3
$\begingroup$

I was looking at this neat page on logarithmic spiral tilings when a question popped up:

http://www.uwgb.edu/dutchs/symmetry/log-spir.htm

It seems that in all of the tilings shown, the area of each tile is exponentially increasing as a function of the distance to the origin. Are there any radial- or spiral-type tilings (or "tiling-like" configurations) in which the area of each tile is a polynomial function of the distance to the origin, say $r^{1/2}$ or $r^2$? I don't require similarity of shapes or anything, just a simple way to fill the plane with shapes possessing (or approximately possessing) this property. I guess it would be nice if the shapes stay reasonable rounded and convex (i.e. not really long and skinny).

Improve this question
$\endgroup$
1
  • $\begingroup$ Obviously, one can do a tiling with parallelograms (or triangle, or anything one can make a parallelogram from) and any rate of growth. But I do not see any other shapes. $\endgroup$
    – ε-δ
    Commented Dec 10, 2011 at 21:30

1 Answer 1

Reset to default
6
$\begingroup$


            Grunbaum and Shephard book cover
See Section 9.5, "Spiral Tilings," p.512ff. These remarkable tilings go back to H. Voderberg in the 1930's.

Improve this answer
$\endgroup$
5
  • $\begingroup$ In these tilings the size does not grow... $\endgroup$
    – Anton Petrunin
    Commented Dec 11, 2011 at 3:25
  • $\begingroup$ Yes, my point was that exponential growth in the OP's examples is not essential. And I guess $f(r)=1$ is technically polynomial growth! :-) $\endgroup$
    – Joseph O'Rourke
    Commented Dec 11, 2011 at 14:10
  • $\begingroup$ No, $f(r)=\exp(cr)$ with $c=0$ is exponential growth :) $\endgroup$
    – Piero D'Ancona
    Commented Dec 11, 2011 at 16:17
  • $\begingroup$ @Piero: Ha! Touché! $\endgroup$
    – Joseph O'Rourke
    Commented Dec 11, 2011 at 17:22
  • 1
    $\begingroup$ The cover of this book is frequently mistaken as a Voderberg tiling. In fact, it it not. The Voderberg tile is shown clearly in Fig. 3.2.4 and the tiling is in Fig. 9.5.1. $\endgroup$
    – Cye Waldman
    Commented Dec 22, 2017 at 17:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .