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

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more