3
$\begingroup$

when I set plot range on 300, polar coordinates as is shown in figure 1 is completely covered but figure range is so far away from outer degree circle and When I set plot range on 200, polar coordinate is not shown in plot clearly,I am thankful if you guide me to fix it

Q = PolarPlot[
      {(100*Cos[t]*Cos[t])/(1.5*(1-0.49)^(1/2)), 
       (100*Sin[t]*Sin[t])/(1.5*(2-0.49)*(1 - 0.49)^(3/2))}, 
      {t, 0, 2*Pi},
      PolarAxes -> True,
      PlotRange -> 300,
      PolarGridLines -> {Automatic, None},
      PolarTicks -> {"Degrees", Automatic}
    ]

plot with PlotRange set to 300

plot with PlotRange set to 200

$\endgroup$
4
  • 4
    $\begingroup$ Along with PlotRange -> 200, use the option PlotRangePadding -> 25 $\endgroup$
    – Bob Hanlon
    May 2, 2020 at 15:49
  • $\begingroup$ More hacky solution, draw an invisible circle with PolarPlot[170, {\[Theta], 0, 2 Pi}, PlotStyle -> {Directive[Opacity[0]]}] and then combine that with Show with your polarplot. $\endgroup$
    – Max1
    May 2, 2020 at 15:54
  • $\begingroup$ @Max1, thank you bro. $\endgroup$
    – Arian
    May 3, 2020 at 7:48
  • $\begingroup$ @Bob Hanlon, thanks a lot $\endgroup$
    – Arian
    May 3, 2020 at 7:49

1 Answer 1

3
$\begingroup$

In addition to setting a PlotRangePadding as Bob Hanlon suggested in comments, you may also want to play with PolarAxesOrigin to fine-tune the appearance of the plot:

PolarPlot[
  {(100*Cos[t]*Cos[t])/(1.5*(1 - 0.49)^(1/2)), 
   (100*Sin[t]*Sin[t])/(1.5*(2 - 0.49)*(1 - 0.49)^(3/2))},
  {t, 0, 2*Pi},
  PolarAxes -> True,

  (*----------*)
  PolarAxesOrigin -> {Pi/4, 150},
  PlotRange -> 130, PlotRangePadding -> Scaled[0.15],
  (*----------*)

  PolarGridLines -> {Automatic, None},
  PolarTicks -> {"Degrees", Automatic}
]

polar plot with fine tuning of position of axes and range padding

PolarAxesOrigin allows you to control the angular position of the polar axis as well as the distance from the origin at which the radial axis crosses the angular axis.

$\endgroup$
1
  • $\begingroup$ thank you so much, it was very helpful. $\endgroup$
    – Arian
    May 3, 2020 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.