plotting - Modular surface of tri-focal Cassini curve ContourPlot3D missing feet

I am wondering why the following fails to cover the surface at points near $k = 0$.

c[z_] := (z + 1) (z - 1) (z + 1 + I);


ContourPlot3D[Abs[c[x + I y]] == k^3, {x, -2.5, 2}, {y, -2, 2}, 
{k, 0, 1.75}, Background -> White, AxesLabel -> {"x", "y", "k"}]

Mathematica graphics

Furthermore, when the left hand side is complex expanded and the equation is rearranged with $2xy+k^3$ on the right, the curve looks significantly different, with one large and one small foot, both touching at zero. Why is that?

Answer

ContourPlot3D is not very good at resolving thin features, because it only knows that the feature exists when one of the sampling points happens to land inside it. In general, one thing you can do is to increase PlotPoints, which improves the plot but takes a very long time.

ContourPlot3D[
 Abs[c[x + I y]] == k^3, {x, -2.5, 2}, {y, -2, 2}, {k, 0, 1.75}, 
 Background -> White, AxesLabel -> {"x", "y", "k"}, PlotPoints -> 20]

enter image description here

In this particular case, though, your plot is equivalent to $k = |c(x+iy)|^{1/3}$, so you could just use Plot3D instead. This is much faster because it only has to sample the two-dimensional $xy$ plane rather than the three-dimensional $xyk$ space. Then you can afford to make MaxRecursion quite large and it's still really quick to plot.

Plot3D[Abs[c[x + I y]]^(1/3), {x, -2.5, 2}, {y, -2, 2}, PlotRange -> {0, 1.75},
 Background -> White, AxesLabel -> {"x", "y", "k"}, ClippingStyle -> None,
 BoxRatios -> 1, MeshFunctions -> {#1 &, #2 &, #3 &}, MaxRecursion -> 5]

enter image description here

(I've added a few options to make it look like your original plot.)

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

Blog

Search This Blog