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> On 11/21/2016 4:23 PM, clipka wrote:
>>
>> So if I understand you correctly, the shape you want is exactly the
>> volume for which R, G and B are all within the range from 0 to 1.
>>
>
> Yes, exactly!
>
>
>> That sounds like a job for an isosurface to me.
>>
>> Variant A: Create six isosurfaces, each representing one of the
>> boundaries (R<0, G<0, B<0, R>1, G>1, B>1), and intersect them.
>>
>> Variant B: Refactor the functions from variant A, so that they have the
>> same threshold, combine them using max() or min() to achieve the
>> intersection effect, and turn that into a single isosurface.
>>
>
> I don't know how to do either variant. :(
>
> Mike
Forgot to add that I want to use polar coordinates (if that makes a
difference).
Mike
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On 11/21/2016 4:23 PM, clipka wrote:
> Am 21.11.2016 um 21:56 schrieb Mike Horvath:
>
>>>> Actually, I can create a pigment already. But I don't know how to use
>>>> the functions in a isosurface or a parametric to make the 3D object the
>>>> pigment gets applied to, if possible.
>>>>
>>>> Mike
>>>
>>> Oops! You'll need to know these as well.
>>>
>>> #declare correctRGB1 = function(R,G,B) {min(max(R,0),1)}
>>> #declare correctRGB2 = function(R,G,B) {min(max(G,0),1)}
>>> #declare correctRGB3 = function(R,G,B) {min(max(B,0),1)}
>>>
>>>
>>
>> That is wrong. The RGB values should *not* be corrected. In fact, if any
>> of R, G or B are outside the range of 0 to 1, then the coordinate/color
>> should be discarded. Sorry!
>
> So if I understand you correctly, the shape you want is exactly the
> volume for which R, G and B are all within the range from 0 to 1.
>
> That sounds like a job for an isosurface to me.
>
> Variant A: Create six isosurfaces, each representing one of the
> boundaries (R<0, G<0, B<0, R>1, G>1, B>1), and intersect them.
>
> Variant B: Refactor the functions from variant A, so that they have the
> same threshold, combine them using max() or min() to achieve the
> intersection effect, and turn that into a single isosurface.
>
I am trying random things now.
isosurface
{
function {max(convertLCH2RGB_R(x,y,z), 1)}
contained_by
{
box {2,+2}
}
// threshold FLOAT_VALUE
accuracy 0.01
max_gradient 5
// [evaluate P0, P1, P2]
// open
// [max_trace INTEGER]  [all_intersections]
pigment {rgb 1}
}
This takes a long time to render, but there's nothing visible happening.
Mike
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I tried this as well.
parametric
{
function { convertLCH2RGB_R(u,100,v) }
function { convertLCH2RGB_G(u,100,v) }
function { convertLCH2RGB_B(u,100,v) }
<0,0>, <100,360>
contained_by
{
box {2,+2}
}
max_gradient 5
accuracy 0.01
precompute 10 x,y,z
pigment {rgb 1}
}
The render is too slow. The pps is below 100 and still dropping.
Mike
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Am 22.11.2016 um 01:46 schrieb Mike Horvath:
> I am trying random things now.
>
> isosurface
> {
> function {max(convertLCH2RGB_R(x,y,z), 1)}
throw away the max() function, and take care of the polarish coordinate
system by replacing:
convertLCH2RGB_R(x,y,z)
with:
convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x/z)*pi/180)
(presuming you want L up and I got the math right)
> contained_by
> {
> box {2,+2}
> }
> // threshold FLOAT_VALUE
... and use threshold 1. That should give you something visible.
> accuracy 0.01
> max_gradient 5
> // [evaluate P0, P1, P2]
> // open
> // [max_trace INTEGER]  [all_intersections]
> pigment {rgb 1}
> }
>
> This takes a long time to render, but there's nothing visible happening.
>
> Mike
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On 11/22/2016 2:30 AM, clipka wrote:
> Am 22.11.2016 um 01:46 schrieb Mike Horvath:
>
>> I am trying random things now.
>>
>> isosurface
>> {
>> function {max(convertLCH2RGB_R(x,y,z), 1)}
>
> throw away the max() function, and take care of the polarish coordinate
> system by replacing:
>
> convertLCH2RGB_R(x,y,z)
>
> with:
>
> convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x/z)*pi/180)
>
> (presuming you want L up and I got the math right)
>
>> contained_by
>> {
>> box {2,+2}
>> }
>> // threshold FLOAT_VALUE
>
> ... and use threshold 1. That should give you something visible.
>
>> accuracy 0.01
>> max_gradient 5
>> // [evaluate P0, P1, P2]
>> // open
>> // [max_trace INTEGER]  [all_intersections]
>> pigment {rgb 1}
>> }
>>
>> This takes a long time to render, but there's nothing visible happening.
>>
>> Mike
>
THANK YOU!!
However atan2 requires two parameters and your math only has one.
Mike
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Am 22.11.2016 um 08:33 schrieb Mike Horvath:
>> convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x/z)*pi/180)
> THANK YOU!!
>
> However atan2 requires two parameters and your math only has one.
Sorry, try atan2(x,z).
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On 11/22/2016 2:44 AM, clipka wrote:
> Am 22.11.2016 um 08:33 schrieb Mike Horvath:
>
>>> convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x/z)*pi/180)
>
>> THANK YOU!!
>>
>> However atan2 requires two parameters and your math only has one.
>
> Sorry, try atan2(x,z).
>
Here is the whole file again with your changes. The shape I get when I
render is that of a pointy cap. I don't know if that's correct or not.
Mike
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Am 22.11.2016 um 08:45 schrieb Mike Horvath:
> On 11/22/2016 2:44 AM, clipka wrote:
>> Am 22.11.2016 um 08:33 schrieb Mike Horvath:
>>
>>>> convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x/z)*pi/180)
>>
>>> THANK YOU!!
>>>
>>> However atan2 requires two parameters and your math only has one.
>>
>> Sorry, try atan2(x,z).
>>
>
> Here is the whole file again with your changes. The shape I get when I
> render is that of a pointy cap. I don't know if that's correct or not.
How would I know  I'm not familiar with that colour model.
Now you need to intersect that shape with the five others  or use the
combined function:
min(
convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
convertLCH2RGB_G(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
convertLCH2RGB_B(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
1convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
1convertLCH2RGB_G(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
1convertLCH2RGB_B(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180)
)
should do the job, I think.
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On 11/22/2016 3:31 AM, clipka wrote:
> min(
> convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> convertLCH2RGB_G(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> convertLCH2RGB_B(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> 1convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> 1convertLCH2RGB_G(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> 1convertLCH2RGB_B(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180)
> )
I think the last parameter should be atan2(x,z)*180/pi because atan2
returns radians and convertLCH2RGB_R expects degrees.
Also, here's a rough approximation of what the surface should look like.
https://commons.wikimedia.org/wiki/File:Cielch_color_solid_cylinder.png
But with smooth curves instead of blocky sections.
Mike
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On 11/22/2016 3:31 AM, clipka wrote:
> min(
> convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> convertLCH2RGB_G(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> convertLCH2RGB_B(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> 1convertLCH2RGB_R(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> 1convertLCH2RGB_G(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180),
> 1convertLCH2RGB_B(y*200100,sqrt(x*x+z*z)*100,atan2(x,z)*pi/180)
> )
Forgot to say that the above just results in a cube.
Mike
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