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

[parisa khademagha]

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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