Re: [math-fun] tetraroller volume
DanA wrote:
<< was able to figure out the face structure of the four equal cylinders (which could be said to intersect as do the great diagonals of a cube). Same method should apply to the higherhedra.
Here's the method (but I've patented it, and you may need to pay me royalties):
Yeah but you didn't say <below> is the projection of a cuboctahedron onto a concentric sphere, so I don't need no steenking license agreement!
Suppose n solid infinite cylinders of unit radius are arranged in 3-space so their axes each pass through the origin.
Then the solid K where they all intersect has boundary surface N, where N has "faces" that are maximal portions of 2D unit cylinders.
DanA> Imagine the intersection of N with the unit
sphere S^2 about the origin. That will be a number of great circles of S^2.
This isn't right; rather it's just each 2D cylinder that intersects the unit sphere in a great circle.
Now look at how each of these great circles on S^2 is cut by the others; each resulting arc corresponds to one "face" of the boundary of the intersection solid.
I hope that was clearer and truer.
--Dan These circles cut each other up into (uncut) great circle arcs. Each of these occurs in just one face of N. So # faces of N = # largest uncut great circle arcs on S^2.
The faces will resemble Voronoi regions of the great circle arcs. In a symmetrical situation, e.g., where the axes are the diagonals of a higherhedron, this will be exact (if the faces of N are radially projected onto S^2).
E.g., for 4 main diagonals of a cube, we can pretend the cube is a unit sphere, and the great circles will be equators w.r.t. opposite pairs of cube vertices -- giving us 4 regular hexagons inscribed on the cube. Each hexagon is cut in sixths by the other hexagons, giving us 24 "great circle" arcs, hence 24 faces of the intersection solid K. The Voronoi regions on the sphere are each a quadrilateral with one axis of symmetry.
--Dan I believe that's a deltoidal icositetrahedron with faces bulged out to cylindrical. It would need a pretty shallow grade and some friction not to start rolling like a sphere. --rwg _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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DanA's great-circle/Voronoi argument seems to give 4(n choose 2) as the number of faces created by intersecting n cylinders. n 2 3 4 6 10 F 4 12 24 40 60 84 112 144 180 A 180-hedron! Is there any limit to the number of congruent faces on a polyhedron? E.g., recursively erecting pyramids of the right shape gives the sequence tetrahedron, cube, rhombic dodecahedron, then our recent friend, the deltoidal icositetrahedron. But the succeeding 48-hedron has faces of two types. Perhaps Eric's Rhombic Dodecahedron article http://mathworld.wolfram.com/RhombicDodecahedron.html could mention that if you place a nitrogen at one of the eight vertices of valence three (which form a cube) and hydrogens at the adjacent vertices, you get an anatomically correct ammonia molecule. I.e., those four vertices form the center and three vertices of a regular tetrahedron. I wonder if anyone has synthesized C6N8 with (strained) carbons at the vertices of valence 4. Somehow removing those carbons would leave the elusive (and explosive) octaazocubane. --rwg AMERICAN CLOTHS - ACHROMATIC LENS
DanA wrote:
<< was able to figure out the face structure of the four equal cylinders (which could be said to intersect as do the great diagonals of a cube). Same method should apply to the higherhedra.
Here's the method (but I've patented it, and you may need to pay me royalties):
Yeah but you didn't say <below> is the projection of a cuboctahedron onto a concentric sphere, so I don't need no steenking license agreement!
Suppose n solid infinite cylinders of unit radius are arranged in 3-space so their axes each pass through the origin.
Then the solid K where they all intersect has boundary surface N, where N has "faces" that are maximal portions of 2D unit cylinders.
DanA> Imagine the intersection of N with the unit
sphere S^2 about the origin. That will be a number of great circles of S^2.
This isn't right; rather it's just each 2D cylinder that intersects the unit sphere in a great circle.
Now look at how each of these great circles on S^2 is cut by the others; each resulting arc corresponds to one "face" of the boundary of the intersection solid.
I hope that was clearer and truer.
--Dan These circles cut each other up into (uncut) great circle arcs. Each of these occurs in just one face of N. So # faces of N = # largest uncut great circle arcs on S^2.
The faces will resemble Voronoi regions of the great circle arcs. In a symmetrical situation, e.g., where the axes are the diagonals of a higherhedron, this will be exact (if the faces of N are radially projected onto S^2).
E.g., for 4 main diagonals of a cube, we can pretend the cube is a unit sphere, and the great circles will be equators w.r.t. opposite pairs of cube vertices -- giving us 4 regular hexagons inscribed on the cube. Each hexagon is cut in sixths by the other hexagons, giving us 24 "great circle" arcs, hence 24 faces of the intersection solid K. The Voronoi regions on the sphere are each a quadrilateral with one axis of symmetry.
--Dan I believe that's a deltoidal icositetrahedron with faces bulged out to cylindrical. It would need a pretty shallow grade and some friction not to start rolling like a sphere. --rwg _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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On Wednesday 12 November 2008, rwg@sdf.lonestar.org wrote:
A 180-hedron! Is there any limit to the number of congruent faces on a polyhedron?
No. Take a skew prism -- lots of congruent triangular faces -- and put a cap on each end made out of triangles of the same dimensions. -- g
DanA's great-circle/Voronoi argument seems to give 4(n choose 2) as the number of faces created by intersecting n cylinders.
n 2 3 4 6 10 F 4 12 24 40 60 84 112 144 180
A 180-hedron! Is there any limit to the number of congruent faces on a polyhedron? E.g., recursively erecting pyramids of the right shape gives the sequence tetrahedron, cube, rhombic dodecahedron, then our recent friend, the deltoidal icositetrahedron. But the succeeding 48-hedron has faces of two types.
Perhaps Eric's Rhombic Dodecahedron article http://mathworld.wolfram.com/RhombicDodecahedron.html could mention that if you place a nitrogen at one of the eight vertices of valence three (which form a cube) and hydrogens at the adjacent vertices, you get an anatomically correct ammonia molecule. I.e., those four vertices form the center and three vertices of a regular tetrahedron.
Tetrahedral minifact: Resting on a flat surface, the (dihedral) angle between the surface and a (sloping) face equals the bond angle (between rays from the center to two vertices), namely pi - asec 3. Is this equality geometrically obvious? --rwg REPLICATILE PERCIATELLI
On 11/14/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
Tetrahedral minifact: Resting on a flat surface, the (dihedral) angle between the surface and a (sloping) face equals the bond angle (between rays from the center to two vertices), namely pi - asec 3. Is this equality geometrically obvious?
Consider a "stella octangula" comprising a pair of interpenetrating regular tetrahedra with common centre, and edges bisecting one another perpendicularly in pairs. The joins of the centre to any two vertices of one tetrahedron, together with the perpendiculars from the corresponding edge mid-point along the faces of the other tetrahedron, form a plane cyclic quadrilateral --- the other two angles being right-angles, by symmetry. Hence the sum of the (interior) dihedral and bond angles equals \pi; and of course sum of the former and exterior dihedral --- which presumably you had in mind above --- also equals \pi. WFL
On 11/14/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
Tetrahedral minifact: Resting on a flat surface, the (dihedral) angle between the surface and a (sloping) face equals the bond angle (between rays from the center to two vertices), namely pi - asec 3. Is this equality geometrically obvious?
Consider a "stella octangula" comprising a pair of interpenetrating regular tetrahedra with common centre, and edges bisecting one another perpendicularly in pairs.
Before learning this term, our Esteemed Moderator used the logically incongruous but delightfully intuitive "tetrahedron of David". By now everyone must know that iterating the Snowflake recursion on this figure produces (something whose closure is) merely a cube instead of a fractal. But the boundary of the construction using open tetrahedra is an interesting Cantor set, (strongly?) resembling eight corners cut off a Menger sponge. (The reason this limit, unlike a "spacefilling" polygon sequence, is defined is that it is a convergent, cumulative union.)
The joins of the centre to any two vertices of one tetrahedron, together with the perpendiculars from the corresponding edge mid-point along the faces of the other tetrahedron, form a plane cyclic quadrilateral --- the other two angles being right-angles, by symmetry.
Hence the sum of the (interior) dihedral and bond angles equals \pi; and of course sum of the former and exterior dihedral --- which presumably you had in mind above --- also equals \pi.
WFL
Very nice.
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