How is the visible gamut bounded?

问题

Tho coordinates of a color in CIE XYZ space represent the stimulation of one each of the cone types in our eyes – hence tristimulus values. Not all coordinates (x, y, z) make sense (for example you can't have negative stimulation in any of the cones), so the domain of all possible combinations in XYZ space will be a blob. Sometimes you'll find that blob depicted, for example on Wikipedia:

(The colors on the blob are rather meaningless since they cannot actually be displayed on a standard RGB monitor.)

I'm now asking myself why this blob is bounded. Couldn't I just pick any point (x, y, z) in the blob and scale it with alpha*(x,y,z,), just like I would crank up a light source, and still be in the visible space? What exactly constitutes the upper boundary here?

回答1:

Context Definitions

The outer surface depicted in your Wikipedia image represents the volume bounded by the Colour Matching Functions (CMFS). The CMFS are defined by the CIE as follows:

tristimulus values of monochromatic stimuli of equal radiant power

Another name for the equal radiant power illuminant in the CIE system is Equal Energy or E.

The CIE XYZ tristimulus values are themselves defined by the CIE:

amounts of the 3 reference colour stimuli, in a given trichromatic system, required to match the colour of the stimulus considered

This is directly related to Colour Matching Experiments from Maxwell and Wright and Guild (1931) and probably out of the scope of this answer.

Outer Surface Generation

The outer surface is likely built using square wave spectral power distributions (SPDs), i.e. slices of the the Equal Energy illuminant. Bruce Lindbloom has an explanation on this page.

The methodology is to convert the square wave SPDs with various bandwidth to CIE XYZ tristimulus values, this will form the outer surface.

Assuming 5 bins, a first set of SPDs would be like that:

1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1

The second one:

1 1 0 0 0
0 1 1 0 0
0 0 1 1 0
0 0 0 1 1
1 0 0 0 1

The third:

1 1 1 0 0
0 1 1 1 0
0 0 1 1 1
1 0 0 1 1
1 1 0 0 1

etc...

Here a Python snippet using Colour to generate the points for the outer surface:

import colour
import numpy as np


class NearestNeighbourInterpolator(colour.KernelInterpolator):
    def __init__(self, *args, **kwargs):
        kwargs['kernel'] = colour.kernel_nearest_neighbour
        super(NearestNeighbourInterpolator, self).__init__(*args, **kwargs)


def generate_square_waves(samples):
    square_waves = []
    square_waves_basis = np.tril(np.ones((samples, samples)))[0:-1, :]
    for i in range(samples):
        square_waves.append(np.roll(square_waves_basis, i))
    return np.vstack((np.zeros(samples), np.vstack(square_waves),
                    np.ones(samples)))


def XYZ_outer_surface(samples):
    XYZ = []
    wavelengths = np.linspace(colour.DEFAULT_SPECTRAL_SHAPE.start,
                            colour.DEFAULT_SPECTRAL_SHAPE.end, samples)

    for wave in generate_square_waves(samples):
        spd = colour.SpectralPowerDistribution(
            wave, wavelengths).align(
                colour.DEFAULT_SPECTRAL_SHAPE,  
                interpolator=NearestNeighbourInterpolator)
        XYZ.append(colour.spectral_to_XYZ(spd))

    return np.array(XYZ).reshape(len(XYZ), -1, 3)


# 43 is picked as number of samples to have integer wavelengths.
colour.write_image(XYZ_outer_surface(43), 'CIE_XYZ_outer_surface.exr')

And the output:

来源：https://stackoverflow.com/questions/48390558/how-is-the-visible-gamut-bounded