The Rigidity of Compound Spatial Grids

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