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1 Copyright © 2007 byASME

Proceedings of the ASME 2007 International Design Engineering Technical Conferences & Computers and

Information in Engineering Conference

IDETC/CIE 2007

September 4-7, 2007, Las Vegas, Nevada, USA

DETC2007-34249

CLOSED-FORM EQUILIBRIUM ANALYSIS

OF PLANAR TENSEGRITY STRUCTURES

Jahan Bayat

Center for Intelligent Machines and Robotics

Department of Mechanical andAerospace Engineering

University of Florida, Gainesville, Florida 32611, USA

Carl D. Crane III

Professor and author of correspondence, Phone: (352) 392-9461, Fax: (352) 392-1071, Email: ccrane@ufl.edu.

ABSTRACT

This paper presents a closed-form analysis of a series of

planar tensegrity structures

equilibrium configurations for each device when no external

forces or moments are applied.

determined by identifying the configurations at which the

potential energy stored in the springs is a minimum. The degree

of complexity associated with the solution was far greater than

expected. For a two-spring system, a 28th degree polynomial

expressed in terms of the length of one of the springs is

developed where this polynomial identifies the cases where the

change in potential energy with respect to an infinitesimal

change in the spring length is zero.

systems are also analyzed. These more complex systems were

solved using the Continuation Method.

are presented.

to determineall possible

The equilibrium position is

Three and four spring

Numerical examples

1. INTRODUCTION

The word tensegrity is a combination of the words tension

and integrity (Edmondson, 1987 and Fuller, 1975). Tensegrity

structures are spatial structures formed by a combination of

rigid elements in compression (struts) and connecting elements

that are in tension (ties). In a classic definition, no pair of struts

touch and the end of each strut is connected to three non-

coplanar ties (Yin et al, 2002). The entire configuration stands

by itself and maintains its form solely because of the internal

arrangement of the struts and ties (Tobie, 1976).

The development of tensegrity structures is relatively new

and the works related have only existed for approximately

twenty five years. Kenner, 1976, established the relation

between the rotation of the top and bottom ties. Tobie, 1976,

presented procedures for the generation of tensile structures by

physical and graphical means. Yin, 2002, obtained Kenner’s

results using energy considerations and found the equilibrium

position for unloaded tensegrity prisms. Stern, 1999, developed

generic design equations to find the lengths of the struts and

elastic ties needed to create a desired geometry for a symmetric

case. Knight, 2000, addressed the problem of stability of

tensegrity structures for the design of deployable antennae.

Duffy, 2000 and 2002, presents static analysis and equilibrium

condition for selected tensigrity structures. Roth, 1981,

establishes some basis for flexibility and rigidity of tensigrity

frameworks. Skelton, 2002, reduces the static model of some

tensigritystructuresto linear

characterizing the problems in vector space. Tibert, 2002, uses

the tensegrity structure in deployable space applications. Festl,

2003, explains the adjustable tensigrity structures and their

applications.

algebra equations by

2. TWO-SPRING SYSTEM

A planar tensegrity system is shown in Figure 1.

device consists of two rigid struts and four ties. For the two-

spring system, two of the ties are compliant (the ties between

points 4 and 2 and points 1 and 3) and two of the ties are non-

compliant. The objective is to determine all equilibrium

configurations for the system when given the lengths of the

struts and non-compliant ties as well as the free length and

spring constant for each of the compliant ties.

The specific problem statement for the two-spring system is

written as follows:

given: a12, a34

lengths of struts,

a23, a41

lengths of non-compliant ties,

k1, L01

spring constant and free length of compliant

tie between points 4 and 2,

The

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2 Copyright © 2007 byASME

k2, L02

spring constant and free length of compliant tie

between points 3 and 1.

length of spring 1 at equilibrium position,

length of spring 2 at

corresponding to length of spring 1, i.e. L1.

It should be noted that the problem statement could be

formulated in a variety of ways, i.e. a different variable (such as

the relative angle between strut a34and tie a41) could have been

selected as the generalized parameter for this problem.

Attempts at using this alternate approach did not successfully

yield a closed form solution for the equilibrium positions.

find: L1

L2

equilibrium position

2.1 Development of Geometry and Energy Equations

Figure 2 shows the nomenclature that is used. L1and L2

are the extended lengths of the compliant ties between points 4

and 2 and points 3 and 1. A cosine law for the triangle with

sides a34, a23, and L1can be written as

a

2

Solving for cos4' yields

2

23

4

L2

A cosine law for the triangle with sides a41, a12, and L1can

be written as

a

2

Solving for cos4" yields

2

12

4

2

A cosine law for the triangle with sides a41, a34, and L2can

be written as

a

2

Solving for cos4yields

2

a

' cosθ

aL

2

L

2

23

4 341

2

34

2

1

. (1)

341

2

34

2

1

a

aLa

'

θ

cos

. (2)

2

a

" cosθ

aL

2

L

2

12

4 411

2

41

2

1

. (3)

411

2

41

2

1

aL

aLa

" cosθ

. (4)

2

L

cosθ

aa

2

a

2

2

4 4134

2

41

2

34

. (5)

4134

2

41

2

34

2

2

4

aa2

aaL

cosθ

. (6)

From Figure 2 it is apparent that

4+ 4' = + 4" (7)

Equating the cosine of the left and right sides of (7) yields

cos (4+ 4') = cos ( + 4")

and expanding this equation yields

cos4cos4' – sin4sin4' = - cos4" .

Rearranging (9) yields

cos4cos4' + cos4" = sin4sin4' .

Squaring both sides of (10) gives

(cos4)2(cos4')2+ 2 cos4cos4' cos4" + (cos4")2

= (sin4)2(sin4')2.

Substituting for (sin4)2and (sin4')2in terms of cos4 and

cos4' gives

(cos4)2(cos4')2+ 2 cos4cos4' cos4" + (cos4")2

= (1-cos24) (1-cos24') .

Equations (2), (4), and (6) are substituted into (12) to yield

a single equation in the parameters L1and L2which can be

written as

AL2

where

A= L1

B = L1

and where

B2= - (a23

B0= (a12– a41) (a12+ a41) (a23– a34) (a23+ a34) ,

C2= (a34– a41) (a34+ a41) (a23– a12) (a23+ a12) ,

C0= (a41a23+ a34a12) (a41a23- a34a12) (a41

(8)

(9)

(10)

(11)

(12)

4+ B L2

2+ C = 0(13)

2,

4+ B2L1

2+ B0, C = C2L1

2+ C0

(14)

2+ a34

2+ a41

2+ a12

2),

2+ a23

2– a12

2– a34

2).

(15)

The potential energy of the system can be evaluated as

1

U

2

0222

2

0111

)L (Lk

2

1

)L (Lk

2

. (16)

Figure 1: Planar Tensegrity System

4

a41

a12

a34

2

3

a23

struts

ties

1

Figure 2: Two-Spring Planar Tensegrity Structure

4

3

2

1

θ4

θ'4

θ"4

a41

a23

a34

a12

L1

L2

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3 Copyright © 2007 byASME

At equilibrium, the potential energy will be a minimum. This

condition can be determined as the configuration of the

structure whereby the derivative of the potential energy taken

with respect to the length L1equals zero, i.e.

dU

2 0111

1

The derivative dL2/dL1 can be determined via implicit

differentiation from equation (13) as

aaL2L(L[L

dL

41232112

1

0

dL

dL

)L (Lk)L (Lk

dL

1

2

022

. (17)

))]

(18)

aa )(a a ()aaaaL2L(L[L

)]aa )(a a ()

aa dL

2

34

2

23

2

41

2

12

2

12

2

34

22222

2

34

2

41

2

23

2

12

2

12

2

34

2

41

2

23

2

1

2

2

2

21

2

.

Substituting (18) into (17) and regrouping gives

D L2

where

D = D1L1,

E = E1L1,

F = F3L1

G = G3L1

H = H5L1

J = J1L1

and where,

D1= k2, E1= -k2L02,

F3= 2 (k2– k1), F2= 2 k1L01,

F1= - k2(a12

G3= -2 k2L02, G1= k2L02(a12

H5= - k1, H4= k1L01, H3= k1(a12

H2= - k1L01(a12

H1= - k1(a34

(a23

H0= k1L01(a34

J1= k2L02(a34

5+ E L2

4+ F L2

3+ G L2

2+ H L2+ J = 0(19)

3+ F2L1

3+ G1L1,

5+ H4L1

2+ F1L1,

4+ H3L1

3+ H2L1

2+ H1L1+ H0,

(20)

2+ a23

2+ a34

2+ a41

2),

2+ a23

2+ a34

2+ a23

2),

2) + k2(a34

2+ a41

2+ a34

2),

2+ a41

2),

2+ a23

2) (a41

2),

2– a23

2– a41

2+ a34

2– a12

2+ a41

2– a23

2– a41

2)

2– a12

2) (a41

2) (a12

2– a12

2– a23

2),

2) . (21)

2.2 Solution of Geometry and Energy Equations

Equations (19) and (13) represent two equations in the two

unknowns L1and L2. These equations can be solved by using

Sylvester’s variable elimination procedure by multiplying

equation (13) by L2, L2

L2

matrix form as

B0A0000

2, L2

3, and L2

4and equation (19) by L2,

2, L2

3to yield a total of nine equations that can be written in

0

0

0

0

0

0

0

0

0

1

L

L

L

L

L

L

L

L

0000C0B0A

000JHGFED

000C0B0A0

00JHGFED0

00C0B0A00

0JHGFED00

0C0B0A000

C0

JHGFED000

2

2

2

3

2

4

2

5

2

6

2

7

2

8

2

. (22)

A solution to this set of equations can only occur if the

determinant of the 9×9 coefficient matrix is equal to zero.

Expansion of this determinant yields a 30thdegree polynomial

in the variable L1.When the determinant was expanded

symbolically, it was seen that the two lowest order coefficients

were identically zero. Thus the polynomial can be divided

throughout by L1

coefficients of the 28thdegree polynomial were obtained

symbolically in terms of the given quantities, but are not

presented here due to their length and complexity.

Values for L2that correspond to each value of L1can be

determined by first solving (13) for four possible values of L2.

Only one of these four values also satisfies equation (19).

2to yield a 28thdegree polynomial.The

2.3 Numerical Example

The following parameters were selected to show the results

of a numerical example:

strut lengths:

a12= 3 in.a34= 3.5 in.

non-compliant tie lengths:

a41= 4 in.a23= 2 in.

spring 1 free length & spring constant:

L01= 0.5 in.k1= 4 lbf/in.

spring 2 free length & spring constant:

L02= 1 in.k2= 2.5 lbf/in.

Eight real and twenty complex roots were obtained for L1.

The real values for L1and the corresponding values of L2are

shown in Table 1.

Table 1: Eight Real Solutions

CaseL1, in.

1 -5.485

2 -5.322

3 -1.741

4 -1.576

51.628

6 1.863

75.129

8 5.476

L2, in.

2.333

-2.901

-1.495

1.870

1.709

-1.354

-3.288

2.394

The values of L1 and L2 listed in Table 1 satisfy the

geometric constraints defined by equation (13) and the energy

condition defined by equation (19). Each of these eight cases

was analyzed to determine whether it represented a minimum or

maximum potential energy condition and cases 3, 4, 5, and 6

were found to be minimum states. Afree body analysis of struts

a12and a34was performed to show that these bodies were indeed

in equilibrium.Figure 3 shows the four equilibrium

configurations for the numerical case under consideration.

Efforts were undertaken to obtain a numerical example that

would yield more than eight real roots. One example of each

type of Grashof and non-Grashof four-bar mechanism was

analyzed, yet for all these cases a maximum of eight real roots

were found.The complex values of L1 were analyzed and

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4 Copyright © 2007 byASME

corresponding complex values of L2were determined. In every

case it was possible to obtain complex pairs of L1and L2that

satisfied the geometric constraint equation, (13), and the

derivative of potential energy equation (19).

extraneousroots wereintroduced

elimination procedure.

Thus no

variable during the

3. THREE-SPRING SYSTEM

Figure 4 shows a three-spring tensegrity system.

parameters must be specified, in addition to the constant

mechanism parameters, in order to define the configuration of

the device. These two parameters will be referred to as the

descriptive parameters for the system.

descriptive parameters are the angles θ4 and θ1. Considering the

non-compliantmembera41

specification of θ4will define the location of point 3. Similarly,

specification of θ1will define the location of point 2.

Two approaches to solve this problem have been analyzed.

Both aim to find a set of descriptive parameters that minimize

the potential energy in the system. In the first approach, the

lengths of the compliant members L1and L2are chosen as the

descriptive parameters.Derivatives of the potential energy

equation are obtained with respect to L1and L2and values for

the descriptive parameters are obtained such that these

derivatives are zero, corresponding to either a minimum or

maximum potential energy state. In the second approach, the

cosines of the angles θ4 and θ1were chosen as the descriptive

parameters. The cosines of the angles were chosen rather than

the angles themselves in the hope that the resulting equations

would be of lesser degree in that, for example, a single value of

cosθ4accounts for the obvious symmetry in solutions that will

occur with respect to the fixed member a41.

Two

One obvious set of

asbeing fixed toground,

3.1 Approach 1 – Descriptive Parameters L1and L2

3.1.1 Problem Formulation

The problem statement can be explicitly written as:

given: length of struts (a12, a34) ; length of the non-

compliant tie (a41) ; spring constants and free lengths

of three springs (k1, L01; k2, L02; k3, L03)

find:length of springs 1 and 2 (L1, L2) and corresponding

length of spring 3 (L3) at equilibrium

The analysis for this case can proceed in a manner similar

to that presented for the two-spring system. Specifically, the

term a23in (15) can be replaced by L3and equation (13) can be

factored into the form

G1L3

where

G1= a41

G2= G2aL1

G3= G3aL1

G4= G4aL1

G5= G5aL1

G6= G6aL1

and where the remaining coefficients are written in terms of the

constant mechanism parameters as

G2a= -1,

G2b= a12

G3a= (a34

G3b= a41

G4a= 1,

G5a= 1,

G5b= (-a12

G5c= a34

G6a= a12

4+ (G2L2

2+ G3) L3

2+ (G4L2

4+ G5L2

2+ G6) = 0 (23)

2,

2+ G2b,

2+ G3b,

2,

4+ G5bL1

2+ G6b

2+ G5c,

(24)

2– a41

2– a41

2(a41

2,

2),

2– a12

2– a34

2) – a12

2a34

2,

2– a34

2(a41

2(a41

2– a41

2– a12

2– a34

2),

2),

2),G6b= a12

2a34

2(a12

2+ a34

2– a41

2) .

(25)

The potential energy of the system can be written as

1

)L (Lk

2

At equilibrium, the potential energy will be a minimum. This

condition can be determined as the configuration of the

2

0333

2

0222

2

0111

)L (Lk

2

1

)L (Lk

2

1

U

.(26)

4

44

41

11

1

2

2

2

2

3

3

3

3

Case3 Case 4

Case 5 Case 6

spring in compression with

a negative spring length

spring in tension

Figure 3: Four Equilibrium Configurations

for Two-Spring Tensegrity System

4

3

2

1

4

a41

a12

a34

L1

L2

L3

1

Figure 4: Three-Spring Tensegrity System

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5 Copyright © 2007 byASME

structure whereby the derivative of the potential energy taken

with respect to the descriptive parameters L1and L2both equal

zero. The geometric constraint equation, equation (23),

contains three unknown terms, L1, L2, and L3.

equation, L3can be considered as a dependent variable of L1

and L2. The following two expressions may be written:

U

3 0111

1

U

3 0222

2

The derivatives δL3/δL1and δL3/δL2can be determined via

implicit differentiation from Equation 23 as

LLGLL2[L

L

21a2313

1

LLGLL [2L

L

21a2313

2

Substituting (29) into (27) and rearranging gives

(D1L2

(D5L2

where

D1= D1aL1,

D2= D2aL1+ D2b,

D3= D3aL1,

D4= D4aL1,

D5= D5aL1,

D6= D6aL1

D7= D7aL1

D8= D8aL1,

D9= D9aL1

D10= D10aL1

and

D1a= - G2ak3,

D2a= 2 G1k1– G3ak3, D2b= - 2 G1k1L01,

D3a= G2ak3L03,

D4a= G3ak3L03,

D5a= - k3,

D6a= G2ak1– 2 k3,

D6b= -G2ak1L01, D6c= G2bk1– G5bk3, D6d= -G2bk1L01,

D7a= G3ak1, D7b= -G3ak1L01, D7c= G3bk1– G6ak3,

D7d= -G3bk1L01,

D8a= k3L03,

D9a= 2 k3L03, D9b= G5bk3L03

D10a= G6ak3L03

Substituting (30) into (28) and rearranging gives

(E1L2+ E2) L3

+ E5L2

where

E1= E1aL1

E2= - 2 G1k2L02,

E3= E3aL1

E4= E4aL1

E5= E5aL1

From this

0

L

L

)L (Lk)L (Lk

L

1

3

033

, (27)

0

L

L

)L (Lk)L (Lk

L

2

3

033

. (28)

]GLGLGLLGLG [2L

]GL

GL

LG

L

b3

2

1a3

2

2b

L

2

222

a6

2

2b5

4

2

2

3a3

2

3

2

2a2

2

2

2

113

(29)

]GLGLGLLGLG [2L

]GL

GL

G

L

b3

2

1a3

2

2b2

222

c5

2

1b5

4

1

2

3b2

2

3

2

1a2

2

2

2

123

(30)

2+D2) L3

2+D7) L3+ (D8L2

3+ (D3L2

2+D4) L3

4+D9L2

2+

2+D10) = 0

4+D6L2

(31)

3+ D6bL1

3+ D7bL1

2+ D6cL1+ D6d,

2+ D7cL1+ D7d,

3+ D9bL1

(32)

(33)

3+ (E3L2) L3

2+ (E4L2

3+ E9L2) = 0

3

2+ E6L2+ E7) L3+ (E8L2

(34)

2+ E1b,

2+ E3b,

2+ E4b,

2+ E5b,

E6= E6aL1

E7= E7aL1

E8= E8aL1

E9= E9aL1

4+ E6bL1

2+ E7b,

2,

4+ E9bL1

2+ E6c,

2+ E9c

(35)

and

E1a= -G2ak3,

E3a= G2ak3L03,

E4a= G2ak2– 2 k3,

E5a= -G2ak2L02,

E6a= - k3, E6b= G3ak2– G5bk3,E6c= G3bk2– G5ck3,

E7a= -G3ak2L02,E7b= -G3bk2L02,

E8a= 2 k3L03,

E9a= k3L03, E9b= G5bk3L03, E9c= G5ck3L03.

E1b= -G2bk3+ 2 G1k2,

E3b= G2bk3L03,

E4b= G2bk2,

E5b= -G2bk2L02,

(36)

3.1.2Solution of Three Simultaneous Equations in Three

Unknowns – Sylvester’s Method

Equations (23), (31), and (34) are three equations in the

three unknowns L1, L2, and L3.

applied in order to obtain sets of values for these parameters

that simultaneously satisfy all three equations. In this solution,

the parameter L1is embedded in the coefficients of the three

equations to yield three equations in the apparent unknowns L2

and L3. Determining the condition that these new coefficients

(which contain L1) must satisfy such that the three equations can

have common roots for L2and L3will yield a single polynomial

in L1.

Equation (23) was multiplied by L2, L3, L2L3, L3

L2L3

L2

L2

Equation (34) was multiplied by L2, L3, L2L3, L3

L2

L2

equations that can be written in matrix for as

M λ = 0 .

The vector λ is written as

λ = [L2

L2

L2L3

L2

L2

L2

The coefficient matrix M is a 52×52 matrix whose

elements are the coefficients G1through G6, D1through D10,

and E1 through E9 which are polynomials in terms of the

variable L1. This matrix is not presented here due to its size.

Since the set of 52 simultaneous equations represented by (37)

must be linearly dependent, the determinant of the matrix M

must equal zero. This will yield an equation in terms of the

single variable L1.

It was not possible to symbolically expand the determinant

of matrix M. A numerical case was analyzed and a polynomial

of degree 158 in the variable L1 was obtained.

possible to numerically solve this high degree polynomial for

Sylvester’s method can be

2, L2

2, and

2, L3

2,

2, L2

3L3

2, L2L3

2L3, L2

2L3

2, L3

3, L2

3, L2L3

3, L2

2L3

3, L2

3L3, L2

2, L2L3

3L3

2, L2

3L3, L2

3L3

3. Equation (31) was multiplied by L3, L2, L3

3, L2

3,

4.

2,

4,

2L3, L2

2L3

2, L2

3, L3

4, L2

3L3, L2

2, and L2L3

2, L2L3

3L3

2L3, L2

4L3, L2

2L3

4L3

2, L3

2, L2L3

3, L2

4, and L2

3, L2L3

3, L2

2L3

2L3

3, L2

2, L2

4, L3

4. This resulted in a set of 52

(37)

7L3

3L3

3, L2

6, L2

7, L2

6, L2

3L3

3, L2

5L3

2L3

7, L2

5, L2

7, L2

6L3, L2

4L3

3, L2L3

2, L3

3L3

7L3, L2

5L3

2, L2

4, L3

3, L2

7, L2

7L3

6L3

2, L2

3L3

5, L2

2, L2L3, L3

2, L2

2, L2

4L3

2L3

4, L2

6L3

5L3

3, L2

4, L2L3

3L3, L2

2, L2, L3, 1]T.

3, L2

3, L2

3L3

5L3

4L3

4, L2

5, L3

2L3

4, L2

4, L2

2L3

6, L2

2, L2L3

4L3

3L3

5, L2L3

5, L2

5,

5, L2

2L3

6,

7,

6, L3

4L3,

3, L3

5L3, L2

2, L2

2L3, L2L3

3, L2

2L3

4,

(38)

It was not

Page 6

6Copyright © 2007 byASME

the values of L1, although several commercial and in-house

written algorithms were attempted. Because of this, a different

method was attempted to solve the set of equations (23), (31),

and (34).

3.1.3Solution of Three Simultaneous Equations in Three

Unknowns – Continuation Method

The continuation method (Garcia and Li, 1980, Morgan,

1983, 1986, 1987, Wampler et al., 1990) is a numerical

technique to solve a set of equations in multiple variables. This

is as opposed to Sylvester’s method which would lead to a

symbolic solution of the problem.

A concise description of the continuity method is presented

by Tsai, 1999. Suppose one wishes to solve the set of equations

F(x) which are defined by

,x,(xf

:F(x)

0)x ,,x ,

1

(xf

0)x,

0) x ,,x ,

1

(xf

n2n

n212

n21

. (39)

F(x) is called the target system.

The continuation method begins by first estimating the total

number of possible solution sets (sets of values for L1, L2, and

L3for our case) that satisfy the given equations. For example,

Bezout’s theorem states that a polynomial of total degree n has

at most n isolated solutions in the complex Euclidean space.

Including solutions at infinity, the Bezout number of a

polynomial system is equal to the total degree of the system.

Next, an initial system, G(x) =0, is obtained, whose

solution will be of the same degree as that of F(x), but whose

solution set is known in closed form.

maintains the same polynomial structure as F(x).

Finally, a homotopy function H(x, t) is prepared such as

H(x, t) = γ (1-t) G(x) + t F(x)

where γ is a random complex constant. When t=0, the

homotopy function equals the initial system, G(x). When t=1,

the homotopy function equals the target system, F(x). Recall

that the solutions to G(x) are known. As the parameter t is

increased in small steps from 0 to 1, the solutions of H(x, t) can

be tracked (referred to as path tracking) and when t =1, these

solutions will be the solutions to the original target system. If

the degree of the solution set was overestimated, some of the

solutions will track to infinity and these can easily be discarded.

In other words, G(x)

(40)

3.1.4 Numerical Example

The following information is given:

strut lengths:

a12= 14 in.a34= 12 in.

non-compliant tie lengths:

a41= 10 in.

spring 1 free length & spring constant:

L01= 8 in.k1= 1 lbf/in.

spring 2 free length & spring constant:

L02= 2 in.k2= 2.687 lbf/in.

spring 3 free length & spring constant:

L03= 2.5 in. k3= 3.465 lbf/in.

Based on these values, the coefficients in equations (23),

(31), and (34) were evaluated to yield the three equations

100 L3

+ (L1

4+ [(-L1

4– 440 L1

2+ 96) L2

2-13824) L2

2+ 44 L1

2– 8624 L1

2– 52224] L3

2+ 6773760 = 0

2+ L1

2L2

4

(41)

(2.5 L1L2

+ 381.165 L1) L3

+ 1196 L1– 768) L2

+ 417792] L3+ 8.663 L1L2

– 3811.651 L1) L2

2+ 90 L1– 1600) L3

2+ [-2.5 L1L2

2+ 44 L1

3+ (-8.663 L1L2

4+ (-6 L1

3– 352 L1

4+ (17.326 L1

2– 74708.354 L1= 0

2

3+ 8 L1

2– 30664 L1

2

3

(42)

[(2.5 L1

– 8.663 L1

+ 5.373 L1

– 236.420 L1

+ (-119755.135 + 8.663 L1

The continuation method was run on this set of three

equations in three unknowns to obtain all solution sets for the

three spring lengths, L1, L2, and L3, for the particular numerical

example. The software PHCpack (Verschelde, 1999) was used

to implement the method.

The PHCpack software estimated the number of possible

solutions to be 136. Seven real solutions were obtained and

these are listed in Table 2.

Table 2: Seven Real Solutions for

Three-Spring Planar Tensegrity System

CaseL1, in.

1 13.000

2 -11.376

3 -7.585

4 -11.029

5 13.969

6 14.248

7 13.181

2+ 160) L2– 1074.637] L3

2) L2L3

2) L2

2+ 280609.161] L3+ 17.326 L1

4– 3811.651 L1

3+ (831.633

3+ (-515.826

2– 69888) L2

2+ [(-7 L1

2+ (-2.5 L1

2+ 192) L2

4+ 1188 L1

2L2

3

2) L2= 0 . (43)

L2, in.

8.000

-10.371

9.097

12.557

-5.800

-9.373

11.599

L3, in.

7.017

-5.333

10.106

-3.044

9.164

-4.774

-2.488

An equilibrium analysis was conducted for the seven cases

and only the first case was indeed in equilibrium, i.e. it

4

1

2

3

a41

a12

a34

L1

L2

L3

Figure 5: Case 1 ; Equilibrium Solution

Page 7

7 Copyright © 2007 byASME

corresponds to a minimum potential energy configuration. Case

1 is shown in Figure 5.

3.2 Approach 2 – Descriptive Parameters cosθ1and

cosθ4

3.2.1 Problem Formulation

The problem statement is written as

given: a41, a12, a34(strut lengths and non-compliant tie length)

k1, k2, k3, L01, L02, L03(spring constants and spring free

lengths)

find: cosθ4 and cosθ1when the system is in equilibrium

Figure 4 shows the nomenclature used. Springs L1, L2, and

L3are the extended lengths of the compliant ties between points

4 and 2, 2 and 3, and 1 and 3 respectively.

lengths of springs are given by L01, L02, and L03. A cosine law

for the quadrilateral 1-2-3-4 can be written as:

L

Z

The unloaded

2

2

3

41

(44)

where,

2

a

Z)cYs (XaZ

2

12

41414 1241

(45)

and where

X4= a34s4

Y4= - (a41+ a34c4)

2

34

22

(46)

(47)

4 4134

2

41

4

caa

aa

Z

. (48)

Substituting (45) into (44) and rearranging gives

2

L

2

a

ZcYasXa

2

3

2

12

414 1214 12

. (49)

Substituting (46), (47), and (48) into (49), then squaring it, and

substituting for s4

entire equation by 4 yields

L3

where

A= A1c4+A2c1+A3c1c4+A4,

B = B1c1

+ B6c4

and where

A1= - 4 a34a41

A2= - 4 a12a41

A3= - 4 a12a34

A4= - 2 (a12

B1= 4 a12

B2= 8 a12

B3= 4 a12a41(a41

B4= 4 a12a34(a34

B5= 8 a12a41a34

B6= 4 a34

B7= 4 a34a41(a41

B8= (a41

+ 2 a12a34+ a12

2= 1- c4

2and s1

2=1-c1

2and multiplying the

4+AL3

2+ B = 0(50)

(51)

2+ B2c1

2+ B7c4+ B8

2c4+ B3c1+ B4c1c4+ B5c1c4

2

(52)

2+ a34

2(a34

2a34a41

2+ a41

2)

2)

2+ a41

2+ a12

2+ a12

2+ a34

2+ 3 a41

2)

2)

2

2(a12

2+ a41

2+ a34

2– 2 a12a34+ a12

2) .

2)

2+ a12

2)

2+ a34

2) (a41

2+ a34

2

(53)

Equation (50) expresses L3as a function of c4and c1. A

cosine law for triangle 4-1-2 may be written as

a

2

Acosine law for the triangle 3-4-1 may be written as

a

2

Equations (50), (54), and (55) define the three spring lengths in

terms of the variables cosθ4 and cosθ1.

equations, each of the three spring lengths L1, L2, and L3can be

determined for a given set of parameters cosθ4 and cosθ1.

Lastly, equation (50) can be regrouped into the format

(r1c1+ r2) c4

+ (r6c1

where

r1= 8 a12a41a34

r2= 4 a34

r3= 8 a34a41a12

r4= (-4 a12a34) L3

r5= (-4 a34a41) L3

r6= 4 a12

r7= (-4 a12a41) L3

r8= L3

The total potential energy stored in all three springs is given by,

U = ½ k1(L1-L01)2+ ½ k2(L2-L02)2+ ½ k3(L3-L03)2. (58)

Differentiating the potential energy with respect to c4and c1and

then evaluating values for c4and c1that cause the derivative of

the potential energy to equal zero, will identify configurations

of either minimum or maximum potential energy.

derivatives may be written as,

dU/dc4= k1(L1– L01) dL1/dc4+ k2(L2– L02) dL2/dc4

+ k3(L3– L03) dL3/dc4

dU/dc1= k1(L1– L01) dL1/dc1+ k2(L2– L02) dL2/dc1

+ k3(L3– L03) dL3/dc1

Since from (54), L1 is not a function of c4, dL1/dc4 = 0.

Similarly, from (55), L2is not a function of c1and thus dL2/dc1

= 0. Equations (59) and (60) reduce to

dU/dc4= k2(L2– L02) dL2/dc4+ k3(L3– L03) dL3/dc4, (61)

dU/dc1= k1(L1– L01) dL1/dc1+ k3(L3– L03) dL3/dc1. (62)

The term dL2/dc4is evaluated from (55) as

a

dc

and the term dL1/dc1is evaluated from (54) as

a

dc

Implicit differentiation of (50) for L3with respect to c4and c1

yields

cBL)cA(A

dc

4133

4

2

L

caa

2

a

2

1

1 4112

2

41

2

12

.(54)

2

L

caa

2

a

2

2

4 41 34

2

41

2

34

. (55)

From these three

2+ (r3c1

2+ r7c1+ r8) = 0

2+ r4c1+ r5) c4

(56)

2

2(a12

2+ a41

2

2)

2+ 4 a12a34(a34

2+ 4 a34a41(a34

2+ a41

2+ 4 a12a41(a41

4– 2 (a34

2+ a12

2+ a12

2+ 3 a41

2+ a41

2)

2)

2(a34

2)

2+ a34

2.

2+ a12

2)

2+ a12

2+ a41

2) L3

(57)

These

(59)

(60)

2

4134

L

4

2

a dL

(63)

1

41 12

L

1

1

a dL

(64)

)AccAcAcAL (2L2

BcB2ccB2

cBdL

441312

2

74641514

2

12

2

31313

(65)

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8 Copyright © 2007 byASME

)AccAcAcAL(2L2

cBcB

Bc

cB2cB2

L)cA (A

dc

Substituting (63) and (65) into (61) and (64) and (66) into (62)

and equating to zero yields

(M1L3

(N1L3

where the coefficients Miand Ni, i=1..6, are functions of c4and

c1as

M1=-k3(A3c1+A1) + 4 k2a34a41

M2=k3L03(A3c1+A1)

M3=-k3B2c1

k3B4) c1+ 2 (k2a34a41A1– k3B6) c4– k3B7+ 2 k2a34

a41A4

M4= k3L03B2c1

L03B6c4+ k3L03B7

M5= -4 k2a34a41L02

M6= -2 k2(a34a41L02A2c1+ a34a41L02A3c1c4

+ a34a41L02A1c4+ a34a41L02A4)

N1= - k3(A3c4+A2) + 4 k1a12a41

N2= k3L03(A3c4+A2)

N3= 2 k1a12a41(A1c4+A2c1+A3c1c4+A4) – k3(B4c4

+ B5c4

N4= k3L03(B4c4+ B5c4

N5= - 4 k1a12a41L01

N6= -2 k1a12a41L01(A4+A1c4+A2c1+A3c1c4) .

Equations (67) and (68) may be written as

(M1L3

= - M5L3

(N1L3

= - N5L3

Squaring both sides of both equations gives

(M1L3

= (M5L3

(N1L3

= (N5L3

Using (54) and (55) to substitute for L1

equations in the parameters c1, c4, and L3. These two equations

can be arranged as

(p1c1

+ (p8c1

+ (p13c1

(q1c1+ q2) c4

+ q7c1

+ q13) c4+ (q14c1

where

p1= p1aL3

p2= p2aL3

p3= p3aL3

p4= p4aL3

dL

44131241

2

33

2

4544341211

2

3432

1

3

.(66)

3+ M2L3

3+ N2L3

2+ M3L3+ M4) L2+ M5L3

2+ N3L3+ N4) L1+ N5L3

3+ M6L3= 0 (67)

3+ N6L3= 0 (68)

2+ [2 (k2a34a41A3– k3B5) c4+ 2 k2a34a41A2–

2+ k3L03B4c1+ 2 k3L03B5c1c4+ 2 k3

(69)

2+ 2 B1c1+ 2 B2c1c4+ B3)

2+ 2 B1c1+ B2c1c4+ B3)

(70)

3+ M2L3

2+ M3L3+ M4) L2

3– M6L3,

2+ N3L3+ N4) L1

3– N6L3.

(71)

3+ N2L3

(72)

3+ M2L3

2+ M3L3+ M4)2L2

3+ M6L3)2,

2+ N3L3+ N4)2L1

3+ N6L3)2.

2

(73)

3+ N2L3

2

(74)

2and L2

2will yield two

2+ p2c1+ p3) c4

4+ p9c1

4+ p14c1

3+ (p4c1

2+ p11c1+ p12) c4

3+ p15c1

4+ (q3c1

2+ q8c1+ q9) c4

3+ q15c1

3+ p5c1

2+ p6c1+ p7) c4

2

3+ p10c1

2+ p16c1+ p17) = 0

2+ q4c1+ q5) c4

2+ (q10c1

2+ q16c1+ q17) = 0

(75)

3+ (q6c1

2+ q12c1

3

3+ q11c1

(76)

2+ p1bL3+ p1c

2+ p2bL3+ p2c

2+ p3bL3+ p3c

2+ p4bL3+ p4c

p5= p5aL3

p6= p6aL3

p7= p7aL3

p8= p8aL3

p9= p9aL3

p10= p10aL3

L3+ p10g

p11= p11aL3

L3+ p11g

p12= p12aL3

L3+ p12g

p13= p13aL3

p14= p14aL3

p15= p15aL3

L3+ p15g

p16= p16aL3

L3+ p16g

p17= p17aL3

L3+ p17g

q1= q1aL3

q2= q2aL3

q3= q3aL3

q4= q4aL3

q5= q5aL3

q6= q6aL3

q7= q7aL3

q8= q8aL3

+ q8fL3+ q8g

q9= q9aL3

+ q9fL3+ q9g

q10= q10aL3

q11= q11aL3

q12= q12aL3

L3+ q12g

q13= q13aL3

L3+ q13g

q14= q14aL3

q15= q15aL3

q16= q16aL3

L3+ q16g

q17= q17aL3

L3+ q17g.

The coefficients p1a through q17g have been expressed

symbolically in terms of the given mechanism parameters and

as such can be considered as known quantities. The coefficients

are not listed here due to the length of the expressions.

4+ p5bL3

4+ p6bL3

4+ p7bL3

2+ p8bL3+ p8c

4+ p9bL3

6+ p10bL3

3+ p5cL3

3+ p6cL3

3+ p7cL3

2+ p5dL3+ p5e

2+ p6dL3+ p6e

2+ p7dL3+ p7e

3+ p9cL3

5+ p10cL3

2+ p9dL3+ p9e

4+ p10dL3

3+ p10eL3

2+ p10f

6+ p11bL3

5+ p11cL3

4+ p11dL3

3+ p11eL3

2+ p11f

6+ p12bL3

5+ p12cL3

4+ p12dL3

3+ p12eL3

2+ p12f

2+ p13bL3+ p13c

4+ p14bL3

6+ p15bL3

3+ p14cL3

5+ p15cL3

2+ p14dL3+ p14e

4+ p15dL3

3+ p15eL3

2+ p15f

6+ p16bL3

5+ p16cL3

4+ p16dL3

3+ p16eL3

2+ p16f

6+ p17bL3

5+ p17cL3

4+ p17dL3

3+ p17eL3

2+ p17f

(77)

2+ q1bL3+ q1c

2+ q2bL3+ q2c

2+ q3bL3+ q3c

4+ q4bL3

4+ q5bL3

2+ q6bL3+ q6c

4+ q7bL3

6+ q8bL3

3+ q4cL3

3+ q5cL3

2+ q4dL3+ q4e

2+ q5dL3+ q5e

3+ q7cL3

5+ q8cL3

2+ q7dL3+ q7e

4+ q8dL3

3+ q8eL3

2

6+ q9bL3

5+ q9cL3

4+ q9dL3

3+ q9eL3

2

2+ q10bL3+ q10c

4+ q11bL3

6+ q12bL3

3+ q11cL3

5+ q12cL3

2+ q11dL3+ q11e

4+ q12dL3

3+ q12eL3

2+ q12f

6+ q13bL3

5+ q13cL3

4+ q13dL3

3+ q13eL3

2+ q13f

2+ q14bL3+ q14c

4+ q15bL3

6+ q16bL3

3+ q15cL3

5+ q16cL3

2+ q15dL3+ q15e

4+ q16dL3

3+ q16eL3

2+ q16f

6+ q17bL3

5+ q17cL3

4+ q17dL3

3+ q17eL3

2+ q17f

(78)

3.2.2 Solution of Three Simultaneous Equations in Three

Unknowns

Equations (75), (76), and (56) are factored such that the

parameter L3 is embedded in the coefficients p1 through r8.

Sylvester’s method was used in an attempt to solve these

equations for all possible sets of values of L3, c1, and c4.

Equation (75) was multiplied by c1, c1

c1c4

2, c4, c4

2, c4

3, c1

4, c12c4,

2, and c1

2c4

2to obtain 10 equations (including itself).

Page 9

9Copyright © 2007 byASME

Equation (76) was multiplied by c1, c1

c1

Equation (56) was multiplied by c1, c1

c1c4, c1

c1

This resulted in a set of 43 “homogeneous” equations in 43

unknowns.

The condition for a solution to this set of equations was

that they are linearly dependent, i.e. the determinant of the

43×43 coefficient matrix must equal zero. This would yield a

polynomial in the single variable L3. Due to the complexity of

the problem, it was not possible to expand this determinant

symbolically.Similarly, it was not possible to root the

polynomial that resulted from a particular numerical example.

Because of this, the Continuation Method was used again to

determine solutions for the same numerical example presented

in Section 3.1.4. The continuation method was run on this set

of three equations in three unknowns to obtain all solution sets

for the three variables, L3, c1, and c4, for the particular

numerical example. The software PHC pack (Verschelde [26])

was again used to implement the method.

2, c1

3, c4, c4

2, c1c4, c1

2c4,

3c4, c1c4

2and c1

2c4

2to obtain 11 equations (including itself).

2, c1

4c4,c1c4

2c4

3, c1

2, c1

4, c4, c4

4c4

2, c4

3, c1

3, c4

2c4

4,

3,

2c4, c1

4, c1c4

3c4, c1

4and c1

2, c1

2c4

2, c1

3c4

2, c1c4

3c1

4to obtain 22 equations (including itself).

3.2.3 Numerical Example

The same input parameters used in the numerical example

in Section 3.1.4 were used here. The coefficients of equations

(75), (76), and (56) were numerically determined and the

PHCpack software, which utilizes the continuation method,

estimated the number of possible solutions to be 136. Atotal of

7 real solutions and 129 complex solutions were obtained. The

seven real solutions are shown in Table 3.

number one is the only real solution case that corresponds to a

minimum potential energy configuration.

The lengths of the springs L1and L2were calculated for

each of the seven cases and it was determined that the seven

real solutions presented in Table 3 correspond with the seven

real solutions that were determined from the first problem

formulation and listed in Table 2. It is an important verification

that that the equilibrium configurations determined from the two

solution approaches are identical.

Of these, case

Table 3: Real solutions;

cosine of θ1, θ4in radians and L3in inches.

cos θ1

Case

cos θ4

L3

1 -0.45357 -0.749997.0166

2-0.59496 -0.56850 -5.3333

3 -0.85169-0.67187 10.1050

4-0.62272-0.35968 -3.0440

5-0.36020-0.876519.1639

6-0.33212-0.65064 -4.7738

7 -.043667-0.45612-2.4876

4. FOUR-SPRING SYSTEM

4.1 Problem Formulation

Afour-spring planar tensegrity system is shown in Figure 6.

The problem statement can be explicitly written as:

Given:

a12, a34

lengths of struts,

k1, L01

spring constant and its free length between

points 4 and 2,

k2, L02

spring constant and its free length between

points 3 and 1,

k3, L03

spring constant and its free length between

points 3 and 2,

k4, L04

spring constant and its free length between

points 4 and 1.

Find at equilibrium position:

L1

length of spring 1,

L2

length of spring 2,

L3

length of spring 3,

L4

length of spring 4.

It should be noted that as before, the problem statement

could be formulated in a variety of ways.

presented here uses L1, L2, and L3 as the three generalized

parameters, knowledge of which completely describes the

system (i.e. the value for L4, the last remaining parameter, can

be deduced). The geometry equation that relates L4in terms of

the three generalized parameters is derived first.

this, a derivative of the potential energy equation is taken with

respect to each of the three generalized parameters to yield an

additional three equations.

The equation that relates the lengths of the four springs

can be obtained directly by substituting the variable L4for the

length a41in equation (25). The potential energy equation can

then be written as

U = ½ k1(L1-L01)2+ ½ k2(L2-L02)2

+ ½ k3(L3-L03)2+ ½ k4(L4-L04)2.

Taking a derivative of the potential energy with respect to

L1, L2, and L3and equating to zero will yield three additional

equations that must be satisfied at equilibrium.

these equations are not presented here, however at this point

four equations now exist in the parameters L1, L2, L3, and L4. It

The solution

Following

(79)

For brevity,

4

3

2

1

L4

a12

a34

L1

L2

L3

Figure 6: Two Strut, Four-Spring Planar Tensegrity Device

Page 10

10Copyright © 2007 byASME

was not practical to attempt to use Sylvester’s method to solve

this set of equations. The continuation method, however, was

used to solve numerical cases.

4.2 Numerical Example

The following parameters were selected as a numerical

example:

Strut lengths: a12= 14 in., a34= 12 in.

Spring free lengths and spring constants:

L01= 8 in.k1= 1 lbf/in.

L02= 2.6866 in.k2= 2.0 lbf/in.

L03= 3.46513 in.k3= 2.5 lbf/in.

L04= 7.3083 in.k4= 1.5 lbf/in.

The geometry equation and the three potential derivatives

of the potential energy equation can now be expressed in terms

of the given quantities as

L2

– 13824.0) L2

+ 6773760.0 – 52224.0 L3

– 4.0376 L2

+ 80.0 L1

+ 80.0 L3

– 1577.6530 L3

– 15680.0 L1

(–28.0751 L1

+ 53.7318 L3

+ 5696.5300 L1

– 4.0375 L1

+ 0.1517 107– 53.7318 L3

+ 53.7318 L3

– 494742.0331 L3+ 12127.9787 L3

+ 86.6284 L3

+ 4722.3470 L3L1

+ (–20.9624 L3L1

– 25.0 L3

+ 0.24450 107= 0

The Polynomial Continuation Method was used to find

simultaneous roots for the four equations.

solutions and 490 complex solutions were found. All the real

and complex solutions were shown to satisfy Equations (80)

through (83) with negligible residuals.

Nine of the eighteen real solutions were not physically

realizable. For example, L1in one of the solutions was greater

than the sum of a12 and L4, and hence point 2 cannot be

constructed. The remaining real solutions can correspond to

either maximum or minimum potential energy configurations.

The second derivative of the potential energy with respect to the

variables L1, L2, and L3was evaluated to identity the minimal

potential energy cases.Each of these solutions was then

analyzed by calculating forces in the struts and elastic ties to

4L1

2+ (L1

2– 8424.0 L1

4– L3

2L1

2– 440.0 L1

2– 8624.0 L1

2+ 96.0 L3

2+ 44.0 L3

2+ L3

2

2L1

2

4= 0(80)

4L1+ (–1.1520 104– 5.9624 L3

2– 18.0751 L1

2+ 11200.0 L3

2L1+ 10.0 L1L3

2– 247420.0110 L1+ 80. L3

2+ 2880.0 – 20.0 L3

2+ 53.7318 L1

2– 15.9624 L3

4– 508664.6558) L2+ 7522.4588 L3

4– 10531.4424 L1

2L1

2L1

3+ 3216.5300 L1

2+ 1960.0 L1

4– 80.0 L3

2) L2

3–10 L3

4+ 0.2257 107

2L1

3+ (–7737.3862

2+ (-3187.6065 L3

2L1

2L1

3

2= 0 (81)

2) L2

2) L2

2

2+ 20.0 L3

4

2

2

2= 0

2– 16979.1702 L1

4– 4307.5137 L3

2+ 25.0 L3

2+ 86.6284 L1

3– 12474.4924 + 86.6284 L3

(82)

2

2L1

2– 86.6284 L3

3

5– 25.0 L3

2+ 3212.3934 L3

3L1

2

2) L2

2

(83)

Eighteen real

determine

configuration. This analysis indicates that only four of the real

solutions were indeed in equilibrium.

configurations are shown in Figure 7.

if ittruly correspondstoanequilibrium

The four equilibrium

5. CONCLUSIONS

The two-spring planar tensegrity system represents the

simplest tensegrity device. It was remarkable that such a simple

device would have such a complicated solution, i.e. a 28th

degree polynomial was obtained in the variable L1. It is true

that conditions corresponding to maximum as well as minimum

potential energy states are obtained in this formulation. The

coefficients of this polynomial were obtained symbolically.

Two solution approaches were presented for the three-

spring system. The complexity of both cases was such that they

could not be solved symbolically. Further, an attempt to solve a

numerical example using Sylvester’s solution method was not

successful.The numerical example was solved using the

continuation method. However, of the seven real solutions for

the three spring lengths, only one solution was found that was in

equilibrium.

Lastly, a four-spring planar tensegrity system was analyzed.

The continuation method estimate of the solution was degree

508. Four real equilibrium solutions for a specific numerical

example were obtained.

The surprising result to the authors was the degree of

complexity of the solutions that were obtained for two-, three-,

and four-spring planar tensegrity systems. It was not apparent

at the outset that such simple planar structures would have such

a high degree for the solution polynomial. Other solution

44

44

11

11

2

2

2

2

3

3

3

3

case 2case 3

case 13case 15

Figure 7: Equilibrium Configurations

for Four-Spring Tensegrity Device

Page 11

11 Copyright © 2007 byASME

approaches that would yield only minimum potential energy

solutions should be addressed to see if a simpler result can be

obtained.

ACKNOWLEDGMENTS

The authors would like to gratefully acknowledge the

support of the Department of Energy, grant number DE-FG04-

86NE37967.

REFERENCES

[1]Duffy, J., Rooney, J., Knight, B., and Crane, C., (2000), A

Review of a Familly of Self-Deploying Tensegrity

Structures with Elastic Ties, The Shock and Vibration

Digest, Vol. 32, No. 2, March 2000, pp. 100-106.

[2]Duffy, J., Crane, C. D., Correa, J. C., (2002), “Static

Analysis of Tensegrity Structures PART 1. Equilibrium

Equations,” Proceedings of DETC’02ASME 2002 Design

Engineering Technical Conferences and Computers and

Information in Engineering Conference Montreal, Canada,

pp. 1-10.

[3]Duffy, J., Crane, C. D., Correa, J. C., (2002), ”Static

Analysis of Tensegrity Structures PART 2. Numerical

Examples,” Proceedings of DETC’02 ASME 2002 Design

Engineering Technical Conferences and Computers and

Information in Engineering Conference Montreal, Canada,

pp. 1-10.

[4]Edmondson, A., (1987) A Fuller Explanation: The

SynergeticGeometry of

Birkhauser, Boston.

[5]Fest1, E., Shea, K., Domer, B., and Smith, I. “Adjustable

Tensegrity Structure,” journal of structural engineering,

pp. 515-526 ©ASCE ,April 2003.

[6]Fuller, R., (1975), Synergetics: The Geometry of Thinking,

MacMillan Publishing Co., Inc., NewYork.

[7]Garcia, C.B. and Li, T.Y., (1980), “On the Number

Solutions to Polynomial Systems of Equations,” SIAM J.

Numer.Anal., Vol. 17, pp. 540-546.

[8]Kenner, H., (1976), Geodesic Math and How to Use It,

University of California Press, Berkeley and Los Angeles,

CA.

[9]Knight, B.F., (2000), Deployable Antenna Kinematics

usingTensegrityStructure

University of Florida, Gainesville, FL.

R. BuckminsterFuller,

Design,

Ph.D.thesis,

[10] Morgan, A.P., (1983), “A Method for Computing All

Solutions to Systems of Polynomial Equations,” ACM

Trans. Math. Software, Vol. 9, No. 1, pp 1-17.

[11] Morgan,A.P.(1986),

Polynomial Systems,” Appl. Math. Comput., Vol. 18, pp.

87-92.

[12] Morgan, A.P., (1987), “Solving Polynomial Systems

UsingContinuationfor

Problems”, Prentice-Hall, Englewood Cliffs, NJ.

[13] Roth,B.,Whiteley,W.,

Transactions American Mathematics Society, Vol.265, pp.

419-446, 1981.

[14] Skelton, R. E., Williamson, D., and Han, J. H.,

“Equilibrium Conditions

Proceedings of the Third World Conference on Structural

Control (3WCSC) Como, Italy, pp. 1-18, April 7-12,

2002.

[15] Stern, I.P., (1999), Development of Design Equations for

Self-Deployable N-Strut Tensegrity Systems, M.S. thesis,

University of Florida, Gainesville, FL.

[16] Tibert, G., “Deployable Tensegrity Structures for Space

Applications,” Ph.D. Thesis.

Technology, Stockholm 2002.

[17] Tobie, R.S., (1976), A Report on an Inquiry into The

Existence, Formation and Representation of Tensile

Structures, Master of Industrial Design thesis, Pratt

Institute, NewYork.

[18] Tsai, L., (1999), “Robot Analysis; The Mechanics of

Serial and Parallel Manipulators,” John Wiley.

[19] Verschelde, J. (1999), “PHCpack: a General-Purpose

Solver forPolynomial

Continuation,” Algorithm 795 in ACM Trans. Math.

Softw.,

http://www.math.uic.edu/~jan/PHCpack/phcpack.html .

[20] Wampler, C., Morgan, A., and Sommese, A., (1990),

“Numerical Continuation Mehtods for Solving Polynomial

Systems Arising in Kinematics,” ASME J. Mech. Des.,

Vol. 112, pp. 59-68.

[21] Yin, J., Duffy, J., and Crane, C., (2002), An Analysis for

the Design of Self-Deployable Tensegrity and Reinforced

Tensegrity Prisms with Elastic Ties, International Journal

of Robotics and Automation, Special Issue on Compliance

and Compliant Mechanisms, Volume 17, Issue 1.

“AHomotopyforSolving

Scientific and Engineering

“Tensegrity Framework,”

of Tensegrity Structure,”

Royal Instituteof

Systems by Homotopy