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SphericalPlot3D Manipulate

sean roubion

Posted 10 years ago

a = ((3.14.005) (1.887 - 1.792))/((.000522) Cos[-.7454Sin[\[Theta]]Cos[\[Phi]] + .6626Cos[\[Theta]]]); b = (1 - Cos[.7454Sin[\[Theta]]Cos[\[Phi]] + .6626 Cos[\[Theta]]]^2)(1 - Cos[-.7454Sin[\[Theta]]Cos[\[Phi]] + .6626Cos[\[Theta]]])^2; c = ab; d = (Cos[c])^2; SphericalPlot3D[d, {\[Theta], 0, \[Pi]}, { \[Phi], 0, 2\[Pi]}] I have spherical 3D plot of some functions. Is there a way that I can manipulate this plot so that I vary theta from 0, pi/4 while I hold phi constant in the plot? Thank you in advance.

a = ((3.14*.005) (1.887 - 1.792))/((.000522)*
     Cos[-.7454*Sin[\[Theta]]*Cos[\[Phi]] + .6626*Cos[\[Theta]]]);
b = (1 - Cos[.7454*Sin[\[Theta]]*Cos[\[Phi]] + .6626*
         Cos[\[Theta]]]^2)*(1 - 
      Cos[-.7454*Sin[\[Theta]]*Cos[\[Phi]] + .6626*Cos[\[Theta]]])^2;
c = a*b;
d = (Cos[c])^2;
SphericalPlot3D[d, {\[Theta], 0, \[Pi]}, { \[Phi], 0, 2*\[Pi]}]

I have spherical 3D plot of some functions. Is there a way that I can manipulate this plot so that I vary theta from 0, pi/4 while I hold phi constant in the plot? Thank you in advance.

POSTED BY: sean roubion

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Hans Milton

Posted 10 years ago

SphericalPlot3D Manipulate

POSTED BY: Hans Milton

Hans Milton

Posted 10 years ago

Just to be clear, since there are more than one convention: In SphericalPlot3D the first angle argument is the "latitudinal" angle. And the second angle argument is the "longitudinal" angle. In the SphericalPlot3D case, the Mathematica documentation does not follow the most common uses of the greek letters theta and phi. f[\[Theta]_, \[Phi]_] := Module[ {a, b, c, d}, a = ((3.14.005) (1.887 - 1.792))/ ((.000522)Cos[-.7454Sin[\[Theta]]Cos[\[Phi]] + .6626Cos[\[Theta]]]); b = (1 - Cos[.7454Sin[\[Theta]]Cos[\[Phi]] + .6626Cos[\[Theta]]]^2)* (1 -Cos[-.7454Sin[\[Theta]]Cos[\[Phi]] + .6626Cos[\[Theta]]])^2; c = ab; d = (Cos[c])^2 ] SphericalPlot3D[f[polar, azimuthal], {polar, 0, \[Pi]}, {azimuthal, 0, 5}]

Just to be clear, since there are more than one convention: In SphericalPlot3D the first angle argument is the "latitudinal" angle. And the second angle argument is the "longitudinal" angle. In the SphericalPlot3D case, the Mathematica documentation does not follow the most common uses of the greek letters theta and phi.

f[\[Theta]_, \[Phi]_] := Module[
    {a, b, c, d},
    a = ((3.14*.005) (1.887 - 1.792))/
        ((.000522)*Cos[-.7454*Sin[\[Theta]]*Cos[\[Phi]] + .6626*Cos[\[Theta]]]);
    b = (1 - Cos[.7454*Sin[\[Theta]]*Cos[\[Phi]] + .6626*Cos[\[Theta]]]^2)* 
        (1 -Cos[-.7454*Sin[\[Theta]]*Cos[\[Phi]] + .6626*Cos[\[Theta]]])^2;
    c = a*b;
    d = (Cos[c])^2
  ]

  SphericalPlot3D[f[polar, azimuthal], {polar, 0, \[Pi]}, {azimuthal, 0, 5}]

enter image description here

POSTED BY: Hans Milton

suvadip mandal

suvadip mandal, Indian Institute of Science Education and Research, Kolkata

Posted 10 years ago

Perhaps the attached file will help. Attachments:

POSTED BY: suvadip mandal

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