SV: Re: Lux-values for points in a plane

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Thank you very much for your reply: I really appreciate your help. But it still troubles me why I cannot get the correct ratio between the output and the entered LT-value for the window. Please refer to the below.
   
  > 6.216 0.00 1.8 0 -1 0 (window)
  > 6.216 0.05 1.8 0 -1 0 (black wall located 0.05 m behind the window
  > - in fact all walls are black)
  >
  > The Window (LT=0.72):
  > void glass window_glass
  > 0
  > 0
  > 3 .78 .78 .78
  >
  > cat IllumKoordinater.pts | rtrace -I -ab 4 -h -w -oov kontor.oct |
  > rcalc -e '$1=$2;$2=179*(.265*$4+.670*$5+.065*$6)' > lux.dat
  >
  > Output
  > -1 4883.69
  > -0.95 3049.70
   
  > If I understand the above correctly you are calculating two points:
> one in front of the window, one behind it. The black walls are used
> to swallow all light reflections on the inside so you will only have the
> transmitted light left.

  Yes I only want to evaluate on the transmitted light through the window. I am interested in, later on, to calculate the transmitted light through the window with external shading based on a CIE-standard overcast sky.
  
> Two improvements:

> 1) Move the wall further back or place your second point at ie. 0.04. If
> the calculation point coincides with the wall everything is possible.

  When I calculate the transmitted light at any other point in the room e.g. 0.04 (away from the black wall) the calculation coincides with the window (LT=1).
  
> 2) Don't use a box at all. At the perimeter there is a dark horizon
> (black walls) which is taken in into the calculation. You could
> create a sphere of the material around your calculation point instead.

I’m not sure what you mean here.

  > Two things worth noticing here. The output coordinates are not
  > listed correctly which gives me the suspicion that lux-values are
  > is not calculated for the correct coordinates. Furthermore the
  > ratio between the output are not according to the entered LT-value
  > for the window (0,62 vs. 0,72)
  
> The first is the result of rtrace's interpretation of the '-I' switch
> (uppercase 'I'!) rtrace creates a 'virtual surface' at the calculation
  > point and starts the calculation from the point of the view vector.
  > In your case these are (6.216+0, 0+-1, 1.8+0) and
  > (6.216+0, 0.05+-1, 1.8+0). Your rcalc command prints only the y-
> coordinate.
   
  Okay, so the calculation is carried out for the correct input coordinates.
   
  > The second problem I can't explane but depending on the size of your
  > box the 'dark horizon' might have some influence here.
   
  My reference point in the room should coincide with the center of the window. I appreciate any help on solving why the ratio between the output and the entered LT-value for the window (0,62 vs. 0,72) differ.

Best regards

Per Haugaard

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> 2) Don't use a box at all. At the perimeter there is a dark horizon
> (black walls) which is taken in into the calculation. You could
> create a sphere of the material around your calculation point instead.

I’m not sure what you mean here.

I would suggest to split your verification of glazing material and shading system
into two separate steps to identify possible problems more easily. For the material
part I would create a simple scene like this:

#### simple test scene ####

## overcast sky with 10000 lux horizontal diffuse
!gensky 3 21 12:00GMT -c -o 0 -a 51 -m 0 -B 55.866

skyfunc glow skyglow 0 0 4 1 1 1 0
skyglow source sky 0 0 4 0 0 1 180

## material for glass sphere
void glass window_glass
0
3 .78 .78 .78

## glass sphere around origin
window_glass sphere glass_sphere
0
4 0 0 0 1

#### end of test scene ####

Then use two points (0,0,0) and (0,0,2) to calculate the lux values
inside and outside of the sphere. The direction has to be (0,0,1)
because the scene does not include a ground glow.

If you get around 10000 lux outside of fhe sphere and with the
above material around 7200 lux inside the scene setup is correct
and you can now start modifying the values for the glass material.
The ratio of ouside/inside will give you the transmittance of the
glazing (which is not equal to the values of the glass material!).

When you have verified your glass material do the next step and
use it in a scene. You can now be sure that the material is correct
and any unexpected values are the result of the scene geometry.

> The second problem I can't explane but depending on the size of your
> box the 'dark horizon' might have some influence here.

My reference point in the room should coincide with the center of the window. I appreciate any help on solving why the ratio between the output and the entered LT-value for the window (0,62 vs. 0,72) differ.

Your reference point might be in the centre of the window but
rtrace samples the whole hemisphere around this point. So if you
have (black) reveals this will influence the calculations even if
you move your calculation point only by 0.05 units.

Without knowing more about your scene it's hard to tell what
exactly causes your values. As a quick help calculate the scene
without glazing and with different glazing values but always
from the same point inside of the room. If these values are correct
it's your geometry that's creating the problems.

Thomas

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On 14 May 2008, at 08:54, Per Haugaard wrote: