# splitting the tetrahedron into convex bricks

• Jul 14th 2009, 05:50 AM
vlnikolic
splitting the tetrahedron into convex bricks
dear sirs

i would like that somebody help me to solve the next question:

i would like to have proof, that when splitting the tetrahedron in the manner that is depicted on this picture:
Imageshack - tetra

the four 8-nodal elements (usually brick like elements) that are produced must be convex, no matter of how much differs the starting tetrahedron.

i have empirical proof already because i developed software for that purpose, the only thing i need is mathematical proof

thank you for your interest, and sorry on not-so-good english
• Jul 14th 2009, 08:11 AM
Laurent
Quote:

Originally Posted by vlnikolic
dear sirs

i would like that somebody help me to solve the next question:

i would like to have proof, that when splitting the tetrahedron in the manner that is depicted on this picture:
Imageshack - tetra

the four 8-nodal elements (usually brick like elements) that are produced must be convex, no matter of how much differs the starting tetrahedron.

i have empirical proof already because i developed software for that purpose, the only thing i need is mathematical proof

thank you for your interest, and sorry on not-so-good english

The picture doesn't seem quite correct (for instance, the line between I-II centers of gravity should go through the vertices of the tetrahedron)

If it were, you should be convinced that the four brick-like elements are defined as intersections of 6 half-spaces (defined by 6 planes). Therefore they must be convex: any intersection of convex subsets is convex.

Thus there is almost nothing to prove. The convexity comes from the very definition of these elements.
• Jul 15th 2009, 04:46 AM
vlnikolic
hello and much thanks for reply

i am splitting the tetrahedron just in the way i depicted on the picture (although not so precisely)

i split tetrahedron facets (triangles) so that firstly i find the CoG for 3 nodes in triangle, then i find CoG for edges and afterwards i connect CoG of edges and facet - just like in the picture.

i have no pleasure to understand this:
***
If it were, you should be convinced that the four brick-like elements are defined as intersections of 6 half-spaces (defined by 6 planes). Therefore they must be convex: any intersection of convex subsets is convex.
***

so if possible please explain me more

> The convexity comes from the
> very definition of these elements.

8 nodal element (brick-like) does not need to be convex in 3D

thanks once more, i would be very happy if i solve this problem
• Jul 15th 2009, 01:39 PM
Laurent
Quote:

Originally Posted by vlnikolic
hello and much thanks for reply

i am splitting the tetrahedron just in the way i depicted on the picture (although not so precisely)

i split tetrahedron facets (triangles) so that firstly i find the CoG for 3 nodes in triangle, then i find CoG for edges and afterwards i connect CoG of edges and facet - just like in the picture.

Not "just like in the picture". This picture could be drawn by hand as follows:

- Mark the middle of the edges (I).
- Then on each face draw the line from a vertex of the face to the middle of the opposite edge; the three lines cross at the center of gravity of the triangle (II).
- Finally, draw the four lines from a vertex of the tetrahedron to the center of the opposite face; the four lines meet at the center of gravity of the tetrahedron (III).

This construction is not correctly done on the picture, and it is important to understand it, because it shows that the brick-like elements are defined by 6 planes, each defined by 4 points. In general, 4 points don't define a plane, but this is the case here because of the construction above (and this is the main reason for the convexity). For instance, one such plane is the plane going through two vertices of the tetrahedron and the middle of the edge they don't touch. This plane goes through the type-III center and two type-II centers.

Once you have come with this visualisation of the situation, you are almost done. Indeed, any subset defined by a conjonction of conditions like "being on this side of this plane" (here, there are 6 such conditions) is convex. More mathematically, an intersection of half-spaces is convex. It is called a convex polytope.

Writing a formal proof may be a bit tedious, but at least you have the elements here that show you why this is true.

n.b.: I wrote "middle" of edges to simplify the writing, but you could put other weights on the vertices of the tetrahedron and the centers of gravity of type I could be elsewhere. This wouldn't affect the above construction about concurring lines.
• Jul 16th 2009, 02:30 AM
vlnikolic
hello once more

> This construction is not correctly done on the picture <

i am aware of how can i get CoGs (I, II, and III)

easy method when programming for set of n nodes is to do the:

Xcog = (X1 + X2 + ... + Xn) / n
Ycog = (Y1 + Y2 + ... + Yn) / n
Zcog = (Z1 + Z2 + ... + Zn) / n

after that i get 15 nodes on one tetrahedron as follows:

4 starting nodes
6 CoG of edges (I cog)
4 Cog of facets (II cog)
1 CoG of tetrahedron

(as shown on the picture)

after connecting theese nodes as presented on the picture i get four 8-nodal 3D elements

there is nothing wrong with that :)

only thing i need is writing a formal proof that when i interconnect nodes as i explained the 8 nodal elements i get are convex :)
• Jul 21st 2009, 05:21 AM
vlnikolic
hello again

here is one another example of how i split the tetrahedron

http://www.math.umass.edu/~qchen/SVP...eD1stOrder.jpg

i hope there will b more re:

thanks