[Radiance-general] illum examples

[Greg Ward]

There is a special case for "glow" sources to take care of this.   
Quoting from the Radiance reference manual <http://radsite.lbl.gov/radiance/refer/ray.html#Materials 
 >:

Glow
Glow is used for surfaces that are self-luminous, but limited in their  
effect. In addition to the radiance value, a maximum radius for shadow  
testing is given:
         mod glow id
         0
         0
         4 red green blue maxrad
If maxrad is zero, then the surface will never be tested for shadow,  
although it may participate in an interreflection calculation. If  
maxrad is negative, then the surface will never contribute to scene  
illumination. Glow sources will never illuminate objects on the other  
side of an illum surface. This provides a convenient way to illuminate  
local light fixture geometry without overlighting nearby objects.


> From: Jia Hu <hujia06 at gmail.com>
> Date: July 20, 2010 10:23:31 AM PDT
>
> Hello experts:
>
>  Even if mkillum is used for window, the shadow ray in a point will  
> be still sent to sun (just pass through the window) to calculate the  
> contribution of sun (direct calculation). The reason is "illum  
> surface do not interfere with the shadow testing of others, non  
> illumn sources" (Page 572 in the book Rendering with Radiance).
>
> In Page 568-569 of the book, a decorative luminaries is illustrated  
> to show why to use illum. One reason of using illum is that the  
> shadow ray in a point may hit the fixture of the lamp and get a  
> shadow which is incorrect (it is incorrect because other parts of  
> the lamp may illuminate the point). Using illum will make the lamp a  
> uniform radiator and therefore a shadow ray hit the illum will  
> return a nonzero value.
>
> In the above first paragraph , as the book said, the shadow ray  
> (that hitting the sun) will not be interfered with by the illum  
> (window, in this example) . But for the decorative luminaries  
> example (in the second paragraph), My problem is that I do not know  
> which of the following is right as to the ways of treatment of  
> Shadow ray test of this example.
>
> (1)  In this way, we do not need to consider the shadow ray test of  
> the lamp. In orther words, we can think of the new illum enclosing  
> the lamp as a special light, and there is no light (lamp) inside the  
> illum, and therefore, only one shadow ray is sent to the illum and  
> NO shadow ray is sent to the lamp (inside illum)
>
> (2) It is similar with the example (window illum) in the above first  
> paragraph,  In this case, one shadow ray is sent to hit the lamp and  
> one shadow ray is sent to hit the illum. Both of them will calculate  
> the direct part of light source. The illum just shows the luminance  
> distribution of lights reflected by the lamp fixture rather than  
> lights directly from lamp.
>
> Thank you,
>
> Jia
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