[Radiance-general] Lux-values for points in a plane

[Thomas Bleicher]

On 13 May 2008, at 13:51, Per Haugaard wrote:

> 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.

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.

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.

> 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.

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

Regards,
Thomas