With the digital file on your computer you know the RGB values of the color. When you print and use a refection densitometer to measure the print. For a given RGB value what kind of values should you expect to read on the densitometer?
Without knowing the Color Space involved, the values are meaningless.If a patch of color in PS is 117,117,117 RGB then what kind of density should the print be using a reflection densitometer?
With the digital file on your computer you know the RGB values of the color. When you print and use a refection densitometer to measure the print. For a given RGB value what kind of values should you expect to read on the densitometer?
For a given RGB value what kind of values should you expect to read on the densitometer?
It's not a simple thing!
If a patch of color in PS is 117,117,117 RGB then what kind of density should the print be using a reflection densitometer?
I was looking for a LAB meter a couple of years back that I could see how far off my prints were to the actual LAB readings in my computer. Apparently there is a paint
chip reader that can record details for paint matching purposes- but not what I was envisioning of using.
To be clear - I am not trying to do this now . It was pie in the sky thinking years back when I was working through colour mangagement, profile, settings , paper issues of the past and how they all related to LAB numbers and my thinking that a portable device like a spot meter that could measure such values would be practical..The cheaper Colormunki Photo from X-Rite can do spot readings as well. I am not sure you can point up and get the values for the blue of the sky or the white of the clouds. However, another way to do what you were looking to do would be to take a picture of a ColorChecker Passport (also made by X-Rite) at the same time and use color correction software that would reproduce your sky or cloud color (relatively) more faithfully than simply using the color temperature setting in the camera or the raw processing.
Exactly. Without knowing more details about the system, this can't be answered.
Let me do a quick work through on this. Assuming that your stated RGB pixel values (117, 117, 117) are in the color space known as sRGB, this can be converted to CIELAB values of L* = 49, a* = 0, and b* = 0. And it can also be seen as roughly equivalent to 18% reflected light; I say this last part because 1) we know the "color" is "neutral" because both a* and b* are both zero, and 2) in CIEXYZ, the Y component, aka visual luminance is about 0.18. Now, IF YOU KNOW THAT THE SPECTRAL REFLECTANCE IS FLAT, as with b&w silver materials, then you can realize that reading any spectral range with a densitometer should correspond to that amount of reflectance, namely a density reading of about 0.74 or 0.75 (per the definition of optical density).
But... if the grey material is not spectrally "flat," as is normal with color materials then the density readings might vary, depending on where the (spectral) dye peaks are, what their shape is, and what part of them the densitometer "sees."
If you're using a conventional color paper, it was most likely made with the intention of being used with the standard "Status M" densitometer used for process control, so the dye peaks line up well with the narrow spectral zones that the densitometer can see. And the densitometer will probably give roughly equal RGB densities for a visual neutral.
But if you're using a media with different dye behavior, all bets are off. Especially if it's a wide color gamut system. Consider that the way to get "stronger" colors is to have higher, narrower dye peaks, and more of 'em, and you can see that the narrow spectral range of a densitometer just isn't a good way to predict how a human will see that material.
I hope I'm explaining decently.
Thanks Bill. Can you show me how to convert the R,G,B value of 117,117,117 to a density of 0.75?
Hi, first realize that the conversion to density is only legit because we think there is a flat spectral reflectance curve, otherwise...?
To get there from RGB numbers, there must be a known color space assigned, otherwise the RGB numbers don't have a firm meaning.
The easiest way is to use an online calculator, such as found on the excellent (and never wrong) website brucelindbloom.com. Start with the "calc" button, then the "CIE Color Calculator." Near the bottom, select the appropriate "RGB Model" (for the example I used sRGB, but you need to use the specific one that your data is in). Also, check the box, "scale RGB" to allow direct entry of your numbers.
Next, in the RGB row, enter your numbers, 117, 117, and 117. Then click on the RGB button at the beginning of the row - this will populate everything in the grid.
Now, there's no direct link to density, but the top row - "XYZ," uses linear data, normalized to a 100% reflective scene element. It's based on either 1 or 100, depending on whether you checked the "scale..." box. Now, when the CIE (International Commission on Illumination) set the imaginary "primaries" they made the "Y" correlate to the human sense of luminance, which is why I used it. I rounded it to 0.18, but it should display to something like 0.177888. Anyway, you should see it as rough equivalent to 0.18 reflectance.
Finally, to get from 0.18 reflectance to density (associated the neutral image with flat spectral response) convert according to the definition. It's something like density = the inverse log of transmission (or reflection). Without verifying (like I normally would do, but not today) I think you would raise 10 to the power of the density, and then invert that number to get reflectance. To crosscheck, a density value of 0.301 should reflect 50% (rounded) of the light, and density of 0.602 reflects ~25% of the light. (To get from reflectance to density, reverse the flow of the math.)
I'm speaking strictly off the top of my head, so if anything disagrees with the formal literature, I'm sure I would be the one in error, but please raise the question.
For the most part it's not a good idea to try to correlate a densitometer reading to these colorimetric data, except in limited cases.
Some how I couldn't get the percentage right.
When I put in 117,117,117 in the RGB row I got something in the 80000 in the XZY row.Hi, in case you don't have it ...
If reflectance = 0.18... invert it, meaning that 1/0.18 = 5.5555. Then log(5.5555) = 0.74, the density.
If you have density of 0.74, but want reflectance... 10^0.74 = 5.495... then 1/5.495 = 0.182 reflectance.
"Season to taste" with the proper amount of digits - I tend to go to overkill on the calcs so I don't have to worry whether I used enough.
When I put in 117,117,117 in the RGB row I got something in the 80000 in the XZY row.
these are values from my experience and measurements over he years:With the digital file on your computer you know the RGB values of the color. When you print and use a refection densitometer to measure the print. For a given RGB value what kind of values should you expect to read on the densitometer?
Now I get it to work. Do you have the formula available? I don't want to rely on the existing of the web and I need to modify it to give me density directly. thanksYou might be aided by the AMB system that you find on old Kodak reflection/transmission greyscales for reproduction photography.
A=Highlight 5%, M=Midtone 50% & B=Shadow 95% and these had predefined values as indicated previously for your "ideal" reproduction curve.
117 / 2.56 = 45.7 so I would say in a scale from 0-100% it is 46% roughly almost an ideal midtone value.
Do you have the formula available?
Thank you very much! Most of my questions are rather academic (I just want to know) than for real use.Hi, you quoted ced, but it seems like you are asking me?
Bruce Lindbloom's site has another button on the top row, "Math," that leads to various formulas. But they're probably more complicated than you are hoping. It might be easier for you to run a large number of values through his calculator and just record the data. Assuming that you are only working with one "flavor" of RGB.
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