Why does Finite Elements Methods (FEM) tile domains with triangles?

Why is it the case that the Finite Elements Method (FEM) tiles domains with triangles? With so many geometrical shapes, is there anything special triangles have to offer?

For starters, a surprisingly small number of regular polygons can even be used to tile a surface. If we want to avoid leaving any gaps in the space, the condition that needs to be fulfilled is that the internal angle of the polygon has to evenly divide 360. Evenly is the keyword here. The internal angle of regular polygons is $(n-2)*\frac{180}{n}$, where $n$ is the number of sides. Here below is the list of regular polygons of up to 100 sides with the corresponding internal angle:

(3, 60.0)
(4, 90.0)
(5, 108.0)
(6, 120.0)
(8, 135.0)
(9, 140.0)
(10, 144.0)
(12, 150.0)
(15, 156.0)
(18, 160.0)
(20, 162.0)
(24, 165.0)
(30, 168.0)
(36, 170.0)
(40, 171.0)
(45, 172.0)
(60, 174.0)
(72, 175.0)
(90, 176.0)

Of these, only the triangle (60), the square (90) and the hexagon (120) fit well into 360. All the others simply don’t. This first evaluation is enough to shift the initial question from “why the triangle” to “why not the square or the hexagon“? All three of them are in fact theoretically equally capable of tiling whatever “infinite surface”, and it’s not a case that beehives are hexagonal in structure.

Now the practical constraints come in:

  1. we don’t deal with boundless surfaces – we always have boundaries, and the meshing needs to follow them as accurately as possible. How much can you bend a square to make it fit a curvy boundary? Not much. With a hexagon the problem is pretty much the same. Triangles, although still not perfect, have one pointy end that can be arranged to follow a curly outline better.
  2. in FEM, every vertex of every triangle is a degree of freedom of the linear system that is then solved. Put in twice as many triangles and you’ll get almost twice as many degrees of freedom (not exactly twice as many, but hey, you get the idea). What happens if we use hexagons? Now every tile will yield 6 degrees of freedom! The problem is that these 6 vertexes are not well distributed, since they are all concentrated on the perimeter of the small hexagon – no degree of freedom covers the interior of the hexagon, which will leave quite some non-computational space while still doubling the computational cost of the simulation.

All in all, the triangle is a much less constrained shape. Think of being given a random cloud of 2D points, with the instructions of linking them with a regular polygon of your choice. Obviously no shape will work on the cloud of points as it is, so you are allowed to move the points to fit your tiling. However, the metric that defines your success is how much, overall, you have had to move individual points to manage the tiling with the polygon of your choice. Triangles will inevitably require less adjustment, while hexagons more.

All this discussion gives for granted that we only allow regular polygons. If you want to allow irregular polygons, go on and see what happens, but good luck building a theory where, from the very first piece, “everything can happen”.

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