• CNRS

    From Francois LE COAT@21:1/5 to All on Sat Oct 19 18:00:02 2019
    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,

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

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  • From =?UTF-8?Q?LE_COAT_Fran=c3=a7ois?=@21:1/5 to Francois LE COAT on Thu Oct 24 16:26:03 2019
    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/

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  • From =?iso-8859-1?Q?Vincent_Bela=EFche?=@21:1/5 to lecoat@atari.org on Thu Nov 14 22:55:53 2019
    LE COAT FranÁois <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)≤+(y+2*b)≤-1)≤-c+2*z≤+d*(y+2*b)≤))*((((x+2*a)≤+(y-2*b)≤-1)≤-c+2*z≤+d*(y-2*b)≤))*((((x-2*a)≤+(y/(2*b+1))≤-1)≤-c+2*z≤+d*(y/(2*b+1))≤))-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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  • From =?iso-8859-1?Q?Vincent_Bela=EFche?=@21:1/5 to lecoat@atari.org on Thu Nov 14 22:53:53 2019
    Copy: vincent.belaiche@gmail.com (=?iso-8859-1?Q?Vincent_Bela=EFche?=)

    LE COAT FranÁois <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)≤+(y+2*b)≤-1)≤-c+2*z≤+d*(y+2*b)≤))*((((x+2*a)≤+(y-2*b)≤-1)≤-c+2*z≤+d*(y-2*b)≤))*((((x-2*a)≤+(y/(2*b+1))≤-1)≤-c+2*z≤+d*(y/(2*b+1))≤))-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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  • From =?UTF-8?Q?LE_COAT_Fran=c3=a7ois?=@21:1/5 to All on Fri Nov 15 14:30:15 2019
    Hi,

    Vincent Bela√Įche 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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    * Origin: fsxNet Usenet Gateway (21:1/5)