It’s been fun. Let me know if you get up to any other geometrical woodwork adventures.

]]>Amazing Ben, I really appreciate all the time you took to think about this, and your wealth of knowledge. I was sort of assuming it wouldn’t work, because like you said, there’d be a name for it, or I’d be able to find images. If I get around to making my art piece that I have in mind, I will be happy to share it with you. Thanks so much!

]]>I’ve had a good think about your question, and although I haven’t been able to reach a cast-iron conclusion, I believe that what you hope to do isn’t possible. Here’s the main line of reasoning (it’s a bit complicated I’m afraid):

Any polyhedron has faces (the flat bits), edges (the ridges where 2 faces meet) and vertices (the peaks where the corners of 3 or more faces meet). There’s a formula that is true for all polyhedra:

F + V – E = 2

where F, V and E are the numbers of faces, vertices, and edges in the polyhedron. For example, a cube has 6 faces, 8 vertices, and 12 edges:

6 + 8 – 12 = 2 as predicted.

Now consider a polyhedron where all the faces are quadrilaterals (4-sided shapes). Each face touches 4 edges, but each edge is shared with the neighbouring face, so the number of edges will only be twice the number of faces. So E = 2 times F.

Let’s suppose, just for the sake of argument, that the polyhedron has 40 four-sided faces. From the above, that means there’ll be 80 edges. Putting these number into the formula gives us

40 + V – 80 = 2

which means that V must 42, ie this polyhedron must have 42 vertices. (In general, for a polyhedron with 4-sided faces, the formula tells us that the number of vertices must be 2 greater than the number of faces).

Just for now, let’s ignore the extra 2 vertices and suppose that we’ve got 40 faces and 40 vertices. Now each face has 4 corners, and so must take part in 4 vertices. But because there are as many vertices as faces, this means that there must be 4 faces meeting at each vertex. Now in a flat tiling like this there’s no problem having 4 squares meeting at a point everywhere.

[Imagine a bathroom wall tiled in the usual way with square tiles. I didn’t realise I couldn’t get an image in the comment]

We could also imagine a distorted version where the tiles are all different non-square quadrilaterals. But there aren’t any peaks (vertices) – it’s flat. Now imagine trying to distort the tiles to make a peak in one place. The angles of the tiles meeting at this peak will have to be made less than 90 degrees. That’s not a problem in itself, but we want every tile to have a vertex at *each* of its corners, which means that the shapes of the faces need to be 4-sided and with angles less than 90 degrees at *every*corner. Maybe this is too strict – perhaps there can be a bit of give-and-take, with each face having some faces bigger than 90 degrees compensated by some being much smaller, such that the average size of the angles is less than 90 degrees. But it turns out that the internal angles of any quadrilateral add up to exactly 360 degrees. So the average angle at each corner must be 90 degrees. It can’t be less. So it’s impossible to have a polyhedron (or even part of one) made of 4-sided faces where each vertex is the meeting of 4 faces.

Now in fact we have 2 more vertices than have faces. This means that 2 of the vertices must have 3 faces meeting at them, not 4. So it lets us off the hook at 2 vertices but doesn’t solve the problem elsewhere.

For this reason, I think what you’re hoping to do isn’t possible. We can’t make the angles at the corners of each face small enough.

I hope that makes some sort of sense. It’s a rather elaborate argument!

There is another reason why I don’t think it’s possible: I think I would have heard of it. The polyhedron would surely have a name of its own, and I don’t know of one.

Having said all that, if in the future you send me a photo of a wooden polyhedron, the faces of which all have 4 sides, I will be delighted to have been proved wrong and will write a special blog post on the subject!

]]>I’ve been trying to design an icosahedron with random flat sides, and came across your 83 sided icosahedron, which is really cool, and basically what I’m envisioning to do. I’m a woodworker, and I’m trying to create a sculpture where the base of the project starts with an icosahedron like yours, and I then continue to build on from there. I was wondering how you can generate software to do that, as I’m not very experienced with digital design. I can use google sketchup, but my knowledge is fairly limited. What I’d like to do is make something like yours, but only using four sides per shape. Most icosahedron images I find, use either all triangles or polygons, but I’d like to make irregular squares. Would your software be able to do that? I’d like to create a sphere about roughly 10-12″ in diameter (304mm) where the shapes vary in size from like 1-3 inches, and print it all out on paper, and then recreate it in wood. If you have any advice or suggestions, please let me know. Thanks!

]]>I’m a bit puzzled about the units here. You mention the ridge push as being 2*10^12 Nm-1. The units here make sense to me as a force per unit of ridge length. Later on you give the same number but with different units: 2,000,000,000,000 Nm (per metre of ridge). This is (as you say in words) a *moment* per unit length of ridge. Is one of these an error or have I misunderstood?

And if it really is a moment per unit length of ridge, is the moment reckoned about an axis passing through the centre of the Earth? If that’s the case, then the force per unit length of the ridge would be about 300,000 N, which in the general context seems very small.

You mention that the main driving force is the pull of the subducting slab, by an order of magnitude. So is the plate overall under tension? Or at least, is a large part of it under tension? Because of the huge resisting forces on the plate, my intuition is that the plate would be under compression near the ridge, and under tension near the subduction zone, and that there would be a line somewhere between the two where the material was unstressed. Is this somewhere near the truth?

]]>Hi David,

Thanks for your comment and for taking the trouble to discuss the issues in such detail.

1) You are right – I’m guilty, if not of flat-earth assumptions, at least of unspoken flat-earth approximations. To see how much my result is in error, we could view my flat-earth linear calculation as a rotational one by considering the plate to be moving around an Earth-radius cylinder in an exactly circumferential direction at the same omega as the real plate. The use of a cylinder ensures that all parts of the plate are one Earth-radius from the axis of rotation. For a plate that is very thin compared to its distance from the axis of the cylinder, the KE would come out the same whether you considered it as a problem in rotational dynamics or linear dynamics. So my calculated KE will be in error by a factor which is the ratio of the MOI of the imaginary plate on the cylinder to the MOI of the real plate on the Earth with its compound curvature. I’m not sure that the very small value of omega excuses the error: all KEs, no matter how calculated, will scale as omega squared. I see what you mean about the moment of inertia being a pig to calculate; certainly not one for the faint of heart. I’m not even going to try, but my intuition is that my conclusion – that the KE of a drifting continent is laughably small – will remain unscathed. One other (ignored) complication of the spherical case is that the linear speed of the plate is different at different parts of the plate.

2) The lack of a reference frame has nagged at me in a vague sort of way for years. Except for the poles, we have no points of reference. But the using the poles leaves the whole rotational symmetry to deal with. And even the poles wander don’t they? But relative to what? I’ve just read that their wandering is regarded relative to the crust, in which case using them as reference points would be circular. Interesting stuff.

Interesting. However, there are a couple of problems.

1) you are dealing with a rigid body rotation (this is the definition of plate tectonics, using Euler’s theorem). The kinetic energy you want to calculate is thus the rotational kinetic energy which is 0.5 * I * omega ^ 2. Omega is the angular velocity vector of the plate, I is the moment of inertia. I is a pig to calculate, and means the equation can only be evaluated numerically. The number that results may not be much different from what you obtained however, since omega is a tiny number of degrees per million years in most cases.

2) The angular velocity needs to be defined relative to a global fixed reference frame independent of plate tectonics. This means something not attached to any moving plate, for instance, markers in the deep mantle. This is difficult stuff. We get large variances in the results for various methods of calculating “absolute” plate motions. A recent paper defined a particular set of deep hot spots in the mantle and fixed plate motions relative to these. The result for Eurasia interestingly, was that it is barely moving at all. This reference frame is probably the most “valid” kinetic reference frame you could derive. The fastest moving plate in this case is the Pacific (omega = 59.79°S, 97.344 E, 0.8023 °/Myr). By contrast, Eurasia has omega = (63.926°N, 6.837°W, 0.0638 °/Myr). Not that the kinetic energy depends not just on the rotation rate, but also the offset of the angular velocity vector from the plate’s centre of mass (because I = sigma (i) (m_i * r_i^2) where each r_i is the distance of some part of the plate of mass m_i to the axis of rotation (omega). ]]>

Hi Paul,

I’m delighted that you like my post and have made good use of the idea 🙂 If your students (or you!) come up with any spin-off ideas it’d be great to hear about them.

Cheers

Ben

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