Related
I have a function of 2 variables - F(x,n).
I need to plot it as a fuction of x for several n's at the same axes system.
I understand I can use Show function in this way (for F(x,a)=x^n):
Show[Table[Plot[x^n, {x, 0, 100}, PlotStyle -> ColorData[1][n]], {n, 10}],
PlotRange -> {All, 10^14}]
But I couldn't figure out how to add lengeds using the parameter n.
I am trying to avoide writing each legend "by hand"' because I need it for different sets of n, so I want it to be generated semi-automatically: I want to set the beginng of the string while the end of the string is depaned on the n. For the example I gave here it should be "F = x^n"...
I have tried using Table with ToString, but it didn't work (maybe I put it on the wronge places inside the Show function)
Thank you!
Is this what you want?
Show[Table[Plot[x^n,{x,0,100},PlotStyle->ColorData[1][n],
PlotLegends->{"F=x^"<>ToString[n]}],{n,10}],PlotRange->{All,10^14}]
You can apply legends without Show like so.
Using automatic expressions
Plot[Evaluate[Array[{x^#} &, 10]], {x, 0, 100},
PlotStyle -> ColorData[1], PlotRange -> {All, 10^14},
PlotLegends -> "Expressions"]
or with custom labels: "F = x^n"
legend = LineLegend[
Array[ColorData[1], 10],
Array["F = x^" <> ToString[#] &, 10]];
Plot[Evaluate[Array[{x^#} &, 10]], {x, 0, 100},
PlotStyle -> ColorData[1], PlotRange -> {All, 10^14},
PlotLegends -> legend]
I need help. I have many variables, that I use in my Graphics[] command, that are dependent of one variable (H in my example). I want to manipulate my graphic so that by changing value of H graphic changes accordingly. But it is not as easy as I've thought.
If you have any idea on how to acomplish this, I would be grateful.
(*This variables are dependent on H that I want to change in
manipulate*)
R = 10;
\[Alpha] = ArcSin[H/R];
p = H/Tan[\[Alpha]];
n = 1.5;
\[Beta] = ArcSin[n Sin[\[Alpha]]];
\[Theta] = \[Beta] - \[Alpha];
l = H/Tan[\[Theta]];
(*This is the graphic I want to make manipulated*)
Graphics[{(*Incident ray*)Line[{{-2, H}, {p, H}}],(*Prism*)
Circle[{0, 0}, R, {0, Pi/2}],
Line[{{0, 0}, {0, 10}}],(*Refracted ray*)
Line[{{p, H}, {p + l, 0}}],(*Surface*)
Line[{{0, 0}, {p + l + 10, 0}}]}]
Here's one of my solutions but it's really messy. What I did is just manually pluged in those values. Is there any more appropriate way to acomplish this:
R = 10;
n = 1.5;
Manipulate[
Graphics[{(*Incident ray*)
Line[{{-2, H}, {H/Tan[ArcSin[H/10]], H}}],(*Prism*)
Circle[{0, 0}, R, {0, Pi/2}],
Line[{{0, 0}, {0, 10}}],(*Refracted ray*)
Line[{{H/Tan[ArcSin[H/10]],
H}, {H/Tan[ArcSin[H/10]] +
H/Tan[ArcSin[n Sin[ArcSin[H/10]]] - ArcSin[H/10]],
0}}],(*Surface*)
Line[{{0,
0}, {H/Tan[ArcSin[H/10]] +
H/Tan[ArcSin[n Sin[ArcSin[H/10]]] - ArcSin[H/10]] + 10,
0}}]}], {H, 0.0001, 10, Appearance -> "Labeled"}]
And also how to make my graphic not to change it's size constantly. I want prism to have fixed size and incident ray to change its position (as it happens when H gets > 6.66 in my example above / this solution).
The question is maybe confusing, but if you try it in Mathematica, you'll see what I want. Thank you for any suggestions.
I think your solution is not bad in general, Mark already noticed in his reply. I loved simplicity of Mark's solution too. Just for the sake of experiment I share my ideas too.
1) It is always a good thing to localize your variables for a specific Manipulate, so their values do not leak and interfere with other dynamic content. It matters if you have additional computation in your notebook - they may start resetting each other.
2) In this particular case if you try to get read of extra variables plugging expressions one into each other your equations became complicated and it is hard to see why they would fail some times. A bit of algebra with help of functions TrigExpand and FullSimplify may help to clarify that your variable H has limitations depending on refraction index value n (see below).
3) If we are aware of point (2) we can make variable n dynamic too and link the value H to n (resetting upper bound of H) right in the controls definition, so always it should be H<10/n . If[..] is also necessary so the controls will not “pink”.
4) If your formulas would depend on R we could also make R dynamic. But I do not have this information, so I localized R via concept of a “dummy“ control (ControlType -> None) – which is quite useful concept for Manipulate.
5) Use PlotRange and ImageSize to stop jiggling of graphics
6) Make it beautiful ;-)
These points would be important if you’d like, for example, to submit a Demonstration to the Wolfram Demonstration Project. If you are just playing around – I think yours and Mark’s solutions are very good.
Thanks,
Vitaliy
Manipulate[If[H >= 10/n, H = 10/n - .0001]; Graphics[{
{Red, Thick, Line[{{-2, H}, {Sqrt[100 - H^2], H}}]},
{Blue, Opacity[.5], Disk[{0, 0}, R, {0, Pi/2}]},
{Red, Thick, Line[{{Sqrt[100 - H^2], H},
{(100 n)/(Sqrt[100 - H^2] n - Sqrt[100 - H^2 n^2]), 0}}]}},
Axes -> True, PlotRange -> {{0, 30}, {0, 10}},
ImageSize -> {600, 200}], {{R, 10}, ControlType -> None},
{{n, 1.5, "Refraction"}, 1.001, 2, Appearance -> "Labeled"},
{{H, 3, "Length"}, 0.0001, 10/n - .0001, Appearance -> "Labeled"}]
I think your first batch of code looks fine and is easy to place into a Manipulate. I would recommend use of the PlotRange option in Graphics.
R = 10;
n = 1.5;
Manipulate[
\[Alpha] = ArcSin[H/R];
p = H/Tan[\[Alpha]];
\[Beta] = ArcSin[n Sin[\[Alpha]]];
\[Theta] = \[Beta] - \[Alpha];
l = H/Tan[\[Theta]];
Graphics[{
Line[{{-2, H}, {p, H}}],(*Prism*)
Circle[{0, 0}, R, {0, Pi/2}],
Line[{{0, 0}, {0, 10}}],(*Refracted ray*)
Line[{{p, H}, {p + l, 0}}],(*Surface*)
Line[{{0, 0}, {p + l + 10, 0}}]},
PlotRange -> {{-1,33},{-1,11}}],
{H,0.0001,6,Appearance->"Labeled"}]
Here is a set of structured 3D points. Now we can form a BSpline using these points as knots.
dat=Import["3DFoil.mat", "Data"]
fu=BSplineFunction[dat]
Here we can do a ParametricPlot3D with these points.
pic=ParametricPlot3D[fu[u,v],{u, 0, 1}, {v, 0, 1}, Mesh -> All, AspectRatio ->
Automatic,PlotPoints->10,Boxed-> False,Axes-> False]
Question
If we carefully look at the 3D geometry coming out of the spline we can see that it is a hollow structure. This hole appears in both side of the symmetric profile. How can we perfectly (not visually!) fill up this hole and create a unified Graphics3D object where holes in both sides are patched.
What I am able to get so far is the following. Holes are not fully patched.
I am asking too many questions recently and I am sorry for that. But if any of you get interested I hope you will help.
Update
Here is the problem with belisarius method.
It generates triangles with almost negligible areas.
dat = Import[NotebookDirectory[] <> "/3DFoil.mat", "Data"];
(*With your points in "dat"*)
fd = First#Dimensions#dat;
check = ParametricPlot3D[{BSplineFunction[dat][u, v],
BSplineFunction[{dat[[1]], Reverse#dat[[1]]}][u, v],
BSplineFunction[{dat[[fd]], Reverse#dat[[fd]]}][u, v]}, {u, 0,
1}, {v, 0, 1}, Mesh -> All, AspectRatio -> Automatic,
PlotPoints -> 10, Boxed -> False, Axes -> False]
output is here
Export[NotebookDirectory[]<>"myres.obj",check];
cd=Import[NotebookDirectory[]<>"myres.obj"];
middle=
check[[1]][[2]][[1]][[1(* Here are the numbers of different Graphics group*)]][[2,1,1,1]];
sidePatch1=check[[1]][[2]][[1]][[2]][[2,1,1,1]];
sidePatch2=check[[1]][[2]][[1]][[3]][[2,1,1,1]];
There are three Graphics groups rest are empty. Now lets see the area of the triangles in those groups.
polygonArea[pts_List?
(Length[#]==3&)]:=Norm[Cross[pts[[2]]-pts[[1]],pts[[3]]-pts[[1]]]]/2
TriangleMaker[{a_,b_,c_}]:={vertices[[a]],vertices[[b]],vertices[[c]]}
tring=Map[polygonArea[TriangleMaker[#]]&,middle];
tring//Min
For the middle large group output is
0.000228007
This is therefore a permissible triangulation. But for the side patches we get zero areas.
Map[polygonArea[TriangleMaker[#]] &, sidePatch1] // Min
Map[polygonArea[TriangleMaker[#]] &, sidePatch2] // Min
Any way out here belisarius ?
My partial solution
First download the package for simplifying complex polygon from Wolfram archive.
fu = BSplineFunction[dat];
pic =(*ParametricPlot3D[fu[u,v],{u,0,1},{v,0,1},Mesh->None,
AspectRatio->Automatic,PlotPoints->25,Boxed->False,Axes->False,
BoundaryStyle->Red]*)
ParametricPlot3D[fu[u, v], {u, 0, 1}, {v, 0, 1}, Mesh -> None,
AspectRatio -> Automatic, PlotPoints -> 10, Boxed -> False,
Axes -> False, BoundaryStyle -> Black];
bound = First#Cases[Normal[pic], Line[pts_] :> pts, Infinity];
corners = Flatten[Table[fu[u, v], {u, 0, 1}, {v, 0, 1}], 1];
nf = Nearest[bound -> Automatic]; {a1, a2} =
Union#Flatten#(nf /# corners);
sets = {bound[[2 ;; a1]], bound[[a1 ;; a2]],bound[[a2 ;; a2 + a1]]};
CorrectOneNodeNumber = Polygon[sets[[{1, 3}]]][[1]][[1]] // Length;
CorrectOneNodes1 =
Polygon[sets[[{1, 3}]]][[1]][[1]]; CorrectOneNodes2 =
Take[Polygon[sets[[{1, 3}]]][[1]][[2]], CorrectOneNodeNumber];
<< PolygonTriangulation`SimplePolygonTriangulation`
ver1 = CorrectOneNodes1;
ver2 = CorrectOneNodes2;
triang1 = SimplePolygonTriangulation3D[ver1];
triang2 = SimplePolygonTriangulation3D[ver2];
Show[Graphics3D[{PointSize[Large], Point[CorrectOneNodes1]},Boxed -> False,
BoxRatios -> 1], Graphics3D[{PointSize[Large], Point[CorrectOneNodes2]},
Boxed -> False, BoxRatios -> 1],
Graphics3D[GraphicsComplex[ver1, Polygon[triang1]], Boxed -> False,
BoxRatios -> 1],
Graphics3D[GraphicsComplex[ver2, Polygon[triang2]], Boxed -> False,
BoxRatios -> 1]]
We get nice triangles here.
picfin=ParametricPlot3D[fu[u,v],{u,0,1}, {v,0,1},Mesh->All,AspectRatio->Automatic,PlotPoints->10,Boxed->False,Axes->False,BoundaryStyle->None];pic3D=Show[Graphics3D[GraphicsComplex[ver1,Polygon[triang1]]],picfin,Graphics3D[GraphicsComplex[ver2,Polygon[triang2]]],Boxed->False,Axes->False]
Now this has just one problem. Here irrespective of the PlotPoints there are four triangles always appearing that just shares only one edge with any other neighboring triangle. But we expect all of the triangles to share at least two edges with other trangles. That happens if we use belisarius method. But it creates too small triangles that my panel solver rejects as tingles with zero area.
One can check here the problem of my method. Here we will use the method from the solution by Sjoerd.
Export[NotebookDirectory[]<>"myres.obj",pic3D];
cd=Import[NotebookDirectory[]<>"myres.obj"];
polygons=(cd[[1]][[2]]/.GraphicsComplex-> List)[[2]][[1]][[1,1]];
pt=(cd[[1]][[2]]/.GraphicsComplex-> List)[[1]];
vertices=pt;
(*Split every triangle in 3 edges,with nodes in each edge sorted*)
triangleEdges=(Sort/#Subsets[#,{2}])&/#polygons;
(*Generate a list of edges*)
singleEdges=Union[Flatten[triangleEdges,1]];
(*Define a function which,given an edge (node number list),returns the bordering*)
(*triangle numbers.It's done by working through each of the triangles' edges*)
ClearAll[edgesNeighbors]
edgesNeighbors[_]={};
MapIndexed[(edgesNeighbors[#1[[1]]]=Flatten[{edgesNeighbors[#1[[1]]],#2[[1]]}];
edgesNeighbors[#1[[2]]]=Flatten[{edgesNeighbors[#1[[2]]],#2[[1]]}];
edgesNeighbors[#1[[3]]]=Flatten[{edgesNeighbors[#1[[3]]],#2[[1]]}];)&,triangleEdges];
(*Build a triangle relation table.Each'1' indicates a triangle relation*)
relations=ConstantArray[0,{triangleEdges//Length,triangleEdges//Length}];
Scan[(n=edgesNeighbors[##];
If[Length[n]==2,{n1,n2}=n;
relations[[n1,n2]]=1;relations[[n2,n1]]=1];)&,singleEdges]
(*Build a neighborhood list*)
triangleNeigbours=Table[Flatten[Position[relations[[i]],1]],{i,triangleEdges//Length}];
trires=Table[Flatten[{polygons[[i]],triangleNeigbours[[i]]}],{i,1,Length#polygons}];
Cases[Cases[trires,x_:>Length[x]],4]
Output shows always there are four triangles that shares only one edges with others.
{4,4,4,4}
In case of belisarius method we don't see this happening but there we get triangles with numerically zero areas.
BR
Import the data and construct the BSpline function as before:
dat = Import["Downloads/3DFoil.mat", "Data"];
fu = BSplineFunction[dat]
Generate the surface, making sure to include (only) the boundary line, which will follow the edge of the surface. Make sure to set Mesh to either All or None.
pic = ParametricPlot3D[fu[u, v], {u, 0, 1}, {v, 0, 1}, Mesh -> None,
AspectRatio -> Automatic, PlotPoints -> 10, Boxed -> False,
Axes -> False, BoundaryStyle -> Red]
Extract the points from the boundary line:
bound = First#Cases[Normal[pic], Line[pts_] :> pts, Infinity]
Find the "corners", based on your parameter space:
corners = Flatten[Table[fu[u, v], {u, 0, 1}, {v, 0, 1}], 1]
Find the edge points best corresponding to the corners, keeping in mind that ParametricPlot3D doesn't use the limits exactly, so we can't just use Position:
nf = Nearest[bound -> Automatic];
nf /# corners
Figure our which range of points on the boundary correspond to the areas you need to fill up. This step involved some manual inspection.
sets = {bound[[2 ;; 22]], bound[[22 ;; 52]], bound[[52 ;; 72]],
bound[[72 ;;]]}
Construct new polygons corresponding to the holes:
Graphics3D[Polygon[sets[[{1, 3}]]], Boxed -> False, BoxRatios -> 1]
Show[pic, Graphics3D[Polygon[sets[[{1, 3}]]]]]
Note that there is probably still a hole that can't be seen where the edge runs between the holes you mentioned, and I haven't tried to fill it in, but you should have enough information to do that if needed.
Your data set looks like this:
Graphics3D[Point#Flatten[dat, 1]]
It consists of 22 sections of 50 points.
Adding a mid-line in each end section (which is actually the end section flattened):
dat2 = Append[Prepend[dat,
Table[(dat[[1, i]] + dat[[1, -i]])/2, {i, Length[dat[[1]]]}]
],
Table[(dat[[-1, i]] + dat[[-1, -i]])/2, {i, Length[dat[[-1]]]}]
];
Graphics3D[{Point#Flatten[dat, 1], Red, Point#dat2[[1]], Green, Point#dat2[[-1]]}]
Now add some weights to the wingtip rim:
sw = Table[1, {24}, {50}];
sw[[2]] = 1000 sw[[1]];
sw[[-2]] = 1000 sw[[1]];
fu = BSplineFunction[dat2, SplineWeights -> sw];
Show[
ParametricPlot3D[fu[u, v], {u, 0, 1}, {v, 0, 1}, Mesh -> All,
AspectRatio -> Automatic, PlotPoints -> 20, Boxed -> False,
Axes -> False, Lighting -> "Neutral"
],
Graphics3D[{PointSize -> 0.025, Green, Point#dat2[[-1]], Red,Point#dat2[[-2]]}]
]
Note that I increased the PlotPoints value to 20.
(*With your points in "dat"*)
fu = BSplineFunction[dat[[1 ;; 2]]];
Show[{ParametricPlot3D[fu[u, v], {u, 0, 1}, {v, 0, 1},
Mesh -> All, AspectRatio -> Automatic, PlotPoints -> 30],
ListPlot3D[dat[[1]]]}]
And with
InputForm[%]
you get the "unified" graphics object.
Edit
Another way, probably better:
(*With your points in "dat"*)
fu = BSplineFunction[dat];
Show[
{ ParametricPlot3D[fu[u, v], {u, 0, 1}, {v, 0, 1},
Mesh -> All, AspectRatio -> Automatic,
PlotPoints -> 10, Boxed -> False, Axes -> False],
ParametricPlot3D[
BSplineFunction[{First#dat, Reverse#First#dat}][u, v], {u, 0, 1}, {v, 0, 1},
Mesh -> None, PlotStyle -> Yellow],
ParametricPlot3D[
BSplineFunction[{dat[[First#Dimensions#dat]],
Reverse#dat[[First#Dimensions#dat]]}]
[u, v], {u, 0, 1}, {v, 0, 1}]}]
In just one structure:
(*With your points in "dat"*)
fd = First#Dimensions#dat;
ParametricPlot3D[
{BSplineFunction[dat][u, v],
BSplineFunction[{dat[[1]], Reverse#dat[[1]]}] [u, v],
BSplineFunction[{dat[[fd]], Reverse#dat[[fd]]}][u, v]},
{u, 0, 1}, {v, 0, 1},
Mesh -> All, AspectRatio -> Automatic,
PlotPoints -> 10, Boxed -> False, Axes -> False]
Edit
You can check that there are small triangles, but they are triangles indeed and not zero area polygons:
fu = BSplineFunction[dat];
check = ParametricPlot3D[{BSplineFunction[{First#dat, Reverse#dat[[1]]}][u, v]},
{u, 0, 1}, {v, 0, 1}, Mesh -> All,
PlotStyle -> Yellow, Mesh -> All, AspectRatio -> Automatic,
PlotPoints -> 10, Boxed -> False, Axes -> False];
pts = check /. Graphics3D[GraphicsComplex[a_, b__], ___] -> a;
m = check[[1]][[2]][[1]][[1]] /. {___, GraphicsGroup[{Polygon[a_]}]} -> a;
t = Replace[m, {a_, b_, c_} -> {pts[[a]], pts[[b]], pts[[c]]}, {1}];
polygonArea[pts_List?(Length[#] == 3 &)] :=
Norm[Cross[pts[[2]] - pts[[1]], pts[[3]] - pts[[1]]]]/2;
t[[Position[Ordering[polygonArea /# t], 1][[1]]]]
(*
->{{{-4.93236, 0.0989696, -2.91748},
{-4.92674, 0.0990546, -2.91748},
{-4.93456, 0.100181, -2.91748}}}
*)
I have already checked all the examples and settings in the Mathematica documentation center, but couldn't find any example on how to choose the numbers that will be shown on the axes.
How do I change plot axis numbering like 2,4,6,.. to PI,2PI,3PI,...?
Howard has already given the correct answer in the case where you want the labels Pi, 2 Pi etc to be at the values Pi, 2 Pi etc.
Sometimes you might want to use substitute tick labels at particular values, without rescaling data.
One of the other examples in the documentation shows how:
Plot[Sin[x], {x, 0, 10},
Ticks -> {{{Pi, 180 \[Degree]}, {2 Pi, 360 \[Degree]}, {3 Pi,
540 \[Degree]}}, {-1, 1}}]
I have a suite of small custom functions for formatting Ticks the way I want them. This is probably too much information if you are just starting out, but it is worth knowing that you can use any number format and substitute anything into your ticks if desired.
myTickGrid[min_, max_, seg_, units_String, len_?NumericQ,
opts : OptionsPattern[]] :=
With[{adj = OptionValue[UnitLabelShift], bls = OptionValue[BottomLabelShift]},
Table[{i,
If[i == max,
DisplayForm[AdjustmentBox[Style[units, LineSpacing -> {0, 12}],
BoxBaselineShift -> If[StringCount[units, "\n"] > 0, adj + 2, adj]]],
If[i == min,
DisplayForm#AdjustmentBox[Switch[i, _Integer,
NumberForm[i, DigitBlock -> 3,
NumberSeparator -> "\[ThinSpace]"], _, N[i]],
BoxBaselineShift -> bls],
Switch[i, _Integer, NumberForm[i, DigitBlock -> 3,
NumberSeparator -> "\[ThinSpace]"], _, N[i]]]], {len, 0}}, {i,
If[Head[seg] === List, Union[{min, max}, seg], Range[min, max, seg]]}]]
And setting:
Options[myTickGrid] = {UnitLabelShift -> 1.3, BottomLabelShift -> 0}
SetOptions[myTickGrid, UnitLabelShift -> 1.3, BottomLabelShift -> 0]
Example:
Plot[Erfc[x], {x, -2, 2}, Frame -> True,
FrameTicks -> {myTickGrid[-2, 2, 1, "x", 0.02, UnitLabelShift -> 0],
myTickGrid[0, 2, {0.25, .5, 1, 1.8}, "Erfc(x)", 0.02]}]
You can find an example here:
Ticks -> {{Pi, 2 Pi, 3 Pi}, {-1, 0, 1}}
Ticks also accepts a function, which will save you the trouble of listing the points manually or having to change the max value each time. Here's an example:
xTickFunc[min_, max_] :=
Table[{i, i, 0.02}, {i, Ceiling[min/Pi] Pi, Floor[max/Pi] Pi, Pi}]
Plot[Sinc[x], {x, -5 Pi, 5 Pi}, Ticks -> {xTickFunc, Automatic},
PlotRange -> All]
If you want more flexibility in customizing your ticks, you might want to look into LevelScheme.
I am looking to plot something like the whispering gallery modes -- a 2D cylindrically symmetric plot in polar coordinates. Something like this:
I found the following code snippet in Trott's symbolics guidebook. Tried running it on a very small data set; it ate 4 GB of memory and hosed my kernel:
(* add points to get smooth curves *)
addPoints[lp_][points_, \[Delta]\[CurlyEpsilon]_] :=
Module[{n, l}, Join ## (Function[pair,
If[(* additional points needed? *)
(l = Sqrt[#. #]&[Subtract ## pair]) < \[Delta]\[CurlyEpsilon], pair,
n = Floor[l/\[Delta]\[CurlyEpsilon]] + 1;
Table[# + i/n (#2 - #1), {i, 0, n - 1}]& ## pair]] /#
Partition[If[lp === Polygon,
Append[#, First[#]], #]&[points], 2, 1])]
(* Make the plot circular *)
With[{\[Delta]\[CurlyEpsilon] = 0.1, R = 10},
Show[{gr /. (lp : (Polygon | Line))[l_] :>
lp[{#2 Cos[#1], #2 Sin[#1]} & ###(* add points *)
addPoints[lp][l, \[Delta]\[CurlyEpsilon]]],
Graphics[{Thickness[0.01], GrayLevel[0], Circle[{0, 0}, R]}]},
DisplayFunction -> $DisplayFunction, Frame -> False]]
Here, gr is a rectangular 2D ListContourPlot, generated using something like this (for example):
data = With[{eth = 2, er = 2, wc = 1, m = 4},
Table[Re[
BesselJ[(Sqrt[eth] m)/Sqrt[er], Sqrt[eth] r wc] Exp[
I m phi]], {r, 0, 10, .2}, {phi, 0, 2 Pi, 0.1}]];
gr = ListContourPlot[data, Contours -> 50, ContourLines -> False,
DataRange -> {{0, 2 Pi}, {0, 10}}, DisplayFunction -> Identity,
ContourStyle -> {Thickness[0.002]}, PlotRange -> All,
ColorFunctionScaling -> False]
Is there a straightforward way to do cylindrical plots like this?.. I find it hard to believe that I would have to turn to Matlab for my curvilinear coordinate needs :)
Previous snippets deleted, since this is clearly the best answer I came up with:
With[{eth = 2, er = 2, wc = 1, m = 4},
ContourPlot[
Re[BesselJ[(Sqrt[eth] m)/Sqrt[er], Sqrt[eth] r wc] Exp[I phi m]]/.
{r ->Norm[{x, y}], phi ->ArcTan[x, y]},
{x, -10, 10}, {y, -10, 10},
Contours -> 50, ContourLines -> False,
RegionFunction -> (#1^2 + #2^2 < 100 &),
ColorFunction -> "SunsetColors"
]
]
Edit
Replacing ContourPlot by Plot3D and removing the unsupported options you get:
This is a relatively straightforward problem. The key is that if you can parametrize it, you can plot it. According to the documentation both ListContourPlot and ListDensityPlot accept data in two forms: an array of height values or a list of coordinates plus function value ({{x, y, f} ..}). The second form is easier to deal with, such that even if your data is in the first form, we'll transform it into the second form.
Simply, to transform data of the form {{r, t, f} ..} into data of the form {{x, y, f} ..} you doN[{#[[1]] Cos[ #[[2]] ], #[[1]] Sin[ #[[2]] ], #[[3]]}]& /# data, when applied to data taken from BesselJ[1, r/2] Cos[3 t] you get
What about when you just have an array of data, like this guy? In that case, you have a 2D array where each point in the array has known location, and in order to plot it, you have to turn it into the second form. I'm partial to MapIndexed, but there are other ways of doing it. Let's say your data is stored in an array where the rows correspond to the radial coordinate and the columns are the angular coordinate. Then to transform it, I'd use
R = 0.01; (*radial increment*)
T = 0.05 Pi; (*angular increment*)
xformed = MapIndexed[
With[{r = #2[[1]]*R, t = #2[[1]]*t, f = #1},
{r Cos[t], r Sin[t], f}]&, data, {2}]//Flatten[#,1]&
which gives the same result.
If you have an analytic solution, then you need to transform it to Cartesian coordinates, like above, but you use replacement rules, instead. For instance,
ContourPlot[ Evaluate[
BesselJ[1, r/2]*Cos[3 t ] /. {r -> Sqrt[x^2 + y^2], t -> ArcTan[x, y]}],
{x, -5, 5}, {y, -5, 5}, PlotPoints -> 50,
ColorFunction -> ColorData["DarkRainbow"], Contours -> 25]
gives
Two things to note: 1) Evaluate is needed to ensure that the replacement is performed correctly, and 2) ArcTan[x, y] takes into account the quadrant that the point {x,y} is found in.