|
Author |
Message |
Christopher J. Henrich science forum Guru Wannabe
Joined: 03 May 2005
Posts: 107
|
Posted: Fri Jun 09, 2006 1:03 am Post subject:
Re: Multiple Reflections (Kaleidoscope) problem
|
|
|
In article <e6ad8q$8d6$1@reader2.panix.com>, Paul Ciszek
<nospam@nospam.com> wrote:
Quote: | My apollogies if this appears more than once; I am having
trouble posting it.
There must be a theorem of group theory somewhere that addresses
this issue.
Someone is trying to model an infinite 2-D grid of identical
features by modeling one of the features, and "reflecting" it
about each of the sides of a square. This is not quite the
same thing as periodic boundary conditions, but we hope it
will yield similar results. I realized though, that not all
2-D grids fare equally well under this treatment.
Imagine this: You draw a lower-case "b" on a piece of paper
and then surround it with four mirrors arranged in a square.
You will see something that looks like this:
b d b d b
p q p q p
b d b d b
p q p q p
b d b d b
This grid is self consistent; whether you look at the relection
of the north mirror in the west mirror, or the reflection of the
west mirror in the north mirror, you will see a "q" in the square
to the northwest of the original "b". Next, I looked at creating
a hexagonal grid by the same method. I think that a hexagon of
mirrors creates a hexagonal grid of reflections, but I'm not
absolutely sure. I do know, though, that the grid would not
be self-consistent--you can get to some cells by either an even
or an odd number of reflections, depending on the "path" taken.
Is there a simple way to determine whether a given "kaleidoscope",
i.e., a prism without ends constructed of mirrors, will result
in a self-consistent infinite grid or not? Is the grid produced
Yes, there is. |
In the plane, two mirrors (represented by half-lines) at an angle of
Pi/m will generate a "dihedral group" of order 2m. This is the symmetry
of many flowers and flwer-like figures.
Three mirrors, as the sides of a triangle, will generate a discrete
group of transformations if the angles between them are Pi/2, Pi/3,
Pi/4, or Pi/6. There are three possibilities: equilateral triangle,
right isosceles triangle, and half-of-an-equilateral triangle.
four mirrors, surrounding a rectangle, generate a discrete group.
A very good reference for this theory (which goes on into more
dimensions) is James E. Humphreys, _Reflection_ _Groups_ _and_
_Coxeter_ _Groups_ , Cambridge University Press.
Quote: | by an equilateral triangle of mirrors self-consistent?
|
--
Chris Henrich
http://www.mathinteract.com
God just doesn't fit inside a single religion. |
|
Back to top |
|
 |
Mike Amling science forum Guru
Joined: 05 May 2005
Posts: 525
|
Posted: Thu Jun 08, 2006 11:52 pm Post subject:
Multiple Reflections (Kaleidoscope) problem
|
|
|
My apollogies if this appears more than once; I am having
trouble posting it.
There must be a theorem of group theory somewhere that addresses
this issue.
Someone is trying to model an infinite 2-D grid of identical
features by modeling one of the features, and "reflecting" it
about each of the sides of a square. This is not quite the
same thing as periodic boundary conditions, but we hope it
will yield similar results. I realized though, that not all
2-D grids fare equally well under this treatment.
Imagine this: You draw a lower-case "b" on a piece of paper
and then surround it with four mirrors arranged in a square.
You will see something that looks like this:
b d b d b
p q p q p
b d b d b
p q p q p
b d b d b
This grid is self consistent; whether you look at the relection
of the north mirror in the west mirror, or the reflection of the
west mirror in the north mirror, you will see a "q" in the square
to the northwest of the original "b". Next, I looked at creating
a hexagonal grid by the same method. I think that a hexagon of
mirrors creates a hexagonal grid of reflections, but I'm not
absolutely sure. I do know, though, that the grid would not
be self-consistent--you can get to some cells by either an even
or an odd number of reflections, depending on the "path" taken.
Is there a simple way to determine whether a given "kaleidoscope",
i.e., a prism without ends constructed of mirrors, will result
in a self-consistent infinite grid or not? Is the grid produced
by an equilateral triangle of mirrors self-consistent?
--
Please reply to: | "Any sufficiently advanced incompetence is
pciszek at panix dot com | indistinguishable from malice."
Autoreply is disabled | |
|
Back to top |
|
 |
Google
|
|
Back to top |
|
 |
|
The time now is Tue Apr 24, 2018 4:34 am | All times are GMT
|
Copyright © 2004-2005 DeniX Solutions SRL
|
Other DeniX Solutions sites:
Electronics forum |
Medicine forum |
Unix/Linux blog |
Unix/Linux documentation |
Unix/Linux forums |
send newsletters
|
|
Powered by phpBB © 2001, 2005 phpBB Group
|
|