geometrymath-softwaretiling
Does anyone know if there's a client or web app that generates Penrose patterns which can then be converted to a tileable rectangular background image for web site?
I found this http://stephencollins.net/penrose but I can't seem to get a rectangular shape out of it for web use. Thanks.
I struggled with this one, but it finally became clear. I hope this answer helps someone else along the way.
The comment around the original MathForum question that suggested that one could calculate the volume of the kinds of solids described above by taking the hyperarea of the base and multiplying it by the average of the heights, while technically on the mark, its wording sent me down the wrong road.
To find the volume of these kinds of solids one needs to multiply the volume formula for a regular simplex (also shown in the original post):
times the average of the heights of the cap.
Calculating a solid with 3 heights to the cap will need a 2 dimensional regular simplex as a base (an equilateral triangle). In this case the formula for the volume of a regular simplex will give the triangle's area, hence the confusion about hyperarea.
When calculating for heights of n dimensions, the base simplex will have n-1 dimensions.
So, a little Mathematica function will do the trick:
volSimplexWithHyperCap[heights_List] := Module[{n, simplexVolume},
n = Length[heights] - 1;
simplexVolume = Sqrt[n + 1]/(n! * Sqrt[2^(n)]);
N[simplexVolume * Mean[heights]]]
volSimplexWithHyperCap[3, {1, 1, 1}]
0.433013
This corresponds correctly to the prism volume in the question. It also, readily extends to higher dimensions:
volSimplexWithHyperCap[{1, 3, 2, 3, 5}]
0.0652186
This has the advantage over integration of being relatively simple and very fast to calculate. Not a Cholesky decomposition solution like I imagined at first, but it looks serviceable.
Maybe @MvG will think of a proof.
Thanks to everyone for their patients with all the edits.
Best Answer
I believe, since Penrose tilings are aperiodic (lacking translational symmetry), there isn't such a rectangular shape.