สมการดอกกุหลาบ

Rose[x_,theta_]:=Module[{phi=(Pi/2)Exp[-theta/(8 Pi)],X=1-(1/2)((5/4)(1-Mod[3.6 theta,2 Pi]/Pi)^2-1/4)^2},
y=1.95653 x^2 (1.27689 x-1)^2 Sin[phi];
r=X(x Sin[phi]+y Cos[phi]);
{r Sin[theta],r Cos[theta],X(x Cos[phi]-y Sin[phi])}
];

ParametricPlot3D[Rose[x,theta],{x,0,1},{theta,-2 Pi,15 Pi},PlotPoints->{25,576},PlotStyle->Red,Mesh->None,Axes->False,Boxed->False]

rose

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