Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
481 views
in Technique[技术] by (71.8m points)

colors - What is the math behind the Colour Wheel

I want to create a pie with 12 slices, with each slice a different colour.

Pretty much every colour wheel seems to follow the same format; eg: http://www.tigercolor.com/color-lab/color-theory/color-theory-intro.htm .

But what algorithms are there for generating the colours? What is the math behind RGB(theta)? Surely there must be some established science on this, but Google is not giving me any clues.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Have a look at http://www.easyrgb.com it has the algorithms behind many color conversions. Here's the RGB -> HSV one.

var_R = ( R / 255 )                     //RGB from 0 to 255
var_G = ( G / 255 )
var_B = ( B / 255 )

var_Min = min( var_R, var_G, var_B )    //Min. value of RGB
var_Max = max( var_R, var_G, var_B )    //Max. value of RGB
del_Max = var_Max - var_Min             //Delta RGB value 

V = var_Max

if ( del_Max == 0 )                     //This is a gray, no chroma...
{
   H = 0                                //HSV results from 0 to 1
   S = 0
}
else                                    //Chromatic data...
{
   S = del_Max / var_Max

   del_R = ( ( ( var_Max - var_R ) / 6 ) + ( del_Max / 2 ) ) / del_Max
   del_G = ( ( ( var_Max - var_G ) / 6 ) + ( del_Max / 2 ) ) / del_Max
   del_B = ( ( ( var_Max - var_B ) / 6 ) + ( del_Max / 2 ) ) / del_Max

   if      ( var_R == var_Max ) H = del_B - del_G
   else if ( var_G == var_Max ) H = ( 1 / 3 ) + del_R - del_B
   else if ( var_B == var_Max ) H = ( 2 / 3 ) + del_G - del_R

   if ( H < 0 ) H += 1
   if ( H > 1 ) H -= 1
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...