Hexagonal Geodesic Domes

Triangulation

It is possible to describe algorithms to categorise structures according to whether they are fully triangulated or not. Such classifications often have some utility when determining the stability of systems.

Here is one such algorithm. It tries to classify as many stable objects as it can easily manage as being "fully triangulated":

An algorithm that categorises systems according to whether they are fully triangulated

Define a strut as being a solid object.

• For each solid object a in the system consider if it is attached to two other solid objects b and c in such a way that b is also attached to c.

  If so, a, b and c form a single new soild object.

• Otherwise for each solid object a in the system consider if it is attached to another other solid object b in at two (or more) places.

  If so, a and b form a single new soild object.

At the end of the process there will be zero or more solid objects in the system. If there is precisely one solid object the system is defined as being fully triangulated.

Illustrations:

Three composite solid objects forming a triangle

Two composite solid objects joined in two places

A composite solid object forming a triangle with two struts

Fully triangulated systems

All such structures are stable. They all have linear stability.

Structures which are not fully triangulated may be either stable or unstable.

Here are some examples of 2D structures which are not fully triangulated - but are still stable:

Not fully triangulated - but still stable

Not fully triangulated - but still stable

Further extensions

Probably.

The next most obvious extension to the existing definition states that if two solid objects are connected together by three solid objects, where each of the three objects has an associated line through their respective "points of attachment"; and those lines are not all parallel; and those lines don't all intersect at the same point - then the resulting composite object is also stable.

The implications of this rule need further exploration.

3D

Much the same algorithm can be applied to three dimensional structures.

However it is no longer true that all fully trigangulated structures are stable.

As an example of a fully-triangulated structure that lacks stability, consider the flexahedron:

Flexahedron - fully triangulated, but unstable

As with the 2D case, structures which are not fully triangulated may also be stable or unstable.

More useful concepts for use in 3D

One approach is to consider structures with clearly-defined surfaces.

If such a surface exists, and it forms a convex hull, then full triangulation of the surface implies stability of the resulting structure.

There are other results along similar lines.

Another approach involves concepts such as tetrahedralization.

Instead of considering if an object is joined to two other objects that are themselves linked, tetrahedralization requires an object be linked to three other solid objects - that are themselves all linked to each other.

Constraints are needed to ensure linear stability - no three points of attachment should lie in a straight line.

All fully tetrahedralized structures are linearly stable.

A flexahedron does not qualify as tetrahedralized under such a definition.

The notion of tetrahedralized may be extended by permitting extensions using octahedra, icosahedra, decahedra - and the other primitive stable three dimensional structures.

However that exercise is beyond the scope of this essay.

As with lack of triangulation, lack of tetrahedralization does not imply anything about stability.

Here is a stable structure which is not triangulated or tetrahedralized. Nor is it clearly a composite of other stable polyhedra:

This structure is stable