scott wrote:
>> The way to do it is to integrate over hemisphere at that point
>> acording to light
>> transport equation. In 2D scene it would be I(1.0 * cos x  dx) and we
>> have to do
>> it from -pi/2 to pi/2. Right?
>>
>> The value of that is sin (pi/2) - sin(-pi/2) = 2. Hmm. Not right. Or
>> is it? Most
>> likely I'm integrating incorrectly or the wrong thing.
> 
> You calculated the total light power collected at that point from all
> direction (and hence the amount emitted assuming 100% reflectance).  The
> correct answer is 2.  You'll get the same answer if you try to work out
> how much light power is emitted from a point on the sky, in fact the
> integral is the same.

Ok. And in 3D the result is? Basically revolving "y=cos x" around y-axis
and calculating the volume. I have forgotten it but I'll figure it out
later...

If irradiance in above case is 2, what about the radiance than is
reflected from that point to a certain direction? (That is equal no
matter what the direction as it is lambertian).

Thanks for your help!

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