# Weighting factors used in the irradiance calculation

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
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.

Parisa

Hi Parisa,

in fact these are weighting functions derived from v(lamda). So you should use them only when all your quantities are in photometric units. Irradiance "usually" implies the solar spectrum, but in RADIANCE this is used also only considering the visible spectrum (and then using the weighting). I'm not sure if a color weighting makes sense when talking about radiometric quantities using the solar spectrum. Usually I calculate everything in "grey" (r=g=b) when using the radiometric quantities, so this problem doesn't occur. Be aware, that you usually calculate everything in photometric quantities in RADIANCE, so if you want to calculate the solar spectral behavior you have to take care of using the related values(solar transmittance for glazing, solar reflectance of surfaces and generate sky and sun with radiometric quantities). As far as I know only if you use gendaylit with the -O 1 option, then sky and sun are automatically in radiometric quantities (which means your results are in W/m*m solar spectrum).

Jan

···

Am 6/22/15 um 5:50 PM schrieb parisa khademagha:

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 I_R + 0.670 I_G + 0.065 I_B ) 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.

Parisa

_______________________________________________
[email protected]

--
Dr.-Ing. Jan Wienold
Ecole Polytechnique Fédérale de Lausanne (EPFL)
EPFL ENAC IA LIPID

http://people.epfl.ch/jan.wienold
LE 1 111 (Office)
Phone +41 21 69 30849

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 <[email protected]> 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" <[email protected]<mailto:[email protected]>> 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.

Parisa

_______________________________________________
[email protected]

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

···

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 <[email protected]> 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" <[email protected]<mailto:[email protected]>> 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.

Parisa

Hi Jan,

It is true that if these weighting functions are from V-lambda they are to be used only in calculation of photometric quantities. Doing so the summation of the triad irradiance values should be multiplied to 179 efficacy function to give illuminance ( I = 179*(0.265 IR+ 0.670 IG + 0.065 IB)). What I don’t understand is why these weighting functions are used in calculation of radiometric quantities like irradiance ( I = 0.265 IR+ 0.670 IG + 0.065 IB). You said you calculate everything in grey (r=g=b) to avoid this problem. Shall I ask what do you mean by everything? What is the luminous reflectance of the grey you use? Moreover, is it possible to measure the effective irradiance with respect to C-lambda using adjusted weighting functions? Should the summation of the weighting functions be again equal to 1 as they are for V-lambda?

Cheers,
Parisa

···

On 22 Jun 2015, at 18:24, Jan Wienold <[email protected]> wrote:

Hi Parisa,

in fact these are weighting functions derived from v(lamda). So you should use them only when all your quantities are in photometric units. Irradiance "usually" implies the solar spectrum, but in RADIANCE this is used also only considering the visible spectrum (and then using the weighting). I'm not sure if a color weighting makes sense when talking about radiometric quantities using the solar spectrum. Usually I calculate everything in "grey" (r=g=b) when using the radiometric quantities, so this problem doesn't occur. Be aware, that you usually calculate everything in photometric quantities in RADIANCE, so if you want to calculate the solar spectral behavior you have to take care of using the related values(solar transmittance for glazing, solar reflectance of surfaces and generate sky and sun with radiometric quantities). As far as I know only if you use gendaylit with the -O 1 option, then sky and sun are automatically in radiometric quantities (which means your results are in W/m*m solar spectrum).

Jan

Am 6/22/15 um 5:50 PM schrieb parisa khademagha:

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.

Parisa

_______________________________________________
[email protected]

--
Dr.-Ing. Jan Wienold
Ecole Polytechnique Fédérale de Lausanne (EPFL)
EPFL ENAC IA LIPID

http://people.epfl.ch/jan.wienold
LE 1 111 (Office)
Phone +41 21 69 30849
_______________________________________________
[email protected]

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
my question is how can I spectrally model the sky?

Cheers,

Parisa

···

On 22 June 2015 at 23:57, Greg Ward <[email protected]> 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 <[email protected]>
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 < > [email protected]> 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).
>>
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)
>>
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" <[email protected] > <mailto:[email protected]>> 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)
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.
>>
>> Parisa

_______________________________________________
[email protected]

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:

Best,
-Greg

···

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 <[email protected]> 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 <[email protected]>
> 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 <[email protected]> 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" <[email protected]<mailto:[email protected]>> 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.
>>
>> Parisa

Hi Parisa,

when we worked on the spectral evaluations back then, we were testing our
and has not been updated since then...). Our idea back then was to check
whether there are RGB coefficients that provide reasonable results for
calculating spectrally weighted values. With the coefficients as given in
the paper you can definitely get an estimation, but there are already
deviations from the correct spectral calculations if you only look at the
light source spectra themselves. This might even get worse if
interreflections significantly contribute to the illumination. This is also
true for illuminances and color, which in fact are nothing else but special
spectral weighting functions (see therefore Greg's paper "Picture Perfect
RGB Rendering Using Spectral Prefiltering and Sharp Color Primaries" at
http://www.anyhere.com/gward/papers/egwr02/index.html or Arne's and mine
"Color-rendering indices in global illumination methods" at
http://dx.doi.org/10.1117/1.3274623)

So, if you really need to be sure about your values (as far as this is
possible within a simulation...), I would always suggest to use spectral
rendering and use the c(lambda) weighting function (as also stated in the
conclusion of the paper). Following the proposal by Ruppertsberg&Bloj you
put them together to you spectral result.

Best,
David

···

2015-06-24 19:15 GMT+02:00 Greg Ward <[email protected]>:

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
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:

Best,
-Greg

*Subject: *Re: [Radiance-general] Weighting factors used in the

*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
In this case my question is how can I spectrally model the sky?

Cheers,

Parisa

On 22 June 2015 at 23:57, Greg Ward <[email protected]> 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 <[email protected]>
> Subject: Re: [Radiance-general] Weighting factors used in the
> 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 < >> [email protected]> 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).
>>
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)
>>
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" <[email protected] >> <mailto:[email protected]>> 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)
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.
>>