Hello,

I have a boundary surface discretized using triangular elements. I would like to compute the outward normal to each triangular element. Although, I was able to compute the normal I am not able to check whether the normal points inwards or outwards. How does one check this normally ?

Thanks, Prem

This is determined by the order of the vertices of the volumetric cell to which this trinagular face belongs. To make it more concrete, suppose you deal with a tetrahedron, with its boundary surface defined by vertices labeled a, b, c and the off-boundary vertex labeled q. Let us define Vab as the vector from a to b, etc. Also assume that the order of vertices is such that q is located at the same direction of face a-b-c as the vector product Vab x Vbc (which is also twice the face area). The the outward normal is therefore the opposite direction, i.e., -Vab x Vbc (and you may wish to normilize it by its magnitude).

The same argument holds for other cell types (e.g., prisms).

I hope this helps.

Thanks for the information, Rami. So without the off boundary node I have no-chance of determining whether the normal points inwards or outwards ? The data that I have contains vertex information for ONLY the boundary nodes. Also for a closed surface, shouldn't the off-boundary node be located inside the surface ?

Thanks, Prem

Hi,

I guess you could also use centroid of the triangle /tetrahedron. In my case what I did was to check the sign of dot product of unit normal and unit vector along line joining centroid and mid pt. of the boundary face. It works in all, but one case... both the vectors could be perpendicular....does'nt happen in 2D but maybe in 3D.

Regards,

Abhijit

Unless the surface is closed, I think it is undetermined what is "inside" and "outside" when only surface verices are given.

If you have only surface vertices and the surface is closed, you may find algorithms for finding the outwards normal. I never did it myself, though, so I can just think of a very simple and inefficient solution. If you are that desparate, I'll put it forward at your request. I hope you'll come up with better ideas.

Rami,

Could you tell me your solution, please ? I am desperate....

Thanks, Prem

OK, if you are that desperate...

Take the candidate normal you wish to examine. Find all its intersections with all other facets of the boundary surface and count them. If the number of intersections is even (including 0), the normal is outwards.

As I mentioned, this is meaningful only for closed surface and probably very inefficient. Maybe someone else has a better idea?