Color Bias in characteristic curve in negative films

Bill Burk · Nov 30, 2017

Ted Baker,

I assume you know about color correction that is necessary during printing because cyan and magenta inks/dyes are flawed?

You could study some of the basic masking techniques from the printing industry used when making color separations. This can help you understand how the ink imperfections are dealt with.

In graphic arts color separation, the masks are considered photomechanical, and they are fairly easy to understand and make.

In photographic color negative film, the same color corrections "have to happen" but instead of a simple-to-understand procedure, it all has to happen in a multilayer film with one exposure processed at once... The story of how Dr. Bunny Hanson came up with the "orange mask" is worth seeing. And as PE has explained, interactions between layers as they develop refine the important color corrections.

I think if you "model" the necessary color corrections, you will have a better chance of matching color.

Dr. Bunny Hanson, who came up with it, explains it in these videos.
https://www.photrio.com/forum/threa...sley-hanson-kodak-director-of-research.94784/

Ted Baker · Nov 30, 2017

Bill Burk said:
Ted Baker,
I assume you know about color correction that is necessary during printing because cyan and magenta inks/dyes are flawed?

Not really, but I think that's a different problem as we are talking about an image for display.

That's an interesting video thanks!, I will watch it properly. The orange mask is not difficult to understand, the details of the "bias in gamma" which appears to only exist in the negative is the mystery to me.

Reviewing PE original answer to my question:

Photo Engineer said:
The gamma of a yellow (blue) curve is different in a single color exposure than in a neutral due to interimage effects which are there to correct for unwanted absorption in the dyes. This is therefore translated into a paper print as a corrected color image with true colors. (the bold is my edit)

My current working theory is the bias exists because of the lack of saturation in the dyes used in the negative, in comparison to those used a print. The cyan(Red) and magenta(Green) dyes being less saturated, and "bleed into" the yellow(Blue) dye spectrum. This would account for the bias and why it is a power function and not a simple coefficient. The dyes in the paper being more saturated would also make sense. A variation on this theory is the dyes in the paper 'bleed' into each others spectrum in a more 'balanced' way and thus the bias is not needed.

Anyway perhaps I am completely off

The purpose of this to have a clearer understanding of how the negative and positive interact, which is important if you remove the analogue positive from the process.

Photo Engineer · Nov 30, 2017

Ted, get a copy of this book: https://www.amazon.com/Principles-Color-Photography-Ralph-Evans/dp/B0000CINFZ

It is pretty good. BTW, I called him Dr. Hanson, not Bunny because he was a VP of Kodak and Director of Research.

PE

MattKing · Nov 30, 2017

Ted Baker said:
Not really, but I think that's a different problem as we are talking about an image for display.

Not really.
Displays are essentially just a different type of paper. Just like paper, they are calibrated to respond in different ways to different inputs. The only difference is that their calibration may be more easily changed.
While the colour response built into film may be calibrated to take into account the developing of, and dyes incorporated in, colour paper, that calibration gives a response that is easy to make use of with a digital display. It is just a matter of making the right choices. There is no inherent difference between that and making the right choices in order to make use of the output from digital devices.

Ted Baker · Nov 30, 2017

MattKing said:
Not really.

You have misunderstood what I tried to say, what ever the the reason is, either the problem or the solution it is not present in the finished product. i.e. the print. There is no bias in the gamma of a current Kodak Endura Paper, (fuji don't provide a characteristic curve). The question is really why the gamma bias exists in the film, and not in the paper, and the correct answer is not because of the mask, or at least as I understand how the mask works.

Photo Engineer said:
Ted, get a copy of this book: https://www.amazon.com/Principles-Color-Photography-Ralph-Evans/dp/B0000CINFZ

In the process of ordering it thanks, I am sure with everything rattling around in my head, something will eventually rattle out that makes sense...

Photo Engineer · Nov 30, 2017

Ted, I drew this set of curves a while back and thought it might help just a bit.

Ted Baker · Nov 30, 2017

Photo Engineer said:
Ted, I drew this set of curves a while back and thought it might help just a bit.

Thanks yes it might help the discussion a little. I have done that numerically with spreadsheets using portra 400 as the film and endura as a paper, also I attempted a digital simulation of a film camera, and paper, but the simulation is probably overkill so I put it aside.

I will superimpose portra 160 and endura onto that graph

Bill Burk · Nov 30, 2017

Ted I looked at an example graph in Bob Shanebrook’s Making Kodak Film. I see the Blue (yellow) curve has the highest contrast index. I think what PE was saying about single color exposures is that those curves are created by exposures to red, green and blue light separately so that the individual color readings can be taken.

But color correction kicks in when you expose to light that exposes two or more colors. The color correction would effectively raise the contrast of the green (magenta) and red (cyan) layers so much that it probably makes all three have the same contrast index under white light.

Photo Engineer · Nov 30, 2017

Ummm, not exactly Bill.

Bill Burk · Nov 30, 2017

So the curves are read from white light sensitometric exposures?

Ted Baker · Dec 1, 2017

Here is the diagram redrawn to represent porta 400 and endura paper.

It is illustrative, with reasonably correct proportions, it represents an exposure with filtration to bring the toes on the film together. It shoes the different gamma of each layer of the film. The paper has roughly the same gamma and toes for each layer, however the maximum densities are different.

So what is still puzzling me is why the different gamma exists? and how does it disappear in the final print.

Photo Engineer · Dec 1, 2017

Bill Burk said:
So the curves are read from white light sensitometric exposures?

I said "not exactly". It is done by dozens of complex exposures to white light, then R/G/B and varying color flashes. Sometimes there are hundreds.

PE

Photo Engineer · Dec 1, 2017

Ted Baker said:
Here is the diagram redrawn to represent porta 400 and endura paper.
View attachment 191199

It is illustrative, with reasonably correct proportions, it represents an exposure with filtration to bring the toes on the film together. It shoes the different gamma of each layer of the film. The paper has roughly the same gamma and toes for each layer, however the maximum densities are different.

So what is still puzzling me is why the different gamma exists? and how does it disappear in the final print.

You have left out the masking, and that is part of the equation here. To print from a masked negative, there must be a speed mismatch in the paper equal to the density of the mask and this is just a minimum. So, the red layer is taken to be the zero point, and the green and blue are faster in the paper by the densities of each mask. Then an addition 50Red is added to keep exposures from needing cyan filtration in most enlargers. This is one key point. The others are very complex and relate to the fact that all emulsions are sensitive to blue light, among other things.

PE

Ted Baker · Dec 1, 2017

Photo Engineer said:
To print from a masked negative, there must be a speed mismatch in the paper equal to the density of the mask and this is just a minimum. So, the red layer is taken to be the zero point, and the green and blue are faster in the paper by the densities of each mask. Then an addition 50Red is added to keep exposures from needing cyan filtration in most enlargers. This is one key point.

OK but I believe I accounted for that exact point, i.e. the different exposure, that is why I brought the toes on the film to the same point, I could adjust the exposure/filtration to make the top of the curves come together, but the mismatch of gamma would still be there?

Photo Engineer said:
The others are very complex and relate to the fact that all emulsions are sensitive to blue light, among other things.

Which is perhaps the part that I am missing?

Photo Engineer · Dec 1, 2017

Ok, I get the first part, but you must remember that there is some blue sensitivity to all 3 layers, and to that extent, they have some exposure when exposed to white light. In fact, the red layer has some green sensitivity for that matter. So, all 3 have overlaps. Using separation filters gives a huge boost in color reproduction, both in the taking portion and printing portion. That overlap muddies some colors and increasing blue contrast alleviates some of that. You cannot mask the blue layer, ie. you cannot have 3 masks.

See here: http://www.wiley.com/WileyCDA/WileyTitle/productCd-047119459X.html for more helpful information and also papers by Sir Robert Hunt of Kodak. I've taken courses from all 3 authors, Saltsman, Billmeyer and Hunt.

PE

Bill Burk · Dec 1, 2017

When I see straight lines with all three colors coinciding - as you show in your graph, Ted it looks to me as if you have solved the riddle.
I see the original where blue (yellow) is steeper than the others... What did you do to pull them together? Was it the paper curves?

Ted Baker · Dec 1, 2017

No the riddle is not yet solved, at least yet for me, You can make the three lines coincide for the film at any point, by changing the filtration (or the exposure of each individual color). (That's the basic process of removing the mask). As this graph has logarithmic scales, an exposure change is shown as a simple subtraction or addition. i.e. move the curve up or down, but it does not change the slope (ignoring the fact that film has a toe and shoulder).

The fact that the slopes are different and how they disappear in the print is the question I posed in the very beginning.

Though I am still chewing through PE's comment which I think maybe the answer to my question.

Photo Engineer said:
Ok, I get the first part, but you must remember that there is some blue sensitivity to all 3 layers, and to that extent, they have some exposure when exposed to white light. In fact, the red layer has some green sensitivity for that matter. So, all 3 have overlaps. Using separation filters gives a huge boost in color reproduction, both in the taking portion and printing portion. That overlap muddies some colors and increasing blue contrast alleviates some of that. You cannot mask the blue layer, ie. you cannot have 3 masks.

I hope this makes some sense, it is not easy to explain.

Bill Burk · Dec 1, 2017

So Ted, you drew paper curves as-is? I thought the print on PE's original graph was the tone reproduction result of the film and paper curves when they are combined.

(So when I see the print curves coincide, I thought you solved it)... No?

Photo Engineer · Dec 1, 2017

Ted, Bill, you have to remember that I have coated both paper and film and played with these curves and the masking and IIE. I have worked for years with it. There are mathematicians that have solved it all at EK and at many major scanner companies. It is not simple and involves matrix algebra and calculus.

PE

Bill Burk · Dec 1, 2017

Yes, it's remarkable PE. I can wrap my head around process camera color separations. But I can't wrap around the film. All I can do is believe that you did the same functions with chemicals, dye couplers and all the layers that add up to make color film with accurate color!

WilmarcoImaging · Dec 2, 2017

To PE’s point that this has been solved by major scanner companies: is there anyone in this forum community or the other analog fora, with career experience in scanner companies, who can comment in a similar way that PE contributes here?

The wheel has been invented, just need to talk to the people who did the inventing.

Ted Baker · Dec 2, 2017

Bill Burk said:
(So when I see the print curves coincide, I thought you solved it)... No?

No, I took PE's detailed graph, that has probably been done very carefully to scale, which among other things shows what happens with the toe and shoulder, and I just got my digital crayons out in an attempt to illustrate my question. I have been doing this numerically, but a picture is worth a thousand numbers...

My version is sort of a yet unresolved question, i.e the filtration/different speed points of the paper have removed the mask, by adjusting the curves of the film up or down relative to the paper, this should not modify the slopes of the film (at least without some other mechanism). Yet the paper at least (endura) does not have any such mismatched gamma to balance this out, it does have however have different maximum densities. What the final density of each the dyes in the final print is not yet known to me, so treat the line for the final result as an unknown.

I am looking at all of this from a certain perspective, i.e. what mathematical functions need to be applied to get the correct result. What really is amazing is all this stuff is done using chemistry. I have read on these very pages that 'when color film goes it's gone for good'

Photo Engineer said:
There are mathematicians that have solved it all at EK and at many major scanner companies.

Exactly!

WilmarcoImaging said:
with career experience in scanner companies, who can comment in a similar way that PE contributes here?

Unfortunately I think we are currently at a low point on that one, I am not going to comment on why that is, but I will comment on how it can be improved...

I am grateful for the help and interest here..

Photo Engineer · Dec 2, 2017

I am sure that the formulas for scanners as with film are highly proprietary.

I have found that my Epson does the best job across the board with all color films.

PE

Bill Burk · Dec 4, 2017

Ted Baker said:
No the riddle is not yet solved, at least yet for me... but it does not change the slope.

It still seems to me that color correction could have something to do with it. Yellow's the purest color. Magenta and Cyan need to be held back whenever they print where there is also the other colors. To hold back the print ink, you would increase the density on the negative. So I think in cases where color correction is happening, the density at any given point will be increased compared to the density that negative would have in a pure color exposure.

Ted Baker · Dec 4, 2017

Bill Burk said:
It still seems to me that color correction could have something to do with it.

True, but it is not an obvious solution, and there are many variations on that theme that could provide the answer or part of the answer. The one that a like the most at the moment is this:

Photo Engineer said:
So, all 3 have overlaps. Using separation filters gives a huge boost in color reproduction, both in the taking portion and printing portion. That overlap muddies some colors and increasing blue contrast alleviates some of that. You cannot mask the blue layer, ie. you cannot have 3 masks.

But I am still working on it...

The reason I am most interested in this, is probably answered by why doesn’t the paper have a bias in gammas as well? The endura paper is designed to print most any c-41 stock, what about the fuji crystal paper, I am sure it works with Kodak stock as well? Basically I am trying to understand in more detail how the negative and paper interact.

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