Center of Mass of a Tetrahedron
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I am working on a problem, I have four points (x1,y1,z1),(x2,y2,z2),(x3,y3,z3),(x4,y4,z4) which can form tetrahedron. I have checked it with tetramesh. I need to find the Center of Mass of these four points. Little background of the points; these are x-easting, y-Northing and z-elevation of ground, I have thousands of points like this.
I am looking for a function file which could give me output as Center of mass of tetrahedron.
Answers (1)
Roger Stafford
on 28 May 2015
1 vote
((x1+x2+x3+x4)/4,(y1+y2+y3+y4)/4,(z1+z2+z3+z4)/4)
2 Comments
Image Analyst
on 28 May 2015
Everything I've seen says that Roger is correct (as usual). For example: http://en.wikipedia.org/wiki/Centroid#Of_triangle_and_tetrahedron
Roger Stafford
on 29 May 2015
Edited: Roger Stafford
on 29 May 2015
It's a simple problem in integration, Sairavichand. Take triangular slices of a solid tetrahedron parallel to one of its faces. It is easy to show that the volume would be one third the area of the face multiplied by the orthogonal altitude of the fourth vertex from the face. If you compute the first moment along this altitude as measured from the fourth vertex, it will give a value of one-fourth the area multiplied by the altitude squared. Hence the ratio will be three-quarters the altitude, again as measured from that fourth vertex. This will be the answer no matter which face is used, and the only point that can satisfy such a requirement is the centroid of the four individual vertices. QED
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