Hexayurt dome details and models

Edit 4/8/12: Andrew Maxwell, Tracy Suskin, Ying Yang, students at SAIT polytechnic in Canada, have put together the engineering details for the tri-dome.

People are now starting to build my tri-dome and quad-dome versions of the hexayurt, so it is time to give some of the technical details. To start, however, here is an application of the intermediate value theorem!

When I started working on the details for the tri-dome I realised I had made a bad assumption (thinking that the form was geometrically pure). This means that some of the details in my original write up were wrong. All a little embarrassing. Ironically, I might have missed a form that does actually work, had I not made the bad assumption. The shape, like the hexayurt, starts with a hexagonal based pyramid. In a traditional hexayurt this lies on top of a hexagon of vertical walls. Instead of this we attach a square to three of the edges and the classic hexayurt triangle (isocoles triangle with base and height the same length) to the other three. We can look at what happens as the pyramid is moved away from the ground, while the edges of the shapes rest on it:

This does not give a great building; there are holes. The holes are triangles and two of the sides have a fixed length. The final edge changes length, starting long, and ending short. We know we can fill the holes with classic hexayurt triangles. Two of the edges are the right length we just need the third. The length changes smoothly as we raise the roof, and starts shorter and ends longer than we want. Here we can apply the intermediate value theorem, so the correct length must be passed. As a mathematician I would stop there, the system works; however people are building the things…

So to calculate the correct angle for the square sides of the model we can look vertically down, calling the angle of the square face θ, (and assuming that the boards we are using are 8′ by 4′) needing as the classic maths problem asks to “find x”.In this case

$x = 4 \sqrt{4 \cos(\theta)^2+1+2\sqrt{3}\cos(\theta)}$,

we want $x = 4\sqrt{5}$ so:

$4\sqrt{5} = 4 \sqrt{4 \cos(\theta)^2+1+2\sqrt{3}\cos(\theta)}$

$5 = 4 \cos(\theta)^2+1+2\sqrt{3}\cos(\theta)$

$0 = 2 \cos(\theta)^2 + \sqrt{3}\cos(\theta) - 2$

$\cos(\theta) = \frac{-\sqrt{3} \pm \sqrt{19}}{4}$
Which gives an angle of about 49°, and the height of the roof (assuming 4’x8′ panels) is $8 \sin(\theta)$, just over 6′ at the edge and 10′ in the centre. We can use these, and useful facts about general tetrahedra to calculate all the angles between faces by using the lengths of their edges. If you want to follow the details yourself, you need to add vectors to get some of the edge lengths, then use the Cayley-Menger determinant to find the volume of the tetrahedron, and then the generalised Sine rule to (halfway down this page) to give the angle.