So now that we've got blurring reflections licked, how do we fade
out reflections?  Maybe media?  (Which I still haven't figured out
how to make work the way I want it to...)

-- 
Tim Cook
http://empyrean.scifi-fantasy.com

-----BEGIN GEEK CODE BLOCK-----
Version: 3.12
GFA dpu- s: a?-- C++(++++) U P? L E--- W++(+++)>$
N++ o? K- w(+) O? M-(--) V? PS+(+++) PE(--) Y(--)
PGP-(--) t* 5++>+++++ X+ R* tv+ b++(+++) DI
D++(---) G(++) e*>++ h+ !r--- !y--
------END GEEK CODE BLOCK------

Post a reply to this message

On Mon, 22 Oct 2001 12:47:57 -0400, Timothy R. Cook wrote:
> So now that we've got blurring reflections licked, how do we fade
> out reflections?  Maybe media?  (Which I still haven't figured out
> how to make work the way I want it to...)

Try this.  It's not perfect, and I'm sure others will improve on it, but
it demonstrates the idea.  (3.5 only, I'm afraid, due to the no_image)
Unfortunately, this may be due to an undocumented and perhaps incorrect
behavior of no_image...

Notice that if you comment out or remove the line that says "reflective 
sphere location" the reflective sphere ceases to have faded reflections; 
this is the part that might be considered a bug.

difference {
  box {-100 100}
  sphere {x-2*z .001} //camera location
  sphere {10*z+x 5.001} // reflective sphere location
  
  hollow 
  pigment {color rgbt 1}
  interior {
    media {
      absorption .3
      density {rgb 1}
    }
  }  
  no_image
} 
 
sphere {10*z+x 5
  pigment {color red 1}
  finish {reflection .9 ambient 0 diffuse .5 specular 0}
}

light_source {<-20,20,-20> rgb 1 shadowless}
 
camera {
  location x-2*z 
  look_at 0
}

#declare I=0;
#while (I<5)     
  cylinder {I*z-y I*z+y .05 pigment {color green 1} finish {ambient 1}}
  #declare I=I+1;
#end



-- 
plane{-z,-3normal{crackle scale.2#local a=5;#while(a)warp{repeat x flip x}rotate
z*60#local a=a-1;#end translate-9*x}pigment{rgb 1}}light_source{-9red 1rotate 60
*z}light_source{-9rgb y rotate-z*60}light_source{9-z*18rgb z}text{ttf"arial.ttf"
"RP".01,0translate-<.6,.4,.02>pigment{bozo}}light_source{-z*3rgb-.2}//Ron Parker

Post a reply to this message

What about fresnel reflection?

-tgq


"Timothy R. Cook" <tim### [at] scifi-fantasycom> wrote in message
news:3BD### [at] scifi-fantasycom...
> So now that we've got blurring reflections licked, how do we fade
> out reflections?  Maybe media?  (Which I still haven't figured out
> how to make work the way I want it to...)
>
> --
> Tim Cook
> http://empyrean.scifi-fantasy.com
>
> -----BEGIN GEEK CODE BLOCK-----
> Version: 3.12
> GFA dpu- s: a?-- C++(++++) U P? L E--- W++(+++)>$
> N++ o? K- w(+) O? M-(--) V? PS+(+++) PE(--) Y(--)
> PGP-(--) t* 5++>+++++ X+ R* tv+ b++(+++) DI
> D++(---) G(++) e*>++ h+ !r--- !y--
> ------END GEEK CODE BLOCK------

Post a reply to this message

"Timothy R. Cook" wrote:
> 
> So now that we've got blurring reflections licked, how do we fade
> out reflections?

What physical property of materials and light would this model?

_____________
Kari Kivisalo

Post a reply to this message

I think I understand what you mean here now, that as objects get further
away from the reflecting object, they fade in brightness as opposed to
blurring out of visibilty, aka light attenuation.  POV does have light
attenuation, but it is only applied to direct illumination and not diffuse
or reflected light, but this problem exists in looking directly at as well
as reflections. i.e., when using light attenuation,  as an object gets
further from a light source, it is illuminated less because the direct
lighting gets faded, but as an equally illuminated object gets further from
the camera, it does not fade (which in reality, it should if direct light
gets attenuated, then so should indirect, but POV light attenuation doesn't
work that way). As Ron shows, one way to simulate this is with absorbing
media.  As a quicker method, you could use constant fog with a black
pigment, as the features of media aren't really required here. Adding this
to your scene (make sure your camera is not in a solid object, that includes
inverted planes!)

fog {
  fog_type 1
  distance 10
  color rgb 0
}


You will need to play with the distance value to get what you need.

Note that sky_sphere and backgroud won't work with these, as they are an
infinite distance away and will get faded to black, use a large hollow
sphere with a high ambient finish to replace sky_sphere.

-tgq

Post a reply to this message

On Mon, 22 Oct 2001 20:59:53 +0300, Kari Kivisalo wrote:
> "Timothy R. Cook" wrote:
>> 
>> So now that we've got blurring reflections licked, how do we fade
>> out reflections?
> 
> What physical property of materials and light would this model?

It's supposed to resemble a poor man's reflection blur.

-- 
#local R=rgb 99;#local P=R-R;#local F=pigment{gradient x}box{0,1pigment{gradient
y pigment_map{[.5F pigment_map{[.3R][.3F color_map{[.15red 99][.15P]}rotate z*45
translate x]}]#local H=pigment{gradient y color_map{[.5P][.5R]}scale 1/3}[.5F
pigment_map{[.3R][.3H][.7H][.7R]}]}}}camera{location.5-3*z}//only my opinions

Post a reply to this message

In article <3bd46719$1@news.povray.org>, Tin### [at] hotmailcom 
says...
> lighting gets faded, but as an equally illuminated object gets further from
> the camera, it does not fade (which in reality, it should if direct light
> gets attenuated, then so should indirect,

No, this is not true. Just look at the Moon. It has roughly the same 
brightness as a landscape in daylight has.
The reason objects further away from a lightsource are less illuminated 
then those near it, is NOT that the brightness of the lightsource 
decreases over distance (this is only possible if there really is some 
absorbing media in between), but because the area (in degrees^2) it 
occupies when viewed from the object is smaller.

The light attenuation in Pov is only necessary because raytraced 
lightsources are infinitly small.

Lutz-Peter

Post a reply to this message

To follow up on this I posted two images in p.b.i showing how it would work.
In one image I used standard light attenuation, and in the second I have
added a black fog with the distance set the same as the attenuation distance
of the light sources, note that the light sources have a white sphere with a
high ambient value applied to them, so they do not fade as much as the red
spheres.

-tgq

Post a reply to this message

On Mon, 22 Oct 2001 14:35:57 -0400, Trevor Quayle wrote:
> work that way). As Ron shows, one way to simulate this is with absorbing
> media.  As a quicker method, you could use constant fog with a black
> pigment, as the features of media aren't really required here. Adding this

You have greatly misunderstood the point of my demonstration.

That absorbing media is funny.  Notice how it doesn't do anything to the
first five green cylinders, but the other five (the reflections of the 
first five in the surface of the sphere) fade out?  There's something special
going on there that you can't do with fog.  

For more fun, try adding another reflective sphere to the scene somewhere and
notice how it doesn't behave the same as the one that's there now.  So 
there's something about that absorbing media that's not only strange from 
the camera's point of view, but also from that one sphere's point of view.

The answer, of course, lies in the two spheres that are differenced out of
the media, and in the no_image keyword.  It's hard to state the exact rules
that dictate when the media appears and when it doesn't, but I'll give it
a shot:

- a camera ray that originates inside the media container is affected.  This
  might very well be a bug, depending on whether you believe that the no_image
  keyword should affect the media in the interior of the object.  Luckily,
  even if this bug is fixed it won't matter for this technique.

- a camera ray that originates outside the media container is not affected,
  because its intersections with the media container are ignored due to the
  no_image.

- When the camera is outside the media container, a reflected ray that 
  originates inside the media container is not affected.  This is because 
  we only add the media to the list of things to consider when the ray 
  enters the media container, which doesn't happen in this case unless 
  the ray happens to intersect a "bubble" in the container.  This fact 
  is probably also a bug.  If it's fixed, it will limit the technique
  considerably (requiring all reflective objects in your scene to fade, 
  and requiring them all to fade to the same color.)  It probably should
  be fixed, but fixing it seems like it would cause a performance hit,
  and it's not as though it's really a real-world feature anyway.  The
  part about the "bubbles" has some effect on what sort of objects and
  scene geometry you can use, but not too much. 

- When a reflected ray originates outside the media container, it is 
  always affected by the media.  This is the only part that really makes
  sense about the whole thing. :)

--
#macro R(L P)sphere{L __}cylinder{L P __}#end#macro P(_1)union{R(z+_ z)R(-z _-z)
R(_-z*3_+z)torus{1__ clipped_by{plane{_ 0}}}translate z+_1}#end#macro S(_)9-(_1-
_)*(_1-_)#end#macro Z(_1 _ __)union{P(_)P(-_)R(y-z-1_)translate.1*_1-y*8pigment{
rgb<S(7)S(5)S(3)>}}#if(_1)Z(_1-__,_,__)#end#end Z(10x*-2,.2)camera{rotate x*90}

Post a reply to this message

On 22 Oct 2001 16:23:35 -0400, Ron Parker wrote:
> On Mon, 22 Oct 2001 14:35:57 -0400, Trevor Quayle wrote:
>> work that way). As Ron shows, one way to simulate this is with absorbing
>> media.  As a quicker method, you could use constant fog with a black
>> pigment, as the features of media aren't really required here. Adding this
> 
> You have greatly misunderstood the point of my demonstration.
> 
> That absorbing media is funny.  Notice how it doesn't do anything to the
> first five green cylinders, but the other five (the reflections of the 
> first five in the surface of the sphere) fade out?  There's something special
> going on there that you can't do with fog.  

Oh, and just for fun, try adding a sky_sphere.  One that's not black.  I'll
wait here while you try that...

sky_sphere {pigment {color blue 1}}

So, uh... shouldn't the background still be black, or at least darker than
blue 1?  There really is something funny going on here.

--
#macro R(L P)sphere{L __}cylinder{L P __}#end#macro P(_1)union{R(z+_ z)R(-z _-z)
R(_-z*3_+z)torus{1__ clipped_by{plane{_ 0}}}translate z+_1}#end#macro S(_)9-(_1-
_)*(_1-_)#end#macro Z(_1 _ __)union{P(_)P(-_)R(y-z-1_)translate.1*_1-y*8pigment{
rgb<S(7)S(5)S(3)>}}#if(_1)Z(_1-__,_,__)#end#end Z(10x*-2,.2)camera{rotate x*90}

Post a reply to this message

Actually I was commenting on how to fade reflections, not the funniness with
media you pointed out, I had just mentioned your post in regards to the
using of media for this.

-tgq

"Ron Parker" <ron### [at] povrayorg> wrote in message
news:slr### [at] fwicom...
> On Mon, 22 Oct 2001 14:35:57 -0400, Trevor Quayle wrote:
> > work that way). As Ron shows, one way to simulate this is with absorbing
> > media.  As a quicker method, you could use constant fog with a black
> > pigment, as the features of media aren't really required here. Adding
this
>
> You have greatly misunderstood the point of my demonstration.
>
> That absorbing media is funny.  Notice how it doesn't do anything to the
> first five green cylinders, but the other five (the reflections of the
> first five in the surface of the sphere) fade out?  There's something
special
> going on there that you can't do with fog.
>
> For more fun, try adding another reflective sphere to the scene somewhere
and
> notice how it doesn't behave the same as the one that's there now.  So
> there's something about that absorbing media that's not only strange from
> the camera's point of view, but also from that one sphere's point of view.
>
> The answer, of course, lies in the two spheres that are differenced out of
> the media, and in the no_image keyword.  It's hard to state the exact
rules
> that dictate when the media appears and when it doesn't, but I'll give it
> a shot:
>
> - a camera ray that originates inside the media container is affected.
This
>   might very well be a bug, depending on whether you believe that the
no_image
>   keyword should affect the media in the interior of the object.  Luckily,
>   even if this bug is fixed it won't matter for this technique.
>
> - a camera ray that originates outside the media container is not
affected,
>   because its intersections with the media container are ignored due to
the
>   no_image.
>
> - When the camera is outside the media container, a reflected ray that
>   originates inside the media container is not affected.  This is because
>   we only add the media to the list of things to consider when the ray
>   enters the media container, which doesn't happen in this case unless
>   the ray happens to intersect a "bubble" in the container.  This fact
>   is probably also a bug.  If it's fixed, it will limit the technique
>   considerably (requiring all reflective objects in your scene to fade,
>   and requiring them all to fade to the same color.)  It probably should
>   be fixed, but fixing it seems like it would cause a performance hit,
>   and it's not as though it's really a real-world feature anyway.  The
>   part about the "bubbles" has some effect on what sort of objects and
>   scene geometry you can use, but not too much.
>
> - When a reflected ray originates outside the media container, it is
>   always affected by the media.  This is the only part that really makes
>   sense about the whole thing. :)
>
> --
> #macro R(L P)sphere{L __}cylinder{L P __}#end#macro P(_1)union{R(z+_
z)R(-z _-z)
> R(_-z*3_+z)torus{1__ clipped_by{plane{_ 0}}}translate z+_1}#end#macro
S(_)9-(_1-
> _)*(_1-_)#end#macro Z(_1 _
__)union{P(_)P(-_)R(y-z-1_)translate.1*_1-y*8pigment{
> rgb<S(7)S(5)S(3)>}}#if(_1)Z(_1-__,_,__)#end#end Z(10x*-2,.2)camera{rotate
x*90}

Post a reply to this message

On Mon, 22 Oct 2001 16:50:28 -0400, Trevor Quayle wrote:
> Actually I was commenting on how to fade reflections, not the funniness with
> media you pointed out, I had just mentioned your post in regards to the
> using of media for this.

But the point is, what you suggested fades everything, not just reflections.

-- 
#local R=rgb 99;#local P=R-R;#local F=pigment{gradient x}box{0,1pigment{gradient
y pigment_map{[.5F pigment_map{[.3R][.3F color_map{[.15red 99][.15P]}rotate z*45
translate x]}]#local H=pigment{gradient y color_map{[.5P][.5R]}scale 1/3}[.5F
pigment_map{[.3R][.3H][.7H][.7R]}]}}}camera{location.5-3*z}//only my opinions

Post a reply to this message

Trevor Quayle wrote:
>
> as an equally illuminated object gets further from
> the camera, it does not fade (which in reality, it should if direct light
> gets attenuated

No, this is not what happens in reality. The energy arriving
from an object when viewed through a constant space angle stays
the same because the observed area (and energy) on the object
increases when viewed from greater distance. The 1/r^2 attenuation
and increase of observed area cancel each other out. The brightness
(pixel values) of surfaces is not a function of distance.


_____________
Kari Kivisalo

Post a reply to this message

What I was trying to show here was that reflections don't fade as a result
of reflective properties (not counting variable reflection, whereby the
reflection fades with the viewing angle).  The intensity of light does fade
over distance (1/r^2), the energy stays the same but it is spread over a
larger area:
http://imagine.gsfc.nasa.gov/YBA/M31-velocity/1overR2-more.html,
http://www2.ncsu.edu/ncsu/pams/science_house/CBLable/resources.html
Thus, making objects further away appear less bright. POVRay can do light
fading, but, as the docs state, it is only applied to direct illumination
from light sources, reflected or diffuse light does not get attenuated.  If
there is any fading of a reflection as the objects distance increases, this
is why, it is not a reflective property, but a light property and should be
extended to the whole scene, not just reflected images.  I was trying to use
a black fog to simulate this effect, I haven't put too much thought into it
yet as to the realism, but perhaps I will and post some of my results and
conclusions at a later time.
But, as always, feel free to comment on or dispute this if you feel I am
mistaken in my assumptions,  as I promote the spreading of knowledge, not
ignorance and I do appreciate any input or illumination on the matter, pun
intended ;).

-tgq

"Kari Kivisalo" <ray### [at] engineercom> wrote in message
news:3BD488F1.AA0F3F21@engineer.com...
> Trevor Quayle wrote:
> >
> > as an equally illuminated object gets further from
> > the camera, it does not fade (which in reality, it should if direct
light
> > gets attenuated
>
> No, this is not what happens in reality. The energy arriving
> from an object when viewed through a constant space angle stays
> the same because the observed area (and energy) on the object
> increases when viewed from greater distance. The 1/r^2 attenuation
> and increase of observed area cancel each other out. The brightness
> (pixel values) of surfaces is not a function of distance.
>
>
> _____________
> Kari Kivisalo

Post a reply to this message

Ron Parker wrote:
> > What physical property of materials and light would this model?
> 
> It's supposed to resemble a poor man's reflection blur.

Actually I was thinking more along the lines of linoleum floors,
where you only have a few inches of reflection.

-- 
Tim Cook
http://empyrean.scifi-fantasy.com

-----BEGIN GEEK CODE BLOCK-----
Version: 3.12
GFA dpu- s: a?-- C++(++++) U P? L E--- W++(+++)>$
N++ o? K- w(+) O? M-(--) V? PS+(+++) PE(--) Y(--)
PGP-(--) t* 5++>+++++ X+ R* tv+ b++(+++) DI
D++(---) G(++) e*>++ h+ !r--- !y--
------END GEEK CODE BLOCK------

Post a reply to this message

Trevor Quayle wrote:
>
> over distance (1/r^2), the energy stays the same but it is spread over a
> larger area:

Yes, but the observed area increases by r^2 so they cancel out.
How the energy got on or leaves the surface (direct, radiosity, transmission,
reflection) doesn't matter, the emitted energy to observer's fixed
space angle is not a function of distance.

Do your own experiment. Look at a wall from 1 m distance. Note the
brightness. Go to 3 m distance and compare the brightness. It's
the same!

A pixel gathers light from fixed space angle so the area it gathers
the energy from increases by r^2. The energy arriving at the pixel
from a fixed area on the surface decreases by 1/r^2. You don't have
to be a rocket scientist to figure this one out.

_____________
Kari Kivisalo

Post a reply to this message

Kari Kivisalo escribió:
> 
> No, this is not what happens in reality. The energy arriving
> from an object when viewed through a constant space angle stays
> the same because the observed area (and energy) on the object
> increases when viewed from greater distance. The 1/r^2 attenuation
> and increase of observed area cancel each other out. The brightness
> (pixel values) of surfaces is not a function of distance.

  Yes, you are absolutely correct. But we are missing one point here.
That's for theorically perfect surfaces! We were talking about fading on
blurred reflection, and that's another thing...

  When some reflective object shows blurred reflections, it's due to the
very small irregularity of the surface, which spreads out the rays. So,
as
greater is the distance to the surface from the light source (or
object),
and less rays hit the surface, then smaller is the chance to get rays
reflected on that small irregularities towards the camera. The visible
effect is that blurred reflections seems to disapear with distance. Note
I said "seems". Because in fact the reflections of distant objects are
highly and very softly distributed along all the surface irregularities.
Isn't?

  So, I agree that a faked reflection fading will make some artistic
tasks easier to do... like the linoleum floor example.

--
Jaime Vives Piqueres

La Persistencia de la Ignorancia
http://www.ignorancia.org/

Post a reply to this message

I think I understand what you are saying here and can see how it applies to
something that covers the entire field of vision such as a wall, but can you
explain how it would apply to a small object against a dark background?  To
me the area that the pixel is gathering light from is increasing by r^2, but
since the object is fully in the field of vision and the background is dark,
there is no additional light to be gathered than what is reflected by the
object.  If you look at small object, say a red sphere (I just love those
things, gotta get me some real ones) under a dim light in a dark room, woul
you expect the apparent intensity of the ball's colour to be the same
whether you are 1m away or 10m away?

Please dispell any myths I have here as I need to know now.

-tgq


"Kari Kivisalo" <ray### [at] engineercom> wrote in message
news:3BD58A14.D268FC06@engineer.com...
> Trevor Quayle wrote:
> >
> > over distance (1/r^2), the energy stays the same but it is spread over a
> > larger area:
>
> Yes, but the observed area increases by r^2 so they cancel out.
> How the energy got on or leaves the surface (direct, radiosity,
transmission,
> reflection) doesn't matter, the emitted energy to observer's fixed
> space angle is not a function of distance.
>
> Do your own experiment. Look at a wall from 1 m distance. Note the
> brightness. Go to 3 m distance and compare the brightness. It's
> the same!
>
> A pixel gathers light from fixed space angle so the area it gathers
> the energy from increases by r^2. The energy arriving at the pixel
> from a fixed area on the surface decreases by 1/r^2. You don't have
> to be a rocket scientist to figure this one out.
>
> _____________
> Kari Kivisalo

Post a reply to this message

Jaime Vives Piqueres wrote:
> 
> Kari Kivisalo escribió:
> >
> > No, this is not what happens in reality. The energy arriving
> > from an object when viewed through a constant space angle stays
> > the same because the observed area (and energy) on the object
> > increases when viewed from greater distance. The 1/r^2 attenuation
> > and increase of observed area cancel each other out. The brightness
> > (pixel values) of surfaces is not a function of distance.
> 
>   Yes, you are absolutely correct. But we are missing one point here.
> That's for theorically perfect surfaces!

No. How the energy get's emitted or how it's distributed spatially
doesn't matter. Surface properties are irrelevant, only the emitted
energy matters. The emission can be from gas, laser, wood, water
fog or anything that emits light via radiosity, transmission, reflection
or diffuse emission from direct lighting.

Put a sample of linoleum in a test bench and measure it's photometric
brightness at various distances. It will take a material considerably
more exotic than linoleum to break the laws of physics :)


_____________
Kari Kivisalo

Post a reply to this message

Further to what I am trying to figure out here, we can also look at faded
reflections as it is the root of the problem.  For things like linoleum
floors that have a low reflection (let's consider a smooth one with no
reflection blurring), objects that are a large distance away can be seen
less clearly than objects that are very close, if they can be seen at all,
however brighter objects can be seen more clearly such as light bulbs or a
bright window.  To me it is logical that the reflective properties don't
change as objects move further away, the surface can only reflect a
percentage of the light it receives, so this gives me the impression that
less reflected light is being recieved from objects as they move further
away.  It is a lot easier to see in this case because only a small
percentage of the light is reflected, showing the contrast a lot better.
Am I wrong in what I am assuming here?

-tgq

Post a reply to this message

Kari Kivisalo wrote:
> Surface properties are irrelevant, only the emitted energy matters.

It's the surface properties which determine the emitted energy.

> Put a sample of linoleum in a test bench and measure its
> photometric brightness at various distances. It will take a
> material considerably more exotic than linoleum to break the
> laws of physics :)

Take a chair, set it on a linoleum floor, then look at its
reflection.  You don't see the entire chair.

-- 
Tim Cook
http://empyrean.scifi-fantasy.com

-----BEGIN GEEK CODE BLOCK-----
Version: 3.12
GFA dpu- s: a?-- C++(++++) U P? L E--- W++(+++)>$
N++ o? K- w(+) O? M-(--) V? PS+(+++) PE(--) Y(--)
PGP-(--) t* 5++>+++++ X+ R* tv+ b++(+++) DI
D++(---) G(++) e*>++ h+ !r--- !y--
------END GEEK CODE BLOCK------

Post a reply to this message

On Tue, 23 Oct 2001 12:31:16 -0400, Trevor Quayle wrote:
> Am I wrong in what I am assuming here?

Yes.  The effect you see in Linoleum[1] is *exactly* reflection blurring.  If
you really had a perfectly smooth Linoleum with no blurring, you'd find 
that it didn't exhibit the effects you associate with Linoleum.

However, there's still a place for faded reflection in the POV toolbox.  It's
definitely the case that no matter how we do it, real blurred reflection takes
a lot of horsepower.  If faded reflection is good enough, there's no reason
it shouldn't be used.  However, I don't think it actually works.  To work,
it would need to fade based on the distance normal to the surface, not the
total distance.  That's just not practical.

[1] Linoleum is - or at least was - somebody's trademark.  It's only fair to
    capitalize it.

--
#macro R(L P)sphere{L __}cylinder{L P __}#end#macro P(_1)union{R(z+_ z)R(-z _-z)
R(_-z*3_+z)torus{1__ clipped_by{plane{_ 0}}}translate z+_1}#end#macro S(_)9-(_1-
_)*(_1-_)#end#macro Z(_1 _ __)union{P(_)P(-_)R(y-z-1_)translate.1*_1-y*8pigment{
rgb<S(7)S(5)S(3)>}}#if(_1)Z(_1-__,_,__)#end#end Z(10x*-2,.2)camera{rotate x*90}

Post a reply to this message

On Tue, 23 Oct 2001 12:32:19 -0400, Timothy R. Cook wrote:
> Kari Kivisalo wrote:
>> Surface properties are irrelevant, only the emitted energy matters.
> 
> It's the surface properties which determine the emitted energy.
> 
>> Put a sample of linoleum in a test bench and measure its
>> photometric brightness at various distances. It will take a
>> material considerably more exotic than linoleum to break the
>> laws of physics :)
> 
> Take a chair, set it on a linoleum floor, then look at its
> reflection.  You don't see the entire chair.

Some of that is Fresnel reflection.  Some of it is blurred reflection.  
None of it is the inverse-square law.

-- 
plane{-z,-3normal{crackle scale.2#local a=5;#while(a)warp{repeat x flip x}rotate
z*60#local a=a-1;#end translate-9*x}pigment{rgb 1}}light_source{-9red 1rotate 60
*z}light_source{-9rgb y rotate-z*60}light_source{9-z*18rgb z}text{ttf"arial.ttf"
"RP".01,0translate-<.6,.4,.02>pigment{bozo}}light_source{-z*3rgb-.2}//Ron Parker

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Trevor Quayle wrote:
>
> explain how it would apply to a small object against a dark background?

You just look at the object surface and ignore the rest. One mesure
of brightness (pixel values) could be W/m^2. It doesn't change
when the size (area) of the object changes. A big object which emits
100 W/m^2 is as bright as a small object which emits 100 W/m^2.
The total energy emitted changes but not the energy/area (brightness).
This is all seen from a fixed camera distance of course.


_____________
Kari Kivisalo

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"Kari Kivisalo" <ray### [at] engineercom> wrote in message
news:3BD5A384.940229DF@engineer.com...
> Trevor Quayle wrote:
> >
> > explain how it would apply to a small object against a dark background?
>
> You just look at the object surface and ignore the rest. One mesure
> of brightness (pixel values) could be W/m^2. It doesn't change
> when the size (area) of the object changes. A big object which emits
> 100 W/m^2 is as bright as a small object which emits 100 W/m^2.
> The total energy emitted changes but not the energy/area (brightness).
> This is all seen from a fixed camera distance of course.

I actually meant the same sized object at varying distances, I can see how
what you are saying works for varying object sizes.

-tgq


>
>
> _____________
> Kari Kivisalo

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"Timothy R. Cook" wrote:
> 
> Take a chair, set it on a linoleum floor, then look at its
> reflection.  You don't see the entire chair.

Yes, the reflection is blurred, we know that already.

_____________
Kari Kivisalo

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Timothy R. Cook <tim### [at] scifi-fantasycom> wrote:
:> It's supposed to resemble a poor man's reflection blur.

: Actually I was thinking more along the lines of linoleum floors,
: where you only have a few inches of reflection.

  That is, what he said.

-- 
#macro N(D,I)#if(I<6)cylinder{M()#local D[I]=div(D[I],104);M().5,2pigment{
rgb M()}}N(D,(D[I]>99?I:I+1))#end#end#macro M()<mod(D[I],13)-6,mod(div(D[I
],13),8)-3,10>#end blob{N(array[6]{11117333955,
7382340,3358,3900569407,970,4254934330},0)}//                     - Warp -

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