Skip to main content

plotting - Placing a ContourPlot under a Plot3D


I would like to combine a 3-dimensional graph of a function with its 2-dimensional contour-plot underneath it in a professional way. But I have no idea how to start.


I have a three of these I would like to make, so I don't need a fully automated function that does this. A giant block of code would be just fine.


The two plots I would like to have combined are:


potential1 = 

  Plot3D[-3600. h^2 + 0.02974 h^4 - 5391.90 s^2 + 0.275 h^2 s^2 + 0.125 s^4, 
    {h, -400, 400}, {s, -300, 300}, PlotRange -> {-1.4*10^8, 2*10^7}, 
    ClippingStyle -> None, MeshFunctions -> {#3 &}, Mesh -> 10,
    MeshStyle -> {AbsoluteThickness[1], Blue}, Lighting -> "Neutral",
    MeshShading -> {{Opacity[.4], Blue}, {Opacity[.2], Blue}}, Boxed -> False,
    Axes -> False]

and


contourPotentialPlot1 = 
  ContourPlot[-3600. h^2 + 0.02974 h^4 - 5391.90 s^2 + 0.275 h^2 s^2 + 0.125 s^4, 

   {h, -400, 400}, {s, -300, 300}, PlotRange -> {-1.4*10^8, 2*10^7},
   Contours -> 10, ContourStyle -> {{AbsoluteThickness[1], Blue}}, Axes -> False, 
   PlotPoints -> 30]

These two plots look like:


The 3-D plot The contour plot


I would also love it if I could get 'grids' on the sides of the box like in http://en.wikipedia.org/wiki/File:GammaAbsSmallPlot.png



Update The new plotting routine SliceContourPlot3D was introduced in version 10.2. If this function can be used to achieve the task above, how can it be done?



Answer




Strategy is simple texture map 2D plot on a rectangle under your 3D surface. I took a liberty with some styling that I like - you can always come back to yours.


contourPotentialPlot1 = ContourPlot[-3600. h^2 + 0.02974 h^4 - 5391.90 s^2 + 
   0.275 h^2 s^2 + 0.125 s^4, {h, -400, 400}, {s, -300, 300}, 
 PlotRange -> {-1.4*10^8, 2*10^7}, Contours -> 15, Axes -> False, 
 PlotPoints -> 30, PlotRangePadding -> 0, Frame -> False, ColorFunction -> "DarkRainbow"];

potential1 = Plot3D[-3600. h^2 + 0.02974 h^4 - 5391.90 s^2 + 0.275 h^2 s^2 + 
    0.125 s^4, {h, -400, 400}, {s, -300, 300}, 
   PlotRange -> {-1.4*10^8, 2*10^7}, ClippingStyle -> None, 
   MeshFunctions -> {#3 &}, Mesh -> 15, MeshStyle -> Opacity[.5], 

   MeshShading -> {{Opacity[.3], Blue}, {Opacity[.8], Orange}}, Lighting -> "Neutral"];

level = -1.2 10^8; gr = Graphics3D[{Texture[contourPotentialPlot1], EdgeForm[], 
Polygon[{{-400, -300, level}, {400, -300, level}, {400, 300, level}, {-400, 300, level}}, 
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}, {0, 1}}]}, Lighting -> "Neutral"];

Show[potential1, gr, PlotRange -> All, BoxRatios -> {1, 1, .6}, FaceGrids -> {Back, Left}]

enter image description here


You can see I used PlotRangePadding -> 0 option in ContourPlot. It is to remove white space around the graphics to make texture mapping more precise. If you need utmost precision you can take another path. Extract graphics primitives from ContourPlot and make them 3D graphics primitives. If you need to color the bare contours - you could replace Line by Polygon and do some tricks with FaceForm based on a contour location.



level = -1.2 10^8;
pts = Append[#, level] & /@ contourPotentialPlot1[[1, 1]];
cts = Cases[contourPotentialPlot1, Line[l_], Infinity];
cts3D = Graphics3D[GraphicsComplex[pts, {Opacity[.5], cts}]];

Show[potential1, cts3D, PlotRange -> All, BoxRatios -> {1, 1, .6}, 
 FaceGrids -> {Bottom, Back, Left}]

enter image description here


Comments

Post a Comment

Popular posts from this blog

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}]
Post a Comment
Read more

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}]
Post a Comment
Read more

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.
Post a Comment
Read more