Hello all,

I want a formula for 3D grids which relates the # of nodes to the # of tetrahedra, i.e.

Nt = fnc(Nv)

where Nv is the # of vertices(nodes) and Nt is the # of tetrahedra.

Also, the # of triangles on boundaries may be involved.

Does anybody know whether or not it is possible to get such a formula? (even an approximation)

Thank you,

Yoshikiyo

Hi,

let V = number of nodes,

E = number of edges,

F = number of faces

T = number of tets

_i : inner

_a : boundary

Then (for a geometry without holes):

V + F = E + H + 1

4 * T = 2 * F_i + F_a

F_a + V_a = E_a + 2

4 * F_a = 2 * E_a

On the average, each internal node has 24 tets adjacent.

Regards,

Robert

Dr. 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/

Thank you for your information, Dr. Schneiders.

I understand the four formulas, but it seems impossible to eliminate both E_i and F_i in the first formula so that I have V=fnc(T,Va).

How did you deduce that each internal node has 24 tets adjacent on average?

Thank you

Dear Mr. Yoshikiyo,

I don't know whether it's possble to get the kind of formula that you want.

Your second question: A heuristic argument is:

- 1D: Each node has 2 = 1*2 edges adjacent - 2D: Each node has 6 = 1*2*3 triangles adjacent ==> 3D: Each node has 24 = 1*2*3*4 tets adjacent.

Best regards,

Robert Schneiders

in your formulae what is 'H'?

"H" should be "T", sorry.

Regards,

Robert

For a good Delaunay mesh, T approx=6*V