Conversion from RGB to HSI

Given the intensities of the three primaries RGB of a color, we can find its HSV representation using different models. Here we use the RGB plane of the cube to find the corresponding HSV. The three vertices are represented by

and

, and the three components of the given color is represented by a 3D point

. We also assume the intensities are normalized so that the

and

values are between 0 and 1, so that point

is inside or on the surface of the color cube.

Determine the intensity I:

One of the definitions of intensity is

$\begin{displaymath}I\stackrel{\triangle}{=}\frac{1}{3}(R+G+B) \end{displaymath}$

Determine the hue H:

First find the intersection of the color vector with the RGB triangle :

$\begin{displaymath}\left\{ \begin{array}{l} r\stackrel{\triangle}{=}R/(R+G+B)=R... ... b\stackrel{\triangle}{=}B/(R+G+B)=B/3I \end{array} \right. \end{displaymath}$

This point

is on the RGB triangle as

. Here we assume the point

is inside the triangle formed by points

, and

. The hue is the angle $\angle H$ formed by the vectors $\overline{pw}$ and $\overline{P_rw}$ . Consider the dot product of these two vectors:

$\begin{displaymath}\overline{P_r w} \cdot \overline{pw} =(P_r-w)\cdot(p-w)=\vert P_r-w\vert\;\vert p-w\vert\;cos \angle H \end{displaymath}$

where

and

, and

$\begin{displaymath}(P_r-w)=(1,0,0)-(1/3,1/3,1/3)=(2/3, -1/3, -1/3) \end{displaymath}$

$\begin{displaymath}(p-w)=(r-1/3, g-1/3, b-1/3) \end{displaymath}$

$\begin{displaymath} (P_r-w)\cdot(p-w)=\frac{2}{3}(r-\frac{1}{3}) -\frac{1}{3}(g... ...)-\frac{1}{3}(b-\frac{1}{3}) =2r-g-b=\frac{2R-G-B}{3(R+G+B)} \end{displaymath}$

$\begin{displaymath}\vert P_r-w\vert=\sqrt{(1-\frac{1}{3})^2+(0-\frac{1}{3})^2+(0-\frac{1}{3})^2} =\sqrt{\frac{2}{3}} \end{displaymath}$

$\displaystyle \vert p-w\vert$	$\textstyle =$	$\displaystyle \sqrt{(r-\frac{1}{3})^2+(g-\frac{1}{3})^2+(b-\frac{1}{3})^2} =\sqrt{\frac{9(R^2+G^2+B^2)-3(R+G+B)^2}{9(R+G+B)^2}}$
	$\textstyle =$	$\displaystyle \frac{\sqrt{6(R^2+G^2+B^2-RG-GB-BR)}}{3(R+G+B)}$

Now the hue angle can be found to be

$\displaystyle \angle H$	$\textstyle =$	$\displaystyle cos^{-1}[\frac{(P_r-w)\cdot(p-w)}{\vert P_r-w\vert\;\vert p-w\vert}] =cos^{-1}[\frac{2R-G-B}{\sqrt{2/3}\sqrt{6(R^2+G^2+B^2-RG-GB-BR)}}]$
	$\textstyle =$	$\displaystyle cos^{-1}[\frac{3R-(R+G+B)}{2\sqrt{R^2+G^2+B^2-RG-GB-BR}}] =cos^{-1}[\frac{(R-G)+(R-B)}{2\sqrt{(R-G)^2+(R-B)(G-B)}}]$

, then $\angle H=360-\angle H$ .

Determine S:

The saturation of the colors on any of the three edges of the RGB triangle is defined as 1 (100% saturated), and the saturation of is zero. Denote as the intersection of the extension of line $\overline{wp}$ with the edge. If the normalized color is , , and if , . The saturation of any color point between and is defined as

$\begin{displaymath}S_p=\frac{\vert wp\vert}{\vert wp'\vert} =\frac{\vert wp'\ve... ...1-\frac{\vert pp'\vert}{\vert wp'\vert} =1-\frac{b}{1/3}=1-3b \end{displaymath}$

Here it is assumed that point

is inside the triangle $\triangle P_rwP_g$ so that

. In general

$\begin{displaymath}S_p=1-3\;min(r,g,b) =\left\{ \begin{array}{cl} 0, & min(r,g,... ...the edges} \\ 0<S_p<1, & 0<min(r,g,b)<1/3 \end{array} \right. \end{displaymath}$