Luminous Efficacy

radiance-general
1

Dear All

As a part of my Ph.D research I am doing a study about the influence of the colors of surfaces on the illuminance at any arbitrary point of the room under daylighting condition. So I need to know about the exact relationship between the color of a surface and its luminance. For that reason I am going to do some simulation with radiance. As I have just common knowledge of using radiance, sorry if my questions are too basic.

The calculation method used in “Radiance” for computing the luminance value is:

Luminance = 179 * (0.265*R + 0.670*G + 0.065*B) (1)

Where the 179 value has been defined so: The value of 179 lm/w is the standard luminous efficacy of equal energy white light that is defined and used by Radiance specifically for this conversion.

1. Unfortunately I could not find this value at any other document about lighting and colour measurement. Does anybody know how the value of 179 has been derived? I assume it is the average of all the luminous efficacies in visible light spectrum (380 – 780 nm). Isn't it ?

2. As I know, the method for calculating the luminance is:

L = 683 ∫ I (λ) * ρ (λ) * V (λ) dλ (2)

Where: 0=<V (λ) < =1 and has a maximum of 1 at 555 nm and I (λ) is the Spectrum power distribution at any wavelength of arbitrary spectrum. If I have understood correctly, in the file “src/common/color.h” the value of (0.265*R + 0.670*G + 0.065*B) will give us the tristimulus value “Y”, which is the same as: Y= ∫ I (λ) * ρ (λ) * ‾y (λ) dλ (3)

Since ‾y (λ) = V (λ) for computing the luminance value (according to the methods (2), (3)) what we should do is to multiply the Y value at 683 which gives:

Luminance = 683 * (0.265*R + 0.670*G + 0.065*B)

I suppose if we use the method (1), it means that we have calculated luminance so: L = 179 ∫ I (λ) * ρ (λ) * V (λ) dλ

That means the value of 179 lm/w has been used as maximum luminous efficacy instead of 683 lm/w at 555 nm??

And if 179 has been used as average, I can not understand why we shoud average again , whereas behind the 683 ∫ V (λ) dλ (380nm- 78nm) there is the meaning of averaging the luminous efficacy over the visible light spectrum.

I hope I could make my questions understandable.

If I have gotten the concept wrong, I would be thankful if somebody could tell me where my mistake is, it would be really important and helpful for the result of my research.

Thank you in advance

Niloofar

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Niloofar Moghbel
Chair for Building Physics
University of Stuttgart
Pfaffenwaldring 7
70569 Stuttgart, Germany
Tel.: ++49 (0)711-685-6301
Fax: ++49 (0)711-685-6583
[email protected]

2

Hello Niloofar,

I hope I can clarify this. We seem to be caught in some kind of International Month of Color Calculations...

The calculation method used in “Radiance” for computing the luminance value is:
Luminance = 179 * (0.265*R + 0.670*G + 0.065*B) (1)
Where the 179 value has been defined so: The value of 179 lm/w is the standard luminous efficacy of equal energy white light that is defined and used by Radiance specifically for this conversion.

1. Unfortunately I could not find this value at any other document about lighting and colour measurement. Does anybody know how the value of 179 has been derived? I assume it is the average of all the luminous efficacies in visible light spectrum (380 – 780 nm). Isn't it ?

That's correct, but the actual value is very sensitive to where you set the endpoints of what constitute "visible," since the tails represent a greater proportion of energy relative to what's visible. In this sense, the value is somewhat arbitrary, and as long as you divide by 179 on the way in and multiply by 179 on the way out, the value cancels and is therefore irrelevant. Integrating equal-energy white from 380 to 780 gets you something very close to 179, though I can't promise it's exact. I can promise it doesn't matter.

2. As I know, the method for calculating the luminance is:
L = 683 ∫ I (λ) * ρ (λ) * V (λ) dλ (2)

Where: 0=<V (λ) < =1 and has a maximum of 1 at 555 nm and I (λ) is the Spectrum power distribution at any wavelength of arbitrary spectrum. If I have understood correctly, in the file “src/common/color.h” the value of (0.265*R + 0.670*G + 0.065*B) will give us the tristimulus value “Y”, which is the same as:

Y= ∫ I (λ) * ρ (λ) * ‾y (λ) dλ (3)

Since ‾y (λ) = V (λ) for computing the luminance value (according to the methods (2), (3)) what we should do is to multiply the Y value at 683 which gives:

Luminance = 683 * (0.265*R + 0.670*G + 0.065*B)

How do you figure? For this to work, the luminous efficacy of the white corresponding to our RGB color space would have to be 683 lumens/watt. The only light that matches this efficacy is a monochromatic source at 555 nm. Let's hope we don't have an RGB color space with a monochromatic green white point.

The white point defined in Radiance (different from most other RGB spaces) is equal-energy white, hence the luminous efficacy of 179, which is just a conversion factor as discussed above.

I suppose if we use the method (1), it means that we have calculated luminance so: L = 179 ∫ I (λ) * ρ (λ) * V (λ) dλ
That means the value of 179 lm/w has been used as maximum luminous efficacy instead of 683 lm/w at 555 nm??

No, because Eq. (1) assumes not a spectrum but a tristimulus color space. They are different beasts, and you can't go freely from one to the other like they represent the same information. They do not. Data is irretrievably lost when you convert from a continuous spectrum to any tristimulus space (eg., RGB or XYZ).

And if 179 has been used as average, I can not understand why we shoud average again , whereas behind the 683 ∫ V (λ) dλ (380nm- 78nm) there is the meaning of averaging the luminous efficacy over the visible light spectrum.

You don't average again. Equations (1) and (2) are correct. Eq. (3) is missing a normalization constant, and the subsequent equations are confused because of that. Y = L from Eq. (2).

If you can understand why the coefficients 0.265 + 0.670 + 0.065 add up to one, whereas the *maximum* value of V(lambda) is 1.0, I think you will resolve your mystery.

-Greg

3

Hello Greg
Thank you very much for your quick and comprehensive reply.

Niloofar