Re: Eikosany tumbling in 5D

Sorry Folks, that should have been

http://dkeenan.com/Music/EikosanyRotation.xls

-- Dave Keenan
http://dkeenan.com

David, it looks like you picked one particular set of lines and fail to ever
rotate into the configuration where that set becomes a donut. I see two
points that are always on the horizontal median and are always at opposite
positions relative to the center.

How about including _all_ the lines?

-----Original Message-----
From: David C Keenan [mailto:D.KEENAN@UQ.NET.AU]
Sent: Friday, October 20, 2000 11:57 PM
To: tuning@egroups.com
Subject: [tuning] Re: Eikosany tumbling in 5D

Sorry Folks, that should have been

http://dkeenan.com/Music/EikosanyRotation.xls

-- Dave Keenan
http://dkeenan.com

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I'm guessing that this has something to do with an Eikosany:

http://members.tripod.com/vismath2/ogawa/3da.htm#f4

Paut Erlich wrote:

>David, it looks like you picked one particular set of lines and fail to ever
>rotate into the configuration where that set becomes a donut. I see two
>points that are always on the horizontal median and are always at opposite
>positions relative to the center.

Yes. I realised that I'm only doing 4 rotations, when I should be doing 9
(out of 10).

>How about including _all_ the lines?

Is the eikosany the 5D generalised octahedron? Am I correct in supposing
that it has 60 edges, and so I "only" have to add another 30?

-- Dave Keenan
http://dkeenan.com

Dave Keenan wrote,

>Yes. I realised that I'm only doing 4 rotations, when I should be doing 9
>(out of 10).

Cool. Looking forward to the fix.

>Is the eikosany the 5D generalised octahedron?

No. I've been baffled by this for hours. Apparantly the 5D generalized
octahedron has 10 vertices, not 20. See

http://www.friesian.com/polyhedr.htm

But maybe this has something to do with it -- remember, the Eikosany has 30
tetrads.

http://members.tripod.com/vismath2/ogawa/3da.htm#f4

>Am I correct in supposing
>that it has 60 edges, and so I "only" have to add another 30?

It has 90 edges. Each note (vertex) is connected to 9 others, not 3 others,
so you need to triple the number of edges. Can you use different colors or
dotted lines?

Paut Erlich wrote:

>>Yes. I realised that I'm only doing 4 rotations, when I should be doing 9
>>(out of 10).
>
>Cool. Looking forward to the fix.

Don't hold your breath. I need a break from this stuff.

>>Is the eikosany the 5D generalised octahedron?
>
>No. I've been baffled by this for hours. Apparantly the 5D generalized
>octahedron has 10 vertices, not 20. See

>But maybe this has something to do with it -- remember, the Eikosany has 30
>tetrads.
>
>http://members.tripod.com/vismath2/ogawa/3da.htm#f4

Maybe. But I can't see how.

>>Am I correct in supposing
>>that it has 60 edges, and so I "only" have to add another 30?
>
>It has 90 edges. Each note (vertex) is connected to 9 others, not 3 others,
>so you need to triple the number of edges.

Damn. Again, don't hold your breath.

>Can you use different colors or dotted lines?

Yes, either or both. If each colour is a separate "series" (in Excel
terminology). Maybe I'll make it 3 series. I've just got to figure out the
vertex sequences now. Sigh. Of course, _you_ could do it if you are
impatient. :-)

Good night, and thanks for all your help.

-- Dave Keenan
http://dkeenan.com

I should have described Oscar Lanzi's matrix as

M = (I + f(n)*E)/sqrt(2)

where
n is the number of dimensions
I is the nxn identity matrix [not 3x3 matrix]
E is the nxn matrix consisting entirely of 1's [not 3x3 matrix]
f(n) = (sqrt(n+1)-1)/n

It is not the same as Keenan Pepper's solution since Pepper's preserves the
X axis (an edge of the simplex (generalised tetrahedron)), while Lanzi's
preserves the line through the origin and the point (1,1,1,....) (an axis
of symmetry of the simplex).

I have not checked either of them.

-- Dave Keenan
http://dkeenan.com

--- In tuning@egroups.com, David C Keenan <D.KEENAN@U...> wrote:
> I should have described Oscar Lanzi's matrix as
>
> M = (I + f(n)*E)/sqrt(2)
>
> where
> n is the number of dimensions
> I is the nxn identity matrix [not 3x3
matrix]
> E is the nxn matrix consisting entirely of 1's [not 3x3
matrix]
> f(n) = (sqrt(n+1)-1)/n
>
> It is not the same as Keenan Pepper's solution since Pepper's
preserves the
> X axis (an edge of the simplex (generalised tetrahedron)), while
Lanzi's
> preserves the line through the origin and the point (1,1,1,....)
(an axis
> of symmetry of the simplex).
>
> I have not checked either of them.

So, Dave, what did you think of my 3)5 dekany movie? Can you help me
make it rotate
through the fourth dimension, such that one of the "squashed"
hexanies becomes the
"big" one, and vice versa? Or can you implement that yourself in
Excel?