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"And" <49341109@ntnu.edu.tw> wrote:
>
> A more subtle details is that the light_color usually not the same.
> the color of POV-Ray light_source describes the color of the illuminated object.
> the color of looks_like object describes the color of (sun) itself. The sun
> color often much brighter than the object it illuminates.
I show you the calculation:
The angular diameter of the Sun is about 1/2 degree, so the solid angle of it is
about 0.00006 steradian viewed from Earth. If the luminance of the Sun is
sun_color, than the illuminace of the sun light is sun_color*0.00006, then the
luminance of the reflect surface will be sun_color*0.00006*surface_albedo/pi, so
you should set the color of the POVRay's light_source sun_color*0.00006/pi.
On the other hand, if you set the light_source color A, than the looks_like
color will be A*pi/0.00006 ~= A*52333.
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On 9/12/2016 3:57 AM, And wrote:
> On the other hand, if you set the light_source color A, than the looks_like
> color will be A*pi/0.00006 ~= A*52333.
>
Is this even possible? Isn't there a limit on how bright you can make
something?
Mike
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Am 12.09.2016 um 14:44 schrieb Mike Horvath:
> On 9/12/2016 3:57 AM, And wrote:
>> On the other hand, if you set the light_source color A, than the
>> looks_like
>> color will be A*pi/0.00006 ~= A*52333.
>>
>
> Is this even possible? Isn't there a limit on how bright you can make
> something?
Yes, absolutely: Brightness values cannot exceed 3.4e+38 ;)
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On 9/12/2016 3:57 AM, And wrote:
> On the other hand, if you set the light_source color A, than the looks_like
> color will be A*pi/0.00006 ~= A*52333.
>
Do I apply this to the pigment or the finish emission?
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Am 13.09.2016 um 23:15 schrieb Mike Horvath:
> On 9/12/2016 3:57 AM, And wrote:
>> On the other hand, if you set the light_source color A, than the
>> looks_like
>> color will be A*pi/0.00006 ~= A*52333.
>>
>
> Do I apply this to the pigment or the finish emission?
That's up to you, but personally I'd recommend the finish emission. The
pigment is also used in the other finish features, which are related to
reflection of light from somewhere else, where values above 1 make no
physical sense. But when it comes to emitting light on its own, there's
no limit for an object's material.
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On 9/13/2016 6:11 PM, clipka wrote:
> Am 13.09.2016 um 23:15 schrieb Mike Horvath:
>> On 9/12/2016 3:57 AM, And wrote:
>>> On the other hand, if you set the light_source color A, than the
>>> looks_like
>>> color will be A*pi/0.00006 ~= A*52333.
>>>
>>
>> Do I apply this to the pigment or the finish emission?
>
> That's up to you, but personally I'd recommend the finish emission. The
> pigment is also used in the other finish features, which are related to
> reflection of light from somewhere else, where values above 1 make no
> physical sense. But when it comes to emitting light on its own, there's
> no limit for an object's material.
>
So, like this?
texture
{
pigment {color srgb light_color}
finish
{
emission pi/0.00006
ambient 0
diffuse 0
}
}
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Mike Horvath <mik### [at] gmail com> wrote:
> On 9/13/2016 6:11 PM, clipka wrote:
> > Am 13.09.2016 um 23:15 schrieb Mike Horvath:
> >> On 9/12/2016 3:57 AM, And wrote:
> >>> On the other hand, if you set the light_source color A, than the
> >>> looks_like
> >>> color will be A*pi/0.00006 ~= A*52333.
> >>>
> >>
> >> Do I apply this to the pigment or the finish emission?
> >
> > That's up to you, but personally I'd recommend the finish emission. The
> > pigment is also used in the other finish features, which are related to
> > reflection of light from somewhere else, where values above 1 make no
> > physical sense. But when it comes to emitting light on its own, there's
> > no limit for an object's material.
> >
>
> So, like this?
>
>
> texture
> {
> pigment {color srgb light_color}
> finish
> {
> emission pi/0.00006
> ambient 0
> diffuse 0
> }
> }
That should work.
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clipka <ano### [at] anonymousorg> wrote:
The
> pigment is also used in the other finish features, which are related to
> reflection of light from somewhere else, where values above 1 make no
> physical sense.
Does that include fluorescence, such as the widespread use of optical
brighteners in laundry detergents and printer paper?
(I'm not trying to claim that the actual reflected light intensity would exceed
the incident light in practice, in those two examples, but in theory I suppose
it might be possible?
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Le 14/09/2016 à 16:09, Bald Eagle a écrit :
> clipka <ano### [at] anonymousorg> wrote:
> The
>> pigment is also used in the other finish features, which are related to
>> reflection of light from somewhere else, where values above 1 make no
>> physical sense.
>
> Does that include fluorescence, such as the widespread use of optical
> brighteners in laundry detergents and printer paper?
Azurant... convert UV-radiation (invisible) into visible spectrum
(usually by doubling the wavelength). Black light is using the same system.
IMHO, you would need a non-diagonal matrix as pigment instead of a
vector to model such things (one column/row per wavelength) and a wider
vector for the light itself... and at the end, the camera collapse the
wider vector into the RGB value for the picture format.
Our traditional pigment would just be a pure diagonal matrix (everything
out of the main diagonal is 0).
>
> (I'm not trying to claim that the actual reflected light intensity would exceed
> the incident light in practice, in those two examples, but in theory I suppose
> it might be possible?
>
>
>
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Am 14.09.2016 um 16:09 schrieb Bald Eagle:
> clipka <ano### [at] anonymousorg> wrote:
> The
>> pigment is also used in the other finish features, which are related to
>> reflection of light from somewhere else, where values above 1 make no
>> physical sense.
>
> Does that include fluorescence, such as the widespread use of optical
> brighteners in laundry detergents and printer paper?
>
> (I'm not trying to claim that the actual reflected light intensity would exceed
> the incident light in practice, in those two examples, but in theory I suppose
> it might be possible?
Even with fluorescence, technically values above 1 make no sense. But to
model this, colours would have to be matrices rather than vectors, and
one of the channels will be UV light, which doesn't show up in the image.
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