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Re: ListPolarPlot questions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg124071] Re: ListPolarPlot questions
  • From: Ashwini <aksingh21 at gmail.com>
  • Date: Sun, 8 Jan 2012 04:21:23 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <je96ru$j8k$1@smc.vnet.net>

On Jan 7, 10:30 am, "Eric Michielssen" <emich... at eecs.umich.edu>
wrote:
> Mathgroup:
>
> I have two questions regarding the use of ListPolarPlot.
>
> 1. The function I'd like to plot is periodic.  Suppose I have samples of it
> at equispaced angles, the first one for angle zero, and the last one for 2
> Pi; the data for 0 and 2 Pi is identical.  I want to plot the function for
> all angles, from 0 to 2 Pi, connected by a smooth line.  Choosing options
> joined -> True and  InterpolationOrder -> 2 almost does the trick, except
> that there is a kink in the graph at phi = 0.  As the interpolator does not
> know the function is periodic.  Is there a way around that?
>
> Example: g1 =  ListPolarPlot[Table[phi = (i - 1) 2 Pi/10; Cos[phi]^2, {i, 1,
> 11}],
>  DataRange -> {0., 2 Pi}, Axes -> None, Joined -> True,
>  InterpolationOrder -> 2]
>
> (In practice my table is numerically derived, not that of a Cos^2 function)
>
> 2.  Suppose I want to plot a circle centered away from the origin, and then
> the above g1 centered about the same point.  How do I do that?
>
> Example:
>
> g1 = Graphics[
>    ListPolarPlot[Table[phi = (i - 1) 2 Pi/10; Cos[phi]^2, {i, 1, 11}],
>      DataRange -> {0., 2 Pi}, Axes -> None, Joined -> True,
>     InterpolationOrder -> 2]];
> g2 = Graphics[Circle[{1., 1.}, 1]];
> Show[g1, g2]
>
> But now with the center of g1 shifted to {1,1} (and, if possible, smooth
> about phi = 0...).  I presume this can be done using Offset but I cannot get
> that to work.
>
> Thanks,
> Eric Michielssen

1. Increasing  the InterpolationOrder to 3 connects the center.
2. the center of the circle is at {1,1} shouldn't it be {0,0}

Regards,
aks



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