[Radiance-general] illum examples

[Jia Hu]

Hello Greg:

Thank you. I also find a similar thread you replied:
http://www.radiance-online.org/pipermail/radiance-general/2003-September/000963.html

I shall change the luminaires material type to "glow" (with maxrad = 0). In
this case, the shadow rays will only be sent to illum rectangle and not be
sent to the luminaire (inside illum rectangle). This case is just similar
with the window (as sky is "glow" type).

Cheers,
Jia

On Tue, Jul 20, 2010 at 2:58 PM, Greg Ward <gregoryjward at gmail.com> wrote:

> 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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