This method is sort of a clunky way to make
a triangular 3d von Koch using Cuboids.
It doesn't start to look triangular except by level 3.
What I did was take a triangular carpet tetrahedron ( 10 cubes , ratio 3)
and added a central cube and face cubes
to get this at 15 cubes and a ratio of 4.
If I could get the Mathematica Tetrahedron[] graphic primitive to work
like
a Cuboid[] , I would have used it instead.
Essentually what I'm doing if a three relative size tetrahedron with a
one size
tetrahedron on each face ( sort of a 3d Jewish Star) using cubes
instead of tetrahedrons.
Picture:
http://profile.imeem.com/GUmj0c/photo/_BKRMfb9tf/
Mathematica code:
Clear[pieces, menger]
(* 3d tetrahedron/ cuboid triangular von Koch*)
(* Roger Bagula 09 sept 2007©*)
pieces =
{{1, 0, 1}, {-1, 0, 1}, {0, -1, -1}, {0, 1, -1}, {0, 0, 1}, {0, 0, -1},
{0, \
0, 0}, {0, 1.5, 0}, {0, -1.5, 0}, {
1.5, 0, 0}, {-1.5, 0,
0}, {1/2, 1/2, 0}, {1/2, -1/2, 0}, {-1/2, 1/2, 0}, {-1/2, -1/2, 0}};
N[Log[Length[pieces]]/Log[4]]
1.9534452978042591`
menger[cornerPt_, sideLen_, n_] :=
menger[cornerPt + #1*(sideLen/4), sideLen/4, n - 1] & /@[EMAIL PROTECTED]
pieces;
menger[cornerPt_, sideLen_, 0] :=
{EdgeForm[], Cuboid[ cornerPt , cornerPt + sideLen*{1, 1, 1}]};
Show[Graphics3D[Flatten[menger[{0, 0, 0}, 1, 1]]], Boxed -> False]
gr = Show[Graphics3D[Flatten[menger[{0, 0, 0}, 1, 3]]], Boxed -> False]
Show[gr, ViewPoint -> {-0.002, 1.297, 3.125}]
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