Color Bias in characteristic curve in negative films

Lack of crosstalk is not intentional in print films. In fact there is repression of the blue sensitive layer, but we wanted more. The emulsions used are not capable of giving significant crosstalk, and DIR couplers are not used.

PE

Is TS asking about an algorithm to remove the integral mask from scanned color negatives? I have read this thread just quickly, so sorry for my ignorance if I have misunderstood completely!

If this thread *is* about removing the integral mask from scanned negatives, then I might be able to contribute some. I tried to solve this task myself some time ago and studied e.g. "Advanced Photography" by M.J Langford to learn about how integral masks are constructed and from that I tried to do some reverse engineering. To my understanding, the integral mask is actually made up from 2 (or 3 depends on how you look at it), a pinkish mask that is due to the cyan layer in the film, and a yellowish mask due to the magenta layer in the film. Together, this two masks form a gives an overall orange appearance. The key observation is that the mask are not constant density, but depends on the density of the cyan and magenta dye layers in the film. So, to remove the integral mask digitally, one must first estimate how dense the integral masks are. Below in this post is piece of code that i have written. The code is in matlab, but should be understandable for anyone with skills in programming. In this code, it is assumed that the input image 'imRGBin' is *linear* (i.e. no jpeg-file, no gamma encoding etc, just raw data from the scanner) and with all pixel values in the range [0 1]. The constants dMY, dCY, dCM are constants representing the maximum density of the masks, these constants depend on the film at hand and must be tuned for each scanned negative/roll. Good starting points are dMY=0.2, dCY=0.4, dCM=0.4.

I hope this makes sense to someone (TS perhaps) and is not completely out of scope!?

function imRGBout = integralMaskEstimation(imRGBin, dMY, dCY, dCM)
% integralMaskRemovalCore core algorithm to estimate integral mask
%
% Convert RGB ---> CMY (CMY = 1 - RGB)
imCMY = bsxfun(@minus, 1, imRGBin);
% Calculate the pinkish integral mask due to cyan layer
imMaskCY = dCY .* imageInvert( imCMY( :,:,1) );
imMaskCM = dCM .* imageInvert( imCMY( :,:,1) );
% Calculate the yellowish integral mask due to magenta layer
imMaskMY = dMY .* imageInvert( (imCMY( :,:,2) - imMaskCM));
% Build integral mask
imIntegralMask = zeros( size(imCMY) );
imIntegralMask( :,:,2) = imMaskCM;
imIntegralMask( :,:,3) = imMaskMY + imMaskCY;
imIntegralMask(imIntegralMask > 1) = 1;
imIntegralMask(imIntegralMask < 0) = 0;
% Generate output (and convert CMY ---> RGB)
imRGBout = bsxfun(@minus, 1, imCMY - imIntegralMask);
imRGBout(imRGBout > 1) = 1;
imRGBout(imRGBout < 0) = 0;
end%function

I'm not sure about this language, but it looks like a matrix. If it is, well, you should know that our algorithms at EK use matrix algebra and calculus to converge on an answer. We have the advantage of viewing single color coatings for comparison.

PE

Photo Engineer said:

I'm not sure about this language, but it looks like a matrix. If it is, well, you should know that our algorithms at EK use matrix algebra and calculus to converge on an answer. We have the advantage of viewing single color coatings for comparison.

PE

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I always like it when PE reverts to the present tense

The films are still made Matt. I know that they are still being used. Now that I am here again on this damned thread about Kodachrome, I suggest the thread below on PN.

https://www.photo.net/discuss/threads/kodachrome-again.428135/#post-4413230

For reasons I don't understand, I can no longer log onto PN or I would probably have more Kodachrome questions, not that I need any. Here I get enough.

PE

Photo Engineer said:

The films are still made Matt. I know that they are still being used.

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I was referring to "our" "use" and "we have".
I truly understand though - my Dad thought of himself as still being part of Kodak right up to his death - 33 years after retirement!
And I know that you have a much closer relationship with those who are still actively "in the trenches".

I do see or communicate with many of them up to this day. I used the syntax I did because I used that methodology and knew parts of the code. I knew all of the guys involved in creating it who were at KRL.

PE

henpe said:

The key observation is that the mask are not constant density, but depends on the density of the cyan and magenta dye layers in the film.

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My understanding is the mask, a positive orangish image, is designed to form in the film during processing to cancel unwanted dye impurities, which collectively form a negative orangish image. The two images cancel each other and form a UNIFORM orange color all over the negative. Since it is uniform, it can be, and is, simply filtered/balanced out during printing, and along with it, the unwanted effects of the dye impurities. It doesn't seem like it would be that hard to remove during scanning.

Photo Engineer said:

I'm not sure about this language, but it looks like a matrix. If it is, well, you should know that our algorithms at EK use matrix algebra and calculus to converge on an answer.

Click to expand...

Yup, Many years ago , I wrote programs using matrix algebra several times for color space conversions and image color correction.

henpe said:

Is TS asking about an algorithm to remove the integral mask from scanned color negatives?

Click to expand...

NO that's not the question, the mask is trival to remove, as you probably understand. If the mask is difficult to remove, then there is something fundamentally wrong with the algorithm used. My question is very straightforward but obviously difficult to answer.

The questions restated is why does the characteristic curve show a different contrast index or gamma between each color layer, of a color negative emulsion?

In summary, this thread discusses:
1. Status-M used to measure the curve versus paper/CCD which would have a different spectral sensitivity.
2. Different contrast indexes when a single layer is exposed (used to manage saturation)
3. The Mask itself

This is all very complex with lots of similar side effects.

IMHO the reason for the bias as shown in the published curve is designed into the film because their is different speed point for each emulsion layer. That's what the graphs I posted earlier attempts to show. However I think there are also a few other alternative explanations.

Thanks again for posting the code.

The difference vanishes when you print! The film and paper or print films etc.... are balanced to give matches curves when printed.

PE

Photo Engineer said:

Here is a copy of an old aim curve in my files.

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PE what does AIM stand for in this context?

Aim means exactly that. With a standard exposure and process, a film or paper must match the aim for speed and curve shape when exposed to a neutral.

PE

Photo Engineer said:

Aim means exactly that

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Thanks, just that AIM keeps popping up in various papers many times in all caps, so I thought perhaps it was acronym, or had a special meaning. It such a common word so hard to search for it.

aphid said:

I came across this article about motion picture film scanning and thought it might be of some interest to you. http://mjbcolor.com/ScanningMetric.pdf

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Aphid, that avenue of enquiry was really helpful! I spent a fair amount of time researching cineon/DPX, and there is so much more technical information from industry professionals or film manufactures (i.e. Kodak) shared in the public domain about negative scanning, compared to still photography negative scanning which in summary is essentially zilch!

This may be because that part of the industry needs to able to share their scans, so they can be either printed on print stock or digital files, and they need to interact with production houses all around the world.

Mr Bill said:

An important part of this is, how does the paper "see" the film, raising the question (to Ted), where are you getting your info about the contrast of the film? If you are using characteristic curves, these are mostly made with a certain densitometer response (status M, I think, for color neg). These use a narrow spectral response, and almost certainly are not how a paper "sees" them - the results are usually broadly similar, but nothing like exact.

Click to expand...

Back to the original question that I asked...

I think Mr Bill actually nailed it the first time. Researching a number of papers on cineon I believe provides the information I was looking for specifically https://www.kodak.com/uploadedfiles/motion/H-387_Digital_LAD.pdf

You can see in table II it shows how the printing density varies against the status-M measurements. So in summary I believe most of the bias in characteristic charts exists because of the characteristic charts show status-M measurements, the bias would largely disappear if the graph showed the density of each layer using the spectral sensitivity of the paper was factored in.

The other assumption that I had made, is that "effective colorspace" of film, is well behaved such that if the Red, Green, Blue densities remained equal while their values were increased, the color would remain neutral, but that is not the case i.e. the colorspace is not well behaved. So believe small variations in the gamma at different points along CC exist, to manage this.

In summary, I believe the main reason for the variation showed in the CC is use of status-M measurements, but there are also suitable reasons for the variation.