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 Post subject: Edge turning dodecahedron jumbling properties.Posted: Wed Apr 17, 2013 6:45 am

Joined: Fri Nov 05, 2010 2:20 am
Location: Wherever
As I was designing several edge turning dodecahedra and face turning rhombic triacontahedra and such, I had a question.
What is the exact jumbling angle for the Rhombic triacontahedron? Also, if the cuts deepen, another jumbling move is available. What is the angle for this one?

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 Post subject: Re: Edge turning dodecahedron jumbling properties.Posted: Wed Apr 17, 2013 7:39 am

Joined: Sun Oct 08, 2006 1:47 pm
Location: Houston/San Antonio, Texas
Hey I know this! An edge turning dodecahedron, or a face turning rhombic tricontahedron actually has 4 different* types of jumbling!

*up for debate I suppose, but this is how I see it

In order of "deepness" they occur at:

1) arccos(1/sqrt(5)) and 180-that (=approx 63.435 and 116.565)
2) arccos(sqrt(5)/3) and 180-that (=approx 41.810 and 138.190)
3) arccos(1/sqrt(5)) and 180-that (*again, but for a different "reason" so I'm counting it twice)
4) 90 (and yes I would argue that this is a jumbling angle even though its rational)

Since each of these can occur either clockwise or counter clockwise, that means a face on a deepcut face turning rhombic dodecahderon could interact with other faces at 12 different angles including 0 degrees:
0
41.810
63.435
90
116.565
138.190
180
221.810
243.435
270
296.565
318.190

Anything else?
Peace,
Matt Galla

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 Post subject: Re: Edge turning dodecahedron jumbling properties.Posted: Wed Apr 17, 2013 7:40 am

Joined: Fri Nov 05, 2010 2:20 am
Location: Wherever
That was.... quick. wow.

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 Post subject: Re: Edge turning dodecahedron jumbling properties.Posted: Wed Apr 17, 2013 7:44 am

Joined: Sun Oct 08, 2006 1:47 pm
Location: Houston/San Antonio, Texas

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 Post subject: Re: Edge turning dodecahedron jumbling properties.Posted: Wed Apr 17, 2013 8:18 am

Joined: Sat Sep 15, 2012 7:42 am
Out of curiosity, how does one calculate a jumbling angle? The math nerd in me has to know.

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 Post subject: Re: Edge turning dodecahedron jumbling properties.Posted: Wed Apr 17, 2013 12:45 pm

Joined: Wed Jan 07, 2009 6:46 pm
Location: Evanston, IL
I'm also curious about how you calculate the angle. Could the process be generalized to any geometry? (or at least to multiple different axis geometries?)

I can personally vouch for the 90-degree jumbling angle on a Face-Turning Rhombic Triacontahedron. I ran into it on the Big Kahuna during the summer of 2011.

-Eitan

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 Post subject: Re: Edge turning dodecahedron jumbling properties.Posted: Wed Apr 17, 2013 7:03 pm

Joined: Sun Oct 08, 2006 1:47 pm
Location: Houston/San Antonio, Texas
It's pretty simple in theory. I use the identity AB=|A||B|cos b where A and B are vectors in 3-D space originating from the same point and b is the angle between them. The tricky part is finding the exact value of the vectors A and B. Basically you have to construct the necessary vectors by hand, using pencil and paper instead of a computer, which rounds everything off to decimals. If you would like to know exactly what vectors you should be computing, see this post

Of course, seeing as the asnwer will always be the form of:
arccos((a1b1+a2b2+a3b3)/(sqrt((a1^2+a2^2+a3^2)(b1^2+b2^2+b3^2))))
we can simplify this expression to say with certainty the answer will always be of the form:
arccos((a+sqrt(b))/c)
And, assuming we can always write the coordinates of te vectors in the form (sqrt(a),sqrt(b),sqrt(c)) which seems to be likely but not guaranteed, then we can simplify the expression further to say that the answer will be of the form:
arccos(sqrt(a/b))

In many cases, it boils down to the second option so you can actually cheat and get the answer quickly by assuming this form. Find the angle in decimal, take the cosine, square it. This miraculously gives you a rational number that you can plug into the above expression and simplify. This at least works for Rhombic Dodecahedra, Rhombic Tricontahedra, and Triangular Dipyramids (Meteor Madness, More Madness), but it does not work for Icosahedra, where the jumbling angle is arccos((3sqrt(5)-1)/8). That one requires some actual work (or Mathematica )

Peace,
Matt Galla

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