Hi,

Is it possible to translate the following equation with SciLab ?

<https://imaginary.github.io/80-cnrs/>

I don't know how to begin ... It's an interesting problem :-)

Best regards,

--
François LE COAT
Author of Eureka 2.12 (2D Graph Describer, 3D Modeller) http://eureka.atari.org/

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* Origin: fsxNet Usenet Gateway (21:1/5)

Hi,

Francois LE COAT writes:

Is it possible to translate the following equation with SciLab ?

    <https://imaginary.github.io/80-cnrs/>

I don't know how to begin ... It's an interesting problem :-)

This "80" 3D shape was design for October 19th 2019, for the eighty
anniversary of the CNRS (centre national de la recherche scientifique)
the French institute for scientific research.

I also tested modelling the "80" shape from the CNRS anniversary, with
MathMod, with the help from its author Taha Abderrahman and it gives :

<http://eureka.atari.org/mathmod.png>

The eighty anniversary of the institute was celebrated worldwide :-)
Is it possible to model the "80" shape with SciLab, I'm wondering ?...

Best regards,

--
François LE COAT
Author of Eureka 2.12 (2D Graph Describer, 3D Modeller) http://eureka.atari.org/

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

LE COAT François <lecoat@atari.org> writes:

Hi,

Francois LE COAT writes:

Is it possible to translate the following equation with SciLab ?

    <https://imaginary.github.io/80-cnrs/>

I don't know how to begin ... It's an interesting problem :-)

This "80" 3D shape was design for October 19th 2019, for the eighty anniversary of the CNRS (centre national de la recherche scientifique)
the French institute for scientific research.

I also tested modelling the "80" shape from the CNRS anniversary, with MathMod, with the help from its author Taha Abderrahman and it gives :

<http://eureka.atari.org/mathmod.png>

The eighty anniversary of the institute was celebrated worldwide :-)
Is it possible to model the "80" shape with SciLab, I'm wondering ?...

Best regards,

Just for my plain understanding, please confirm the following you have
some x y z polynomial

((((x+2*a)²+(y+2*b)²-1)²-c+2*z²+d*(y+2*b)²))*((((x+2*a)²+(y-2*b)²-1)²-c+2*z²+d*(y-2*b)²))*((((x-2*a)²+(y/(2*b+1))²-1)²-c+2*z²+d*(y/(2*b+1))²))-10*e

ie x, y and z are variables, and all the other letters (a, b, c, d, and e) are constants.

If, so, in parctical terms, what are the values of a, b, c, d, e to get
the 80 shape ?

And I understand that the shape is the locus where this polynomial is
equal to zero, is that correct ?

So on the picture, the intensitiy of the color tells the value of z.

I would say that here we have two questions, first question is how to
make a 3D plot of a surface in Scilab, and second question is how to
determine a surface from a polynomial equation in Scilab.

You have polt3D to make 3D plots, and you can use genfac3d to get some
tiling of the surface from some equation z = f(x,y). You can also make
the f function from the polynomial using some solver (x and y are fixed,
and you ask the solver to find the set of z zeroing the polynomial, then
you take the max of this set, if not empty, or %nan otherwise).

For the solver, the optim scilab function can be used, just the cost
function is the square of the polynomial, and one can use also Scilab polynomial manipulation function to derive formally its gradient.

V.

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Copy: vincent.belaiche@gmail.com (=?iso-8859-1?Q?Vincent_Bela=EFche?=)

LE COAT François <lecoat@atari.org> writes:

Hi,

Francois LE COAT writes:

Is it possible to translate the following equation with SciLab ?

    <https://imaginary.github.io/80-cnrs/>

I don't know how to begin ... It's an interesting problem :-)

This "80" 3D shape was design for October 19th 2019, for the eighty anniversary of the CNRS (centre national de la recherche scientifique)
the French institute for scientific research.

I also tested modelling the "80" shape from the CNRS anniversary, with MathMod, with the help from its author Taha Abderrahman and it gives :

<http://eureka.atari.org/mathmod.png>

The eighty anniversary of the institute was celebrated worldwide :-)
Is it possible to model the "80" shape with SciLab, I'm wondering ?...

Best regards,

Just for my plain understanding, please confirm the following you have
some x y z polynomial

((((x+2*a)²+(y+2*b)²-1)²-c+2*z²+d*(y+2*b)²))*((((x+2*a)²+(y-2*b)²-1)²-c+2*z²+d*(y-2*b)²))*((((x-2*a)²+(y/(2*b+1))²-1)²-c+2*z²+d*(y/(2*b+1))²))-10*e

ie x, y and z are variables, and all the other letters (a, b, c, d, and e) are constants.

If, so, in parctical terms, what are the values of a, b, c, d, e to get
the 80 shape ?

And I understand that the shape is the locus where this polynomial is
equal to zero, is that correct ?

So on the picture, the intensitiy of the color tells the value of z.

I would say that here we have two questions, first question is how to
make a 3D plot of a surface in Scilab, and second question is how to
determine a surface from a polynomial equation in Scilab.

You have polt3D to make 3D plots, and you can use genfac3d to get some
tiling of the surface from some equation z = f(x,y). You can also make
the f function from the polynomial using some solver (x and y are fixed,
and you ask the solver to find the set of z zeroing the polynomial, then
you take the max of this set, if not empty, or %nan otherwise).

For the solver, the optim scilab function can be used, just the cost
function is the square of the polynomial, and one can use also Scilab polynomial manipulation function to derive formally its gradient.

V.

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* Origin: fsxNet Usenet Gateway (21:1/5)

Hi,

Vincent Belaïche writes:

LE COAT François writes:

Is it possible to translate the following equation with SciLab ?

    <https://imaginary.github.io/80-cnrs/>

I don't know how to begin ... It's an interesting problem :-)

This "80" 3D shape was design for October 19th 2019, for the eighty
anniversary of the CNRS (centre national de la recherche scientifique)
the French institute for scientific research.

I also tested modelling the "80" shape from the CNRS anniversary, with
MathMod, with the help from its author Taha Abderrahman and it gives :

<http://eureka.atari.org/mathmod.png>

The eighty anniversary of the institute was celebrated worldwide :-)
Is it possible to model the "80" shape with SciLab, I'm wondering ?...

Best regards,

Just for my plain understanding, please confirm the following you have
some x y z polynomial

((((x+2*a)²+(y+2*b)²-1)²-c+2*z²+d*(y+2*b)²))*((((x+2*a)²+(y-2*b)²-1)²-c+2*z²+d*(y-2*b)²))*((((x-2*a)²+(y/(2*b+1))²-1)²-c+2*z²+d*(y/(2*b+1))²))-10*e

ie x, y and z are variables, and all the other letters (a, b, c, d, and e) are constants.

If, so, in parctical terms, what are the values of a, b, c, d, e to get
the 80 shape ?

And I understand that the shape is the locus where this polynomial is
equal to zero, is that correct ?

So on the picture, the intensitiy of the color tells the value of z.

I would say that here we have two questions, first question is how to
make a 3D plot of a surface in Scilab, and second question is how to determine a surface from a polynomial equation in Scilab.

You have polt3D to make 3D plots, and you can use genfac3d to get some
tiling of the surface from some equation z = f(x,y). You can also make
the f function from the polynomial using some solver (x and y are fixed,
and you ask the solver to find the set of z zeroing the polynomial, then
you take the max of this set, if not empty, or %nan otherwise).

For the solver, the optim scilab function can be used, just the cost
function is the square of the polynomial, and one can use also Scilab polynomial manipulation function to derive formally its gradient.

Thinking about the formulae, that is the one of an isosurface, it may
not be representable with SciLab. The representations of this software
in 3D, are only parametric surfaces, so far as I know ...

Thanks for your answer,

Regards,

--
François LE COAT
Author of Eureka 2.12 (2D Graph Decriber, 3D Modeller)
http://eureka.atari.org/

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