Hello,
Any idea what I’m doing wrong here? The data nvf seems to match the structure wanted for an explicit list of vector field points, but nothing plots.
f[x_, y_] := Sin[x y]
normF = {D[f[x, y], x], D[f[x, y], y], -1} // Normalize;
nvecs = Table[{{x, y, f[x, y]}, Normalize[normF]}, {x, 0, 3, .5}, {y,
0, 3, .5}];
nvf = Flatten[nvecs, 1]; nvf // Dimensions;
ListVectorPlot3D[nvf]
You miss a dimension, having 3
In[64]:= Dimensions[nvf]
Out[64]= {49, 2, 3}
on the other hand, from the manual
vectors = Table[{x, y, z}, {x, -1, 1, .1}, {y, -1, 1, .1}, {z, -1, 1, .1}]; ListVectorPlot3D[vectors]
works fine and has 4 dimensions
In[65]:= Dimensions[Table[{x, y, z}, {x, -1, 1, .1}, {y, -1, 1, .1}, {z, -1, 1, .1}]]
Out[65]= {21, 21, 21, 3}
Thanks to you both. But I still find this peculiar. I think my nvf has the format specified in the second entry of the docs for ListVectorPlot3D, which is { {{x1,y1,y2},{vx1,vy1,vz1}}, . . . }, and has rank 3. – which is just a list of field points and vector values.
The unflattened nvecs does plot as you show, Marco, but I think all the field points should lie on the surface f[x,y] = Sin[x y]. Here I plot the field points of nvf on the surface:
Show[Plot3D[
f[x, y], {x, 0, 3}, {y, 0, 3}], {PointSize[0.02],
Point[nvf[[All, 1]]]} // Graphics3D]
But the field points for the unflattened nvecs don’t lie on the surface where they were generated, but seem to be on a grid, as though Mathemetica were interpreting it as some array, which is like the first entry in the documentation for ListVectorPlot3D.
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Dear David,
I think I understand better now what you mean. That behaviour is odd and has been discussed before:
http://forums.wolfram.com/mathgroup/archive/2009/Feb/msg00770.html
http://mathematica.stackexchange.com/questions/27921/problem-with-listvectorplot3d
Is it true that you want something like this:
Show[Graphics3D[{Thick, Arrow[{#[[1]], #[[1]] + #[[2]]}]} & /@ nvf],
ListPlot3D[nvf[[All, 1]]]]
I couldn’t make it work with ListVectorPlot3D. This, however, is similar and does work
scalarField = x^2 - y^2 - z;
vectorField = D[scalarField, {{x, y, z}}];
data = Table[vectorField, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}];
v = ListVectorPlot3D[data, DataRange -> {{-2, 2}, {-2, 2}, {-2, 2}},
VectorPoints -> 25, VectorScale -> {0.1, Scaled[0.5]},
RegionFunction -> Function[{x, y, z}, -0.1 <= scalarField <= 0.1]]
c = ContourPlot3D[
scalarField == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, Mesh -> None,
ContourStyle -> Opacity[0.5, Green]]
Show[v, c]
I know that this does not really answer the question of why ListVectorPlot3D does not do it.
Cheers,
Marco
Hi Marco,
That’s quite useful.
I had not noticed that ListVectorPlot and ListVectorPlot3D are trying to re-interpolate the field and generate a plot on a regular grid. When I noticed that, I tried the option VectorPoints → All, which is supposed to cause the vectors to be plotted only at the explicit points. That still did not work. I suppose that on a 2D manifold the interpolation algorithm is breaking down. This is too bad, since for a list plot I would really rather have a vector field plotted from an explicit list. I suppose I will make my own.
Best,
David
Here is some simple impl to draw the normal vector on a 2D function with rectangular support
Remove[bodeN]; (* bode normal *)
bodeN[fn_ (* function of two arguments *),
{v1_, v10_, v11_} (* range of argument 1 *),
{v2_, v20_, v21_} (* range of argument 2 *),
l0_ (* length of normals *),
p0_Integer: 20 (* plot points *)
] :=
Module[{dv1, dv2, o, dfn1, dfn2, x, y, p, pp, pic0, pic1},
(* Check fn *)
If[FreeQ[fn[v1, v2], v1] || FreeQ[fn[v1, v2], v2],
Print["Function ", fn, " unabhaengig von ", v1, " und/oder ",
v2 "! Bye."];
Return[$Failed]
];
dv1 = (v11 - v10)/p0;
dv2 = (v21 - v20)/p0;
dfn1[x_, y_] := Evaluate[D[fn[x, y], x]];
dfn2[x_, y_] := Evaluate[D[fn[x, y], y]];
(* die Fusspunkte des Normalenfeldes - "in" der Flaeche *)
p = Flatten[Table[{x, y, fn[x, y]}, {x, v10, v11, dv1}, {y, v20, v21, dv2}], 1];
(* die Normalenrichtungen *)
pp = (((l0/Sqrt[#.#]) & /@ #)*#) &[Flatten[Table[{dfn1[x, y], dfn2[x, y], -1},
{x, v10, v11, dv1}, {y, v20, v21, dv2}], 1]];
(* das l0 normierte Normalenrichtungsfeld mit Pfeilen *)
pic1 = Graphics3D[{Arrowheads[.01], Arrow /@ Transpose[{p, p + pp}]}, DisplayFunction -> Identity];
(* die Flaeche z = fn[v1, v2] selbst *)
pic0 = Plot3D[fn[v1, v2], {v1, v10, v11}, {v2, v20, v21}, DisplayFunction -> Identity, PlotPoints->p0];
(* Anzeige *)
Show[{pic0, pic1}, DisplayFunction -> $DisplayFunction]
] /; l0 != 0 && v10 < v11 && v20 < v21 && AtomQ[v1] &&
AtomQ[v2] && ! StringMatchQ[ToString[v1], ToString[v2]]
originating back from 2001; at that times there was a function ListPlotVectorField3D[] which has seemingly gone. bodeN gives Output like
Notebook attached.
Does the following work for you?
f1[x_, y_] := Sin[x y]
vecF = Normalize[{D[f1[x, y], x], D[f1[x, y], y], -1}];
scalarR = f1[x, y] - z;
vecData = Table[vecF, {x, 0, 3, 0.5}, {y, 0, 3, 0.5}, {z, -1, 1, 0.5}];
For this, I created a regularly spaced array of vectors in 3D. The iteration on z is for that purpose. The scalar region function, scalarR, is used as the rule for RegionFunction. BTW, it seems like you have a redundant Normalize in your original code, so I left it out.
vG = ListVectorPlot3D[vecData, DataRange -> {{0, 3}, {0, 3}, {-1, 1}},
VectorPoints -> 20,
RegionFunction -> Function[{x, y, z}, -0.2 <= scalarR <= 0.2]]
cG = ContourPlot3D[scalarR == 0, {x, 0, 3}, {y, 0, 3}, {z, -1, 1},
Mesh -> None, ContourStyle -> Opacity[0.5, Green]]
Show[vG, cG]
