[Radiance-general] vta 2 vth

[Mehlika Inanici]

Hi Greg (and everybody in the group),

I have used the formula Greg provided for a previous discussion 
(http://www.radiance-online.org/pipermail/radiance-general/2009-January/005644.html) 
to convert an angular projection to hemispherical fisheye projection.

I am getting bizarre results. So, just to see what is going on, I have 
generated an overcast sky with vta option (-vta -vp 1.000 1.000 0.000 
-vd 0.000 0.000 1.000 -vu -1 0 0 -vh 180 -vv 180 -vs 0 -vl 0)

I have used your formula:

pcomb -e 'sq(x):x*x' \
	 -e 'ar=190/2*PI/180*sqrt(sq(2/xmax*x-1)+sq(2/ymax*y-1))' \
	-e 'cf=WE*sq(190*PI/180*2/(xmax+ymax))*if(ar-PI/2,0,if(ar-.01,cos(ar) 
*sin(ar)/ar,1))' \
	-e 'lo=cf*li(1)'  input.hdr > corrected.hdr

(I used 181 degrees instead of 190)

and I was hoping to have an image similar to what I would have generated 
using the vth option (-vth -vp 1.000 1.000 0.000 -vd 0.000 0.000 1.000 
-vu -1 0 0 -vh 180 -vv 180 -vs 0 -vl 0).

After the correction, my luminance values dropped significantly (2500 
cd/m2 dropped to 4.3). You can find the original and corrected images 
at:  http://faculty.washington.edu/inanici/vta/

Any ideas why I am having difficulties with this? Thanks in advance...

Cheers,
Mehlika