for the breakout groups, day 2, twilight zone workshop
kirby urner
Exercise 1: The growing V.
Draw a V with a 60 angle.
Take a ruler and place it across the V at 90 degrees such that all three edges (2 radii plus edge opposite the vertex) are equal.
Slide the rule towards and away from the apex of the V.
Note the equilateral triangle grows and shrinks.
Exercise 2: Triangular Lid
Draw or make a "cup" with three triangular sides and a share apex. The triangles are equiangular.
As you slowly fill this cup with water, the horizontal triangle formed by the water's surface is like a lid.
The plane segment P, formed by water, is rising away from apex O, the "origin".
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Teacher Notes:
Exercise 1 demonstrates 2nd power growth as a function of the ruler's distance from the opposite apex.
Exercise 2 demonstrates 3rd power growth as a function of the water's depth above O.
2nd powering associates with a growing triangle (special case: equilateral).
3rd powering associates with a growing tetrahedron (special case: equilateral)
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Exercise 3: Four Balls
Glue them together at tangent points to form a tetrahedron.
Define the edges of a tetrahedron that connects adjacent ball centers.
Exercise 4: Paper Tetrahedron
Make a paper tetrahedron of the same size as that define by the four ball centers.
Exercise 5: 12-around-1
Glue 12 equi-sized balls around a nuclear ball such that
- these 12 remain inter-tangent to one another and
- all the edges and radii between adjacent sphere centers are of equal length
Start with six-around-1 on a table.
Note the two different ways to orient the triangles of balls above and below, relative to one another.
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Teacher Notes:
These two ways of packing 12-around-1 as a basis for isotropic equi-density sphere packing (on outward), suggest different kinds of symmetry.
When the two triangles align, vertex atop vertex, that's option 1.
When the two triangles point oppositely, that's option 2.
Option 2 is more symmetrical in the sense of being more omni-symmetrical.
Stipulation (b) may be relaxed, while preserving (a) inter-tangency. Show this.
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Exercise 6: 42-around-12-around-1
Add a layer of 42 balls around the previous layer of 12. This may be a group activity with ball sharing.
Introduce the formula 10 * F * F + 2 where F is the number of intervals between ball centers, along any of 12 outward edges.
Activities:
Watch collected videos about CCP, Barlow Packings, HCP, ball agglomeration.
Introduce the Jitterbug Transformation transforming the cuboctahedron into the icosahedron
Challenge:
John Brawley
----- Original Message -----
From: "kirby urner" <kirby...@gmail.com>
To: <r-buckminster-fuller...@googlegroups.com>
Sent: Sunday, February 19, 2012 5:01 PM
Subject: for the breakout groups, day 2, twilight zone workshop
> *Exercise 5: 12-around-1*
>
> Glue 12 equi-sized balls around a nuclear ball such that
> 1. these 12 remain inter-tangent to one another and
>
> 2. all the edges and radii between adjacent sphere centers are of equal
> length
Can't be done.
(*grin*)
--JBw
dbkoski
John Brawley
of the 13 spheres").
2. worded in a copout manner, yes, all the edges between adjacent sphere
centers are equal, but not all spheres that could be intertangent, are
intertangent. A trade-off is in place.
(Intertangency definable as every ball that could touch another ball,
touching that ball. Example: in 2D, a plane of 6-around-1-ness has full
intertangency. Every ball is surrounded by and touching six other balls.
Full intertangency of this sort is impossible in 3D 12-around-none, so the
movement of balls into that one-of-a-kind cuboctahedral ('VE') arrangement
is thought by some to fit this wording Kirby uses, but there's a different
packing of 12-around-1 that has only 9 non-touching sphere-pairs, instead of
12.
(It's an old, old, gripe of mine, this prejudice for the Fullerist-*only*
12-around-1 packing; students will be directed to ignore or dismiss the
nonsymmetrical packing of 12-around-1, which is tighter, denser, and fits
the 1. and 2. descriptions better: more intertangency is present, and all
the edges and radii *between adjacent (touching surfaces) sphere centers*
are equal (even though, as in the cubocta form, some sphere-pairs can't
touch surfaces).
It's a point of contention, caused by there being one and only one
*symmetrical* maximization of partial intertangency (Fuller'sl the cubocta
form). The prejudicial wording Kirby uses (has always, will always, others
always, etc.) defaults only to the cuboctahedral packing of 13, which has no
less than 12 near-neighbor two-ball pairs that lack tangency.
--JBw
(I'd ignore my comment were I you; it's like a 'bad penny', which will
return from time to time. (*g*))