Editing HSL and HSV (section)

====Luma, chroma and hue to RGB====
Given a color with hue {{math|''H'' ∈ [0°, 360°)}}, chroma {{math|''C'' ∈ [0, 1]}}, and luma {{math|''{{prime|Y}}''<sub>601</sub> ∈ [0, 1]}},{{refn|group=upper-alpha |Some points in this cylinder fall out of [[gamut]].}} we can again use the same strategy. Since we already have ''H'' and ''C'', we can straightaway find our point {{math|(''R''<sub>1</sub>, ''G''<sub>1</sub>, ''B''<sub>1</sub>)}} along the bottom three faces of the RGB cube:
: <math>\begin{align}
 H^\prime &= \frac{H}{60^\circ} \\
 X    &= C \times (1 - |H^\prime \bmod 2 - 1|)
\end{align}</math>
: <math>(R_1, G_1, B_1) = \begin{cases}
  (0, 0, 0) &\text{if } H \text{ is undefined} \\
  (C, X, 0) &\text{if } 0 \leq H^\prime \leq 1 \\
  (X, C, 0) &\text{if } 1 \leq H^\prime \leq 2 \\
  (0, C, X) &\text{if } 2 \leq H^\prime \leq 3 \\
  (0, X, C) &\text{if } 3 \leq H^\prime \leq 4 \\
  (X, 0, C) &\text{if } 4 \leq H^\prime \leq 5 \\
  (C, 0, X) &\text{if } 5 \leq H^\prime < 6
\end{cases}</math>

Overlap (when <math>H^\prime</math> is an integer) occurs because two ways to calculate the value are equivalent: <math>X = 0</math> or <math>X = C</math>, as appropriate.

Then we can find ''R'', ''G'', and ''B'' by adding the same amount to each component, to match luma:
: <math>m = Y^\prime_{601} - (0.30R_1 + 0.59G_1 + 0.11B_1)</math>
: <math>(R, G, B) = (R_1 + m, G_1 + m, B_1 + m)</math>