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Synergetics: Explorations in the Geometry of Thinking


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#35 Turtle

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Posted 21 September 2008 - 10:38 PM

Read it long ago, and am trying to catch up now.

I love Uncle Buckster, but I'll tell ya' I often find his penchant for supercilious locquaciousness is obfuscating and sometimes comes across as faux sagacity!

He was seriously smart, so its fun to try to follow along, but sometimes I so feel like Alice having half a cuppa tea!

Speak English! I don't know the meaning of half those long words, and I don't believe you do either! :phones:
Buffy


:hyper: Now that I agree with. :hihi: If you pardon the pun, Fuller seems to have a bit of a chip on his shoulder from bucking the status quo. :doh:

I quite agree with you. And the moral of that is: Be what you would seem to be, or if you'd like it put more simply: Never imagine yourself not to be otherwise than what it might appear to others that what you were or might have been was not otherwise than what you had been would have appeared to them to be otherwise. :turtle:

:lol:
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#36 modest

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Posted 22 September 2008 - 12:55 AM

and being as how the last section introduced the icosahedron...

Posted Image


Very sweet! Can I play with the icosahedron too? :phones:

YouTube - Icosahedron

Did fuller mention the red, blue, and gree squares are golden?

~modest

EDIT: T, I used an unfamiliar computer when rendering that video. I think it used a funky codec encoding it. Let me know if you have trouble playing it and I'll redo it.
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#37 Turtle

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Posted 22 September 2008 - 12:58 AM

Posted Image

Posted Image

Likeable or not, Fuller rigorously adds to Euler's geometric accounting. I have a rather good & hopeful outlook that if Leonhard had seen some of Bucky's work that he would say, "why didn't I think of that!?" (the geodesic dome & the dymaxion map projection for examples) In any case, I have waded through far worse verbiage on less promise before, and already I have some new insights. To whit:

I once before built the A quanta modules from the planar net, but out of folded paper. It wasn't until I made the open straw models that I saw how it was possible to grap one edge & pull it through a face and out pops the same thing but changed right/left. (or the other way. :phones::lol: ) Then I realized that the same is true of the tetrahedron, except it is completely symmetric so you can't tell right-from left, or inside from outside. A little epiphany. :doh:

I'll be back tomorrow night after I do the next reading myself. :hyper: :hihi: :turtle:
1000.00 OMNITOPOLOGY
§ 1044.00 - 1044.13; pgs 409-412; Synergetics volume 2

#38 Turtle

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Posted 22 September 2008 - 05:26 PM

Very sweet! Can I play with the icosahedron too? :hihi:

YouTube - Icosahedron

Did fuller mention the red, blue, and gree squares are golden?

~modest

EDIT: T, I used an unfamiliar computer when rendering that video. I think it used a funky codec encoding it. Let me know if you have trouble playing it and I'll redo it.


Dude!! You are freakin' me out now! :eek: I totally missed this post last night; I must have been posting #37 at the time. Anyway, when I got up today, I too decided to play with my icosahedron. :hihi: I saw them at least 3 times on TV in bed last night too! Freakin' me out I tell ya. :hyper:

So, welcome, welcome, :bow: and yes please play :lol: and no Fuller hasn't mentioned Goldenosity yet. He is soon going to justify not refering to any irrationals however.

But the reason I stopped by so early was to post some eye candy I made this morning of my cored-through* icosahedron rotating in suspension from a perspective through its center & into the ceiling corner of the room.
*'cored-through' is a new term I saw Bucky use in the latest reading, and seems equivalent to my phrase 'straw & string model'.

Have a taste of this pudding. :) :)
YouTube - rotating a cored-through icosahedron

#39 modest

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Posted 22 September 2008 - 06:49 PM

So, welcome, welcome, :) and yes please play :hyper:


It's a plan. I was well-into a game of catch-up last night. I started with
100.01 Introduction: Scenario of the Child
and got up to bisecting the triangle and tetrahedron. My plan was to follow your lead and do some modeling along those lines tonight and gain some insight into how the universe works and whatnot. Unfortunately, I wasn't considering the 2-hour NBC Heroes premiere tonight! :lol:

But, I am excited by what little I read thus far. At the very least we can learn some new and interesting geometry and no doubt some interesting word-isms as well. So, ok, I'll be your huckleberry.

~modest

PS: attached a glitch-free version of my icosahedron.

#40 Turtle

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Posted 22 September 2008 - 10:23 PM

It's a plan. I was well-into a game of catch-up last night. I started with
100.01 Introduction: Scenario of the Child
and got up to bisecting the triangle and tetrahedron. My plan was to follow your lead and do some modeling along those lines tonight and gain some insight into how the universe works and whatnot. Unfortunately, I wasn't considering the 2-hour NBC Heroes premiere tonight! :)

But, I am excited by what little I read thus far. At the very least we can learn some new and interesting geometry and no doubt some interesting word-isms as well. So, ok, I'll be your huckleberry.

~modest

PS: attached a glitch-free version of my icosahedron.


Roger. I'll let you catch up, and just quote the salient feature of the current section as it goes (eventually :hihi:) to the accounting Fuller employs in divisioning 'stuff', and immediately to 'correcting' Euler. :eek: :lol: :hihi: Formatting mine. :hyper:

1044.01 Euler + Synergetics: The first three topological aspects of all minimum systems__vertexes, faces, and edges__were employed by Euler in his formula V + F = E + 2. (See Table 223.64 and Sec. 505.10.) Since synergetics' geometry embraces nuclear and angular topology, it adds four more minimum aspects to Euler's inventory of three:

EULER
vertexes
faces
edges

SYNERGETICS
angles
insideness & outsideness
convexity & concavity
axis of spin

1000.00 OMNITOPOLOGY

#41 Turtle

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Posted 23 September 2008 - 02:51 PM

Very sweet! Can I play with the icosahedron too? :hihi:


YouTube - Icosahedron

Did fuller mention the red, blue, and green squares are golden?

~modest


Okaly-dokaly. So I watched the vid a few times, combined it with my models and what I have read so far, and have these observations.
1) In your vid, you show an orthoganal division, with 3 Golden ratio quadrilaterals. Just keep in mind that the icosahedron has 30 edges, 15 opposite pairs, and so the vid leaves out 12 planes of the possible 15.

2) Bucky says divide space on the 7 axes of the tetrahedron, and so I shoved my tetrahedron model into a face of the icosahedron. This still doesn't give a tetrahedron in the center, but we have skipped ahead on one of the Trimster's parenthetical vectors. :eek2::hihi:

3) In the photo below of my icosahedron with the shoved-in tetrahedron, it is convenient to notice that shoving in another 19 tetrahedrons would fill the icoshedron. :clue: (little epiphany):idea:

4) In the inset, I have marked in green & blue one of the Golden rectangles. The green is actually the longer side; perspective makes it look shorter. ;)

5) All 15 of your rectangle planes pass between all the tetrahedrons! :eek2: So, this is a major 'mistake' the Buckster is harping on; divisoning space cubically (orthogonally?) rather than tetrahedrally. :D :) We have yet to get to Fuller's basic construction of the icosahedron, so this is all my own interpretation so far.

6) I'm all played out thens for now, but the game is still a foot...afoot. :D :) .......:turtle:

Post Script: Just watching my models twist in the wind, and realized that Modest's rectangles are divided lengthwise by 2 equilateral triangles, making the ratio of their altitudes-to-edges each 1/2 phi. I simply don't recall seeing it expressed thata way. Am I seein' that right? Cut...print it. Smoke 'em if ya gott 'em. ;)

Posted Image

#42 Turtle

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Posted 23 September 2008 - 04:39 PM

4) In the inset, I have marked in green & blue one of the Golden rectangles. The green is actually the longer side; perspective makes it look shorter. :eek2:

...
Post Script: Just watching my models twist in the wind, and realized that Modest's rectangles are divided lengthwise by 2 equilateral triangles, making the ratio of their altitudes-to-edges each 1/2 phi. I simply don't recall seeing it expressed thata way. Am I seein' that right? Cut...print it. Smoke 'em if ya gott 'em. :)

Posted Image


On further thought then, Modest's rectangles aren't Golden ratioed at all! ;) By Pythagoras, the altitude of a unit equilateral triangle is 0.86602540378443864676372317075294, which is 1/2 the square root of 3, not 1/2 of Phi. :turtle: (2*0.86602540378443864676372317075294 = 1.7320508075688772935274463415059) The rectangles in Modest's video then, have the ratio sqr3/1. :) Oui/no? :D

#43 modest

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Posted 23 September 2008 - 10:00 PM

Roger. I'll let you catch up, and just quote the salient feature of the current section as it goes (eventually :hihi:) to the accounting Fuller employs in divisioning 'stuff', and immediately to 'correcting' Euler. :eek: :hyper: :read: Formatting mine. :turtle:

1044.01 Euler + Synergetics: The first three topological aspects of all minimum systems__vertexes, faces, and edges__were employed by Euler in his formula V + F = E + 2. (See Table 223.64 and Sec. 505.10.) Since synergetics' geometry embraces nuclear and angular topology, it adds four more minimum aspects to Euler's inventory of three:

EULER
vertexes
faces
edges

SYNERGETICS
angles
insideness & outsideness
convexity & concavity
axis of spin

1000.00 OMNITOPOLOGY


I understand (or, at least I have some affinity for) Fuller's addition of convexity and concavity. Without getting to deep into it yet, there are good reasons to treat non-Euclidean shapes as a separate geometry. So, I think there's something to that. Physical laws such as gravity are represented well with "convexity and concavity".

In the same way, I like his addition of "axis of spin". I think there's something to spin and rotation that isn't yet understood. Why rotating frames have fictitious pseudo forces, and what is this quantum mechanical property of spin. Why does it present like conservation of angular momentum? I think there's something there and maybe Fuller's ideas of incorporating spin into geometry is a good approach.

The first thing I'm reading along those lines is 100.103:

Rational numerical and geometrical values derive from (a) parallel and (B) perpendicular halving of the tetrahedron. (See Fig. 100.103.)

  • The parallel method of tetrahedral bisecting has three axes of spin__ergo, three equators of halving. Parallel equatorial halving is both statically and dynamically symmetric.
  • The perpendicular method of tetrahedral bisecting has six axes of spin__ergo, six equators of halving. Perpendicular equatorial halving is only dynamically symmetric.


which has this pic,

Posted Image

and I have modeled with the property of spin:

YouTube - Rotating bisected tetrahedron

But, I don't get it. I don't get why the axis of spin is so emphasized here. What's he trying to say, T? I have a feeling it could be pretty significant.

Okaly-dokaly. So I watched the vid a few times, combined it with my models and what I have read so far, and have these observations.
1) In your vid, you show an orthoganal division, with 3 Golden ratio quadrilaterals. Just keep in mind that the icosahedron has 30 edges, 15 opposite pairs, and so the vid leaves out 12 planes of the possible 15.


I agree exactly.

3) In the photo below of my icosahedron with the shoved-in tetrahedron, it is convenient to notice that shoving in another 19 tetrahedrons would fill the icoshedron. :clue: (little epiphany):idea:


I tried doing this and couldn't get it to work. The angles were a bit off and there were gaps where I tried to get things to fit. I haven't yet done the numbers to show where either I'm going wrong or to prove it doesn't work, but we should, I think, look closer at this. I apologize for objecting without anything to show as a basis for my objection, but something seems off.

On further thought then, Modest's rectangles aren't Golden ratioed at all! :naughty: By Pythagoras, the altitude of a unit equilateral triangle is 0.86602540378443864676372317075294, which is 1/2 the square root of 3, not 1/2 of Phi. :eek: (2*0.86602540378443864676372317075294 = 1.7320508075688772935274463415059) The rectangles in Modest's video then, have the ratio sqr3/1. :sherlock: Oui/no? :turtle:


When making the icosahedron in studio max I gave it a radius of 95.10565 which by wiki gives it a segment length of
[math]r=a/2 \sqrt{ \phi \sqrt{5}}[/math]
or 100 units. I checked this and verified the segment length once created. The squares I made are 100 x 100(phi) or 100 x 161.8. They do fit very well, I'll try to work out the math.

Posted Image

~modest

EDIT:

Wikipedia's icosahedron article has:

The following Cartesian coordinates define the vertices of an icosahedron with edge-length 2, centered at the origin:

  • (0, ±1, ±φ)
  • (±1, ±φ, 0)
  • (±φ, 0, ±1)
where φ = (1+√5)/2 is the golden ratio (also written τ). Note that these vertices form five sets of three mutually centered, mutually orthogonal golden rectangles.


So, I think we're pretty safe on the golden rectangles.

#44 Turtle

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Posted 23 September 2008 - 10:49 PM

I understand (or, at least I have some affinity for) Fuller's addition of convexity and concavity. Without getting to deep into it yet, there are good reasons to treat non-Euclidean shapes as a separate geometry. So, I think there's something to that. Physical laws such as gravity are represented well with "convexity and concavity".

In the same way, I like his addition of "axis of spin". I think there's something to spin and rotation that isn't yet understood. Why rotating frames have fictitious pseudo forces, and what is this quantum mechanical property of spin. Why does it present like conservation of angular momentum? I think there's something there and maybe Fuller's ideas of incorporating spin into geometry is a good approach.

The first thing I'm reading along those lines is 100.103:



which has this pic,

Posted Image

and I have modeled with the property of spin:

YouTube - Rotating bisected tetrahedron

But, I don't get it. I don't get why the axis of spin is so emphasized here. What's he trying to say, T? I have a feeling it could be pretty significant.


He is saying there are 2 basic ways to cut a tetrahedron exactly in half. The perpendicular way leads to the divisioning that gives the A quanta modules. This is significant because the A quanta module is just one new building block Fuller is showing us.


In the photo below of my icosahedron with the shoved-in tetrahedron, it is convenient to notice that shoving in another 19 tetrahedrons would fill the icoshedron. (little epiphany) :idea:

I tried doing this and couldn't get it to work. The angles were a bit off and there were gaps where I tried to get things to fit. I haven't yet done the numbers to show where either I'm going wrong or to prove it doesn't work, but we should, I think, look closer at this. I apologize for objecting without anything to show as a basis for my objection, but something seems off.


Don't worry about the looseness, and an exact fit is out because these are actual structures and no 2 'real' points can occupy the same space at the same time. You objection is noted, but...

When making the icosahedron in studio max I gave it a radius of 95.10565 which by wiki gives it a segment length of
[math]r=a/2 sqrt{ phi sqrt{5}}[/math]
or 100 units. I checked this and verified the segment length once created. The squares I made are 100 x 100(phi) or 100 x 161.8. They do fit very well, I'll try to work out the math.

Posted Image

~modest

EDIT:

Wikipedia's icosahedron article has:
So, I think we're pretty safe on the golden rectangles.


No. I'm afraid I have to hold my assertion still, that the rectangles aren't Golden in spite of wiki. Their ratio I still find as sqr3 to 1. :shrug: I plan to also show this by taking Fuller's lengths from his A quanta module planar net, as each of the 20 tetrahedrons making an icosahedron is made of 24 A quanta modules. My models use the unit edge, as does Fuller's net. Looking forward to more on it that you find.
There is phi in an icosahedron however, and it is found in the ratio of the diagonal of one of the pentagons formed by the 5 triangles, to the unit edge. The whole proof is given in a back section of Synergetics(1), pg. 829, by Harvard professor Arthur L. Loeb. Standard phi in the pentagon.

I think now that we should return to the 100.00's section whence we came, and get on to a bit of why cartesian coordinates just make divsioning space harder. I'll start with the start. :read: :turtle:
100.00 SYNERGY

100.30;pg13 Omnirational Subdividing

Fig. 100.301 100.301 Omniquadrilaterally interconnecting the mid-edge-points of any dissimilarly- edge-lengthed quadrilateral polygon always produces four dissimilar quadrangles. Omnitriangularly interconnecting the mid-edge-points of any dissimilarly-edge-lengthed triangle always produces four similar triangles. (See Fig. 990.01.) Whereas omniinterconnecting the mid-edge-points of a cube always subdivides the cube into eight similarly equiedged cubes, interconnecting the mid-edge-points of any dissimilarly-edge- lengthed quadrangular-faced hexahedra always subdivides the hexahedron into eight always dissimilar, quadrangular-faced hexahedra. (See Fig. 100.301.)

Fig. 100.301
Fig. 990.01

#45 modest

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Posted 24 September 2008 - 01:07 AM

No. I'm afraid I have to hold my assertion still, that the rectangles aren't Golden in spite of wiki. Their ratio I still find as sqr3 to 1. :) I plan to also show this by taking Fuller's lengths from his A quanta module planar net, as each of the 20 tetrahedrons making an icosahedron is made of 24 A quanta modules. My models use the unit edge, as does Fuller's net. Looking forward to more on it that you find.


There's a lot in your post I want to respond to (which, by the bye, has a lot of good info that makes me realize how much reading I have to do), but before I go to bed tonihgt - I'll post the three sections (projections) my computer gives will the following input:

Green: regular icosahedron with edge length 1
Red: rectangle side a: 1.6180 (phi); side b:1
Blue: rectangle side a: 1.7321 (sqr3); side b:1

Top:
Posted Image

Front:
Posted Image

Left:
Posted Image

We can fill in all the lengths and angles with a bit of trigonometry from this, but I'm a bit tired for trig tonight.

~modest :)

#46 Turtle

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Posted 24 September 2008 - 01:26 AM

There's a lot in your post I want to respond to (which, by the bye, has a lot of good info that makes me realize how much reading I have to do), but before I go to bed tonihgt - I'll post the three sections (projections) my computer gives will the following input:

Green: regular icosahedron with edge length 1
Red: rectangle side a: 1.6180 (phi); side b:1
Blue: rectangle side a: 1.7321 (sqr3); side b:1

We can fill in all the lengths and angles with a bit of trigonometry from this, but I'm a bit tired for trig tonight.

~modest :sleeps:


Acknowledge drawings. They displayed dark for me so I saved them to disk and enhanced for a look. My trig is as rusty as a witch's nail, but I can muddle through Euclidian constructions. I'll leave you to catching up reading while I mull this more... :sherlock: :turtle:

PS OK. I'm convinced you have it because my ***presumption (;))that 20 regular tetrahedron close-pack into an icosahedron is wrong. If I had made all 20 straw-tetrahedrons & tried to jam them in the straw icosahedron at once, then I would have clearly seen the gappage. :doh: :hihi: Geometry in Condensed Matter Physics ... - Google Book Search

Back to Synergetics then. :read: :turtle:
100.30;pg13 Omnirational Subdividing
100.00 SYNERGY
Fig. 100.301
Fig. 990.01

#47 modest

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Posted 24 September 2008 - 10:42 PM

They displayed dark for me so I saved them to disk and enhanced for a look.


Sorry. Rendered on white background and attached if needed.

Also, I found a site that works out the angles and whatnot.

PS OK. I'm convinced you have it because my ***presumption (;))that 20 regular tetrahedron close-pack into an icosahedron is wrong. If I had made all 20 straw-tetrahedrons & tried to jam them in the straw icosahedron at once, then I would have clearly seen the gappage. :) :)


Yeah, that's what happened to me, albeit virtually. Couldn't get the little buggers to fit - and good source on that as well (google book).

Back to Synergetics then.


Yes, exactly what I was thinking.

~modest :)

#48 Turtle

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Posted 25 September 2008 - 01:03 PM

I'm convinced you have it because my asspresumption ()that 20 regular tetrahedron close-pack into an icosahedron is wrong. If I had made all 20 straw-tetrahedrons & tried to jam them in the straw icosahedron at once, then I would have clearly seen the gappage.

Yeah, that's what happened to me, albeit virtually. Couldn't get the little buggers to fit - and good source on that as well (google book).


I think what helped keep this illusion, is that tetrahedrons do close-pack, i.e. they are space filling just like cubes.

Back to Synergetics then.

Yes, exactly what I was thinking.

~modest :(


So we return to a common vertex, the current reading, and I think we'll just idle a while on the 2 diagrams (cool new Hypog feature with links opening little boxies!!! :() until I'm sure you have caught up with the reading. What is a key feature of Synergetics is the division of space triangularly/tetrahedrally, as opposed to squarely & cubically, and these diagrams show the angles distortion of cubic space measuring, and the accounting equity between both systems. :( :doh:
100.30;pg13 Omnirational Subdividing
100.00 SYNERGY
Fig. 100.301
Fig. 990.01


Thanks for joining in folks, :bow:, as I am fully capable of getting into mischief if left unguarded. :hihi:

#49 Turtle

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Posted 26 September 2008 - 03:11 PM

Synergetics is like teaching a monkey to sing; you have to use sign language. ;) :)

Moving on with the reading at hand, I think some deserves quotables. :clue:

100.00 SYNERGY

100.302 A triangle is a microaltitude tetrahedron with its apex almost congruent with one of its base triangle's vertexes. A right-angled triangle, an isosceles triangle, and a scalene triangle are all the same triangle. The seeming difference in edge lengths and angles is a consequence only of shifting the base-plane locus of the observer.
100.303 Most economically intertriangulating the midpoints of any regular equiedged or any dissimilarly edged tetrahedra will always subdivide that tetrahedron into four similar tetrahedra and one octahedron whose volume is always four times that of any of the four similar and equivolumed tetrahedra.
100.304 Cheese Tetrahedron: If we make all the symmetrical Platonic solids of firm cheese, and if we slice the cube parallel to one of its faces, the remaining hexahedron is no longer equiedge-lengthed. So too with all the other Platonic solids__the dodecahedron, the octahedron, or the icosahedron__with one, and only one, exception: the tetrahedron. The cheese tetrahedron may be sliced parallel to any one, or successively all four, of its faces without losing its basic symmetry; ergo, only the tetrahedron's four-dimensional coordination can accommodate asymmetric aberrations without in any way disrupting the symmetrical integrity of the system.
100.310 Two Tetra into Cube: The child wants to get inside things. What is the minimum something it can get inside of? The necklace tetrahedron strung together with long-tube-beads. A child tries the necklace cube, and it collapses. The child then takes the edge tubes of the collapsed cube and reassembles them as an octahedron__which holds its shape. The child also takes two sets of six tubes and makes two tetrahedra producing an omnitriangulated superficially induced cube with eight corners.


With the help of my diminuative hired monkey with teensy hands, I employed Buckster's childlike naivete and nested 2 tetrahedrons so as to have the vertices lie at the corners of a cube. I love getting inside of stuff! :evil: :evil:

Posted Image

#50 Turtle

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Posted 27 September 2008 - 10:25 PM

This reading covers pgs. 15 to 20 in the hardcover, and 3 of the pages have diagrams of some geometry of foldability. Longer than most readings so far, it is still only about 4 minutes long and it hangs together well. Moreover, this will bring us to the end of the Child As Explorer section. :) .:mickmouse:

§100.320 - 100.63

#51 modest

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Posted 28 September 2008 - 12:09 AM

Mmm-k, T. Check my understanding...

Rational numerical and geometrical values derive from (a) parallel and (:) perpendicular halving of the tetrahedron. (See Fig. 100.103.)

  • The parallel method of tetrahedral bisecting has three axes of spin__ergo, three equators of halving. Parallel equatorial halving is both statically and dynamically symmetric.
  • The perpendicular method of tetrahedral bisecting has six axes of spin__ergo, six equators of halving. Perpendicular equatorial halving is only dynamically symmetric.


The parallel method which I have depicted more-or-less isometrically (matching figure 100.103):
Posted Image
means the axis is perpendicular to the 2 relevant segments and the equator (or the bisected slice) is parallel to the relevant segments. There are 3 parallel (to each other) segments, so this can be done 3 ways.

The perpendicular method:
Posted Image
has the axis and relevant segment (singular) parallel while the equator and slice are perpendicular to said segment. Being there are 6 segments, this can be accomplished 6 ways.
Posted Image
It is here bisected twice perpendicularly with axes AB and CD.

The three-way, symmetry-imposed, perpendicular bisecting of each of the tetrahedron's four triangular faces results in an inadvertent thirding. This halving and inadvertent thirding physically isolate the prime number three and its multiples and introduce the 24 A Quanta Modules.


I guess he means like this:
Posted Image
bisected perpendicularly with axes AC, CD, and AD. The tetrahedron's triangular face ACD would then look like figure 100.1041. This makes 6 shapes. Doing the same with the other 3 faces of the tetrahedron should then make 6 x 4 = 24 shapes which I suppose are the "24 A quanta modules", but he doesn't really explain that in this chapter.

So, I guess I should take a detour into 913 now 'n figure out these moduli - is that the idea?

~modest :mickmouse: