SCI 200 Physical Science Lecture 9 Color Mixing Rob Daniell July 28, 2011.

SCI 200 Physical Science Lecture 9

Color Mixing

Rob Daniell

July 28, 2011

SCI200 - Lecture 10 2revised 2011.07.27

Color Mixing

Gilbert & Haeberli: Physics in the ArtsChapter 7 Additive color mixingChapter 8 Subtractive color mixing

Supplementary materials:Malacara, Daniel: Color Vision and

Colorimetry: Theory and Applications,SPIE Press, Bellingham, 2001

Wikipedia articles on Hue, sRGB, & ITU Recommendation 709 (Rec. 709)


Color Mixing

Trichromacy allows fairly strong hue discriminationNevertheless:

The same hue can be produced by different spectral distributions

Response of cones to any particular spectral distribution of light is complex

Is there a (relatively) simple way to describe or specify a particular hue?


Spectral Response of Cones

Two laser pointers (left) at 532 nm and 633 nm give the same hue as a single laser pointer (right) at 570 nm Note that the authors seem to ignore the 2 pulses in the

Type II cone produced by the 633 nm light Gilbert & Haeberli, Physics in the Arts, pp. 86-87.


Idealized Color Wavelength Ranges

Gilbert & Haeberli, Physics in the Arts, p. 83.

400-500 nm 500-600 nm 600-700 nm


Color Mixing

Additive mixingTwo light sources

Combined by adding the intensity at each wavelength

Subtractive mixingFilters, paints, dyes

Combined effect is produced by subtracting colors

Also depends on the light source


Additive Color Mixing (a), (b), & (c): three non-

monochromatic light sources

(d), (e), & (f): three two-color mixtures, each producing a third color

(g): spectral yellow (h): unsaturated yellow

produced by combining spectral green and spectral red

(i): the three non-spectral light sources mixed to produce white


Additive Color Mixing (a): spectral blue + spectral

yellow = white Complementary colors

(b): broad band white Mixture of many colors

metamers: Two different intensity distributions that appear the same to your eye


Additive Color Mixing

•Additive color rules:• R + G + B = W• R + G = Y• G + B = C• R + B = M

• Complementary colors:• R + C = W• G + M = W• B + Y = W

• Can 3 colors be combined to produce any other color?

C = Cyan, M = Magenta, Y = Yellow

W = White



• Red, Green, & Blue can be combined to produce most colors, but some saturated colors cannot be reproduced.• Red, Green, & Blue can be

combined to produce more colors than any other choice of primary colors

C = Cyan, M = Magenta, Y = Yellow

W = White



CIE Color Matching experiments in 1922 Illuminated region subtended a visual angle of 2°

Only the viewer’s fovea is illuminated Matching field produced by 700 nm, 546.1 nm, 435.8 nm

Three spectral lines in mercury vapor



Hue, Saturation, and Luminance of the reference field were to be matched as closely as possible.

For some wavelengths (in fact, nearly all wavelengths) of the monochromatic source, a perfect match was impossible. Except by adding one of the red, green, or blue matching colors to the

reference field.


Additive Color Mixing• Suppose you attempt to match

spectral cyan (490 nm) using• spectral red (650 nm)• spectral green (530 nm)• spectral blue (460 nm)

• Spectral cyan produces• I: 1 pulse• II: 7 pulses• III: 2.5 pulses

• spectral blue + spectral green produce• I: 2 + 0 = 2 1• II: 2.5 + 16.5 = 19 9.5• III: 1 + 9 = 10 5


Additive Color Mixing• Spectral cyan:

• I: 1 pulse• II: 7 pulses• III: 2.5 pulses

• 50% blue + 36% green:• I: 1 + 0 = 1• II: 1 + 6 = 7• III: 0.5 + 3 = 3.5

• Spectral cyan + 33% red:• I: 1 + 0 = 1• II: 7 + 0 = 7• III: 2.5 + 1 = 3.5

• 50% blue + 36% green matches spectral cyan + 33% red


Additive Color Mixing• 50% blue + 36% green

matches spectral cyan + 33% red• But spectral cyan + spectral

red yields white:• So spectral cyan + 33% red is

the same as 67% cyan + 33% white

• That is, unsaturated cyan



• Above: Spectral cyan on left; unsaturated cyan on the right• 50% blue + 36% green matches spectral cyan + 33% red• But spectral cyan + spectral red yields white:

• So spectral cyan + 33% red is the same as 67% cyan + 33% white• That is, unsaturated cyan



• Combine 460 nm (blue), 530 nm (green), and 650 nm (red)• To match the wavelength on the horizontal axis• Negative fractions mean you must ‘unsaturate’ the color you are trying to

match• Standard observer: Based on data from 1931



• Previous figure redrawn so that complementary spectral colors lie opposite each other.• Saturated colors lie along the

outside• Mixtures of two colors are

proportional to the distance between them• along the line joining them

• A line from one spectral color through the white point ends at the complementary color• White point depends on the

precise standard used• The hue of the color at F is found

by extending a line from white through F to the edge



• International Commission for Illumination (CIE)• Did not want to use ‘negative

colors’ to match saturated colors• Defined ‘imaginary colors’ [X], [Y],

and [Z]• Not physical colors, but related

• Defined ‘color matching functions’ x, y, and z• Relative amounts of [X], [Y], and

[Z] needed to match a given spectral color (wavelength)

• Results in “tristimulus” values x, y, and z

• z = 1 - x - y • x and y together are called the

‘chromaticity’ of a color



• CIE Chromaticity Diagram:• Horseshoe shaped curve

represents saturated monochromatic colors• 380 nm - 700 nm

• ‘purples’ (mixtures of blue and red) lie along bottom edge• z = 1 - x - y• Interior of horseshoe

represents unsaturated colors



• CIE Chromaticity Diagram:• Specify a color by three

values• Chromaticity (x and y)• Lightness (or

Brightness)

• No representation is perfect• None can reproduce

the precise color sensitivity of the human eye

• Individual differences



• RGB example:• 3 standard wavelengths:

• 700 nm (red)• 546.1 nm (green)• 435.8 nm (blue)

• These three colors can only reproduce colors in the interior of the triangle



•CIE chromaticity diagram:• Three “primary” colors

can only produce the colors inside the triangle - the “gamut” of the color set.• This triangle

corresponds to the three colors used in a typical color monitor.



•Monochromatic complementary pairs• 700 nm -- 494 nm• 600 nm -- 490 nm• 571 nm -- 380 nm• No complementary pairs

between 494 nm & 571 nm



• Color Triangle (Fig. 7.4) from textbook• Doesn’t cover gamut of the

human eye• Does not conform to the

sRGB standard used for color televisions and computer monitors

• Does come close to maximizing the gamut of representable colors



• Color Triangle inside a chromaticity diagram (Fig. 7.15)• It appears that the text uses

• Red = 700 nm• Green = 530 nm• Blue = 440 nm

• Does not conform to the “sRGB” standard for color television and computer monitors



• sRGB standard• Based on 1931 CIE

chromaticity• Red: x = 0.64, y = 0.33• Green: x = 0.30, y = 0.60• Blue: x = 0.15, y = 0.06• White point:

• x = 0.3127, y = 0.3290

From the diagram, the dominant wavelengths are

Red: 611 nm Green: 549 nm Blue: 464 nm

From Wikipedia

http://en.wikipedia.org/wiki/Rec._709



• AdobeRGB standard• Based on 1931 CIE

chromaticity• Red: x = 0.64, y = 0.33• Green: x = 0.21, y = 0.71• Blue: x = 0.15, y = 0.06• White point:

• x = 0.3127, y = 0.3290

From the diagram, the dominant wavelengths are

Red: 611 nm Green: 535 nm Blue: 464 nm

Green point is main difference with sRGB

From Wikipedia

http://en.wikipedia.org/wiki/Rec._709



• Complementary Colors to the “line of purples” appear to range from c491 nm to c579 nm



• correspondence between wavelength and hue (angle):– complicated


Additive Color Mixing• correspondence

between RGB and hue (angle):



• LEDs approximate monochromatic red, green and blue:• Red = 645 nm• Green = 510 nm• Blue = 465 nm



• LEDs approximate monochromatic red, green and blue:• Red = 645 nm• Green = 510 nm• Blue = 465 nm


Additive Color Mixing Methods

Methods and TechniquesSimple addition

Two or more light sourcesStage lights

Partitive mixingSeparate sources close to each other

Television screens & CRT displaysLCD displays

Rapid successionUses persistence of vision


Subtractive Color Mixing

•Subtractive color combination:• Filters that absorb or block

light of certain colors• Ink or pigments that reflect

only certain colors and absorb the others

• Primary Subtractive Colors:• Cyan, Magenta, Yellow• Supplemented by Black in

“four color printing”C = Cyan, M = Magenta, Y = Yellow


Subtractive Color Mixing: Idealized Filters

Idealized Red Filter Idealized Green Filter Idealized Blue Filter



Idealized Red Filter + Idealized Green Filter = “black” (no transmission)



Idealized Cyan Filter (blocks Red)

Idealized Magenta Filter (blocks Green)

Idealized Yellow Filter (blocks Blue)



Idealized Cyan Filter + Idealized Magenta Filter = Blue filter


Subtractive Color Mixing: Block Diagrams

Subtractive Primaries:

• C = B + G = W - R

• M = B + R = W - G

• Y = G + R = W - B

-G

-R

-B


Subtractive Color Mixing: Block Diagrams

Basic Subtractive Rules:

• C + M = B

• C + Y = G

• M + Y = R

• C + M + Y = Bk (Black)

-R -G

-R -B

-G -B

-R -G -B



Dyes & Pigments:

• Chemicals that absorb and/or reflect selective parts of the visible spectrum

• Ideal dyes & pigments follow the same rules as ideal filters (i.e., subtractive color mixtures)

• Real filters, dyes, and pigments often behave in complicated and (almost) unpredictable ways.



Real filters, dyes & pigments:

• Usually smooth “corners”

• Sometimes complicated structure

• Rarely achieve trans-missions and reflectances of 100%




• Can produce unexpected results

The transmission of two or more filters in series is the product of the transmissions of each individual filter.

For example: one filter with 70% transmission combined with another filter with 50% transmission produces a combined transmission of 35%

1 filter

2 identical filters

10 identical filters




• Can produce unexpected results

Combining filters or dyes with different transmittance functions can result in startling color shifts.

blue dye

yellow dye

one-to-one mixture



With subtractive colors the perceived color depends on the light source

cool white flourescent bulb

reflectance of a magenta object

resulting intensity distribution, the product of (a) and (b).



With subtractive colors the perceived color depends on the light source (example 2)

a gray object

a brown object

In sunlight, the two objects at right have very different colors. Under an incandescent bulb, which emits much less blue light, the objects have very similar colors.



Different light sources have different intensity distributions:

Incandescent light bulb

Deluxe Warm White Fluorescent bulb

High Pressure Sodium Lamp



Gamut of subtractive primaries:

Gamut of color TV or computer monitor:

Gamut of color slides:



Printing inks: color is due to a combination of reflection and transmission:


Homework Discussion

• Translation between Textbook color triangle and 8 bit RGB

• Hue, especially “purples”


Homework DiscussionTranslation between Textbook color

triangle and 8 bit RGB8 bit RGB is the color representation

system used on most computer monitors and other LCD graphic displays

1 byte = 8 bits: 0101100112 = 17910

Smallest number = 010

Largest number = 28 – 1 = 255

Each pixel is represented by 3 bytes1 each for red (R), green (G), and blue (B)


Homework Discussion• Translation between color

triangle and 8 bit RGB• r, g are “given”• b = 1 – r – g • Example: The point

indicated • r = 0.55,• g = 0.2,• b = 1 - 0.55 - 0.2 = 0.25

• Problem: Find the equivalent RGB representation



triangle and 8 bit RGB• Point:

• r = 0.55,• g = 0.2,• b = 1 - 0.55 - 0.2 = 0.25

• 1st Step:• Identify brightest component• r = 0.55 R = 255

• 2nd Step: Calculate the other two components:• G = (g/r) × 255• = (0.2/0.55) × 255 = 93 • B = (b/r) × 255• = (0.25/0.55) x 255 = 116

8 bit representation:R = 255G = 93B = 116



triangle and 8 bit RGB• Practice Problem #1:• Locate this point on your

color triangle• r = 0.15,• g = 0.25,• b = 1 – r – g = ___________

• 1st Step:• Identify brightest component

• 2nd Step: • Calculate the other two

components:

8 bit representation:R = ____G = ____B = ____



triangle and 8 bit RGB• Practice Problem #2:• Determinre r, g, b from the

indicated point:• r = _______,• g = _______,• b = 1 – r – g = ___________

• 1st Step:• Identify brightest component

• 2nd Step: • Calculate the other two

components:

8 bit representation:R = ____G = ____B = ____


Homework Discussion• Hue• Color: r = 0.35, g = 0.55• b = 0.1

• Dominant wavelength = 575 nm

• RGB:• R = 162• G = 255• B = 46• G > R > B


Homework Discussion• Hue• Color: r = 0.35, g =

0.55, b = 0.1• RGB:

• R = 162• G = 255• B = 46

• Lab:• L = luminance• a = red vs. green• b = blue vs. yellow


Homework Discussion• Hue• Color: r = 0.40,

g = 0.15, b = 0.45

• Dominant wavelength = c545 nm


Homework Assignment

• Read Chapters 7 & 8• Homework Packet 10

– Due August 4• Lab on Thursday, July 28 (today)

– Color Addition


Upcoming Labs

• Lab 7: Color Addition– July 28

• Lab 8: Color Subtraction– August 4

• Reminder: Last Exam– August 11

SCI 200 Physical Science Lecture 9 Color Mixing Rob Daniell July 28, 2011.

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