How can I plot a truncated cube with size (sqrt(2)-1)? The code I am using is simply
PolyhedronData["TruncatedCube"]
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1$\begingroup$ And by size you mean what? $\endgroup$– KubaCommented Jun 2, 2016 at 9:03
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1 Answer
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In case size means length of each edge:
edgeLength = PolyhedronData["TruncatedCube", "EdgeLengths"][[1]];
Graphics3D[
GeometricTransformation[
First @ PolyhedronData["TruncatedCube"],
ScalingTransform[{1, 1, 1} (Sqrt[2] - 1)/edgeLength]
]
,
Axes -> True
]
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2$\begingroup$ For people like moi who prefer to avoid the use of
GeometricTransformation[]:Graphics3D[ MapAt[# (Sqrt[2] - 1)/edgeLength &, PolyhedronData["TruncatedCube", "Faces"], 1], Axes -> True]$\endgroup$ Commented Jun 2, 2016 at 10:26 -
$\begingroup$ @J.M. one could simply use
Scaletoo. p.s. Why do you prefer to avoidGeometricTransformation? $\endgroup$– KubaCommented Jun 2, 2016 at 10:28 -
$\begingroup$ What if I need the side of the cube to be
sqrt(2)-1? $\endgroup$– randomalCommented Jun 2, 2016 at 10:31 -
3$\begingroup$ Yes, but
Scale[]is likeGeometricTransformation[]too; in my usual applications, I often apply replacement rules to the coordinates, andGeometricTransformation[]does not often play nice with such replacements. Another point against it for me is thatDiscretizeGraphics[]andBoundaryDiscretizeGraphics[]do not (currently) work withGeometricTransformation[]objects. $\endgroup$ Commented Jun 2, 2016 at 10:33 -
2$\begingroup$ @albus, did you notice the scaling factor used in Kuba's and my snippet? Replace that. $\endgroup$ Commented Jun 2, 2016 at 11:07