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Unique vertices of triangulated surface!

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Old   July 12, 2000, 03:47
Default Unique vertices of triangulated surface!
  #1
Anders Jönson
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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
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Old   July 12, 2000, 04:17
Default Re: Unique vertices of triangulated surface!
  #2
Robert Schneiders
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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/

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Old   July 12, 2000, 16:39
Default Re: Unique vertices of triangulated surface!
  #3
Xiaoming
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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

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Old   July 13, 2000, 02:26
Default Re: Unique vertices of triangulated surface!
  #4
Anders Jönson
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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
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Old   July 13, 2000, 04:18
Default Re: Unique vertices of triangulated surface!
  #5
Robert Schneiders
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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

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Old   July 16, 2000, 11:04
Default Re: Unique vertices of triangulated surface!
  #6
PRAVEEN C
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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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