Title: Rigidity and Tensegrity
1Rigidity and Tensegrity
- Robert Connelly
- Cornell University
2Part I The local theory
3What determines rigidity?
4What determines rigidity?
- The physics of the materials.
5What determines rigidity?
- The physics of the materials.
- The external forces on the structure.
6What determines rigidity?
- The physics of the materials.
- The external forces on the structure.
- The combinatorics/topology of the structure.
7What determines rigidity?
- The physics of the materials.
- The external forces on the structure.
- The combinatorics/topology of the structure.
- THE GEOMETRY OF THE STRUCTURE.
8What model?
9What model?
- My favorite is a tensegrity.
10What sort of rigidity/stablility?
11What sort of rigidity/stablility?
- A very basic choice is prestressability.
- This means that there is potential function Eij
defined for each member ( cable, bar, or strut)
i,j, and the configuration p (p1, , pn) is
at a unique local minimum, modulo congruences, of
the total potential - E(p)Sij Eij(pi - pj2)
- using the Hessian for the second-derivative test.
12What energy function to choose?
13What energy function to choose?
- It does not matter too much.
- It only matters what the first and second
derivatives are of Eij at pi-pj2. - (Eij )(pi-pj2) wij the (pre)stress
coefficient. - (Eij )(pi-pj2) cij gt 0, the stiffness
coefficient.
14The basic stress-stiffness decomposition
- Theorem The Hessian H of the energy potential
is the sum of the quadratic forms - H stress energy stiffness energy,
- and for the displaced pi pi for each i,
- stress energy Sij wij pi - pj2,
- stiffness energy Sij cij (pi - pj) (pi -
pj)2.
15The stiffness matrix
- The matrix of the quadratic form giving the
stiffness energy is - R(p)TDR(p),
- where R(p) is the rigidity matrix, D is the
diagonal matrix with cijs as diagonal entries,
and ()T is the transpose. This stiffness matrix
is always positive semi-definite.
16The stress matrix
- 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 W, where each
off-diagonal entry is -wij-wji, and the row and
column sums are 0.
17An example of a stress matrix
18Equilibrium stresses
- For a tensegrity framework G(p), an equilibrium
stress w is an assignment of a scalar wij wji to
each pair of distinct vertices i,j of G, such
that wij 0 when i,j is not an edge of G, and
for each i, the equilibrium equation - Sj wij(pj-pi)0
- holds.
- When a configuration is at a critical point (e.g.
a local minimum), then the corresponding stress
is in equilibrium. - A cable corresponds to wij gt 0, i.e. tension.
- A strut corresponds to wij lt 0, i.e. compression.
19Properties of the stress matrix
- If the affine span of the points of the
configuration p (p1, , pn) is d-dimensional,
then the rank of W is at most n-d-1. - If the rank of W is n-d-1, and some other
configuration q is such that w is an equilibrium
stress for G(q), then the points of q are an
affine image of the points of p.
20Part II The global theory
- Stress matrices applied again.
21Dominance
- A tensegrity G(q) dominates G(p) and we write
G(p) G(q) if - Each cable ij of G(p) is no longer than the
corresponding cable ij of G(q). - Each strut ij of G(p) is no shorter than the
corresponding strut ij of G(q). - Each bar ij of G(p) is the same length as the
corresponding bar ij of G(q).
22Global Rigidity
- A tensegrity G(p) in Ed is globally rigid, if
G(q) G(p) implies the configuration q in Ed is
congruent to the configuration p.
23How do you tell when a given framework is
globally rigid?
24How do you tell when a given framework is
globally rigid?
- Answer Its hard, even when there are only
bars!! - 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)
25Tools to show global rigidity
- Theorem Suppose that G(p) is a tensegrity in Ed
with an equilibrium stress such that - The stress matrix W is positive semi-definite of
rank n-d-1. - The only affine motions of the configuration p
that preserve the member constraints are
congruences. - Then G(p) is globally rigid in any EN containing
Ed.
26- This tool is sufficient for many of the popular
tensegrities that are built by artists and
others. - You can see several examples of symmetric
tensegrities at my web page - http//mathlab.cit.cornell.edu/visualization/tense
g/tenseg.html - See also Simon Guest for information about how to
use symmetry in rigidity calculations.
27Part III Bar frameworks
- More stress matrices using the generic philosophy.
28Generic configurations
- 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. - p, e, g, are algebraically independent.
29Generic rigidity
- Theorem A bar framework is rigid at a generic
configuration if and only if it is
infinitesimally rigid at any configuration. - This means that generic rigidity is a
combinatorial property of the graph G only. - 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.
30Algorithms
- 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.)
31Generic global rigidity
- A graph G is generically globally rigid in Ed if
for some (every?) generic configuration p, the
framework G(p) is globally rigid in Ed. - The following graphs are generically globally
rigid in the plane
32Generic global rigidity
- A graph G is generically globally rigid in Ed if
for some (every?) generic configuration p, the
framework G(p) is globally rigid in Ed. - The following graphs are not generically globally
rigid in the plane
33When is a graph G generically globally rigid in
Ed?
- Necessary conditions
- G must be vertex (d1)-connected. (This means
d1 or more vertices are needed to disconnect the
vertices of G.) - 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). - Conjecture (Hendrickson) For d2, these
conditions are also sufficient. - For d3, these conditions are not sufficient. (Me
1991)
34Vertex connectivity
- If G is not (d1) vertex connected, then d1
vertices separate G, and reflection of one of the
components of G about the (hyper)-line through
those d1 vertices violates global rigidity.
35Vertex connectivity
- If G is not (d1) vertex connected, then d1
vertices separate G, and reflection of one of the
components of G about the (hyper)-line through
those d1 vertices violates global rigidity.
36When is a graph G generically globally rigid in
Ed?
- Sufficient condition
- For p generic, G(p) has an equilibrium stress w,
and an associated stress matrix W with maximal
rank n-d-1, where n is the number of vertices of
G. (Me, to appear.) - 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 W. - Conjecture If G is generically globally rigid
in Ed, then either G is a simplex or it satisfies
the condition above for some generic
configuration p.
37The rigidity matrix
- For a graph G, the rigidity map f End -gt Ee is
the function that assigns to each configuration p
of n vertices in d-space, the squared lengths of
edges of G, f(p)(. . ., pi - pj2, . . .),
where e is the number of edges of G. - The rigidity matrix R(p) df is the differntial
of f. - Fact The co-kernel of R(p) is the space of
equilibrium stresses of G(p). I.e. w is an
equilibrium stress if and only if wR(p) 0.
38Proof 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 dfp dfq dhp. 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.
39Henneberg transformations
- Suppose that G(p) is a rigid framework in the
plane, with p generic, such that it has an
equilibrium stress w whose stress matrix W 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 (n1)-3.
40Henneberg transformations
- Suppose that G(p) is a rigid framework in the
plane, with p generic, such that it has an
equilibrium stress w whose stress matrix W 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 (n1)-3.
41Henneberg transformations
- Then a perturbation of the new vertex creates a
generic configuration for this transformed graph,
as below. - 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.
42What graphs can be obtained from Henneberg
transformations?
- Conjecture Any redundantly rigid, vertex
3-connected graph G can be obtained from a
sequence of Henneberg operations and edge
insertions, starting from K4, the complete graph
on 4 vertices. - For example
43What graphs can be obtained from Henneberg
transformations?
- Conjecture Any redundantly rigid, vertex
3-connected graph G can be obtained from a
sequence of Henneberg operations and edge
insertions, starting from K4, the complete graph
on 4 vertices. - For example
44What graphs can be obtained from Henneberg
transformations?
- Conjecture Any redundantly rigid, vertex
3-connected graph G can be obtained from a
sequence of Henneberg operations and edge
insertions, starting from K4, the complete graph
on 4 vertices. - For example
45What graphs can be obtained from Henneberg
transformations?
- Conjecture Any redundantly rigid, vertex
3-connected graph G can be obtained from a
sequence of Henneberg operations and edge
insertions, starting from K4, the complete graph
on 4 vertices. - For example
46What graphs can be obtained from Henneberg
transformations?
- Conjecture Any redundantly rigid, vertex
3-connected graph G can be obtained from a
sequence of Henneberg operations and edge
insertions, starting from K4, the complete graph
on 4 vertices. - For example
47Hungarian and student to the rescue
- 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 K4 by
a sequence of Henneberg transformations.
48Hungarian and colleague to the rescue
- Theorem (T. Jordan and W. Jackson, 2003) Let G
be a 3-connected, generically redundantly rigid
graph. Then G can be obtained from K4 by a
sequence of Henneberg transformations and edge
insertions. - Corollary Hendricksons conjecture is true in
the plane, and there is a polynomial time
algorithm to test for generic global rigidity.
49An example of the Berg-Jordan Henneberg
transformations
50An example of the Berg-Jordan Henneberg
transformations
51An example of the Berg-Jordan Henneberg
transformations
52An example of the Berg-Jordan Henneberg
transformations
53An example of the Berg-Jordan Henneberg
transformations
54An example of the Berg-Jordan Henneberg
transformations
55An example of the Berg-Jordan Henneberg
transformations
56Whoops!
- Care must be taken as how to choose the
operations of inverse Hennenberg operations. - The graph to the right is not vertex 2-connected.
57Part IV Existence of realizations
- When does a graph have a realization with
predermined edge lengths?
58The molecule problem
- 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.
59A modified problem
- Suppose that instead that you are given a
realization of the graph G in some possibly high
dimensional Euclidean space EN. For what graphs
can you ALWAYS be sure that there is a
realization p in E3 with the same edge lengths?
60A modified problem
- Suppose that instead that you are given a
realization of the graph G in some possibly high
dimensional Euclidean space EN. For what graphs
can you ALWAYS be sure that there is a
realization p in E3 with the same edge lengths? - Answer See Maria Sloughter. It is related to
the idea of backbones.