It is a given that the most efficient regular polygon to completely fill an area (tessellate) is the hexagon. It has an angle of 120deg.
IIRC, the most efficient regular polyhedron to fill a volume is the dodecahedron
It has an angle of 108deg.: if each vertex meets four others in a tetrahedral angle
109.4712deg…well, that’s close enough for government work.
What polyXXX possessing what angle degree is optimum for 4-space?
What is more efficient about filling a space with a (regular) hexagon, as opposed to a triangle or square? All will fill 100% of a plane.
Also, what’s efficient about the dodecahedron? It won’t fill space without gaps. A tetrahedron or cube will, though.
A 4-cube will fill 4D space efficiently. I’d suppose a “hyperpyramid” would as well, though I’m not certain about that.
[QUOTE=Enola Straight]
It is a given that the most efficient regular polygon to completely fill an area (tessellate) is the hexagon. It has an angle of 120deg.]
[/QUOTE]
Like Dr. Strangelove above, I don’t know what you mean by ‘most efficient’. Triangles, squares, and hexagons will all form a regular tessellation of the plane and thus have packing density 1. In R^3, the only tesselation is by cubes (and not by solid dodecahedra; the solid angle at each vertex of the dodecahedron is not a rational multiple of 4\pi, so they can’t fill the space around a vertex). In R^4, there are three possibilities. (For what it’s worth, there are are other possibilities in H^n, including a regular tesselation by (hyperbolic) dodecahedra in H^3).
Assume you’re going to be paying for each of those lines.
On that basis, I should have thought the square would be best. You could make 100 squares with just 22 lines. The hexagons, at any rate, would involve drawing a heck of a lot more line segments.
Fair enough. In terms of perimeter, the hexagon is best. I guess the further extension is for area and volume for 3D and 4D.
And I stand corrected on 3D: as **Itself ** said, only the cube will tile space; the tetrahedron won’t. So, the answer is easy there.
FWIW, the equivalent of the polyhedra 4D are called polychoron, and a tiling of a pace is called (appropriately enough) a honeycomb.
Bees knew this long before Euclid did. When I was in 7th grade (circa 1964) our math teacher showed us a film. (Note that films are very uncommon in math class – they are much more common in social studies classes.) It was called “The Mathematics of the Honeycomb”, and it showed a whole bunch of ways that hexagonal beehive combs are optimal – most storage space per amount of wax required; greatest strength ditto; whole bunch of other optimal engineering specifications.
The regular old dodecahedron won’t fill space, but the rhombic dodecahedron will. However, it’s not the most efficient way of doing so. Kelvin conjectured (in 1887) that the most efficient cell that could fill 3D space was a version of the truncated octahedron, and this result stood for over 100 years before being improved upon in 1994.
Ok, I’ll play. What was discovered in 1994 that is better than a truncated octohedron?
The Weiare-Phelan structure, which consists of two different types of cells (some 12-sided, some 14-sided, and none of whose faces are regular polyhedra.) I don’t think it’s been improved upon since, but I don’t think it has been proved to be the best possible either.
Oh, and I first learned about it via Larry Gonick’s “Science Classics” cartoons in Discover magazine in the mid-'90s. Here’s the one on the Weaire-Phelan foam: Part 1, Part 2.
Yeah, but back to my OP…
A 1D line, there can only be one angle…180 deg.
a 2D plane, 120 deg for the hexagon
a 3d space, 108/109.4712 for the dodecahedron/tetrahedron.
The angle in 4space is…?
You’re going to have to explain what you mean by “the angle” a little better. As has been noted, regular dodecahedrons (with all angles equal) don’t fill space the way you’ve described them. Rhombic dodecahedrons do fill space—see my post above—but the angles between the edges of the cells vary since the faces are rhombi (about 70.53° for the acute angles, 109.47° for the obtuse.) Similarly, the angles between the edges of the truncated octahedron are all 90° or 120° — again, the angles between the meeting at a given point in space varies.
The point here is that there are many different “foams” that fill n-dimensional space, and that even if you say that you’re interested in the “most efficient” one, the edges that converge at a given vertex are not all at the same angle to each other. Asking what “the angle” is assumes that there’s only one angle worth talking about, which is not true in general.
If you’re looking for the interior angle between the regular N-dimensional simplex (N-dimensional analogue of a regular tetrahedron or equilateral triangle), the formula is
atan(1 / sqrt(N^2 - 1)) + 90 (degrees)
So:
2D: atan(1/sqrt(3))*180/pi+90 = 120
3D: atan(1/sqrt(8))*180/pi+90 = 109.4712206345
4D: atan(1/sqrt(15))*180/pi+90 = 104.4775121859
This approaches 90 degrees in the limit of large dimensions.
:smack:
Or asin(1 / N) + 90 (degrees)
So:
2D: asin(1/2)*180/pi+90 = 120
3D: asin(1/3)*180/pi+90 = 109.4712206345
4D: asin(1/4)*180/pi+90 = 104.4775121859
Now THATS the kind of answer I was looking for!
Booyah!
Just realised something:
do we encounter the 104.4775 deg angle anywhere in nature?
If so, could this be A 4D object intersecting our 3D reality?
The only reference I could find anywhere to that angle is a packing of points on a sphere in four dimensions. Couldn’t find any reference to 3 dimensions.
Update:
The 104.4775 deg angle is found in the common molecule of H2O!