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I want to add some number ticks to a unit circle in PolarPlot like the one attached, but how can I do it?

PolarPlot[Circle[], {t, 0, 1}, PolarAxes -> Automatic, 
     PolarTicks -> {Drop[Table[i, {i, 2*Pi, 0, -Pi/5}], 1], Automatic}]

The effect I want

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closed as off-topic by Bob Hanlon, MarcoB, LLlAMnYP, ilian, Öskå Aug 24 '15 at 16:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, MarcoB, LLlAMnYP, ilian, Öskå
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ So, you have a working code that uses custom ticks, but you are not willing to experiment with it and see if you can solve the problem on your own? I am sorry, but you have to show some effort. $\endgroup$ – Sektor Aug 24 '15 at 12:01
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    $\begingroup$ If all else fails, read the documentation for PolarTicks. $\endgroup$ – Bob Hanlon Aug 24 '15 at 12:31
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PolarPlot[1, {t, 0, 2 Pi}, PolarAxesOrigin -> {0, 1}, 
 PolarAxes -> Automatic, 
 PolarTicks -> 
  Evaluate[{#, If[# == 0, "0/1", ToString[Round[#/(2 Pi), 0.1]]]} & /@
     Range[0, 9/5 Pi, 2 Pi/10]]]
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  • $\begingroup$ Thank you very much. However, the unit circle is within the boundary of the plot. Do you know how to make the boundary of the plot to be the boundary of the unit circle? $\endgroup$ – Math Guy Aug 24 '15 at 12:12
  • $\begingroup$ I updated the code. Added the PolarAxesOrigin option that takes care of your concern $\endgroup$ – Zviovich Aug 24 '15 at 12:56

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