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| From: kurtz le pirate Subject: Re: profile following a parametric curve
 Date:  1 May 2025 09:53:55
 Message: <68137cf3@news.povray.org>
 
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|  |  | On 01/05/2025 14:39, Bald Eagle wrote:
> kurtz le pirate <kur### [at] free fr> wrote:
> 
>> Curves are defined with Cartesian parameters: x=f(t), z=g(t).
>> The profile is a simple polygon.
> 
> Nice job with the profiles.
> 
> What's the arc-length of that lemniscate?   :D
Simple. Here, half-width a = 24 = c√2.
                   2*π
Then, length = --------- * c
                M(1,1/√2)
With M(a,b) the arithmetic–geometric mean function.
;)
-- 
kurtz le pirate
compagnie de la banquise Post a reply to this message
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|  |  | On 5/1/25 06:22, kurtz le pirate wrote:
> My first attempts.
> 
> Curves are defined with Cartesian parameters: x=f(t), z=g(t).
> The profile is a simple polygon.
Cool!
Bill P.
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|  |  | kurtz le pirate <kur### [at] free fr> wrote:
> Simple. Here, half-width a = 24 = c√2.
>
>                    2*π
> Then, length = --------- * c
>                 M(1,1/√2)
>
>
Maybe not so simple?
https://en.wikipedia.org/wiki/Elliptic_function#History
(I only know that because I was trying to hunt down a good overview of
calculating the arc-length of a catenary, and move along the curve at regular
intervals.)
- BW Post a reply to this message
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|  |  | On 01/05/2025 17:37, Bald Eagle wrote:
> kurtz le pirate <kur### [at] free fr> wrote:
> 
>> Simple. Here, half-width a = 24 = c√2.
>>
>>                     2*π
>> Then, length = --------- * c
>>                  M(1,1/√2)
>>
>>
>> With M(a,b) the arithmetic–geometric mean function.
> 
> Maybe not so simple?
> https://en.wikipedia.org/wiki/Elliptic_function#History
That's right.
But my answer is more of a joke and a simple copy/paste from Wiki
<https://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli#Arc_length_and_elliptic_functions>
-- 
kurtz le pirate
compagnie de la banquise Post a reply to this message
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|  |  | kurtz le pirate <kur### [at] free fr> wrote:
> That's right.
>
> But my answer is more of a joke and a simple copy/paste from Wiki
>
<https://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli#Arc_length_and_elliptic_functions>
Nice.  (I suspected)
"It therefore obviously follows that . . ."    ;)
- BW Post a reply to this message
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|  |  | kurtz le pirate <kur### [at] free fr> wrote:
> hello,
>
>
> My first attempts.
>
> Curves are defined with Cartesian parameters: x=f(t), z=g(t).
> The profile is a simple polygon.
>
> Basic texture, lighting with area_light.
> Low resolution (800x600).
>
>
>
>
>
>
>
>
> --
> kurtz le pirate
> compagnie de la banquise
Do you adjust the size of your profile object as you follow the curve, or does
it just have a very tiny thickness at each t step?  When I use this technique, I
struggle with getting the resulting object smooth if the curves are wide.
-- Chris R Post a reply to this message
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|  |  | On 02/05/2025 16:17, Chris R wrote:
> 
> Do you adjust the size of your profile object as you follow the curve, or does
> it just have a very tiny thickness at each t step?  When I use this technique, I
> struggle with getting the resulting object smooth if the curves are wide.
> 
> -- Chris R
> 
Hello Chris.
In fact, my profile has no thickness !
The profile is only defined by points in an a array and doesn't really 
exist.
For example, the profile for a triangle is defined like this:
#declare i = 0;
#while ( i < NumberOfVertices)
   #declare ShapePoints[i] = RotateX(<0, ShapeRadius, 0>, Angle*i);
   #declare i = i +1;
#end
The shape is in the y-z plane.
I position the first profile for t=0;
Then I calculate the position for the second (t=t+epsilon) and build 
triangles* that connect the two profiles.
* : using the true POV object : triangle { p0, p1, p2 }
-- 
kurtz le pirate
compagnie de la banquise
Post a reply to this message
 Attachments:
 Download 'parametrique_limacon de pascal _connect_prof0.png' (124 KB)
 
 
 Preview of image 'parametrique_limacon de pascal _connect_prof0.png'
  
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| From: kurtz le pirate Subject: Re: profile following a parametric curve
 Date:  3 May 2025 11:18:56
 Message: <681633e0@news.povray.org>
 
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|  |  | A little more artistic, but this time using a curve defined by a polar 
equation.
Profile : hexagon.
Low resolution (800x600).
-- 
kurtz le pirate
compagnie de la banquise
 Post a reply to this message
 Attachments:
 Download 'polaire_trefle_connect_0.png' (264 KB)
 
 
 Preview of image 'polaire_trefle_connect_0.png'
  
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|  |  | kurtz le pirate <kur### [at] free fr> wrote:
> On 02/05/2025 16:17, Chris R wrote:
> >
> > Do you adjust the size of your profile object as you follow the curve, or does
> > it just have a very tiny thickness at each t step?  When I use this technique, I
> > struggle with getting the resulting object smooth if the curves are wide.
> >
> > -- Chris R
> >
>
>
> Hello Chris.
>
> In fact, my profile has no thickness !
> The profile is only defined by points in an a array and doesn't really
> exist.
>
>
> For example, the profile for a triangle is defined like this:
> #declare i = 0;
> #while ( i < NumberOfVertices)
>    #declare ShapePoints[i] = RotateX(<0, ShapeRadius, 0>, Angle*i);
>    #declare i = i +1;
> #end
> The shape is in the y-z plane.
>
> I position the first profile for t=0;
>
> Then I calculate the position for the second (t=t+epsilon) and build
> triangles* that connect the two profiles.
>
> * : using the true POV object : triangle { p0, p1, p2 }
>
>
>
>
> --
> kurtz le pirate
> compagnie de la banquise
Thanks for the explanation.  My brain is so focused on isosurfaces and
extrusions that I didn't even think of using triangles or a mesh.
-- Chris R Post a reply to this message
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|  |  | On 05/05/2025 15:32, Chris R wrote:
> 
> Thanks for the explanation.  My brain is so focused on isosurfaces and
> extrusions that I didn't even think of using triangles or a mesh.
more explanations on my website ;)
-- 
kurtz le pirate
compagnie de la banquise
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