Region from points

Question

Is there a built-in way to define a 3D region from an arbitrary list of points. The points will represent vertices of the edges of the region and we can assume that 1) they are not all on the same plane and 2) that there are at least 4 points. The points should define the region that allow the biggest volume given the provided vertices. I imagine something like Polygon except for 3 dimensional regions.

It needs to be a Mathematica region - something that I could use with functions like RegionIntersection or Volume. I have provided a list of example points below along with a (crude) picture of the volume I would like to derive from them, which I made manually. I am not necessarily interested in this volume in particular (it is just an example). I am mainly wondering if there is a generic method for any set of points

 points = {{0, 0, 1}, {5, 0, 0}, {1, 3, 0}, {0, 0, 2}, {4, 3, 0}, 
           {5, 0, 2}, {1, 3, 2}, {4, 3, 2}}

Answer 1

points = {{0, 0, 1}, {5, 0, 0}, {1, 3, 0}, {0, 0, 2}, {4, 3, 0}, {5, 
    0, 2}, {1, 3, 2}, {4, 3, 2}};

reg = ConvexHullMesh @ points

RegionDimension @ reg

3

Volume @ reg
 (* or *)
RegionMeasure @ reg

21.5

p1 = {2.6161, 2.60692, 0.880247};
RegionMember[reg, p1]

True

p2 = {0.915295, 0.507214, 0.367796};
RegionMember[reg, p2]

False

RegionCentroid @ reg

{2.61628, 1.44767, 1.08721}

Integrate[x y z, {x, y, z} ∈ reg]

82.0625

---

RegionIntersection, RegionUnion etc. all have in their documentation (under Possible Issues) that they are "not implemented for MeshRegion objects embedded in 3D" and "nor for BoundaryMeshRegion objects embdedded in 3D". My idea is to triangulate the MeshRegion into Tetrahedrons and RegionUnion them into the desired Region:

tri = TriangulateMesh[reg, MaxCellMeasure -> Infinity];
pts = MeshCoordinates @ tri;
ord = Cases[Show[tri][[1]], Tetrahedron[x_] :> x, Infinity] // First;
tetra = Tetrahedron /@ (pts[[##]] & /@ ord);
Volume /@ tetra // Total

21.5

The volume agrees.

r = RegionUnion[tetra];
RegionPlot3D @ r

Volume @ r

21.4583

The volume slightly diverges from the correct value. Nevertheless:

ball = Ball[{0, 0, 1}];
RegionPlot3D @ (int = RegionIntersection[r, ball])

Volume @ DiscretizeRegion @ int

0.530894

---

I want to check how accurate is the volume measurement of int.

I start with altered points:

points1 = {{0, 0, 0}, {5, 0, 0}, {1, 3, 0}, {0, 0, 2}, {4, 3, 0}, {5, 
    0, 2}, {1, 3, 2}, {4, 3, 2}};

where I changed the first point - {0, 0, 1} - to {0, 0, 0}. Next I played with it forcefully

{b1, t1} = GatherBy[points1, Last];
ord = Drop[#, -1] & /@ (Flatten[#, 1] & /@ 
      FindCurvePath /@ {b1, t1}) // First;

to create the correct ordering for making

hex = Hexahedron @ Join[b1[[ord]], t1[[ord]]];

which has

Volume @ hex

24

I then take manually

b2 = {{1, 3, 0}, {5, 0, 0}, {0, 0, 0}, {0, 0, 1}};

for creating the chunk additially added to the above hex:

tetra = Tetrahedron @ b2;

These two can be handled with

Volume @ (region = RegionDifference[hex, tetra]) // N

21.5

to create the initial region of interest. Because this is a Region, one can

ball = Ball[{0, 0, 1}];
RegionPlot3D[#, PlotPoints -> 100]& @ (inter = RegionIntersection[region, ball])

(and do other Region* functions on it).

Volume @ N @ inter

0.534992

So the two measurements of the volume agree with each other.