![]() You can get arbitrarily close to instantaneous and still expect meaningful chromaticity, within the bounds of precision, so the limit as the sampling bandwidth goes to 0 is the ideal spectral locus, even if it disappears at exactly 0. Instead, they're narrow bands of the spectrum near their wavelengths. The use of sampling means that the spectrums for the monochromatic sources are not taken to be instantaneous values. The simplest explanation is that Y at the base of the shape is actually ever-so-slightly greater than zero. However, that then raises the question: how do they have chromaticity at all, since the other two functions should also be 0? This now makes some sort of sense, since they are monochromatic colors, and their spectrums should consist of a single point, and thus when you take the integral over a single point you'll always get 0. It's identical to the rendering I had produced a few hours earlier, and trying to figure out why it didn't make sense is, in part, what led me here.įor readers: the rendering is what results when you convert from, i.e. If (RGB.x >= 0.0 & RGB.y >= 0.0 & RGB.z >= 0.I happened upon this question while searching for a slightly different but related issue, and what immediately caught my eye is the rendering at the top. For each xy coordinate of my diagram I need to determine the corresponding Y value to obtain the complete xyY representation, from which XYZ tristimulus values can be calculated and finally sRGB values derived. ![]() Mat3 XYZtoRGB = mat3(vec3( 0.032406f, -0.009689f, 0.000557f), Starting with xy coordinates I would like to plot a chromaticity diagram displaying just the sRGB gamut portion of the CIE xy chromaticity diagram. Theme Copy tic format long N 7 color depth per channel in bits Gamma 2.2 Rx 0. But starting with xy coordinates is there a method for determining the Y component ps. Vec3 XYZ = xyY.z * vec3(xyY.x / xyY.y, 1.0, (1.0 - xyY.x - xyY.y) / xyY.y) Using the math on I've managed to get a point cloud of the surface of a color volume (code below) in the CIELAB space, but that's where I get stuck. By starting with the known sRGB Red, Green, Blue and white values: FF0000, 00FF00, 0000FF, FFFFFF respectively I can use the Bradford matrix to deduce their XYZ tristimulus values and thus their xyY representations. Vec3 xyY = vec3(pass_Position.xy, 100.0) Here is some GLSL code that does the job: vec4 color = vec4(0.0, 0.0, 0.0, 1.0) optionally, convert linear (R,G,B) to gamma-corrected sRGB values.otherwise, divide each of R, G and B by the largest of the three.if one of R, G or B is negative, we are out of the sRGB space.So one way to take an (x,y) pair and find out whether it can be obtained from an sRGB value is: Multiplying each of the RGB triplets with a constant doest not affect x or y. Note that if linear RGB values (R,G,B) are transformed to (x,y,Y), then (a*R,a*G,a*B) are transformed to (x,y,a*Y). ![]() Your computer cannot display violet, do not make people thinking that colour with low x,y is violet (it is not). Now you can plot in x,y diagram a RGB colour (and in the correct place). It is possible to know whether there is at least one sRGB colour that will map to a given x,y pair, though. If you want to do properly, select 16x16x16 RGB values, calculate they X,Y,Z and then x,y. The Y component is for you to choose, because for a given x,y pair, depending on Y, the colour may or may not lie within the sRGB volume.
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