1.
Report
No.
No.
3.
Recipient's
Title andSubtitle ADVANCED STRUCTURAL GEOMETRY STUDIES PART I - POLYHEDRAL SUBDIVISION CONCEPTS FOR STRUCTURAL APPLICATIONS
5.
Report
NASA 4.
2.
7. Author(s) 9.
Government
Accession
Joseph
Performing
Organization
Name
D. Clinton and
Address
Southern Illinois University Carbondale, Illinois
No.
Sponsoring
Agency
Name
Performing
Organiza:tion
Code
8.
Performing
Organization
Report
10.
Work
11.
Contract
16.
Abstract
Unit
No.
,or Gran.J No.
Type
of Report
and
Period
Contractor Space
Covered
Report
Administration Sponsoring
Agency
Code
Notes
A study leading to the formulation of computer-oriented mathematical models pertaining to methods of subdividing polyhedra into triangulated spherical space frames. The models perform the truncations and transformations of the polyhedral forms and calculate the geometrical properties of the generated space frames (spheres, hemispheres, and domes).
17.
Key
Words
(Selected
Geodesic
19.
by Author(s))
18.
Domes Subdivision
Structural
Spheres,
Security
Classif.
Distribution
Statement
Domes
Structural Polyhedral
(of
Unclassified
this
report)
Unclassified
- unlimited
Spaceframes 20.
Security
No.
NGR-14-O0_-O02
14.
Supplementary
1971
6.
and Address
National Aeronautics and Washington, D. C. 20546
15.
Date
September
13. 12.
Catalog
CR-1734
Classif.
(of
this
Unclassified
For sale by the National Technical Information
page)
21.
No.
of Pages
iii
Service, Springfield,
22.
Price
$3.00
Virginia
22151
FOREWORD
This
final
report
Technology
at
Illinois
under
was
administered
and
Technology.
Personnel Julian
was
prepared
Southern NASA by
Illinois NGR
the
Office
NASA
Lauchner,
Clinton,
prime
research
consultant;
Michael
Keeling,
Richard
M.
in
the
Moeller,
R.
computer
14-008-002. of
The
Advanced
Mark
Ann B.
programmers.
iii
contract Research
included: Joseph
Buckminster
Boo%h,
Kilty,
of Carbondale,
investigator;
investigator;
Allen
School
research
principal
Wayne
the
University,
Contract
participating H.
by
C.
Fuller, Garrison,
Mabee,
and
D.
PART
POLYHEDRAL FOB
I
SUBDIVISION STRUCTURAL
CONCEPTS
APPLICATIONS
i.i
Introduction
.............
1-2
1.2
Polyhedron
1.3
Structural
1.4
Definitions
1.5
Method
of
1.6
Method
1
1.7
Methods
2
&
3
............
1-76
1.8
Methods
4
&
5
............
1-145
1.9
Methods
6
&
7
............
1-185
..............
Orientation
1-7
........
.............
Subdivision
1-16
........
...............
I-ll
1-18
1-36
v
Computer Software and Information
The documentation and program advanced structural geometry made available to the public
Management Center
developed for the studies will be through COSMIC.
COSMIC (Computer Software Management and Information Center) was established early in 1966 at the University of Georgia to collect and disseminate to the public computer software developed by government agencies. Since that time thousands of computer programs in all areas of aerospace engineering, mathematics, business, and industry have been distributed to requesters throughout the United States. The Technology Utilization Division of NASA, designed to enlarge the return on the public investment in aeronautical and space activities, was the first government agency to participate formally. _In July 1968 the Atomic Energy Commission and in November 1968 the Department of Defense joined in the COSMIC endeavor. With the addition of these two major agencies, the original concept of making taxpaid developments available to the public was expanded to make COSMIC a transfer point between and within government agencies as well. Requests for this program
documentation or information should be directed to: COSMIC The University of Georgia Barrow Hall Athens, Georgia 30601 REF:
HQN-10677
I-i
concerning
I.I
One
of
the
cohtemporary Designs by
the
environment of
system
form and
been
on
structures
which
are of
it
based
will
the
systems spherical
influenced
the
structure, be
in form.
primarily spacial
subjected,
and
the
fabrication.
basic
spherical
structural
purpose to
materials
economical
has
such
ultimate
Two
most
use
for
INTRODUCTION
the
systems for
are
structural
multi-polar
used
for
subdividing
application: system.
Figures
the The
bi-polar
I.I,
1.2.
I
! I
i Bi-polar Figure
I Multi-polar
System
Figure
I.I
Z-2
System 1.2
The bi-polar
system is related
latitude-longitude
approach to subdividing
commonexamples of this 1.3)
and the lamella
to the familiar a sphere.
system are the ribbed
dome (Figure
Ribbed Dome Figure
1.3
T- 3
1.4).
Two
dome (Figure-
0 C:_
w-!
\
\
\
\
I-4
I:::: ,,_1
• rI_1_
The of
multi-polar
polyhedra.
system Fuller.
is
system
Perhaps the
geodesic
Figure
1.5
the dome
Geodesic Figure
is most
related familiar
discoverd
Dome 1.5
to
the
spherical
example by
R.
of
Buckminster
form this
A typical with
the
tion.
design
geometrical
Computer
initiated
to
of
number
of
single
joint, This
aids
handle
determination
geometrical dividing
up
has
great
of
and
will
model
for
form
for
structural
the
three
triangles:
i cosahedron.
I-G
in
at
the
a
other, itself
system
etc. with
the for
applications.
polyhedral the
be
total
determining
multi-polar
regular
each
concern
the
to
may
members,
to
is
configura-
intersectinq
members
report
frame
variables
of
limited of
of of
members of
computer
a spherical
number
frequency
the
properties
completely
octahedron
of
design
the
system
investigation
the
relationship
been
bi-polar
in
number
and
the of
lengths,
portion
with
relationships
the
joints,
a mathematical
model
problem
forms
tetrachedron,
subThe made
1.2
Hoppe, metrical
in
1882,
figure in
polyhedron.*
as
study
Klein,
based
on
regular and
Greeks
with
the
the
the
structural
several
of
of
it
and
of
polytope:
portions
the
Schi_fli
is
studied
in
of
In
polyhedral
three
Others
much
this
to
section
are
specifically
Tetrahedron,
the
of
discussed
forms,
a
over
Euclid.
introduced
polytope.
or
polyhedra
findings
the
planes,
a polygon;
Coxeter,
polyhedra;
a geo-
lines,
configurations
the
(Platonic)
the
Octahedron,
Icosahedron. In
Euclid's
definition
is
to
the
be
regular
are
ancient if
congruent,
Table and
ago
concepts
report:
word
dimensions
years
and
the
two
the
by
However,
thousand
such
coined
bounded
hyperplanes;
two
POLYHEDRON
I.I
list
Icosahedron
regular
writings, give
the
world. each
The have
regular
convex
regular
if
the
properties
which
Elements,
f_ve
and
polyhedral
*Coxeter,
to
The
they
are
are
N.
S.
M.
of of
considered
1
T- 7
solids
polyhedra and
forms.
explanation
egual
regular the
as are
known said
to
if
they
faces, polyhedral
Tetrahedron, as
and
three
angles. Octahedron,
of
the
five
Table Properties Tetrahedron,
1 .I
of the Basic Octahedron,
Tetrahedron V=4, Dihedral
angle#
F=4,
by of
center
to
center
edge
center
to
vertex
of
face
edge mid
edge
to
center
mid
edge
to
vertex
mid
edge
to
opposite
height opposite area vol
of ume
(vertex face) face
to
E=6
of
= 70 °
cos(llV_)
iI_,
44"
I/3)
= 109°28'16
(-iI_,
I/_
O .57735
I/3
0.33333
1
1
2 _IJ-3
1 .63299 0.47140 1 .41421
mid center
edge of
2/_/-3
1.15470
4/3
1 .33333
2/_
1.15470 0.51320
27
T-8
''
(liVe,-llVT,-11v_)
-iI_)
_/-{/3
face
31'
an
llY_,
(-li#_, mid
3
8=
(llV_,
Vertices
to
3
= 2 sin_/3
Angle subtended edge at center polyhedron
Center
Polyhedra Icosahedron
-llJ-%,
llJ-_)
Table
1 .I
Octahedron V=6, Dihedral
F:8,
Angle/3=
Angle substended an edge at center of polyhedron
2_/2-=
109 °
28'
by 8=90
°
o,
o)
(o,
-+l,
o)
(o,
o, -+I)
1.41421
center
to
vertex
center
to
mid
center face
to
center
Mid edge vertex
to
Mid edge vertex
to
Volume
4
(-+I,
edge
area
3 E=I2
tan
Vertices
(cont.)
of
face
1
1
I/v/2
0.70711
I/_
0.57735
_3/2
1.22474
F
1.58114
edge of
near
distant 5/2 v_ 4/3
T- 9
/2
0.86603 1.33333
16"
Table
1 .I
(cont.) 5
Icosahedron V=I2,
3
F=20,
E=30
2/2
_1
Dihedral
angle
6 = _ -(I
sin
Angle subtended by an edge at center of polyhedron 6 = cos
:
Vertices
(f5-/5
= 138o11,22
)
)
1 +v/-5 - = 1.61803 2
= 63o26
'05.818''
1 f4
11
5 J 1
= T-
''
5
1
T
51/4/-_ -
51/4
2
m
= -_ + 1
md/-5-= T + 2
_+ 51/4,/7-
center
to
vertex
center
to
mid
center face
to
center
area volume
114
2/5
edge
of
face
1
51/4V/T
1.05146 1
1 i/4
0.85065
31_5114d-_
o.79465
edge of
, _+
I/_vi3/5
0.47873
(4(5 I 1 4) d_)13
2.53615
I -i0
o)
1.3
The through form
structural
chosen
forms
1.6
only
the
the to
chosen
for
of
x,
the
then
for
y,
the
geometrical z axis this
is
the
in is
1.2
chosen
calculation Figure with the
face
intersection
of
coordinates
PPT.
z
\,
Y P3
Tetrahedron
I-ll
in
(0,0,0)
center
5
1 .6
polyhedral
of
Z
Figure
compu-
form
origin
the
Octahedron
the
The
the
the
the
vertices
the
list as
surface
structure.
the
at
regular
rectangular
used
polyhedral
being
polyhedral
the
of
computations.
Table faces
basis
the
with
point
the
dimensional the
of
located
of
onto
polyhedron
z axis
acquired
faced
existing
of
is
trianqular
tY.:islated
properties
x,
faces
A three
the
the
the
three
chosen
of
sphere. of
is
orientation
y,
cumscribed
the
the
symmetries
face
polyhedron,
vertices
the
geometrical
shows
of
was
to
one
of
sphere.
system Due
respect
the
grid
a circumscribed
tations.
of
one
The
desired
gridding
from
coordinate
ORIENTATION
configuration
a three-way
polyhedra. of
STRUCTURAL
the
of cirof
the
x
I
Icosahedron Figure
1.6
Table Coordinates
of
the
(cont.)
1.2
Principal
Polyhedral
Triangles
t L
Tetrahedron
P_ = (-l/d-3-,-lldT, P2 = (lld-%, P3 :
(-lld_,
lid-3-)
=
(-.57735027,-.57735027, (.57735027,-.57735027,-.57735027)
-IIdY,
-lld3)
=
lld_,
-lld_)
= (-.57735027,
.57735027)
.57735027,-.57735027)
0ctahedron
nl
: (l,
P2 :
o, o)
(o,
I,
o)
P3 : (o,
o,
I) Icosahedron 1
d_-) P1 = ( 0 ,d-_-I5114 , 115 14 P2 = (I/51/4_
0'
P3 =
1/51/4v/-_-
(_-/51/4
= (0,
_/T/51/4)
,
0)
.85065081,
(.52573111, =
(.85065081
1-12
.52573111) 0,
.85065081) 52573111
O)
.,
Throughout division used
of with
the the
polyhedral
examples
derived
using
vertex,
and
discussion
the face.
of
of forms
computer
three
the maps
traditional
Figure
Edge
1.7.
Orientation
Fi qure
the
1. 7
1-13
methods
of
Icosahedron of
the
spherical
orientations:
subwill
be forms edge,
/
Vertex
/
Orientation Figure
1.7
1-14
(cont.)
Face
Orientation
__(_o_./_'_
1-15
1.4
AXIAL
ANGLE
(_)
radius
from
common
point
a vertex CENTRAL
= an
ANGLE
angle
the
center
and
of
the
the (6)
polyhedron
DEFINITIONS
formed
by
of
polyhedron
the
vertex
of
an
the
element
and
a
meeting
axial
in
angle
a
sharing
polyhedron. = an
angle
passing
formed
by
the
end
through
two
radii
points
of of
the
a principal
side. CHORD FACTOR upon
(cf)
= the
a radius
spherical
of
The
may
lengths
calculated
a non-dimensional
form.
structures
element
be
length
found
where
unit
of
any
by
the
cf
x
cf
= chord
r
1 for
element
ANGLE
(8)
in
a common
of
the
To
measure
vertex sides each
= an
line.
The
dihedral
is are face
angle,
the on
angle
r
dihedral the
element
perpendicular of
the
dihedral
formed
planes
angle
to
larger
facto_
by
of the forms
desired
of
element
two
common measure
the the
angle.
1-16
dihedral element
the
planes
themselves the
of
for
= 1
= the radius structural
and
the
equation:
1 = the length sought DIHEDRAL
based
are
meeting the
line
is
the
angle angle
and
lie
faces
the
element. whose
and
whose
one
in
FACE ANGLE in
-
a common
the FACES
(_)
faces = the
an
angle
point
of
formed
and
the
lying
by in
two
elements
a plane
meeting
that
is
one
of
polyhedron.
triangles
making
up
the
"exploded"
structural
form. FREQUENCY
(_)
= the
a principle PRINCIPLE
side
PRINCIPLE principle
is
POLYHEDRAL
equilaterial regular
number
of
parts
or
segments
into
which
subdivided.
TRIANGLE
triangles
(PPT)
which
= any
forms
the
one
of
face
the of
equal the
Polyhedron. SIDE
(PS)
= any
polyhedral
one
of
triangle.
1-17
the
three
sides
of
the
1.5
Upon
using
it
is
readily
in
its
pure
that
hedral the
the apparent
be
can
form
that
the
not
satisfy
geometrically
will
be
form
into
fabrication
and
of
be made
to
limits
Seven
basic
components
erection
conditions
met. the
of
form,
range
reducing
may
unit,
polyhedral
the
number
properties
a structural
structurally
for
a larger
as
basic
and
discussed
geometrical
structural
OF SUBDIVISION
spherical
state,
must
methods
METHODS
polyfrom
remain for
which
within
the
a desired
configuration. Due
to
the
polyhedral for
symmetrical
form
only
calculating
tural
one
or and
The
reflections
its
the
detail
in
Method
1: The
chosen
as
Figure
1.8
is PPT
the
PPT
of
the
of
remaining the
basic
is
used
polyhedron
properties
of
the
faces
of
the
may
be
principal
polyhedral
the
methods
strucfound
by tri-
transformations.
Attention dividina
face
geometrical
configuration.
rotations angle
the
characteristics
given in
equal
to
a broad
following
is
here
sense
seven and
will
be
of
treated
subin
section.
subdivided
into
divisions
n frequency,
along
the
three
with
the
principal
parts sides.
A i
|
!
J
I
NOTE:
Figure
1.8
1-18
A1
= 12
Each line
point
segment
giving
of
subdivision
parallel
a three-way
triangles
are
to grid
is their
so
formed.
then
connected
respective
that
Figure
with
sides
a series
of
a
thereby
equilateral
1.9 I
A
Note:
AB is
parallel
to
B
Figure
Each passing its
on
through
the
respective
sphere. the
vertex
chords
the
PPT
The of
element
is
origin
vertex,
1 .9
then
(0,0,0)
onto
the
of
the
surface
connecting
a three-way
translated
the
great
along polyhedron
of
the
translated
circular
grid.
a line and
circumscribed vertices
form
Fiqure
I.I0
P_
P3
( o,o,o ) Figure
I.I0
1-19
12
Methods
2 & 3:
The
PPT
therein the
as
is
subdivided
equal
arc
polyhedron.
into
n frequency
divisions
Figure
of
the
with
central
the
parts
angles
of
I.II.
A
2 Note:
A-i- # I_
B
(o,o,o) Figure
the
The
points
PPT
are
respective of
occur
subdivision
connected sides.
points
method
of
which of
in
define
grid.
on
with Each
subdivision, the
I.II
line
line
of
small
principal
segments
segment
a qrid
Figure
each
side
parallel
intersects
equilateral
to at
subdivision.
of
Due
their
a number to
trianqular
"windows"
1.12. I
A
NOTE:
b
Q
A--B is parallel to 12
Aa _ ab Windows triangles
B
Figure
the
1.12
1-20
are
equilateral
The
center
of
methods
and
for
the
PPT.
the
circumscribed
respective The
used
They
vertex
element
chords
are
of
these as are
the then
sphere and
connecting a three-way
"windows"
the great
of
translated
translated circular
1.13
1-21
the onto
a line
origin
(0,0,0
Figure
found
vertices
along
the
are
by
of vertices
grid.
of
three-way the
passing
(0,0,0)
one
two grid
surface through
the
of the
polyhedron. form
Figure
the 1.13.
Method
4:
The
PPT
is
subdivided
equal
divisions
chosen
as
Figure
1.14.
into along
n frequency, the
with
three
the
principal
parts sides.
No te :
Figure
Each segments thus right
point
of
giving triangles.
1.14
subdivisions
perpendicular
to
a three-way Fiqure
A--I = I--2
is their
grid
then
connected
respective comprised
with
principal of
equilateral
line side and
1.15.
2
A
Note:
Figure
1 .15
1-22
A-B _L
12
Each
vertex
surface
of
through
the
the
the
form
Figure
1.16
the
the
PPT
is
circumscribed
respective
polyhedron.
vertex
on
then
translated
sphere
vertex
The
elements
chords
of
and
along the
a line
origin
connecting
a three-way
onto
passing
(0,0,0)
the
circular
grid.
I
/ I I / / / /
( o,o,o
)
Figure
1.16
1-23
/
of
translated
great
/
the
Method
5 The
chosen
as
PPT
is
subdivided
equal
polyhedron.
arc
into
divisions
Figure
n frequency of
the
with
central
the
angle
parts of
the
1.17.
NOTE:
AT _ I--2
B ( 0,0,0
)
Figure The PPT
are
However,
points connected the
respective grid
is
of
subdivision with
line
Upon
created.
Due
"windows"
on
line
are
not
completion
to
each
segments
segments
sides.
triangular
1.17
the
occur
similar
the
side to
of of grid.
the
to
>
fI i
>
a
subdivision,
small 1.18.
NOTE:
>
4.
their
connections
Figure
of
Method
perpendicular
method in
principal
A-B-/_T2
Small triangular windows occur
Figure
1.18
1-24
the
The as
the
tices
centers vertices
are
scribed
then
sphere
vertex
and
elements Of
the
joining
a three-way
of
these
"windows"
ef
a three-way
translated along origin the great
onto a line (0,0,0)
of
for
the
surface
and
the
PPT. of
through the
the
the
the
1.19.
PI
P2
I
/ / / / / / /
( 0,0,0
)
Figure
1-29
1.19
used
The
ver-
circum-
The
form
Figure
are
respective
polyhedron.
vertices grid.
found
grid
passing
translated circle
are
chords
Method
6:
The being
PPT
may
be
a reflection
described or
as
rotation
six of
right the
triangles
other.
each
Fiqure
1.20.
Note:
A
B
Fiqure
In ABC.
this The
rotations is the
subdivided central
method
of
remaining and
angle
1.20
subdivision section
reflections into
ABC is a right triangle
of
of of
parts
we the this
chosen
the
shall
treat
PPT may basic
as
unit.
equal
polyhedron.
be
arc
Figure
only found The
divisions
0
B
(o,o,o )
Figure
1.21
1-26
through Line
AB of
1.21.
No te :
A
triangle
A--I _ I--2
Once the
the
points
through
subdivisions of
the
side
AC,
AC.
Figure
division
points
this
are on
of
found
side
A-C and
division
giving
the
they
on
points
C--B.
side
of
are
used
to
find
Perpendiculars
A--B are
subdivision
extended on
to
side
1.22.
Note:
2_IA-B A1 _ 23
1
2
3
4
Figure
The extending side
points
of
a line
AC perpindicular
1 .22
division through to
5
on
the
the
points
side
C_.
side of Figure
CB were subdivision
formed on
1 .23.
Note:
54 _J_ CB 12
C
4
3 2
1 B
Fiqure
1.23
1-27
by
# 34
Having acquired
the points
three
sides of the triangle,
point
on side A-Cto alternate
Figure
of subdivision
diagonals points
along the
are drawn from each of sides A--Band B--C.
1.24. C
A
B
Figure
To points division
complete of
the
subdivision of
side
1.24
three-way of
side
B--C. Figure
qrid
connect
A--@to
alternate
1.25
C
Figure
1.25
1-28
alternate points
of
Through and
its
rotations
and
subdivisions,
PPT may
be
the
found.
reflections entire
Figure
of
the
three-way
basic
unit
gridding
of
the
1.26
J J J J J J A-
B
Figure
The lated line (0,0,0) lated grid.
vertices
to
the
passing of vertices Figure
the
of
the
surface
of
through
the
polyhedron. form
the
1.26
three-way the
grid
are
circumscribed
respective The chords
of
1.27.
1-29
trans-
sphere
vertex element
then
and joining
a three-way
along the the
great
a
origin transcircle
/ ! !
( 0,0,0
)
Figure
1.27
1-30
Method
7 The
being
PPT
is
described
a reflection
or
as
six
rotation
right of
triangles
the
other.
each Figure
1.28.
3
triangle
A B
Figure In
this
triangles into
will
be
a three-way
chosen
as
AC and bieng
method
the
the
of treated
arc
origin center
subdivision as
grid.
equal
The
the
the
only
the line
division of
of
1.28
of
of
basic
unit
AC is
subdivided
an
polyhedron triangle
one
angle with
of
the
for
made the
subdivision.
right
subdivision into up
of
parts
the
origin
triangle
(0,0,0) Figure
NOTE:
1.29
A1
# 23
oc _L _-C 1
2
3
o (o,o,o)
c
Figure
1 .29
1-31
u
Once used
to
lines
the
find
through
perpendicular on
side
subdivisions the
points
the
of
points
to
A--B.
are
side
Figure
division
of _-B,
found
on on
line side
sibudvision this
on
giving
the
AC they AB and side
points
are CB.
The
AC are
taken
of
division
1.30.
C
Note:
2--5 _I_ AB 12
1
2
Figure
The line to
points
through CB.
Figure
of the
division
points
3
4
$
1.30
on of
_ 34
CB are
subdivision
found on
by
extending
a
AC perpendicular
1.31.
C
Note:
5__LC_ 12
A
Figure
1.31
1-32
# 54
Having three
sides
point
on
acquired of
AC to
the
the
points
triangle,
alternate
on
AB to
complete
the
alternate
points
on
on
grid
AB and
connect
BC.
Figure
C
A
B
Figure
are
along drawn BC.
from Figure
the each 1.32.
1 .32
three-way points
subdivision
diagonals
Figure
To
of
1.33
1-33
alternate 1.33.
points
Through found the
and PPT
rotations
it's
may
be
and
subdivison, found.
reflections the
Figure
entire
of
the
basic
three-way
unit
grid
of
1.34.
J _J
J_J J_J J_ J_J A-
Figure
The to
the
passing of
the
tices Figure
vertices
surface
of
of
through
the
the
polyhedron. form
the
the
chords
1.34
three-way
grid
circumscribed
respective The of
elements
are
sphere
vertex
and
joining
a three-way
1.35.
1-34
great
then along
the the
translated a line
origin
(0,0,0)
translated
circle
grid.
ver-
P_
/ / / / / / /
( o,o.o
)
Figure
1-35
1.35
1.6
The for
mathematical
subdividing
METHOD
and
by
a unit
as
to
illustrate
an
example The
icosahedron
rectangul_r PPT
computer
a tetrahedron,
circumscribed
1
coordinate,
was
octahedron,
sphere.
is
model
The
the
system
or
in so
an
icosahedron
geometry
oriented
developed
of
that
was
the
a three
icosahedron chosen
model.
dimensional
the
vertices
of
one
are: (X1'
YI'
(X2'
Y2'
Z1)
Z2)
: (o,
d-T
,
= (0,
.850651,
:(
1
)
.525731)
, o,
= (.525731,
I
O,
dT-_
)
.850651)
o)
(.850651,
=
where
.525731,
O)
1 +d52
with origin
the
intersection (0,0,0)
of
of the
the
axis
icosahedron.
X,
Y, Figure
I136
Z located 1.36.
at
the
(X2,Y2,Z
Figure
This where
PPT
the
XI +
I
is
divided
vertices
X2
-
Xl
+ J
of
X3
N Z2 Zl
+
-
ZI
triangles
N is
Y1
X 2
Z3
-
Z2
\
of
the
+ J N
N the
integers
such
for
each
vertex
Figure
1.37.
in
-
smaller
+
equilateral are
I
N
I
where
1.36
into
the
frequency that and
O
2)
used
Y2
of
the
triangles form
- Y1 + a N
Y3
- Y2 N I.I
) structure
The to
values identify
1-37
of
and
I
and
I
and
J
each
vertex
J are are as
unique shown
(N,1) ( N,N-!
)
o_
•
(N
A
V( o,o )
Figure
1.37
1-38
_/-
To onto of
the
(X 1,Yl,Z
the the
PPT,
find
the
projection
unit
sphere
PPT and
the
is origin.
divided
by Figure
of
along origin the
a
each line
each distance
vertex
of
segment
coordinate between
the
the
PPT
through
the
of
each
vertex,
vertex
PPT
1.38.
z)
r
k( X 2'Y2'z 2)
(0,0,0)
(X3,Y
Figure
1.38
1-39
3,Z 3 )
vertex
and
Using the coordinates, of the elements of elements
(axial
angle _),
(_),
the lengths
the angle between pairs
the angle between the elements
from the origin
angle _),
structure
program finds
of the structure
(face
and a radius
this
to an endpoint
of the element
and the angle between adjacent
(dihedral
angle 8).
Figure
faces
of the
1.39
'"Z (XI,YI,ZI)
,Y2,z2 )
/
/
)
( 0,0,0
3'Y3'Z3)
Figure
To the is
find
the
coordinates a common
of
The
the
same
manner.
the
points and
P2
other
resulting
between
their
endpoint
origin.
P1
angle
1.39
endpoints.
to two
elements
each
element
endpoints
Letting from
(Xl, the
Pl Y_,
1-40
face
The
vertex
and
is
and Zl)
translations
,
the
of
we
the
to
translated
(X2, of
use
angle
translated
P2 are and
_ _,
Y2, the
Z2) endpoints
the in
be
COS
_
=
XlX 2 + Y1Y2
+ Z1Z
dld2
where
is
d2
the
desired
origin
axial
angles
Y22
+
used
with
angle
the
Z 2
above
established the
angle
found
+
angle.
is
desired
+ Z1 2
Y 2 I
2
x_X
vertex is
The is
/_2 1 +
=
find
the
The
=
and
To that
dl
1.2
is
method
at
other
one
is
end
endpoint
used
of
to
an
except
element
define
the
and angle.
_.
between
two
adjacent
faces,
the
dihedral
C
_,
using cos 8
=
A1A 2 +BIB 2
2
A1
+ B1
2
+
ClC
2
+
C1
2
2
2
2
A 2 + B 2 + C2
1.3
where is
AIX one
face
taining The
+ BIY and the
negative The
A,
+
A2X
ClZ
desired
+ DI
+ B2Y
other
angle.
= 0 defines
+ C2Z + D2
the
plane
= 0 defines
containing the
plane
con-
face.
sign B,
the
is
and
used
C fcr Y1
because each
plane
Zl
1
A = Y2
Z2
1
Y3
Z3
1
1-41
the
obtuse are
angle
computed
is as
desired.
the
= X2
Z2
1
X1 X3
Z1 Z3
1 1 1.4
C = X2 X1 X3
where
YI,
(Xl,
Z1)'
the
plane.
In
are
used.
For
above
that
face
and
in
the
the
the
the
origin
same
manner
structural
the
three
the the
doubled.
above
the
Y3,
vertices
Z3) of
the
polyhedron,
bisects
face
(X 3 ,
where
containing
and
and
case
of
plane
Z2)'
the
special
faces
the
1 1 1
Y2'
(X2'
particular
separate
made
Y2 Y1 Y_
two the
lie
each
face
faces
used
assumption
element
common
angle.
This
This
method
polyhedral
in
is
to
angle is
lie
each
is
used
face
is
not
found
by
using
found because generated
properly. The
length
general
of
the
elements,
_,
are
the
equation =
Px I
is To certain
the
_ p x 2 )2
desired
reduce
and
lengths.
one
outputed
symmetries.
The value Figure
(p yl
p Y2
)2
+ (pzl
_ pz z
program
takes
)2 1.5
length.
total
symmetries
+
output, and rest and
this
outputs of can
the
only
a part
values
are
easily
1.40.
1-42
be
found
of the
into the
same using
account
total as the
angles at
least
following
(N,N)
(N,N-1)
(N,I)
J
(N-
,
-
Figure
1.40
1-43
(N,O)
FACE
ANGLES For
away
every
from)
towards
the
(or the
is
to
(N,N-I),
the
If
the
vertex
(I
+ I,
J + l),
(I-I,J).
Also,
or
point
on
a
of
the
The
equal
point
(N,O)
and
(0,0),
(l,O)
with
(N-l,O),
(N,l,
N-l).
towards is
to
lie
(I,J), only
line
Thus, away
at
(I,J),
(I +l,
passing
side
of
the
with
The
dihedral
associated
between
the
element, since are have
the equal
two
faces
there
are
and
be
considered
elements
parallel
midpoint
program
side of
are
symmetric
this
manner.
to
of
the to
will the
a line
opposite one
of
an
of
the
and
put
one
at
circle,
right
the
compute
mid-
lengths
1-44
the
For each
the
angle
each
end,
two case,
we and
around
(X i, Yi,
Zi)
Z1)
and
lengths angles
but
angles
values
other and
one-to-
angles.
elements
(X i, Yi' All
into
is
In this
between
through
the
either
the
dihedral
opposite
side.
on
element.
one.
only
side
be
(I,J),
element
the
correspondence
This
angle
computed.
will
Zi,)
(0,0)
angles
J-l),
be
and
angles,
a cord
a one-to-one
right
axial
is
may
can
with
containing
two
are
Yi,
lengths
element
angles.
the
the
the
face
angle
at
computed.
structure
correspondence
and
falling
(Xi,
For
vertex
the
(I-l,
angles
are
one
angle
or
opening
(N,N).
(0,0)
the
J)
face
only
(or
angles
(N,l)
from
through
opposite
elements
(N,O),
or
the
towards
are
the
(l,l),
directly
there
angle
(N,N), directly
opening
(0,0),
from)
angle
facing
of
angle
point
away
example, equal
face
axial
and
on
the and
computed
angles in
THE
COMPUTER PAGES IS
AVAILABLE
PROGBAM 1-45
DESCBIBED to
FBOM
1-75 COSMIC
ON
1.7
This or
program
icosahedron
works
was
of
program.
chosen
dimensional
as The
an
with
a tetrahedron,
of
one
(x_,
PPT
y_,
(x2'
Y2'
by
a unit
example
to
icosahedron
rectanqular
vertices
2 & 3
circumscribed
hedron the
METHODS
is
coordinate
sphere.
oriented
z)2
z3)
=
(o,
=
(0,
.850651,
=
(
l
=
T
with
the
intersections
the
origin
(0,0,0)
in
system
_
(
_
(850651,
where
the
so
icosa-
qeometry a three
that
the
are
z )
Y3'
The
illustrate
,
=
of of
the
I
)
.525731)
,
(.525731,
(x3'
octahedron,
O,
v/_-T
O,
)
.850651)
,
1
,
.525731,
O)
O)
1 +v_-
the
axis
icosahedron,
I_76
X,
Y,
Z,
Figure
located 1 .43.
at
Z
X2,Y2
,Z2
X
I Figure
This
PPT
is
divided
are
translated
the
desirable Using
the
edqes
(X3'
Y3'
onto space
the
into
1.43
the
smaller surface
triangluar of
units
a sphere
which
constituting
form.
following
of
the
PPT
and
Z3)
are
rotated
formula the from
1.44.
1-77
the
origin 3-space
planes (XI,
consisting
YI,
into
ZI)
of
(X2,
2-space,
Y2,
Figure
Z2)
x
Figure
Where
x,
_,
and
Z'-axis
are
found
u are
=
%1 x
+
_1 y
+
Ul z
y'
=
_2x
+
u2y
+
_2 z
Z'
=
X3X
+
P3W
+
_3
cosines
respectively
with
of
respect
the to
1.6
z
X -axis, the
old
by: %1
=
Xl/_
Pl
=
Yl
_)1
=
Zl
2
Xl V
Xl
//
X3;
X'
direction
//
X 2 ,
1.44
!J2,P3;
/_/
2
+ Yl 2
+
Yl
+
Yl
2 Xl
2
2
+
Z1
+
Zl
+
Z 1
2
and_2,_3
2 2
are
1-78
found
similarly.
and axis
Y'-axis,
and
The following
edge
the
consisting the
the
method,
FIND:
at
of
of
PPT
is
Figures angle
PIP 2,
subdivided 1.45
and
_ contained and
the
into
units
by
the
1.46.
within
origin
the
with
the
rotated vertex
triangle located
origin. =
Arctan
(Py2)
r
1.7
Px 2 where
r
= 1 and
is
considered
constant
Y
•
P2
x
Figure
THEN:
subdivide
the e =
Where
T =
1.45
angle
_ into
_ N
T
Increment
1 to
N angles
1.8
N
Y
°
_
t Figure
1 .46
1-79
e
X
O ÷
e,I X
oo
I
cO 4-
CO
X v
II
._.
II
e4
c_ c,,I 0r--
_ e.I
f c,l
X
14o
t
O
4-
, C
x
X
_ e-
OX
• 0
II
II
_
o
_
co
_
II
"t-
_
"-" "-" %
i
x
x
I
t
4-
"5
II
II
2
e" 4-_ @4
e,l
@4
i
X {_
U I
i
X
X
.,_
_ _
_
I
X_
_ v
v
X v
I
II
II
II
co
x v
co C)
X
×
0 4-
I
0
"5 0r.-
4-_
0r-
or-
0rO °r--
_
_
_
e4
e-
.g _g cI---
I H
,-,
y
Rotate
the
2-spaces
points back
where,, and
a2
c2
intersection
along
_i x'
+
pl.y _ +
_1 z
y
=
_2 x'
+
_2y'
v2Z
z
=
_3
X I
+
v,
are
with
_-1 =
Xl
Pl
Yl
=
_2, the
Zl
_3;
u3Y
+ I
+
PPT
edge
from
1 .II
v3 z
direction
cosines
of
the
old
axis
respect
to
/x
v1 2
+
Zl 2
12 +
//x /X
P2,
co-ordinates Figure
the
3-spaces.
=
Z'-axis
in
cI
X
p,
Retain
aI
of
to
Vl=
shown
=
12
+
Yl
+
Zl
12
+ Yl 2
+
Zl 2
P3;
and along
_2,
v3 the
1.47.
1-81
the
X'-axis, and
are
found
edges
Sl,
are
Y'-axis, found:
similarly. S2
and
S3
as
1
2
N--
3
=
;
$3
4
N
=
_N
3
3
2
2
Figure
After the
PPT,
findinq they
smaller
grid
are
equal,
not
centers final
of 3-way
the
are
unit
connected
network. the
these grid
1.47
measurements through
Since gridding
the will
a grid units create
"windows"
must
network.
Fiqure
1 .48.
P2
P4
Figure
1.48
1-82
along
be
found
the
edges
determining
alonq
the
PPT
"windows" to
establish
of a
edge The the
The
gridding
method: and
From
$3,
the
calculate
intersection PsT6
and
of
and
the
equation
and
solve
P3
coordinates
of
Pz
P2 with
a line
in
x
-
is:
find
takes
the the
window
the
two for
the
x 3
Y
X4-
X3
Y_+-Y3
z4-
z3
x -
xs
Y
z -
zs
z6-
Zs
-
"
Z2-
,Y3
z
-Y5
Y6-Ys
of
PIP2
with
S2
finding
the
-
form
three
of
lines
intersection.
Y2-
Yl
Sl,
point
of
z
of
P2 with
P1
points
following
by
Xl
intersection following
the
edges
Y -Yl
x6-x5
To
using
the
xI
x
is:
the
three-space for
X2-
PsP6
by
by
the
P-3--P-4 and
Ps-_G
is:
P-_4
found
along
simultaneously
PIP2
are
coordinates
P4 with of
windows
zI
1.12
Zl
-
P3P4
z 3
the
equation
form:
(I)
PIP2
is:
x(y2-yl)
+ y(xl-x
2
)
= Yl(Xl-X2)
+ x1(Y2_Yl)
(2)
PIP2
is:
y(z2-zl)
+ z(y2-y
I
)
= zl(Yl-Y2)
+ yl(z2_zl
(3)
P3P4
is:
x(y4-Y3)
+ Y(x3-x4
)
:
(4)
P3P_
is:
y(z4-z3)
+ z(y3-y
)
= z3(Y3,y4)
1-83
4
Y1(X3-X
4)
)
+ xl(y4-y3) + y3(z4_z3
)
For
PI_2
For
P--_4
let:
(y2-yl)
:
(Xl-X2)
= bI
yl(x1_x2
)
let:
aI
+ x1(y2-yl)
(y4-y3)
= a2
(x3-x4)
= b2
y1(x3-x4)
using the
the
formula
intersections Find
For
I,I0
P--_2
the
of z
The
let:
other
two
manner.
Once
are
determined,
two
methods:
METHOD
3)
x and
:
y
c2
coordinates
of
P--_-4"
coordinate:
let:
P--_4
for
with
(Z2-ZI)
= al
(yl-Y2)
= b 1
Zl(Yl-y
For
+ xI(Y4-Y
solve P-_2
= c
+ yl(z2-z3)
(z4-z3)
:
(y_-y4)
= b2
z3(y3-y
4)
of
the
vertices the
2)
a2
+ y3(z4-z3) window
coordinates
for
its
center
is
PPT
Plane
the
= cI
are
the
found
= c2 found
vertices by
one
of
in
a similar
of
the
the
window
following
I: On
the
triangles p_(x3y3z3)
with as
windows
vertices shown
in
appear
P1(xiYlZl) Figure
1-84
1.49.
as
equilateral
, P2(x2Y2Z2
)'
P!
P
"x
Figure
The
center
1.49
C(cx,cy,cz)
is
found
with
the
following
formula: CX = Xz+
x2+
x 3
1.13
3 CY = y1+
y2 + Y3 3
CZ = zz+
z2+
z
3 METHOD
II: The of the then To
coordinates the
PPT
of are
sphere. found find
the
first
The by
the
center
the
window
found
"exploded"
to
of
the
intersection
projection
Z-89
of
each
on the
the
surface
surface
of
exploded
window
of
bisectors.
angle
vertex
of
the
is
window
onto
the
a line each
unit
throuqh
coordinate
the
distance
Figure
sphere, the of
translate
vertex each
between
of
the
vertex,
the
vertex
each PpT PPT, PPT
vertex and is and
the
the
Y
-X
Figure
1.50
I'86
origin;
divided
1.50.
z
along
by
origin,
2
d
=
r
= 1
X
'
1
--
xI
2
2
+ Yl
+ zl
where from
d = distance origin to P
1 .14 l
where r = radius of the sphere to be exploded upon and is considered constant
rx 1
d
1 .15
y ' 1 = ry. 1 d z'
=
rz
1
Translate
"window"
1
with
vertice
P3 at
P2 T
Figure
1.51
1-87
the
origin,
Figure
1 .51.
P1Tx
P1
PiTy
= Ply
Rotate and
PI
P2Tx
P2 x :
-
P2x
P3
P3Txl'
P3Ty t ,
at
P2,
the
origin
of
is the
x
P3 z
PI,
center
Z
P3 x
P2 z
P3 is The
P3
Z
P2Tz
plane
bisectors
P3y
PIT z
P2Ty
1 .16
P3 X
X
= 0
P3 Tzl P3 so
that
using
found
with
triangular
PIP3 equation the
window
will
fall
on
the
X-axis
of
two
angle
1.6
intersection PIP2P3,
Figure
1.52.
Y
x
Z
Figure
1.52
1-88
The
angles
_ and
_ are
Arctan
Y
found: 1 .17
2 =
X
Arctan
-,f
2
Y2
= 6 X 1 -X 2
rotate
P2 about
Y4
1/2
y
PI'
= Y2 cos
_ degrees
origin
xs = Y5
With
Rotate
= Y2
I/2
the
center
previously
-
x 2 sin
I/2
y
then
rotate
8-
Y2
(x 2-
1 .18
P2 about
sin
1/2
Xl)
sin
Pl
toward
a + xI
1.19
I/2
P3P4
of
to
y
origin
intersection C back
1/2
a +
PIP s and
degrees
Y
sin
c0sl/2
cos
P3 at
translate the
at
(x2-xl)
defining
the
1/2
Y + Y2
P3
for
P1,
1/2
the
thus
toward
x 4 = x 2 cos
locate I/2
P3
three C back
formula
1.9
may
line
P3P4,
PIPs
space
using
formula
to
"exploded"
three
space
window),
C'x
= Cx
+ P3 x
C'y
= Cy
+ P3Y
C'z
= Cz +
findinq
("C"
figure
used center
I.II is
to
solve C.
Then located
in
1.53. 1 .20
P z 3
T-89
be
"
P1
P2
C'
P3 T
P2
P3 T
For
Method
"exploded" 1.14
of
to
formula
Using
the
pairs
of
element of
II,
surface
of
of
the
the
centers
sphere
this
the
structure
(face
angle
a radius (axial
the
Method
coordinates,
elements and
or
1.53
found
using
are
formula
1.15
elements
elements
faces
the
and
the
the
I
Fiqure
structure
from angle
(_), _),
the _),
program
the
origin and
(dihedral
the angle
1-90
finds the
the
anqle
lengths
between
angle
between
the
to
endpoint
of
an
angle _),
between Figure
adjacent 1.54.
Figure
To use
the
angle to
the
endpoints
angle
the
of
a common
origin. in
be
the
coordinates is
lated z 2)
find
the
same points
P1 and
between their
endpoint
The
other
1.54
elements
the
endpoints. to two
each
manner.
Letting
resultina
from
P2,
1-91
The
element
endpoints (xi, the
face
}
m,
vertex
of
and
is
Pz
and
P2 are
Yi,
zi)
translations
we the
translated
and of
trans(x2, the
Y2,
1 .21 COS
e¢
=
_XlX 2 + YlY2 dld
where and
is
the To
that
the
the
angle.
axial
+
angles is
origin
is
2
+ Yl
2
2
2
2
Y2
2
+ Zl +
2
z2
angle.
vertex
The
12
v_X
=
desired
the
v//X
=
d 2
find
and
6,
dI
+ zlz
used
the
angle
between
found
at
with
desired
angle
above
established
is The
the
two
is
method one
end
other
is of
used an
endpoint
except
element to
define
_.
adjacent
faces,
the
dihedral
using I 6
cos
=
-AIA
+ C C I 2 I 2I
2 +BIB 2
2
1
B1
+
2//
1 22 2
+ C 1 _/A2
2 +
Be
2 +
C2
where B is
the
AI x + BIY one plane used
face
desired +
and
containinq because The
A,B,
ClZ
+ DI
A2X the
the and
angle.
+ B2Y other
obtuse C for
a
= 0 defines + C2Z face.
the
+ D2 = 0 defines The
angle
is
each
plane
planecontaining
negative
are
computed
Yl
Zl
1
Y
Z
l
2
2
y_
Z3
l
Z2 Z1 Z3
1 1 1
X2 X3
1-92
is
desired.
XI B
sign
the
as 1 .23
C
=
Y2 Y1 Y3
X2
Xl X3
where
(X
1 ,
the
plane.
are
used.
Zl)'
In
The general
Y1,
particular
length
of
is To account
found
the
the
and
three
elements_
(X3,
Y3,
vertices
are
Z3)
of
found
by
lie
each
the
p x2
total
certain
as
)2
desired
reduce
angles
in face
using
the 1 .24
xI
same
Z2),
equation: (P
total
Y2,
(X2'
1 1 1
and
at
least
using
the
+
(P
Yl
P Y2
)2
(P
-
p
)2
zl
z2
takes
into
length. output,
this
program
symmetries
and
outputs
lengths.
The
rest
one
+
outputed
following
value
symmetries,
1-93
only of
the
and
can
Figure
a part values easily 1.55.
of
the
are
the
be
N ,1
v
Figure
0,0
1.55
1-94
N,O
FACE
ANGLES For
away
every
from)
towards
the (or the
is
to
equal
(N,N-I),
(N,N), directly
If
the
vertex
(I
+ I,
or
point
The
(N-I,O),
(N-I,
N-I).
lie
(l,J),
Also,
only
a line
of
the
The
elements
element,
of
two
the
element
only
are
the
face are
angle
computed. be
J-l),
YI,
(0,0)
angles
will
(I-I,
can
the
either
(l,J),
on Z1,)
be
lengths with
the
and
an
right the
mid-
is
angles. the
For
at
circle,
one-
dihedral
element. one
the
into
element
the
of
put
and
angles,
a cord
the
falling
structure
axial
and
at
computed.
containing
two
For
vertex
angle
(X I,
opening
(N,N).
(0,0)
or
(or
angles
(N,I)
angles
associated
is
with
J)
face
with
are
(I,0)
the
+ I,
the
and
from
through
faces
there
since
(l,J),
side
angle
the
at
passing
opposite
dihedral
away
the
equal
(N,O),
or
towards
(N,O)
Thus,
(I
correspondence
between
(0,0),
angle
to
are
point
(I,I),
is
directly
there
the
towards
on
to-one
from)
J + I),
(l-l,J).
opening
(0,0),
angle the
facing
angle
point
away
example,
of
face
each the
angle each
end, two
but
angles
J
are
equal
and
a one-to-one This
parallel
to of
to
one
program the
a line
opposite
side. of
be
considered
correspondence
angles.
side
may
the
side
lengths
between
will
only
opposite
through All
one.
other
compute
Y1,Z1)
lengths
and
angles
I-9_
this
elements
(X 1,
(X I ,
In
case, and
values
on
right
and
and
the
midpoint
computed
in
axial elements
Z1)
angles
have
around
YI,
and
we
the of
are
symmetric
this
manner.
the
The for
computer
the
IBM
language.
7040/7044 The
Octahedron, chosen upon may
be
The example
as
of
takes
advantage as
be
used
the
spherical
a basis
for
determining
input
data and
discussed
of
may
data
is
is
given
be
read
the
within text
upon given
in
Table the
units
and
a six in
coordinates
in
large
material.
1-96
the
form
for
IV
a Tetrahedron,
given
as
symmetries in
for
is
written
FORTRAN
depending
1 for
of
was
utilizing
output
output
sphere,
contained
The
example
hedral
hedron
data. of
used
may
Icosahedron,
input
a radius
here
computer,
program
or
as
program
based
therefore, structures.
Table
1.6.
frenuency 1.7. spherical Figure
The Icosa-
The
output
Icosa1.56
THE
COMPUTER PAGES IS
AVAILABLE
PROGRAM 1-97
DESCRIBED _o
FROM
1-144 COSMIC
ON
1.8
This
mathematical
subdivision
of
and
by
a unit
as
to
illustrate
example
polyhedron
is
coordinate
system (xl
so 'Yl
that 'zl)
(x2,Y2,Z
2 )
the
with origin
the
intersection (0,0,0)
of
of the
of
of
O,
v/T-T,
---
(0,
.850651,
=
1
=
(.525731,
the
was
the
chosen
model.
The
rectangular one
PPT are:
1
.525731)
O,
O,
,/T,
1
(.850651, l
for
icosahedron
dimensional
=
m =
written
Icosahedron
vertices
= where:
was or
geometry
a three
:
(x3 'Y3 ,z
The
the in
model
octahedron,
sphere.
oriented
4 & 5
computer
a tetrahedron,
circumscribed an
METHODS
.850651)
,o
.525731,
O)
+ v/_2
axis
polyhedron.
1-149
x,
1 .25
y,
Figure
z
located 1.58.
at
the
Z
&
/ X I,YI,
Zl
/ X ,Y3, Z3
/
Figure Due
to
only
the
methods
even
in
frequency
which
PPT
vertices
is
take
the
three-way
subdivisions
Subdivision The
1.58
and
subdivided the
may
be
Gridding into
grids
Method
parts
4
so
that
where
I
N is
such and
+
that are
Figure
N
the 0
used
}
,
Yl
frequency I to
the
form: z
xI
generated
used.
for
equal
are
_
N.
identify
+
I
of The
wthe
values each
N
,
Zl
structure of vertex
1 .59.
z-z46
I
+
I and
are for
-
1.26
N I
is
unique each
z
an for
side
interger each
as
shown
vertex in
S2 3
x3 ,Y3, z3
2
X2'Y2
_\
Figure
PPT, way to
found the
grid their
the
points is
unit of
generated
respective
• Z2
/, _ XI,YI
Having
I
1.59
division
along
subdivision with sides.
• Zl
are the
the
connected
lines
Figure
S 2
principal
of
the
1 .60
P2
P_
S3
Sl P1
Figure
1.60
1-147
so grid
side that
of
a three-
perpendicular
t_
v >_ cO
N
>,,
I
I
I
N I
>,
N
_
N
°lI CO
..i.a
+
+
+
+
_J c_ co
cq
X I
.._
_
X I
I ,-4
>_ I o3 _D II II
i
cO
N II
II
c.i.a
II
II
,-4
II
>_
_
_ e4
>_
I H
(xl
,--I
x
>_
I
I c_l
I
/>_.
=
>_
X 1
I
_
4--
II
0
-t-
r_
+
II
c-
+
+
II
°_'-
I
m (1# I:::
.io
c_ >._
_
_
N
cN N
I
I
4-
C_I
°r--
I
X I
II
"'"
_
X
I
>_ I
X ..
I_'_
,°
""
°"
""
•r--
.I--
°r--
._--
.r-"
,_
(1.1
4-_
0
Cq
0.]
._"
_-
13-
0 4I--
_
_
v
I
X
'_ X
X
I
cO
e°r-
_
m
X
°,
+
II e4
I
.._
+
_
v
v
Solve
the
section
equations
for
of
PIP2
with
D =
a
b
i
a
x & y
coordinates
ef
the
inter-
P--_4
1
b 2
X =
the
2
CI
b1
c2
b2
1 .28
D
y
find
:
the
a I
C1
a2
c2
z coordinates For
PIP2
let:
(z2-zl)
= a1
(Yz-
Y2 )
:
zl(Yl-Y2) For
P3P4
let:
+ Y1(Z2
(z4
-
z3)
= a2
(Y3
-
Y4)
= b2
z3(Y 3 The need All found the
intersection not other in surface
be
bl
of
P-_P-6with
of
intersection
Y4)
-
+ Y3(Z_
P--_4&
z 3)
-
F_F 2 are
= c
z 3)
1 .29
= c2
coincident
and
found,
points like
manner of
the
and
of
are stored
circumscribed
the for
sphere,
z-z49
three-way final
grid
translation
are to
Subdivision With the
the
edge
of
3-space,
and
following the
for
equations
PPT
Figure
Gridding
and
the
the
origin
Method planes
5 consisting
(0,0,0)are
of
rotated
from
1.61.
/ y, Y
Figure
1.61 1 .30
Where
x,
Z'-axis
_,
=
XlX
+
PlY
+
VlZ
y
:
x2x
+
p2 y
+
v2z
z
_
x3x
+
u3Y
+
_3 z
are
direction
respectively X1
=
_I
=
D1
X2'
_
x _
X3;
_2'
=
JX
y
_Z
_3;
cosines
with 2
i /
xl
1/
xl
1 /
Xl
and
2
2
92'
of
respect +
Yl
+
Yl
+
Yl
_3
2
2
to zl
+
zl
+
Zl 2
are
the
X'-axis, old
2
+
2
the
2
found
I-zSo
similarly.
axis
Y_-axis, and
are
and found
by:
Due
to
the
only
even
side
of
method frequency
the
following
PPT
the
consisting the
of
which
the
three-way
subdivisions is
may
subdivided
method:
FIND:
at
in
Figure angle
PIP2,
into 1,62
the
be
used.
equal
and
6 contained and
grid
generated
The
arc
units
the
rotated
principal by
the
1.63. within
origin
is
with
the
vertex
triangle located
origin. = Arctan
Where
r
(
= 1 and
Py2 P2 x
)
is
r
l .31
considered
constant
¥
v
Figure TH EN:
subdivide
the =
e where
T =
angle
_ N Increment
1.62
_ into
N angles
T
l .32
l to
N
1-19i
Y
P2
P3 X
Figure
The
points
of
PIP
2
intersection is
Y
-
1.63
of
Yl
O--P3 and =
Y2
PIP2
are
found:
Yl 1 .33
-
0-_ 3
The
is
equation PzP2
X 1
y
-
0
X
-
0
takes is
=
the
x(y
let
X2
2
Yz)
(Y2
-
Yl
(x 2
-
Xl)
-
xy 3 Y._ -X
3
0
=
Y3
-
0
X 3
-
0
following
-
Yl(Xl
-
)
a
=
b 2
form: Y(Xl
=
al
:
b I
x2)
yx 3
=
+
+ =
X 1
Xl(Y2
-
=
x2)
-
0
2
C 2
I-1_2
Yl)
Yl(Xl
=
ci
-
x2)
+
x
(y2
-
yl)
Solve the equations
for
the point
of intersection:
i bbi[
D_
bI
C 1
1 .34 X=
C2
2
b 2
D a
V_
c I
i
a 2
Rotate
the
2-spaces
where and
points back
%, Z'-axis
c
_,
of
to
2
intersection
along
=
%1
x"
+
ply"
+
Vl
z
y
=
%2
x "
+
p2y"
+
_2
z
z
=
_ 3 x"
+
_3Y"
+
_3 z
are
with
direction
cosines
respect
to
the
%1
=
X 1
jJ2
I_I
=
Yl
/
=
Z
//X
]
%2 ,
in
Figure
the
edge
from
1 .35
of
old
X I
+
/x
co-ordinates
1
%3 ;
Yl
axis Zl
+
z
Yl
+
Zl
and
_2,
the
edges
2
2
+
P
along
3
;
1.64.
Z-Z93
X'-axis,
and
are
are
found
Y'-axis, found:
2
+
+Yl
I
1
P2,
the
2
2 1
2
Retain
PPT
3-spaces.
X
_,
the
2
_3
Sl,
S2
and
similarly.
S
as 3
shown
N X3,Y3,
3
2
I Z2
Z3 1
_"
_"
=
/NX2'Y2'
2
3
2
X!
,YI,
Figure
After the a
finding PPT,
grid
not
the
the
points
network. of
centers three-way
equal of
unit
these grid
the the
"windows" on
the
1.64
along
subdivison
Since length,
Z1
divisions of
PPT.
3
are
units
the
connected
along
gridding
will
must
be
Figure
1-19_
found 1.65.
principal
the
sides thereby
principal
create to
creating sides
"windows" establish
of
are The
the
.
The method: calculate of
PIP2
the
two
the
three
gridding From
and the
point lines
1.65
windows
are
coordinates
coordinates with
Figure
of
P3P4
and
form
of
and
along the
PIP2 the
solve
found the
window with
by
PsPG
equation simultaneously
intersection,
I-i55
by
of
the
edlges
following of
finding
and
P3P4
a line for
S I,
the with in the
S2 and
S 3,
intersection PsP_
by
three-space points
using for
of
_O CO
e4-_
_J ('4 r_
I
N I
_ I
N I
>_
N
_
N
o °_
0
x
_
x
>_
-I-
-I-
-l-
e4
-I-
_J
_
O
_
0 *e-
II
ll
iia
c-
x
_
I co N
I
I
I
N
N
I
N
II
_o
N
_
N
II
II
II
II
X I
_ I _._
X I _
X
_ I CO
_ v
X
x
co
I
I
_
r-I I I-I
4-O
X
_
_1
co
I
_
i
tPI
v
N
_
N
-I-
-I-
-I-
x
._
_,
N
_,
N
I
U
I
I
I
I
_
X
_
X ......_
,_ _
_
_ X .--.-..._.
X
-I_
'_ _
.,-
X
_
N
_
N
_
x
°°
x
X
I=::
o°
0 _._
o x el.,.
c-
c"
¢/) °r-
0
_o
_1"
X
eJ
°_
_
X
X I X
o°
II
co
I c_
_
_,
c-
x
II
II cxl
c_
X v
II
',.0
e,I
0
II
-I-
I
t-
X c,,I
v
.if
_j ,la
•i-I-
_,,
I
I
c-I
X
-_
°r--
_D
II
I
N
N
I
I
I
:_,
II
I
CO
x
e,l
e,l
I
CO
_
co
I
N I
._-
c-I N
r-I
x
Iio
N
N
N I
I
I-o co
N
_3J
x
I
•;--
.lII _io
_
*°
°°
°°
•r=-
*r-
.r-
°r-"
°°
.°
°°
X
,_
Q_
SO
• r-
°
4--
40 L_-
Find For
P1P2
the
z
coordinate:
let:
(Z2-Zl)
= aI
(yz-y2)
= b I
Zz(yl-y For
P P 3
4
1 et :
(z4-z3) (y z
The
other
two
manner. are
3
3
the
determined,
the
(w
+ w
I
cw
respective
vertices
surface vertex
= b
the
for is
(z4-z)
= c2
3
are
the
found
found
vertices by
the
in
a similar
of
the
following
window method:
+ w )/3 3
of
w = the x, vertices
y,
of
the
the
the
sphere
and
the
Grid
along
origin =
windows
or z coordinate of the window.
three-way
the
X
3
window
= center
of of
2
= c 1
2
+ y
4
center
Translation
to
)
coordinates
where:
The
4
of
its =
-y
+ yz(Z2-Z3)
= a2
(y3-y)
vertices
Once
cw
2)
for
Method
grid a line (0,0,0)
are
of
4 & 5 then
passing of
the
the
translated through polyhedron
the by:
rx
i
1
d Yl
= rYl 1 .37
d Z
_
=
rz
1
1
d
I-i57
Where:
iX
d
I2
+ Yl
and
d = distance
and
r
= the
r
= 1
where :
Using lengths
of
pairs
of
and
a radius
(axial structure
angle
radius
translated
the
elements
from _),
(dihedral
+ zi 2 from
the
elements
2
and
of
the
to unit
coordinates, of
(face the
origin
the
angle origin angle
angle
8),
1
sphere
this
structure _), to
the
P
the an
Figure
T-i58
(_), angle
endpoint
between
program the
angle
between of
adjacent 1.66
finds
the
the between
the
elements
element
faces
of
the
/ ! /
\
/ /
i
//
\
Figure
To
find
coordinates common The
other
the of
endpoint two
manner.
Letting
resulting
from
angle their to
between
elements
endpoints. each
endpoints (Xz, the
1.66
The
element PI
Yl,
translations
and
and Zl)
of
face
vertex is
P2 are and
the
(xl, the
of
_ the
to
translated zl)
endpoints
in be
the P
use is
the the
the a
origin. same
points and
1
Z-z59
we
angle
translated
Yl,
_,
P 2'
COS
o_
=
XlX
2
--
+
and is
the
desired
To the
find
vertex
origin The
is
The is
found
+
/xx2 2
=
d2
12
2
1 .38
2
Yl
+ Y
2
2
+
2
+ z
2
Zl
2
2
angle.
axial is
angles
the
established
used
desired
v_X
dl=
1
d d 1
where
+z, 1
YlY2
with
angle angle
at
the is
above one
other
method
end
of
endpoint
is
an to
used
except
element
and
define
the
the
dihedral
that the
angle.
_.
between
two
adjacent
faces,
),
B,
using COS
B
=
-I AIA2 2 V/A 1
+ +
BIB2
B
2
CIc21
+
+
C
1
1 .39
J'A2 2
2 1
+
B 2
2
+
C 2
2
where B is
the
AI X + B1Y face the
and other
angle
is The
A2X
desired
angle.
+ CI Z + D 1 = 0 defines
+ B2Y + C2Z
face.
The
the
plane
+ D2 = 0 defines
negative
sign
is
the
used
containing plane
one
containing
because
the
obtuse
desired. A,
B,
and
C for
each y
a
plane
are Z
I
Y2
Z2
1
y
Z
1 8
X
Z 1
_
1 .40 l
1
X
Z
2
1 2
X
Z 3
1 3
1-160
as
1
I
3
B
computed
C
where
(Xl,
the
plane.
are
used.
Yz'
Zl)'
In
The
of
1
X2
Y2
1
X3
Y3
1
Z2),
the
the
and
three
elements
(X 3,
Y3'
vertices
are
Z3) of
found
by
lie
in
each
face
using
the
equation:
=
p
_ p Xl •
is To certain and
Y1
Y2,
particular
length
general
(X2,
Xl
reduce
following
desired total
symmetries
lengths.
least
the
one
The outputed symmetries,
)
+
(p
X2
_ p Yl
)
+
(p
V "2
_
)2
p
Z1
Z2
length.
outDut, and
this
outputs
rest
of
value
the and
Figure
program
takes
only
a part
values
are
can 1.67.
1-161
easily
of the
be
into the
same found
account
total as using
angles at the
1.41
+
I"-',.0
-i-
I Z
S,-
L2
0,1 _0 r-4 I I--I
THE
COMPUTER
PAGES
IS
PBOGP..a_4
1-163
AVAILABLE
to
FROM
DESCRIBED
1-184
COSMIC
ON
1.9
This
mathematical
subdivision
of
and
by
a unit
as
to
illustrate
The
example
polyhedron
tangular PPT
is
model
was
octahedron,
sphere.
The
the
oriented
coordinate
6 & 7
computer
a tetrahedron,
circumscribed an
Methods
system
or
a three
so
that
of
was
the
rec-
vertices
of
are:
(Xl'
Yl'
ZI)
=( o, __g
(x2,
(X3'
Y2,
:
(0,
.850651
=
(,850651,
4v_v__) ,
.525731)
z2)
.525731,
Z3)
Y3'
= (.525731, whe re :
'
T
=
1 + _rg 2
1-189
0,.850651)
chosen
model.
dimensional
the
for
icosahedron
icosahedron
geometry
in
written
0)
one
The intersections the origin
(0,0,0)of
of the axis
X, Y, Z is
the polyhedron.
Figure
located
at
1.68.
(X3,Y3,Z3)
X
\
X2,Y2,Z2 )
\ \y
\
/
\ \
/
Figure
1.68
1-186
I--
O c-
%
e-I
= 0
Cl •r-
_0 c-
N +
k---
E
O
O
m O N
t,,•r-" _
=
°r-O
_-°rN
N F'-
_'O
°t--
N N
:5
_-" _ e-
C'4
>_
t'-
C,l
_
|
°r-
O
N I'-v
X I
c-"
C-I
"IN I-'-"
bCO r--t
O •r--
N Il-
_
°r-
>_
+._
C-I
v t'--
_
c-
l
0
0 _
_ C"
e-"
c-
4_
-I-:' (..)
4._
II
II
II
"N
Ill-V
"tO °° 0 IlL
H
X
m
N "0
I
O
X I
°rZ r_ X
II
o_
X F--
"x
:_ r_
>-
N
N I'--
k-v
I.-v
_
_
0
o I
bv _ 0
X v II
11
II
N
X v
I--v _ .r-
0
CO CO _-_ I H
Note:
ABC
right
triangle
is
a
C 4
A
w
Figure
Subdivision
for
The are
Method
Line
divisions
AB of
following
6
is
subdivied
the
=
x
-
parts
angle
Figure
2[Arcsin 22
into
central
equation.
¢
1.69
of
chosen
the
as
equal
polyhedron
by
the
1.70.
(J(x2-xl)
2
+(Y2-Yl)2/2)]
-x 1 1.46
J(X2-Xl)2 Y2-Y y
+(y2-Yl
)2
+(Y2-Yl
)2
N
l
=
Sin
J(x2-xl)2
where:
0
0
=
i/
=
the
<
I
(xl, points
2_ central
YI)
and on
angle
(x 2,
the
PPT
I-z89
y2) edqe
of
the
represent
polyhedron
any
two
(X1,¥1)
(x2 ,v2)
I
B
/ /
/ / /
/ / --
8 .._/ / /
/ / /
/o,o,o) Figure
Subdivision
for
The
Method
Line
AC is
7
subdivided
arc
divisions
the
origin
of
the
polyhedron
the
center
of
the
triangle
equations
of
are
A:
Arcsin
2 x =
angle
for
[
(x2-x
3 /(x2-x 2
y
an
used
1.70
into
made
up
with of
this
2, 3 /
parts
of
the
the
chose triangle
origin
(0,
subdivision.
The
subdivision.
(xl-x2)2
Figure
+(Yl-Y2
equal
AC and O,
O)
being
following 1.71
)2 ]
I ) I)2
1 .47
Sin(NIA)
+(Y2-Yz)2
(Y2-Yl)
=
3 /(X2-Xl)2
as
+(Y2-Yz
)2
Sin
1-19o
(
N
where:
z_=
the
angle
AO and
between (X1,Y1)
OC
(X2,Y 2)
'_
Le
0 < I
1)
sent
any
PPT
face
and
(x2,y2)
two
repre-
points
on
the
bisector.
0
Figure Gridding
& Projection
The stored each
in
points
of
a matrix
respective
Matrix sphere:
for
Methods
6 & 7
subdivision and
are
method.
1.71
for used
methods
in
Figure
the
6 and
gridding
7 are
process
for
1.72
PT(A,B,C) A = the
type
l-on
the
2-on 3-on
of edge
of
triangle
bisector
of
angle
internal
points
4-external B = line
point
of
triangle
points value
and
x,
y,
z
values
I-3
represents
x,
y,
z
values
of
line
1
4-6
represents
x,
y,
z values
of
line
2
7-9
represents
x,
y,
z
values
of
line
3
on
line.
C = the
number
of
z-z9z
points
the
_ a_
p ao
C',d
r'--"
f JO, i
Ic_ 5-
olin ii
I I--I
Once points
of
triangle
the
divisions by
x
:
subdivisions on
the
(Y1
are the
[M21])
= x(M21 )
where:M21
=-x2-x
-
(
l/3
= y3-y4
X3-X
(neg.
reciprocal
(slope)
2
X4,Y4)
("_,h)
(,,,,v,)
sides
are of
Figures
used the 1.73
to
find
right & 1.74.
-x 3 [M34])
)
Y2-Yl M34
they
M21
+ yl-xz(M2z I
two
equations.
M34 y
other
following
+Xl
found,
(,,_,_'_)
A
Figure
1-193
1.73
of
slope)
1 ,48
and:
x
:
(Y3
+Mix3)
(M 2 y
where
:
=
xM;
+Y3
M1
- (--_--2)
M2
= Y2 X 2
-
(Yl
+M2Xl)
-
M 1)
1 .49
+x3M1
(neg.
reciprocal
-Yz -X
(slope)
1
( X 2_'Y2)
F
(x,_)
(xa,*a)
(x,,,1)
I
i i A
Figure
1.74
1-194
of
slope)
Reflection Figure
of
points
in
two-space
are
found
by;
1.75. x4
=
[(Y3
+Mix3)
-
(Yl
+M2x1) ]
(M 2 -M 1 ) 1
Y4 where
and
the
= x4M1
+
M1 =
_ 1/M 2
M2 =
(Y2
-Yl
(X
-X
2
reflected
/3
+x3M1 1 .50
)
1 )
points
are:
x = x I + 2 (x_
- x 1)
Y = Yl
- Yl )
+ 2 (Y4
(X2,Y
(X1
2 )
,Y1 )
(X,Y)
'x4,Y.)
(X3,Y
Figure
Rotations Figure
of
points
in
3)
1.75
two-space
are
found
by:
1.76. X
=
(X2-Xl)
y
= (x2-xl)
Cos
@
-
(y2-yz)
Sin
_ +x I
Sin
_
+
(y2-Yl)
COS
_
1 .51
1-199
+Yl
in matrix
form
(x,y)
Cos
@ - Sin
Sin
_
(XI,YI)
_
(X, Y) I
Figure
The the
points
equations:
Cos
for Figure
the
1.76
internal
gridding
are
found
by
1.77.
x = (B2C I
-
BIC 2)
(A2B 1 -
A1B 2)
T_196
l .52
y
=
(A2C
AIC 2)
I
(AIB 2 where
:
BIA2)
AI=
Y2
Yl
BI=
x I
-
CI=
Xl
(Yl
A2=
Y4
- y
B2=
x3 - x4
C2=
x3
x 2
-
(Y3
Y2
)
+ Yl
(x2
- Xl)
+ Y3
(x4
-
3
- Y4)
x3)
(x,,Y,)
A
Through of
the
Figure
basic
rotations unit,
the
Figure
1,77
and
reflections
entire
PPT
1.78
z-z9?
three-way
(equations grid
1,50, is
found.
1.51)
,
j
"_JA
p_J
J J_ J
J
B
A
Figure
The
external
1.52
and
1.44
where:
1.43. to
are
rotated the
The
their
angles
external
PPT
their of
are
are
found
respective
rotation
points
of
using :
the
plane
are
rotated
by
found by
equation by
equation
by
equation
equation
1.51
positions.
points
three-space
of
into
respective All
where
points
1.78
the
three-way
equation
grid
are
then
rotated
into
1,45
Tz
= 211
Tz
Ty
= 211 -
Ty
T
= 211-T
1 .53
X
position
are
the
The
origin
X
angles
is
of
then
rotation.
retranslated
by:
z-z98
to
its
original
x
where
:
s
= x+T y
X
y'
:
z'
= z + Tz
(Tx,
+ Ty
Ty,
the
T z)
coordinates
originally All to
the
points
of
surface
1 .54
of
to
which
translated the
the
origin
was
are
then
projected
to.
three,way sphere
the
grid
by:
Figure
1.79
X X
J
=
Dis y'
-
Y 1 .54
Dis Z Z
t
=
Dis
where:
Dis
= j(x)
2 +
(y)2
+
(z)2
/ ! !
Figure 1-199
1.79
For the PPT, the number of: edges
=
3_(3_
-
2)
8 half
edges
= 3_ 2
faces
= 3_(_
-
2) 1 .55
4 half
faces
vertices
= 3_ = 3_(_
+ 2) + 1 8
where:_
= frequency
For
total
spherical
Edges
E_ = 3 4
Faces
= E_
the
and
must
be
form,
even
the
number
of:
2
2
2
Vertices
where:
E = no.
Using the
from and
the
structure
(face
the
(dihedral
_),
origin angle angle
of
edges
coordinates, (_),
angle the
= E_ 4
the
the to
an
between 8),
in the
angle
angle
+ 2
polyhedral lengths
between
between
element
faces
of
the
of
the
adjacent
1-200
pairs
elements
of
calculated.
of
the
endpoint
are
unit.
the
Figure
elements
of
elements and
a radius
(axial
angle
structure 1.80
_),
\
\
Figure
To the
find
the
coordinates
angle of
a common
endpoint
origin.
The
other
the
same
manner.
be
the
points
resulting
P1 and
P2'
COS
=
and
endpoints,
PI (x l,
from
the The
element
Letting
o_
elements,
endpoints.
each
two
in
points
between
their to
1.80
the
is
vertex
Zz)
+ YlY2
_,
of
P2 are and
translations
xlx2
_
+
we
the
translated
and
Yl,
face
use
angle
to
is
the
translated (x 2,
of
Y2'
the
z2) end-
ziz2
did 2 where
d
1
=
_Xl
2
1-201
+
yl
2
+
ZI2
1 .57
and
is
the To
that the
=
desired
angle.
find
the
axial
vertex
origin
angle.
is
d2
is
is
The
used
angles,
angle
found
using
the
above
established with
desired
The
v/x2 2 + Y2 2 + Z2 2
the
angle
between
at
two
one
other
is
8
is
end
of
endpoint
used
an
to
except
element
define
and
the
_.
adjacent
-IA A2 + cos
method
faces,
the
dihedral_
B,
BIB 2 + CIC21
=
l .58 AI2+
B I 2+
C12
J
A22+
B2
2+
C22
where 8 AI x + BIy one
face A2X
the
desired
angle.
+ CIZ
+ D 1 = 0 defines
the
plane
containing
+ C2Z
+ D 2 = 0 defines
the
plane
containing
used
obtuse
and + B2Y
the
other
The
negative The
is
face.
A,
sign B,
and
is
C for
because each
plane
Y A
B
the
are
Z
computed
is
desired.
as
l
1
I
Y2
Z2
1
Y_
Z3
1
Xl
ZI
1
X2
Z2
1
X3
Z3
1
1-202
angle
1 .59
X
Y
1
I C
1
X
Y 2
X
Y 3
where
(X
, Y 1
the
plane.
are
used. The
, Z ), 1
In
general
, Y 2
, Z 2
particular,
length
of
the
, Y
three
, Z ) lie 3
vertices
_ are
found
in
3
of
by
each
using
face
the
equation:
is
At these
1971
(X 3
elements
_ p xl
NASA-Langley,
), and 2
the
/(p
been
l 3
(X
1
l 2
the
time
methods included
--
32
the
had in
0R-1734
desired
of not this
publication, been
)2
+
x2
(p
_ p Y2
)2
+
(p
Y2
_ p z 3
)2 z 3
length
the completed
report.
1-203
computer and
programs
therefore,
for have
not
1 .60