scott wrote:
>> 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).
> 
> No it's not equal, it's proportional to cos(x), that is what lambertian
> means.  It then "looks" the same brightness to a camera/eye/surface from
> any angle, because the area you see for each pixel/cone/rod/patch is
> proportional to 1/cos(x), so the cos factor cancels out leaving constant
> "brightness" from any angle.
> 
> In the case of irradiance of 2, the radiance is actually exactly cos(x),
> because if you integrate cos(x) over -pi/2 to pi/2, you will get 2, as
> you already found out.  So in fact your sky and ground are identically
> looking planes (assuming the sky is not reflective at all), which you
> proved already.
> 
> In 3D your radiance is now in units of power per solid angle per area,
> compared to power per angle per area.  Lambertian still means that the
> radiance is proportional to cos(x) (where x is the angle between the
> surface normal and radiated/reflected light), it's just you need to
> integrate over a hemi-sphere rather than a semi-circle.
> 
> For lambertian surfaces, in reality this just means that your units are
> different and you only integrate from 0 to pi/2.
> 
> So in 3D, if you have a lambertian surface with radiance function
> 1*cos(x) Watts/steradian/metre2, the total irradiance from a point will
> be 1 Watt/metre2.


Damn it. I have to chew on this one a bit. I need my renderer to give
accurate results and I want to be able to define light sources using
real units.

Now ssRay gives good looking results which are or are not accurate. I
have no way to verify it. I'll be back later...

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