没有人知道任何式的调光频率转换为RGB值？
Answer 1:
这里的整个转换过程的详细说明： http://www.fourmilab.ch/documents/specrend/ 。 包含的源代码！
Answer 2:
对于懒惰的人（比如我），这里是在@ user151323的回答（即从发现PASCAL代码只是一个简单的翻译找到的代码Java的实现光谱实验室报告 ）：
static private double Gamma = 0.80;
static private double IntensityMax = 255;
/** Taken from Earl F. Glynn's web page:
* <a href="http://www.efg2.com/Lab/ScienceAndEngineering/Spectra.htm">Spectra Lab Report</a>
* */
public static int[] waveLengthToRGB(double Wavelength){
double factor;
double Red,Green,Blue;
if((Wavelength >= 380) && (Wavelength<440)){
Red = -(Wavelength - 440) / (440 - 380);
Green = 0.0;
Blue = 1.0;
}else if((Wavelength >= 440) && (Wavelength<490)){
Red = 0.0;
Green = (Wavelength - 440) / (490 - 440);
Blue = 1.0;
}else if((Wavelength >= 490) && (Wavelength<510)){
Red = 0.0;
Green = 1.0;
Blue = -(Wavelength - 510) / (510 - 490);
}else if((Wavelength >= 510) && (Wavelength<580)){
Red = (Wavelength - 510) / (580 - 510);
Green = 1.0;
Blue = 0.0;
}else if((Wavelength >= 580) && (Wavelength<645)){
Red = 1.0;
Green = -(Wavelength - 645) / (645 - 580);
Blue = 0.0;
}else if((Wavelength >= 645) && (Wavelength<781)){
Red = 1.0;
Green = 0.0;
Blue = 0.0;
}else{
Red = 0.0;
Green = 0.0;
Blue = 0.0;
};
// Let the intensity fall off near the vision limits
if((Wavelength >= 380) && (Wavelength<420)){
factor = 0.3 + 0.7*(Wavelength - 380) / (420 - 380);
}else if((Wavelength >= 420) && (Wavelength<701)){
factor = 1.0;
}else if((Wavelength >= 701) && (Wavelength<781)){
factor = 0.3 + 0.7*(780 - Wavelength) / (780 - 700);
}else{
factor = 0.0;
};
int[] rgb = new int[3];
// Don't want 0^x = 1 for x <> 0
rgb[0] = Red==0.0 ? 0 : (int) Math.round(IntensityMax * Math.pow(Red * factor, Gamma));
rgb[1] = Green==0.0 ? 0 : (int) Math.round(IntensityMax * Math.pow(Green * factor, Gamma));
rgb[2] = Blue==0.0 ? 0 : (int) Math.round(IntensityMax * Math.pow(Blue * factor, Gamma));
return rgb;
}
顺便说一句，这对我来说工作正常。
Answer 3:
总体思路：
- 使用CEI配色函数波长转换为XYZ色彩 。
- XYZ转换到RGB
- 夹子组件[0..1]和乘以255，以适应无符号字节范围。
步骤1和2可以变化。
有几种颜色匹配函数，可作为表或作为解析近似值（由@Tarc和@Haochen解所建议的）。 表是最好的，如果你需要一个平稳preсise结果。
没有单一的RGB色彩空间。 多个变换矩阵可被用于和不同种类的伽马校正。
下面是我想出了最近的C＃代码。 它使用线性内插在“1964年CIE标准观察者”表和sRGB的矩阵+伽玛校正 。
static class RgbCalculator {
const int
LEN_MIN = 380,
LEN_MAX = 780,
LEN_STEP = 5;
static readonly double[]
X = {
0.000160, 0.000662, 0.002362, 0.007242, 0.019110, 0.043400, 0.084736, 0.140638, 0.204492, 0.264737,
0.314679, 0.357719, 0.383734, 0.386726, 0.370702, 0.342957, 0.302273, 0.254085, 0.195618, 0.132349,
0.080507, 0.041072, 0.016172, 0.005132, 0.003816, 0.015444, 0.037465, 0.071358, 0.117749, 0.172953,
0.236491, 0.304213, 0.376772, 0.451584, 0.529826, 0.616053, 0.705224, 0.793832, 0.878655, 0.951162,
1.014160, 1.074300, 1.118520, 1.134300, 1.123990, 1.089100, 1.030480, 0.950740, 0.856297, 0.754930,
0.647467, 0.535110, 0.431567, 0.343690, 0.268329, 0.204300, 0.152568, 0.112210, 0.081261, 0.057930,
0.040851, 0.028623, 0.019941, 0.013842, 0.009577, 0.006605, 0.004553, 0.003145, 0.002175, 0.001506,
0.001045, 0.000727, 0.000508, 0.000356, 0.000251, 0.000178, 0.000126, 0.000090, 0.000065, 0.000046,
0.000033
},
Y = {
0.000017, 0.000072, 0.000253, 0.000769, 0.002004, 0.004509, 0.008756, 0.014456, 0.021391, 0.029497,
0.038676, 0.049602, 0.062077, 0.074704, 0.089456, 0.106256, 0.128201, 0.152761, 0.185190, 0.219940,
0.253589, 0.297665, 0.339133, 0.395379, 0.460777, 0.531360, 0.606741, 0.685660, 0.761757, 0.823330,
0.875211, 0.923810, 0.961988, 0.982200, 0.991761, 0.999110, 0.997340, 0.982380, 0.955552, 0.915175,
0.868934, 0.825623, 0.777405, 0.720353, 0.658341, 0.593878, 0.527963, 0.461834, 0.398057, 0.339554,
0.283493, 0.228254, 0.179828, 0.140211, 0.107633, 0.081187, 0.060281, 0.044096, 0.031800, 0.022602,
0.015905, 0.011130, 0.007749, 0.005375, 0.003718, 0.002565, 0.001768, 0.001222, 0.000846, 0.000586,
0.000407, 0.000284, 0.000199, 0.000140, 0.000098, 0.000070, 0.000050, 0.000036, 0.000025, 0.000018,
0.000013
},
Z = {
0.000705, 0.002928, 0.010482, 0.032344, 0.086011, 0.197120, 0.389366, 0.656760, 0.972542, 1.282500,
1.553480, 1.798500, 1.967280, 2.027300, 1.994800, 1.900700, 1.745370, 1.554900, 1.317560, 1.030200,
0.772125, 0.570060, 0.415254, 0.302356, 0.218502, 0.159249, 0.112044, 0.082248, 0.060709, 0.043050,
0.030451, 0.020584, 0.013676, 0.007918, 0.003988, 0.001091, 0.000000, 0.000000, 0.000000, 0.000000,
0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000,
0.000000
};
static readonly double[]
MATRIX_SRGB_D65 = {
3.2404542, -1.5371385, -0.4985314,
-0.9692660, 1.8760108, 0.0415560,
0.0556434, -0.2040259, 1.0572252
};
public static byte[] Calc(double len) {
if(len < LEN_MIN || len > LEN_MAX)
return new byte[3];
len -= LEN_MIN;
var index = (int)Math.Floor(len / LEN_STEP);
var offset = len - LEN_STEP * index;
var x = Interpolate(X, index, offset);
var y = Interpolate(Y, index, offset);
var z = Interpolate(Z, index, offset);
var m = MATRIX_SRGB_D65;
var r = m[0] * x + m[1] * y + m[2] * z;
var g = m[3] * x + m[4] * y + m[5] * z;
var b = m[6] * x + m[7] * y + m[8] * z;
r = Clip(GammaCorrect_sRGB(r));
g = Clip(GammaCorrect_sRGB(g));
b = Clip(GammaCorrect_sRGB(b));
return new[] {
(byte)(255 * r),
(byte)(255 * g),
(byte)(255 * b)
};
}
static double Interpolate(double[] values, int index, double offset) {
if(offset == 0)
return values[index];
var x0 = index * LEN_STEP;
var x1 = x0 + LEN_STEP;
var y0 = values[index];
var y1 = values[1 + index];
return y0 + offset * (y1 - y0) / (x1 - x0);
}
static double GammaCorrect_sRGB(double c) {
if(c <= 0.0031308)
return 12.92 * c;
var a = 0.055;
return (1 + a) * Math.Pow(c, 1 / 2.4) - a;
}
static double Clip(double c) {
if(c < 0)
return 0;
if(c > 1)
return 1;
return c;
}
}
结果为400-700纳米范围：
Answer 4:
尽管这是一个老问题，并且已经得到了一把很好的答案，当我试图在我的应用程序来实现这种转换功能我并不满足于在这里已经列出的算法，做我自己的研究，这给了我一些很好的结果。 所以，我要发布一个新的答案。
一些期试验研究后，我碰到本文简单分析逼近的CIE XYZ色彩匹配功能 ，并试图采用引进多叶分段高斯拟合算法在我的应用程序。 本文只描述的功能的波长转化为相应的XYZ值 ，所以实现的功能到XYZ转换为RGB中的sRGB色彩空间和组合它们。 其结果是别出心裁，值得分享：
/**
* Convert a wavelength in the visible light spectrum to a RGB color value that is suitable to be displayed on a
* monitor
*
* @param wavelength wavelength in nm
* @return RGB color encoded in int. each color is represented with 8 bits and has a layout of
* 00000000RRRRRRRRGGGGGGGGBBBBBBBB where MSB is at the leftmost
*/
public static int wavelengthToRGB(double wavelength){
double[] xyz = cie1931WavelengthToXYZFit(wavelength);
double[] rgb = srgbXYZ2RGB(xyz);
int c = 0;
c |= (((int) (rgb[0] * 0xFF)) & 0xFF) << 16;
c |= (((int) (rgb[1] * 0xFF)) & 0xFF) << 8;
c |= (((int) (rgb[2] * 0xFF)) & 0xFF) << 0;
return c;
}
/**
* Convert XYZ to RGB in the sRGB color space
* <p>
* The conversion matrix and color component transfer function is taken from http://www.color.org/srgb.pdf, which
* follows the International Electrotechnical Commission standard IEC 61966-2-1 "Multimedia systems and equipment -
* Colour measurement and management - Part 2-1: Colour management - Default RGB colour space - sRGB"
*
* @param xyz XYZ values in a double array in the order of X, Y, Z. each value in the range of [0.0, 1.0]
* @return RGB values in a double array, in the order of R, G, B. each value in the range of [0.0, 1.0]
*/
public static double[] srgbXYZ2RGB(double[] xyz) {
double x = xyz[0];
double y = xyz[1];
double z = xyz[2];
double rl = 3.2406255 * x + -1.537208 * y + -0.4986286 * z;
double gl = -0.9689307 * x + 1.8757561 * y + 0.0415175 * z;
double bl = 0.0557101 * x + -0.2040211 * y + 1.0569959 * z;
return new double[] {
srgbXYZ2RGBPostprocess(rl),
srgbXYZ2RGBPostprocess(gl),
srgbXYZ2RGBPostprocess(bl)
};
}
/**
* helper function for {@link #srgbXYZ2RGB(double[])}
*/
private static double srgbXYZ2RGBPostprocess(double c) {
// clip if c is out of range
c = c > 1 ? 1 : (c < 0 ? 0 : c);
// apply the color component transfer function
c = c <= 0.0031308 ? c * 12.92 : 1.055 * Math.pow(c, 1. / 2.4) - 0.055;
return c;
}
/**
* A multi-lobe, piecewise Gaussian fit of CIE 1931 XYZ Color Matching Functions by Wyman el al. from Nvidia. The
* code here is adopted from the Listing 1 of the paper authored by Wyman et al.
* <p>
* Reference: Chris Wyman, Peter-Pike Sloan, and Peter Shirley, Simple Analytic Approximations to the CIE XYZ Color
* Matching Functions, Journal of Computer Graphics Techniques (JCGT), vol. 2, no. 2, 1-11, 2013.
*
* @param wavelength wavelength in nm
* @return XYZ in a double array in the order of X, Y, Z. each value in the range of [0.0, 1.0]
*/
public static double[] cie1931WavelengthToXYZFit(double wavelength) {
double wave = wavelength;
double x;
{
double t1 = (wave - 442.0) * ((wave < 442.0) ? 0.0624 : 0.0374);
double t2 = (wave - 599.8) * ((wave < 599.8) ? 0.0264 : 0.0323);
double t3 = (wave - 501.1) * ((wave < 501.1) ? 0.0490 : 0.0382);
x = 0.362 * Math.exp(-0.5 * t1 * t1)
+ 1.056 * Math.exp(-0.5 * t2 * t2)
- 0.065 * Math.exp(-0.5 * t3 * t3);
}
double y;
{
double t1 = (wave - 568.8) * ((wave < 568.8) ? 0.0213 : 0.0247);
double t2 = (wave - 530.9) * ((wave < 530.9) ? 0.0613 : 0.0322);
y = 0.821 * Math.exp(-0.5 * t1 * t1)
+ 0.286 * Math.exp(-0.5 * t2 * t2);
}
double z;
{
double t1 = (wave - 437.0) * ((wave < 437.0) ? 0.0845 : 0.0278);
double t2 = (wave - 459.0) * ((wave < 459.0) ? 0.0385 : 0.0725);
z = 1.217 * Math.exp(-0.5 * t1 * t1)
+ 0.681 * Math.exp(-0.5 * t2 * t2);
}
return new double[] { x, y, z };
}
我的代码是用Java编写的8，但它不应该是很难移植到降低Java等语言版本。
Answer 5:
你说的是从波长转换为RGB值。
看看这里，可能会回答你的问题。 你有一个与源代码以及一些解释这样做的效用。
WaveLengthToRGB
Answer 6:
我想我也可以跟着我的一个正式的答复意见。 最好的办法是使用HSV色彩空间 -虽然色调是波长它不是一个一对一的比较。
Answer 7:
我没有已知的色调值和频率的线性拟合（脱落红色和紫色，因为它们在频率值到目前为止延伸它们歪斜东西位）和我有一个粗略的转换式。
它是这样
频率（在太赫兹）= 474 +（3/4）（色调角（单位：度））
我试着环顾四周，看看是否有人想出了这个等式，但我没有发现任何东西为2010年5月。
Answer 8:
方法1
这是位清理并可进行@浩辰勰的C ++ 11的版本。 我还添加，其将值0到1到在可见光谱中的波长，其与该方法一起使用的功能。 你可以把下面的一个头文件，并使用它没有任何依赖关系。 这个版本将保持在这里 。
#ifndef common_utils_OnlineStats_hpp
#define common_utils_OnlineStats_hpp
namespace common_utils {
class ColorUtils {
public:
static void valToRGB(double val0To1, unsigned char& r, unsigned char& g, unsigned char& b)
{
//actual visible spectrum is 375 to 725 but outside of 400-700 things become too dark
wavelengthToRGB(val0To1 * (700 - 400) + 400, r, g, b);
}
/**
* Convert a wavelength in the visible light spectrum to a RGB color value that is suitable to be displayed on a
* monitor
*
* @param wavelength wavelength in nm
* @return RGB color encoded in int. each color is represented with 8 bits and has a layout of
* 00000000RRRRRRRRGGGGGGGGBBBBBBBB where MSB is at the leftmost
*/
static void wavelengthToRGB(double wavelength, unsigned char& r, unsigned char& g, unsigned char& b) {
double x, y, z;
cie1931WavelengthToXYZFit(wavelength, x, y, z);
double dr, dg, db;
srgbXYZ2RGB(x, y, z, dr, dg, db);
r = static_cast<unsigned char>(static_cast<int>(dr * 0xFF) & 0xFF);
g = static_cast<unsigned char>(static_cast<int>(dg * 0xFF) & 0xFF);
b = static_cast<unsigned char>(static_cast<int>(db * 0xFF) & 0xFF);
}
/**
* Convert XYZ to RGB in the sRGB color space
* <p>
* The conversion matrix and color component transfer function is taken from http://www.color.org/srgb.pdf, which
* follows the International Electrotechnical Commission standard IEC 61966-2-1 "Multimedia systems and equipment -
* Colour measurement and management - Part 2-1: Colour management - Default RGB colour space - sRGB"
*
* @param xyz XYZ values in a double array in the order of X, Y, Z. each value in the range of [0.0, 1.0]
* @return RGB values in a double array, in the order of R, G, B. each value in the range of [0.0, 1.0]
*/
static void srgbXYZ2RGB(double x, double y, double z, double& r, double& g, double& b) {
double rl = 3.2406255 * x + -1.537208 * y + -0.4986286 * z;
double gl = -0.9689307 * x + 1.8757561 * y + 0.0415175 * z;
double bl = 0.0557101 * x + -0.2040211 * y + 1.0569959 * z;
r = srgbXYZ2RGBPostprocess(rl);
g = srgbXYZ2RGBPostprocess(gl);
b = srgbXYZ2RGBPostprocess(bl);
}
/**
* helper function for {@link #srgbXYZ2RGB(double[])}
*/
static double srgbXYZ2RGBPostprocess(double c) {
// clip if c is out of range
c = c > 1 ? 1 : (c < 0 ? 0 : c);
// apply the color component transfer function
c = c <= 0.0031308 ? c * 12.92 : 1.055 * std::pow(c, 1. / 2.4) - 0.055;
return c;
}
/**
* A multi-lobe, piecewise Gaussian fit of CIE 1931 XYZ Color Matching Functions by Wyman el al. from Nvidia. The
* code here is adopted from the Listing 1 of the paper authored by Wyman et al.
* <p>
* Reference: Chris Wyman, Peter-Pike Sloan, and Peter Shirley, Simple Analytic Approximations to the CIE XYZ Color
* Matching Functions, Journal of Computer Graphics Techniques (JCGT), vol. 2, no. 2, 1-11, 2013.
*
* @param wavelength wavelength in nm
* @return XYZ in a double array in the order of X, Y, Z. each value in the range of [0.0, 1.0]
*/
static void cie1931WavelengthToXYZFit(double wavelength, double& x, double& y, double& z) {
double wave = wavelength;
{
double t1 = (wave - 442.0) * ((wave < 442.0) ? 0.0624 : 0.0374);
double t2 = (wave - 599.8) * ((wave < 599.8) ? 0.0264 : 0.0323);
double t3 = (wave - 501.1) * ((wave < 501.1) ? 0.0490 : 0.0382);
x = 0.362 * std::exp(-0.5 * t1 * t1)
+ 1.056 * std::exp(-0.5 * t2 * t2)
- 0.065 * std::exp(-0.5 * t3 * t3);
}
{
double t1 = (wave - 568.8) * ((wave < 568.8) ? 0.0213 : 0.0247);
double t2 = (wave - 530.9) * ((wave < 530.9) ? 0.0613 : 0.0322);
y = 0.821 * std::exp(-0.5 * t1 * t1)
+ 0.286 * std::exp(-0.5 * t2 * t2);
}
{
double t1 = (wave - 437.0) * ((wave < 437.0) ? 0.0845 : 0.0278);
double t2 = (wave - 459.0) * ((wave < 459.0) ? 0.0385 : 0.0725);
z = 1.217 * std::exp(-0.5 * t1 * t1)
+ 0.681 * std::exp(-0.5 * t2 * t2);
}
}
};
} //namespace
#endif
从的375nm到725nm的颜色情节看起来象下面这样：
这个方法的一个问题是，它只能400-700和外部之间的，它急剧下降到黑色。 另一个问题是窄蓝色。
为了便于比较，下面是从视觉FAQ颜色在maxmax.com：
我用这个可视化的深度图，其中每个像素在米代表深度值，这看起来象下面这样：
方法2
这被实现为部分bitmap_image通过Aeash Partow单个文件仅标头库：
inline rgb_t convert_wave_length_nm_to_rgb(const double wave_length_nm)
{
// Credits: Dan Bruton http://www.physics.sfasu.edu/astro/color.html
double red = 0.0;
double green = 0.0;
double blue = 0.0;
if ((380.0 <= wave_length_nm) && (wave_length_nm <= 439.0))
{
red = -(wave_length_nm - 440.0) / (440.0 - 380.0);
green = 0.0;
blue = 1.0;
}
else if ((440.0 <= wave_length_nm) && (wave_length_nm <= 489.0))
{
red = 0.0;
green = (wave_length_nm - 440.0) / (490.0 - 440.0);
blue = 1.0;
}
else if ((490.0 <= wave_length_nm) && (wave_length_nm <= 509.0))
{
red = 0.0;
green = 1.0;
blue = -(wave_length_nm - 510.0) / (510.0 - 490.0);
}
else if ((510.0 <= wave_length_nm) && (wave_length_nm <= 579.0))
{
red = (wave_length_nm - 510.0) / (580.0 - 510.0);
green = 1.0;
blue = 0.0;
}
else if ((580.0 <= wave_length_nm) && (wave_length_nm <= 644.0))
{
red = 1.0;
green = -(wave_length_nm - 645.0) / (645.0 - 580.0);
blue = 0.0;
}
else if ((645.0 <= wave_length_nm) && (wave_length_nm <= 780.0))
{
red = 1.0;
green = 0.0;
blue = 0.0;
}
double factor = 0.0;
if ((380.0 <= wave_length_nm) && (wave_length_nm <= 419.0))
factor = 0.3 + 0.7 * (wave_length_nm - 380.0) / (420.0 - 380.0);
else if ((420.0 <= wave_length_nm) && (wave_length_nm <= 700.0))
factor = 1.0;
else if ((701.0 <= wave_length_nm) && (wave_length_nm <= 780.0))
factor = 0.3 + 0.7 * (780.0 - wave_length_nm) / (780.0 - 700.0);
else
factor = 0.0;
rgb_t result;
const double gamma = 0.8;
const double intensity_max = 255.0;
#define round(d) std::floor(d + 0.5)
result.red = static_cast<unsigned char>((red == 0.0) ? red : round(intensity_max * std::pow(red * factor, gamma)));
result.green = static_cast<unsigned char>((green == 0.0) ? green : round(intensity_max * std::pow(green * factor, gamma)));
result.blue = static_cast<unsigned char>((blue == 0.0) ? blue : round(intensity_max * std::pow(blue * factor, gamma)));
#undef round
return result;
}
从375-725nm波长的情节看起来象下面这样：
因此，这是在400-725nm更可用。 当我想象相同的深度图与方法1中，我得到以下。 还有那些黑色的线，我想表示这个代码，我还没有看更深入轻微错误的一个明显的问题。 也紫罗兰是在该方法中这导致对远处的物体对比度较低位窄。
文章来源: Convert light frequency to RGB?
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