The Butterfly (Foldables 1)

We all know that cardboard cubes are rigid, which is why we get our packages in boxes. We also all know that if we remove two opposite faces from a cube, we can fold it together. This started to interest me when I noticed that the polyhedral approximation of the Schwarz P surface is surprisingly flexible. This summer, I showed this to our local Origami and Paper Folding expert, Jiangmei Wu from our School of Art and Design, and she became interested. A few days later she came with a paper model that looked like this:

ButterflyTriply

She called it a simple variation of the polyhedral P-surface. Hmm. This is a triply periodic polyhedral surfaces tiled with rhombi. To understand it, we build it out of smaller units (which we called butterflies):

Butterbuild

The really cool thing about it is that it can be folded together in two different ways, like so:

Butterdeform

You can find an animation showing the continuous deformation here. We stared at this for a (long) while, until we realized that this has to do with rhombic dodecahedra. The structures up above are composed of the rhomboids from last week that tile a rhombic dodecahedron. The latter has, as the name hints, 12 faces, which occur in opposite pairs. Like the cube, it is rigid per se, but becomes foldable if we remove two pairs of parallel faces, leaving us with four faces to use, which are distinguished by color up above.Fractal0b

Above you can see the four hollow parallelepipeds (which we called hollowpeds). The almost trivial but nevertheless mind bending realization is that everything you build out of these hollowpeds becomes a structure foldable in two different ways. Next week I’ll show Jiangmei’s second model, a foldable fractal… If you can’t wait, check out this.

 

 

 

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Beach Balls

Following Jack’s suggestion, let’s try to visualize holomorphic maps from the Riemann sphere to itself. As a reference, here is the Riemann sphere with polar coordinates, and on the right hand side its cylindrical projection.

Identity

The preimage of this under a quadratic polynomial looks like this:

Square

And here a rational function function of degree 4:

Quartic

One can get quick images using uv-mapping, but there are some rendering artifacts I don’t know how to get rid of yet. Degree 5:

Deg5

Finally, the Gamma function, which has an essential singularity at infinity.

Gamma

Martha says their gypsum printer can print these in color. We’ll see, I hope.

Julianna and Friends

I have written before about Sofia, one of the wonderful cheeses from Capriole Farm.

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Like her, Julianna (up above) is made from goat cheese, but comes with a nice herbal crust. It’s the stronger companion of the Old Kentucky Tome, which you find below to the right.

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There are other things from here I would like to take with me to my next life, whenever this will happen. The bread, for instance. American bread used to be the biggest nightmare in this country. Not anymore. One reason is the Muddy Fork Bakery that produces this Rustic Sourdough with a perfect crust,

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or the beautiful Sesame Spelt

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that goes so well with the goat cheese. All their breads are hand-made and wood fired. Amazing stuff. You can find both bread and cheese at the local Farmer’s Market or at Bloomingfoods.

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Stellated Triacontahedron

If you have mastered the Slidables from last year and had enough of the past gloomy posts, you are ready for this one.

Let’s begin with the rhombic triacontahedron, a zonohedron with 30 golden rhombi as faces. There are two types of vertices, 12 with valency 5, and 20 with valency 3. In the image below, the faces are colored with five colors, one of which is transparent.

Triacontahedron

The coloring is made a bit more explicit in the map of this polyhedron below.

Triagraph 01

We are going to make a paper model of one of the 358,833,072 stellations of it. This number comes from George Hart’s highly inspiring Virtual Polyhedra.

Stellatriacontahedron

In a stellation, one replaces each face of the original polyhedron by another polygon in the same plane, making sure that the result is still a polyhedron, possibly with self intersections.

Newface 01

In our case, each golden (or rather, gray) rhombus becomes a non convex 8-gon. The picture above serves as a template. You will need 30 of them, cut along the dark black edges. The slits will allow you to assemble the stellation without glue. Print 6 of each of the five colors:

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Now assemble five of them, one of each color, around a vertex. Note that there are different ways to put two together, make sure that the original golden rhombi always have acute vertices meeting acute vertices. This produces the first layer.

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The next layer of five templates takes care of the 3-valent vertices of the first layer. Here the coloring starts to play a role.

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The third layer is the trickiest, because you have to add 10 templates, making vertices of valency 5 again. The next image shows how to pick the colors to maintain consistency.

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Below is the inside of the completed third layer.

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Two more to go. Layer 4 is easy:

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The last layer is again a bit tricky again, but just because it gets tight. Here is my finished model. It is quite stable.

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Another Brick in the Wall

When Apple announced in July this year they had sold 1 billion iPhones, I started wondering about another brick maker: How many blocks has Lego made? Their friendly customer service couldn’t tell me how many elements they have made in total, but the yearly production is 19 billion. Scary. Unfortunately, the shape of the standard lego brick is too limited for my needs. For a long time, I had wanted a lego brick in the shape of a rhombic dodecahedron (better would be a four dimensional lego hypercube of which the rhombic dodecahedron is a mere shadow, but let’s not be delusional). As you can see, this polyhedron tiles space as well if not better than the cube.

RhombicDodecahedronTiling

Various companies have produced shapes with more or less cleverly embedded magnets, but keeping track of the polarity on all faces of a 12 sided object is tricky. And this would be a lot of magnets. The actual problem, however, is the enormous amount of choices one has: 12 faces to attach to is just too much. I strongly believe that Lego’s success stems from the fact that they have reduced the number of possible ways how you can attach two lego pieces dramatically. No choice means dictatorship, two choices US capitalism, but more choices sounds like European liberalism or even anarchism, and we see where that leads.

This gave me the idea to replace the complicated rhombic dodecahedron by a simple object that is less attachable. Here is the new brick.

Brick

To make it, take three faces of the rhombic dodecahedron that are symmetrically positioned, and replace each of the three rhombi by its inscribed ellipse. Then take the convex hull of the ellipses. The resulting shape consists of the ellipses, two equilateral triangles in parallel planes, and three intrinsically flat mantel pieces.

You will notice that there are two versions of this brick, a left and a right handed one. This leaves just the right amount of choices.

Hexring

If you alternatingly attach a left to a right brick, you get a hexagonal annulus. Remember that we are still tiling space using slimmed down versions of the rhombic dodecahedron. Due to our imposed limitation of choice, nor every place can be reached anymore. The hexagonal annulus is a little simplistic. What do we get if we just use the left handed brick?

Prestar2

Let’s start with a red central brick, attach a brick on all three sides, and another six at the free faces of the new bricks. We notice that the bricks can occur in four different rotated positions. I have distinguished them by color. Add another 12 bricks:

Prestar

And another 24. No worry, no intersections can occur, because, I insist, we just tile a portion of space with rhombic dodecahedra.

Star

Now we see that the tree like structure we have produced so far does not persist. In the next generation, we obtain closed cycles of length 10, and we finally recognize the Laves graph.

Ball

In the very near future you will see what else one can make with these bricks.

Ragged Rectangles (From the Pillowbook II)

In a ragged rectangle, the sides zigzag diagonally as in the left figure below, which shows a ragged rectangle of dimensions 6⨉7, and within a ragged 3⨉3 square. Note that the boundary changes directions at every unit step. These shapes make interesting candidates for regions to be tiled with polyominoes. The example in this post illustrates nicely how the interplay between making examples and generalization leads to a miniature theory.

Raggedex 01

To tile a shape like this with polyominoes, it will help to know its area in terms of unit squares. This is easy: If you color the squares in a ragged a⨉b rectangle beige and brown, you will get a⨉b squares of one color, and (a-1)⨉(b-1) squares of the other color.

This right away shows that it is hopeless to tile a ragged rectangle with dominoes. The first really interesting case is to use L-trominoes. The area formula implies that we need one dimension of the rectangle to be divisible by 3, and the other to leave remainder 1 after division by 3. Thus the shortest edge that can occur has length 3, and the other them must have length 3n+1. The figure below shows how to tile any ragged rectangle of dimensions 3x(3n+1) with L-trominoes:

ragged3-01.jpg

The next shortest edge possible has length 4, and then the other edge must have length 3n. Again, a few experiments lead to a general pattern which shows that any 4x(3n) ragged rectangle can be tiled with L-trominoes:

Ragged4

This covers the two basic kinds of thin and arbitrarily long rectangles. What about larger dimensions? If we already have a ragged rectangle tiled with L-trominoes, we can put a frame around it that is also tiled with L-trominoes:

Raggedframe

These three constructions together show that a ragged rectangle can be tiled with L-trominoes if and only if its area is divisible by 3. Next time we will see how this helps us to tile curvy rectangles with pillows.

The Helicoid (again!)

In 1760, Leonhard Euler studied the curvature of intersections of a surface with planes perpendicular to the surface, and showed that the maximal and minimal values of their curvature are attained along orthogonal curves. In 1776, Jean Baptiste Marie Charles Meusnier de la Place showed that for minimal surfaces these principal curvatures are equal with opposite sign. He went on to show that both the catenoid and the helicoid satisfy this condition, thus exhibiting the first two non-trivial examples of minimal surfaces. Euler had discussed the catenoid as a minimal surface before, but only in the context of surfaces of revolution.

In its standard representation as a ruled surface, the parameter lines are the asymptotic lines of the helicoid. For a change, here is the helicoid parametrized by its curvature lines:

Helicurvature

The purpose of this note is a little craft, similar to what I explained earlier using Enneper’s surface: A ruled surface that has as directrix a curvature line of a given surface, and as generators the surface normals, will be flat and can thus be constructed by bending a strip of paper. Doing this for an entire rectangular grid of curvature lines results (for the helicoid) in an attractive object like this one:

Helicoid

To make a paper model, one first needs to find planar isometric copies of the ribbons. This is done by computing the geodesic curvature of the curvature lines of the helicoid, and, using the fundamental theorem of plane curves, then finding a planar curve with the same curvature. The (planar) ribbon is then bounded by parallel curves of this plane curve:

Js 01

Using four (due to the inevitable symmetry of things) copies of the template above, carefully cut out & slit, allows you to easily build the model below, which also makes a nice pendant. Print out the template so that the smallest distance between two slits is not much wider then your fingers, otherwise assembling the pieces will be tricky.

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Begin with the largest J-piece and use the four copies to build a frame, by sliding the hook into hook and non-hook into non-hook. Then continue inwards, adding four copies of the second largest J, by placing the hook of a new J next to a hook of the old J.

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