60 degree coordinates

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60 degree coordinates

Postby Dick Fischbeck » Fri May 14, 2004 6:02 pm

Hi

New here. I read at the intro page that, "The universe that we live in has only three spatial dimensions. We are limited to length, width, and height, and we can only travel along three perpendicular paths."

HAs anyone here considered that 4 sixty degree coordinates might work better for polychoron that the standard 3 ninty degree coordinates? Fuller claims the tet is the model to base our measuring on, not the cube.

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Postby Geosphere » Fri May 14, 2004 6:45 pm

It would work different, not necessarily better.
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Re: 60 degree coordinates

Postby pat » Fri May 14, 2004 7:31 pm

Dick Fischbeck wrote:HAs anyone here considered that 4 sixty degree coordinates might work better for polychoron that the standard 3 ninty degree coordinates? Fuller claims the tet is the model to base our measuring on, not the cube.


For polychorons, one needs four ninety-degree coordinates or four sixty-degree coordinates. As Geosphere mentioned, it'd be different, but not necessarily better.

It is true that the tetrahedron is more fundamental than the cube. For that reason, it is tempting to approach things from a tetrahedron-based coordinate system. A tetrahedron-based coordinate system is not as conducive (I don't think) if we intend to think of a polytope as a bounded intersection of half-spaces.

To deal with half-spaces, one pretty much needs the concept of 'orthogonal to a given vector'. Orthogonality calculations are, unsurprisingly, much more convenient in an orthogonal coordinate system than in a tet-based one.
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60 degree coordinates

Postby Dick Fischbeck » Sat May 15, 2004 3:48 pm

Hello Pat

"To deal with half-spaces, one pretty much needs the concept of 'orthogonal to a given vector'. "

Can you please explain more about what this means?

I am exploring what half-space is.

Thanks
Dick

Edit by BobXP: Removed the excess quotation from the end of your post. I assume it was accidentally put there.
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Re: 60 degree coordinates

Postby pat » Sun May 16, 2004 2:21 am

Dick Fischbeck wrote:To deal with half-spaces, one pretty much needs the concept of 'orthogonal to a given vector'.


A half-space in our three-dimensional world would be some plane and all of the volume below it. A typical way to represent a plane in three-dimensions is to specify a vector which is perpendicular to all lines in the plane and then specify a point that is in the plane. Orthogonal and perpendicular mean the same thing (at least in Euclidean geometry).

So, for example, I might divide the Universe into two halves. There is all of the Universe that is above my floor and all of the Universe that is below my floor. Then, I could specify this by saying that I have the vector: < 0, 1, 0 > which points directly up... and then I would also need to specify a point on my floor (so we can distinguish that I was really talking about the floor and not the ceiling or some other parallel plane).

In n-dimensional space, any hyperplane divides the space into two halves. We can specify the hyperplane by telling one point in the hyperplane and giving a vector which is perpendicular to all lines in the hyperplane.

More later if this doesn't clear things up...
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Re: 60 degree coordinates

Postby Dick Fischbeck » Sun May 16, 2004 5:10 pm

A half-space in our three-dimensional world would be some plane and all of the volume below it. A typical way to represent a plane in three-dimensions is to specify a vector which is perpendicular to all lines in the plane and then specify a point that is in the plane. Orthogonal and perpendicular mean the same thing (at least in Euclidean geometry).

Does a cone divide space in half? Or are the 2 spaces different sizes?

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Re: 60 degree coordinates

Postby pat » Mon May 17, 2004 3:39 pm

Dick Fischbeck wrote:Does a cone divide space in half? Or are the 2 spaces different sizes?


It does divide it into two volumes of the same cardinality. But, typically one means "bounded by a hyperplane" when one says half-space. You have probably heard things like "the half-plane below the x-axis" when talking about 2D stuff and such. That's the same thing.
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Postby pat » Mon May 17, 2004 3:45 pm

Further, one cannot construct polytopes as intersections (or unions) of cones.

I'm having trouble with the 60-degree coordinate system things at the moment though. I'm so used to dealing with things in terms of dot-products of vectors. In a 60-degree coordinate system, rather than having <b>a . b = |a| |b| cos θ</b>, you would have what? <b>a . b = |a| |b| cos <sup>3θ</sup>/<sub>2</sub></b>?
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Postby cnelson9 » Mon May 17, 2004 5:11 pm

pat wrote:Further, one cannot construct polytopes as intersections (or unions) of cones.

I'm having trouble with the 60-degree coordinate system things at the moment though. I'm so used to dealing with things in terms of dot-products of vectors. In a 60-degree coordinate system, rather than having <b>a . b = |a| |b| cos ?</b>, you would have what? <b>a . b = |a| |b| cos <sup>3?</sup>/<sub>2</sub></b>?


See:

http://users.adelphia.net/~cnelson9/Lin ... lnk_8.html

A brief description of Synergetics 60-degree coordinates is at:

http://mathworld.wolfram.com/Synergetic ... nates.html

The Mathematica notebook SynergeticsApplication7 at:

http://library.wolfram.com/infocenter/MathSource/600/

or the SynergeticsApplication7 notebook as html at:

http://users.adelphia.net/~cnelson9/

R. Buckminster Fuller's Synergetics 1 and 2 at:

http://www.rwgrayprojects.com/synergeti ... etics.html

Cliff Nelson
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Postby Dick Fischbeck » Wed May 19, 2004 4:33 pm

pat wrote:Further, one cannot construct polytopes as intersections (or unions) of cones.

I'm having trouble with the 60-degree coordinate system things at the moment though. I'm so used to dealing with things in terms of dot-products of vectors. In a 60-degree coordinate system, rather than having <b>a . b = |a| |b| cos ?</b>, you would have what? <b>a . b = |a| |b| cos <sup>3?</sup>/<sub>2</sub></b>?


Pat

I asked about your question and Al wrote the following. I am still unclear about dot products myself but maybe he can help you understand.

Dick

"In response to your earlier message, the dot product is defined the
same way
it always is, just use the vectors you have picked as your coordinate
system
and dot an arbitrary vector with respect to them. I hope this makes
sense."

Al
http://www.signal-science.com
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Postby pat » Thu May 20, 2004 8:14 pm

Dick Fischbeck wrote:I asked about your question and Al wrote the following. I am still unclear about dot products myself but maybe he can help you understand.


Actually, my difficulty was that I thought a point in the plane only had two synergistic coordinates. I hadn't realized that each point had three synergistic coordinates. I'm going to have to re-evaluate all of my claims from the beginning. It does seem possible (even likely) now that the dot-products still work out and that orthogonal vectors have a zero dot-product.
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Postby RQ » Tue May 25, 2004 1:33 pm

Ahem....
a 60 degree angle is not a different dimension, because only perpendicular motions are independent from each other proving that 60 degree angles are slopes of a combination of the two perpendicular dimensions it occupies.
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re:60 degree coordinates

Postby Dick Fischbeck » Tue May 25, 2004 2:10 pm

RQ wrote:Ahem....
a 60 degree angle is not a different dimension, because only perpendicular motions are independent from each other proving that 60 degree angles are slopes of a combination of the two perpendicular dimensions it occupies.


Hi RQ

Isn't precession the behavior of perpendicular motions? What does it mean, 'only perpendicular motions are independent'?

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