[Radiance-general] Zenith Luminance from rvu

[Guglielmetti, Robert]

Hi Vaib,

Falsecolor now has an "auto" option to the –s parameter, which will automatically set the top of the scale to the peak value found in the image. Note the peak value will not necessarily be at the zenith (which, yes, is where the vector 0,0,1 points to in a hemispherical fisheye view).

I also find that for other than overcast skies, using a logarithmic scale (-log 2) is very useful for getting the most value out of a falsecolor representation of a sky's luminous distribution.

On 5/23/14, 12:14 PM, "Vaib" <vaibhavjain.co at gmail.com<mailto:vaibhavjain.co at gmail.com>> wrote:

Hello Everyone,

I am trying to create falsecolor fish-eye image of just the sky. For that I am using:

rpict -vta -vp 0 0 0 -vd 0 0 1 -vu 0 1 0 -vh 180 -vv 180 sky.oct | falsecolor -s XYZ -l cd/m2 > sky.hdr

Dr.Mardaljevic in his dissertation said to set -s argument of falsecolor close to the Zenith luminance, that can be obtained by rvu's trace command.

To get the Zenith luminance, I am tracing the ray (using rvu) that is approximately at the centre of the image created with the above view parameters, and setting this ray's luminance value as the -s option of falsecolor.

Is this the right way to show/cover the whole luminance spectrum in the sky ? (It doesn't look so in the image..)

Or shall I trace the ray at the apparently at the brightest pixel ? (by reducing the exposure, and figuring out that pixel)

I think the term Zenith Luminance may mean the luminance at the centre of the sky, but I must take your opinion before concluding. Thank you!

Best regards,
Vaib