Rigidity And Tensegrity By Connelly

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Rigidity And Tensegrity By Connelly

  1. 1. Rigidity and Tensegrity Robert Connelly Cornell University
  2. 2. Part I: The local theory Statics
  3. 3. What determines rigidity?
  4. 4. What determines rigidity? <ul><li>The physics of the materials. </li></ul>
  5. 5. What determines rigidity? <ul><li>The physics of the materials. </li></ul><ul><li>The external forces on the structure. </li></ul>
  6. 6. What determines rigidity? <ul><li>The physics of the materials. </li></ul><ul><li>The external forces on the structure. </li></ul><ul><li>The combinatorics/topology of the structure. </li></ul>
  7. 7. What determines rigidity? <ul><li>The physics of the materials. </li></ul><ul><li>The external forces on the structure. </li></ul><ul><li>The combinatorics/topology of the structure. </li></ul><ul><li>THE GEOMETRY OF THE STRUCTURE. </li></ul>
  8. 8. What model?
  9. 9. What model? <ul><li>My favorite is a tensegrity. </li></ul>
  10. 10. What sort of rigidity/stablility?
  11. 11. What sort of rigidity/stablility? <ul><li>A very basic choice is prestressability . </li></ul><ul><li>This means that there is potential function E ij defined for each member (= cable, bar, or strut) {i,j}, and the configuration p = (p 1 , … , p n ) is at a unique local minimum, modulo congruences, of the total potential </li></ul><ul><li>E( p )=  ij E ij (|p i - p j | 2 ) </li></ul><ul><li>using the Hessian for the second-derivative test. </li></ul>
  12. 12. What energy function to choose?
  13. 13. What energy function to choose? <ul><li>It does not matter too much. </li></ul><ul><li>It only matters what the first and second derivatives are of E ij at |p i -p j | 2 . </li></ul><ul><li>(E ij )’(|p i -p j | 2 ) =  ij the (pre)stress coefficient. </li></ul><ul><li>(E ij )’’(|p i -p j | 2 ) = c ij > 0, the stiffness coefficient. </li></ul>
  14. 14. The basic stress-stiffness decomposition <ul><li>Theorem : The Hessian H of the energy potential is the sum of the quadratic forms </li></ul><ul><li>H = stress energy + stiffness energy, </li></ul><ul><li>and for the displaced p i + p i ’ for each i, </li></ul><ul><li>stress energy =  ij  ij |p i ’ - p j ’| 2 , </li></ul><ul><li>stiffness energy =  ij c ij [(p i - p j ) (p i ’ - p j ’)] 2 . </li></ul>
  15. 15. The stiffness matrix <ul><li>The matrix of the quadratic form giving the stiffness energy is </li></ul><ul><li>R( p ) T DR( p ), </li></ul><ul><li>where R( p ) is the rigidity matrix, D is the diagonal matrix with c ij ’s as diagonal entries, and () T is the transpose. This stiffness matrix is always positive semi-definite. </li></ul>
  16. 16. The stress matrix <ul><li>Note that the stress energy quadratic form is the sum of d identical forms if the structure is in d-dimensional space. Then each of these forms has its symmetric matrix  where each off-diagonal entry is -  ij =-  ji , and the row and column sums are 0. </li></ul>
  17. 17. An example of a stress matrix
  18. 18. Equilibrium stresses <ul><li>For a tensegrity framework G( p) , an equilibrium stress  is an assignment of a scalar  ij =  ji to each pair of distinct vertices {i,j} of G, such that  ij = 0 when {i,j} is not an edge of G, and for each i, the equilibrium equation </li></ul><ul><li> j  ij (p j -p i )=0 </li></ul><ul><li>holds. </li></ul><ul><li>When a configuration is at a critical point (e.g. a local minimum), then the corresponding stress is in equilibrium. </li></ul><ul><li>A cable corresponds to  ij > 0, i.e. tension. </li></ul><ul><li>A strut corresponds to  ij < 0, i.e. compression. </li></ul>
  19. 19. Properties of the stress matrix <ul><li>If the affine span of the points of the configuration p = (p 1 , …, p n ) is d-dimensional, then the rank of  is at most n-d-1. </li></ul><ul><li>If the rank of  is n-d-1, and some other configuration q is such that  is an equilibrium stress for G( q ), then the points of q are an affine image of the points of p . </li></ul>
  20. 20. Part II: The global theory Stress matrices applied again.
  21. 21. Dominance <ul><li>A tensegrity G( q ) dominates G( p ) and we write G( p )  G( q ) if </li></ul><ul><li>Each cable {ij} of G( p ) is no longer than the corresponding cable {ij} of G( q ). </li></ul><ul><li>Each strut {ij} of G( p ) is no shorter than the corresponding strut {ij} of G( q ). </li></ul><ul><li>Each bar {ij} of G( p ) is the same length as the corresponding bar {ij} of G( q ). </li></ul>
  22. 22. Global Rigidity <ul><li>A tensegrity G( p ) in E d is globally rigid , if G( q )  G( p ) implies the configuration q in E d is congruent to the configuration p . </li></ul>
  23. 23. How do you tell when a given framework is globally rigid?
  24. 24. How do you tell when a given framework is globally rigid? <ul><li>Answer: It’s hard, even when there are only bars!! </li></ul><ul><li>More precisely, if you could find a polynomial time algorithm for this problem, you could solve a huge list of unsolved equivalent problems, and most likely earn a million $$. This problem is non-deterministically polynomially (NP) complete. For example, even for global rigidity in the line, global rigidity is the uniqueness part of the knapsack problem. (Saxe, 1979) </li></ul>
  25. 25. Tools to show global rigidity <ul><li>Theorem : Suppose that G( p ) is a tensegrity in E d with an equilibrium stress such that </li></ul><ul><li>The stress matrix  is positive semi-definite of rank n-d-1. </li></ul><ul><li>The only affine motions of the configuration p that preserve the member constraints are congruences. </li></ul><ul><li>Then G( p ) is globally rigid in any E N containing E d . </li></ul>
  26. 26. <ul><li>This tool is sufficient for many of the popular tensegrities that are built by artists and others. </li></ul><ul><li>You can see several examples of symmetric tensegrities at my web page: </li></ul><ul><li>http://mathlab.cit.cornell.edu/visualization/tenseg/tenseg.html </li></ul><ul><li>See also Simon Guest for information about how to use symmetry in rigidity calculations. </li></ul>
  27. 27. Part III: Bar frameworks More stress matrices using the generic philosophy.
  28. 28. Generic configurations <ul><li>A configuration p is called generic if the coordinates of all of its points are algebraically independent over the rationals. In other words, any non-zero polynomial with rational coefficients will not vanish when the variables are replaced by the coordinates of p . </li></ul><ul><li> , e,  , … are algebraically independent. </li></ul>
  29. 29. Generic rigidity <ul><li>Theorem: A bar framework is rigid at a generic configuration if and only if it is infinitesimally rigid at any configuration. </li></ul><ul><li>This means that generic rigidity is a combinatorial property of the graph G only. </li></ul><ul><li>Theorem (Laman 1972): Suppose that G has n vertices and e = 2n-3 edges. Then G is generically rigid in the plane if and only if for every subgraph of G with n’ vertices and e’ edges e’  2n’-3. </li></ul>
  30. 30. Algorithms <ul><li>Corollary (Lovasz and Yemini + …): Given a graph G with n vertices and e edges, there is an algorithm to determine whether G is generic rigid in the plane in at most O(ne) steps. (Variations: 2-tree condition, the pebble game, etc.) </li></ul>
  31. 31. Generic global rigidity <ul><li>A graph G is generically globally rigid in E d if for some (every?) generic configuration p , the framework G( p ) is globally rigid in E d . </li></ul><ul><li>The following graphs are generically globally rigid in the plane: </li></ul>
  32. 32. Generic global rigidity <ul><li>A graph G is generically globally rigid in E d if for some (every?) generic configuration p , the framework G( p ) is globally rigid in E d . </li></ul><ul><li>The following graphs are not generically globally rigid in the plane: </li></ul>
  33. 33. When is a graph G generically globally rigid in E d ? <ul><li>Necessary conditions: </li></ul><ul><li>G must be vertex (d+1)-connected. (This means d+1 or more vertices are needed to disconnect the vertices of G.) </li></ul><ul><li>G must be generically redundantly rigid. (This means that, for p generic, G( p ) must be rigid, even when any edge of G is removed.) B. Hendrickson (1991). </li></ul><ul><li>Conjecture (Hendrickson ): For d=2, these conditions are also sufficient. </li></ul><ul><li>For d=3, these conditions are not sufficient. (Me 1991) </li></ul>
  34. 34. Vertex connectivity <ul><li>If G is not (d+1) vertex connected, then d+1 vertices separate G, and reflection of one of the components of G about the (hyper)-line through those d+1 vertices violates global rigidity. </li></ul>
  35. 35. Vertex connectivity <ul><li>If G is not (d+1) vertex connected, then d+1 vertices separate G, and reflection of one of the components of G about the (hyper)-line through those d+1 vertices violates global rigidity. </li></ul>
  36. 36. When is a graph G generically globally rigid in E d ? <ul><li>Sufficient condition: </li></ul><ul><li>For p generic, G( p ) has an equilibrium stress  , and an associated stress matrix  with maximal rank n-d-1, where n is the number of vertices of G. (Me, to appear.) </li></ul><ul><li>Unlike the necessary conditions above, this condition is numerical, and it involves solving linear equations, as well as computing the rank of the n-by-n matrix  . </li></ul><ul><li>Conjecture : If G is generically globally rigid in E d , then either G is a simplex or it satisfies the condition above for some generic configuration p . </li></ul>
  37. 37. The rigidity matrix <ul><li>For a graph G, the rigidity map f: E nd -> E e is the function that assigns to each configuration p of n vertices in d-space, the squared lengths of edges of G, f( p )=(. . ., |p i - p j | 2 , . . .), where e is the number of edges of G. </li></ul><ul><li>The rigidity matrix R( p ) = df is the differntial of f. </li></ul><ul><li>Fact : The co-kernel of R( p ) is the space of equilibrium stresses of G( p ). I.e.  is an equilibrium stress if and only if  R( p ) = 0. </li></ul>
  38. 38. Proof that the stress condition implies global rigidity The Tarski-Seidenberg theory of quantifier elimination implies that the situation below cannot happen, since the configuration p is generic. So a neighborhood of p can be mapped to a neighborhood of q by a diffeomorphism h, so that fh = f, and df p = df q dh p . This implies that any equilibrium stress for p is an equilibrium stress for q . If the rank of the associated stress matrix is maximal, this implies that p and q are affine images of each other, and ultimately that they are congruent.
  39. 39. Henneberg transformations <ul><li>Suppose that G( p ) is a rigid framework in the plane, with p generic, such that it has an equilibrium stress  whose stress matrix  has rank n-3. Then the following transformation adds one new vertex to G while the new framework has an equilibrium stress whose stress matrix has rank (n+1)-3. </li></ul>
  40. 40. Henneberg transformations <ul><li>Suppose that G( p ) is a rigid framework in the plane, with p generic, such that it has an equilibrium stress  whose stress matrix  has rank n-3. Then the following transformation adds one new vertex to G while the new framework has an equilibrium stress whose stress matrix has rank (n+1)-3. </li></ul>
  41. 41. Henneberg transformations <ul><li>Then a perturbation of the new vertex creates a generic configuration for this transformed graph, as below. </li></ul><ul><li>The rigidity matrix for all three frameworks has maximal rank, and so the stress space changes continuously, and the stress matrix remains maximal, and we get generic global rigidity for this generic configuration for this transformed graph. </li></ul>
  42. 42. What graphs can be obtained from Henneberg transformations? <ul><li>Conjecture : Any redundantly rigid, vertex 3-connected graph G can be obtained from a sequence of Henneberg operations and edge insertions, starting from K 4 , the complete graph on 4 vertices. </li></ul><ul><li>For example: </li></ul>
  43. 43. What graphs can be obtained from Henneberg transformations? <ul><li>Conjecture : Any redundantly rigid, vertex 3-connected graph G can be obtained from a sequence of Henneberg operations and edge insertions, starting from K 4 , the complete graph on 4 vertices. </li></ul><ul><li>For example: </li></ul>
  44. 44. What graphs can be obtained from Henneberg transformations? <ul><li>Conjecture : Any redundantly rigid, vertex 3-connected graph G can be obtained from a sequence of Henneberg operations and edge insertions, starting from K 4 , the complete graph on 4 vertices. </li></ul><ul><li>For example: </li></ul>
  45. 45. What graphs can be obtained from Henneberg transformations? <ul><li>Conjecture : Any redundantly rigid, vertex 3-connected graph G can be obtained from a sequence of Henneberg operations and edge insertions, starting from K 4 , the complete graph on 4 vertices. </li></ul><ul><li>For example: </li></ul>
  46. 46. What graphs can be obtained from Henneberg transformations? <ul><li>Conjecture : Any redundantly rigid, vertex 3-connected graph G can be obtained from a sequence of Henneberg operations and edge insertions, starting from K 4 , the complete graph on 4 vertices. </li></ul><ul><li>For example: </li></ul>
  47. 47. Hungarian and student to the rescue <ul><li>Theorem (A. Berg, T. Jordan, 2002): Let G be a 3-connected, generically redundantly rigid graph with 2n-2 edges, where n is the number of vertices of G. Then G can be obtained from K 4 by a sequence of Henneberg transformations. </li></ul>
  48. 48. Hungarian and colleague to the rescue <ul><li>Theorem (T. Jordan and W. Jackson, 2003): Let G be a 3-connected, generically redundantly rigid graph. Then G can be obtained from K 4 by a sequence of Henneberg transformations and edge insertions. </li></ul><ul><li>Corollary : Hendrickson’s conjecture is true in the plane, and there is a polynomial time algorithm to test for generic global rigidity. </li></ul>
  49. 49. An example of the Berg-Jordan Henneberg transformations
  50. 50. An example of the Berg-Jordan Henneberg transformations
  51. 51. An example of the Berg-Jordan Henneberg transformations
  52. 52. An example of the Berg-Jordan Henneberg transformations
  53. 53. An example of the Berg-Jordan Henneberg transformations
  54. 54. An example of the Berg-Jordan Henneberg transformations
  55. 55. An example of the Berg-Jordan Henneberg transformations
  56. 56. Whoops! <ul><li>Care must be taken as how to choose the operations of inverse Hennenberg operations. </li></ul><ul><li>The graph to the right is not vertex 2-connected. </li></ul>
  57. 57. Part IV: Existence of realizations When does a graph have a realization with predermined edge lengths?
  58. 58. The molecule problem <ul><li>Suppose that you given distance constraints on the edges lengths of a graph G, exact lengths, or upper and lower bounds. Determine a configuration that satisfies those constraints. </li></ul>
  59. 59. A modified problem <ul><li>Suppose that instead that you are given a realization of the graph G in some possibly high dimensional Euclidean space E N . For what graphs can you ALWAYS be sure that there is a realization p in E 3 with the same edge lengths? </li></ul>
  60. 60. A modified problem <ul><li>Suppose that instead that you are given a realization of the graph G in some possibly high dimensional Euclidean space E N . For what graphs can you ALWAYS be sure that there is a realization p in E 3 with the same edge lengths? </li></ul><ul><li>Answer: See Maria Sloughter. It is related to the idea of “backbones”. </li></ul>

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