
[Sponsors] 
July 12, 2000, 03:47 
Unique vertices of triangulated surface!

#1 
Guest
Posts: n/a

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 

July 12, 2000, 04:17 
Re: Unique vertices of triangulated surface!

#2 
Guest
Posts: n/a

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 D52072 Aachen Kackertstr. 11 Germany Tel.: +492418890113 email: R.Schneiders@magmasoft.de www: http://wwwusers.informatik.rwthaachen.de/~roberts/ 

July 12, 2000, 16:39 
Re: Unique vertices of triangulated surface!

#3 
Guest
Posts: n/a

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 

July 13, 2000, 02:26 
Re: Unique vertices of triangulated surface!

#4 
Guest
Posts: n/a

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 

July 13, 2000, 04:18 
Re: Unique vertices of triangulated surface!

#5 
Guest
Posts: n/a

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 

July 16, 2000, 11:04 
Re: Unique vertices of triangulated surface!

#6 
Guest
Posts: n/a

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 

Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Error : Self intersecting surface mesh, computing intersections & Error : Impossible  velan  Open Source Meshers: Gmsh, Netgen, CGNS, ...  3  October 22, 2015 11:05 
Problem with Gmsh  nishant_hull  Open Source Meshers: Gmsh, Netgen, CGNS, ...  23  August 5, 2015 02:09 
[ICEM] Problems with coedge curves and surfaces  tommymoose  ANSYS Meshing & Geometry  0  August 5, 2011 16:02 
boundaries with gmshToFoam  ouafa  Open Source Meshers: Gmsh, Netgen, CGNS, ...  7  May 21, 2010 12:43 
CFX4.3 build analysis form  Chie Min  CFX  5  July 12, 2001 23:19 