计算亮度的常数k是多少

Question

I want to calculate the luminance of image using RGB value but in the formula there is a constant ki didn't know what it is this is the formula from the document:

"In calculating luminance (L), CIE Y is multiplied by a constant value 'k', where k can be either aconstant for the camera or determined for a scene based on a measurement of a selected region in the scene.

L=k*(0,2127 Red+0,7151 Green+0,0722*Blue) (cd/m²)

this is the link to the document: https://faculty.washington.edu/inanici/Publications/LRTPublished.pdf emphasized text

Answer 1

Short Answer

You are conflating something that is only for a specific piece of software called "Photosphere" and that math is not for general luminance use.

Complete Answer

You said:

calculate the luminance of image using RGB value

What do you mean? The luminance of some given pixel? Or the RMS average luminance? Or???

AND: What colorspace (profile) is your image in?

I am going to ASSUME your image is sRGB, and 8 bit per pixel. But it could be P3 or Adobe98, or any number of others... The math I am going to show you is for sRGB.

1) Convert to float

Each 8bit 0-255 sRGB color channel must be converted to 0.0-1.0

    let rF = sR / 255.0;
    let gF = sG / 255.0;
    let bF = sB / 255.0;

2) Convert to linear

The transfer curve aka "gamma" must be removed. For sRGB, the "accurate" method is:

    function sRGBtoLin(chan) {
       return (chan > 0.04045) ? Math.pow(( chan + 0.055) / 1.055, 2.4) : chan / 12.92
    }

3) Calculate Luminance

Now, multiply each linearized channel by the appropriate coefficient, and sum them to find luminance.

  let luminance = sRGBtoLin(rF) * 0.2126 + sRGBtoLin(gF) * 0.7152 + sRGBtoLin(bF) * 0.0722;

BONUS: Andy's Down and Dirty Version

Here's an alternate that is still usefully accurate, but simpler for better performance. Raise each channel to the power of 2.2, then multiply the coefficients, and sum:

  let lum = (sR/255.0)**2.2*0.2126+(sG/255.0)**2.2*0.7152+(sB/255.0)**2.2*0.0722;

Let me know if you have additional questions...