, I fear. Surely, it can't be merely the discrete step issue (i.e., nothing between filter #1 and #1.5 while split filtering lets you reach that). Folks actually claim to be altering gradation i.e., changing curve SHAPE. I haven't seen any evidence to that but for me sometimes it does make for a more easy way of working. That is, it lets me get to a good print faster. I've used Steve Anchell's method. He suggests making a test strip to find exposure through F1, then making a full print and making a fresh test strip on that print through F2. Once developed, you have a good idea what the exposures should be through each filter. In other words, it lets you home in faster on what the shadows and highlights should look like. Any other ideas? DJ

-- N Dhananjay (ndhanu@umich.edu), October 29, 2001.

The reply for Ted.

You suggest the averaging formulas:

F3 = (F1+F2)/2 T3 = (T1+T2)/2

Both are not correct, it is easy to prove. Lets consider the formula for T3. If it keeps true for any F1 and F2, it should remain true for the particular case when F1=F2. But here we have a simple double exposure with the same filter, thus T3 = T1 + T2. I guess it was a misprint, and you meant the sum, not the average.

The formula for F3 is also objectionable. Lets assume we have a very short T1 relative to T2 (does not matter what are F1 and F2), say T1 = 1 sec, T2 = 100 sec. It is obvious, that total exposure is almost identical to the single second exposure (F2, T2). But your formula gives F3 approximately equal F2/2. It is wrong. You probably had to write something like

F3 = (T1*F1 + T2*F2) / (T1+T2)

-- a weighted average, not the plain average. Hadn't you? But I'm curious are these formulas just an empirical hypothesis or there is a theory behind?

The reply for James.

My experience is close to your. If F1 and F2 are too different I get sometimes results that I'm unable to reproduce using a single filer. But it is neither a proof nor a disproof of the statement (may be I was just not diligent enough to try the enough number of tests). Till the moment I attributed this inability to my negligence and still believed that the equivalence exists (i.e. such a pair F3, T3 does exist). But last week I got an interesting result. I had to decrease contrast in highlights (on the print) and at the same time to increase the exposure latitude. It was a perfect situation for pre- flash technique. I easily estimated amount of pre-flash and after of couple of test I got the desired result. No problem. But then I became interested what if I use different filter for the pre-flash? The main exposure was made with filter#3 (the negative was rather flat) and to emphasize the difference I tried #00 for the pre-flash. The result was shocking: dull, deadly depressing light grays, almost no texture. The microcontrast difference in "normal" tones and in my newly made "lights" was drastic. Of course, the effect differs depending on the pre-flash amount, but in any case it was a radically different effect and it obviously could not be reproduced with filter#3 in the pre-flash. Well, I felt the difference...

Of course the pre-flash is different than the conditions I described in my question (long step wedge as a negative), but now I'm really curious is the theorem true or not.

Till the moment I used the split contrast printing technique only conjunct with burning and dodging (to control the local microcontrast), but now I strongly suspect that there are other applications.

N Dhananjay wrote: "Folks actually claim to be altering gradation i.e., changing curve SHAPE". --- Exactly this is the desire. But I cannot yet figure out how to calculate the effect. If we had a dozen emulsions incorporated in the paper and each emulsion corresponded exactly to "his" filter then the double exposure calculations could be simple: the only thing we had to do is to add densities from different curves. But in the realty there are only 3 emulsions with different spectral sensitivity... I'm lost, at least right now...

My thanks to everybody who replied. I hope it is to be continued.

-- Andrey Vorobyov (AndreyVorobyov@yahoo.com), October 30, 2001.


For those who are interested, Phil Davis's article "Variable Contrast Papers Revealed", appeared in Darkroom and Creative Camera Techniques, sometime in 1994.

I still don't think there is any magic to split filtering - at least I've never seen a theoretical explanation of why. I have heard theoretical explanations of why there is no magic to them. Think of it this way. Each VC filter allows green and blue light to pass through in varying amounts. You start out with some amount of blue and green light, and the filters each block more or less of the blue and green.

If you look at it this way, all that matters is how much blue and green light hits the paper. It doesn't matter (ignoring things like reciprocity departure or intermittency effects) whether the blue light arrives first in one exposure, and then the green light, or the blue and green light is delivered all at once. The end result would be the same.

I don't think that split filter printing can achieve tonal distributions that aren't achievable with a single exposure under the appropriate filtration.

Mostly what the split filter proponents say is that the two exposure method gives a very easy way to determine the appropriate print density and contrast. Like so many things, that is an issue of personal preference, and we can and do disagree.

Preflashing is a different kettle of fish all together, because it does distort the toe of the paper's characteristic curve. And I would expect different distortions in curve shape depending on the filter that was used because the filter actually changes the sensitivity of the paper - remember the part of the emulsion sensitive to green light is much more sensitive than the part sensitive to blue light. So preflashing through two different filters probably amounted to a difference in the amount of preflashing.

Cheers, DJ.

-- N Dhananjay (dhananjay-nayakankuppam@uiowa.edu), October 30, 2001.



Also, for those who are interested in the technique, here is some more recent reading. http://www.users.zetnet.co.uk/ktphotonics/reference.htm Chris Woodhouse has done a fair bit of work and has designed some pretty well thought of products. The two articles related to split grade printing titled "A graph is worth a thousand pictures" are well worth reading. Cheers, DJ.

-- N Dhananjay (dhananjay-nayakankuppam@uiowa.edu), October 30, 2001.

N Dhananjay wrote: "...all that matters is how much blue and green light hits the paper..."

DJ, great hint! I don't know if it was clear from my messages above, but I've chosen an unnecessarily complex approach to the problem: I tried to set up a hypothesis about properties of real 3 emulsions separately, to guess their respond to light of different spectrum and then reconstruct the curves for every filter number… Too complex.

Your approach is simple and immediately solves the problem:

1) The number of the filter controls the (blue:green) ratio in the transmitted light 2)We have a one-parameter family of curves. The parameter is that (b:g) ratio.

It follows that if we take the filter with minimal (b:g) ratio (filter #00) and the filter with maximal (b:g) ratio (filter #5), we are able to produce in the combined exposure any ratio (total b: total g) between that min and max by modifying the T1/T2 ratio (I omit the simple algebra here). I.e. we can arrive at any curve in that family. The theorem is proved. (Although after your hint the word "theorem" is too sumptuous for the statement).

It follows also that in principle we don't need any filter other than the softest and the hardest.

It follows also that if we had formula for that one-parameter family of curves: D = f(H, R) where D is Density, H is exposure, R is (b:g) ratio (the mentioned parameter), we were in principle able to calculate even the effect of preflash with different filters.

Both articles of Chris Woodhouse are excellent. By the way, DJ, how do you find such great links? Could you please share some more links of the same quality from your collection? :)

Thank you!

-- Andrey Vorobyov (AndreyVorobyov@yahoo.com), November 01, 2001.


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