[Radiance-general] Re: trans or glass material, rectangular, luminous, opening problem

[Gregory J. Ward]

Hi Christian,

> thank you very much for your detailed answer, i think the biggest  
> problem i am confronted with are these different terms.
> As i understand:
>  transmissivity is the amount of light that is not absorbed inside  
> an object, varies between 0and 1, not taking in consideration the  
> reflection
> total transmittance the amount of light, that passes through an  
> object, taking in consideration the index of refraction and is less  
> than the transmissivity
> But what is transmission ?

That's my mistake -- I should have said "transmissivity" rather than  
transmission.  A transmission is the thing in your car that shifts  
gears...  In fact, I think the term is used interchangeably in a more  
informal context with total transmittance.

Total transmittance takes into account reflections and  
interreflections (and absorbtion) inside the glass, and is the  
easiest value to measure.

> As i want to apply the color and transmission of this color filter  
> to the glass i got these values from LEE:
> Transmission Y%: 62.4  ; Absorption: 0.21 ; as well as a Spectrum  
> from which i calculated the xy chromaticity values with the mgfilt  
> command: cxy 0.2884 0.3034
> They have precalculated XYZ and Yxy values for different sources,  
> but i want to apply my own source at different dim and color  
> temperature levels.  http://www.leefilters.com/LPFD.asp?PageID=193
> To get my RGB values, i have to calculate the XYZ values and for  
> that i need Y! Can i take 0.624 ?

Actually, the Y value of transmittance can change for different light  
sources.  What you should do is multiply the value you were given  
(0.624) by the Y you get with your source spectrum, divided by the Y  
you get when you use their reference source spectrum.  The confusing  
part is that you need to normalize the two spectra using the Y value  
computed from each WITHOUT your glass transmittance spectrum.

Is this clear enough?  I know it's confusing, but I don't know how to  
explain it any better without adding a bunch of pictures and diagrams.

-Greg