Centroid of a convex polyhedron

Question

I have a closed convex polyhedron, which is defined by an array of convex polygons (faces), which are defined by arrays of vertices in three-dimensional space. I am trying to find the center of gravity of a polyhedron, assuming uniform density. At the moment, I am calculating it using an algorithm in this pseudo-code.

public Vector3 getCentroid() { Vector3 centroid = (0, 0, 0); for (face in faces) { Vector3 point = face.centroid; point.multiply(face.area()); centroid.add(point); } centroid.divide(faces.size()); return centroid; } 

This essentially takes a weighted average of the centroids of the faces. I am not 100% sure, this is correct, since I could not find the correct algorithm online. If someone could confirm my algorithm or direct me to the correct one, I would appreciate it.

Thanks.

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[EDIT]

So, here is the actual Java code that I use to find the centroid. It divides the polyhedron into pyramids converging at an arbitrary point inside the polyhedron. The weighted average for the centroids of the pyramid is based on the following formula.

C _all = SUM _{all pyramids} (C _pyramid * volume _pyramid ) / volume _allsub>

Here (heavily commented code):

  // Compute the average of the facial centroids. // This gives an arbitrary point inside the polyhedron. Vector3 avgPoint = new Vector3(0, 0, 0); for (int i = 0; i < faces.size(); i++) { avgPoint.add(faces.get(i).centroid); } avgPoint.divide(faces.size()); // Initialise the centroid and the volume. centroid = new Vector3(0, 0, 0); volume = 0; // Loop through each face. for (int i = 0; i < faces.size(); i++) { Face face = faces.get(i); // Find a vector from avgPoint to the centroid of the face. Vector3 avgToCentroid = face.centroid.clone(); avgToCentroid.sub(avgPoint); // Gives the unsigned minimum distance between the face and a parallel plane on avgPoint. float distance = avgToCentroid.scalarProjection(face.getNormal()); // Finds the volume of the pyramid using V = 1/3 * B * h // where: B = area of the pyramid base. // h = pyramid height. float pyramidVolume = face.getArea() * distance / 3; // Centroid of a pyramid is 1/4 of the height up from the base. // Using 3/4 here because vector is travelling 'down' the pyramid. avgToCentroid.multiply(0.75f); avgToCentroid.add(avgPoint); // avgToCentroid is now the centroid of the pyramid. // Weight it by the volume of the pyramid. avgToCentroid.multiply(pyramidVolume); volume += pyramidVolume; } // Average the weighted sum of pyramid centroids. centroid.divide(volume); 

Please feel free to ask me any questions you may have about this or point out any errors you see.

Answer 1

This usually depends on the structure of your polyhedron. 4 cases are possible:

• Only the vertices have weight, i.e. your polyhedron is a point system. Then you can simply calculate the average of the weighted sum of all points:

   r_c = sum(r_i * m_i) / sum(m_i) 

  Here r_i is a vector representing the ith vertex, m_i is its mass. The case of equal masses leaves us with a simpler formula:

   r_c = sum(r_i) / n 

  Where n is the number of vertices. Please note that both amounts are vectorized.

• Only the edges have weight, and the polyhedron is essentially a frame. This case can be reduced to the previous one, substituting each edge in a vertex located right in the middle of the edge and having the weight of the entire edge.

• Only individuals have weight. This case can be reduced to the first. Each face is a two-dimensional convex figure from which you can find a centroid. Substitution of each face of its centroid leads this case to the first.

• Solid polyhedron (your case deriving from the "assumed uniform density"). This problem requires a more complex approach. The first step is to divide the polyhedron into tetrahedra. Here is a brief description of how to do this. For a centroid, the tetrahedron is at the point where all its medians intersect. (The median of the tetrahedron is the line connecting its vertex and the center of gravity of the opposite side.) The next step is to replace each tetrahedron in the section with its centroid. And the last step is to find the centroid of the resulting set of weighted points, which is exactly the first case.

Answer 2

For the hard case, there is a much simpler method than trying to tetrahedronize a polyhedron (which is a trap ).

Here's the java-ish pseudo-hash code (assuming a decent implementation of Vector3):

 // running sum for total volume double vol = 0; // running sum for centroid Vector3 centroid = (0, 0, 0); for each triangle (a,b,c) { // Compute area-magnitude normal Vector3 n = (ba).cross(ca); vol += a.dot(n)/6.; // Compute contribution to centroid integral for each dimension for(int d = 0;d<3;d++) centroid[d] += n[d] * ((a[d]+b[d])^2 + (b[d]+c[d])^2 + (c[d]+a[d])^2); } // final scale by inverse volume centroid *= 1./(24.*2.*vol); 

Note that if you have higher degree faces than triangles, you can trivially using a fan is trivial, and this will work anyway.

This works conveniently even if the polyhedron is not convex.

I also posted matlab code