Re: ListPolarPlot questions

*To*: mathgroup at smc.vnet.net*Subject*: [mg124086] Re: ListPolarPlot questions*From*: "David Park" <djmpark at comcast.net>*Date*: Sun, 8 Jan 2012 04:26:36 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <3193533.81836.1325933394495.JavaMail.root@m06>

Eric, Make a radius interpolation function for your curve using the PeriodicInterpolation option. And use a higher InterpolationOrder. datapoints = Table[phi = (i - 1) 2 Pi/10; Cos[phi]^2, {i, 1, 11}]; radiusFunc = ListInterpolation[datapoints, {{0, 2 \[Pi]}}, InterpolationOrder -> 4, PeriodicInterpolation -> True] Then use PolarPlot with the radiusFunc. To translate to the circle use Translate or GeometricTransformation with a Show statement but it is confusing because of the necessary graphics level jumping. With the Presentations Application from my web site it is easy. We just draw the circle and translate the polar curve to the center of it. << Presentations` Draw2D[ {Circle[{1, 1}, 1], PolarDraw[radiusFunc[\[Theta]], {\[Theta], 0, 2 \[Pi]}] // TranslateOp[{1, 1}]}] David Park djmpark at comcast.net http://home.comcast.net/~djmpark/index.html From: Eric Michielssen [mailto:emichiel at eecs.umich.edu] 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