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Unique vertices of triangulated surface!
Hi,
Does anyone know if one can state the maximum number of unique (not sure of the spelling :) ) vertices for a triangulated surface? That is, if I have 300 surface triangles, what is the maximum number of vertices? Any good sites for such questions and other mesh topology related information out there? Regards Anders |

Re: Unique vertices of triangulated surface!
Anders,
the number of vertices is roughly half the number of faces. This can be concluded from Euler's theorem: number of faces + number of vertices = number of edges + 1 (for simply connected bodies without holes). For the surface mesh, the following equation holds: 3 * number of faces = 2 * number of edges It follows: number of nodes = 0.5 * number of faces + 1 Best regards, Robert Robert Schneiders MAGMA Giessereitechnologie GmbH D-52072 Aachen Kackertstr. 11 Germany Tel.: +49-241-88901-13 email: R.Schneiders@magmasoft.de www: http://www-users.informatik.rwth-aachen.de/~roberts/ |

Re: Unique vertices of triangulated surface!
Dear Robert,
I think you must have committed a small error: Euler 's fomula is: number of faces + number of vertices = number of edges + 2 (instead of 1) So the final answer is: number of nodes = 0.5 * number of faces + 2 Best, Xiaoming |

Re: Unique vertices of triangulated surface!
Hi Robert,
Thanx for the fast answer. However, I am a bit confused about the last formulas. They does not seem to match a simple testcase. I agree on the first one number of faces + number of vertices = number of edges + 1 However, if you take a square and draw a line diagonally, then you will end up with 2 triangular faces, 4 vertices and 5 edges. Thus Eulers formula: 2 + 4 = 5 + 1 OK However, the last formula does not add up. While writing this I also read Xiaoming answer, and it seemed that his Euler variant must be wrong! Any ideas? Regards Anders |

Re: Unique vertices of triangulated surface!
Hi,
the Xiaoming comment is right. The test case you give is not a valid solid, and it does not fulfil the correct equation 2 + 4 = 5 + 2 If you a one quadrilateral face, you have a valid solid model (at least from a topological point of view), and the formula gives 3 + 4 = 5 + 2 Best regards, Robert |

Re: Unique vertices of triangulated surface!
The simplest solid is a triangular based pyramid.
Faces(F) = 4 Vertices(V) = 4 Edges(E) = 6 So the correct formula is F + V = E + 2 |

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