I was trying to figure out how big an area light should be to simulate the  
sun when the light is a lot closer than the sun actually is. Looking at a  
wall about 3m high I noticed the shadow on the ground blurred about 3cm. I  
came to the conclusion that the diameter of my light should be about 1% of  
the distance from the object. To test my theory I looked up the diameter  
of the sun and the distance from the sun to the earth. Close enough! The  
diameter of the sun is 1 391 000 km and the distance from the earth to the  
sun is 149,669,180 km. That's 0.929% :)


-- 
-Nekar Xenos-

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> I was trying to figure out how big an area light should be to simulate
> the sun when the light is a lot closer than the sun actually is. Looking
> at a wall about 3m high I noticed the shadow on the ground blurred about
> 3cm. I came to the conclusion that the diameter of my light should be
> about 1% of the distance from the object. To test my theory I looked up
> the diameter of the sun and the distance from the sun to the earth.
> Close enough! The diameter of the sun is 1 391 000 km and the distance
> from the earth to the sun is 149,669,180 km. That's 0.929% :)
>
>

It's interesting to note that the Moon have the same apparant size as 
the sun.
So, you can use the same settings for both the Sun and the Moon, with 
the only difference been the actual brightness.

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On 23-8-2012 21:27, Nekar Xenos wrote:
> I was trying to figure out how big an area light should be to simulate
> the sun when the light is a lot closer than the sun actually is. Looking
> at a wall about 3m high I noticed the shadow on the ground blurred about
> 3cm. I came to the conclusion that the diameter of my light should be
> about 1% of the distance from the object. To test my theory I looked up
> the diameter of the sun and the distance from the sun to the earth.
> Close enough! The diameter of the sun is 1 391 000 km and the distance
> from the earth to the sun is 149,669,180 km. That's 0.929% :)
>
>

I use the following, originally proposed by Cousin Ricky iirc, which 
seems fairly correct:

#declare SunDis = vlength(SunPosition)*2/215;

   light_source{
     SunPosition
     color SunColor
     area_light
     SunDis*x, SunDis*z,
     5, 5
     adaptive 2
     jitter
     circular
     orient
     parallel
     point_at <0, 0, 0>
   }

Thomas

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Alain <kua### [at] videotronca> wrote:
> It's interesting to note that the Moon have the same apparant size as 
> the sun.
> So, you can use the same settings for both the Sun and the Moon, with 
> the only difference been the actual brightness.

Only during full moon.

-- 
                                                          - Warp

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On Fri, 24 Aug 2012 09:22:17 +0200, Thomas de Groot <tho### [at] degrootorg>  
wrote:

> On 23-8-2012 21:27, Nekar Xenos wrote:
>> I was trying to figure out how big an area light should be to simulate
>> the sun when the light is a lot closer than the sun actually is. Looking
>> at a wall about 3m high I noticed the shadow on the ground blurred about
>> 3cm. I came to the conclusion that the diameter of my light should be
>> about 1% of the distance from the object. To test my theory I looked up
>> the diameter of the sun and the distance from the sun to the earth.
>> Close enough! The diameter of the sun is 1 391 000 km and the distance
>> from the earth to the sun is 149,669,180 km. That's 0.929% :)
>>
>>
>
> I use the following, originally proposed by Cousin Ricky iirc, which  
> seems fairly correct:
>
> #declare SunDis = vlength(SunPosition)*2/215;
>
>    light_source{
>      SunPosition
>      color SunColor
>      area_light
>      SunDis*x, SunDis*z,
>      5, 5
>      adaptive 2
>      jitter
>      circular
>      orient
>      parallel
>      point_at <0, 0, 0>
>    }
>
> Thomas
>
>

Yes, that works out to 0.93% closer than my 1% guesstimate :)

-- 
-Nekar Xenos-

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Alain <kua### [at] videotronca> wrote:
> So, you can use the same settings for both the Sun and the Moon, with
> the only difference been the actual brightness.

Which is factor of approximately half a million when the Moon is full.

Gotta love that non-linear perception.

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On 24-8-2012 22:53, Nekar Xenos wrote:

> Yes, that works out to 0.93% closer than my 1% guesstimate :)
>
Perihelion / Aphelion difference ;-)

Thomas

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