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Referring to this page,

Plotting implicitly-defined space curves

How can I project the curves of intersection onto the x-y plane?

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  • 2
    $\begingroup$ This kind of questions usually receives much more attention if you show that you've tried something before asking. Post some code ... $\endgroup$ – Dr. belisarius Dec 29 '13 at 5:52
  • $\begingroup$ Add ViewPoint -> {0, 0, \[Infinity]}, ViewVertical -> {0, 1, 0}, Ticks -> {True, True, None}, I suppose it is a duplicate but I can't find a good link :). Also, maybe it is not, you want 3D or 2D plot as an output? $\endgroup$ – Kuba Dec 29 '13 at 15:31
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The same as David Park's answer but with 2D output

cp3d = ContourPlot3D[(x^2 + y^2 + z^2 + 8)^2 == 
    36 (x^2 + y^2), {x, -4, 4}, {y, -4, 4}, {z, -2, 2}, 
   MeshFunctions -> {Function[{x, y, z}, y^2 + (z - 2)^2 - 4]}, 
   Mesh -> {{0}}, ContourStyle -> None, PlotPoints -> 30, 
   BoxRatios -> Automatic];

Graphics[Cases[cp3d, GraphicsComplex[p___, OptionsPattern[]] :> 
   (GraphicsComplex[p] /. {x_Real, y_Real, z_Real} :> {x, y})], Axes -> True]

enter image description here

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Using the example from Plotting implicitly-defined space curves

ContourPlot3D[(x^2 + y^2 + z^2 + 8)^2 == 36 (x^2 + y^2), {x, -4, 
   4}, {y, -4, 4}, {z, -2, 2}, 
  MeshFunctions -> {Function[{x, y, z}, y^2 + (z - 2)^2 - 4]}, 
  Mesh -> {{0}}, ContourStyle -> None, PlotPoints -> 30, 
  BoxRatios -> Automatic] /. {x_Real, y_Real, z_Real} -> {x, y, 0}

enter image description here

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