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

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kurtz le pirate <kur### [at] freefr> 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

- BE

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On 01/05/2025 14:39, Bald Eagle wrote:
> kurtz le pirate <kur### [at] freefr> 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

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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] freefr> 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

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On 01/05/2025 17:37, Bald Eagle wrote:
> kurtz le pirate <kur### [at] freefr> 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

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kurtz le pirate <kur### [at] freefr> 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

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kurtz le pirate <kur### [at] freefr> 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

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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

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