I would like to do some radiosity testing with outdoors scenes and
for this I need to know how much energy comes just from the sun
to the surface and how much from the rest of the sky. All I need is
the ratio maybe for a clear sky and 50% cloudy.

I tried following Ken's links and altavista but didn't find hard data.

_______________________________________________________________________
Kari Kivisalo                                          www.kivisalo.net

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Kari Kivisalo wrote:
> 
> I would like to do some radiosity testing with outdoors scenes and
> for this I need to know how much energy comes just from the sun
> to the surface and how much from the rest of the sky. All I need is
> the ratio maybe for a clear sky and 50% cloudy.

That depends on how high in the sky the sun is.  At sunrise or sunset,
the sun is scattered through a maximum thickness of atmosphere, and half
of the lit sky is blocked.  At local noon, the sun is brightest, but a
larger proportion of the sky is scattering light as well.  Also,
remember that the sky is typically brighter closer to the sun, even at
noon.  It's not realistic for the sky to have uniform brightness unless
it's either uniformally dark or overcast.

Sorry if this makes life more complicated without answering your
question.

-Mark Gordon

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Mark Gordon wrote:
> remember that the sky is typically brighter closer to the sun, even at
> noon.  It's not realistic for the sky to have uniform brightness unless
> it's either uniformally dark or overcast.

I was planning to make an image map for the sky and calculate
it's average brightness. If the ratio is 5 I would use color*5
for sun and ambient 1/average_brightness for sky.

All I need is a rough figure for the ratio.
Is it 1,2,5,10 or 20?

K.K.

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"Kari Kivisalo" <kar### [at] kivisalonet> wrote in message
news:39587BE1.162461B9@kivisalo.net...
| Mark Gordon wrote:
| > remember that the sky is typically brighter closer to the sun, even at
| > noon.  It's not realistic for the sky to have uniform brightness unless
| > it's either uniformally dark or overcast.
|
| I was planning to make an image map for the sky and calculate
| it's average brightness. If the ratio is 5 I would use color*5
| for sun and ambient 1/average_brightness for sky.
|
| All I need is a rough figure for the ratio.
| Is it 1,2,5,10 or 20?

I'm not sure of the actual idea but if you could go by magnitudes then it
might be plausible.  The Sun is around -25 mag. while the clear blue sky (not
near the sun anyway, but daylight sky) should be around -2 mag.  I'm guessing
this right now from what the planet Venus mag. can be while still seeing it in
the daytime, meaning the sky would be about the same or a little less.
So, looking up astronomical magnitudes might yield some useful info.
Hmm, I just looked up what magnitude the Sun has and I was close, -26.7.
Since Venus varies in brightness it averages probably about -3 (I must've been
thinking of Jupiter before :-) Sure I was...) but I think it's daytime showing
is only when at around -4 or greater mag., which I looked up as well (the
magnitude, not the specifics).  So daylight sky must average about that, -4.
Now, in case you don't know this already, a magnitude of +1 is 2.5 times
dimmer than magnitude 0 and each preceding or proceeding magnitude is likewise
2.5X the adjacent one.
Maybe that will help you, but if I were you I'd look it up someplace.

Bob

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I wrote:
> 
> I would like to do some radiosity testing with outdoors scenes and
> for this I need to know how much energy comes just from the sun
> to the surface and how much from the rest of the sky.

I found that the ratio is from 8 to 9. Thanks to aquatic plant
hobbyists :)

______________________________________________________________________
Kari Kivisalo                                  http://www.kivisalo.net

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"Kari Kivisalo" <kar### [at] kivisalonet> wrote in message
news:399D983B.26C0137F@kivisalo.net...
|
| I found that the ratio is from 8 to 9. Thanks to aquatic plant hobbyists
:)

Oh, that's good, because I multiplied out the number due to magnitude
differences (which I was suggesting before) and it was around 3 and a half
billion times.  By comparison I think your "8 or 9" can be done and get some
renderable results.  ha ha

Bob

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Direct sunlight produces illumination level of 8000-9000 foot candelas
and clear sky obscured 50% by a tall building produces 500 foot candelas.



______________________________________________________________________
Kari Kivisalo                                  http://www.kivisalo.net

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