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5

This answer replaces an earlier one that deleted only some of the intersecting cylinders. (My thanks to paw for pointing this out.) It also is much faster. The square of the distance between points at p1 and p2 is p1.p1 + p2.p2 - 2 p1.p2 and a cylinder axis can be parameterized by pi + dp t, where pi is one end of the axis, dp is the vector from pi to ...

5

How about adding `ColorFunction' to your Image3D: m = RandomChoice[{1/4, 1/2, 3/4, 1}, {4, 4, 4}]; Image3D[m, ColorFunction -> Hue] Play with the Colorfunction to get different color schemes: m = RandomChoice[{1/4, 1/2, 3/4, 1}, {5, 10, 10}]; Image3D[m, ColorFunction -> "TemperatureMap"]

4

1) Simply amazing: http://intothecontinuum.tumblr.com/tagged/Mathematica 2) Jeff Bryant: http://members.wolfram.com/jeffb/visualization/3d.shtml 3) http://vqm.uni-graz.at/notebooks/index.html 4) http://www.vis.uni-stuttgart.de/~kraus/LiveGraphics3D/tutorial/tutorial.html#Basic%20Graphics 5) http://ieng9.ucsd.edu/~ma155f/Handouts/index.html 6) ...

3

In drawing these Steinmetz solids I tried to use as many of the coding points as I could from Paul Bourke's page where your images come from. He uses PovRay, but the code is human readable even if you can't use that program. Module[{l = 1.75, viewpoint, cylinders1, cylinders2}, viewpoint = 1.2 {-1, -1, 1}; cylinders1 = {Specularity[White, 40], ...

3

The best I can do: Graphics3D[{Specularity[White, 10], Darker[Red, 3/4], Cylinder[{{-5/2, 0, 0}, {5/2, 0, 0}}], Specularity[White, 10], Darker[Blue, 3/4], Cylinder[{{0, -5/2, 0}, {0, 5/2, 0}}]}, Boxed -> False, Lighting -> {{"Point", GrayLevel[1/2], {2, -2, 2}}, {"Ambient", White}}, Method -> {"CylinderPoints" -> 1000}] Just ...

3

In Mathematica tongue x^2 - y^2 = 1 is pronounced as x^2 - y^2 == 1 x^2-y^2=1 It is a hyperbola, Wolfram|Alpha is verry helpfull for first findings, The Documentation Center (hit F1) is helpfull as well, see Function Visualization, Plot3D[x^2 - y^2 == 1, {x, -5, 5}, {y, -5, 5}] ContourPlot3D[x^2 - y^2 == 1, {x, -5, 5}, {y, -5, 5}, {z, -5, 5}] ...

2

Unfortunately, on version 10.3 the solutions proposed by Jens do not seem to work in at least some situations (see edit on this other question). A workaround that I found to work well in my case is simply using GraphicsRow, which seems to automatically correctly tweak the options of Inset in a way that produces the expected result. If the 3D graphics to ...

2

I will use the following data for ListPlot3D data = Table[Sin[j^2 + i], {i, 0, Pi, 0.1}, {j, 0, Pi, 0.1}]; Some initialization to be used later {w, h} = {300, 300}; pt = Scaled[{0.5, 0.5}]; {{l, r}, {b, t}} = {{80, 40}, {60, 10}}; Now make the ViewPoint dynamic as shown here vp = Options[Graphics3D, ViewPoint][[1, 2]]; p1 = ListPlot3D[data, ViewPoint ...

2

Taking Rahul's idea: ContourPlot3D[x^2 - y^2 == 1, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]

1

I understand the solution proposed in the comments by @YvesKlett has bee sufficient. However, I gave it a go out of curiousity and this seemed to work fine: Code: (*Dummy function*) f[x_, y_] := Sin[x^2 + y^2] Exp[-x^2] + Cos[x^2 + y^2] (*Sample data*) g2 = Plot3D[f[x, y], {x, -2, 2}, {y, -2, 2}, PlotPoints -> 50, Axes -> False, Boxed -> ...

1

Show[ ParametricPlot3D[{u,Sqrt[u^2-1],v},{u,-2,2},{v,-2,2}], ParametricPlot3D[{u,-Sqrt[u^2-1],v},{u,-2,2},{v,-2,2}] ]

1

In my opinion the easiest approach is to use Polygon[] 3D graphics and and texture it with the BarLegend[]. The sample code is: myplotfcn = RandomReal[{0, 1}, {10, 3}]; x = Max[myplotfcn[[All, 1]]]; y = Max[myplotfcn[[All, 2]]]; z = Max[myplotfcn[[All, 3]]]; Show[ListPlot3D[myplotfcn, ViewPoint -> {-(N[Pi]/8.), -(N[Pi]/2.), 0.45}, AxesLabel -> ...

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