I want to rotate some figure. For example

fig0 = Plot[0.001 Sin[x], {x, 0, Pi}, Axes -> False]
Show[fig0 /. prim : _Line | _Point | _Polygon :>
   GeometricTransformation[prim, RotationTransform[Pi/3, {0, 0}]], 
 PlotRange -> All]

The sine curve after rotation becomes a straight line. How can I make the figure remain the original shape after rotation?


2 Answers 2


It is because amplitude of your Sin function is very small relative to the length of its arc (you picked 0.001 Sin[x]), so it looks almost flat when you rotate it. You can see it non-flat initially because Mathematica auto-rescales Y-axes to zoom in on your small amplitude. When you rotate by Pi/3 curve that is Pi-long in base, its end lifts by Pi Tan[Pi/3] or N[Pi Sqrt[3]] or ~ 5.4414. Relative to this hight .001 will not bee seen. You can see that there is nothing wrong with rotation if you pick amplitude comparable to the length arc:


enter image description here

  • 1
    $\begingroup$ Thanks for your answer. Is these a way to rotate the figure with original shape even though the y range is relatively small compared with x range, as if the canvas is rotated. $\endgroup$
    – Ice0cean
    Mar 13, 2020 at 5:28
  • $\begingroup$ @Ice0cean I added explanation in reply. $\endgroup$ Mar 13, 2020 at 5:35

You can replace RotationTransform[Pi/3, {0, 0}]] with

ScalingTransform[{1, 10^-3}] @* 
  RotationTransform[Pi/3, {0, 0}] @* ScalingTransform[{1, 10^3}]

to avoid the issue caused by different scales (as explained by Vitaliy).

sRs = ScalingTransform[{1, 1/1000}] @* RotationTransform[#, {0, 0}] @*
  ScalingTransform[{1, 1000}] &;

    MapAt[GeometricTransformation[#, sRs[t]] &, fig0, {1}] /. l_Line -> {Red, l}, 
  PlotRange -> {{-3, 3}, {-.003, .003}}, Axes -> True, Frame -> True, AspectRatio -> 1],
 {{t, Pi/3.}, 0., 2. Pi, Appearance -> {"Labeled", "Open"}}]

enter image description here


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