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

[Greg Ward]

Hi Parisa,

If you are only comparing one set of C-lambda values to another, then the scaling does not matter.  If, on the other hand, the RGB coefficients from David & Arne's paper were somehow computed to correlate with other aspects of irradiance, such as photopic adaptation, then you probably should use them as given.  I guess I need to read their paper....

A full spectral calculation probably is not needed in this case, but it would help to have a good estimate of the sky color either way.  You should check out Mark Stock's implementation of the Utah sky model using Radiance:

	http://markjstock.org/radiance

Best,
-Greg

> From: parisa khademagha <p.khademagha at gmail.com>
> Subject: Re: [Radiance-general] Weighting factors used in the irradiance calculation
> Date: June 24, 2015 6:04:12 AM PDT
> 
> Hi Greg and Jan,
> 
> Thank you for your detailed replies!
> 
> Here is my precise application: I want to use Radiance to estimate/simulate the effective irradiance with respect to the C-lambda.
> 
> First option is to use different (-0.034 IR + 0.323 IG + 0.558 IB) weighting functions for C-lambda as suggested by Geisler-Moroder & Dur (2010) in their paper (entitled ‘Estimating melatonin suppression and photosynthesis activity in real-world scenes from computer generated images’). My question is whether using these C-lambda weighting functions is a correct way of getting effective irradiance with respect to C-lambda in W/m2 or not?
> 
> Second option is to use the spectral rendering as suggested by Ruppertsberg & Bloj (2008) in their paper (entitled ‘Creating physically accurate visual stimuli for free: Spectral rendering with RADIANCE’) . In this case my question is how can I spectrally model the sky?
> 
> Cheers,
> 
> Parisa
> 
> 
> On 22 June 2015 at 23:57, Greg Ward <gregoryjward at gmail.com> wrote:
> 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
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