3
$\begingroup$

I am trying to redraw this illustration for the Poincaré section using Mathematica:

poincare illustration

For which I used this code as a start:

fullData = 
  Table[With[{\[Theta] = 2 \[Pi] u, \[CurlyPhi] = 
       2 \[Pi] v}, {(2 + Cos[\[CurlyPhi]]) Cos[\[Theta]], 
      Sin[\[CurlyPhi]], (2 + Cos[\[CurlyPhi]]) Sin[\[Theta]]}], {u, 0,
      1, 0.001}, {v, 0, 1, 0.003}] // Flatten[#, 1] &;
tol = 0.02;
secData = Select[fullData, Abs[#[[2]]] < tol &];
fullPlot = Graphics3D[{Blue, PointSize[Small], Point[fullData]}];

sectionPlot = Graphics3D[{Green, PointSize[Small], Point[secData]}];

planePlot = 
  Graphics3D[{Yellow, Opacity[0.4], 
    Polygon[{{-3, 0, -3}, {3, 0, -3}, {3, 0, 3}, {-3, 0, 3}}]}];

Show[fullPlot, sectionPlot, planePlot, Axes -> True, 
 AxesLabel -> {Style["x(\[Tau])", 14], Style["x'(\[Tau])", 14], 
   Style["x(\[Tau]-\[Tau]₀)", 14]}, BoxRatios -> {1, 1, 1}, 
 Lighting -> "Neutral", ImageSize -> Large]

Is there a better way by which I can draw a similar image, or any improvement I can make on this starting code to get a close image like the above figure?

Improve this question
$\endgroup$
1
  • $\begingroup$ If you can formulate your process as a differential equation, then you can use NDSolve and its WhenEvent features to detect crossings of the section plane. Simple example: NDSolve[{y[0] == 0, y'[0] == 1, y''[t] == -y[t], WhenEvent[y[t] == 0, Sow[t]]}, y[t], {t, 0, 10}] // Reap $\endgroup$ Commented Jun 23 at 19:44

1 Answer 1

Reset to default
2
$\begingroup$

You may want to use ParametricPlot3D to highlight the envelope of the cross section (dropping the axes labels here for easier code):

expr = {a  (Cos[t] + b) * Cos[phi], a  (Cos[t] + b) * Sin[phi], 
   0.3 a  Sin[t] + (a  (Cos[t] + b) * Sin[phi])*0.3};
params = {a -> .1, b -> 3, c -> 8 Pi, z -> .2};

fullData = 
  Table[Evaluate[expr /. phi -> c  t /. params], {t, 0, 40 Pi, 
    2 Pi/2000}];

Show[
 ParametricPlot3D[
  Evaluate[expr /. params], {t, 0, 2 \[Pi]}, {phi, 0, 2 \[Pi]}, 
  RegionFunction -> 
   Function[{x, y, z, u, v}, 0 < -y < 0.02 && x > 0], 
  PlotStyle -> Green, Mesh -> None, PlotTheme -> "Detailed", 
  PlotLegends -> None,
  PlotRange -> 
   Evaluate[{{-2  a  b, 2  a  b}, {-2  a  b, 2  a  b}, {-z, z}} /. 
     params]],
 Graphics3D[{Blue, PointSize[.001], Point@fullData, Yellow, 
   Opacity[.95], 
   Polygon[{{0, 0, -z}, {b*a*2, 0, -z}, {b*a*2, 0, z}, {0, 0, z}} /. 
     params]}]
 ]

Poincare Crosssection

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.