Original Post
When they talk about BRDFs having to be energy conserving, in that the reflected light cant be more than the incoming light, so integrating the BRDF over the hemisphere should return <=1.
If the BRDF for Lambert diffuse is 1/pi, then integrating that over the hemisphere, you get 2. You only get 1 if you include the cos(theta) weight, giving, cos(theta)/pi
So, within direct lighting, you usually have
expanding that out of Lambert which reflects all wavelengths, wouldnt that be
So when they say the brdf must integrate to <=1 over the hemisphere, do they mean
or
If the BRDF for Lambert diffuse is 1/pi, then integrating that over the hemisphere, you get 2. You only get 1 if you include the cos(theta) weight, giving, cos(theta)/pi
So, within direct lighting, you usually have
Lo = BRDF() * Li() * cos(theta);expanding that out of Lambert which reflects all wavelengths, wouldnt that be
Lo = (cos(theta)/pi) * Li() * cos(theta);So when they say the brdf must integrate to <=1 over the hemisphere, do they mean
BRDF() * cos(theta), over the hemisphereor
BRDF() , over the hemisphere