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--

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X

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4-

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X

_ e-

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II

_

o

_

co

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II

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_

"-" "-" %

i

x

x

I

t

4-

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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

×

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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



=

_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=::



0 _._

o x el.,.

c-

c"

¢/) °r-

0

_o

_1"

X

eJ

°_

_

X

X I X



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

Joe Clinton Paper - Signal Science, LLC

IBM 7040/7044 computer, utilizing. FORTRAN. IV language. The program may be used for a Tetrahedron,. Octahedron, or Icosahedron, depending upon the ..... iX. 2. 2. 2 d. I. + Yl. + zi d = distance from origin to. P. 1 r = the radius of the unit sphere r = 1. Using the translated coordinates, this program finds the lengths of the.

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