La igidi 6 des 6seaux spa iaux compos6s
pa Alain Dandu and
*
Topologie
s uc u ale
#lo,
1984
In oduc ion
Ce documen se di ise en deux pa ies; p emi emen , la dCpendance linkai e des d oi es,
e deuxi&memen , la igidi 6 des beaux spa iaux. I1 en e de dk ini une mk hodologie
pe me an de ai e l'in es iga ion d'un kau spa ial a in &en d e mine
la
igidi 6
ou
la
non igidi k. Le Cseau C an cons i u de liens (d oi es) e de nceuds ( o ules) don la
con igu a ion elie deux ClCmen s igides. Au emen di , le kau o me une a ache
en e les deux Clhen s e il s'agi de dC e mine
si
ce e a ache
es
igide ou si elle n'es pas
igide (mou emen mkanique
ou
in ini bimal).
La dbpendance linhi e des d oi es
Deux d oi es dans l'espace son , soi concou an es, soi gauches. Si les deux d oi es ab e
bc son
coacou an es,
les d oi es du
aimu
de cen e b son di es lin ai emen dbpen-
dan es de ab e
bc.
Si
Ies
deux d oi es son gauches, les seules d oi es d pendan es son
les
deux d oi es elles-memes.
T ois d oi es gauches
dans
l'espace
db inissen un hype boloide
A
une
nappe
ou
une
pa abololde hype bolique
(Figu e
1).
Tou es
les
gCn a ica de
la
mhe amille su ce e
su ace, appelk
le
dgnhm,
son in ai emm d-ndan es aux
ois
d oi es donn es.
1
I
S uc u al Topology
#lo,
1984
I
The
Rigidi y
o
Compound Spa ial
G ids
I
c
I
In oduc ion
This a icle is in wo sec ions: i s ,
on
linea dependence o lines, and second,
on
he
igidi y o spa ial amewo ks.
Ou
pu pose
is
o de ine
a
me hodology which will pe mi
us
o in es iga e
a
spa ial amewo k, o de e mine i s igidi y o non- igidi y. The
amewo ks in ques ion consis o ba s (line segmen s) joining wo igid bodies, he ba s
being a ached o he bodies a uni e sal join s. In o he wo ds, he amewo k o ms a
link be ween wo igid bodies, and i is
a
ques ion o de e mining i his linkage is igid,
o
else
is
non
igid and pe mi s ei he a mechanical o in ini esimal ela i e mo ion o he wo
bodies.
Linea Dependence
o
Lines
Two lines in space a e ei he concu en
o
skew. I he
wo
lines ab and
bc
a c
commen ,
he lines in he la
pencil
wi h cen e b a e said o be linea ly dependen
on
ab and
bc.
I
he !wo ines a e skew, he only dependen lines a e he
wo
lines hemsel es.
Th ee skew lines in space ddine
a
hype boloid o one shee
o
a
hype bolic pa aboloid
(Figll e
1).
All
he lines in one amily o gene a o s o his su ace, which amily is called a
g9h ,
a e
linea ly dependen on any gi en
se
o h ee o hose lina.
el
---
--
42
Topologie
s uc u ale
#lo,
1984
G n alemen , une con igu a ion den d oi es, aucune d‘elles n’ an dkpendan es des n
-
1
au es, son di es lin ai emen indkpendan es.
E an donne un ensemble
B
de d oi es e son sous-ensemble A, nous pou ons di e que le
sous-ensemble
A
engend e I’ensemble
B
si ou es les d oi es de
B
son lin ai emen
dkpendan es
I
la con igu a ion o m e pa les d oi es de
A.
Le ang de l’ensemble
B
es le
nomb e minimum de d oi es pou an engend e I’ensemble
B.
L’ensemble ide es de
ang
0.
L’ensemble de ou es les d oi es de l’espace es de ang
6.
Ainsi ous les ensembles
de d oi es posseden une aleu de ang de numk o
R
en e
0
e
6.
On appelle a ib un ensemble de d oi es qui con ien :
a) a ec les deux d oi es concou an es, aussi les d oi es du aisceau g n ;
b) a ec les ois d oi es gauches, aussi les d oi es du kgulus g n k.
Ceci an , un ensemble de d oi es es une a ik si aucune d oi e ex k ieu e
B
I’ensemble
n’es dependan e des d oi es de I’ensemble.
Nous
a ons d ji men ionn les deux a ie s de d oi es de ang
2
e une des a i s de
ang 3. Au o al,
il
exis e qua e a ik s de ang 3:
3a) le gulus,
3b) I’union de deux aisceaux non coplanai es e de cen es di en s mais ayan une
d oi e commune,
3c) ou es
les
d oi es passan pa un poin ,
3d) ou es les d oi es dans un plan.
Les a ik ks de ang 4 son appelkes cong uences.
I1
en exis e qua e:
4a) la di usion lin ai e, la a i de d oi es con enan exac emen une d oi e pa chaque
poin de I’espace (engend e pa un ensemble de qua e d oi es gauches e ind pendan es),
4b) ou es les d oi es concou an es
B
deux d oi es gauches,
4c)
I’ensemble des aisceaux
i
un pa ad e, ous ayan une d oi e commune e o man
une a ik ,
4d)
ou es
les
d oi es dans un plan
ou
passan pa un poin de ce plan.
Les a i s de ang
5
son appelkes complexes.
I1
en exis e deux:
5a) non singuli e:
la
a ik de d oi es con enan exac emen un aisceau de d oi es pa
chaque poin de I’espace (engend ke pa un ensemble de cinq d oi es gauches e
indipendan es),
5b) singulihe: ou es les d oi es concou an es
B
une d oi e donn e.
Ces a i s son illus kes dans le Tableau
1.
Le Tableau 2dkc i Ie cheminemen c k pa
I’addi ion d’une d oi e ind pendan e
I
lacon igu a ion minimum. Pa exemple, il
ya
ois
aqons dis inc es d’ajou e une qua ihe d oi e ind pendan e
I
un gulus o m6 de ois
d oi es ind pendan es. Dans la p emii e, la d oi e ne p ne e pas le kgulus. Dans la
seconde,
la
d oi e p n e le gulus e le pe ce en deux poin s. E dans la oisi me, la
d oi e es angen e en un poin du gulus. Les p op ik s de g nk a ion de la nou elle
con igu a ion d penden de la acon don la d oi e ind pendan e es ajou Ce. Ainsi
chacune de ces addi ions mhe
B
une a i k di en e de ang
4.
Le Tableau
3
illus e des
con igu a ions poss dan les memes p op i ks de g n a ion que ce aines des con igu a-
ions du Tableau
1
mais posskdan des p op ik ks combina oi es di k en es.
Mo e gene ally, a con igu a ion o n lines, no one
o
which
is
dependend on he o he n
-
1
lines, is said o be linea ly independen .
Gi en a se
B
o lines and a subse A o
B,
we say ha he subse A spans he se
B
i all he
lines in
B
a e linea ly dependen upon he con igu a ion o med by he lines in
A.
The
ank
o he se
B
is he minimum numbe o lines which span he se
B.
The emp y se has ank
0.
The se o all lines in space has ank
6.
Thus any se o lines has as i s alue
R
o ank,
some numbe be ween
0
and
6.
A a ie y is a se o lines which con ains:
a) along wi h any wo concu en lines, also he lines o he la pencil hey gene a e;
b) along wi h any h ee skew lines, also he lines o he egulus hey gene a e.
This being he case,
a
se o lines is a a ie y i no line ou side he se is dependen upon he
lines in he se .
We ha e al eady men ioned he wo a ie ies o ank
2,
and one o he a ie ies o ank 3.
In all, he e a e ou di e en ypes o a ie ies o ank 3:
3a) he egulus,
3b) he union o wo la pencils, no coplana , wi h di e en cen es, ha ing a line in
common,
3c) all he lines passing h ough a poin ,
3d) all he lines lying in a plane.
The a ie ies o ank
4
a e called cong uences. The e a e ou ypes:
4a) a linea sp ead, ha ing exac ly one line passing h ough each poin in space (gene a ed
by ou independen skew lines),
4b) all he lines concu en wi h wo skew lines,
4c)
a one-pa ame e amily o la pencils, ha ing one line in common, and o ming a
a ie y,
4d)
all he lines in a plane, o passing h ough one poin in ha plane.
Va ie ies o ank
5
a e called complexes. The e a e wo
ypes:
5a) non singula : a a ie y o lines con aining exac ly one la pencil h ough each poin in
space (gene a ed by i e independen skew lines),
5b) singula : all he lines mee ing one gi en line.
These a ie ies a e illus a ed in Table
1.
Table 2 desc ibes he consequence o adding lines
one by one s a ing wi h
a
single line. Fo example, he e a e h ee dis inc ways o adding
a ou h independen line o a se o h ee skew lines gene a ing a egulus. In he i s case,
he added line does no mee he egulus. In he second, i pene a es he egulus, mee ing
i in wopoin s. In he hi d, he line is angen o he egulusa one poin . The p ope ies
o he esul ing con igu a ion gene a ed by he addi ion o one line depend on he choice
o he added line. Each o he abo e ways o adding a line esul s in a di e en
con igu a ion o ank 4. Table
3
shows combina o ially dis inc igu es o lines which a e
equi alen o he gene a ing sys ems shown in Table
1.
1
2
3
'[XI
4'[X]
5
6
'[I
'[XI
"773-1
B[m]
'[,I
B
S uc u al
Topology
#lo,
I984
43
D[j77]
1
..
.
Tableau
1
-Table
1
44
Topologie
smc u ale
#lo,
1984
Tablmu
2
-
Tde
2
S we anal
Topology
#
10.
I984
45
0
4c
0000
00.0
Tabkau
3
-
Tabk
3
46
Topologie
s uc u ale
#lo,
1984
Les cons uc ions sui an es dk e minen les mb hodes pou ou e
il
pa i d’une con i-
gu a ion de d oi es independan es une d oi e dependan e
il
la con igu a ion.
Cons uc ion
1
(Figu e
2).
T ou e la d oi e concou an e aux d oi es
1
e 2pa le poin p.
1. Le poin p e la d oi e 1 o men le plan opq.
2.
Le poin p e la d oi e
2
o men le plan s .
3.
La d oi e che chk es la d oi e d’in e sec ion des plans opq e s :
xy.
Cons uc ion
3A1
(Figu e
3).
T ou e une d oi e dependan e au kgulus 1-2-3.
1. T ou e ois d oi es quelconques concou an es
il
1, 2 e
3:
ab, cd e e .
2. La d oi e che chk doi & e concou an e
21
ab, cd e e
xy.
The ollowing cons uc ions de e mine he me hods by which, s a ing wi h
a
con igu a-
ion
o
independen lines, a line dependen on he con igu a ion can be ound,
Cons uc ion
1
(Figu e
2).
1. The poin p and he line 1 o m a plane
opq.
2. The poin p and he line
2
o m a plane s .
3.
The equi ed line is he line o in e sec ion o he planes
opq
and s , ha is: xy.
Find he line h ough he poin
p
which mee s lines 1 and 2.
Cons uc ion
3A1
(Figu e
3).
1. Find h ee a bi a y lines mee ing lines 1,2 and
3,
say ab, cd and e .
2. The equi ed line will
be
concu en wi h ab, cd and
e ,
say xy.
Find a line in he egulus gene a ed by lines 1-2-3.
Figu e
2.
Cons uc ion
I
Figu e
3.
Cons uc ion
3A1
S uc u al Topology
#lo,
1984
47
Cons uc ion
4A1
(Figu e
4).
T ou e la d oi e d pendan e
A
la cong uence 1-2-3-4 (de
ype 4A) pa le poin p.
1.
Pa p ou e une d oi e concou an c aux d oi es 1 e 2: ab (en se se an de
1).
2. Les d oi es 2 e ab o men le plan
opq.
3.
Les
d oi es 3 e 4 pe cen
opq
en c e d.
4.
La
d oi e cd coupe ab en
i.
5.
T ou e deux d oi es concou an es aux d oi es 2,
3
e 4: e e
gh
(en se an de 1).
6.
T ou e pa i la d oi e concou an e
A
e e gh: ij (en se se an de 1).
7.
T ou e deux d oi cs concou an es
d
1,2 e ij: kl e mn (en se se an de 1).
8.
La d oi e che chke es
la
d oi e passan pa p e concou an es aux d oi es kl e mn: xy.
Cons uc ion
4A2
(Figu e
5).
T ou e la d oi e d pendan e
B
la cong uence 1-2-3-4 (de
ype 4A) dans le plan opq.
1.
Les d oi es
1
e
2
pe cen
opq
en
a
e b.
2. Pa un poin quelconque
C
de la d oi e ab ou e une d oi e dkpendan e
A
la
cong uence 1-2-3-4:
cd
(en se se an de 4A1).
3. T ou e deux d oi es concou an es
A
cd,
1
e 2: e e gh (en se se an de 1).
4. Les d oi es e e gh pe cen opq en x e
y.
5.
La d oi e che chk es la d oi e passan pa x e
y:
xy.
Cons uc ion
4A1
(Figu e
4).
Find he line h ough a poin p, in he cong uence gene -
a ed by lines 1, 2, 3,4
(o
ype 4A).
1.
Th ough p, ind a line concu en wi h lines
1
and 2, say ab (using cons uc ion 1).
2. The lines 2 and ab o m a plane opq.
3. The lines
3
and 4 pie ce
opq
a poin s c and d.
4. The line
cd
in e sec s line ab a poin
i.
5.
Find wo lines e
and
gh which mee lines 2, 3 and 4 (using cons uc ion
1).
6.
Th ough poin i, ind he line ij mee ing lines e and gh (using cons uc ion
1).
7.
Find wo lines kl and mn mee ing lines 1, 2 and ij (using cons uc ion 1).
8.
The equi ed line is he line xy h ough p mee ing lines kl and mn.
Cons uc ion
4A2
(Figu e
5).
Find
a
line in he plane opq which is in he cong uence
o
ype 4A gene a ed by lines 1,2, 3, 4.
1. The lines 1 and 2 pie ce
opq
a poin s a and b.
2. Th ough an a bi a y poin con he line ab ind a dependen line in he cong uence 1,2,
3, 4
say
cd
(using cons uc ion 4A1).
3. Find wo lines e and gh mee ing cd, 1, and 2 (using cons uc ion
1).
4. The lines e and gh pie ce he plane opq a x and
y.
5.
The equi ed line is he line xy h ough he poin s
x
and
y.
k
Figu n
5.
Cons uc ion
4A2
48
Topologie
s uc u ale
U
10,
1984
Cons uc ion
4B1
(Figu e
6).
La
d oi e
4
pe ce le gulus
1-2-3
en a e c.
1.
Pa les poin s a e c passen les d oi es ab e
cd
appa enan au Cgulus
1-2-3.
2.
Pa les poin s a e c passen les d oi es ab’ e
cd’
appa enan au Cgulus conjugu deb
3.
Pa le poin a
il
exis e un aisceau de d oi es d pendan es
B
ab e ac qui elie ous
les
poin s de
cd
’
B
a.
4.
Pa le poin c il exis e un aisccau de d oi es d pendan es
B
cd
e ac qui elie ous les
poin s de ab’
A
c.
5.
Ainsi ou es les d oi es concou an es
A
ab’ e cd‘ son dependan es
A
la
cong uence
1-2-3-4
(de ype
4B).
Cons uc ion
4B2
(Figu e
7).
T ou e la d oi e d pendan e
B
la cong uence
1-2-3-4pa
le
poin p. La d oi e che ch e es la d oi e passan pa p e concou an e
B
ab e cd: xy (en se
se an de
1).
Cons uc ion
4B3
(Figu e
8).
T ou e la d oi e d pendan e
B
la cong uence
1-2-3-4 dans
le plan opq. Les d oi es ab e cd pe cen opq en
x
e
y.
La d oi e che chke es la d oi e
passan pa
x
e
y:
xy.
1-2-3.
Cons uc ion
4B1
(Figu e
6).
The line
4
pie ces he egulus 1,
2,3
a a and c.
1.
Th ough he poin s a and c he e pass lines ab and cd belonging o he egulus
1,2,3.
2.
Th ough he poin s a and c he e pass lines ab’ and cd’ belonging o he egulus
conjuga e o
1,
2,
3.
3.
Th ough he poin a he e is a la pencil o lines dependen upon ab and ac which join
he poin s on line
cd’
o a.
4.
Th ough he poin c he e
is
a la pencil o lines dependen upon
cd
and ac which join
he poin s on line ab’ o c.
5.
Thus all he lines mee ing lines
ab’
and
cd’
a e in he cong uence (o ype
4B)
gene a ed
by lines
1,
2, 3.4.
Cons uc ion
4B2
(Figu e
7).
Find he line h ough a poin
p,
in
he cong uence gene -
a ed by lines
1,2,3,4.
The equi ed line is he line
xy
h ough he poin p, mee ing lines ab
and cd (ob ained by cons uc ion
1).
Cons uc ion
4B3
(Figu e
8).
Find he line in he plane opq which is in he cong uence
gene a ed by lines
1,2,3,4.
The lines ab and cd pie ce opq a
x
and
y.
The equi ed line
is
he line
xy
h ough hese poin s
x
and
y.
Figu e
7.
Cons uc ion
4B2
Figu e
6.
Cons uc ion
4B1
S uc u al Topology
#lo,
1984
49
Cons uc ion
4C1
(Figu e
9).
La
d oi e
4
es angen e au kgulus
1-2-3
en
a.
1.
Pa le poin
a
passe la d oi e ab appa enan au kgulus
1-2-3.
2.
Pa le poin a il exis e un aisceau de d oi es dipendan es
B
4
e ab.
3.
La d oi e ab’ appa enan au kgulus conjuguk de
1-2-3
ai pa ie de ce aisceau.
4.
La d oi e ab’ es concou an e
B
ou es
les
d oi es du kgulus
1-2-3.
5.
Ainsi chaque in e sec ion de la d oi e ab’ e des d oi es du kgulus
1-2-3
engend e un
aisceau de d oi es dkpendan es
B
la cong uence
1-2-3-4
(de ype
4C).
Cons uc ion
4C2
(Figu e
10).
T ou e
la
d oi e dbpendan e
B
la
cong uence
1-2-3-4
pa
le poin p.
1. Le poin p e la d oi e
1
o men le plan opq.
2.
T ou e deux d oi es concou an es aux ois d oi es
2,
3
e
4:
ab e
cd
(en se se an
de
1).
3.
Les d oi es ab e cd pe cen le plan opq en e e .
4.
La d oi e e coupe la d oi e
1
en
x.
5.
La d oi e che chke es la d oi e passan pa p e x: xy (en se se an de
1).
Cons uc ion
4C1
(Figu e
9).
The line
4
is angen o he egulus gene a ed by lines
1,2,3
a a.
1.
Th ough he poin a he e passes a line ab belonging o he egulus
1,2, 3.
2.
Th ough he poin
a
he e
is
a la pencil o lines dependen upon lines
4
and ab.
3.
The line ab’ belonging o he egulus conjuga e o
1,2, 3
is in ha la pencil.
4.
The line ab’ mee s all he lines
in
he egulus
1,
2, 3.
5.
Thus whe e e he line ab’ mee s
a
line
o
he egulus
1,2,3
i gene a es a la pencil o
lines which belong o he cong uence (o ype
4C)
gene a ed by lines
1,2,3,4.
Cons uc ion
4C2
(Figu e
10).
Find he line h ough he poin p in he cong uence
gene a ed by lines
1,
2, 3,4.
1.
The poin p and he line
1
o m a plane opq,
2.
Find wo lines ab and cd which mee lines
2, 3,
and
4
(using cons uc ion 1).
3.
The lines ab and cd pie ce he plane opq a e and .
4.
The line e in e sec s he line
1
a x.
5.
The equi ed line is he line xy h ough poin s
p
and x (by cons uc ion
1).
Figu e
9.
Cons uc ion
4CI
Figww
10.
Cons uc ion
4C2
56
Topologie
s uc u ale
#lo,
1984
I
ELEMENTS
POINT
POINT
POINT
DROllE
POINT
CORPS
DROITE
DROITE
DRolTE
CORPS
CORPS
1
CORPS
LIENS
c =
1
ang
1:
ang
2:
C=l
C ='
ang
1:
ang
2:
C
=1
ang
3:
C,5=1
c
=2
ang
1:
ang
2:
c
=1
ang
3:
C?=4
ang
4:
C =1
c
=a
c
=I7
ang
1:
ang
2:C
=1
ang
3:
Ca=m
ang
4:
C
=is
ang
5:
c!=
;
ang
6:C
=
c
=a
CAS
TYPES
NON
RIGIDES
00
0
n
Tableau
4
-
Tlbk
4
