Skip to main content

performance tuning - Faster way to compute the distance from a point to a surface in 3D


I am trying to compute the shortest distance between a point and a triangle in 3D


distance[point_, {p1_, p2_, p3_}] := Module[{p, s, t, sol},
p = s*p1 + (1 - s)*(t*p2 + (1 - t)*p3);
MinValue[{(point - p).(point - p),
0 <= s <= 1, 0 <= t <= 1}, {s, t}]];

but it seems to be quite slow, is there any way to make it faster?




Answer



Well, you can use the undocumented RegionDistance which does exactly this as follows: (This answer, as written, only works for V9 as noted by Oska, for V10 see update below)


here is a triangle in 3D


region = Polygon[{{0, 0, 0}, {1, 0, 0}, {0, 1, 1}}];

Graphics3D[region]

Mathematica graphics


Now suppose you want to find the shortest distance from the point {1, 1, 1} in 3D to this triangle just do the following:


Load the Region context



Graphics`Region`RegionInit[];

Then


RegionDistance[region, {1, 1, 1}]

Mathematica graphics


As a bonus, you can get the exact point on the triangle that is closest to the given point as follows:


RegionNearest[region, {1, 1, 1}]

Mathematica graphics



Visualize it


Graphics3D[{region, Darker@Green, PointSize[0.03], Point[{1, 1, 1}], 
Red, PointSize[0.03], Point[{1/3, 2/3, 2/3}]}]

Mathematica graphics


Update for Version 10


The above undocumented functions used in this answer now works out of the box in V10 so no need to load the Region context as I did above. Otherwise everything works as is. Also, now you can use the new Triangle function in place of Polygon above.


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.