color - Create colour with given lightness value

How can I create a colour of my chosen hue with a given fixed lightness level?

For example, how can I create a reddish colour, col, so that ColorConvert[col, "GrayScale"] will return precisely GrayLevel[0.6]?

Does Mathematica have something built-in to assist in this, or do I need to read up on how colours are converted between various representations? Note that the b value in Hue[h, s, b] does not correspond to the GrayLevel value. I need to have precise control over what certain colours I use will look like when converted to greyscale.

Answer

Conversion Formula

ColorConvert uses the following formula for "Grayscale" conversion:

$ \mathrm{Grayscale} = 0.299 R + 0.587 G + 0.114 B$

where $R$, $G$, and $B$ are normalized.

Interactive Example

The following manipulate example will help you finding (and confirming) the conversion based on a fixed $\mathrm{Grayscale}$ and $R$ values.

f[g_, gs_, r_] := (gs - .299 r - .587 g)/.114;

Manipulate[
 Column[{
   Show[
    Graphics[{Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}},
       VertexColors -> {RGBColor[r, 0, 0], RGBColor[r, 1, 0], 
         RGBColor[r, 1, 1], RGBColor[r, 0, 1]}]},
     PlotRange -> {{0, 1}, {0, 1}}, PlotRangeClipping -> True, 
     AspectRatio -> 1,

     ImageSize -> 400,
     Frame -> True, FrameLabel -> {"Green", "Blue"},
     LabelStyle -> {FontFamily -> "Arial", FontSize -> 14}],
    Plot[f[g, gs, r], {g, 0, 1}],
    Graphics[{
      Locator[
       Dynamic[pt, (With[{b = f[#[[1]], gs, r]}, 
           If[0 <= b <= 1, pt = {#1[[1]], b}, {.5, .5}]]) &]]
      }]
    ],

   "",
   Row[{"ColorConvert[RGBColor[", Prepend[pt, r], 
     ",\"Grayscale\"]==\n", 
     ColorConvert[RGBColor[Prepend[pt, r]], "Grayscale"]}]
   }],
 {{pt, {.5, .5}}, ControlType -> None},
 {{gs, .5, "Grayscale"}, 0, 1, Appearance -> "Labeled"},
 {{r, 1, "Red"}, 0, 1, Appearance -> "Labeled"},
 SaveDefinitions -> True]

grayscale

The blue line indicates the colors which has the exact grayscale value. If you move the locator, it tries to snap in to the line and show you RGB value as well as converted graysale value (which should be the same as the slider value).

Discussion

belisarius asked, "Then what is the yellowest color whose grayscale value is .3?".

There are several problems. First, what is "yellow"-ish colors? Are these colors close to {1,1,0} in RGB space? Then the solution is the point on the plane 0.299 x + 0.587 y + 0.114 z = 0.3 whose distance is the shortest from {1,1,0} which can be easily solvable. But you will quickly realize that it is not really "yellow". For instance, {1,0,0} and {0,1,0} are both exactly the same distance from the yellow, but none perceive them as yellowish. Then, is it a point with the proportion of R and G is 1:1? It is better but not entirely. Red and green have difference luminescence (thus different weights in grayscale conversion). Now, should we consider the weight? Well, then how about gamma and gamut? Then Euclidean distance becomes useless... I wouldn't even start the fact that human color perception is not linear but more of quadratic space... And then there is a whole story about chromaticity and visible spectrum (then different "yellow" there).

Getting the closest color for some color can be surprisingly hard.

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],

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