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MathGroup Archive 2012

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ListPolarPlot questions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg124066] ListPolarPlot questions
  • From: "Eric Michielssen" <emichiel at eecs.umich.edu>
  • Date: Sat, 7 Jan 2012 05:25:21 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com

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






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