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Math behind mixing two colors

by MM
Tags: colors, math, mixing
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MM
#1
Aug25-08, 05:00 AM
P: n/a
Hi,

I am stuck at the math one needs to do to mix colors. Here is the
exact problem statement:
I am provided the intensities (a.u.) of all the wavelengths in the
visible range (400 to 670 nm). I know the RGB values (between 0-1) at
all the wavelengths. I need to mix all the wavelengths and find out
the color I would get. For example: If the intensity at 650nm is 1 and
rest are all 0 I would get a shade of red. If I have intensities of
0.5 at 530 nm (green) and 670 nm (red) then I should get yellow.

To simplify the above problem:
If I am given the intensity and the RGB value (between 0.0 and 1.0) at
all 3 channels, for two colors. How do I "mix" them so as to obtain a
new color. Remember the math should be able to satisfy all the basic
rules like:
1. 1 Red + 1 Red = 2 Red
2. Red + Blue + Green = White
3. The RGB value for the final color should stay between 0 and 1.
4. 0.5 Red + 0.5 Green = 1.0 Yellow BUT 0.99 Red + 0.01 Green NOT= 1.0
Yellow

Sorry for the long post. Someone please HELP!!

-mm

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Redbelly98
#2
Aug25-08, 07:43 AM
Mentor
Redbelly98's Avatar
P: 12,068
Colors are characterized by their x and y chromaticity coordinates. You can start with the Wikipedia article,
http://en.wikipedia.org/wiki/CIE_1931_color_space

A google search on the following:
x y chromaticity
(or variations on those terms) might turn up more info.
CompuChip
#3
Aug25-08, 09:41 AM
Sci Advisor
HW Helper
P: 4,300
Err, how do I read
1 Red + 1 Red = 2 Red
?

Shouldn't 1 red + 1 red be 1 red (allowing all coefficients to range from 0 to 1 separately) or undefined (e.g. only allowing coefficients which sum to 1)

Ian Parker
#4
Aug27-08, 05:00 AM
P: n/a
Math behind mixing two colors

On 24 Aug, 21:07, MM <extra.mem...@gmail.com> wrote:
> Hi,
>
> I am stuck at the math one needs to do to mix colors. Here is the
> exact problem statement:
> I am provided the intensities (a.u.) of all the wavelengths in the
> visible range (400 to 670 nm). I know the RGB values (between 0-1) at
> all the wavelengths. I need to mix all the wavelengths and find out
> the color I would get. For example: If the intensity at 650nm is 1 and
> rest are all 0 I would get a shade of red. If I have intensities of
> 0.5 at 530 nm (green) and 670 nm (red) then I should get yellow.
>
> To simplify the above problem:
> If I am given the intensity and the RGB value (between 0.0 and 1.0) at
> all 3 channels, for two colors. How do I "mix" them so as to obtain a
> new color. Remember the math should be able to satisfy all the basic
> rules like:
> 1. 1 Red + 1 Red = 2 Red
> 2. Red + Blue + Green = White
> 3. The RGB value for the final color should stay between 0 and 1.
> 4. 0.5 Red + 0.5 Green = 1.0 Yellow BUT 0.99 Red + 0.01 Green NOT= 1.0
> Yellow
>
> Sorry for the long post. Someone please HELP!!
>
> -mm


What you have to do is to sum up sensitivities at all your
wavelengths, and then you will get an apparent color. These cureves
are given here.

http://en.wikipedia.org/wiki/Color_model

The red curve is quite interesting. Blue is in fact a mixure of blue
and red as the red curve has 2 humps. It does not therefore work quite
as you are suggesting. You need to work out at each wavelength what
you need to get the color. You have to solve 3 simultaneous equations
in each case.

- Ian Parker

Gerard Westendorp
#5
Aug28-08, 05:00 AM
P: n/a
MM wrote:
> Hi,
>
> I am stuck at the math one needs to do to mix colors. Here is the
> exact problem statement:
> I am provided the intensities (a.u.) of all the wavelengths in the
> visible range (400 to 670 nm). I know the RGB values (between 0-1) at
> all the wavelengths. I need to mix all the wavelengths and find out
> the color I would get.


This is the same problem a color camera has to solve.
I think they use 3 filters, one for each of {RGB}. So you just integrate
the spectrum weighted by a filter function for each component.

Gerard

Thomas Smid
#6
Aug28-08, 05:00 AM
P: n/a
Just add the corresponding R,G,B values together for all the colours
you want to mix, then divide everything by the highest resulting
value. So in mathematical terms for n colours

R=R1+R2 +...Rn
G=G1+G2 +...Gn
B=B1 +B2+...Bn

r=R/max(R,G,B)
g=G/max(R,G,B)
b=B/max(R,G,B)


Thomas

MM
#7
Aug29-08, 05:00 AM
P: n/a
On Aug 26, 9:43 am, Ian Parker <ianpark...@gmail.com> wrote:
> On 24 Aug, 21:07, MM <extra.mem...@gmail.com> wrote:
>
>
>
> > Hi,

>
> > I am stuck at the math one needs to do to mix colors. Here is the
> > exact problem statement:
> > I am provided the intensities (a.u.) of all the wavelengths in the
> > visible range (400 to 670 nm). I know the RGB values (between 0-1) at
> > all the wavelengths. I need to mix all the wavelengths and find out
> > the color I would get. For example: If the intensity at 650nm is 1 and
> > rest are all 0 I would get a shade of red. If I have intensities of
> > 0.5 at 530 nm (green) and 670 nm (red) then I should get yellow.

>
> > To simplify the above problem:
> > If I am given the intensity and the RGB value (between 0.0 and 1.0) at
> > all 3 channels, for two colors. How do I "mix" them so as to obtain a
> > new color. Remember the math should be able to satisfy all the basic
> > rules like:
> > 1. 1 Red + 1 Red = 2 Red
> > 2. Red + Blue + Green = White
> > 3. The RGB value for the final color should stay between 0 and 1.
> > 4. 0.5 Red + 0.5 Green = 1.0 Yellow BUT 0.99 Red + 0.01 Green NOT= 1.0
> > Yellow

>
> > Sorry for the long post. Someone please HELP!!

>
> > -mm

>
> What you have to do is to sum up sensitivities at all your
> wavelengths, and then you will get an apparent color. These cureves
> are given here.
>
> http://en.wikipedia.org/wiki/Color_model
>
> The red curve is quite interesting. Blue is in fact a mixure of blue
> and red as the red curve has 2 humps. It does not therefore work quite
> as you are suggesting. You need to work out at each wavelength what
> you need to get the color. You have to solve 3 simultaneous equations
> in each case.
>
> - Ian Parker


Hi Ian,

Thanks for our reply. The wikipedia link that you referred me to
describes how one can express colors in numbers. There are different
ways (RGB, CIE, HSV etc) and I think one can convert from one format
to another as well. If I sum up all the intensities then I will have
the total intensity but of what color? For example: if if the
intensity of blue light if 1 (a.u.) and that of red is 0.05 then then
the total intensity is 1.05 but I dont have anything that tells me
that it should have a bluish shade. Is the weighted average of RGB
values the right way to go?

-mm

Chalky
#8
Aug29-08, 05:00 AM
P: n/a
On Aug 26, 2:43 pm, Ian Parker <ianpark...@gmail.com> wrote:
> On 24 Aug, 21:07, MM <extra.mem...@gmail.com> wrote:
>
>
>
> > Hi,

>
> > I am stuck at the math one needs to do to mix colors. Here is the
> > exact problem statement:
> > I am provided the intensities (a.u.) of all the wavelengths in the
> > visible range (400 to 670 nm). I know the RGB values (between 0-1) at
> > all the wavelengths. I need to mix all the wavelengths and find out
> > the color I would get. For example: If the intensity at 650nm is 1 and
> > rest are all 0 I would get a shade of red. If I have intensities of
> > 0.5 at 530 nm (green) and 670 nm (red) then I should get yellow.

>
> > To simplify the above problem:
> > If I am given the intensity and the RGB value (between 0.0 and 1.0) at
> > all 3 channels, for two colors. How do I "mix" them so as to obtain a
> > new color. Remember the math should be able to satisfy all the basic
> > rules like:
> > 1. 1 Red + 1 Red = 2 Red
> > 2. Red + Blue + Green = White
> > 3. The RGB value for the final color should stay between 0 and 1.
> > 4. 0.5 Red + 0.5 Green = 1.0 Yellow BUT 0.99 Red + 0.01 Green NOT= 1.0
> > Yellow

>
> > Sorry for the long post. Someone please HELP!!

>
> > -mm

>
> What you have to do is to sum up sensitivities at all your
> wavelengths, and then you will get an apparent color. These cureves
> are given here.
>
> http://en.wikipedia.org/wiki/Color_model



Since the OP is given the RGB values for each wavelength, I don't
think he needs to worry too much about sensitivities, as these are
implied in the RGB values. Furthermore, since there are more colours
than names for them, it seems to me that he would be better employed
by displaying the resultant colours on a computer screen. This can be
done very easily in HTML using COLOR=(hash) xxyyzz where xx,yy and zz
are the RGB intensities translated into hexadecimal.

Bear in mind though that different screens display colour slightly
differently, as do different printers, unless carefully calibrated
using specialised equipment.

arivero@unizar.es
#9
Aug29-08, 05:00 AM
P: n/a
Hmm while this thread is interesting, it is not physics. The OP could
get best help in some biology forums, or even in art forums. It is
about the perception of retinal cells.


On 26 ago, 15:43, Ian Parker <ianpark...@gmail.com> wrote:
> On 24 Aug, 21:07, MM <extra.mem...@gmail.com> wrote:
>
>
>
> > Hi,

>
> > I am stuck at the math one needs to do to mix colors. Here is the
> > exact problem statement:
> > I am provided the intensities (a.u.) of all the wavelengths in the
> > visible range (400 to 670 nm). I know the RGB values (between 0-1) at
> > all the wavelengths. I need to mix all the wavelengths and find out
> > the color I would get. For example: If the intensity at 650nm is 1 and
> > rest are all 0 I would get a shade of red. If I have intensities of
> > 0.5 at 530 nm (green) and 670 nm (red) then I should get yellow.

>
> > To simplify the above problem:
> > If I am given the intensity and the RGB value (between 0.0 and 1.0) at
> > all 3 channels, for two colors. How do I "mix" them so as to obtain a
> > new color. Remember the math should be able to satisfy all the basic
> > rules like:
> > 1. 1 Red + 1 Red = 2 Red
> > 2. Red + Blue + Green = White
> > 3. The RGB value for the final color should stay between 0 and 1.
> > 4. 0.5 Red + 0.5 Green = 1.0 Yellow BUT 0.99 Red + 0.01 Green NOT= 1.0
> > Yellow

>
> > Sorry for the long post. Someone please HELP!!

>
> > -mm

>
> What you have to do is to sum up sensitivities at all your
> wavelengths, and then you will get an apparent color. These cureves
> are given here.
>
> http://en.wikipedia.org/wiki/Color_model
>
> The red curve is quite interesting. Blue is in fact a mixure of blue
> and red as the red curve has 2 humps. It does not therefore work quite
> as you are suggesting. You need to work out at each wavelength what
> you need to get the color. You have to solve 3 simultaneous equations
> in each case.
>
>   - Ian Parker


Tom Roberts
#10
Aug29-08, 05:06 AM
P: n/a
MM wrote:
> The wikipedia link that you referred me to
> describes how one can express colors in numbers. There are different
> ways (RGB, CIE, HSV etc) and I think one can convert from one format
> to another as well. If I sum up all the intensities then I will have
> the total intensity but of what color? For example: if if the
> intensity of blue light if 1 (a.u.) and that of red is 0.05 then then
> the total intensity is 1.05 but I dont have anything that tells me
> that it should have a bluish shade. Is the weighted average of RGB
> values the right way to go?


It's considerably more complicated than that, as people's retina
pigments vary (e.g. color blindness). For instance, IIRC most people's
"red" pigment has a secondary peak in the blue. The brain is HEAVILY
involved, and not all 3 colors are required for you to see a color image
-- the brain's image processing will infer colors that are not actually
present.

I remember a striking demonstration at a colloquium by Dr.
Land. He projected a red-and-white and a black-and-white
image of the same scene next to each other. He then carefully
merged them by aiming the projectors, and when the alignment
was correct the image suddenly "jumped" into full color.

The best way to get a basic understanding is to play with a color wheel
-- on most computers selecting colors has an option to use such a wheel.
Certainly my Mac does. Or you can use RGB sliders and see the resulting
color, or CMYK sliders. But, of course, this will all be dependent on
the responses of YOUR eyes, and the pigments of YOUR monitor, and the
color processing of YOUR computer; other peoples' perceptions and
equipment will vary.

I believe forums about digital photography discuss this in detail.

Tom Roberts



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