title: Afterpiece
description: Part of Synergetics - Explorations in the Geometry of Thinking
author: R. Buckminster Fuller, E. J. Applewhite
tags:
- synergetics
- geometry
- mathematics
- systems theory
- buckminster fuller
published: true
date: 1975-01-01
editor: markdown
¶ 0 Afterpiece
The phenomenon lag is simply due to the limited mechanism of the brain.
We have to wait for the afterimage in order to realize.
The norm of Einstein is absolute speed instead of at rest. ‘‘At rest’’ was what we called instantaneous in our innocence of yesterday. We evolute toward ever lesser brain-comprehension lags—ergo, toward ever diminishing error; ergo, ever diminishing misunderstandings; ergo, ever diminishing fear, and its brain-lagging painful errors of objectivity; wherefore we approach eternal instantaneity of absolute and total comprehension.
The eternal instantaneity of no lag at all. However, we have now learned from our generalizations of the great complexity of the interactions of principles—as we are disembarrassed of our local, exclusively physical chemistry of information-sensing devices—that what is approached is eternal and instant awareness of absolute reality of all that ever existed.
All the great metaphysical integrity of all the individuals, which is potential and inherent in the complex interactions of generalized principles, will always and only coexist eternally.
Fig. 450.10 The Twenty-five Great Circles of the Vector Equilibrium: (1) One of the three square-face-centered great circles of the vector equilibrium’s 25 great circles. (2) One of the four triangle-/ace-centered great circles of the vector equilibrium’s 25 great circles. (3) One of the six vertex-centered great circles of the vector equilibrium’s 25 great circles. (4) One of the 12 edge-centered great circles of the vector equilibrium’s 25 great circles. 3 + 4 + 6 + 12 = 25. The drawing shows the axes of rotatability.
Fig. 450.11 Edge Lines Described by Twenty-five Great Circles of Vector Equilibrium: Each system corresponds with illustration 450.10.
Fig. 450.12 Three Great Circles of the Vector Equilibrium.
Fig. 453.02 Inside-Outing of Triangle: This illustrates the inside-outing of a triangle, which transformation is usually misidentified as ‘‘left vs. right’’ or ‘‘positive and negative’’ or as ‘‘existence vs. annihilation’’ in physics. The inside-outing is four-dimensional and often complex. The inside-outing of the rubber glove explains ‘‘annihilation’’ and demonstrates complex intro-extroverting.
Fig. 454.01 The six great circles of the vector equilibrium disclose the six square faces of the spherical cube facets whose eight vertexes are centered in the areal centers of the vector equilibrium’s eight spherical triangles.
Fig. 454.02 The six great circles of the vector equilibrium disclose the spherical tetrahedra and the spherical cube and their chordal, flat-faceted, polyhedral counterparts.
Fig. 454.03 The six great circles of the equilibrium disclose the 12 rhombic di; facets (cross-hatching) of the rhombic icosahedron, whose centers are coincident with the 12 vertexes (dots) of the vector equilibrium.
Fig. 459.01 Great Circle Foldabilities of Icosahedron.
Fig. 464.01 Triangle in Cube as Energetic Model: The rotating shaft is labeled AB. The model demonstrates that there are six vector wave moves per cycle.
Fig. 464.02 Rotation of Triangle in a Cube: Oscillating Effect. (Rendering from author's sketch by S. Sadao.)
Fig. 541.00 Energy Separated Out into Tetrahedral Photon Packages.
Fig. 541.30H Circular Cornucopia Assembled Around Interior Points to Form a Spherical Array. The tangent circle areas as well as concave triangle interstices constitute the total spherical surface.
Subdivide the Total
Fig. 541.30M Gradually the four tetrahedron-defining, vertexial components of the photon package’s spiralling results in an equilibrium-seeking interaction of their four separate interattractions, which generates the four great circles of the vector equilibrium and establishes its tactical energy center as the four planes of the zerophase tetrahedron.
Fig. 541.301 Three-Sided, Triangular Cornucopia Sphere.
Fig. 901.03 The Basic Disequilibrium 120 LCD Triangle: 12 vertexes surrounded by 10 converging angles 20 vertexes surrounded by 6 converging angles 30 vertexes surrounded by 4 converging angles
360 converging angles
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12 x 10 = 120
20x6=120
30x4=120
The 360 convergent angles must share the 720°reduction from absolute sphere to chorded sphere: 720 = 2°per each corner; 6°per each triangle.
360
All of the spherical excess 6°has been massaged by the irreducibility of the 90°and 60°corners into the littlest corner. . . 30--36.
In reducing 120 spherical triangles described by the 15 great circles to planar faceted polyhedra, the spherical excess 6°would be shared proportionately by the 90°-60°-30°flat triangle relationship = 3:2:1.
The above tells us that freezing the 60-degree center of the icosa triangle and sharing the 6-degree spherical excess find the A Quanta Module angles exactly congruent with the icosa’s 120 interior angles.
Fig. 943.OOA Quanta Module Orientations as Neutron and Proton 1 /24-Sphere Centers: A and B Quanta Modules.
Fig. 943.OOB Hierarchy of Quanta Module Orientations: A and B Quanta Modules.
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Fig. 954.00B Mites and Couplers
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Fig. 953.10 Positive and Negative Mites Constituted of Two A. Quanta Modules and One B Quanta Module.
Fig. 970.20 Basic Vector Equilibrium Concentric Shell Structure: The legend at the bottom illustrates the interstitial between-sphere spaces.
Fig. 971.01 Chart:
Table of Concentric, Sphere-Shell Growth Rates
Fig. 982.58 Nuclear Sphere of Volume 5 Enclosing the Vector Equilibrium of Volume 2xlz with the Vector Equilibrium’s Vertexes Congruent with the Nuclear Sphere: Shown are 15 of the Basic Disequilibrium 120 LCD triangles per sphere which transform as A Quanta Module tetrahedra. In the 25-great-circle subdividing of the vector equilibrium’s sphere, the three great-circles produce the spherical octahedron, one of whose eight spherical triangles is shown here. As was shown on the icosahedron, the 120 triangles of the 15 great circles divide the sphere in such a way that the spherical octahedron’s triangle can be identified exactly with 15 of the Basic Disequilibrium 120 LCD Triangles. Here we show the 15 disequilibrium triangles on the spherical octahedron of the vector equilibrium: 8x15 = 120 spherical right triangles which tangentially accommodate —closely but
not exactly—the 120 A Quanta Modules folded into tetrahedra and inserted, acute
corners inward to the sphere’s center, which could not be exactly accommodated in the
shallower icosahedral phase because of nuclear collapse and radius shortening in the
icosahedron.
Fig. 982.61 Synergetics Isometric of the Isotropic Vector Matrix: See text for full legend. Note the twelve-around-one, closest-packed spheres.
Fig. 1012.14A Indig Octave System of Four Positive, Four Negative and Zero-Nine Wave Pattern of Experiential Number: This basic discontinuous wave disclosure is intimately related to inherent octavization through tension-chord halving discovered by the Pythagoreans, and the major-minor-mode ‘‘fifthing’’ obtained by tensionchord thirding of length. These inherent additive-subtractive, alternate pulsing effects of number produce positive waves, but not continuously as had been misassumed. Zero—or ‘‘No-tet, None, Nine’’ intrudes. Waves are discontinuous and confirm unit quantation, one tetrahedron inherently constituting the basic structural system of Universe. The star Sun’s combining of four hydrogen atoms into helium atoms generates quanta radiation.
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Evolution of Synergetics: a portfolio of six drawings made by the author in 1948.
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Evolution of Synergetics: a portfolio of six drawings made by the author in 1948.
Transformations of Dymaxion Comprehensive (8 February 1948).
Dymaxion Nuclear Growth (10 June 1948).
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FULLER RESEARCH FOUNDATION
FULLER RESEARCH FOUNDATION
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¶ Contribution to Synergetics
Arthur L. Loeb
HARVARD UNIVERSITY
A Introduction
B Proof of Euler’s Equation
C Doubly Connected Polyhedra
D The Sum of the Surface Angles of a Singly Connected Polyhedron
E Duals
F Close Packing
G The Icosahedron and the Golden Section
H Remarks on Some Elementary Volume Relations Between Familiar Solids
I Space-Filling Polyhedra
J The A and B Modules and the Trirectangular Tetrahedron
K The Rectangular Cross Section of the Regular Tetrahedron
L Symmetry and Great Circles of Solids
M Proof of the Equation N = 10v2 + 2
N Parallel and Series Connections
O The Number of Connections Between Events
P Vector Equilibrium in Crystals
Q Bibliography
¶ A Introduction
This portion of Synergetics is intended to complement R. Buckminster Fuller’s text in several ways. The most extensive section, ‘‘Vector Equilibrium in Crystals,’’ summarizes the author’s own investigation of spatial order on a microscopic scale, and discloses an architecture of crystal structures that has many features in common with Fuller’s macroscopic coordinate systems. The question has been raised whether such structural similarities on such different scales are fortuitous, or whether the organizing forces are comparable over such different distances.
Such a question ignores the geometrical properties of space: space is not a passive
vacuum, but has properties which impose powerful constraints on any structure that
inhabits it. These properties of space, unlike the various specific interactive forces
between atoms or dome joints, are the same on every scale, and override any specific
interactions.
Fuller’s experimental mathematical approach is characterized by an uncanny sensitivity to significant numerical as well as geometrical patterns. Sometimes the underlying origin of these patterns is easily established, whereas in other cases it is still obscure. In a number of cases I have presented a proof or derivation from fundamental principles. While aware of R.B.F.’s disdain of proofs I am of the opinion that these proofs constitute important bridges between Fuller’s discoveries and established bodies of information and knowledge. I have attempted in these proofs to adhere to Fuller’s experimental intuitive style without sacrificing rigor, stressing the concrete and geometric rather than the abstract and symbolic.
These sections were originally planned as interpolations in Fuller’s principal exposition. The difference in style and trains of thought made such interruptions inappropriate. These contributions do not, for this reason, pretend to great continuity, but should be read as marginal notes to Fuller’s exposition.
¶ B Proof of Euler’s Equation
Euler’s equation is V + F = E + 2 where V = number of vertices, F = number of faces, E = number of edges. For a tetrahedron (not necessarily regular), V = 4, F = 4, E = 6; hence Euler’s equation is satisfied. From this tetrahedron, we can ‘‘grow’’ any other triangulated polyhedron by the following procedure. Choose a point outside a given face of the tetrahedron, and join that point to the vertices of that face by three straight lines. The polyhedron thus generated has one more vertex than did the tetrahedron and three additional edges. The original face was replaced by three new faces, so that the new polyhedron has two
faces more than did the tetrahedron. The same procedure can be repeated indefinitely,
each time growing a tetrahedron from a triangular face. With each step the number
of vertices is increased by one, the number of faces by two, the number of edges by
three.
A. polyhedron having V vertices can thus be created by adding to the tetrahedron (four vertices) in succession (V— 4) vertices. The resulting polyhedron has a number of edges equal to that of the original tetrahedron (6) plus 3 (7—4); the number of its faces equals that of the original tetrahedron (4) plus 2(V—4). Therefore, E = 6 + 3(V − 4) = 37 − 6, F = 4 + 2(7 − 4) = 27 − 4 − E + 2 = 3V − 4; V + F = 37 − 4; −7 + F = E + 2, q.e.d. We have thus proven Euler’s equation for a polyhedron that has only triangular faces. Suppose that we have a polyhedron having a face that is a polygon with n vertices, called an ‘‘n-gon’’; all its other faces are triangles. Call the number of vertices of this polyhedron v, its number of faces f, and its number of edges e. Choose a point outside the n-gon, and join it by straight lines to the n vertices of the n-gon. The polyhedron so created has only triangular faces; we have already proven that Euler’s equation is valid for it. In creating the new polyhedron from the original one we added a single vertex, so that this new polyhedron has (v + 1) vertices. We replaced the original n-gon by n triangular faces; as a result, the total number of faces of the new polyhedron is (f+n— 1). The new number of edges is (e + n). Since, for the new polyhedron, Euler’s equation holds,
(v + 1) + (f+n—l) = e+n+2
v + f = e + 2
We see, therefore, that Euler’s equation also holds for the polyhedron that had the n-gonal face. Since each n-gonal face can be replaced in the same fashion by triangular faces, we have proven Euler’s equation for polyhedra having any kind of polygonal faces.
¶ C Doubly Connected Polyhedra
A doughnut is a doubly connected body (Fig. Al). By this we mean that we can choose two points in a doughnut, connect them by two ropes, and pull the ropes taut without having the ropes coincide.
Doubly connected body
Singly connected body
Fig. A1. Doubly and singly connected figures.
The creation of a polyhedron from a tetrahedron, as described above, will generally result in a singly connected polyhedron (Fig. A1). However, in some cases, multiply connected bodies can result, and in this case Euler’s equation must be modified. Let us suppose that a polyhedron has a horseshoe-like outline. This might be thought of as a horseshoe magnet, with a very small gap between north and south poles and a faceted surface (Fig. A2). Suppose, now, that the faces on both sides of the gap are identical n-gons. For this polyhedron, Euler’s equations are valid: it could be grown from a tetrahedron by the procedure just discussed. Now if we let the gap width go to zero, the north and south poles annihilate each other, and a doubly connected body results. When this happens, the two faces on both sides of the gap
Fig. A2. Polyhedron with horseshoe-like outline.
vanish. Moreover, the n vertices of one face merge with the n vertices of the opposite face, so that 2n vertices merge into only n vertices, with a resulting loss of n vertices. Similarly, there is a loss of n edges. Thus the balance sheet is as follows:
| Horseshoe-Shaped Polyhedron | Doubly Connected Body | |
| No. of vertices: | V | V = v-n |
| No. of faces: | f | F=f-2 |
| No. of edges: | e | E =e—n |
From this balance sheet it follows that for the doubly connected body: V +F —E—v—n +f— 2—e +n =v +f—e—2. Since for the horseshoe-shaped polyhedron v +/=e + 2, V + F—E = 0. For a doubly connected body, Euler’s equation is therefore: V +F =E. Fuller would then associate the ‘‘2’’ in the equation for singly connected polyhedra with the two faces on either side of the gap, which are destroyed when the doubly connected body is created. The argument also holds in reverse; any multiply connected body can be turned into a singly connected one by ‘‘sawing through’’ each connection and thus creating a pair of faces.
¶ D The Sum of the Surface Angles of a Singly Connected Polyhedron
Consider a polyhedron whose faces are general n-gons. In each face, choose a point, and connect this point by straight lines to the vertices of the n-gon. The surface of the polyhedron is then triangulated.
Two triangles meet at each edge; the sum of the angles of the triangles in the surface of the entire polyhedron is therefore 2E x 180°. However, this sum includes the angles that meet at the points chosen in each face, which add up to 360°in each face. The sum of the surface angles of the polyhedron therefore equals (2E—2F)x 180°.
For singly connected polyhedra, E—F = V -2, so that the sum of the surface angles is: V X 360°—720°.
For doubly connected polyhedra, E − F = V , so that the sum of the surface angles is: V X 360°.
¶ E Duals
If we choose a single point in each face of a polyhedron, then these points may be considered as vertices of a new polyhedron. Two polyhedra so related are called each other’s duals. The number of faces of a polyhedron equals the number of vertices of its dual. The sum of the number of faces and the number of vertices is therefore the same for any polyhedron as for its dual. From Euler’s equation, it follows that dual polyhedra have identical numbers of edges.
To find the dual of a regular polyhedron, we choose points at the geometric centers of its faces. Let us consider first the regular, triangulated solids, which Fuller considers fundamental. The simplest of these is the tetrahedron: V = 4, F = 4, E = 6. Since the number of vertices of the tetrahedron equals its number of faces, its dual is again a tetrahedron. The tetrahedron is its own dual; it is unique in this respect. Note that Fuller’s positive and negative tetrahedra are each other’s duals.
The octahedron has K = 6, F = 8, E = 12. Its dual must have V = 8, F = 6, E = 12. This is the cube.
The icosahedron has V = 12, F = 20, E = 30. Its dual has V = 20, F = 12, E = 30; it is the pentagonal dodecahedron. We see, therefore, that the two non-triangulated Platonic solids are just the duals of two triangulated solids. The numbers of vertices, edges, and faces for the Platonic solids are thus summarized as follows:
Tetrahedron
Octahedron
Icosahedron
V E F
4 6 4
6 12 8
12 30 20
Tetrahedron
Cube
Pentagonal
Dodecahedron
¶ F Close Packing
Consider for a moment a planar array of closely packed identical circular disks. Imagine a bee inside each circular cell, producing a wax cylinder. If the bees all produce at equal rate but do not stop at the point when the cylinders just touch, then the well-known honeycomb pattern is produced (Fig. A3).
We call the hexagonal cell around each bee his ‘‘domain’’; the plane is divided equally between all bees. Every point within a cell is closer to its central bee than to any other bee.
Fig. A3. Honeycomb pattern.
The edges of the hexagonal domains perpendicularly bisect the lines joining neighbor bees; every point of such an edge is equidistant between these bees. Edges meet at vertices that are equidistant from three bees located at the corners of an equilateral triangle.
Let us apply the same idea to a three-dimensional, cubically close-packed array of identical spheres. Suppose there to be a creature called a ‘‘cee,’’ which spins wax spheres around itself at a steady radial rate. Suppose that these cees locate themselves most efficiently—namely, at the centers of cubically close-packed identical spheres. Like the bees, the cees continue their activity beyond the point where their spheres just make contact. What is the resulting domain for each cee? It is the so-called ‘‘rhombohedral dodecahedron’’ (cf. Fig. A19), which has 12 diamond-shaped faces. Because each face has one acute and one obtuse angle, this is not a regular polyhedron, the latter
having all edges and all angles equal. The rhombododecahedron is a space filler (cf. Sec. I). For the rhombododecahedron, K=14, F=12, E = 24. Its dual has V = 12, F = 14, E = 24; it is our old friend the cuboctahedron, or in Fuller’s terms, the ‘‘vector equilibrium.’’ The duality between ‘‘vector equilibrium’’ and rhombododecahedron is easily visualized as follows: Each ‘‘cee’’ in the center of a cubically close-packed sphere has 12 neighbor cees at the corners of a ‘‘vector-equilibrium’’ polyhedron. Each ‘‘radial’’ from a cee to one of its 12 neighbors is perpendicularly bisected by a face of its domain. Therefore, to each vertex of the ‘‘vector equilibrium,’’ there corresponds a face of the rhombododecahedron, and vice versa; this observation establishes the duality relation between the two polyhedra.
Being only a ‘‘semi-regular’’ polyhedron, the rhombododecahedron has two different kinds of vertices. There are six vertices at which four acute surface angles meet. In addition, there are eight vertices at which three obtuse surface angles meet. The six vertices form the vertices of an octahedron around each cee in the center of the rhombododecahedron and are themselves in turn octahedrally surrounded by cees. The eight vertices form a cube around the cee at the center of the rhombododecahedron, and are themselves tetrahedrally surrounded by cees. Any creature desirous of maintaining a maximum distance from as many cees as possible would locate itself preferably on one of the octahedrally
surrounded vertices of the domain, with second choice going to the tetrahedrally surrounded
vertices. Although ‘‘cees’’ are only fictional animals, we find that the configuration of
ions in many crystalline solids actually do obey these geometrical principles (cf. Sec.
P).
¶ G The Icosahedron and the Golden Section
The icosahedron is the most complex of the three triangulated Platonic solids. Its thirty edges belong to five distinct sets of six edges each, such that all six edges within one set are mutually parallel or perpendicular. We say, therefore, that the 30 edges belong to five orthogonal sets. The centers of the six edges belonging to the same orthogonal set form the vertices of a regular octahedron. Therefore, five regular octahedra can be inscribed in an icosahedron, having their vertices on the centers of the edges of the icosahedron.
A straight line joining any vertices that are not adjacent is called a ‘‘diagonal.’’ The icosahedron has diagonals of only two different lengths. Thelonger diagonals pass through the center of the icosahedron, joining vertices on opposite sides of the center. The shorter diagonals join vertices that are second-nearest neighbors; going along the surface of the icosahedron, it would require two edge lengths to go from one end of a ‘‘shorter diagonal’’, to the other.
If we make a cross section (Fig. A4) through the icosahedron along a plane perpendicular to
a long diagonal and containing five vertices, a pentagon results. The edges of this pentagon are
all edges of the icosahedron, its vertices are vertices of the icosahedron, and its diagonals are
shorter diagonals of the icosahedron. The edges and shorter diagonals of the icosahedron are
related by the golden section, as will be shown below.
e = edge length
d = diagonal length
Fig. A4. Pentagonal cross section through an icosahedron.
The golden section is the division of a whole into two unequal parts such that the smaller part is related to the larger as the larger is to the whole. This section has a perfection that has appealed to designers for centuries; its relation to the most perfect of Platonic solids, the icosahedron, is therefore worth investigating. Below (Fig. A5), we have drawn a circle around the pentagonal cross section of the icosahedron, as well as a second diagonal that intersects the first at point X. The five vertices divide the circumference of the circle into five equal arcs, with the result that each angle subtended at the circumference of the circle by each arc equals 36 degrees. From the equality of the angles BXC and Z.XBC, it follows thatXC =BC, so that the point X divides the diagonal AC into two portions, of which the larger equals the edge length. It similarly divides diagonal BE into two unequal portions, of which the larger equals the edge length.
Fig. A5. Regular pentagon and the golden section.
As a result of the similarity of IXAXB and ABC:
AX + AB=AB+AC
Therefore, (d—e) be =e 4-d. This proves that the shorter diagonals of the icosahedron intersect at points that divide each diagonal according to the golden section, and that the larger portion of each diagonal just equals the edge length of the icosahedron.
If we draw in all the diagonals of the pentagon (Fig. A6), a new pentagon is generated whose vertices are U, V, W, X, Y. This pentagon can be considered a cross section through an icosahedron whose edges are portions of the shorter diagonals of the original icosahedron. There is also a ‘‘magical’’ relation between the edge lengths of the larger and smaller icosahedra: because of the similarity of SBXY and T\BED, XY+ED =BX-bBE. Therefore, XY-be = (d-e)+d. The ratio of the edge lengths of the smaller and larger icosahedra is therefore the same as that of the smaller portion to the whole in a golden section.
Fig. A6. Generation of smaller pentagon from larger one by the golden section.
Diagonals of the pentagon UVWXY generate a smaller pentagon, just as UVWXY was generated from ABCDE, and the process can be repeated indefinitely; each time the edge length is decreased by A factor (d—e)ld.
The polygon AXBYCUDVEW is a pentagonal star, whose remarkable geometrical properties derived above were often thought to give it magic power.
The age of reason rejected the magical power of geometrical forms, and rightly so. However, in rejecting this power, it lost interest in the forms per se; we are now discovering that there is distinct physical significance in many geometrical forms.
¶ H Remarks on Some Elementary Volume Relations Between Familiar Solids
The volume of a regular octahedron equals exactly four times the volume of a regular tetrahedron of the same edge length. A tetrahedron inscribed in a cube such that the edges of the tetrahedron constitute face diagonals of the cube occupies exactly one third of the volume of the cube. As we are, in the calculation of areas and volumes, concerned basically with problems in affine geometry, these relations are not limited to regular solids, but apply equally well to irregular tetrahedra and octahedra having congruent faces, and to an irregular tetrahedron inscribed in a general parallelepiped. When these relations are derived with the aid of the usual formula for the volume of a pyramid, V = 1 ∕ 3Ah, a good many irrational numbers are involved, and the simple integral ratios emerge almost incidentally. Somehow these simple integral values of the volume
ratios of common solids are not part of our scientific culture, and a lack of familiarity
with them frequently leads to unnecessarily cumbersome computations. It appears
that a bias of our culture to orthogonal Cartesian coordinates has obscured these
relations.
Instead of accepting and mechanically applying some geometric formulas, let us follow the common line of reasoning of physics, namely, to generalize from some physical observations. We shall then find that the common geometrical formulas are based on an arbitrary assignment of units that happen to simplify calculations in orthogonal Cartesian coordinates, but complicate work in other coordinates.
In Fig. A7 we show two geometrically similar pairs of figures; in Fig. A7tz these figures are squares, while in Fig. Mb they are irregular triangles. It is observed that in both cases the areas of these pairs of figures are in ratio 4:1, whereas their edge lengths are in ratio 2:1.
t Cf. Fig. 950.31. Cf. also Loeb and Pearsall, Am. J. Phys. 31, 191.
Fig. A7a. Square ABCD is similar to square A 'B'C'D.
Fig. A7b. Triangle ABC is similar to triangle A'B'C'.
Lemma I: The areas, A, of geometrically similar figures are proportional to the square of the length of corresponding lines in these figures, d:
A=kd2 (1)
where K is a constant depending on the shape of the figure.
Equation (1) has fundamental significance; the numerical value of the proportionality constant K depends on the choice of units for A and d. In orthogonal Cartesian coordinates it is convenient to choose units such that a square of unit length has unit area. There is no fundamental objection, though, to choosing a system of units such that an equilateral triangle of unit edge length has a unit area. In fact, someone working with the geometry of close-packed identical spheres might consider such a system of units ‘‘rationalized’’!
Similar considerations lead to:
Lemma II: The volumes, V, of geometrically similar solids are proportional to the cube of the length of corresponding lines in these solids, d:
V=Kd3 (2)
where A’’ is a constant depending on the shape of the solid.
It should be observed that Loeb and Pearsall’s derivation of the volume ratios of an octahedron and a tetrahedron was based on comparing volumes having the same ‘‘shape constant’’ K, so that assignment of units did not enter the problem at all. Application of the ‘‘familiar’’ formula V = VsAh involves the numerical determination of the constant K, which is irrational for a regular tetrahedron, and eventually cancels out again.
The twelve lines joining the centers of the edges of a (generally irregular) tetrahedron constitute the edges of a (generally irregular) octahedron (Fig. A8). The volume of the tetrahedron, VT, equals that of the octahedron, V , plus the combined volume of four tetrahedra similar to the original tetrahedron but having half its edge length. If the volume of each smaller tetrahedron is Vt:
From Lemma II:
(3)
Fig. A8. The tetrahedron ABCD (volume VT) consists of:
In Fig. A9 there is a parallelepiped with an inscribed (generally irregular) tetrahedron. The parallelepiped contains, besides the inscribed tetrahedron, four octants of an octahedron whose faces are congruent with those of the inscribed tetrahedron. If the volume of the parallelepiped is denoted by Vp:
V=V +-V .
p ' 8°
Substitute Equation 3:
(4)
Fig. A9. Tetrahedron EBDG inscribed in parallelepiped ABCDEFGH. (Another tetrahedron, ACFH, enantiomorphic with EBDG, can be inscribed in the same parallelepiped.)
Hence, a tetrahedron inscribed in a parallelepiped occupies exactly one- third the volume of that parallelepiped. In general, there are two enantiomorphic tetrahedra of equal volumes that can be inscribed in a parallelepiped.
Professor Philippe LeCorbeiller has pointed out that one of the proofs 2 only; four tetrahedra
/\ABC ~/SACD ~sCBD.
Considering the hypotenuse as a ‘‘corresponding line’’ in each of these triangles, one concludes from Lemma I:
Area (J\ABCF)=kAB2 Area (A/fCD) = L42 Area (JSCBDkCB2
Fig. A10. Proof of Pythagoras’ theorem.
Since Area (AABC) =Area (Az4C>) + Area (ACB>), AB2=AC2+CB2, q.e.d.
Conclusion
Uncritical acceptance of geometrical formulas as fundamental laws, particularly in systems that do not naturally fit orthogonal Cartesian coordinates, frequently leads to unnecessarily clumsy calculations and tends to obscure fundamental relationships. It is well to avoid instilling too rigid a faith in the orthogonal system into students of tender and impressionable age!
¶ I Space-Filling Polyhedra
Fig. Al 1. Filling space with cubes.
and an octahedron are necessary to produce a tetrahedron having eight times the volume (twice the linear dimension) of the original tetrahedron (Fig. A12). Conversely, six octahedra and eight tetrahedra are necessary to produce an octahedron having eight times the volume (twice the linear dimension) of the original octahedron (Fig. A13). In turn, the larger octahedra and
tetrahedra can be combined in the appropriate proportions to produce even larger
tetrahedra and octahedra until all of space can be filled. Therefore, whereas octahedra and
tetrahedra cannot by themselves fill all of space, they can be packed together to do
so.
In Fig. A14, we place two tetrahedra on opposite faces of an octahedron. The result is a parallelepiped that has six identical faces, each being a rhomb or diamond with angles 60 degrees and 120 degrees. Such a particular parallelepiped is called a rhombohedron. Identical rhombohedra can be placed next to
Fig. A12. Scaling the tetrahedron by a linear factor of two.
Octahedron
Fig. A13. Scaling the octahedron by a linear factor of two.
each other, in a fashion analogous to that shown in Fig. All, to fill all of space. Since the rhombohedron is a space filler, and can be built from an octahedron and two tetrahedra (in opposite orientations), we may conclude that in order to fill space, each octahedron must be accompanied by a pair of tetrahedra in opposite orientation. We shall later test several polyhedra and combinations of polyhedra for space filling. Since these polyhedra can
Tetrahedron
Fig. A14. Generation of rhombohedron from an octahedron and a pair of tetrahedra. frequently be expressed as combinations of tetrahedra and octahedra, we shall require that each time, as a necessary condition for space filling, each octahedron occur in conjunction with a pair of oppositely oriented tetrahedra.
As an example of this necessary condition, let us check whether the large tetrahedron formed in Fig. A12 and the large octahedron formed in Fig. A13 can indeed fill space. Suppose that we did not know the ratio in which they must be mixed. Let us try A tetrahedra and B octahedra, and determine the ratio of a to B. We then have the following balance sheet:
No. of Small No. of Small
Tetrahedra Octahedra
A Tetrahedra: 4a a
B Octahedra: 8b 6b
—4- A Tetrahedra + B Octahedra: 4a + 8b a + 6b
The necessary condition for space filling is then: 4a + 8b = 2(a + 6b). Hence a = 2b, so that the large tetrahedron and octahedron also must mix in the 2 : 1 ratio, as would be expected.
Since the cube is a space filler, it too should be expressible in terms of two tetrahedra per octahedron. This is done in Fig. A15, where a tetrahedron has one octant of an octahedron attached to each of its four faces to form a cube. Hence:
Cube = Tetrahedron + 4/8 Octahedron = Tetrahedron + 1 ∕ 2 Octahedron
Therefore, the cube contains a tetrahedron and a half-octahedron, hence satisfies the necessary condition for space filling.
We have now surveyed all of the regular solids, except the icosahedron and its dual, the pentagonal dodecahedron. These polyhedra are not space fillers because they have fivefold rotational symmetry: a fundamental theorem in symmetry theory precludes the existence of parallel fivefold axes of rotational symmetry4.
Fig. A15. The cube expressed in terms of octahedra and tetrahedra.
We next turn our attention to semiregular solids, i.e., solids having all edges of the same length but having various surface angles. The first is a truncated octahedron. Eight of its faces are regular hexagons, six are squares. It is obtained from the cube by bisecting each of its octants, as shown in Fig. A16a. Since this solid is obtained by a simple bisection of the cube, it is itself a space filler.
Fig. A16a. One octant of the cube: the plane of the hexagon perpendicularly bisects the body diagonal.
Figure A 16c shows how the truncated octahedron can be related to octahedra and tetrahedra. A ‘‘supercube’’ is bisected as before: eight small cubes (such as shown in Fig. A16&) here produce one octant of a truncated superoctahedron. The octahedra and tetrahedra are differentiated by shading, as indicated in the drawing. Six of the eight small cubes are cut by the bisecting plane; one cube lies entirely on the reader’s side of the plane; the remaining one lies beyond that plane. The bisecting plane cuts a single octant of an octahedron off each of six cubes: in three of these cubes, this octant lies on the reader’s side of the plane; in the remaining three, it lies beyond this plane. Therefore, there are beyond the plane:
1 tetrahedron and 4/s octahedron in a single cube.
1 tetrahedron and 3/s octahedron in three cubes.
Vs octahedron in three cubes.
Fig. A16b. The entire cube with inscribed truncated octahedron.
Therefore, the portion of the supercube beyond the plane contains in total (1 + 3) tetrahedra and (4/s + 9/s + 3/s) octahedra, i.e., 4 tetrahedra and 2 octahedra. Of course, the portion of the cube on the reader’s side of the plane also contains 4 tetrahedra and 2 octahedra. From this construction we conclude:
Fig. A16c. Relation of truncated octahedron to octahedra and tetrahedra.
Octant of octahedron
Tetrahedron
- a)
- The supercube contains 8 tetrahedra and 4 octahedra.
- b)
- The truncated superoctahedron contains 32 tetrahedra and 16 octahedra.
It should be noted that in each of these figures the condition for space filling is obeyed.
Next we consider the cuboctahedron, Buckminster Fuller’s ‘‘vector equilibrium.’’ This solid has six square faces and eight triangular ones; its twelve vertices are the midpoints of the twelve edges of a cube (Fig. A17). In terms of octahedra and tetrahedra, each square face is the equatorial square of an octahedron, and each triangular face is the base of a tetrahedron. There are, therefore, eight tetrahedra and six half-octahedra, or a ratio of 8 tetrahedra to 3 octahedra. The cuboctahedron is therefore one octahedron short of meeting the necessary space-filling condition. When an octant of an octahedron is added to each of its eight triangular faces, the cuboctahedron is transformed into a cube (Fig. A17), hence a space filler. Cuboctahedra and octahedra thus together fill space in ratio 14--1.
Fig. A17. Cuboctahedron.
Next is the truncated tetrahedron (Fig. Al 8a). It has four triangular and four hexagonal faces. It can be constructed by placing an octahedron on each of the four faces of a tetrahedron; the triangular faces of the truncated tetrahedron are faces of these octahedra (Fig. A18b). The dotted edges in Fig. A18b are supplied by tetrahedra: one at the center and six at the dotted edges. The ratio is therefore 7 tetrahedra 4--4 octahedra, 1 tetrahedron short of satisfying the space-filling condition. Therefore, truncated tetrahedra together with tetrahedra fill space in a ratio 14 : 1.
Fig. A18a. The truncated tetrahedron with equal edges.
Since truncated tetrahedra are lacking in tetrahedra, and the cuboctahedron has an excess of tetrahedra for meeting the space-filling condition, it might be interesting to see whether they could be admixed appropriately to fill space. The following would be the balance sheet:
No. of No. of
Tetrahedra Octahedra
A cuboctahedra 8a 3a
B truncated tetrahedra lb 4b
A cuboctahedra + B truncated tetrahedra 8a + lb 3a + 4b
The space-filling condition therefore requires: 8a + lb = 2(3a + 46). . B = 2a. Thus the necessary condition is fulfilled if there are two truncated tetrahedra per cuboctahedron. However, this condition is not sufficient, for the two solids do not have compatible faces; the square faces of the cuboctahedron cannot be adjacent to the hexagonal faces of the truncated tetrahedron. What
Fig. A18b. The truncated tetrahedron expressed in terms of octahedra and tetrahedra. we need is yet another solid, which provides hexagonal and square faces, as a bridge between the other two solids. Such a solid is the truncated octahedron considered above (Fig. A16). We make up two balance sheets: one for the ratio of tetrahedra to octahedra, the other for matching faces:
| No. of | No. of | |
| Tetrahedra | Octahedra | |
| A cuboctahedra | 8a | 3a |
| B truncated tetrahedra | lb | 4b |
| c truncated octahedra | 1 2 C | J_ 4 C |
| 8a + lb + Tc | 3a + 4b +~4C | |
.8a + lb + 4c = 2(3a + 4b + Tc), ,-.b = 2a. The balance for the truncated octahedra follows from the fact that the truncated octahedron is obtained by bisecting the cube. It is observed that the number c cancels out in the balancing equation; this happens because the truncated octahedron is itself a space filler, and may be admixed without disturbing the balance. The balance sheet for matching faces is as follows:
| No. of | No. of | No. of Hexa- | ||
| Square Faces | Triangular Faces | gonal Faces | ||
| A | cuboctahedra | 6a | 8a | — |
| B | truncated tetrahedra | — | 4b | 4b |
| c | Truncated octahedra | 6c | — | 8c |
Since the number of square faces of the cuboctahedra must equal the number of square faces of the truncated octahedra (assuming that no two cuboctahedra have square faces in contact), a—c. Similarly, the balance of triangular faces requires 8a = 4b, b =2a=2c. We have already seen that the condition b=2a satisfies the tetrahedra 4-octahedra balance. The equation b—2c gives the appropriate balance of hexagonal faces. Therefore, both conditions are satisfied if cuboctahedra, truncated tetrahedra, and truncated octahedra are combined in the ratio 14 − 2 − t − 1. We have here an example of three semiregular solids combining to fill all of space.
A regular solid can be subdivided into congruent pyramids, the base of each being a polygonal face of the solid, each apex the center of the solid. For example, the regular tetrahedron can be divided into four congruent pyramids, each having as its base an equilateral triangle whose corners subtend angles of approximately 109 degrees at the apex. An octahedron can be divided into eight pyramids, each also having an equilateral triangle as base, but the angles around the apex are here 90 degrees. For a cube there are six pyramids, each having a square base. The pyramids so produced can be combined in interesting ways to produce semiregular solids.
One particularly important solid, already encountered in connection with close packing, is the rhombohedral dodecahedron, or briefly rhom- bododecahedron (Fig. A19). This solid is the dual of the cuboctahedron. It can be generated in two different ways. First take two identical cubes. Divide one of these cubes into its six component pyramids. Place each of these pyramids with its square face against a face of the second cube. The result is the rhombododecahedronj which has a volume twice that of each cube from which it is generated. Secondly, take an octahedron and two tetrahedra. Divide both tetrahedra into their four component pyramids. Place each of the
resulting eight pyramids against one of the faces of the octahedron. The result is again
the rhombododecahedron. This solid has eight corners where three angles of about
109 degrees meet, and six corners at each of which four angles of about 71 degrees
meet.
From both constructions, it is obvious that the rhombododecahedron is a space filler; the component tetrahedra and octahedra occur in the desired 2 4-1 ratio, and all faces are identical lozenges that adjoin in space filling. It should be noted that whereas a cube is a space filler; its dual the octahedron is not.
Fig. A19. Rhombododecahedron.
OCTAHEDRON
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ojpaqojoo jo on
Fig. A20. Regular and semiregular solids expressed in terms of tetrahedra and octahedra.
The rhombododecahedron fills space; its dual the cuboctahedron does not. Generally, then, the duals of space fillers do not fill space. It should also be noted that only one of the regular (Platonic) solids is also a space filler, and that is the cube.
Fig. A20 summarizes the proportions of octahedra and tetrahedra in each of the solids discussed here. The straight line represents the condition of space filling. Such a diagram suggests that each polyhedron can be represented by a two-dimensional vector whose components express the ‘‘tetrahedral and octahedral natures’’ admixed in each polyhedron.
When several polyhedra are packed together, a necessary condition for space filling is that the sum of their vectors lies along the solid line. A polyhedron possessing too much ‘‘octahedrality’’ for space filling lies above the solid line; one possessing too much ‘‘tetrahedrality’’ (too little octahedrality) lies below the solid line.
¶ J The A and B Modules and the Trirectangular Tetrahedron
The feeling for balance of space between octahedrality and tetrahedrality led Fuller to devise his so-called A and B modules. These particles (cf. Fig. 951.10) are derived from the regular octahedron and regular tetrahedron as follows: Take an octant of the octahedron, and subdivide the face of the octahedron symmetrically into six mutually equivalent triangles (Fig. A21).
Fig. A21. Subdivision of triangular face into six equal parts.
The tetrahedron whose base is one sixth of the octahedral face and whose apex is the center of the octahedron (corner of an octant) is called the asymmetric unit of the octahedron; its volume equals 1/g X 1/s = V48 of that of the octahedron.
The asymmetric unit of the tetrahedron is obtained similarly: the tetrahedron is first subdivided into four equivalent parts, each having as base a face of the tetrahedron and as apex its center of gravity. Next the base is subdivided into six equivalent parts according to Fig. A21: each asymmetric unit of the tetrahedron has one of the right triangles as base, and the center of the original tetrahedron as apex. Its volume is 1/g X V4 of that of the octahedron.
The volume ratio of the octahedral and tetrahedral asymmetric units equals:
(x/48 volume of octahedron) 4-(V24 volume of tetrahedron) =
nA, volume of octahedron A ~
24/48 x - - = 24/48 x 4 = 2
vol. of tetrahedron
The A module is defined as the asymmetric unit of the regular tetrahedron; the B module is the portion of the asymmetric unit of the octahedron that is left over after the/1 module has been scooped out of it. These, then, are the units in terms of which Fuller expresses his space. Since they cannot be symmetrically subdivided, they are true modular quanta. The A and B modules have equal volumes, but are not commensurable: they are not expressible in terms of a common unit.
It should be noted that there are actually four fundamental units, two being the mirror images of the other two. We shall denote the mirror images by asterisks: A* and B*. Since the regular and semiregular polyhedra all have mirror symmetry, we can expect their constituent A and B modules always to be paired to their respective mirror images A* andB*.
In Fig. A22 we have redrawn Fig. A20 in terms of the basic units A and 7?: this plot is somewhat comparable to a plot of imaginary numbers, in which case the units are also non-commensurable quantities of equal magnitude, namely 1 andV —1. The volume of each polyhedron plotted equals (in terms of the volume of a single A of B module) the sum of its absissa and its ordinate in Fig. A22.
Buckminster Fuller would not fail to remark on the fact that all polyhedra represented in Fig. A22 are composed of exact dozens of As and Bs, the number 72 having special significance in the cubic system. (All polyhedra discussed have so-called cubic symmetry, i.e., the same symmetry as the cube.) In this system every asymmetric unit is repeated three times around each axis perpendicular to triangular faces or cross-sections, and four times around each axis perpendicular to a square face or cross-section, hence at least twelve times.
OCTAHEDRON
I L_L
O CO (0 o oo r* ro CM 04
J L
cm
<0 IO
CM CM
OU>stCMOCO<0"4- CMOlL??>TtCMOCD<0M’CM rf CM—O O> CO in <f rOCMOOror-aJM-CM—
CMCMCMcj—-—————-
—g )O -ON
Fig. A22. Polyhedra expressed in terms of .4 andB units.
Fig. A23. Space filling with AA*B and A *AB* modules.
The trirectangular tetrahedron modules shown in Fig. 951.10 both consist of an A module, its mirror image, and a single# module. If an asterisk is used to denote a mirror image, then one module can be denoted AA*B, its mirror image A*AB*. These two modules are shown in Fig. A23 and again in Fig. A24; these two figures demonstrate the two ways in which these modules may be stacked to fill space. In Fig. A23, triangles QRS and Q*R*S are brought in contact, whereas in Fig. A24, triangles PQS and p*Q*s* are brought in contact for space filling. Since# has no mirror symmetry, # and #* are oppositely congruent, so that/L4*# would be oppositely congruent with /IM#*.
In Fuller’s demonstration that/L4*# and A*AB* together can fill space in two distinct ways, something quite remarkable happened, namely that the two modules exchanged places (cf. Figs. A23 and A24). Since one module can exactly occupy the space previously occupied by the other, albeit in a rotated position, the following logical conclusions may be drawn:
Fig. A24. The alternate arrangement of space filling with/l/l*# and/lM5* modules.
- 1.
- /L4*5 is directly congruent with A *AB*.
- 2.
- Since the two modules are directly congruent, only one of them is required to fill all of space.
- 3.
- The mirror-symmetrical module AA* can combine with the asymmetrical module B in 1 : 1 ratio to fill all of space. The module /L4* could equally well combine with B* to fill all of space.
These conclusions are quite remarkable because so far we have only encountered space fillers that have minor symmetry, such as octahedra plus tetrahedra, cubes, or rhombododecahedra. Fuller himself uses AA*B in conjunction with its enantiomorph (mirror image) to fill space. Nevertheless, we shall presently see that the conclusions listed above are entirely correct, and hinge on the direct congruency of AA*B and A*AB*.
We already know that AA*B and A*AB* are oppositely congruent (Figs. A23 and A24). Two objects that are simultaneously directly and oppositely congruent are necessarily each mirror symmetrical. The three above-mentioned conclusions can therefore only be valid if we locate a plane of mirror symmetry in each of the modules AA*B and A*AB*. To this purpose we have, in Fig. A25, folded the faces of/L4*B out, and laid them flat beside each other. Module A A *B *, being oppositely congruent to/l/l*B, would yield the same layout. The triangle PQR is a face of the B module. Since AB was defined as the asymmetrical unit of a regular octahedron, f\PQR represents just one quadrant of an equilateral plane of such an octahedron. Therefore, PQ =RQ, so that &PQS is congruent with &RQS. From this congruency, we conclude (cf. Fig. A23 or A24) that module AA*B has indeed a plane of mirror symmetry; this plane contains the edge QS, and bisectsZpQR. Module A*AB* is similarly bisected by a mirror plane through edge Q*S*.
Fig. A25. The/L4*B or/l*AB* module unfolded.
Space filling with only the module/L4 *B is illustrated in Fig. A26: here, /\QRS of one module is brought into coincidence with /\QPS of an adjacent
Fig. A26. The module AA*B in two orientations, ready for space filling.
module. This arrangement is, in fact, a third alternative to the two already proposed by Fuller (Figs. A23 and A24); it is, of course, complemented by an entirely equivalent packing of modules A *AB*.
There is a direct relation between the AA *B module and the rhombododecahedron. It was shown (see p. 813) that the rhombododecahedron can be thought of as a regular octahedron with a quarter-tetrahedron attached to each of its eight faces. The rhombododecahedron can be divided into eight octants, each of which consists of an octant of an octahedron plus a quartertetrahedron. One such octant is shown in Fig. A27. In turn, the octant can be subdivided into six equivalent modules, of which three are enantiomorphic to the other three. As a matter of fact, three are/L4*2? modules, the other three are Zl*242?* modules. Of the AA*B modules, the AB portion constitutes the asymmetric unit of the octahedron, while theZl* portion constitutes the asymmetric unit of the quarter-tetrahedron. Thus the rhombododecahedron can be constructed from 24 AA*B modules and 24X*/15* modules. Since, however, these modules were shown to be directly congruent with each other, 48 of either kind of module will also form a rhombododecahedron.
Fig. K17. Octant of a rhombododecahedron exploded into an octant of an octahedron plus a quarter-tetrahedron.
The rhombododecahedron can be constructed in two ways from 24 AA *B and 24 A*AB* modules, which correspond to Fuller’s two space-filling modes. In Fig. A28, we labeled some vertices to correspond to the orientation of the modules shown in Fig. A23. The orientation of Fig. A24 might equally well have been used, however. In the first case, the R-vertices of the modules all coincide at the center of the rhombododecahedron, whereas in the second, the R-vertices constitute the acute vertices of the rhombododecahedron. The
p*
A* A B* module
Q*
Pair to form R—centered
A A* B module
R
Q
Pair to form P—centered rhombododecahedren
rhombododecahedren.
R
P
Fig. A28. Ambivalence of the modes of Figs. A23 and A24 in space filling.
orientation of Fig. A26 leads to the construction of a rhombododecahedron from 48/L4*B modules; we shall return to this construction later on.
I want to raise first the question whether or not the two modes represented by Figs. A23 and A24 respectively are indeed distinct modes of space filling. There is no doubt about their representing two different modes of constructing a rhombododecahedron. As the rhombododecahedron is a space filler, the problem would appear to be solved right here. However, when rhom- bododecahedra are stacked together to fill all of space, the centers of the rhombododecahedra form a close-packed cubic lattice, as do their acute vertices (cf. section F). One might
therefore fill space with rhombododecahedra having R-vertices at their centers, but
slice space up again into equivalent rhombododecahedra whose centers are P-vertices.
This ambivalence is illustrated by Fig. A28, in which an AAB module is juxtaposed
with two adjacent AA *B modules in a space-filling array. One might equally well
pair
* with one AA*B neighbor to form an 7?-centered rhombododecahedron as with the other to form a P-centered rhombododecahedron: in actual space filling, it does not matter at all which vantage point one takes. We conclude, therefore, that the orientations of Figs. A23 and A24 lead to different finite polyhedra, but are equivalent in filling all of space. However, the orientation of Fig. A26, which uses B without B*, is fundamentally different, so that there are two fundamentally different arrangements for space filling, namely the one in which AA*B and A*AB* are used together, and the one in which only/L4*B or A*AB* is used.
¶ K The Rectangular Cross Section of the Regular Tetrahedron
A cross section through a regular tetrahedron made parallel to a pair of nonintersecting edges is a rectangle. Every rectangle so generated from a given tetrahedron has the same circumference. This latter observation is proven conveniently by unfolding the tetrahedron, keeping the trace of the rectangular cross section on the faces of the tetrahedron (Fig. A29). The faces of the tetrahedron can be unfolded into a parallelogram having one pair of edges as long as the edge length of the tetrahedron, the second pair twice as long. The trace of the rectangular cross section on the surface of the tetrahedron appears in the unfolded figure as a line parallel to the longer edge of the parallelogram. The length of this line equals twice the edge length of the tetrahedron regardless of where the cross section is made. Therefore, the circumference of any rectangular cross section of a regular tetrahedron equals twice the edge length of the tetrahedron.
¶ L Symmetry and Great Circles of Solids
A solid has rotational symmetry if it does not appear changed after rotation around a straight line through an angle smaller than 360 degrees. The straight line is called an axis of rotational symmetry, or simply a symmetry axis.
Tetrahedron with Rectangular Crossection
Trace of Rectangular Crossection
Tetrahedron Unfolded
Fig. A29. Cross sections of tetrahedra.
When a solid is rotated 360 degrees around a symmetry axis, the same pattern makes its appearance several times: the number of times it appears through a 360 degrees rotation is called the symmetry value of the axis. A body diagonal of a cube is a three-fold symmetry axis; lines joining centers of opposite faces of a cube are four-fold symmetry axes, and lines joining centers of opposite edges of a cube are two-fold axes.
If the geometrical center of a polyhedron is collinear with and equidistant from a pair of vertices, this polyhedron has inversion symmetry. A cube and an octahedron have inversion symmetry, but a tetrahedron does not.
If a solid can be divided by a plane into two halves that are each other’s mirror image, then that solid has reflection symmetry; the plane is called a mirror plane. When a solid possesses several mirror planes, any line of intersection of these planes is an axis of rotational symmetry.
The cuboctahedron (‘‘vector equilibrium’’) has the following axes of rotational symmetry:
Six two-fold axes joining the six pairs of opposite vertices.
Four three-fold axes joining the centers of opposite triangular faces.
Three four-fold axes joining the centers of opposite square faces.
In addition, there are lines joining the centers of opposite edges; these have a novel symmetry property. When the cuboctahedron is rotated through 180 degrees around such a line, the original pattern reappears, but upside down rather than in its original orientation. These lines are called roto-reflection axes: rotation through such an axis followed by reflection in a plane perpendicular to the axis reproduces the original polyhedron. There are twelve such roto-reflection axes.
Each of the 25 great circles of the cuboctahedron is defined by a plane perpendicular to one of the symmetry axes. The six circles perpendicular to the two-fold axes lie in reflection planes of the cuboctahedron. The four great circles perpendicular to the three-fold axes are not mirror planes: one half of the cuboctahedron, upon being reflected into the plane of such a great circle, must also be rotated 60 degrees to reproduce the other half. Rotation of 60 degrees of one half of the cuboctahedron with respect to the other half produces a polyhedron associated with hexagonal close packing; such a polyhedron has not four, but only a single three-fold axis, and a mirror plane perpendicular to it.
The three great circles perpendicular to the four-fold axes lie in mirror planes. The 12 great circles perpendicular to the roto-reflection axes do not lie in mirror planes.
The icosahedron has 30 edges; the centers of opposite edges are pairwise joined by 75 two-fold axes. There are 15 great circles perpendicular to these axes, each lying in a mirror plane of the icosahedron. There are 20 (triangular) faces, whose centers are pairwise joined by ten three-fold axes. The 10 great circles perpendicular to these axes do not lie in mirror planes. There are 12 vertices, pairwise joined by six five-fold axes; the six great circles perpendicular to the five-fold axes do not lie in mirror planes. A cross section made through the icosahedron perpendicular to a five-fold axis equidistant from opposite vertices makes a trace on the surface of the icosahedron that has the shape of a regular decagon. This cross section divides the icosahedron into two equivalent halves, which are oriented 36 degrees from mirror symmetry with respect to each other.
¶ M Proof of the Equation N = 10v2 + 2
The cuboctahedron (‘‘vector equilibrium,’’ cf. Fig. 222.01) has six square faces and eight triangular ones, totaling 14 faces. There are 24 edges and 12 vertices; Euler’s equation is satisfied. To find the total number of spheres in each layer, we must add up the total number of spheres in all faces, but allow for the fact that each sphere in an edge is counted twice (two faces meet at an edge), and each sphere at a vertex is counted four times (four faces meet at each vertex). Thus the total number of spheres in each layer, N, is given by:
N = (no. of square faces) x (number of spheres in a square face) +
+ (no. of triangular faces) x (number of spheres in a triangular face) 4-
- (no. of edges) X (number of spheres on an edge but not on a vertex) +
- 3 (no. of vertices) =
= 6 (number of spheres in a square face) +
+ 8 (number of spheres in a triangular face) +
- 24 (number of spheres on an edge but not on a vertex) + -36.
Here it is assumed that the frequency is at least unity. If the frequency is denoted by v, the number of spheres in a square face is (t + I)2: a unit-frequency square has four spheres, a square of frequency two has nine spheres, etc. The number of spheres in a triangular face is certainly a quadratic function of frequency, expressible in the general form (av2 +bv + c). We
can easily determine experimentally that the unit-frequency triangle contains three
spheres, that the triangle of frequency two contains six spheres, and that the triangle of
frequency three has 10 spheres. From these data it follows that A = V2, b—3li, and c = 1.
Therefore:
N = 6(y + I)2 + 8(1/2V2 + 3/2v + 1)—24(f—1)—36 = IOp2 + 2, q.e.d.
¶ N Parallel and Series Connections
The statement ‘‘A chain is as strong as its weakest link’’ represents the so- called series connection of a set of links (Fig. A30a). When a single link gives way, the loading capacity of the entire chain is destroyed. Each link here carries the full load of the chain.
Conversely, the use of links in parallel connection (Fig. A30Z?) could be expressed as,‘‘This system is as strong as its strongest link,’’ for here the load is shared by the links, and the strongest link alone might be able to bear the entire load. A series connection is very economical, but not very safe, a parallel connection is safe, but not very economical. In general, an optimum combination can be found (Fig. A30c).
Series Connection
Parallel Connection
The Number of Connections Between Events
Series-Parallel Connection
Fig. A30. Connections.
¶ O The Number of Connections Between Events
Fuller uses the expression 1/2(N2 —N) for the number of connections between N ‘‘events.’’ This formula can be proven by the so-called method of mathematical induction. Accordingly, we must prove: (a) the formula is valid for at least one value of N; (b) if the formula is valid for a single integer value of A, say No, then it is also valid for Ni No + 1. Once (a) and (b) are both proven, then the expression is necessarily proven for every integer value of N.
Part (a) has already been proven by Fuller, e.g., for TV = 3 (three connections). Part (b) is proven as follows: The addition of a single event to No existing events necessitates the addition of No connections, for each of the No original events must be connected to the new corner. If the number of connections between No events is (No2 -No), then the number of connections between Ni events, where Ni=No + 1, must be:
1/2(No2—No) +No = 1/2(No2 +No) = (No + l)No = 1/2Nx(Ni - 1) = 1/2(Ni2 —N)
Therefore we have shown that the expression 1/2(N2 —N) holds for Ni=No+ 1 if it is valid for No, q.e.d.
¶ P Vector Equilibrium in Crystals
It is interesting to note the similarity between the arrays formed by rods in tension-compression equilibrium and those formed by identical efficiently packed spheres. The former form the basis for the Fuller domes, the latter provide a model for crystal structures. Because rods and spheres have opposite geometrical properties, it is difficult to imagine a fundamental connection between Fuller’s macrostructures and nature’s microstructures. Yet Fuller’s sensitivity to patterns gave him an interest in sphere packing in connection with his tensegrities. Conversely, this author has discovered that a tension-compression model for crystals has broader applicability than does the
usual sphere-packing model. We shall see that the geometrical shapes encountered in both the
macrostructures and the microstructures are less the result of properties of the individual
components than of the geometrical constraints on the system of components as a
whole.
A sphere has infinite symmetry: any direction is equivalent to every other direction. A rod, on the other hand, has two unique (oppositely directed) directions. The most stable configuration of three identical spheres is one in which the spheres are in contact with each other, their centers the vertices of an equilateral triangle. Three identical rods are in a stable configuration when they form the sides of an equilateral triangle. Accordingly, both the spheres and the rods stabilize in a triangular array, even though the individual components have totally different geometries. The reason is that the geometrical constraints are the same: the centers of the sphere have a minimum distance of approach equal to the sphere diameter, and the ends of the rods are rigidly separated by the length of the rods. We see that, although the sphere has infinite symmetry, the symmetry of the aggregate of spheres is not higher than that of the aggregate of rods.
The highest number of identical spheres that can be in contact with each other and with an identical central sphere is 12. The centers of these 12 spheres constitute the vertices of a cuboctahedron (cf. section I), i.e., of Fuller’s ‘‘vector equilibrium.’’ The compression strength of such a ‘‘close-packed’’ array of spheres is due to the forces acting between the spheres at the points of contact. These forces are directed toward the centers of the spheres, i.e., along the directions of the ‘‘vector equilibrium.’’ The problem with a model of packed spheres is that it has compression strength but not tensional strength; we shall see presently that this lack of tensional strength is a weakness of the sphere-packing model of crystal structures.
In many metals, e.g., copper, the atoms arrange themselves as if they were closely packed spheres. They form layers with a triangular configuration, and these layers stack in such a way that any atom is cuboctahedrally surrounded by its nearest neighbors (Fig. A31). Because the cuboctahedron has the same symmetry as the cube, such an array is called ‘‘cubically close-packed’’ (ccp). Equally efficient packing of the spheres is obtained when the lower half of the cuboctahedron, as seen in Fig. A31, is twisted 60 degrees with respect to the upper half; the resulting array is called hexagonally close-packed (hep). Here each sphere is also surrounded by 12 identical spheres, but less symmetrically than in the ccp case. Many metals, e.g., cobalt, are
hexagonally close-packed. For all of these metals, the sphere-packing model is very
satisfactory. A fly in the ointment, though, is the fact that more metals assume a so-called
body-centered cubic configuration than the ccp one. This body-centered array will be
discussed presently; we shall first continue to follow the successes of the sphere-packing
model.
Fig. A31. Cuboctahedron configuration viewed perpendicularly to triangular layers.
Many crystals are represented by a model in which large, closely packed spheres correspond to larger ions, and small spheres in the voids between the closely packed ones correspond to smaller ions. There are two kinds of voids between closely packed spheres: those bounded by eight spherical surfaces (octahedral voids) and those bounded by four spherical surfaces (tetrahedral voids). For each close-packed sphere, there are two tetrahedral voids and a single octahedral one. In Fig. A32, there is a schematic representation of the most common minerals whose structure can be represented as a distribution of small ions over the voids between closely packed large ions.
This author has invented four models called Moduledra Crystal Building Blocks * to represent structures such as those of Fig. A32. Two of these blocks are tetrahedra, two are octahedra (Fig. A33); they represent the voids between spheres. Their vertices represent the centers of closely packed ions,
Fig. A32. Common minerals that can be represented as close-packed large spheres, with smaller spheres distributed over the voids between the large spheres.
Trade Mark Registered and a sphere in the center represents occupancy of a void. The closely packed spheres are not shown explicitly.
We have already seen that octahedra and tetrahedra together can fill all of space in a ratio l-j-2. Different permutations of the four blocks produce models of about 80 percent of all inorganic crystals (Fig. A34).
The success of the sphere-packing model is thus surprisingly great when one considers how different an atom is from a rigid sphere. We have already noted, however, that the sphere model has compression strength but no tension strength. For this reason, it works well for the most compact structures, but it totally fails to represent less compact but just the same very prevalent structures. One of such is the body-centered cubic (bcc) structure, which is almost twice as prevalent as the ccp structure among metals. In this structure, each atom is surrounded by eight equivalent atoms at the corners of a cube. Since the number of near neighbors is here smaller than that in the ccp and hep structures, it is less compact, and not accounted for by the sphere-packing model. Some examples where the sphere-model is almost absurdly successful are presented by several alloys where metal ions are tetrahedrally or octahedrally surrounded without being small enough to fit in voids between the other metal ions. Although such alloys can be conveniently represented by the Moduledra Blocks, it makes no sense to think here of close-packed spheres: the framework is the same, but the explanation must be different.
Fig. A33. Moduledra Crystal Building Blocks.
Fig. A34. Some structures modeled with Modulera Crystal Building Blocks.
Curious also is the fact that the array of metal ions in the copper-magnesium alloy known as a ‘‘Laves phase’’ (after the German-Swiss crystallographer F. Laves) is exactly the same as that of the metal ions in the mineral spinel. In the latter structure, the metal ions do fit in the voids between closely packed oxygen ions, but in the Laves phase, there is no closely packed framework guiding the location of these ions.
We are therefore faced with the problem of devising a model that is consistent with the results of the sphere-packing model where the latter is applicable, but that will also apply to less compact structures such as the bcc arrays. Let us first imagine two atoms or ions that exert a force on each other, a force that depends only on the distance between their centers and is independent of the particular shape or orientation of the atoms. In the case of rigid spheres, this force would be infinite for distances less than the sphere
diameter, zero for distances greater than the sphere diameter (Fig. A35a). In the case of a
tension-compression equilibrium, the force is positive (repulsive) for small distances, negative
(attractive) for large distances. At a certain distance, the forces just balance (Fig.
A35b).
The two atoms will form a stable duo if .there is a distance at which the repulsive and attractive forces just balance, just as in the case of tensioncompression equilibrium. If the forces between the atoms are repulsive for all distances, the atoms will fly apart, never to meet again. However, if they are attractive for all distances, the atoms will move toward each other and become one forevermore. In the case of a stable equilibrium, the two atoms will remain at (or oscillate around) an equilibrium distance. Conversely, stable systems must have forces similar to those of Fig. A35??: the exact nature of the repulsive and attractive forces is immaterial, as long as they balance at a certain distance.
When a third identical atom is added to the first two, we can assume that the force between individual pairs is unaffected by the presence of the third one. In this case, the three identical atoms will arrange themselves at the corners of an equilateral triangle, so that the environment of each is identical with that of the other two. Any attempt to place one of the atoms in a unique position, e.g., on a line with the other two, disturbs the equilibrium, and the atoms will rearrange themselves to form the equilateral triangle once again. Four atoms will prefer a tetrahedral arrangement to a square one, because in the tetrahedral array, each atom has the same distance from each of the other three, whereas in the square array, one atom (the one diagonally across) is farther away than the other two. If the nearer two in the square are at the equilibrium distance, the two pairs of atoms diagonally across from each other will experience an attraction, and the square will pucker to form a tetrahedron. Conversely, if the diagonally opposite atoms are at the equilibrium distance, the adjacent ones are closer together, hence in the repulsive-distance region. As a result, the square will buckle to form a tetrahedron. Basic to the new model is this fact that identical atoms or ions want as much as possible to establish identical environments so that a maximum number of them can be at equilibrium distance from each other. This requirement is put in the form of a postulate:
The Vector Equilibrium Postulate
Crystal structures tend to assume configurations in which a maximum number of identical atoms or ions are equidistant from each other. If more than a single type of atom or ion is present, then each atom or ion tends to be equidistant from as many as possible of each type of atoms or ions.
This postulate (to be referred to as the VEP) is purely phenomenological, but therein lies its power. It establishes a condition for equilibrium configurations, assuming only that a distance of equilibrium exists for each pair of atoms.
Having substituted this postulate for the rigid-sphere model, we must investigate: (a) whether it produces results consistent with this sphere model where the latter is applicable, and (b) whether it agrees with experimental observations where the sphere model does not apply, in particular whether it is applicable to the bcc array.
The cubically and hexagonally close-packed structures conform ideally to the VEP: every atom is identically surrounded by the maximum of 12 identical atoms. Every atom also has six next-nearest neighbors, at a distance 41 percent greater than that of its nearest neighbors. In ccp structures, these six next-nearest neighbors form the vertices of a regular octahedron; in hep ones, they are located at the vertices of a triangular prism. The relative distances of nearest and next-nearest neighbors in ccp structures are shown in Fig. A36a.
Fig. A35b. Tension-compression equilibrium.
By comparison, the bcc structure, which does not fit the sphere-packing model, fares quite well by the VEP. As shown in Fig. A36Z?, an atom in a bcc array has eight nearest neighbors, compared to 12 in ccp and hep arrays, but its six next nearest neighbors are only 15 percent farther away, instead of 41 percent in ccp. Thus a bcc atom has 14 rather near neighbors. When the forces between atoms have appropriate dependence on distance, the relative scarcity of nearest neighbors in the bcc structure may be offset by the relative proximity of next-nearest neighbors, resulting in a stable bcc structure.
Because we have abandoned the spheres, we have also lost the voids between them. The VEP must therefore provide a substitute, which must be based on a sufficiently general concept that we can find something equivalent to voids in the bcc arrays as well.
Starting with a ccp, hep, bcc, or any other regular array of points, we must locate positions of vector equilibrium between them. Instead of concentrating separately on ccp, hep, and bcc arrays, we shall first approach the problem generally, and apply the result to each of these arrays later on. In general, we shall call the regular array, as exemplified by the ccp, hep, and bcc structure, a Point Complex. The points of vector equilibrium are points equidistant from a maximum of constituents of the Point Complex.
Next-nearest neighbor
Q ) fee: unit cell chosen with its center on a close-packed atom
Fig. A36a. Neighbors of a cubically close-packed atom.
Next-nearest neighbor
Fig. A36b. Neighbors of an atom in a body-centered cubic array.
The locus of all points in space equidistant from two given points is the plane that perpendicularly bisects the line segment joining these points. The locus of all points equidistant from three given points is a line perpendicular to the plane of these points, passing through that plane at the intersection of the perpendicular bisectors of the sides of the triangle formed by the
three points (cf. Fig. 401.01). The locus of the points equidistant from any four given points in
space is the center of a sphere passing through these four given points. There is thus an
infinity of points equidistant from two or three points, but in general only a single point
equidistant from any four points not in the same plane. To locate vector-equilibrium
points in a Point Complex, one needs, therefore, to specify at least four points of the
Complex.
To arrive at a general construction of vector-equilibrium locations, the concept of a Domain is helpful. A Domain of a Point Complex is a region within which all points are closer to a given constituent of the Point Complex than to any other constituent of that Complex. The given constituent is at the center of the Domain. Consider, for example, the bees constructing their honeycomb (cf. section F). These bees were located at the centers of closely packed identical cylinders or circular disks, which expanded beyond the point of tangency to form hexagonal cells. These cells are two-dimensional Domains in the plane of the honeycomb: each point inside the cell is closer to the bee at its center than to any other bee. The boundaries of this Domain are equidistant from two bees; the vertices of the hexagonal cells would be vector-equilibrium positions (in the plane of the honeycomb) in the Point Complex formed by the bees. We shall find out presently which would be the vector-equilibrium points in the Point Complex formed by the hypothetical creatures we called ‘‘cees.’’
For the general construction of a Domain, perpendicularly bisect all line segments joining the center of the Domain to all other points in the Complex. The region enclosed by the innermost polyhedron formed by the bisector planes is the Domain. Ordinarily, only the nearest neighbors and at most next- nearest neighbors actually contribute to the Domain, because more distant points give rise to planes outside the innermost polyhedron. The vertices of the Domain, being the intersection of three perpendicular bisector-planes, are equidistant from four constituents of the Point Complex, hence vector equilibrium points in the Point Complex!
It is concluded, therefore, that the vertices of the Domain of a Point Complex constitute vector-equilibrium positions in that Complex. Since each face of a Domain perpendicularly bisects a line segment joining the center of that Domain to another constituent of the Point Complex, each face of the Domain corresponds to a neighboring point in the Point Complex. These neighbors in the Point Complex form what is called a Coordination Polyhedron around the central point.
In the ccp array, each constituent is surrounded by twelve points at the corners of a cuboctahedron: this cuboctahedron is the Coordination Polyhedron for ccp arrays. The Domain for the ccp array has accordingly 12 faces: it is the rhombododecahedron (see sec. I) shown in Fig. A37. The rhombododecahedron has two kinds of vertices: the ‘‘acute’’ ones, where four acute angles meet, and the ‘‘obtuse’’ ones, where three obtuse angles meet. We have already seen that this polyhedron is a space filler. When packed together to fit all of space, four of those dodecahedra meet at the obtuse vertices and six meet at the acute vertices. Accordingly, each obtuse vertex is symmetrically surrounded by four, each acute vertex by six constituents of the ccp Point Complex. Each obtuse vertex is therefore exactly equivalent to a tetrahedral interstice, each acute vertex to an octahedral interstice in the sphere-packing model. We have thus proven that the VEP leads to results entirely equivalent to those of the sphere-packing model in the ccp case.
In the hep array, the Domain is one that can be visualized as a rhombododecahedron that is first bisected by a plane perpendicular to a diagonal joining opposite obtuse vertices, after which the two halves are twisted 120 degrees with regard to each other. This procedure does not alter either the space-filling property or the vertices, with the result that again the VEP and sphere-packing model are in agreement.
Fig. A37. Rhombododecahedron.
With confidence gained by this agreement between VEP and sphere-packing model, we can now proceed to the bcc array, where the sphere-packing model utterly failed. Here the Domain involves nearest and next-nearest neighbors: the eight nearest neighbors give rise to eight hexagonal faces, and the six next-nearest neighbors to six square faces. The Domain of the bcc array is accordingly the truncated octahedron (see sec. I), as shown in Fig. A38. The vertices of this polyhedron would be points of vector equilibrium in the bcc array. Fig. A39 shows that, indeed, a very important structure, ‘‘/3- tungsten’’ is made up of a bcc array of atoms accompanied by an array of atoms that occupy one half of the vector-equilibrium positions! These vector equilibrium sites are occupied symmetrically, in accordance with the VEP: any other distribution of atoms over the available sites would bring pairs of atoms too close together.
Fig. A38. Domain of a bcc Point Complex.
bcc Point Complex
The /3-tungsten structure is usually described as a bcc array with linear chains, perpendicular to each other but never intersecting, weaving in between the bcc points. That these chains do in fact exist can be seen by stacking cubes identical to Fig. A38 together in three dimensions, all in identical orientations. The existence of these chains was not understood, and the connection of the chains with the bcc array (they always seem to occur together) was even more of a mystery. The VEP accounts for both phenomena and establishes a fundamental relation between the ccp and hep systems on the one hand, and the bcc one on the other hand. It is interesting to compare the /3-tungsten structure (Fig. A39) with the six-strut tensegrity icosahedron (Figs. 717.01 and 740.21): both structures can now be understood on the basis of vector equilibrium.
In /3-uranium hydride and in a gold-zinc alloy, one type of atom (uranium, resp. gold) forms a bcc array and also occupies half of the vector-equilibrium positions, in effect forming a ??-tungsten structure. The other type of ions occupy sites that on closer investigation turn out to be exactly equidistant from one bcc point and three of its vector-equilibrium positions. This means that the ??-tungsten structure, which is itself made up of a bcc Point Complex plus half its vector-equilibrium positions, can be in turn considered as a Point Complex in which some of its own vector-equilibrium positions are occupied!
Interesting also is the above-mentioned Laves phase. Here copper ions form a diamond Point Complex, so called because, just as in diamond, each copper atom is surrounded by four equivalent copper atoms at the corners of a regular tetrahedron. Such ‘‘low coordination’’ (i.e., low number of neighbors) is not easy to understand in the light of the VEP. In the diamond structure, the VEP does not apply because the assumption that the force between atoms depends on distance and not on orientation is not applicable here. On the contrary, each carbon atom in diamond has four chemical bonds directed toward the corners of a regular tetrahedron: in diamond, these bonds lock in place, resulting in the tetrahedral coordination. The Domain in the diamond structure is a truncated tetrahedron plus four quarter-tetrahedra. It should be recalled that the truncated tetrahedron has four hexagonal faces
and four triangular ones. It is not a space filler, but it can combine with tetrahedra
in 1 4-1 ratio to fill all of space (cf. sec. I and Figs. A 14a and A14??). By its very
definition, a Domain must be a space filler: every point in space must be closer to
some
A vector-equilibrium positions occupied constituent than to any other constituent
of the Point Complex. The truncated tetrahedron and tetrahedron can fill space by
stacking in such a way that the tetrahedron shares each of its faces with a triangular face
of a truncated tetrahedron (Fig. A40). The Domain of the diamond Point Complex
consists of the truncated tetrahedron, with a quarter-tetrahedron attached to each
triangular face. This Domain has two kinds of vertices: 12 vertices of the truncated
tetrahedron, and four vertices that were the centers of the tetrahedra between the truncated
tetrahedra.
Fig. A39. Relation of the ??-tungsten structure to the Domain of the bcc Point Complex.
bcc Point Complex
We can now understand how a diamond Point Complex can occur in metallic alloys, where there are no chemical directed bonds as in diamond. In the Laves phase, the copper atoms are not each other’s nearest neighbors, but rather next nearest neighbors. The nearest neighbors of copper atoms are 12 magnesium atoms at the vertices of a truncated tetrahedron! The magnesium atoms occupy vector-equilibrium positions in the Point Complex occupied by copper atoms, thus stabilizing the system. The two types of vertices of the Domain of the diamond Point Complex are too close together to be occupied simultaneously. The vertices at the center of the tetrahedra together form a diamond Point Complex congruent to the one formed by the Domain centers. The two diamond complexes together form a bcc array!
Fig. A40. Domain of the diamond complex; shaded faces represent quarters of a tetrahedron, which are added on to the triangular faces of a truncated tetrahedron.
In conclusion, then, we find that there are geometrical relations between the various polyhedra that are more profound than the simple, though very serviceable, sphere-packing model. The definition of concepts such as vector equilibrium, domains, and coordination polyhedra are useful in establishing these relationships. They help us recognize patterns (such as the one formed by
hydrogens in ??-uranium hydrides), and explain the natural co-occurrence of arrays that can be
interpreted as Point Complexes together with their vector-equilibrium positions. The
relationship between Point-Complex and vector-equilibrium positions is actually a mutual
one, but the simpler one of the two is usually taken as the Point Complex starting
point.
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