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:
> 

> >
> > 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

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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------

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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}

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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

Post a reply to this message

"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

Post a reply to this message

"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

Post a reply to this message

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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