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Christopher James Huff wrote:
> In article <4081ceec$1@news.povray.org>,
> Dan P <dan### [at] yahoo com> wrote:
>
>
>>>The Earth doesn't really add anything by radiosity alone so I'll add a light
>>>source to it.
>>
>>Right; it's too far away.
>
> Distance has little to do with it, it is the area of sky covered and
> brightness per unit area that counts. The diameter is about 3.6 times
> that of the moon, the sky area covered is almost 13.5 times greater. The
> moon's albedo is less than Earth's, between 7% and 12% compared to 30%.
> (The lunar surface reflects light preferentially in the direction it
> came from, so it appears brighter when nearly full.)
I must be misunderstanding radiosity -- I figured because it was so far
away, it wasn't close enough to the moon to reflect any light to it
because of fall-off.
<snip/>
--
Respectfully,
Dan P
http://<broken link>
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"Dan P" <dan### [at] yahoocom> wrote in message
news:408498ee$1@news.povray.org...
>
> I must be misunderstanding radiosity -- I figured because it was so far
> away, it wasn't close enough to the moon to reflect any light to it
> because of fall-off.
Nice to see people discussing this here. ;-)
Well, um, hey... isn't sky_sphere considered infinitely remote? And yet that
still affects everything else when it comes to radiosity. Okay, wait, bad
example there; considering it usually affects things globally anyhow without
having a finish property, that might not have any true distance.
I've used looks_like (sphere) for making the Sun before, on the assumption
radiosity of a very diffuse object will add to the other objects no matter
how far away. Likewise, false window panes meant to be oversaturated with
outdoor lighting, or light panels in the ceiling, even though closer in to
the scene subjects. Thing is, without skylight the blackness of space can
overwhelm the scene due to darkness also being factored into the radiosity.
Or at least I always thought of it in that way.
I just never considered there to be a distance limitation for radiosity,
other than proximity having weight in the calculation, perhaps because it
isn't really a light source and is more of a color effect which could be
seen from any distance based on it's size and diffuseness and/or ambience.
Much more related to object presence rather than illumination.
--
Bob H.
http://www.3digitaleyes.com
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"Dan P" <dan### [at] yahoocom> wrote in message
news:407f1de5@news.povray.org...
> Oh YEAH -- fantastic!
> It reminds me of Wallace and Grommit.
> "Cheese, Grommit! We need cheese! Cheese!!!!"
Heheh... Yes, I can see it. :)
>
> Another parallel light at low intensity pointing down from behind
the
> object will bring out the details of the shadow and add depth.
You're doin' my head in you are, Mr. DP! How do you know all this
stuff?! ;)
Anyways, I'm glad people like you and Bob are around.
~Steve~
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St. wrote:
> "Dan P" <dan### [at] yahoocom> wrote in message
> news:407f1de5@news.povray.org...
>
>>Oh YEAH -- fantastic!
>>It reminds me of Wallace and Grommit.
>>"Cheese, Grommit! We need cheese! Cheese!!!!"
>
> Heheh... Yes, I can see it. :)
God, I loved that penguin too. :-)
>>Another parallel light at low intensity pointing down from behind
>
> the
>
>>object will bring out the details of the shadow and add depth.
>
> You're doin' my head in you are, Mr. DP! How do you know all this
> stuff?! ;)
I learn it from the guys like Tek and Warp and Chris and all the rest.
If it weren't for this newsgroup, I'd still just be making logos! :-)
I'm one of the least knowledgeable on the group, really; guys like Tek,
Warp, the Chris's, Thorsten, Ken -- the list goes on; all those are far
knowledgeable than I. I just like to post a lot -- they're really busy
right now with the next version of POV-Ray coming soon or making scene
files instead of writing papers about topics like www.hibernate.org for
hungry profs :-)
> Anyways, I'm glad people like you and Bob are around.
>
> ~Steve~
Right back atchya!!
--
Respectfully,
Dan P
http://<broken link>
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In article <408498ee$1@news.povray.org>,
Dan P <dan### [at] yahoocom> wrote:
> I must be misunderstanding radiosity -- I figured because it was so far
> away, it wasn't close enough to the moon to reflect any light to it
> because of fall-off.
Light doesn't magically become less intense with distance, it spreads
out. The inverse square law comes from the area of the surface of a
sphere around the origin point. The moon reflects light from many points
on its sunlit surface. At the distance from it of Earth, the inverse
square law approximates the falloff pretty well, but nearer the surface,
most of the surface area is hidden, and you can only be really near a
small area, unlike a true point source where all light comes from a
single point. Yes, the earth only intercepts a small portion of the
reflected light, but the moon reflects a lot of light. Haven't you ever
been outside on a night with a full moon?
In the case of an object like the moon, an object twice as far away
would have to be twice the radius to cover the same area of sky. At
twice the radius, it would have four times the visible area, exactly
compensating for the fact that one fourth the reflected light per unit
of surface area will reach an observer on Earth. From the lunar surface,
the Earth is much bigger and brighter, and it lights the surface
brightly enough to be seen on the dark side from earth. (That is, enough
light reflects off Earth onto the moon, and back to earth again, to be
visible with the naked eye.)
--
Christopher James Huff <cja### [at] earthlinknet>
http://home.earthlink.net/~cjameshuff/
POV-Ray TAG: <chr### [at] tagpovrayorg>
http://tag.povray.org/
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In article <4084c0a7$1@news.povray.org>,
"Hughes, B." <omn### [at] charternet> wrote:
> Well, um, hey... isn't sky_sphere considered infinitely remote? And yet that
> still affects everything else when it comes to radiosity. Okay, wait, bad
> example there; considering it usually affects things globally anyhow without
> having a finish property, that might not have any true distance.
Whether it has a distance or not does not matter...but it does not have
a finite distance. The sky_sphere is just a background effect, not an
object. In a given direction, you see a given color, no matter what
point you are viewing from. Features on it only have angular area.
> I've used looks_like (sphere) for making the Sun before, on the assumption
> radiosity of a very diffuse object will add to the other objects no matter
> how far away. Likewise, false window panes meant to be oversaturated with
> outdoor lighting, or light panels in the ceiling, even though closer in to
> the scene subjects. Thing is, without skylight the blackness of space can
> overwhelm the scene due to darkness also being factored into the radiosity.
> Or at least I always thought of it in that way.
It's not that the dark is factored in, it's that there isn't any light
to be included. A white wall is much better at reflecting light than
empty space.
> I just never considered there to be a distance limitation for radiosity,
> other than proximity having weight in the calculation, perhaps because it
> isn't really a light source and is more of a color effect which could be
> seen from any distance based on it's size and diffuseness and/or ambience.
> Much more related to object presence rather than illumination.
No...radiosity simply calculates diffusely scattered light. There is
nothing special about this light. Brightness and the visible size are
the important quantities.
--
Christopher James Huff <cja### [at] earthlinknet>
http://home.earthlink.net/~cjameshuff/
POV-Ray TAG: <chr### [at] tagpovrayorg>
http://tag.povray.org/
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Christopher James Huff wrote:
> [...]
>
> In the case of an object like the moon, an object twice as far away
> would have to be twice the radius to cover the same area of sky. At
> twice the radius, it would have four times the visible area, exactly
> compensating for the fact that one fourth the reflected light per unit
> of surface area will reach an observer on Earth. From the lunar surface,
> the Earth is much bigger and brighter, and it lights the surface
> brightly enough to be seen on the dark side from earth. (That is, enough
> light reflects off Earth onto the moon, and back to earth again, to be
> visible with the naked eye.)
>
Note that the same effect causes the moon to be visible during a lunar
eclipse as well, just that in this case the earth faces its dark side to
the moon so the light reaching the moon and making it visible to us is
mostly scattered light. Therefore the appearance of the moon during a
lunar eclipse strongly varies depending on the amount of dust in the
atmosphere.
Christoph
--
POV-Ray tutorials, include files, Sim-POV,
HCR-Edit and more: http://www.tu-bs.de/~y0013390/
Last updated 21 Mar. 2004 _____./\/^>_*_<^\/\.______
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Christopher James Huff wrote:
> In article <408498ee$1@news.povray.org>,
> Dan P <dan### [at] yahoocom> wrote:
>
>>I must be misunderstanding radiosity -- I figured because it was so far
>>away, it wasn't close enough to the moon to reflect any light to it
>>because of fall-off.
>
> Light doesn't magically become less intense with distance, it spreads
> out.
Yes, in this case,
you speak of space,
but in the air,
you must take care,
to remember the particle,
(he's quite the article!),
his friends and foes,
and others he knows,
absorbs the light,
with all its might,
its effort made,
light seems to fade!
> The inverse square law comes from the area of the surface of a
> sphere around the origin point.
Bah, I consider that more a guideline than a law :-)
> The moon reflects light from many points
> on its sunlit surface. At the distance from it of Earth, the inverse
> square law approximates the falloff pretty well, but nearer the surface,
> most of the surface area is hidden, and you can only be really near a
> small area, unlike a true point source where all light comes from a
> single point. Yes, the earth only intercepts a small portion of the
> reflected light, but the moon reflects a lot of light. Haven't you ever
> been outside on a night with a full moon?
What you are describing is called the "albedo" of an object. This
website[1] defines albedo as, "the fraction of light that is reflected
by a body or surface." This site contains some albedos that illustrate
your point:
(higher = brighter)
Venus: 0.76
Earth: 0.33
Moon: 0.12
The Earth is 21% brighter than the moon (as you've said), probably
because of all the water on the surface and in the clouds. Venus is 43%
brighter because it has a lot more clouds to reflect the light. Venus
has been so bright to us that people have mistaken it for the light of
an on-coming train!
> In the case of an object like the moon, an object twice as far away
> would have to be twice the radius to cover the same area of sky. At
> twice the radius, it would have four times the visible area, exactly
> compensating for the fact that one fourth the reflected light per unit
> of surface area will reach an observer on Earth. From the lunar surface,
> the Earth is much bigger and brighter, and it lights the surface
> brightly enough to be seen on the dark side from earth. (That is, enough
> light reflects off Earth onto the moon, and back to earth again, to be
> visible with the naked eye.)
Lemme get out my slide-rule (hmmm... carry the three)... sounds good to
me! :-)
[1] http://zebu.uoregon.edu/~js/glossary/albedo.html
--
Respectfully,
Dan P
http://<broken link>
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From: Jellby
Subject: Re: Cavorite Sphere (off the shelf) [~105K JPG]
Date: 22 Apr 2004 09:08:51
Message: <4087c3e2@news.povray.org>
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Among other things, Dan P wrote:
>> The moon reflects light from many points
>> on its sunlit surface. At the distance from it of Earth, the inverse
>> square law approximates the falloff pretty well, but nearer the surface,
>> most of the surface area is hidden, and you can only be really near a
>> small area, unlike a true point source where all light comes from a
>> single point. Yes, the earth only intercepts a small portion of the
>> reflected light, but the moon reflects a lot of light. Haven't you ever
>> been outside on a night with a full moon?
>
> What you are describing is called the "albedo" of an object. This
> website[1] defines albedo as, "the fraction of light that is reflected
> by a body or surface."
There are different factors: the albedo, the apparent size, the distance to
the sun (or the apparent size of the sun as seen from the body)...
The body gets some light -> distance from the sun
The body reflects some of this light -> albedo
The observer sees a fraction of this light -> apparent size
... and there might be some losses in between due to cosmic dust, atmosphere
interaction, etc.
> Venus
> has been so bright to us that people have mistaken it for the light of
> an on-coming train!
It may be seen in daylight (if you know exactly where to find it), and on a
night without moon, it can cast shadows.
--
light_source{9+9*x,1}camera{orthographic look_at(1-y)/4angle 30location
9/4-z*4}light_source{-9*z,1}union{box{.9-z.1+x clipped_by{plane{2+y-4*x
0}}}box{z-y-.1.1+z}box{-.1.1+x}box{.1z-.1}pigment{rgb<.8.2,1>}}//Jellby
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Dan P wrote:
>
> What you are describing is called the "albedo" of an object. This
> website[1] defines albedo as, "the fraction of light that is reflected
> by a body or surface." This site contains some albedos that illustrate
> your point:
>
> (higher = brighter)
> Venus: 0.76
> Earth: 0.33
> Moon: 0.12
>
> The Earth is 21% brighter than the moon (as you've said), probably
> because of all the water on the surface and in the clouds. Venus is 43%
> brighter [...]
What?
I don't know where you learned math but this can't be meant seriously.
If earth reflects a fraction of 0.33 of the incoming light and the moon
0.12 that means the earth surface reflects 2.75 times the amount of
light the moon surface reflects per surface area. I don't even want to
try to imagine how you came to '21% brighter'...
Christoph
--
POV-Ray tutorials, include files, Sim-POV,
HCR-Edit and more: http://www.tu-bs.de/~y0013390/
Last updated 21 Mar. 2004 _____./\/^>_*_<^\/\.______
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