I'm trying to find an algorithm to generate equidistant points along a hexagonal spiral for a 3d printing project and am curious if there's a simple algorithm for this. I've managed this with a circular spiral such as:

circular equidistant spiral

But would like to achieve the same in the format of:

hex spiral

With similar equidistant center points/vertices along the path.

  • 1
    $\begingroup$ does this help? $\endgroup$ Commented Aug 16, 2021 at 18:50

1 Answer 1

PyramidSprl=Table[k{Cos[k 2Pi/3.],Sin[k 2Pi/3.],- 0.9},{k,1,40,.5}]

Export["G:\ PyramidSprl.stl",%]

enter image description here

If you do not want 3d remove the Z-coordinate (-0.9) for flat spiral.

  • $\begingroup$ I need actual (x, y) coordinates for the calculations of each point - I've got to convert this into OpenSCAD to apply a bunch of other transformations to it. Is that just k*(cos(k*2*(pi/3))), k? What iterates k and how? $\endgroup$
    – CoryG
    Commented Aug 16, 2021 at 23:00
  • $\begingroup$ For reference, I'm essentially going for this in a hexagonal shape: i.imgur.com/nO6oaWr.png $\endgroup$
    – CoryG
    Commented Aug 16, 2021 at 23:04
  • $\begingroup$ Is there a way to get equidistant points from this? It doesn't appear to have enough parameters - but it needs to be more than just along the corners as well - and to account for offsets from the edges based on distance radially from point-to-point to be equidistant. $\endgroup$
    – CoryG
    Commented Aug 16, 2021 at 23:06
  • $\begingroup$ You need both (x,y) terms. For hexagon I took 3, half of 6 in a full rotation. I gave STL, you can also give OpenSCAD perhaps to 3d print. It is an Archimedean spiral. For spacing of sides multiply by $\sqrt{3}/2$ $\endgroup$
    – Narasimham
    Commented Aug 16, 2021 at 23:12
  • $\begingroup$ I may not have explained it properly, the result should be a series of points akin to the first image, but in the hexagonal spiral of the second one. $\endgroup$
    – CoryG
    Commented Aug 18, 2021 at 0:11

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