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This question is related to this answer.

What should I change in the following code to replace black color with blue color in the second image?

PolarPlot[Evaluate @ Table[n + ChebyshevT[n, t/Pi - 1], {n, 0, 40, 2}], {t, 0, 2 Pi}]
Graphics @ FilledCurve @ Cases[%, _Line, -1]

enter image description here

enter image description here

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Put the styling directives before the graphics taken from the plot:

PolarPlot[
 Evaluate@Table[n + ChebyshevT[n, t/Pi - 1], {n, 0, 40, 2}],
 {t, 0, 2 Pi}];
Graphics[{Blue, FilledCurve@Cases[%, _Line, -1]}]

enter image description here

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  • $\begingroup$ Is it possible to change the background color as well? $\endgroup$ – Peđa Terzić Aug 3 at 12:00
  • $\begingroup$ @PeđaTerzić Have you looked up the Background option to Graphics? Just add it. $\endgroup$ – Michael E2 Aug 3 at 13:04
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plrplt = PolarPlot[Evaluate@Table[n + ChebyshevT[n, t/Pi - 1], {n, 0, 40, 2}], {t, 0, 2 π}];

To get different colors between layers you can sort the lines by the area of minimal bounding disk and replace Lines with Polygons:

sorted = SortBy[-Area[BoundingRegion[#, "MinDisk"]] &] @ Cases[plrplt, Line[x_] :> x, All];

colors = { Yellow, Red, White, Blue};

Graphics[{First[colors = RotateLeft@colors], Polygon @ #} & /@ sorted]

enter image description here

An alternative way to get the same picture is to partition sorted lines and use FilledCurve for each pair:

Graphics[{First[colors=RotateLeft@colors], FilledCurve[Line/@#]}&/@ Partition[sorted, 2, 1]]

same picture

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