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Color matching experiment 1

Slide credit: W. Freeman

Slide 22

Color matching experiment 1

Color matching experiment 1

p1 p2 p3

Slide credit: W. Freeman

Slide 23

Color matching experiment 1

Color matching experiment 1

p1 p2 p3

Slide credit: W. Freeman

Slide 24

Color matching experiment 1

Color matching experiment 1

p1 p2 p3

Slide credit: W. Freeman

Slide 25

Color matching experiment 2

Color matching experiment 2

Slide credit: W. Freeman

Slide 26

Color matching experiment 2

Color matching experiment 2

p1 p2 p3

Slide credit: W. Freeman

Slide 27

Color matching experiment 2

Color matching experiment 2

p1 p2 p3

Slide credit: W. Freeman

Slide 28

Color matching experiment 2

Color matching experiment 2

p1 p2 p3

p1 p2 p3

We say a “negative” amount of p2 was needed to make the match, because we added it to the test color’s side.

The primary color amounts needed for a match:

Slide 29

Color matching

Color matching

What must we require of the primary lights chosen?

How are three numbers enough to represent entire spectrum?

Slide 30

Metamers

Metamers

If observer says a mixture is a match  receptor excitations of both stimuli must be equal.

But lights forming a perceptual match still may be physically different

Match light: must be combination of primaries

Test light: any light

Metamers: pairs of lights that match perceptually but not physically

Slide 31

Forsyth & Ponce, measurements by E. Koivisto

Forsyth & Ponce, measurements by E. Koivisto

Metamers

Slide 32

Grassman’s laws

Grassman’s laws

If two test lights can be matched with the same set of weights, then they match each other:

Suppose A = u1 P1 + u2 P2 + u3 P3 and B = u1 P1 + u2 P2 + u3 P3. Then A = B.

If we scale the test light, then the matches get scaled by the same amount:

Suppose A = u1 P1 + u2 P2 + u3 P3. Then kA = (ku1) P1 + (ku2) P2 + (ku3) P3.

If we mix two test lights, then mixing the matches will match the result (superposition):

Suppose A = u1 P1 + u2 P2 + u3 P3 and B = v1 P1 + v2 P2 + v3 P3. Then A+B = (u1+v1) P1 + (u2+v2) P2 + (u3+v3) P3.

Here “=“ means “matches”.

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