[Radiance-general] Re: Radiance-general digest, Vol 1 #40 - 1 msg

[Georg Mischler]

Zack Rogers wrote:

> First, the efficacy radiance uses for a 32W T8 is 179lumens/W?  I'm assuming
> that efficacy is for light energy and not power consumed.  Is this the same
> for all lamps?  Most of the time the efficacy of a lamp refers to the power
> consumed, a typical T8 having an efficacy of 2900lumens/32W or 90lumens/W.

I'm not sure if efficacy is really the best word here. The value
of 179 is simply the factor that Radiance normally uses for
converting between photometric and radiometric units (lumens per
Watt of radiation in the visible spectrum), taking into account the
response curve of the human eye for a "uniform white" light color.

Since Radiance knows nothing about electricity, the electrical
power consumption of your lamp has no relevance in the
calculation at all. For all we know, you could just as well be
working with a magic oil lamp!


> Also, I agree that the luminous surface is 0.0979m2 (I had missed a length of
> 4' in my calc).  Then you calculated the radiance with
>
>   16.2W / (0.0979m2 * Pi) = 52.7 W/m2/sr
>
> This is the total radiant output per steradian per area.  However, it seems
> like the total steradians you should be dividing by is the entire sphere that
> the 16.2W of radiant energy is filling, a sphere being 4*pi steradians.
> This gives;
>
>  16.2W / (0.0979m2 * 4 * Pi) = 13.17 W/m2/sr

Why do we divide by Pi? Good question...
I don't have a degree in math myself, so I can't explain this
very well either. The best hint might be that we're looking at a
discrete direction (a point out of a continuum), and not at a
fraction of the total [hemi]sphere (a subset of that continuum).
Our direction has a solid angle of zero, while your line of
thought assumes a solid angle of 1.

That doesn't really explain the division by Pi, but at least it
shows why a division by 4*Pi for the full sphere (or 2*Pi for the
hemisphere) would be wrong. The factor of Pi can probably be
found somewhere in the derivation of the integral that stands
between radiance and radiant exitance, but I'll leave that as an
excercise for the readers... ;)


> So, I'm still a little confused with the math here,

Don't worry, so is everybody else!


-schorsch

-- 
Georg Mischler  --  simulations developer  --  schorsch at schorsch.com
+schorsch.com+  --  lighting design tools  --  http://www.schorsch.com/