It’s that time of the year when I like to evade all the hectic and retreat to a little craft. One of my favorites to make is Alan Schoen’s uniform gyroid, which he discovered on his way from the Laves Graph to the minimal gyroid.

This is a triply periodic uniform polyhedron of type 3-3-6-3-4, meaning that all vertices are equivalent by a motion of space, and at each vertex meet triangle, triangle, hexagon, triangle, square, always in this order.

The triangles and hexagons group together to planar six pointed stars, which I prefer for model making, because it makes the model more sturdy and is less work… If you want to make your own, you need to print the templates on colored card stock. Make sure not to scale the pdfs below so that squares and stars have the same edge lengths. I use two different colors for the stars so that touching stars get different colors.

Take a star, attach a square to every other edge, bending the squares alternatingly up and down. Then attach six more stars to the free edges of the first star, fitting them to one free edge of one of the squares each, as shown above.

Two copies of this piece (without the downward pointing stars and and squares) make a translational fundamental piece of the uniform gyroid. Then you keep going until you get tired or run out of paper. If you have access to a a hyperbolic universe, you can also make a hyperbolic version (which would make a wonderful carpet design):


Printable PDF templates of stars and squares

2 thoughts on “Make Your Own Gyroid

  1. Dear Jayadev,

    Glad to hear you like it. I don’t think that taking the 12th power of the cone metric makes it automatically holomorphic, there can still be holonomy. I suspect that you need a 24th power to eliminate that. My count is that you need 8 hexagons, 12 squares, and 48 triangles for a fundamental piece. I want a large steel sculpture of this thing…

  2. Dear Matthias,

    Thanks for this lovely post! I’m going to make a wooden version of this using a laser cutter. I also am having a bit of trouble figuring out the genus of the quotient- at each vertex the total angle is 13pi/6, which from my perspective makes the naive Euclidean structure that of a 12-differential and each vertex yields a simple zero. Then the number of vertices V has to satisfy V = 12 (2g-2) = 24g-24. My current count is that there are 24 vertices, yielding g=2 for the quotient.


    P.S. Your student Dami Lee is currently working with me as a postdoc at UW

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