The tetrahelix of
Fuller's Synergetics
consists of face
bond regular tetrahedra. The mathematics for this spiraling
structure is quit interesting. Dispite the tetrahelix being composed
of regular tetrahedra (the "simplest" polyhedron), I have not been
able to find a simple
way to calculate the information for the tetrahelix.
From the Zheng paper (see References below): "The tetrahedral helix is also called the 'Bernal spiral' in association
with discussions of liquid structures in the physics literature."
The vertices of the regular tetrahedra of the tetrahelix all lay
on the surface of a cylinder. Let us visualize this cylinder
to be along the z-axis.
The radius of the cylinder will be
r = (3 sqrt(3) / 10) EL
where EL is the edge length of the tetrahedra used to build the
tetrahelix.
Let us put a vertex (call it V0) of one of the tetrahedra on
the x-axis. That is
V0 = (r, 0, 0)
Then the next vertex of the tetrahelix (V1) will be at the coordinates
V1 = (r cos(theta), r sin(theta), h)
where theta is the angle around the z-axis and is given by
and where h is the distance in the z-axis direction and is given by
h = (1/sqrt(10)) EL
In the above figures, the yellow band connects a vertex to the "next" vertex, while the
distance h is the distance between the 2 blue bands around the cylinder.
In general, the coordinates for the vertices of a Counter Clockwise tetrahelix
Vn (n = 0, 1, 2, 3, ...) are given by
Vn = (r cos(n*theta), r sin(n*theta), n*h)
The coordinates for the vertices of a Clockwise tetrahelix
Vn (n = 0, 1, 2, 3, ...) are given by
Vn = (r cos(n*theta), - r sin(n*theta), n*h)
Note that cos(theta) = -2/3 and that sin(theta) = sqrt(5)/3. You can
calculate exact expressions for the vertex coordinates by using these
relations together with the following trig identities:
One of the reasons that deriving the above information is
difficult is that the axis of symmetry of the cylinder (the axis
through the center of the cylinder) does not pass through the
center of volume of the tetrahedra. The distance from the
z-axis to the tetrahedron center of volume is given by the equation
dist. = (sqrt(2)/10) EL
Therefore, all the Tetrahelix cylinder axis of symmetry pass tangentially by a sphere of
radius (sqrt(2)/10)EL centered at the Tetrahedron's center of volume.
Comments
Tetrahelix come in 2 orientations; a right-handed spiral and
a left-handed spiral.
The vertices of the tetrahelix never line up. That is, no two vertices
will ever be directly above one another. This is because the theta angle
above is an irrational number.
It looks like you might be able to "nest" 3 additional tetrahelix
about the original tetrahelix in a nice tight bundle. But this is
not possible. The reason is that face binding 5 tetrahedra together
about a single edge/axis leaves a gap (which Fuller calls the
unzipping angle.) The dihedral angle of the regular tetrahedron is
5 times this amount is approx 352.643895 degrees which is
approx 7.356105 degrees (the unzipping angle) short of 360 degrees. So
the tetrahedra of multiple helixes can not pack together without gaps.
There are 12 possible Tetrahelix passing through a single Tetrahedron:
6 Clockwise and 6 Counter clockwise.
As the figure shows, the spiral/coil associated with a Tetrahelix going through
a Tetrahedron passes through all 4 vertices of the Tetrahedron sequentially. Since the
spiral has the same symmetry axes as the Tetrahelix, we can count the number of possible
different spirals to count the number of Tetrahelix.
We label the Tetrahedron's vertices by numbering them 1, 2, 3, and 4. Then
there are 4x3x2x1 = 24 combinations for the order in which the spiral can pass through
the vertices. However, 1/2 of these are simply reversals: (1,2,3,4) has the same
spiral symmetry axis as the spiral (4,3,2,1). That leaves 12 spirals/Tetrahelix.
For more details on the way 12 Tetrahelix axes pass through a single Tetrahedron,
see the next web page.
I find that there is a Clockwise and a Counterclockwise spiral associated with every
Tetrahelix.
Note that the spacing between adjacent loops in the spiral is different for the CCW
spiral versus the CW spiral.
In both cases, the spirals pass through all the vertices in a Tetrahedron sequentially.
In both cases, the spiral travels a distance along the z-axis by an amonut
h = (1/sqrt(10)) EL
But in one case, the spiral travels around the Tetrahelix by an angular amount of
(My notes on this (dated 1997) suggests that I am not the originator of this. Unfortunately,
they do not indicate who suggested this originally.)
If the tetrahedra used to build the tetrahelix are 10-frequency (have 11 vertices per
edge) then the axis of the surrounding tetrahelix
cylinder will always pass through the tetrahedra faces at an inner
triangular face vertex. The "triangular face coordinate" which
the symmetry axis passes through is (7,3).
For a finite length Tetrahelix, all of the Tetrahelix's vertices can be given
rational (x, y, z) coordinates. Then by scaling, the coordinates' x, y, z, components
can be made to be integers.
I have found that if you allow the vertices to be flexible and allow some of the
tetrahedra egdes to expand in length, the
tetrahelix can fold up into another (shorter) tetrahelix. When it does, it passes
through an Octahedron phase.
Here is a sequence of images to help visualize the transformation.
First, consider the Tetrahelix to be made up of a number of 3-Tetrahedra units. By folding up
each of these units, the Tetrahelix is reduced.
We can fold up a 3-Tetra unit as shown in the following figures.
Not that only the C-to-D edge needs to change its length. All other edges of the
Tetrahedra remain the same.
At this point, the 3 Tetrahedra are very close to defining an Octahedron. They actually
do form an Octahedron as the vertices "A" and "B" are brought closer togther.
When vertices "A" and "B" are brought together, a double Tetrahedron is formed. That is
two Tetrahedra face bound togther. (This figure only shows one of the 2 Tetrahedra. The
other is hidden behind this one.)
The transformation from the 3 Tetrahedra to the Octahedron Fuller calls the
"Richter Transformation". See Color plate 6 in Fuller's book
Synergetics.
I am not aware that Fuller continued the transformation described
here to transform a Tetrahelix.
Joe Matto suggested to me (October, 2004) that 2 Tetrahelix can intersect each other
at 90 degrees. If we define "intersect" to mean that the axis of symmetry of 2
Tetrahelix pass through a common Tetrahedron and
pass through a common point at 90 degrees, then this statement is not
true. The 2 Tetrahelix "intersect" each other at
2 arcsin(1/sqrt(10)) = 36.86989765... degrees.
However, if we define "intersect" to mean that the 2 Tetrahelix pass through a
common Tetrahedron and thier symmetry axes pass by each other at 90 degrees, then
this is true.
There are 2 ways to place the end of a Tetrahelix flat on the top of a table such that
the Tetrahelix riases above the table.
See the calculations here.
The angles which the symmetry axis makes with
the table top are:
90 - arccos(sqrt(1/15)) = 14.96321744... degrees
and
90 - arccos(sqrt(3/5)) = 50.76847952... degrees
References
Here is a list of references. However, I did not use any of these references
for my calculations.
H. S. M. Coxeter, Introduction to Geometry, Wiley: New York, 1969, p. 412.
Chong Zheng, Roald Hoffman, David R. Nelson, "A Helical Face-Sharing Tetrahedron
Chain with Irrational Twist, Stella Quadrangula, and Related Matters",
J. Am. Chem. Soc., 112, pp. 3784-3791, 1990.
I have not see the following references. These references are given in the Zheng paper.
A. H. Boerdjik, Phillips Res. Rep., 7, p. 303, 1952.
J. D. Bernal, Proc. R. Soc. London Ser. A, 280, p. 299, 1964.
P. J. Steinhardt, D. R. Nelson, M. Ronchetti, Phys. Rev. Lett., 47, p. 1297, 1981.
P. J. Steinhardt, D. R. Nelson, M. Ronchetti, Phys. Rev. B, 28, p. 784, 1983.