[Date Index]
[Thread Index]
[Author Index]
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
Prev by Date:
**Re: SortBy won't order irrationals**
Next by Date:
**Re: Default sort of vector is by "complexity" of expression!!?????**
Previous by thread:
**Column widths in Grid**
Next by thread:
**Re: ListPolarPlot questions**
| |