HSV三角形中的公式 [英] The formulars in HSV triangle

问题描述

我不知道以下两个公式是如何得出的.请解释一下.我的名声太低，以至于我不能问那个公式的编写者.

I do not know how the following two formulas are derived. Please kindly explain. My reputation point is so low that I can't ask the person who wrote the formula.

C#中的HSV三角形

var sat = (1 - 2*y1) / (sqrt3*x1 - y1 + 2);
var val = (sqrt3*x1 - y1 + 2) / 3;

推荐答案

转换后，我需要三角形的角触摸单位圆的<0,-1>"top"，<-√3/2,1/2>"bottom-left "和<√3/2,1/2>右下".我选择左下角为黑点( val = 0)，当 val = 1时，顶部为色点( val = 1和 sat = 1)，右下角是白点( val = 1和 sat = 0)

After the transformation, I needed the corners of the triangle to touch the unit circle at the points <0,-1> "top", <-√3/2,1/2> "bottom-left" and <√3/2,1/2> "bottom-right". I picked bottom-left to be the black point (val = 0), and when val = 1, the top would be the color point (val = 1 and sat = 1), and the bottom-right would be the white point (val = 1 and sat = 0).

为此，我首先将右边缘定义为方程式

To achieve this, I first defined the right edge as the equation

<x,y> = <0,-1>*sat + <√3/2,1/2>*(1-sat)
        ^            ^-- When sat = 0, the result is this point
        '-- When sat = 1, the result is this point

然后，当 val 变为零时，我将这条线向<-√3/2,1/2>缩放

I then scaled this line towards <-√3/2,1/2> when val goes to zero

<x,y> = ( <0,-1>*sat + <√3/2,1/2>*(1-sat) )*val + <-√3/2,1/2>*(1-val)
<x,y> = <0,-1>*sat*val + <√3/2,1/2>*(1-sat)*val + <-√3/2,1/2>*(1-val)
        ^                ^                        ^-- When val = 0, the result is this point
        |                '-- When val = 1 and sat = 0, the result is this point
        '-- When val = 1 and sat = 1, the result is this point

从矢量形式转换:

x = 0*sat*val + √3/2*(1-sat)*val + -√3/2*(1-val)
y = -1*sat*val + 1/2*(1-sat)*val + 1/2*(1-val)

扩展:

x = -√3/2*val*sat + √3*val - √3/2
y = -3/2*val*sat + 1/2

重新排列:

(√3*x - y + 2)/3 = val
(1 - 2*y)/3 = sat*val

求解 sat 和 val :

sat = (1 - 2*y)/(√3*x - y + 2)
val = (√3*x - y + 2)/3

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