5' Diameter Trussed Frame Model Dome

Figure 1

This trussed frame model, about 5' diameter,  was made without using any chord factor data, beyond deciding on a radius, and arbitrarily picking an approximate chord length for the struts around  the base circumference of the inner frame.

Figure 2

Fig.2: In this case, the base consists of 29 struts. They are positioned by means of a rotating tripod-crane mounted at the center.  The crane can be adjusted to the inner frame radius, or extended to the outer frame radius, which in this case is about 2" greater than the inner radius. A hub is attached to the end of the crane, attached to adjacent hubs, and then the crane is retracted and repositioned.

Figure 3

The 2" distance between the inner and outer frames is an arbitrary distance. If it had been greater- say 3 or 4", then the model would have been stiffer.  

Although the base is made of approximately even chords,  something¡¯s got to give when triangulating up the curve. Either all the chords have to get shorter by an equal amount, or some chords have to get smaller and some stay the same, or some of the apexes of the triangles around the base need to be higher. In this model, the locations of the apexes for the base row of triangles were picked in a hit-or-miss manner.

Figure 4

Once a triangle in the inner frame is complete, the crane is placed more or less in the center of this triangle, and extended out to the outer shell position. Bracing us used to support the structure while it is being built (Fig.1).

All the hubs in the outer shell attach to 6 struts: three triangulate to the inner shell, and three connect to adjacent outer shell hubs.  The inner shell hubs usually connect 12 struts, (Fig.4) alternately connecting to adjacent inner shell hubs, and to the outer shell.  Pentagon hubs connect 10 struts.

I did not have a logical framework for making decisions, or knowing what the result would be. I was aware that the 12 pentagons in an icosahedron breakdown are somehow crucial to getting a smooth curve, and also that the lengths of struts get longer toward the centers of the spherical triangles of the icosahedron.  In this model, everything was played by ear, except for constant radii for the inner and outer frames. 

At a certain point, I found that there were four struts connected to a hub, and a resulting angle which was clearly less than 180 degrees.  So at that point it seemed advisable to make this a pentagonal hub, and attach only one additional strut, to make a 5-strut pentagon hub, rather than two, which would give a 6-strut, hexagon hub.  

It is hard to explain the criteria for choosing where to place hubs, and so I began thinking of it as ¡°spider math¡±  Every once in a while it seemed appropriate to have a pentagon. At one point, I found that I had a hub with 5 struts attached, and a very obtuse angle remaining, and so instead of adding a single strut to make a hexagonal vertex,  I added two more hubs,  making a heptagon, wondering if this would create major problems elsewhere.  It seemed not to matter, except that the decision to create heptagons seemed to lead to the need for more pentagons in the vicinity.  

The system of folded connectors shown elsewhere on this web site makes it possible to add or subtract connectors to any hub. 

Some questions come to mind. 

-The model seems ¡°quite strong¡±.  How strong? Would this approach make any sense for a full-scale structure? Could it possibly be ¡°engineered¡±?

-Why consider using the spider-math approach rather than going with mathematically-derived chord factors?

- Is there a covering method which would be adaptable to the irregular polyhedra of the outer frame?