Color Bias in characteristic curve in negative films

[Photo Engineer]

Here is a copy of an old aim curve in my files. Does this help?

And actually, the speeds are matches but offset by mask density. Draw parallel vertical lines on the curve(s) and you will see that they can actually print neutrally when balanced for the mask density.

PE

 

[Ted Baker]

Here is a copy of an old aim curve in my files. Does this help?

And actually, the speeds are matches but offset by mask density. Draw parallel vertical lines on the curve(s) and you will see that they can actually print neutrally when balanced for the mask density.

PE

Thanks I was hoping for that kind of detail/accuracy for current films. However I am correct that the speed point for blue is typically higher in colour negative? I think I figured out the answer to my riddle BTW.

 

[Photo Engineer]

The threshold speed of the blue layer is slightly higher, but the mid scale speeds are the same.

PE

 

[aphid]

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

 

[Ted Baker]

might be of some interest to you

Thanks, yes that is very interesting, I will go through that carefully.

I think I may have solved the puzzle, well maybe...

This first image, represents a positive, where the negative has a blue layer that has a higher speed point than red or green. the gamma of each is identical and has been adjusted to 2.4. Filtration has been applied to remove the orange mask for a neutral black,

Note: X and Y are both linear scales, normalised to 0-1, so become intensity ratios. This 'film and paper' has a negligible toe and shoulder.

This second image, is identical to the first but the filtration has been applied to remove the orange mask for a neutral white. Note that the slope of each layer should change because of the filtration however the slopes will never be same between all three layers as the blue has higher speed point. (I have not adjusted the slopes for that reason)

This final image, is like the second in that filtration has been used to remove the orange mask for a neutral white, but the blue layer of the negative now has a different gamma (2.2), so both a neutral highlight and midtone has been achieved.

I have done this in more detail tracing the whole process through from scene, to negative, to positive and this seems to be the best explanation thus far.

 

[Photo Engineer]

Using my curve alone, I have to say that you are missing something.

PE

 

[Ted Baker]

I have to say that you are missing something.

That's very possible Though thus far this seems to me to the most correct answer, the math seems to add up, unless I have made a mistake... At some point I will post a more detailed analysis from which covers scene->negative->print.

 

[Photo Engineer]

Do you see the vertical line drawn measuring the "real" speed? Superimpose all of the 3 curves by moving them to lie atop each other.

PE

 

[alanrockwood]

Just a general comment about color reproduction: When you consider the difficulty of the problem, it is amazing that the C41 process reproduces color as well as it does. Consider the problem from the following perspective. There are at least twelve spectral curves that need to line up sufficiently well to give good color reproduction, four in each of the three layers.

Considering just a single layer, there is what I would call the "activation spectrum". (I don't know the correct name for it.) When the layer absorbs photons it activates silver crystals so they can be developed. The efficiency of activation depends on the wavelength, and this wavelength dependence is what I am calling the "activation spectrum". Once the development and dye coupling takes place there is another spectrum to consider, which is the absorption spectrum of the dye in that layer. The whole picture repeats itself in the paper, i.e. an activation spectrum for activating development of a particular color layer, and the absorption spectrum of the dye in the same layer after then end of processing.

Multiply this by three to account for the three layers and you have twelve different spectra that all have to combine in a way to give accurate color reproduction. (And I am sure that this is only a small part of all of the factors that need to be aligned, such as characteristic curves and such.) Given this, it seems almost a miracle that color films/prints do as well as they do. All of the scientists and engineers who developed these processes deserve high praise indeed!

The same goes for color transparencies, although there are only six spectra to align, which probably simplifies things a bit and may be a factor in making it easier;, but there is also no chance to adjust color like there is in the color print process, which would tend to make the problem harder, so again the scientists and engineers deserve high praise.

 

[Photo Engineer]

Alan, I am quite familiar with this aspect. As a team member, my job was the Magenta layer, ie, coupler, colored coupler and DIR coupler and spectral sensitizer. As the person who might change the spectral sensitizer, I was also the leader of the team for overall color reproduction. So, as I fiddled, each of the others fiddled (Cyan and Yellow), and I had to balance all as I contemplated a change from dyes G2 to G84 in our terminology. I oversaw a similar change in color paper about 10 years earlier.

This is not an easy task. I made about 10 coatings every other week changing the Magenta while using the latest Yellow and Cyan from my teammates. The intervening weeks, we evaluated and compared data.

PE

 

[MattKing]

my job was the Magenta layer

Which makes some of your APUG experience kind of ironic, wouldn't you say?

 

[DREW WILEY]

What's wrong with magenta? Without it, everyone would be Jolly Green Giant!

 

[alanrockwood]

Alan, I am quite familiar with this aspect. As a team member, my job was the Magenta layer, ie, coupler, colored coupler and DIR coupler and spectral sensitizer....
PE

PE, I thought you might have been involved. It sounds incredibly difficult.

 

[Photo Engineer]

Which makes some of your APUG experience kind of ironic, wouldn't you say?

Yes Matt, very much so.

Alan, it was. It was more than a years work by 3 people coating and 3 making and sensitizing the emulsions. That would be 6 people coating every other week amounting to a lot of silver just for tests. Admittedly, about 1/2 of them were what we called single layers or single layer coatouts. The former refers to a true single layer, and the latter to a coating with all components of a layer or typically all 9 emulsions.

PE

 

[Adrian Bacon]

Thanks, yes that is very interesting, I will go through that carefully.

I think I may have solved the puzzle, well maybe...

This first image, represents a positive, where the negative has a blue layer that has a higher speed point than red or green. the gamma of each is identical and has been adjusted to 2.4. Filtration has been applied to remove the orange mask for a neutral black,

Note: X and Y are both linear scales, normalised to 0-1, so become intensity ratios. This 'film and paper' has a negligible toe and shoulder.

View attachment 192023

This second image, is identical to the first but the filtration has been applied to remove the orange mask for a neutral white. Note that the slope of each layer should change because of the filtration however the slopes will never be same between all three layers as the blue has higher speed point. (I have not adjusted the slopes for that reason)

View attachment 192024

This final image, is like the second in that filtration has been used to remove the orange mask for a neutral white, but the blue layer of the negative now has a different gamma (2.2), so both a neutral highlight and midtone has been achieved.

View attachment 192025

I have done this in more detail tracing the whole process through from scene, to negative, to positive and this seems to be the best explanation thus far.

I’m working on a similar program (it’s called simple image tools, if you search for that tag on the various image sharing sites you’ll find images made with it in varying stages of development) and ran across this thread while doing some research. Ted, you’re about where I was about a year ago here in this post.

I thought I’d provide some input to hopefully move you forward a bit, if you’ve already gotten there, then great.

As others have alluded, digital is just another type of paper. It’s a nearly infinitely malleable piece of paper, but it’s a type of paper nonetheless. With that being said, there are a couple of truths that apply to that paper: all the math and colorspace stuff works best if your image data is in scene referred linear light space, meaning if something in a scene is twice as bright, it has twice the sample value.

This means that in order to really correctly digitize film, you must first correctly linearize it. Once you’ve done that, it isn’t really any different than digital image data from any other source, and you deal with it the same way.

Looking at your charts, you’re trying to get there and you’re close, but I suspect what you’re missing is that with color film the contrast curve of each color channel should be parallel, but isn’t. They’re close, but not totally parallel.

Look at the published characteristic curves for Portra 160 (E-4051, pub feb. 2016), at first blush, they look parallel, but upon closer inspection, the blue channel is slightly steeper, and the red channel is slightly less steep relative to the green channel. Your charts above largely reflect that. Portra 400 and 800 show an even greater difference than 160, and Ektar 100 is kind of wonky where the lines aren’t particularly straight and not particularly parallel.

So what do you do about it? Well, when I look at Portra’s published curves, I see three bw contrast curves offset from each other, one assigned to red, one assigned to blue, and one assigned to green. When I develop and scan bw film, I develop to a given contrast index, scan it in as a raw positive, then invert and linearize it with that contrast index. With color negative film each of those channel curves will have a contrast index. You need to linearize each one separately just like you would bw. Once you’ve done that, you still need to take that now linear positive image data, white balance it, conform it to a color space, and optionally do per hue angle hue and saturation adjustments so that a Macbeth chart looks correct, though the hue and saturation adjustments will mostly remove the look of that emulsion, so it depends on how agressive you want to get with that. All the post linearization stuff doesn’t really have anything to do with film as you’re now dealing with linear digital data, and should probably be a whole nother topic in hybrid or digital.

What I was searching for which caused me to stumble across this thread is what are the known contrast indexes for the various currently manufactured color negative films for in AIM development. Kodak does publish them in the film tech sheets, but those charts are small and not very high resolution, which makes using them to map out to digital linear data somewhat inprecise and difficult. I’ve manually exposed and mapped out the contrast curves of a couple of films that I regularly shoot, but that is extremely time consuming and if I’m to get simple image tools to a place where it can reasonably handle a relatively large number of films and be useful to more people than me, then I need to find either higher resolution published characteristic curves, or actual CI numbers, or ask the community to help me shoot and make high resolution in AIM characteristic curves.

 

[Photo Engineer]

Adrian, what you are missing is the fact that in the absence of a blue and green curve, the red curve is much higher in contrast due to interimage effects. The same is true of a single green or blue curve. So, simple consideration of R/G/B contrasts alone is not entirely correct.

PE

 

[Adrian Bacon]

Adrian, what you are missing is the fact that in the absence of a blue and green curve, the red curve is much higher in contrast due to interimage effects. The same is true of a single green or blue curve. So, simple consideration of R/G/B contrasts alone is not entirely correct.

PE

PE, through my own experiments, I’ve found that to be true. The published curves CI have tended to be much higher than what I’ve ended up using to linearize, however, the published relationships of each channel still comes through. For example, if I linearize EKTAR 100 with the same CI for all three channels, the mid tones are mostly OK, but the highlights and shadows show that the blue channel has more contrast than the red and green channel. If I change the blue channel curve out to a different CI linearization, this largely tames that. The same goes for the other channels.

The contrast index numbers I’m after are what the scanner RGB channels would see if the emulsion where illuminated with a high CRI light source (ideally a strobe), which would take the presence of the other channels into account.

 

[Photo Engineer]

Make a pure red exposure through a WR70 or equivalent, and compare it to a neutral. The single color contrast will be found to be significantly higher. Conversely, make a full Dmax exposure to blue and green light (WR98 and WR99) and a red step (WR70) and you will find a much lower red contrast.

PE

 

[Adrian Bacon]

Make a pure red exposure through a WR70 or equivalent, and compare it to a neutral. The single color contrast will be found to be significantly higher. Conversely, make a full Dmax exposure to blue and green light (WR98 and WR99) and a red step (WR70) and you will find a much lower red contrast.

PE

I’ve done something kind of like that, but with Rosco Calcolor gels. Don’t know what that maps in WR terms, I did see less contrast. How much less contrast needs more exploration and would probably be helpful.

 

[Ted Baker]

I’m working on a similar program (it’s called simple image tools, if you search for that tag on the various image sharing sites you’ll find images made with it in varying stages of development) and ran across this thread while doing some research.

Thanks, for you detailed reply. I have implemented a couple of approaches, including what you have described here but are still researching and improving my understanding. I am still undecided how much the bias exists when scanning, as opposed to when it measured using status-m. But I think the reason for the bias is ultimately because the three emulsion do not have the same speed point. This is what the graphs that I produced show. But it is just a working theory.

 

[Ted Baker]

Make a pure red exposure through a WR70 or equivalent, and compare it to a neutral. The single color contrast will be found to be significantly higher. Conversely, make a full Dmax exposure to blue and green light (WR98 and WR99) and a red step (WR70) and you will find a much lower red contrast.

PE

PE are these camera exposures. i.e. put the filters on the lens, take a picture, develop and measure?

 

[Photo Engineer]

Either, as long as you can get a good step scale. Remember though that the camera exposure includes flare.

PE

 

[Ted Baker]

Make a pure red exposure through a WR70 or equivalent, and compare it to a neutral. The single color contrast will be found to be significantly higher. Conversely, make a full Dmax exposure to blue and green light (WR98 and WR99) and a red step (WR70) and you will find a much lower red contrast.

PE

PE, I understand this actually intentional, and this would not necessary apply when using a film for making an inter positive?

 

[Photo Engineer]

This is intentional and does apply when making a print or an inter-positive.

PE

 

[Ted Baker]

This is intentional and does apply when making a print or an inter-positive.

PE

Yes that is kind of what I meant. It is my understanding that duping film or film for making cine inter positives with a gamma of 1.0, specifically does NOT have this crosstalk effect, so as to preserve, the ORIGINAL and INTENTIONAL effect within the original negative.