Skip to main content

graphics3d - Strange behavior of `BoundaryMeshRegion`


BoundaryMeshRegion is new function of version 10. I am not familiar with the function. I want to decide whether a point is inside or outside in the Boundary using RegionMember. This code is not able to work. why is it?


BoundaryMeshRegion[{{4, 4, 4}, {4, 4, 6}, {4, 6, 4}, {4, 6, 6}, {6, 4,
4}, {6, 4, 6}, {6, 6, 4}, {6, 6, 6}},
Polygon[{{2, 3, 1}, {6, 8, 7}, {2, 5, 6}, {1, 7, 5}, {4, 7, 8}, {2,
6, 4}, {2, 3, 4}, {6, 5, 7}, {1, 2, 5}, {1, 3, 7}, {3, 4, 7}, {8,
4, 6}}]]



BoundaryMeshRegion[{{4, 4, 4}, {4, 4, 6}, {4, 6, 4}, {4, 6, 6}, {6, 4, 4}, {6, 4, 6}, {6, 6, 4}, {6, 6, 6}}, Polygon[{{2, 3, 1}, {6, 8, 7}, {2, 5, 6}, {1, 7, 5}, {4, 7, 8}, {2, 6, 4}, {2, 3, 4}, {6, 5, 7}, {1, 2, 5}, {1, 3, 7}, {3, 4, 7}, {8, 4, 6}}]]



But this is able to work. What is different.


BoundaryMeshRegion[{{4, 4, 4}, {4, 4, 6}, {4, 6, 4}, {4, 6, 6}, {6, 4,
4}, {6, 4, 6}, {6, 6, 4}, {6, 6, 6}},
Polygon[{(*{2,3,1},{6,8,
7},*){2, 5, 6}, {1, 7, 5}, {4, 7, 8}, {2, 6, 4}, {2, 3, 4}, {6, 5,
7}, {1, 2, 5}, {1, 3, 7}, {3, 4, 7}, {8, 4, 6}}]]



Blockquote



Are these bugs too?


Case 1


Graphics3D[{Opacity[0.5], tmp, Opacity[1], Red, PointSize[0.05], 
Point[{2, 0, 0}]}, Boxed -> False]


Blockquote




tmp = RevolutionPlot3D[{2 + Cos[t], Sin[t]}, {t, 0, 2 Pi},
PlotPoints -> 2]; tmp =
GraphicsComplex[tmp[[1, 1]], tmp[[1, 2, 1, 1, 5, 1]]];
r1 = BoundaryDiscretizeGraphics@tmp


Blockquote



RegionQ[r1]



True



RegionMember[r1, {2, 0, 0}]


False



Case 2


BoundaryDiscretizeGraphics[

GraphicsComplex[{{4, 4, 4}, {4, 4, 6}, {4, 6, 4}, {4, 6, 6}, {6, 4,
4}, {6, 4, 6}, {6, 6, 4}, {6, 6, 6}},
Polygon[{(*{2,3,1},{6,8,
7},*){2, 5, 6}, {1, 7, 5}, {4, 7, 8}, {2, 6, 4}, {2, 3, 4}, {6, 5,
7}, {1, 2, 5}, {1, 3, 7}, {3, 4, 7}, {8, 4, 6}}]]]


Blockquote




Answer




This is a bug, I think, and I filed it as such: The second region should not evaluate to a RegionQ BoundaryMeshRegion. A BoundaryMeshRegion is valid if it contains a closed surface. The subtle point about BoundaryMeshRegion is that this closed surface is a (sparse) representation of the entire region the surface encloses. Why the first one does not work, I must admit, I do not know. At least it's not obvious to me.


You can use:


bmr = BoundaryMeshRegion[{{4, 4, 4}, {4, 4, 6}, {4, 6, 6}, {4, 6, 
4}, {6, 4, 4}, {6, 4, 6}, {6, 6, 6}, {6, 6, 4}},
Polygon[{{1, 2, 3, 4}, {1, 2, 6, 5}, {2, 3, 7, 6}, {3, 4, 8,
7}, {4, 1, 5, 8}, {5, 6, 7, 8}}]];
rmf = RegionMember[bmr]

To generate the RegionMemberFunction. I just perturbed the coordinates a bit. You could try to replace the quadrilaterals with triangles and see if that works.


Comments

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}]

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.