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December 6

Breaking down a gyroid into layers

I'm trying to figure out how a gyroid surface, expressed as

$\sin x\cos y+\sin y\cos z+\sin z\cos x=0$

can be sliced horizontally.

This is a solved problem, as illustrated by the figures below. The first image shows a cross section of a cube generated by a 3D printing slicer to fill a cube with a gyroid infill, clearly showing that the slicer software figured out how to break the surface down into horizontal layers. The second image shows an animation of a gyroid surface being built up vertically in steps. The third is a unit cell, which would be ideal if I could figure out how to create it with some thickness, but I thought maybe the slicing approach would be simpler. I could be wrong though.

• 
  A 5% gyroid infill cross section from 3D slicer software
• 
  A gyroid surface built up in horizontal steps
• 
  A gyroid unit cell (3D interactive)

My goal here is to see if I can create a model of a gyroid surface in OpenSCAD, which I can then use as I please as a component of a 3D object rather than structural infill. I don't know yet if that is even feasible, but if I can generate coordinates of triangular facets, I can create any object.

The equation is simple enough. To generate a layer, I just fix the z height to some value, and generate y values as a function of x. Using Wolfram Alpha to solve the equation above for y, I get:

$y=2\pi n+2{\tan }^{-1}\left(\frac{\cos z\pm \sqrt{{\sin }^{2}x+{\cos }^{2}z-{\cos }^{2}x{\sin }^{2}z}}{\sin x-\cos x\sin z}\right)$

The problem is, when I try to plot this, I get curves with discontinuities that look sort of like a gyroid cross section in a piecewise fashion, but clearly aren't the same.

I've been searching for days and have not found any algorithm for generating vertices or facets or cross sections of a gyroid. And in spite of the seeming simplicity of the equation I am feeling somewhat stumped. How should I approach this? ~Anachronist (talk) 02:46, 6 December 2020 (UTC)

I have not looked into your solution, but here is how I'd solve the gyroid equation for $y.$

First, determine $\rho \ge 0$ and $\alpha$ (modulo $2\pi$) such that $\rho \cos \alpha =\cos z$ and $\rho \sin \alpha =\sin x.$ This can be done by taking

$\rho =\sqrt{{\cos }^{2}z+{\sin }^{2}x}$ and

$\alpha =\mathrm{a}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{2}(\sin x,\cos z).$

Then the equation can be rewritten as

$\rho \sin (y+\alpha )+\sin z\cos x=0.$

From this equation you can see a numerical problem arising when $\rho =0$ (or very close to it). The values of $\rho$ and $\sin z\cos x$ cannot simultaneously be zero, however; otherwise, all values for $y$ would solve the equation. But assume $|\sin z\cos x|<{\epsilon }^2$. Then $\mathrm{m}\mathrm{i}\mathrm{n}(|\sin z|,|\cos x|)<\epsilon$, so

$\rho \ge \mathrm{m}\mathrm{a}\mathrm{x}(|\cos z|,|\sin x|)=\sqrt{\mathrm{m}\mathrm{a}\mathrm{x}(1-{\sin }^{2}z,1-{\cos }^{2}x)}=\sqrt{1-\mathrm{m}\mathrm{i}\mathrm{n}(|\sin z|,|\cos x|{)}^2}>\sqrt{1-{\epsilon }^2}.$

The solution set for $y$ is empty whenever $\rho <|\sin z\cos x|,$ and the inequation implies that not attempting to solve such cases also will avoid dividing by values of $\rho$ that are (close to) zero.

Put $\eta =-\arcsin ({\rho }^{-1}\sin z\cos x).$ Then the solutions for $y$ are given by:

$y=\eta -\alpha +2\pi n,$ $y=\pi -\eta -\alpha +2\pi n.$

Taking the Minkowski sum of the set of triples $(x,y,z)$ $+$ a small ball will give this some thickness. Caveat lector: I have not tested this.  --Lambiam 13:12, 6 December 2020 (UTC)