[Radiance-general] Weighting factors used in the irradiance calculation

[Greg Ward]

If I may jump in, here...

Ultimately the 179 lumens/watt conversion derives from the peak defined efficacy of 683 lumens/watt at 555nm and my particular choice of endpoints for the visible spectrum in integrating an equal energy source.  Since the power-to-photopic conversion is very sensitive to the choice of endpoints (for the very reason that viewers are *not* sensitive to those endpoints), I thought to standardize it at some point and end the confusion.  Divide by 179 going from photometric units and multiply by 179 going to and the two factors cancel, problem solved.

That said, the factor is somewhat arbitrary and Radiance does not really care what lighting units it works in.  This is why Jan's trick of substituting solar reflectances and solar radiances works just fine.  If you dig through most of the Radiance code, you will find very few references to wavelength, and in the end you can compute in infrared and ultraviolet and nothing will go wrong, so long as you can get by with 3 channels or don't mind doing multiple runs.

The RGB coefficients for luminance (ignoring the 179 conversion factor) add up to 1.0 again as a convention, but it's a common one for conversion between color spaces.  In this case, we're converting RGB to one channel of CIE XYZ, the Y channel.  The other matrix coefficients can be found in ray/src/cal/cal/xyz_rgb.cal, or ray/src/common/spec_rgb.c if you prefer C code.

If you would tell us your precise application, what you have as input and what you expect as output, it would help us to better answer your question.

Cheers,
-Greg

> From: Parisa Khademagha <p.khademagha at gmail.com>
> Subject: Re: [Radiance-general] Weighting factors used in the irradiance calculation
> Date: June 22, 2015 2:14:24 PM PDT
> 
> Hi Rob,
> 
> Thank you for your reply! Is the formula incomplete for calculation of the irradiance value or it is missing 179 luminous efficacy value for conversion of irradiance to illuminance? What if we decide to measure the effective irradiance with respect to another curve (for instance C-lambda)? Should the summation of the C-lambda weighting functions be also equal to 1? Is this a rule in Radiance or it can be violated?
> 
> Cheers,
> Parisa
> 
> On 22 Jun 2015, at 18:10, Guglielmetti, Robert <Robert.Guglielmetti at nrel.gov> wrote:
> 
>> Hi Parisa,
>> 
>> Yes, these values are the so-called V-lambda or photopic weighting function, and yes these three will always equal 1 (naturally your derived result will generally be > 1).
>> 
>> *Be advised*, your formula as printed in your post is incomplete; when converting irradiance as computed by Radiance (the software) to illuminance, you need to multiply the whole thing by 179 which is the luminous efficacy value to be used _with Radiance_. In other words:
>> 
>> I = 179 * (0.265 IR+ 0.670 IG + 0.065 IB)
>> 
>> There are scores of posts in the archives about this value and how it came to be, and why it works even though it's different from any luminous efficacy value you've ever seen in a textbook on light. =)
>> 
>> - Rob
>> 
>> On 6/22/15, 9:50 AM, "parisa khademagha" <p.khademagha at gmail.com<mailto:p.khademagha at gmail.com>> wrote:
>> 
>> Dear all,
>> 
>> I have two questions regarding the weighting factors (0.265, 0.670, 0.065) that are used in the formula ( I = 0.265 IR+ 0.670 IG + 0.065 IB) with which one can convert the spectral irradiance triad to irradiance. My first question is: where these weighting factor come from? Do they incorporate the spectral sensitivity of the human eye (so called V(λ)) in the irradiance calculation? My second question is: should the summation of these weighting factors be always equal to 1.
>> 
>> Thank you in advance,
>> Parisa