This is the second iteration of a geodesic ice dome, the first version was Ice Station Thuban.

The spherical photograph is actually the eye of a krill.

http://en.wikipedia.org/wiki/File:Krilleyekils.jpg

It's analogous to what is going on with the Thuban Prime dome, and as you can see, makes a pretty good sphere, with small units of mostly hexagonal shape.

If you use an icosahedron as your basis, twenty equal triangles can be projected onto a sphere. An icosahedron is a pretty poor approximation of a sphere, which is why Ice Station Thuban used a rhombic triacontahedron as its basis, which is a bit better. However, an icosahedron has the benefit of having very few vertices (twelve) per sphere, and this is actually a benefit for a very practical reason.

Consider the triangle of the icosahedron. If you were to subdivide it as per the following diagram, you would get a hexagon in the middle, and the triangles in red would join with the triangles from the adjacent icosahedral faces, to form pentagons. The result would be a polyhedron with twelve pentagons and twenty hexagons. This composite polyhedron approximates a sphere more closely than an icosahedron - in fact, this pattern is quite similar to the very familiar Telstar soccer ball (proportions slightly different).

If you decided to double the linear precision, and instead of dividing the icosahedral edge into three, you mensurated by six, you could tessellate the triangle as per the following diagram:

As you can see, you will now have five and a half hexagons per icosahedral face, instead of only one. But you still only have three red triangles, which form the twelve pentagons. A projected sphere will now have 110 hexagons, and this will be much more spherical than an icosahedron. All hexagons are not quite the same, they are distorted once you project the vertices on to a sphere, as per the following diagram, which is a bit exaggerated for illustration (and somewhat poorly - the red triangles really should all be equivalently proportioned).

For a dome, to do this precisely, you would now have to create a lot of different, uniquely shaped forms, We don’t want to do that. (Or at least I don’t want to - that is not an optimized procedure, it is labor intensive and highly susceptible to derailment due to practical concerns. With Ice Station Thuban, I had only four unique shapes, and it was more than enough. Too much, really.)

But let’s go further. In the following diagram the icosahedral edge has been mensurated by twelve (double again!), leading to a tessellation that has 23.5 hexagons per face.

This will give you an approximate sphere with 470 hexagons… and still only 12 pentagons. Projected onto a sphere, the hexagons will still be very different, a lot of unique shapes.

The thing is, if you continue this process, you will get more and more tiny hexagons, and exactly twelve pentagons every time. At a certain point, the size of the hexagon is within your manufacturing tolerance. This effectively makes all the hexagons statistically “the same” - plus or minus.

This is what you see in the krill eye. Nature has a certain amount of time to grow these units, so they are all going to be just about the same size. But in order to make a sphere, they will jam together and naturally from hexagons (primarily) - except for some number of dislocations to correct for sphericity. If nature did things exactly repeatably, there would need to be twelve pentagons to account for the dislocations. (Note I did not say perfectly, because nature is the gold standard of perfection - perfection does not necessarily mean repeatable. It means robust, optimized for production quality per energy usage.) In the krill eye, you can see a handful of dislocations. They are not perfect pentagons, but I am guessing, if you average all krill eyes ever to have existed, there will be pretty close to twelve per idealized full-sphere eye. (Obviously an eye is not an entire sphere, it must attach, etc.)

Ice Station Thuban is only a hemisphere, so we can expect six dislocations, plus or minus. (Actually it’s only plus, error does not de-accrue somehow.) But we can expect only six or maybe as many as eight or nine (?) dislocations in an imperfect hemisphere. Even if it is a lot more, it is against a number in the thousands, of regular hexagons. This amount of adjustment seems quite practical.

A twenty foot diameter (OD) sphere will have a great circle circumference of about sixty three feet (62.83….). One projected icosahedral edge at that diameter would be about eleven and a half feet. If the icosahedron was mensurated by thirty-six, we’d end up with hexagons with an edge a little less than four inches. For a sphere, this means 215.5 hexagons per icosahedral face, or, 4310 hexagons per sphere.

Yup, that’s a lot.

But, they are small, and we are only doing a hemisphere, so we would need approximately 2155 hexagons (and six pentagons). The thing is, the manufacturing tolerance with ice at this size, and with this manufacturing method, is around half an inch or so, which is quite a large percent of a less-than-four inch side length hexagon. It means that ALL hexagons can be the same, and lie within an ideal, average-error hexagon.  Error can be compensated manually and incrementally over the assembly process with snow mortar, and in six or so spots with a pentagonal piece.

So all we need to do is calculate a hexagon, and a bevel angle into the center of the sphere, to create a tapered hexagonal form that is universal, easy to make, small and easy to manipulate. Constructing the dome out of these units should result in six(-ish) spots where we need to account for error with a pentagonal piece, or just a bunch of snow mortar.

The optimization problem now becomes quite different. We have to make thousands of these bricks, and are dependent on the weather to do so. Their small size helps - they will freeze fast - and even better, they will freeze faster than the large triangles of the first dome, because there is a much larger surface area to air-exposed volume ratio.

In fact this dome will use a mensuration of 42, which will need about 3000 hexagon bricks to complete a hemisphere. A little less, minus the bricks for a doorway as well. Still, 3000 is a lot of bricks.

They are easily made at this size, because a one-gallon water container easily encompasses the volume of a brick. The optimization question now becomes: how fast can one make 3000 bricks of that shape? The answer is, it all depends on how many you do at once.

If we have a set of molds, and can count on having enough time with the weather to freeze things ten times, we will need 300 molds. If we have 600 molds, we can complete the task in only five freezings. One freezing cycle will take what it takes, several hours. So this is the variable we can optimize. If one person with no sleep spends all of the time filling and releasing molds, maybe 1000 molds is the practical limit. But considering needing to eat and sleep and rest (remember, that is still a lot of water, and it is heavy!), at minimum, 300 to 400 is probably the maximum conservative number. So we will need a run of about a week of cold weather, minimum.

We will also need 300-400 molds.

Using hundreds of recycled water containers is my answer. Polyethylene terephthalate bottles can be easily re-formed into hexagonal buckets with the right bevel angle. They can all be the same, and they nest, so they can be stored in a very small space. The process to make one takes about 5 minutes per.bucket, so if you do one per day, it is barely noticeable, and after a year you have 365 buckets. Reasonable.

I had about 80 when I had a run of cold weather last winter, and was able to do two freezings. I saw what worked and what didn't with these containers, so could adjust the design slightly. I was able to extremely rapidly assemble a small section of dome, but was not able to construct a full dome with 3000 bricks.

I’m all set for the coming winter.