Conversion from RGB to HSI

Given the intensities of the three primaries $R$, $G$, and $B$ 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 $P_r=(1,0,0)$, $P_g=(0,1,0)$ and $P_b=(0,0,1)$, and the three components of the given color is represented by a point $P=(R,G,B)$ in the RGB 3-D space. As we assume the intensities are normalized so that the $R$, $G$ and $B$ values are between 0 and 1, so that point $P$ is inside or on the surface of the color cube.

• Determine the intensity $I$:

  The intensity $I$ can be defined as:
  

  $$I\stackrel{\triangle}{=}\frac{1}{3}(R+G+B),\;\;\;\;\mbox{i.e.} \;\;\;\;\;R+G+B=3I$$
  
  

• Determine the hue $H$:

  First find the intersection of the color vector $(R,G,B)$ with the RGB plane (triangle) $R+G+B=1$:
  

  $$\left\{ \begin{array}{l} r\stackrel{\triangle}{=}R/(R+G+B)=R... ... b\stackrel{\triangle}{=}B/(R+G+B)=B/3I \end{array} \right.$$
  
  
  This point $P=(r,g,b)$ is on the RGB triangle as $r+g+b=1$. The hue is the angle $H$ formed by the vectors $\overline{pw}$ and $\overline{P_rw}$. Consider the dot product of these two vectors:
  
  $$\overline{P_r W} \cdot \overline{PW} =(P_r-W)\cdot(P-W)=\vert P_r-W\vert\;\vert P-W\vert\;\cos H$$
  
  
  where $P_r=(1,0,0)$, $P=(r,g,b)$, and $W=(1/3,1/3,1/3)$, and
  
  $$\overline{P_rW}=P_r-W=(1,0,0)-\left(\frac{1}{3},\frac{1}{3},\... ...}\right) =\left(\frac{2}{3}, -\frac{1}{3}, -\frac{1}{3}\right)$$
  
  
  
  
  $$\overline{PW}=P-W=\left(r-\frac{1}{3}, g-\frac{1}{3}, b-\frac{1}{3}\right)$$
  
  
  We assume $G>B$; i.e., $0\le H \le \pi$, and find the hue angle to be:
  
  $$H=\cos^{-1}\left(\frac{(P_r-W)\cdot(P-W)}{\vert P_r-W\vert\;\vert P-W\vert}\right)$$
  
  
  where
  
  $$(P_r-W)\cdot(P-W)=\frac{2}{3}\left(r-\frac{1}{3}\right) -\fra... ...}{3}\left(b-\frac{1}{3}\right) =2r-g-b=\frac{2R-G-B}{3(R+G+B)}$$
  
  
  
  
  $$\vert P_r-W\vert=\sqrt{\left(1-\frac{1}{3}\right)^2+\left(0-\... ...3}\right)^2 +\left(0-\frac{1}{3}\right)^2}=\sqrt{\frac{2}{3}}$$
  
  
  
  

  $$\vert P-W\vert = \sqrt{\left(r-\frac{1}{3}\right)^2+\left(g-\frac{1}{3}\right)^2 +\left(b-\frac{1}{3}\right)^2} =\sqrt{\frac{9(R^2+G^2+B^2)-3(R+G+B)^2}{9(R+G+B)^2}}$$

  $$= \frac{\sqrt{6(R^2+G^2+B^2-RG-GB-BR)}}{3(R+G+B)}$$

  
  
  Substituting, we get the hue angle:
  

  $$H = \cos^{-1}\left(\frac{(P_r-W)\cdot(P-W)}{\vert P_r-W\vert\;\vert P... ... =\cos^{-1}\left(\frac{2R-G-B}{\sqrt{2/3}\sqrt{6(R^2+G^2+B^2-RG-GB-BR)}}\right)$$

  $$= \cos^{-1}\left(\frac{3R-(R+G+B)}{2\sqrt{R^2+G^2+B^2-RG-GB-BR}}\right) =\cos^{-1}\left(\frac{(R-G)+(R-B)}{2\sqrt{(R-G)^2+(R-B)(G-B)}}\right)$$

  
  
  If $B>G$, then $H=360-H$.

• 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 $w=(1/3,1/3,1/3)$ is zero. Denote as $p'$the intersection of the extension of line $\overline{WP}$ with the edge. If the normalized color is $p=w$, $S=0$, and if $p=p'$, $S=1$. The saturation of any color point $p$ between $w$ and $p'$ is defined as
  

  $$S=\frac{\vert WP\vert}{\vert WP'\vert} =\frac{\vert WP'\vert... ...1-\frac{\vert PP'\vert}{\vert WP'\vert} =1-\frac{b}{1/3}=1-3b$$
  
  
  Here it is assumed that point $p$ is inside the triangle $\triangle P_rWP_g$ so that $b=min(r,g,b)$. In general
  
  $$S=1-3\;min(r,g,b) =\left\{ \begin{array}{cl} 0, & min(r,g,b)... ...f the edges} \\ 0<S<1, & 0<min(r,g,b)<1/3 \end{array} \right.$$