What could be the equation of this curve?












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Does anyone have an idea what (parametric) equation might give a plot like this?



enter image description here



This is in the XY plane. I want to render it nicely on Mathematica, hence why I need the equation.



So far I came up with ${t*sin[t/2], cos[t/2]}$ which seems to be on the right track but doesn't quite do what I want...










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    0












    $begingroup$


    Does anyone have an idea what (parametric) equation might give a plot like this?



    enter image description here



    This is in the XY plane. I want to render it nicely on Mathematica, hence why I need the equation.



    So far I came up with ${t*sin[t/2], cos[t/2]}$ which seems to be on the right track but doesn't quite do what I want...










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Does anyone have an idea what (parametric) equation might give a plot like this?



      enter image description here



      This is in the XY plane. I want to render it nicely on Mathematica, hence why I need the equation.



      So far I came up with ${t*sin[t/2], cos[t/2]}$ which seems to be on the right track but doesn't quite do what I want...










      share|cite|improve this question









      $endgroup$




      Does anyone have an idea what (parametric) equation might give a plot like this?



      enter image description here



      This is in the XY plane. I want to render it nicely on Mathematica, hence why I need the equation.



      So far I came up with ${t*sin[t/2], cos[t/2]}$ which seems to be on the right track but doesn't quite do what I want...







      functions graphing-functions






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      asked Dec 18 '18 at 16:51









      SuperCiociaSuperCiocia

      285213




      285213






















          2 Answers
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          $begingroup$

          ParametricPlot[(1 + .1 Sin[θ/2 - π/4]) {Cos[θ], Sin[θ]}, {θ, 0, 4 π}]


          enter image description here






          share|cite|improve this answer









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            2












            $begingroup$

            If you don't need perfect circles, maybe:
            Figure2Figure2






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              2 Answers
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              2 Answers
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              active

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              $begingroup$

              ParametricPlot[(1 + .1 Sin[θ/2 - π/4]) {Cos[θ], Sin[θ]}, {θ, 0, 4 π}]


              enter image description here






              share|cite|improve this answer









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                3












                $begingroup$

                ParametricPlot[(1 + .1 Sin[θ/2 - π/4]) {Cos[θ], Sin[θ]}, {θ, 0, 4 π}]


                enter image description here






                share|cite|improve this answer









                $endgroup$
















                  3












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                  3





                  $begingroup$

                  ParametricPlot[(1 + .1 Sin[θ/2 - π/4]) {Cos[θ], Sin[θ]}, {θ, 0, 4 π}]


                  enter image description here






                  share|cite|improve this answer









                  $endgroup$



                  ParametricPlot[(1 + .1 Sin[θ/2 - π/4]) {Cos[θ], Sin[θ]}, {θ, 0, 4 π}]


                  enter image description here







                  share|cite|improve this answer












                  share|cite|improve this answer



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                  answered Dec 18 '18 at 17:04









                  David G. StorkDavid G. Stork

                  11.5k41533




                  11.5k41533























                      2












                      $begingroup$

                      If you don't need perfect circles, maybe:
                      Figure2Figure2






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        If you don't need perfect circles, maybe:
                        Figure2Figure2






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          If you don't need perfect circles, maybe:
                          Figure2Figure2






                          share|cite|improve this answer









                          $endgroup$



                          If you don't need perfect circles, maybe:
                          Figure2Figure2







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 18 '18 at 17:23









                          Shubham JohriShubham Johri

                          5,462818




                          5,462818






























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