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