[CIG-SHORT] Direction of traction-normal for Neumann BC

[Satoshi Okuyama]

Hello,

Recently I started using pylith and I already love it. However, I have
an question about Neumann boundary condition;

What determines the direction of positive traction-normal?

or

What determines the order of the vertices when pylith construct faces
from a group of vertices for boundary condition?


Here is my story,

I am trying to simulate the deformation caused by a pressure source. I
created a mesh with spherical source and put all the vertices on source
surface into a group, then applied Neumann BC with just traction-normal.

However, the deformation of the source was far from isotropic. I checked
the initial traction and found that deflation (traction toward source
center) is applied to some faces, while inflation is applied to the others.

Following is an example of initial-traction output. I placed 5 vertices 
on a plane of z=0 and formed 4 triangle face. Then I applied +1Pa of
traction-normal to this group.

#######################################################################
# vtk DataFile Version 2.0
Simplicial Mesh Example
ASCII
DATASET UNSTRUCTURED_GRID
POINTS 5 double
-1.000000e+00 -1.000000e+00 0.000000e+00
1.000000e+00 -1.000000e+00 0.000000e+00
1.000000e+00 1.000000e+00 0.000000e+00
-1.000000e+00 1.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00
CELLS 4 16
3  2 1 4
3  3 0 4
3  3 2 4
3  4 1 0
CELL_TYPES 4
5
5
5
5
CELL_DATA 4
VECTORS initial_traction double
0.000000e+00 0.000000e+00 -1.000000e+00
0.000000e+00 0.000000e+00 1.000000e+00
0.000000e+00 0.000000e+00 -1.000000e+00
0.000000e+00 0.000000e+00 -1.000000e+00
#######################################################################

As you see, 2nd cell (or face) receives traction of (0,0,1) while other
cells receives (0,0,-1). I noticed that if I consider 2 vectors - 1st
vertex to 2nd, and 1st to 3rd - the direction of the traction vector is 
equal to the cross product of them. 

cell #1: 
  v1: #2 -> #1 = ( 0,-2,0)
  v2: #2 -> #4 = (-1,-1,0)
  v1 x v2 = (0,0,-2)

cell #2:
  v1: #3 -> #0 = ( 0,-2,0)
  v2: #3 -> #4 = ( 1,-1,0)
  v1 x v2 = (0,0,2)

One step closer to the answer, I believe. But I have no idea how this
order is determined. The order of the vertices for 2nd cell is 3-0-4,
not 3-4-0. But why?
 

Regards,
----
Satoshi Okuyama