I have a set of points in a nx3 matrix and I would like to convert them into a surface, so that I may calculate its surface area.

The function ListSurfacePlot3D creates the surface how I want it. How do I compute the surface area from there? enter image description here

I have tried other functions, such as DelaunayMesh. However, DelaunayMesh creates a closed region defining a volume, causing the surface area to be greater than expected. Which function do I use? If this surface is created correctly, I would use RegionMeasure to get the area. enter image description here

Edit: Here is my set of data points https://www.dropbox.com/s/db79ewahx24it3r/Data.txt?dl=0

Using Geomagic Control allows me to get the area of the wrapped points, at about 11522 mm2. I don't need the visual, I only want the area from these points. enter image description here

I'll be running the area calculation through a routine of about 6000 samples.

  • 3
    $\begingroup$ See DiscretizeGraphics and TriangulateMesh. You should also provide a sample data if you want people to pay attention and provide help. $\endgroup$
    – RunnyKine
    Commented Jun 17, 2015 at 22:34
  • $\begingroup$ I have added a sample of my data. It appears triangulate mesh only triangulates an existing surface, so I couldn't use that. $\endgroup$
    – Symphos
    Commented Jun 18, 2015 at 15:59

1 Answer 1


It turns out ListSurfacePlot3D does a terribly poor job of approximating the surface in the OP, otherwise one will just apply DiscretizeGraphics to the output obtained from ListSurfacePlot3D and be done with it. But since that's not applicable here, we present an approach that uses alpha shapes to approximate the shape of the given point set by tuning a parameter. First a helper function for computing the circum-radius of a tetrahedron (we Compile it and give it the Listable Attribute for speed gains):

circumRadius = 
 Compile[{{v, _Real, 2}}, 
  With[{a = v[[1]] - v[[4]], b = v[[2]] - v[[4]], c = v[[3]] - v[[4]]},
   With[{a1 = Plus @@ (a^2), b1 = Plus @@ (b^2), c1 = Plus @@ (c^2),
     α1 = b[[2]] c[[3]] - b[[3]] c[[2]], 
     α2 = b[[3]] c[[1]] - b[[1]] c[[3]], 
     α3 = b[[1]] c[[2]] - b[[2]] c[[1]],
     β1 = c[[2]] a[[3]] - c[[3]] a[[2]], 
     β2 = c[[3]] a[[1]] - c[[1]] a[[3]], 
     β3 = c[[1]] a[[2]] - c[[2]] a[[1]],
     γ1 = a[[2]] b[[3]] - a[[3]] b[[2]], 
     γ2 = a[[3]] b[[1]] - a[[1]] b[[3]], 
     γ3 = a[[1]] b[[2]] - a[[2]] b[[1]]},
    Norm[a1 {α1, α2, α3} + b1 {β1, β2, β3} + c1 {γ1, γ2, γ3}] / 
         (2 Norm[Plus @@ (a[[1 ;; 3]] {α1, α2, α3})])]
  CompilationTarget -> "C", RuntimeOptions -> "Speed", 
  RuntimeAttributes -> {Listable}, Parallelization -> True]

And now the code to compute the alpha shape:

alphaShapes[points_, crit_] := 
 Module[{alphacriteria, del = Quiet @ DelaunayMesh @ points, tetras, 
         tetcoords, tetradii, selectExternalFaces}, 
  alphacriteria[tetrahedra_, radii_, rmax_] := 
         Pick[tetrahedra, UnitStep @ Subtract[rmax, radii], 1]; 
  selectExternalFaces[facets_] := MeshRegion[points, facets]; 
  If[Head[del] === EmptyRegion, del, 
      tetras = MeshCells[del, 3]; 
      tetcoords = MeshPrimitives[del, 3][[All, 1]]; 
      tetradii = Quiet@circumRadius@tetcoords /. ComplexInfinity -> $MaxMachineNumber; 
      selectExternalFaces @ alphacriteria[tetras, tetradii, crit]

For the plot, setting crit to $45$ seems to do a good job, but there appear to be holes in the surface. The area appears to be close to the value posted by OP from Geomagic.

reg = RegionBoundary @ alphaShapes[data, 45]; (* data is your nx3 matrix *)
HighlightMesh[reg, {Style[2, FaceForm[None]], Style[1, Darker@Green]}]

Mathematica graphics

Here is the approximate area:

0.5 Area @ reg


Which is very close to 11522 (the value provided by the OP). Note that I'm dividing the total area by $2$ since the Area function will compute the areas of the cells on both sides of the image.

For the Image Gurus here:

I would like to know if Mathematica can produce something as nice as the image posted in the question obtained from GeoImage.

  • $\begingroup$ Unfortunately, ListSurfacePlot3D does a pretty poor job of approximating this surface. Try running Show[ListSurfacePlot3D[data], ListPointPlot3D[data]] to see how bad. $\endgroup$ Commented Jun 18, 2015 at 16:25
  • $\begingroup$ @MichaelSeifert. Indeed, it does a poor job. $\endgroup$
    – RunnyKine
    Commented Jun 18, 2015 at 16:27
  • $\begingroup$ Maybe your code is computing the total area of a two-sided surface? $\endgroup$
    – user484
    Commented Jun 18, 2015 at 21:43
  • $\begingroup$ @Rahul, That's what I thought, I guess that makes sense. Thanks. $\endgroup$
    – RunnyKine
    Commented Jun 18, 2015 at 21:55
  • 2
    $\begingroup$ @g3kk0. I'm glad you found my code useful. You can look up Edelsbrunner's Algorithm, which is what I implemented. $\endgroup$
    – RunnyKine
    Commented Jul 9, 2015 at 14:18

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