The Dimensions of Colour - a colour theory discussion thread - Page 9

# Thread: The Dimensions of Colour - a colour theory discussion thread

1. Registered User Level 1 Gladiator: Andabatae
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hi david,
at the site it said,
"Take a moment at this point to also ensure that you have completely eradicated from your mind every trace of the primary school notion that green is "made of" yellow and blue. Subtractive mixing doesn't work like that. We get a green mixture from yellow and cyan because our components are both in part "made of" green. If any colour can be said to be "made of" yellow and blue, it's white."

"If any colour can be said to be "made of" yellow and blue, it's white" did you mean it black? cause yellow and blue has not wavelength in common.
thanks before david.

2. Registered User Level 1 Gladiator: Andabatae
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is the additive mixing mean the mixing of two color of light, and subtractive mixing is the mixing of color of light with the color of surface (irrespective with the surface's material)?

if the subtractive mixing of the light and the surface which the color of them doesn't have the same wavelength is it really result black?
what if a blue skin creature (like the avatar movie) get pure yellow (R+G) light ? is it result black? i'm confuse with how come if one surface with the light can do subtractive color if the color of light and surface has no similar wavelength... how come it became darker as black? even there is a light...

thanks before for the helping, david.

3. Registered User Level 1 Gladiator: Andabatae
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hi
again david
now i include the sample of my painting about subtractive and additive mixing
it says if the red subtract with green surface or blue light subtract with green surface resulting "black" color did you mean by "neutral or gray" color?

i want to ask, which picture is the correct one sir?

or

and why sir?
thanks before

4. Ah, trick question! Both additive and subtractive processes are involved. Can you follow this? -

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6. Registered User Level 1 Gladiator: Andabatae
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hi david, thanks for the explanation
did you mean by this?

at no 1. i see the green became more yellowish at the diffuse reflection, is that because the RED + GREEN? also the number two, the green became more blue (lets say cyan) is that because G + B?

then what happened to the subtractive mixing theory that say if between two color that has no wavelength in common will appear black? EX. R + G = Black or G + B = Black?
or is there any example in real object where we can see this "black" happen between light and surface color?

what if a yellow surface got blue light?

how the subtractive mixing going on?
the hue of yellow value is higher than the hue of blue light, is the yellow surface became lower in value if it got blue light?

could you please include the formula for me like the one u've did before at the upper of this picture?

thanks before david for your helping. i really appreciate that.

7. With subtractive mixing you get the wavelengths that are in common to the two colours, so the result depends on the actual spectral distributions. With a non-monochromatic blue light and a typical yellow surface you're likely to get a green reflection (first example). With a desaturated red light and a typical green surface you might get a slightly more yellowish green than the local colour (second example).

You might like to try out other combinations yourself with the online applet I used to make these diagrams. Choose the option you want (e.g. "Run with browser Java) and then click and drag on the diagrams to create your spectra.

http://www.cs.brown.edu/exploratorie...ion_guide.html

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9. Registered User Level 1 Gladiator: Andabatae
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thanks dave, but i'm not really understand how to used the diagram, the incoming light, reflectance, and product.

10. No problem g. In each set of three graphs, "Incoming Light" represents the spectrum of the coloured light source (I don't know why they didn't add the spectrum colours to that graph - it's a bit confusing), "Reflectance" represents the wavelengths of the spectrum that the colored object can reflect, and "Product" is the result of multiplying those two graphs together, wavelength by wavelength, and represents how much of that coloured light will be reflected by that coloured object. (For example, in the second diagram, for a middle green wavelength we get incoming light (0.4) times reflectance (1.0) gives a product of 0.4, while for an extreme violet wavelength we get incoming light (0.4) times reflectance (0.0) gives a product of 0.0).

I just made up the reflectance spectra, but they represent rather ideal bright yellow and bright green objects.

Last edited by briggsy@ashtons; April 25th, 2012 at 08:00 AM.

11.

12. I don't know what it is, but I want one! It's described in the article both as "stone" and "glass" but it looks very pale for a volcanic glass, so it seems more likely to be artificial.

Very cool site!

13. I suspect it is, although I've seen similar, but less dramatic, effects with moonstone.

14. Registered User Level 1 Gladiator: Andabatae
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yes, the incoming light got no color
do we identify the color by seeing the highest mount at the diagram?
let say, what monochromatic blue light beam into monochromatic yellow surface

does the diagram look like this?

then what the result of product's diagram saying?

thanks for helping dave.

Last edited by grapholic; April 27th, 2012 at 03:16 AM.

15. Wow, this is great, thanks for sharing!!!

16. Registered User Level 1 Gladiator: Andabatae
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## Thanks, Thanks and again Thanks

I'm just on page 1 of the web site and I'm a very happy person because I have allways thought of colour as spatial and have created representational paintings based on these theories-once I've fully read the site will post images with theoretical explaination.
thanks David Biggs

17. Registered User Level 1 Gladiator: Andabatae
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a very special thanks to Doctor Briggs for give me such a wonderful color study

now my artwork is getting better and better.
now i'm more understand about the subtractive and additive mixing
all this time i was confuse because i was miss the monochromatic intensity about color... when i understand it more... all the mixing suddenly magically fit into the color stimulation.

i feel that My feelings can not be uttered in words now,
all my artwork is always dedicated for you doctor briggs
my respect.

18. Thanks for the encouragement guys. Hopefully you'll be pleased to know I'm planning to work through the site this year, adding some new images and trying to improve the text. I've recently finished revising the two introductory pages:

http://www.huevaluechroma.com/011.php
http://www.huevaluechroma.com/012.php

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20. Registered User Level 1 Gladiator: Andabatae
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## Question

Thanks again for your fantastic website its excellent!!!!!!!!!!!

Q.1.What is the logic of a painter "muting" a colour by mixing it with its complementry instead of GRY especially if its NOT to change hue i.e. mixing YEL and BLU for GRN but rather to desaturate it?
Q.2.Putting aside overly literal abberation of actual pigment and generalising theoretically is it true that you could arrive at a reasonable approximation of say yellow/blue mixed red/violet just by looking for the appropriate hue (if the colour wheel was "blended infinitely")and adding a GRY?

P.S.I think there is a yellow/orange light from the blue glass stone because the light entering the stone is refracted in internal dispersal therefore the concentrated beam of light that eventually emerges is of a different wavelength and that also means colour difference.But I might be wrong.

21. C.M.

Glad you like the site. I'm not sure if this answers your first question, but it is possible to find or mix a pigment-mixing complementary for any coloured paint that will neutralize it, but this sometimes will not be the same colour as the additive complementary. For a yellow, for example it will be not blue but a particular violet paint or paint mixture.

I'm afraid I can't even guess what you mean by your second question, so for me to answer it you'll need to rephrase it completely.

22. Registered User Level 1 Gladiator: Andabatae
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Wonderful & helpful info.
Thank you very much

23. Registered User Level 1 Gladiator: Andabatae
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hello briggsy
does this site precise? as you give me the one before?
http://graphics.stanford.edu/courses...lormixing.html

24. Overall the applets and explanations on that site are very good, but just on that last page you linked to there are a couple of things that could be confusing to students. I don't think it is stressed enough that all actual yellow, cyan amd magenta pigments are of the second, "broad C,M,Y" type, and I doubt that the narrower, essentially theoretical CMY colours in the first applet would be as bright as they are shown. (In fact the displayed colours are definitely not accurate, because when I add light from the red and orange parts of the spectrum to the narrow "yellow" spectrum, the displayed colour gets darker). Also in the summing up at the end, while it is correct to say that the additive/subtractive primaries "don't need to be" R,G,B/C,M,Y respectively, it is potentially misleading not to emphasize that those primaries do give the maximum gamut.

25. I'd like some help to see if a diagram I'm working on is looking right for most observers/monitors.

Please click on the link and observe each of the two finely-lined squares at the bottom of the diagram from a distance of a couple of metres or so, so that the vertical lines are no longer visible. Please check that your browser is set to 100% zoom, and very importantly, position yourself so that your line of sight to the square you are looking at is at exactly 90 degrees to the screen. If your browser resizes the diagram, please click on it to return it to full size. Do the squares look grey or coloured?

http://www.huevaluechroma.com/pics/4.3.3.jpg

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27. Originally Posted by briggsy@ashtons
I Do the squares look grey or coloured?

http://www.huevaluechroma.com/pics/4.3.3.jpg
Square (A) looks colored (light purple) Square (B) looks grey to me.

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29. Thanks first and foremost for your website, it was extremely informative and has helped me no end in my artistic growth. I am very much grateful.

I have this idea, I want to desaturate the pure (100% saturation) colours in order to find out their values in black and white as a reminder.

Would this be possible using the desaturate or greyscale function in ps? Would it be inaccurate or misleading?

30. You're very welcome, and thanks for asking your questions here, Care.

The answer to your first question is that you can either use the desaturate command in Lab mode (not RGB mode), or convert to greyscale mode, among other methods that achieve the exact same result, i.e. conversion of each colour to a grey of the same "L".

My answer to your second question is that it seems this conversion may sometimes be somewhat inaccurate or misleading for some colours (mainly high saturation blues), but there may not be any better general solution. Here's why:

In the attachment each coloured square is on a background of a grey of the same value according to these methods.

The red square may look at first glance to be lighter than the grey background, but if I squint I can see that it is actually similar, or at least much more similar in value than it first seems. So I think that this is probably just an example of the impression of extra lightness that strong colours seem to possess, which been called "chromatic luminance" or the Helmholtz–Kohlrausch effect.

However, no matter how much I squint, the blue square still looks lighter to me than the grey surround, so I'd have to say that it actually is lighter. A possible explanation is that the blue "phosphor" of my laptop is less violet/less saturated than that assumed for the conversion, which would make pure blue (B255) relatively bright on my screen, and also pure yellow (white minus blue) a little less bright. So maybe there's no conversion that will be perfectly accurate for all monitors.

If anyone has any ideas please comment!

A separate issue is that on most laptops the relative values of the colours and the greys change alarmingly with viewing angle; my observations were made looking flat on to the screen.

Last edited by briggsy@ashtons; December 6th, 2012 at 06:52 PM.

31. thanks briggs you always come through, i really enjoy how u manage to verbalize things that i suppose we must unconsciously recognize, everytime you explain something i get that aaaa now it makes sense moment...

like i feared, there might not be an all encompassing way to accurately determine this. but its no big deal. i guess its just a case of learning the exceptions to the rule etc

32. This year my regular classes are (so far) all on late afternoons, evenings and weekends, which means I'll be available to run a few 5-day colour workshops more or less on demand at my studio in Sydney. The focus can be either on theory, or on putting the theory into practice using traditional or digital media. If anyone is interested please email me to discuss suitable dates.

Here's a still life demo from the first one last week.

33. Originally Posted by briggsy@ashtons
In the attachment each coloured square is on a background of a grey of the same value according to these methods.

The red square may look at first glance to be lighter than the grey background, but if I squint I can see that it is actually similar, or at least much more similar in value than it first seems. So I think that this is probably just an example of the impression of extra lightness that strong colours seem to possess, which been called "chromatic luminance" or the Helmholtz–Kohlrausch effect.

However, no matter how much I squint, the blue square still looks lighter to me than the grey surround, so I'd have to say that it actually is lighter. A possible explanation is that the blue "phosphor" of my laptop is less violet/less saturated than that assumed for the conversion, which would make pure blue (B255) relatively bright on my screen, and also pure yellow (white minus blue) a little less bright. So maybe there's no conversion that will be perfectly accurate for all monitors.
The value of that blue looks pretty good to me when I squint (at least equivalent to the red in terms of matching the ground), so I'd say your suspicion about the phosphors is probably right.

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35. Not sure if this is appropriate place to say it, but I appreciate the feedback on my SB and all the help you've given this community towards questions on color and light, sir.

36. Originally Posted by Elwell
The value of that blue looks pretty good to me when I squint (at least equivalent to the red in terms of matching the ground), so I'd say your suspicion about the phosphors is probably right.
Just a followup to that, I'm checking this thread today from a school computer, and the blue looks fine, but the red is appreciably darker than its corresponding gray.

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