Cuboctohedral Fourteen-sided Die

Cuboctohedral Fourteen-sided Die 3d printed Download
Cuboctohedral Fourteen-sided Die 3d printed Download
Cuboctohedral Fourteen-sided Die 3d printed Download
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About this Product

This is a fourteen-sided die with cuboctohedral symmetry. The hexagonal and square faces are exactly the same area, so they are equally likely landing positions. I've verified that it's fair to within 0.5% with 600 rolls of this die.

I'm making it available as a co-creator template so that others can use it as a base for their fourteen-sided die ideas.

There's a 0.1mm hole in one of the square faces to ensure that Shapeways' software sees it as hollow, as described here: . The hole is too small to be visible in the printed version.


IN: 0.726 w x 0.726 d x 0.726 h
CM: 1.844 w x 1.844 d x 1.844 h
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In many modeling programs (such as Blender) there is a function to display the surface area of each polygon, which is an easy way to verify exactly. Also, this model does have a 0.1 mm hold, so the software thinks it is hollow. However actually removing all the powder after printing would be all but impossible, as the hole that small would probably close up. See the design guidelines for more detail on how large the holes need to be to prevent this. A die partially filled with powder would not roll evenly (not to mention Shapeways would not be happy, as they want to recycle the unused powder).
May 12, 2012, 1:47 am
Yes, you can figure out the area of any polygon by breaking it down into triangles, that's how I made sure the irregular hexagonal sides were the same area as the squares. I have no idea how to figure out whether a given three-dimensional shape rolls fairly, though. It probably has to do with the face area, face perimeter, edge-to-edge angles, and, if the shape is irregular, the specific face configuration too. Since this shape is so regular, I just guessed that equal area would do the trick and verified it experimentally.
December 6, 2010, 9:37 pm
This is a nice shape. I wanted to do something like this (different shapes but with the same surface areas) but hadn't a clue how to work it out. Do you have to break the areas down into triangles first?
December 6, 2010, 6:32 pm
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