Volume 112, Advances in the Astronautical Sciences

Spaceflight Mechanics 2002

AAS 02-177

EQUILIBRIUM CONDITIONS OF A CLASS I TENSEGRITY

STRUCTURE

Robert E. Skelton 1 , Darrell Williamson 2 , JeongHeon Han 3

ABSTRACT

Static models of tensegrity structures are reduced to linear algebra problems, after firstcharacterizing the problem in a vector space where direction cosines are not needed. That is, wedescribe the components of all member vectors as opposed to the usual practice of characterizingthe statics problem in terms of the magnitude of tension vectors. While our approach enlarges(by a factor of 3) the vector space required to describe the problem, the computational space isnot increased. The advantage of enlarging the vector space makes the mathematical structureof the problem amenable to linear algebra treatment. Using the linear algebraic techniques,many variables are eliminated from the final existence equations. This paper characterizes theexistence conditions for all tensegrity equilibria.

Key Words: tensegrity, structure, statics, equilibrium conditions, linear algebra.

INTRODUCTION

The Tensegrity structures introduced by Kenneth Snelson pose a wonderful blend of geometry andmechanics. In addition, they have engineering appeal in problems requiring large changes is struc-tural shape. We define “class I” tensegrity as a stable connection of bars and strings with onlyone bar connected to any given node. Nodes connecting more than one bar or other constraintsform “class II” tensegrity. Thus, class I structures are unconstrained, whereas, class II may haveboundary constraints or bar to bar connections. Most existing smart structure methods are limitedto small displacements. Since class I tensegrity structures [8] have no bar-to-bar connections, thecontrol of tendons allows very large shape changes. Therefore, an efficient set of analytical toolscould be the enabler to a hoist of new engineering concepts for deployable and shape controllablestructures.

This paper characterizes the static equilibria of class I tensegrity structures. Furthermore, we usevectors to describe each element( bars and tendons), eliminating the need to use direction cosinesand the subsequent transcendental functions that follow their use.

It is well known in a variety of mathematical problems that enlarging the domain in which theproblem is posed can often simplify the mathematical treatment. For example, nonlinear Riccatiequations are known to be solvable by linear algebra in a space that is twice the size of the originalproblem statement. Many nonlinear problems admit solutions by linear techniques by enlarging thedomain of the problem. The purpose of this paper is to show that by enlarging the vector space in

1 University of California, San Diego, Dept. of Mechanical and Aerospace Engineering, La Jolla, CA, USA. email:

bobskelton@ucsd.edu2 University of Wollongong, Faculty of Informatics, Wollongong, NSW 2522 Australia email: dar-

rell williamson@uow.edu.au3 University of California, San Diego, Dept. of Mechanical and Aerospace Engineering, La Jolla, CA, USA. email:

jehan@mae.ucsd.edu

1

which we characterize the tensegrity statics problem, the mathematical structure of the equationsadmit treatment by linear algebra methods, for the most part.

Our results characterize the equilibria conditions of both Class I tensegrity structures in terms ofa very small number of variables, since the necessary and sufficient conditions of the linear algebratreatment allows the elimination of several of the original variables. In the future, these resultswill be programmed into object-oriented software to design and simulate a large class of tensegritystructures. This paper provides the enabling technology for efficient algorithms to design tensegritystructures, which have been around for fifty years without efficient design procedures.

The paper is laid out as follows. Section 2 introduces the network representations of tensegritystructures as an oriented graph in real three dimensional space. Geometric connectivity, equilibrium,and a coordinate transformation will be introduced. Section 3 introduces the algebraic equilibriumconditions for a class I tensegrity structure. After we derive necessary and sufficient conditions forthe existence of an unloaded tensegrity structure in equilibrium, we write the necessary and sufficientconditions for the externally loaded structure in equilibrium. A couple of examples will show howto construct a tensegrity structure that concludes the paper.

Notation

We let In define the n×n identity matrix, and 0 define an n×m matrix of zeros. (The dimensions of0 will be clear from the context.) We also let ρ(A) define the rank of the matrix A. Let A ∈ Rm×nand B ∈ Rp×q, then the Kronecker product of A and B is defined as

A⊗

B =

a11B a12B · · · a1nBa21B a22B · · · a2nB

......

...am1B am2B · · · amnB

∈ Rmp×nq

NETWORK REPRESENTATION OF STRUCTURES

In this paper, we choose to represent a tensegrity structure as an oriented graph in real threedimensional space R3 defined in terms of np nodes and ns + nb directed branches which are allrepresented as vectors in R3. A loop is any closed path in the graph. As we shall see, the advantageof this approach is that both the magnitude and the direction of the forces are contained in vectorswhich can be solved using linear algebra. Thus linear algebra plays a larger role in this approachcompared to the usual approach in mechanics and finite element methods using direction cosines.

In the oriented graph of a class I tensegrity structure, the nodes consist of the ends of the bars asrepresented by the np nodes (or vectors) {pk}. Hence if there are nb bars, then there are np = 2nbnodes. We choose to identify two different types of directed branches; the ns string branches (orvectors) {sn} and the nb bar branches (or vectors) {bm}.

Geometric ConnectivityEach directed branch can undergo a displacement in reaching its equilibrium state. String vectors

can change both their length and orientation while rigid bar vectors can only change their orientation.Node vectors can change both their length and orientation but subject to a Law of GeometricConnectivity which we state as follows:

The vector sum of all branch vectors in any loop is zero. (1)

These equations are in the form of a set of linear algebraic equations in the branch vectors.

Force EquilibriumIn our study of tensegrity structures, we are concerned with structures in which bars sustain

compressive forces. We therefore choose to distinguish between the string (or tensile) forces {tn}

2

and the bar (or compressive) forces {fm} which are defined in terms of the string and bar vectorsrespectively as follows.

Definition 1 Given the tensile force tn in the string characterized by the string vector sn and thecompressive force fn in the bar characterized by the bar vector bn, the tensile force coefficient γm ≥ 0and the compressive force coefficient λn ≥ 0 are defined by

tn = γnsn ; fm = λmbm (2)

The forces of the tensegrity structure are defined by the external force vector w ∈ R3np , thecompression vector f ∈ R3nb , and the tension vector t ∈ R3ns where

wT = [wT1 ,wT2 , ... ,w

Tnp

] ; fT = [fT1 , fT2 , ... , f

Tnb

] ; tT = [tT1 , tT2 , ... , t

Tns

] (3)

Force Convention:Suppose each node pk is subjected to compressive vector forces {fmk}, tensile vector forces {tnk}

and external force wk. Then the Law for Static Equilibrium may be stated as follows:

∑

n

tnk −∑

m

fmk − wk = 0 (4)

where a positive sign is assigned to a (tensile, compressive, external) force leaving a node, and anegative sign is assigned to a (tensile, compressive, external) force entering a node. The negative signin (4) is a consequence of the fact that we choose to define nonnegative compressive force coefficientsλn.

Consider a class I tensegrity structure consisting of np nodes, nb bars and ns strings. Supposethe positions of the nodes are described by the np vectors {p1,p2, ... ,pnp}, the positions of thebars are described by the nb vectors {b1,b2, ... ,bnb}, and the positions of the strings are describedby the ns vectors {s1, s2, ... , sns}.

Definition 2 The geometry of the tensegrity structure is defined by the tensegrity node vector p ∈R3np , the tensegrity bar vector b ∈ R3nb , and the tensegrity string vector s ∈ R3ns where

pT = [pT1 ,pT2 , ... ,p

Tnp

] ; bT = [bT1 ,bT2 , ... ,b

Tnb

] ; sT = [sT1 , sT2 , ... , s

Tns

] (5)

From the network, it follows that components of the string vector s and the bar vector b can bewritten as a linear combination of components of the node vector p. Also if branch k is a bar whichleaves node i and enters node j, then bk = pj − pi, whereas if branch k is a string which leavesnode i and enters node j, then sk = pj − pi. Hence we have ATp = [sT ,bT ]T where the matrix Aconsists only of block matrices of the form {0,±I3}.

In particular, if we consecutively number the ns + nb branches {s1, s2, ... , sns , b1,b2, ... ,bnb}as {1, 2, .... , ns, ns + 1, ... , ns + nb}, then the 3np × (3ns + 3nb) matrix A = [Aij ] is defined by

Aij =

I3 if tension (compression) j leaves (enters) node i−I3 if tension (compression) j enters (leaves) node i

0 if tension (compression) j is not incident with node i(6)

Also: (i) each column of A has exactly one block I3 and one block −I3 with all other columnblocks 0, and (ii) for any row i there exists a column j such that Aij = ±I3. Specifically, the ”barconnectivity” matrix B and the ”string connectivity” matrix S form the matrix A as follows.

[

sb

]

=

[

ST

BT

]

p = ATp (7)

3

FIG. 1. 2-Bar 4-String Class I Tensegrity

In network analysis, the matrix A is known as the incidence matrix. This matrix is not thereduced incidence matrix since we have included the datum node which means that one block rowof equations in (8) are dependent on the other rows. This fact does not cause any difficulties insubsequent developments. On the contrary, some symmetry is preserved in the algebraic equations.From Network analysis and the law for static equilibrium, we have the following result.

Lemma 1 Consider a tensegrity structure as described by the geometric conditions given by (7).Then the equilibrium conditions for a class I tensegrity structure under the external load w are

A

[

t−f

]

= w ; A∆= [S,B] (8)

or equivalently

St = Bf + w. (9)

Example 1 Consider the 2-bar 4-string class I tensegrity structure illustrated in Figure 1 with ten-sile force vectors {t1, t2, t3, t4} and compressive force vectors {f1, f2}. The Geometric Connectivityconditions are:

−p1 + p3 = s1 ; p2 − p3 = s2−p2 + p4 = s3 ; p1 − p4 = s4 (10)

Also, in terms of the stated force convention (4), the conditions for Static Equilibrium at nodes 1-4are:

t1 − t4 − f1 − w1 = 0 ; −t2 + t3 + f1 − w2 = 0−t1 + t2 − f2 − w3 = 0 ; −t3 + t4 + f2 − w4 = 0 (11)

4

The static equilibrium conditions and the geometric conditions can be written in the form (6),(7), (8), where

S =

−I3 0 0 I30 I3 −I3 0I3 −I3 0 00 0 I3 −I3

; B =

−I3 0I3 00 −I30 I3

(12)

We now derive a canonical network representation in which the bar vectors are a subset of thetransformed network nodes.

Lemma 2 Given the tensegrity node vector p, define the coordinate transformation

p = Pq (13)

for some nonsingular matrix P ∈ R3np×3np where

P = [PT1 ,PT2 ] ; P1,P2 ∈ R3nb×3np (14)

Then:

(i) In terms of the transformed tensegrity node q, the tensegrity geometry is given by

STq q = s ; BTq q = b (15)

where

STq = [ST1 ,S

T2 ] ; B

Tq = [B

T1 ,B

T2 ] (16)

with

S1 = P1S ; S2 = P2S ; B1 = P1B ; B2 = P2B (17)

(ii) The tensegrity force equilibrium is given by

S1t = B1f + P1w ; S2t = B2f + P2w (18)

Proof : Part (i) follows directly from the definition of P. Part (ii) follows from the expansionof the equilibrium condition Sqt = Bqf + P

T w.

Since each bar has two end points, np = 2nb. Without any loss of generality, we label the nodesof the bar bm to be p2m and p2m−1 for class I tensegrity structures, hence

bm = −p2m−1 + p2m , m = 1, 2, ... , nb.BT = blockdiag{

[

−I3 I3]

, · · · ,[

−I3 I3]

} (19)In particular, the following Lemma gives three special choices for coordinate transformations

based on the special bar connectivity matrix given by (19).

Lemma 3 The transformed equilibrium conditions (18) for all three coordinate transformations aregiven by

Sqt − Bqf = PT w ; STq = [ST1 ,ST2 ] ; BTq = [Inb ,0nb ] (20)where S1 ∈ R3nb×3ns ; S2 ∈ R3nb×3ns are given by (17).

S1t = f + P1w ; S2t = P2w (21)

5

The transformed coordinate q is given by

qT = [bT ,qTe ] (22)

where b is the bar vector and qe is given in (26), (29), and (32).

(i) Let the coordinate transformation matrix P ∈ R3np×3np in (14) be given by

P =[

PT1 PT2

]

=1

2

[

B (Ie − Io)B]

(23)

where B is given by (19) and odd and even node selection matrix Io, Ie ∈ R3np×3np are definedby

Io = blockdiag{I3,03, · · · , I3,03} ; Ie = blockdiag{03, I3, · · · ,03, I3}. (24)

Then the inverse transformation is

P−1 = [PT1 ,PT2 ]

−1 = 2

[

P1P2

]

=

[

BT

BT (Ie − Io)

]

(25)

qe ∈ R3nb is the vector of the mass center of each of the bars given by

qTe = [pTc1

,pTc2 , ... ,pTcnb

] (26)

where pcj4= 12 (p2j + p2j−1).

(ii) Let the coordinate matrix P ∈ R3np×3np in (14) be given by

P =[

IoB (Ie − Io)B]

(27)

where B is given by (19) and odd and even node selection matrix Io, Ie ∈ R3np×3np are givenby (24). Then the inverse transformation is

P−1 = [PT1 ,PT2 ]

−1 =

[

BT

BT Ie

]

(28)

qe ∈ R3nb is a vector of the even nodes given by

qTe = [pT2 ,p

T4 , ... ,p

T2nb

] (29)

(iii) Let the coordinate matrix P ∈ R3np×3np in (14) be given by

P =[

IeB (Io − Ie)B]

(30)

where B is given by (19) and odd and even node selection matrix Io, Ie ∈ R3np×3np are givenby (24). Then the inverse transformation is

P−1 = [PT1 ,PT2 ]

−1 =

[

BT

BT Io

]

(31)

qe ∈ R3nb is a vector of the odd nodes given by

qTe = [pT1 ,p

T3 , ... ,p

T2nb−1

] (32)

Proof : Proof of PP−1 = P−1P = I follows directly from identities

BT IeB = BT IoB = I ; IeIo = 0 ; I

2

e= Ie ; I

2

o= Io.

6

Example 2 The equations (10), (11) of the 2-bar 4-string tensegrity introduced in Example 1 canbe written in the form (8) and (7) where

BT =

[

−I3 I3 0 00 0 −I3 I3

]

; ST =

−I3 0 I3 00 I3 −I3 00 −I3 0 I3I3 0 0 −I3

Then from part (i) in Lemma 3, we have

P1 =

[

−I3 0 0 00 0 −I3 0

]

; P2 =

[

I3 I3 0 00 0 I3 I3

]

(33)

S1 =

[

I3 0 0 −I3−I3 I3 0 0

]

; S2 =

[

−I3 I3 −I3 I3I3 −I3 I3 −I3

]

Or from part (ii) in Lemma 3, we have

P1 =1

2

[

−I3 I3 0 00 0 −I3 I3

]

; P2 =1

2

[

I3 I3 0 00 0 I3 I3

]

S1 =1

2

[

I3 I3 −I3 −I3−I3 I3 I3 −I3

]

; S2 =1

2

[

−I3 I3 −I3 I3I3 −I3 I3 −I3

]

ANALYSIS OF THE TRANSFORMED EQUILIBRIUM CONDITIONS FOR A CLASS I

TENSEGRITY STRUCTURE

Definition 3 A tensegrity structure with tensile force coefficients {γn ≥ 0}, compressive force coef-ficients {λn ≥ 0}, node vector p, string vector s and bar vector b is said to be in equilibrium if theelement relationships (2), the force equations (9) and the geometric equations (7) are all satisfied.

For the remainder of this paper, we choose to use the coordinate transformation derived in part (i)Lemma 3.

Requirements for Equilibrium: Given an external force vector w, the problem of determiningthe geometric and force configuration of a tensegrity structure consisting of ns strings and nb barsin equilibrium is therefore equivalent to finding a solution b,qe ∈ R3nb of the equations:

s = ST1 b + ST2 qe (34)

t = Γs ; Γ∆= diag{γ1I3, γ2I3, ... , γnsI3} (35)

S2t = P2w (36)

f = S1t− P1w (37)f = Λb ; Λ

∆= diag{λ1I3, λ2I3, ... , λnbI3} (38)

for given matrices

S1,S2 ∈ R3nb×3ns ; P1,P2 ∈ R3nb×3np (39)

where

np = 2nb ; ns > nb (40)

7

Beyond equilibrium requirements, one might require shape constraints by requiring p to take on aspecific set of values p = p̄, where p = Pq = PT1 b+P

T2 qe. Our focus in this paper is characterizing

equilibria, hence the freedom in choosing the nodal vector p will appear as free variables in thevector qe, as the sequel shows. The diagonal matrices {Γ,Λ} shall be referred to as the tensile forcematrix and compressive force matrix respectively.

Prestressed Equilibrium( Unloaded Structure )

We now proceed to derive necessary and sufficient conditions for the existence of a structure inequilibrium that is prestressed in the absence of any external load (i.e. w = 0) in (34) - (38). Ourstrategy for the examination of the conditions (34) - (38) is as follows. The solution of the linearalgebra problem (36) yields nonunique t which lies in the right null space of S2. The existencecondition of qe for linear algebra problem (34) yields a condition on the left null space of S2. Hence(34) and (36) can be combined to obtain a unique expression for t in terms of b. This is key to themain results of this paper.We now establish necessary and sufficient conditions for a solution of equations (34)-(40) in theabsence of external forces (i.e. w = 0) by examining each of these equations in turn beginning withthe solution of (34), (36).

Lemma 4 Suppose

r∆= ρ(S2) ≤ min{3np, 3ns} (41)

and let S2 have the singular value decomposition {U,Σ,V} given by

S2 = UΣVT ∈ R3nb×3ns (42)

where

U = [U1,U2] ; Σ =

[

Σ11 00 0

]

; V = [V1,V2]

U1 ∈ R3nb×r ; U2 ∈ R3nb×(3nb−r) ; V1 ∈ R3ns×r ; V2 ∈ R3ns×(3ns−r)

Then a necessary and sufficient condition for (34) to have a solution qe ∈ R3nb is given by

VT2 (s− ST1 b) = 0 (43)Furthermore, when (43) is satisfied, all solutions qe are of the form

qe = U1Σ−111 V

T1 (s − ST1 b) + U2ze (44)

where ze ∈ R3nb−r is arbitrary.

We now consider the solution of (36) when w = 0.

Lemma 5 When w = 0, all solutions t of (36) which guarantee (43) are of the form

t = V2M−1VT2 S

T1 b ; M

∆= VT2 Γ

−1V2 (45)

Proof : From (42), (43) , we have t = V2zt where zt is the free solution of (36). Then from (35)

VT2 (s− ST1 b) = VT2 (Γ−1V2zt − ST1 b)

8

Since V2 has full column rank, the matrix M = VT2 Γ

−1V2 is invertible if it exists (that is, ifγn 6= 0, n = 1, · · · , ns). Hence (43) is satisfied when zt = M−1VT2 ST1 b, and this gives (45).

We now consider the solution of (37), (38) when w = 0.

Lemma 6 When w = 0, a necessary and sufficient condition for (37), (38) to have a solutionb ∈ R3nb is given by

(X − Λ)b = 0 (46)where

X∆= (S1V2)M

−1(S1V2)T ; M

∆= VT2 Γ

−1V2 (47)

In particular, define

rb∆= ρ(X − Λ) (48)

Then:

(i) When rb = 3nb, then b = 0 is the only solution of (46),(ii) When rb = 0, any b ∈ R3nb is a solution of (46), and(iii) When 0 < rb < 3nb, all solutions b satisfy the equation

[I− VX1Σ−1X11UTX1Λ]b = VX2zb (49)where zb ∈ R3nb−rb is free, and where {UX ,ΣX ,VX} is the singular value decomposition of thematrix X ∈ R3nb×3nb ; that is

UX = [UX1,UX2] ; ΣX =

[

ΣX11 00 0

]

; VX = [VX1,VX2] ; rX = rank(X)

with UX1 ∈ R3nb×rX , VX2 ∈ R3nb×(3nb−rX ).

Proof : From (45), (38) and (37)

S1t − f = S1V2M−1VT2 ST1 b− Λb = (X − Λ)bThe result then follows from the singular value decomposition of X after writing (46) in the formXb = f , f = Λb.

For the establishment of a non-trivial tensegrity structure, it is necessary that the node com-ponent vectors pk are not required to lie on any line for a planar tensegrity structure and it isnecessary that the node component vectors pk are not required to lie on any plane for a spatialtensegrity structure. Hence we establish the following Corollary.

Corollary 1 When w = 0, a necessary and sufficient condition for (37), (38) to have a nontrivialsolution b ∈ R3nb is given by

(X − Λ)b = 0 rb ∆= ρ(X − Λ) ≤ (nb − 2)d

where dimension d is 2 for planar tensegrity structure, and 3 for spatial tensegrity structure.

Proof : When rb = nb × d, b = 0 is the only solution. When rb = (nb − 1) × d, all bar vectorsare parallel.

Corollary 2 If Γ = γI, and Λ = λI, then all unforced tensegrity equilibria are characterized by themodal data of the matrix X̄ = S1V2V

T2 S

T1 . That is, all admissible values of

λγ

are eigenvalues of

X̄, and all bar vectors are eigenvectors of X̄.

9

Proof : From (46) and Γ = γI,Λ = λI, X = γS1V2VT2 S

T1 = γX̄. Then

(

γX̄ − λI)

b = 0 or(

X̄ − ρI)

b = 0, where ρ = λγ. Hence ρ and b are the eigenvalues and eigenvectors of X̄.

Lemma 7 Consider the partitioned matrix

A =[

A BC D

]

.

Then

(i) detA = det[

A BC D

]

= det[A]det[D − CA−1B] if det[A] 6= 0.

(ii) detA = det[

A BC D

]

= det[D]det[A − BD−1C] if det[D] 6= 0.

Now Γ positive definite implies M in (47) is positive definite. Then M positive definite impliesX in (47) is positive semi-definite with at least one non-zero eigenvalue. Since (46) has a nonzerosolution b if and only if det(X − Λ) = 0, the following Lemma is important.

Lemma 8 Given a nonzero symmetric positive semi-definite matrix X and a positive definite diag-onal matrix Λ, the following statements are equivalent.

(i) det(X − Λ) = 0, where X = S1V2(

VT2 Γ−1V2

)−1VT2 S

T1 .

(ii) det

{[

Λ S1V2VT2 S

T1 V

T2 Γ

−1V2

]}

= 0.

(iii) det

{[

I 00 VT2

] [

Λ S1ST1 Γ

−1

] [

I 00 V2

]}

= 0.

(iv) det{

VT2(

Γ−1 − ST1 Λ−1S1)

V2}

= 0.

Proof : Since VT2 Γ−1V2 is a positive definite matrix, the solutions of det {X − Λ} = 0 are not

affected by multiplying by det {Λ}. This yields

det{

VT2 Γ−1V2

}

det {X − Λ} = 0.

Now we can apply (ii) in Lemma 7 by setting A4= Λ, B

4= S1V2, C = V

T2 S

T1 , and D

4= VT2 Γ

−1V2,which proves part (ii). Part (iii) can be easily proved by the equality given by

A 4=[

Λ S1V2VT2 S

T1 V

T2 Γ

−1V2

]

=

[

I 00 VT2

] [

Λ S1ST1 Γ

−1

] [

I 00 V2

]

.

Since Λ is a positive definite matrix, applying (i) in Lemma 7 to A yields (iv).

Remark 1 For 0 < rb < 3nb, (X − Λ)b = 0 always has at least one solution b 6= 0 of the form(49) where the freedom in the choice of b is available in the choice of a free vector zb.

Theorem 1 Consider a class I tensegrity structure as defined by the geometry and force equationsin the absence of external load as described by the geometric conditions

BTp = b ; STp = s ; p ∈ R3np , b ∈ R3nb , s ∈ R3ns

with np = 2nb, and the equilibrium conditions

St = Bf ; t = Γs, f = Λb

10

where for {γm ≥ 0, λn ≥ 0},

Γ = diag{γ1I3, γ2I3, ... , γnsI3} ; Λ = diag{λ1I3, λ2I3, ... , λnbI3}.

Then given any tensile force coefficients {γm ≥ 0, 1 ≤ m ≤ ns}, there exits compressive forcecoefficients {λn ≥ 0, 1 ≤ m ≤ nb} and b, which define an equilibrium structure, satisfy the condition:

(X − Λ)b = 0 ; X ∆= (S1V2)M−1(S1V2)T, M ∆= VT2 Γ−1V2, (50)

where S1 is given by (17), and V2 is given by (43).

Moreover, for any b and Γ satisfying (50), the nodal vector p is of the form

p = PQ[bT , zTe ]T (51)

where ze may be arbitrarily chosen and

Q∆=

[

I3nb 0L U2

]

; L∆= U1Σ

−111 V

T1 [Γ

−1V2M−1VT2 − I3ns ]ST1 . (52)

The corresponding tensegrity tension vector t, string vector s, and compression vector f are givenin terms of the bar vector b by

t = V2M−1VT2 S

T1 b ; s = Γ

−1t ; f = Λb (53)

Externally Loaded Structures

Under the action of an external force vector w with component vectors {wj} given by

wT = [wT1 ,wT2 , ... ,w

Tnp

] (54)

suppose that the new equilibrium structure is assumed to be given by node vector p, bar vector b,string vector s, compressive force vector f , tensile vector t, compressive force coefficient matrix Λand tensile force matrix Γ as described by (34)-(38). Note that all force coefficients together withall node geometry will normally change. We now seek necessary and sufficient conditions for theexternally loaded structure to be in equilibrium. An extension of Lemma 5 and Lemma 6 gives usthe following result.

Theorem 2 (i) All solutions t of (36) which guarantee (43) are of the form

t = V2M−1VT2 S

T1 b + D1w ; M

∆= VT2 Γ

−1V2

D1 = (I3ns − V2M−1VT2 Γ−1)V1Σ−111 UT1 P2 (55)

(ii) A necessary and sufficient condition for (37), (38) to have a solution b ∈ R3nb is given by

(X − Λ)b = DwUT2 P2w = 0 (56)

where X is given by (47), and

D∆= P1 − S1D1 (57)

11

Proof : Since (34) is not directly dependent on w, Lemma 4 applies for w 6= 0. Now consider thesolution of (36) for w 6= 0. A necessary condition for the existence of a solution t is UT2 P2w = 0,and in this case, all solutions t are of the form

t = V2zt + V1Σ−111 U

T1 P2w

for any zt ∈ R3nb−r where as in (41), r is the rank of S2. Now in order that condition (43) issatisfied, zt must be selected such that

VT2 Γ−1(V2zt + V1Σ

−111 U

T1 P2w) − VT2 ST1 b = 0

That is

zt = M−1VT2 S

T1 b − M−1VT2 Γ−1V1Σ−111 UT1 P2w

which gives (55). From (55), (38) and (36)

0 = S1t − P1w − f = S1V2M−1VT2 ST1 b + S1D1w − P1w − Λb

which gives (56).

The first condition in (56) is a nonhomogeneous equivalent of condition (46). However it is

unlikely (although not impossible) that Λ = X for w 6= 0. Instead, is more likely that rb ∆= ρ(X−Λ)satisfies 0 < rb ≤ 3nb. If rb = 3nb, then b = (X − Λ)−1Dw is unique.

When 0 < rb < 3nb, suppose {Ub,Σb,Vb} is the singular value decomposition of the matrixX − Λ ∈ R3nb×3nb ; that is

Ub = [Ub1,Ub2] ; Σb =

[

Σb11 00 0

]

; Vb = [Vb1,Vb2]

with Ub1 ∈ R3nb×rb , Vb2 ∈ R3nb×(3nb−rb). Then when UT2 Dw = 0, the solution b is of the form

b = Vb2zb + Vb1Σ−1b11U

Tb1Dw ; U

T2 Dw = 0

where zb ∈ R3nb−rb is arbitrary. The particular equilibrium obtained will depend on the way inwhich the external load w is introduced. The structural implications of the null space conditionUT2 Dw = 0 on the external load w would then also require a physical interpretation.

The existence of an equilibrium solution however requires the second condition in (56) on theexternal force w to be satisfied. In this regard, we have the following result.

Lemma 9 For all structures {S,B}, the (3nb − r) × 3ns matrix product UT2 P2 is of the form

UT2 = E[I3, I3, ... , I3] (58)

for some nonsingular matrix E. Hence UT2 P2w = 0 if and only if

np∑

k=1

wk = 0 (59)

Proof : It follows from svd(S2) in Lemma 4 that UT2 S2 = 0 which from (17) implies U

T2 P2S = 0.

Now from Lemma 1, each column of S has exactly one block I3 and exactly one block −I3 with allother column elements 0. Furthermore, for every ith row of S, there exists a column j such that theijth component of S is ±I3. These properties of S then imply that UT2 P2 is of the form (58) forsome (but not necessarily nonsingular) matrix E, and so

12

UT2 P2w = E

np∑

k=1

wk

Now if the full row rank matrix U2 is partitioned into the block form UT2 = [Z1,Z2, ... ,Znb ] it

follows from (23) that

UT2 P2 = [Z1,Z1,Z2,Z2 ... ,Znb ,Znb ]

which then guarantees that UT2 P2 also has full row rank, and consequently that the matrix E isinvertible.

Condition (59) expresses the requirement that for an externally loaded tensegrity structure tobe in equilibrium, it is necessary (but not sufficient) that the sum of the external forces is zero.

COMPUTATIONAL ALGORITHM FOR EQUILIBRIA

One suggested procedure for construction of a class I tensegrity structure in equilibrium is providedas follows.

Design Algorithm A :

step 1. Given the connectivity matrices S and B from the network topology, find a nonsingularmatrix P = [PT1 ,P

T2 ] such that B

Tq = B

TP = [Inb ,0nb ], and calculate {S1 = P1S, S2 =P2S}. (Note that if B is defined as in (19), then {P1,P2} are given by (23) ,(27), or (30).

step 2. Choose {γm > 0} and {λn > 0} such that det (X− Λ) = 0 .step 3. Select suitable bz and compute b = (X− Λ)⊥ bz based on a desired bar vector.step 4. Choose the free vector ze in (51), (52) to give a suitable node vector p based on a desired

shape.step 5. calculate {t, s, f} from (53).

Design Algorithm for Given Bar Vector b

In the design of tensegrity structures, an alternative approach for the selection of the node vectorp and bar vector b may be useful. From (50), it is easy to eliminate the variable Λ if b is known,since

Λb = b̃λ

where b̃4= blockdiag [b1, · · · ,bnb ] ; λ

4= [λ1, · · · , λnb ]

T. Hence, we can establish the following

Lemma.

Lemma 10 For any tensegrity equilibrium, λ is given uniquely by

λ = b̃+Xb, (60)

where b̃+ =(

b̃T b̃)−1

b̃T and b and Γ for all tensegrity structures must satisfy

b̃LXb = 0. (61)

where b̃L is the left null space basis of b̃ and satisfies

b̃Lb̃ = 0 ; b̃Lb̃T

L> 0.

13

Proof : From (50), note that

Xb = Λb = b̃λ

has a solution λ if and only if (61) holds. In this case, the unique solution is given by (60) assumingb̃ has full column rank. This assumption is guaranteed for bi 6= 0, i = 1, · · · , nb.

Note that (61) is equivalent to

(

I− b̃b̃+)

Xb = 0 (62)

since(

I − b̃b̃+)

forms the left null space of b̃.

Design Algorithm B :

step 1. Same as in Design Algorithm A.step 2. Choose b and then compute Γ to satisfy (61).step 3. Compute λ from (60).step 4. Choose the free vector ze in (51), (52) to give a suitable node vector p based on a desired

shape.step 5. calculate {t, s, f} from (53).

Design Algorithm for Given Nodal Vector p

One can first select the node vector p as in (51), (52) subject (if necessary) to other design constraints.The corresponding bar vector b is then determined from the first 3nb components of the vector P

−1p.When the node position vector p is given, we can establish following lemma.

Lemma 11 A given node position vector p corresponds to a class I tensegrity equilibrium if andonly if there exists a nonzero vector z such that

[

γ

λ

]

=[

Ss̃ −Bb̃]⊥

z > 0, (63)

where

b̃4= blockdiag [b1, · · · ,bnb ] ; s̃

4= blockdiag [s1, · · · , sns ] .

Proof : Using f = Λb = b̃λ and t = Γs = s̃γ leads to the equilibrium condition

St− Bf =[

Ss̃ −Bb̃]

[

γ

λ

]

= 0. (64)

The solutions of (64) for positive γ and λ yields (63).When we premultiply (64) by the coordinate transformation matrix PT given by (13), we can

further simplify (64). Hence we establish the following Corollary.

Corollary 3 A given node position vector p corresponds to a class I tensegrity equilibrium if andonly if there exists a nonzero vector z such that

[

γ

λ

]

=

[

I

b̃+S1s̃

]

(S2s̃)⊥

[

b̃LS1s̃ (S2s̃)⊥

]⊥

z > 0. (65)

Proof :

14

Premultiplying by the coordinate transformation matrix PT given by (23) to (64) and using (20)yields

PT (St− Bf) = Sqt − Bqf =[

Sq s̃ −Bqb̃]

[

γ

λ

]

=

[

S1s̃ −b̃S2s̃ 0

] [

γ

λ

]

= 0.

Hence the existence condition for γ and λ require

(S2s̃)⊥ 6= 0 ; b̃LS1s̃γ = 0. (66)

Then, all admissible force coefficients are

γ = (S2s̃)⊥

zr (67)

λ = b̃+S1s̃γ = b̃+S1s̃ (S2s̃)

⊥zr.

Substituting γ in (67) into the existence condition (66) yields

b̃LS1s̃ (S2s̃)⊥

zr = 0.

Hence zr is zr =[

b̃LS1s̃ (S2s̃)⊥

]⊥

z, which completes the proof.

Design Algorithm P :

step 1. Same as in Design Algorithm A.step 2. Choose nodal vector p.step 3. Choose the free vector z in (63) and Compute λ and γ.step 4. calculate {t, s, f} from (53).

ILLUSTRATED EXAMPLES

We now illustrate the construction procedure for a simple tensegrity structure.

2-bar 4-string planar tensegrity structure

A general force configuration for the class I tensegrity structure in Example 1 and 2 will be inves-tigated in this section. Suppose the force coefficient matrices are given by Λ = diag {λ1, λ2} andΓ = diag {γ1, γ2, γ3, γ4} . We will demonstrate two procedures using Design Algorithms A and B.

Design Algorithm A :

step 1. The connectivity matrices S, B and the coordinate transformation P = [PT1 ,PT2 ] are

given in Example 1 and 2. Since V2 spans the null space of S2, we compute

V2 =

1 −1 11 0 00 1 00 0 1

⊗

I2.

step 2. Choose {γm > 0} and {λn > 0} such that det (X− Λ) = 0 , where

X = S1V2(

VT2 Γ−1V2

)−1VT2 S

T1

=1

γ1 + γ2 + γ3 + γ4

[

(γ2 + γ3) (γ4 + γ1) (−γ3γ1 + γ2γ4)(−γ3γ1 + γ2γ4) (γ3 + γ4) (γ2 + γ1)

]

⊗

I2.

15

In all choices for {λn} that led to the rank of (X−Λ) having rank 1, the 4× 2 matrixVX2 is of the form V

TX2 = [A

T1 ,±AT1 ]; that is, in (49), the two bar vectors {b1,b2} are

always parallel, so the equilibrium structure is one dimensional with {p1 = p3,p2 = p4}.Hence for a two-dimensional structure, X − Λ must have rank zero. This requires

γ4 =γ3γ1

γ2

λ1 =(γ2 + γ3) (γ4 + γ1)

γ2 + γ3 + γ4 + γ1

λ2 =(γ3 + γ4) (γ2 + γ1)

γ2 + γ3 + γ4 + γ1, (68)

where γ1, γ2, and γ3 are free positive constants. If we choose {γk = 1; k = 1, 2, 3}, thenΓ = I4. It follows that X = I4 and Λ = I2 satisfy condition (50) in Theorem 1.

step 3. Select suitable bz and compute b = (X− Λ)⊥ bz based on a desired bar vector.Since we choose γ and λ such that X−Λ = 0, the bar vector is arbitrary. Let’s choose

b1 = [2, 0]T ,b2 = [0, 2]

T .step 4. Choose the free vector ze in (51), (52) to give a suitable node vector p based on a desired

shape.The nodes {p1 = [−1, 0]T ,p2 = [1, 0]T ,p3 = [0,−1]T ,p4 = [0, 1]T} define an equilib-

rium solution from (51) setting ze is zero. When ze = [1, 1]T , we get

p =[

−0.6464 0.3536 1.3536 0.3536 0.3536 −0.6464 0.3536 1.3536]T

.

This choices of ze only translate the geometric center of the structure from [0, 0]T to

0.3536[1, 1]T , since the force coefficients and b have been specified.

Design Algorithm B :

step 1. Connectivity matrices and the coordinate transformation matrix is the same as the step1 in design algorithm A.

step 2. Let’s choose b1 = [2, 0],b2 = [0, 2]. From (61),

b̃L =

[

0 1 0 00 0 1 0

]

; b̃LXb =2

∑4i=1 γi

[

−γ1γ3 + γ2γ4−γ1γ3 + γ2γ4

]

= 0.

Hence the equilibrium condition is reduced to

γ4 =γ3γ1

γ2. (69)

Let’s choose γ1 = γ2 = γ3 = 1, and then γ4 = 1 given by (69).step 3. Compute λ from (60).

λ = b̃+Xb =1

γ2 + γ3 + γ4 + γ1

[

(γ2 + γ3) (γ4 + γ1)(γ3 + γ4) (γ2 + γ1)

]

=

[

11

]

(70)

step 4. When we choose the free vector ze = [1,−1]T in (51), we get the node vector

p =[

−0.6464 −0.3536 1.3536 −0.3536 0.3536 −1.3536 0.3536 0.6464]T

.

Note that the existence conditions (70),(69) for λ and γ give the same results as (68).

Design Algorithm P :

16

(A) Class I Design (B) Class II Design (z = 1)

(C) Class II Design (z = −1)FIG. 2. 2-Bar 4-String Tensegrity Structure

step 1. Same as in design algorithm A.step 2. We choose p1 = [1, 0]

T ,p2 = [3, 0]T ,p3 = [0,−1]T ,p4 = [0, 1]T .(See Figure 2.(A) )

step 3. There is no choice of ze which satisfies (63). The choice of z = 1 yields

γ =[

3 −1 −1 3]T

; λ =[

−3 1]T

.

The choice of ze = −1 obviously reverses all signs of this γ and λ.

This proves that the configuration in Figure 2A can only be stabilized with class II tensegrity withnegative γ2, γ3 and λ1. Negative γ2, γ3 means that compression is required in these members, hencebars must replace these strings and negative λ1 means that tension is required in this member. Thisyields the structure in Figure 2B, where thick lines are bars and thin lines are strings. In conclusion,the only class I tensegrity that exists for the topology of Figure 1 requires overlap of the bars. Forthe nodal configurations in Figure 2A, the bars do not overlap and no class I tensegrity exists. Twoconfigurations of class II tensegrities are possible as shown in Figure 2B and 2C.

Symmetrical Force Configuration for 3-bars 9-strings Tensegrity Structure

A (3; 9; 3) class I tensegrity [9] consisting of one stage of nb = 3 bars (so np = 6 nodes) with ns = 9strings has topology as illustrated in Figure 3

17

FIG. 3. 3-Bar 9-String (3; 9; 3) Class I Tensegrity

The corresponding geometric matrices {S,B} are given by

S =

−I 0 0 0 I 0 −I 0 00 −I 0 0 0 I 0 I 0I 0 −I 0 0 0 0 −I 00 I 0 −I 0 0 0 0 I0 0 I 0 −I 0 0 0 −I0 0 0 I 0 −I I 0 0

; B =

−I 0 0I 0 00 −I 00 I 00 0 −I0 0 I

A symmetrical force configuration will be investigated with equal bar force coefficients {λ1 = λ2 =λ3 = λ}, equal ‘base’ string force coefficients {γ1 = γ2 = γ3 ∆= γb}, equal ‘top’ string coefficients{γ4 = γ5 = γ6 ∆= γt}, and equal vertical string coefficients {γ7 = γ8 = γ9 ∆= γv}. Then

X = S1V2(

VT2 Γ−1V2

)−1VT2 S

T1

=1

(2 γb2 + 8 γb γt + 6 γb γv + 6 γv γt + 2 γt2 + 3 γv 2) γ3t γ3b γ

3v

[

X1 X2 X3]

⊗

I3,

where

X1 =

(γt + γv + γb)(

γv2 + 3 γv γt + 3 γb γv + 4 γb γt + γb

2 + γt2)

−γb γv γt − γb γt2 − 4 γb2γt − 3 γb2γv − γb3 + γv γt2 + 2 γt γv 2 + γv 3

−4 γb γt2 − γb γv γt − γb2γt + 2 γb γv 2 + γb2γv − 3 γv γt2 − γt3 + γv 3

X2 =

−γb γv γt − γb γt2 − 4 γb2γt − 3 γb2γv − γb3 + γv γt2 + 2 γt γv 2 + γv 3

10 γb γv γt + 4 γb γt2 + 7 γb

2γt + 5 γb γv2 + 6 γb

2γv + γb3 + 2 γv γt

2 + 3 γt γv2 + γv

3

−3 γb γt2 − γb γv γt − γv γt2 + γt γv 2 − 3 γb2γt + γb γv 2 − γb2γv + γv 3

X3 =

−4 γb γt2 − γb γv γt − γb2γt + 2 γb γv 2 + γb2γv − 3 γv γt2 − γt3 + γv 3

−3 γb γt2 − γb γv γt − γv γt2 + γt γv 2 − 3 γb2γt + γb γv 2 − γb2γv + γv 3

6 γv γt2 + 5 γt γv

2 + 10 γb γv γt + 7 γb γt2 + 4 γb

2γt + γt3 + γv

3 + 3 γb γv2 + 2 γb

2γv

18

Now we have to solve the force coefficients based on the equilibrium conditions given by Lemma8. If we apply (i) condition in Lemma 8,

det (X− Λ) = − (λ − γv )2 γb2 + 8 γb γt + 6 γb γv + 6 γv γt + 2 γt2 + 3 γv 2

λsecond = 0,

where

λsecond4= ( − 2 γt3λ + 2 γv 2γt2 + 2 γv γt3 + 6 γb3γt + 15 γb2γt2 + 2 γb2γv 2

+ 6 γb γt3 + 2 γb

3γv − 6 γv 2γt λ − 10 γv γt2λ − 16 γb2γt λ+ 2 γb

2λ2 − 2 γb3λ − 10 γb2γv λ + 6 γb λ2γv − 6 γb λγv 2

+ 8 γb λ2γt − 16 γb λγt2 − 22 γb λγv γt + 2 λ2γt2 + 3 λ2γv 2

+ 6 λ2γv γt + 16 γb γt2γv + 8 γb γt γv

2 + 16 γb2γv γt

)

.

Since smaller rank of (X − Λ) yields more freedom for the choice of b, we choose γv = λ. Nextevaluating second term when γv = λ, we get

λsecond|γv=λ = −6 λ2γt2 − 6 γb2λ2 + 6 γb γt3 + 6 γb3γt + 15 γb2γt2 − 6 γb λ2γt + 3 λ4

=(

λ2 − γtγb − 2γ2b) (

λ2 − γtγb − 2γ2t)

.

We conclude λ =√

γtγb + 2γ2b or λ =√

γtγb + 2γ2t , since λ > 0.

When we apply λ =√

γtγb + 2γ2t , we get

X − Λ = (γb − γt) (γb + γt )2 γb2 + 6 γb γbt + 11 γb γt + 8 γt2 + 6 γt γbt

X̄

where γbt4=

√

γt (γb + 2 γt) and

X̄ =

γb + 2 γbt + 3 γt −γb − 4 γt − 3 γbt γt + γbt−γb − 4 γt − 3 γbt γb + 4 γbt + 6 γt −2 γt − γbt

γt + γbt −2 γt − γbt γt

⊗

I3

Note that the rank of the matrix X̄ is 1. An interesting case is when γt = γb. In this case, anequilibrium solution with Λ = X is provided by λ = γv =

√3γt for all choices of {γt}. When

γt = γb = 1 and λ = γv =√

3, the structural shape is the prism given in Figure 3.

Symmetrical Force Configuration for 6-bars 24-strings Tensegrity Structure

A (6;24;3;3) class I tensegrity structure consisting of two stage of nb = 6 bars ( sonp = 12 nodes ) with ns = 24 strings has topology as illustrated in Figure 4.Since the connectivity matrix for this example is quite big, we introduced another nota-tion ’Bar’ and ’Tendon’ which is defined by Bar (Bar Id#, { Node Id #1, Node Id #2}) andTendon (Tendon Id#, { Node Id #1, Node Id #2}). For simplicity, a symmetrical force configu-ration will be investigated in this example. Bars are defined by

Bar(1, {1, 2}); Bar(2, {3, 4}); Bar(3, {5, 6});

Bar(4, {7, 8}); Bar(5, {9, 10}); Bar(6, {11, 12});

with equal bar force coefficients {λ1 = λ2 = λ3 = λ4 = λ5 = λ64= λ} . Similariliy base and top

strings are defined by

Tendon(7, {1, 3}); Tendon(9, {3, 5}); Tendon(11, {5, 1});

Tendon(8, {8, 10}); Tendon(10, {10, 12}); Tendon(12, {12, 8});

19

FIG. 4. 6-Bar 24-String Class I Tensegrity

with equal ’base’ and ’top’ string force coefficients {γi4= γt} for all i = 7 · · · 12. Saddle strings are

defined by

Tendon(13, {2, 7}); Tendon(14, {2, 11}); Tendon(15, {4, 9});

Tendon(16, {4, 7}); Tendon(17, {6, 11}); Tendon(18, {6, 9});

with equal ’saddle’ string force coefficients {γi4= γs} for all i = 13 · · · 18. Diagonal strings are

defined by

Tendon(19, {1, 11}); Tendon(20, {3, 7}); Tendon(21, {5, 9});

Tendon(22, {2, 8}); Tendon(23, {4, 10}); Tendon(24, {6, 12});

with equal ’diagonal’ string force coefficients {γi4= γd} for all i = 19 · · · 24. Vertical strings are

defined by

Tendon(25, {1, 6}); Tendon(26, {3, 2}); Tendon(27, {5, 4});

Tendon(28, {7, 10}); Tendon(29, {9, 12}); Tendon(30, {11, 8});

with equal ’vertical’ string force coefficients {γi4= γd} for all i = 25 · · · 30.

For simplicity of calculation, we introduce scaled variables as following

γ̄s4=

γs

γt; γ̄v

4=

γv

γt; γ̄d

4=

γd

γt; λ̄

4=

λ

γt.

Then the equilibrium condition (ii) in Lemma 8 is given by

det

[

Λ S1V2VT2 S

T1 V

T2 Γ

−1V2

]

= 0.

This expression yields by using our scaled variables

−6γ̄6vγ̄6sγ̄6d

(

− γ̄d + λ̄ − γ̄v) (

λ̄ γ̄d − γ̄s γ̄d + γ̄s λ̄ − γ̄v γ̄d − γ̄s γ̄v)

(

λ̄ γ̄d − 3 γ̄d − γ̄s γ̄d − γ̄v γ̄d − 3 γ̄v + 3 λ̄ + 3 γ̄v λ̄ − 3 γ̄s − γ̄s γ̄v + γ̄s λ̄)2

(

λ̄ γ̄d − γ̄d − γ̄s γ̄d − γ̄v γ̄d − γ̄v + λ̄ + γ̄v λ̄ − 3 γ̄s − γ̄s γ̄v + γ̄s λ̄)2

= 0 (71)

20

There are many solutions satisfying this condition. Since we are interested in three dimensionalstable equilbria, we seek those combinations that generate stable equilibria. After simulation toverify the stability, the following combinations were found to generate stable equilibria.

λ̄ γ̄d − γ̄s γ̄d + γ̄s λ̄ − γ̄d γ̄v − γ̄s γ̄v = 0 (72)λ̄ γ̄d − γ̄d − γ̄s γ̄d − γ̄d γ̄v − γ̄v + λ̄ + γ̄v λ̄ − 3 γ̄s − γ̄s γ̄v + γ̄s λ̄ = 0 (73)

Now we analyze the equilibria (72) and (73). We solve (72) for λ̄ to obtain

λ̄ =γ̄s γ̄d + γ̄d γ̄v + γ̄s γ̄v

γ̄s + γ̄d. (74)

Since every variable is positive, this equation yields positive λ̄ as required. Substituting (74) into(73) yields

−γ̄d γ̄s γ̄v + γ̄s γ̄d + γ̄2d − γ̄d γ̄2v − γ̄s γ̄2v + γ̄2s = 0. (75)

Requiring positiveness of γ̄v, (75) can be solved by

γ̄v =1

2

−γ̄s γ̄d +√

γ̄2dγ̄2s

+ 8 γ̄2sγ̄d + 8 γ̄s γ̄2d + 4 γ̄

3s

+ 4 γ̄3d

γ̄s + γ̄d. (76)

With the use of original variables, (74) and (76) can be expressed as following

λ =γs γd + γd γv + γs γv

γs + γd(77)

γv =1

2

−γsγd +√

γ2dγ2s + 4γt (γs + γv) (γ

2s + γ

2v + γsγv)

γs + γd. (78)

CONCLUSION

This paper characterizes the static equilibria of a class I tensegrity structure. Analytical expres-sions are derived for the equilibrium condition of a tensegrity structure in terms of member forcecoefficients and string and bar connectivity information. We use vectors to describe each element(bars and tendons), eliminating the need to use direction cosines and the subsequent transcendentalfunctions that follow their use. By enlarging the vector space in which we characterize the problem,the mathematical structure of the equations admit treatment by linear algebra methods, for themost part. This reduces the study of a significant portion of the tensegrity equilibria to a series oflinear algebra problems. Our results characterize the equilibria conditions of tensegrity structures interms of a very small number of variables since the necessary and sufficient conditions of the linearalgebra treatment has eliminated several of the original variables. This formulation offers insightand identifies the free parameters that may be used to achieve desired structural shapes. Sinceall conditions are necessary and sufficient, these results can be used in the design of any tensegritystructure. Special insightful properties are available in the special case when one designs a tensegritystructure so that all strings have the same force per unit length(γ), and all bars have the same forceper unit length(λ). In this case, all admissible values of λ

γare the discrete set of eigenvalues of a

matrix given in terms of only the string connectivity matrix. Furthermore the only bar vectors whichcan be assigned are eigenvectors of the same matrix. Future papers will integrate these algorithmsinto software to make these designs more efficient.

ACKNOWLEDGEMENTS

This research were conducted by Dynamic Systems Research Inc. under the sponsorship of theJet Propulsion Laboratory, technical monitors Dr. Marco Quadrelli and Fred Hadaegh.

21

(i) Equilibrium Configuration when γt = 0.1 (ii) Equilibrium Configuration when γt = 1

(iii) Equilibrium Configuration when γt = 10

FIG. 5. Two-stage Shell Class Tensegrity Structure when γs = γd = 1

References

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[2] R. Connelly, Rigidity in: P.M. Gruber and J.M. Wills, ed., Handbook of Convex Geometry,Elsevier Publishers Ltd, pp 223-271, 1993.

[3] R. Connelly, Tensegrity Structures: Why are they stable?, in: M.F. Thorpe and P.M. Duxbury,ed., Rigity Theory and Applications , Plenum Press, pp47-54, New York, 1999.

[4] R. Connelly and A. Bach, Mathematics and Tensegrity, American Scientist, March, 1998.

[5] D. E. Ingber, Tensegrity: The Architectural Basis of Cellular Mechanotransduction, Ann. Rev.of Physiology, 59, pp575-599, 1997.

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