Kinematic Analysis of a Planar Tensegrity Mechanism with Pre-Stressed Springs

KINEMATIC ANALYSIS OF A PLANAR TENSEGRITY MECHANISM WITH PRE-STRESSED SPRINGS

By VISHESH VIKAS

A THESIS PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE

UNIVERSITY OF FLORIDA

2008

1 °c 2008 Vishesh Vikas

2 Vakratunda mahakaaya Koti soorya samaprabhaa Nirvighnam kurume deva Sarva karyeshu sarvadaa.

3 TABLE OF CONTENTS page LIST OF TABLES ...... 5

LIST OF FIGURES ...... 6

ACKNOWLEDGMENTS ...... 7

ABSTRACT ...... 8

CHAPTER 1 INTRODUCTION ...... 9 2 PROBLEM STATEMENT AND APPROACH ...... 12

3 BOTH FREE LENGTHS ARE ZERO ...... 17 3.1 Equilibrium Analysis ...... 17 3.2 Numerical Example ...... 19 4 ONE FREE LENGTH IS ZERO ...... 21

4.1 Equilibrium Analysis ...... 21 4.2 Numerical Example ...... 24 5 BOTH FREE LENGTHS ARE NON-ZERO ...... 28 5.1 Equilibrium Analysis ...... 28 5.2 Numerical Example ...... 31

6 CONCLUSION ...... 36

APPENDIX A SHORT INTRODUCTION TO THEORY OF SCREWS ...... 37 B SYLVESTER MATRIX ...... 40

REFERENCES ...... 44 BIOGRAPHICAL SKETCH ...... 45

4 LIST OF TABLES Table page 3-1 Six solutions for Case 1 ...... 20

4-1 List of operations to obtain Sylvester Matrix for case 2 ...... 23

4-2 Twenty solutions for Case 2 ...... 25

5-1 List of operations to obtain Sylvester Matrix for Case 3 ...... 34

5-2 Twenty-four solutions for Case 3 ...... 35

5 LIST OF FIGURES Figure page 1-1 Biological model of the knee ...... 11

1-2 Tensegrity based model of cross-section of knee ...... 11

2-1 Tensegrity mechanism ...... 12

3-1 Four real solutions for Case 1 ...... 20

4-1 Bifurcation diagram for solution of x1 and varying parameter L01 ...... 26 4-2 Eight real solutions for Case 2 ...... 27 5-1 Twenty-four real solutions for Case 3 (cases 1 to 12) ...... 32

5-2 Twenty-four real solutions for Case 3 (cases 13 to 24) ...... 33

6 ACKNOWLEDGMENTS I would like to express profound gratitude to my advisor, Prof. Carl Crane, for his invaluable support, encouragement, supervision and useful suggestions throughout this research work. His moral support and continuous guidance enabled me to complete my work successfully. I would like to thank Prof. John Schueller and Prof. Warren Dixon for serving on my committee. I am also grateful to Prof. Jay Gopalakrishnan and Dr. Jahan Bayat for listening to my queries and answering my questions regarding the research work. I would like to acknowledge the support of the Department of Energy under grant number

DE-FG04-86NE37967. I am especially indebted to my parents, Dr. Om Vikas and Mrs. Pramod Kumari

Sharma, for their love and support ever since my childhood. I also wish to thank my brother, Pranav, for his constant support and encouragement. I thank my fellow students at the Center for Intelligent Machines and Robotics. From them, I learned a great deal and found great friendships. I would also like to thank Nicole, Piyush, Rakesh, Rashi and Sreenivas for their friendships.

7 Abstract of Thesis Presented to the Graduate School of the University of Florida in Partial Fulﬁllment of the Requirements for the Degree of Master of Science KINEMATIC ANALYSIS OF A PLANAR TENSEGRITY MECHANISM WITH PRE-STRESSED SPRINGS By Vishesh Vikas August 2008 Chair: Carl Crane Major: Mechanical Engineering

This thesis presents the equilibrium analysis of a planar tensegrity mechanism. The device consists of a base and top platform that are connected in parallel by one connector leg (whose length can be controlled via a prismatic joint) and two spring elements whose linear spring constants and free lengths are known. The thesis presents three cases: 1) the spring free lengths are both zero, 2) one of the spring free lengths is zero and the other is nonzero, and 3) both free lengths are nonzero. The purpose of the thesis is to show the enormous increase in complexity that results from nonzero free lengths. It is shown that six equilibrium conﬁgurations exist for Case 1, twenty equilibrium conﬁgurations exist for

Case 2, and no more than sixty two conﬁgurations exist for Case 3.

8 CHAPTER 1 INTRODUCTION The human musculoskeletal system is often described as combinations of levers and

pulleys. However, at a number of places (particularly the spine) this lever-pulley-fulcrum model of the musculoskeletal system calculates such extreme amount of forces that will tear muscles oﬀ the bones and shear bones into pieces. This, however, does not happen in real life and can be explained by the concept of ‘tensegrity’. Tensegrity (abbreviation

of ‘tensional integrity [1], [2]), is synergy between tension and compression. Tensegrity structures consist of elements that can resist compression (e.g., struts, bones) and elements that can resist tension (e.g., ties, muscles). The entire conﬁguration stands by itself and maintains its form (equilibrium) solely because of the internal arrangement of the

struts and ties ([3],[4]). No pair of struts touch and the end of each strut is connected to three non-coplanar ties ([5]). More formal deﬁnition of tensegrity is given by Roth and Whiteley([6]), introducing a third element, the bar, which can withstand both compression and tension. Tensegrity structures can be broadly classiﬁed into two categories, prestressed and geodesic, where continuous transmission of tensional forces is necessary for shape

stability or single entity of these structures([7]). ‘Prestressed tensegrity structures’, hold their joints in position as the result of a pre-existing tensile stress within the structural network. ‘Geodesic tensegrity structures’, triangulate their structure members and

orient them along geodesics (minimal paths) to geometrically constrain movement. Our bodies provide a familiar example of a prestressed tensegrity structure: our bones act like struts to resist the pull of tensile muscles, tendons and ligaments, and the shape stability (stiﬀness) of our bodies varies depending on the tone (pre-stress) in our muscles. Examples of geodesic tensegrity structures include Fullers geodesic domes, carbon-based buckminsterfullerenes (buckyballs), and tetrahedral space frames, which are of great interest in astronautics because they maintain their stability in the absence of gravity

and, hence, without continuous compression. Idea of combining several basic tensegrity

9 structures to form a more complex structure has been analyzed([8], [9]) and diﬀerent methods to do so have also been studied([10]). There has been a rapid development in static and dynamic analysis of tensegrity

structures in last few decades ([11]). This is due to its beneﬁts over traditional approaches in several ﬁelds such as architecture ([12]), civil engineering, art, geometry and even bi-

ology. Beneﬁts of tensegrity structures are examined by Skelton et al.([13]). Tensegrity structures display energy eﬃciency as its elements store energy in form of compression or tension; as a result of the energy stored in the structure, the overall energy required to activate these structures will be small([14]). Since compressive members in tensegrity structures are disjoint, large displacements are allowed and it is possible to create deployable structures that can be stored in small volumes. Deployable antennas and masts are notable space applications ([15],[16]). Kenner established the relation between the rotation of the top and bottom ties. Tobie ([3]), presented procedures for the generation of tensile

structures by physical and graphical means. Yin ([5]) obtained Kenner’s ([17]) results using energy considerations and found the equilibrium position for unloaded tensegrity prisms. Stern ([18]) 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 ([19]) addressed the problem of stability of tensegrity structures for the design of deployable antennae. On macro level, tensegrity structures are used to model human musculoskeletal system, deployable antennae, architecture structures, etc.; on cellular level, Donald

Ingber ([7]) proposes Cellular Tensegrity Theory in which the whole cell is modelled as a prestressed tensegrity structure, although geodesic structures are also found in the cell

at smaller size scales. Stephen Levin ([14]) proposed a truss-tensegrity model of the spine mechanics. Beneﬁts of tensegrity structures make them interesting for designing mobile robots. Aldrich ([20]) has built and controlled robots based on tensegrity structures and Paul et al. ([21]) have built triangular prism based mobile tensegrity robot.s

10 Most of the papers in the field of tensegrity assume zero free length of springs. In real life systems, esp. biological systems, such assumption does not hold true. Following research shows that this assumption is not trivial, complexities to find solutions increase tremendously and the number of static equilibrium configurations also increase. Cross- section modeling of the human knee joint (Figure 1-1, 1-2) is the biological motivation of the mechanism in the following research. In the following research, a planar pre-stressed tensegrity structure is examined. The structure consists of three struts connected by two ties.

Figure 1-1. Biological model of the knee

Figure 1-2. Tensegrity based model of cross-section of knee

11 CHAPTER 2 PROBLEM STATEMENT AND APPROACH

The mechanism analyzed here is shown in Figure 2-1. The top platform (indicated by points 4, 5, and 6) is connected to the base platform (indicated by points 1, 2, and 3) by two spring elements whose free lengths are L01 and L02 and by a variable length connector whose length is referred to as L3. Although this does not match the exact deﬁnition of tensegrity, the device is prestressed in the same manner as a tensegrity mechanism. The exact problem statement is as follows: Given:

L12 distance between points 1 and 2

p3x, p3y coordinates of point 3 in coordinate system 1

L45 distance between points 4 and 5

p6x, p6y coordinates of point 6 in coordinate system 2

L3 distance between points 1 and 4

k1,L01 spring constant and free length of spring 1

k2,L02 spring constant and free length of spring 2 Find: All static equilibrium conﬁgurations

Figure 2-1. Tensegrity mechanism

12 A solution approach of satisfying force and moment conditions for equilibrium is

considered. It is apparent that since the length L3 is given, the device has two degrees of freedom. Thus, there are two descriptive parameters that must be selected in order to

deﬁne the system. For this analysis, the descriptive parameters are chosen as the angles

γ1 and γ2 which are respectively the angle between X axis of coordinate system attached to the base and line deﬁned by points 1 and 4 and angle between X axis of coordinate

system attached to the base and line deﬁned by points 4 and 5 as shown in Figure 2-1. Other parameters were investigated, but none yielded a less complicated solution than is presented here.

The co-ordinates of points 1 to 6 are        0   L12   p3x        B   B   B   P1 =  0  P2 =  0  P3 =  p3y  (2–1)       0 0 0        0   L45   p6x        T   T   T   P4 =  0  P5 =  0  P6 =  p6y  (2–2)       0 0 0

Where the superscripts B, T denote the co-ordinate systems ﬁxed on the reference frames of the bottom platform and top platforms respectively. It is also known that    c1    B   P4 = L3  s1  (2–3)   0    c2 −s2 0    B   T R =  s2 c2 0  (2–4)   0 0 1

13 where ci, si are abbreviations for cos γi, sin γi (i = 1, 2) respectively. Points 5, 6 may be calculated as

B B B T Pj = P4 + T R Pj j = 5, 6 . (2–5)

Calculating coordinates of points 5, 6 gives      L3c1 + L45c2   L3c1 + p6xc2 + p6ys2      B   B   P5 =  L3s1 − L45s2  , P6 =  L3s1 − p6xs2 + p6yc2  . (2–6)     0 0

The Pl¨ucker coordinates of the three lines are        S01   S02   S03  $1 =   , $2 =   , $3 =   (2–7) S01L S02L S03L

and

B B B B B P5 − P2 L1 B P2 × P5 S01 = B B = , S01L = P2 × S01 = (2–8) | P5 − P2| d1 d1 B B B B B P6 − P3 L1 B P3 × P6 S02 = B B = , S02L = P3 × S02 = (2–9) | P6 − P3| d2 d2 B B P4 P4 S03 = B = , S03L = 0 . (2–10) | P4| L3

Forces in the three legs are f1, f2, f3 which may be expressed as

f1 = k1(d1 − L01) (2–11)

f2 = k2(d2 − L02) (2–12)

2 B B 2 d1 = || P5 − P2|| (2–13)

2 B B 2 d2 = || P6 − P3|| . (2–14)

Forces in the springs and connector are wrenches with zero pitch (i.e., pure forces along the direction of the respective lines). For static equilibrium, the sum of the three wrenches

must be equal to zero.

f1$1 + f2$2 + f3$3 = 0 . (2–15)

14 Substituting Equations 2–7 to 2–12 into Equation 2–15 gives

B B f3 B k1(1 − λ1) L1 + k2(1 − λ2) L2 + P4 = 0 (2–16) L3 B B B B k1(1 − λ1)( P2 × P5) + k2(1 − λ2)( P3 × P6) = 0 (2–17)

where λ1 = L01/d1, λ2 = L02/d2. As the problem is planar, vector Equation 2–16 is equivalent to two scalar equations and vector Equation 2–17 is equivalent to one scalar

equation. Therefore, there are three unknowns (d1, d2, f3) and three equations. One

scalar equation in d1, d2 can be obtained from Equation 2–16 by performing a cross

B product operation with P4 to both sides of the equation to yield

B B B B k1(1 − λ1) P4 × L1 + k2(1 − λ2) P4 × L1 = 0 . (2–18)

It is desired to solve for γ1, γ2 from Equations 2–13, 2–14, 2–17 and 2–18. Substituting Equations 2–1, 2–3 and 2–6 into these equations yields.

k1(1 − λ1)(c1s2L45 − s1c2L45 + s1L12) + = 0 (2–19) k2(1 − λ2)(c1(s2p6x + c2p6y − p3y) − s1(c2p6x − s2p6y − p3x))

k1(1 − λ1)L12 (s1L3 + s2L45) + = 0 (2–20) k2(1 − λ2)(p3x(s1L3 + s2p6x + c2p6y) − p3y(c1L3 + c2p6x − s2p6y))

2 d2 + ((2p3y − 2s2p6x − 2c2p6y)s1 + (2p3x − 2c2p6x + 2s2p6y)c1)L3 = 0 (2–21) 2 2 2 2 −(p3x + p3y + p6y + p6x) + (2p6xp3y − 2p6yp3x)s2 + (2p6xp3x + 2p6yp3y)c2 2 2 2 2 d1 − L12 − L3 + ((2L12 − 2c2L45)c1 − 2s1s2L45)L3 + 2c2L45L12 − L45 = 0 . (2–22)

The equations can now be analyzed for three diﬀerent cases -

1. Both Free lengths are zero. (i.e., λ1 = λ2 = 0)

2. One Free lengths is zero, another non-zero. (i.e., λ1 = 0, λ2 6= 0)

3. Both Free lengths are non-zero. (i.e., λ1 6= 0, λ2 6= 0)

Equations 2–19 to 2–22 are nonlinear functions of sin γ1, cos γ1, sin γ2 and cos γ2. The concept of ‘tan-half angles’, ([22]) converts these four equations into four nonlinear

15 polynomial equations. It deﬁnes

γ x = tan i for i = 1, 2 (2–23) i 2

thus, 2 2xi 1 − xi si = 2 , ci = 2 for i = 1, 2 . (2–24) 1 + xi 1 + xi Each of the four cases will be analyzed in the chapters that follow.

16 CHAPTER 3 BOTH FREE LENGTHS ARE ZERO

3.1 Equilibrium Analysis

Zero free length for both the springs (L01 = L02 = 0) implies

λ1 = λ2 = 0 . (3–1)

It is important to observe that Equations 2–19 and 2–20 are decoupled from Equations 2–

22, 2–21 (i.e., terms containing d1, d2 vanish). Substituting Equation 3–1 into Equations 2–19 and 2–20 gives

k2p6y(c1c2 + s1s2) + (k1L45 + k2p6x)(s2c1 − s1c2) + (k2p3x + k1L12)s1 − k2p3yc1 = 0 (3–2)

(k1L3L12 + k2L3p3x)s1 + (k1L12L45 + k2p3yp6y + k2p3xp6x)s2 − k2L3c1p3y = 0 . (3–3) +(k2p3xp6y − k2p3yp6x)c2

Applying ‘tan-half angle concept’,(Equation 2–24) to Equations 3–2 and 3–3 yields

2 2 2 2 (A1x2 + A2x2 + A3)x1 + (A4x2 + A5x2 + A6)x1 + (A7x2 + A8x2 + A9) = 0 (3–4)

2 2 2 2 (B1x2 + B2x2 + B3)x1 + (B4x2 + B5x2 + B6)x1 + (B7x2 + B8x2 + B9) = 0 (3–5)

where Ai,Bi(i = 1, 2, ..., 9) are deﬁned in terms of the given parameters as

A1=k2(p3y + p6y) B1=k2(p3yp6x − p3xp6y + L3p3y)

A2=−2(k2p6x + k1L45) B2=2(k1L12L45 + k2(p3yp6y + p3xp6x))

A3=k2(p3y − p6y) B3=k2(p3xp6y + L3p3y − p3yp6x)

A4=2((L45 + L12)k1 + (p6x + p3x)k2) B4=2k1L3L12 + 2k2L3p3x

A5=4k2p6y B5=0

A6=2((L12 − L45)k1 + (p3x − p6x)k2) B6=B4

A7=−(p6yp3y)k2 B7=−B3

A8=2(k1L45 + k2p6x) B8=B2

A9=(p6y − p3y)k2 B9=−B1 .

17 Equations 3–4, 3–5 can be rewritten as

2 P1x1 + P2x1 + P3 = 0 (3–6)

2 Q1x1 + Q2x1 + Q3 = 0 (3–7)

where

2 2 2 P1 = (A1x2 + A2x2 + A3),P2 = (A4x2 + A5x2 + A6),P3 = (A7x2 + A8x2 + A9)(3–8)

2 2 2 Q1 = (B1x2 + B2x2 + B3),Q2 = (B4x2 + B5x2 + B6),Q3 = (B7x2 + B8x2 + B9) .(3–9)

We form the Sylvester’s Matrix(Appendix B) by multiplying Equations 3–6, 3–7 with x1

and write       P P P 0 x3 0  1 2 3   1             2     0 P1 P2 P3   x1   0      =   . (3–10)        Q1 Q2 Q3 0   x1   0       

0 Q1 Q2 Q3 1 0 The determinant of Sylvester matrix is zero for common roots of Equations 3–6, 3–7 and

thus ¯ ¯ ¯ ¯ ¯ ¯ ¯ P1 P2 P3 0 ¯ ¯ ¯ ¯ ¯ ¯ 0 P1 P2 P3 ¯ f (x ) = ¯ ¯ = 0 . (3–11) ZFL 2 ¯ ¯ ¯ Q1 Q2 Q3 0 ¯ ¯ ¯ ¯ ¯ ¯ 0 Q1 Q2 Q3 ¯

Expansion of Equation 3–11 (fZFL(x2)) yields an eighth degree polynomial in x2. It was found that this eighth degree polynomial could be divided symbolically by the

2 term (1 + x2) without any remainder resulting in a sixth degree polynomial in x2. The coeﬃcients of this polynomial have been obtained symbolically, but are not listed here due to their length.

18 Values of x1 that correspond to each of the six solutions of x2 can be determined from observing    −1   3  x1   P1 P2 P3   0         2  =     . (3–12)  x1   0 P1 P2   −P3        x1 Q1 Q2 Q3 0

The corresponding value for x1 is the 3rd value of the solution vector on the left side of

Equation 3–12. Unique corresponding values for γ1 and γ2 are calculated for each value of

x1 and x2 from Equation 2–23

−1 γi = 2 tan (xi) i = 1, 2 (3–13)

3.2 Numerical Example

The following values were selected for a numerical example

L12 = 10 m

p3x = −3 m, p3y = 7 m

L45 = 5 m

p6x = −1.1990 m, p6y = −2.2790 m

L3 = 7.56 m

k1 = 1.5 N/m, L01 = 0 m

k2 = 3.7 N/m, L02 = 0 m

Coeﬃcients Ai,Bi(i = 1,... 9) are evaluated numerically and a sixth degree poly-

2 nomial in x2 is obtained by expanding Equation 3–11 and dividing it by (1 + x2). Six

solutions of γ1, γ2 are listed in Table 3-1. The four real solutions are shown in Figure 3-1. The two complex solutions are shown to satisfy Equations 3–4 and 3–5.

19 Table 3-1. Six solutions for Case 1

γ1 (radians) γ2 (radians) 1 +0.66861 −0.63297 2 −1.16166 −0.20769 3 −0.51037 −0.05764 −1.7079i −1.0903i 4 −0.51037 −0.05764 +1.7079i +1.0903i 5 +0.81733 +1.88221 6 −1.57259 +2.54351

Figure 3-1. Four real solutions for Case 1

20 CHAPTER 4 ONE FREE LENGTH IS ZERO

4.1 Equilibrium Analysis

The free length of one spring is zero (L01 = 0) implies

λ1 = L01/d1, λ2 = 0 . (4–1)

It is important to observe that Equations 2–19, 2–20 are coupled only to Equation 2–22

(i.e., the terms containing d2 vanish) and these equations may be written as

((k1L45 + k2p6x)(s2c1 − s1c2) + k2p6y(c1c2 + s1s2) = 0 (4–2) +k1L12s1 + k2(p3xs1 − p3yc1))d1 + k1L01(L45(s1c2 − c1s2) − L12s1)

(L3(k1L12 + k2p3x − k2L3p3yc1)s1 + k2(p3xp6y − p3yp6x)c2 = 0 (4–3) +(k1L12L45 + k2(p3yp6y + p3xp6x))s2)d1 − k1L12L01L3s1 − k1L12L01s2L45 2 2 2 2 d1 + (2L3L12 − 2L3c2L45)c1 − L12 + 2c2L45L12 − 2L3s1s2L45 − L45 − L3 = 0.(4–4)

Applying the ‘tan-half angle concept’ (Equation 2–24), to Equations 4–5, 4–6 and 4–7, the following equations are obtained

A1d1 + A2 = 0 (4–5)

B1d1 + B2 = 0 (4–6)

2 C1d1 + C2 = 0 (4–7) where

2 2 A1 = D1x2 + D2x2 + D3 A2 = D4x2 + D5x2 + D6 (4–8)

2 2 B1 = E1x2 + E2x2 + E3 B2 = E4x2 + E5x2 + E6 (4–9)

2 2 C1 = F1x2 + F2x2 + F3 C2 = F4x2 + F5x2 + F6 (4–10)

21 2 D1 = k2(p3y + p6y)x1 + 2(k1(L45 + L12) + k2(p6x + p3x))x1 − k2(p6y + p3y)

2 D2 = −2(k1L45 + k2p6x)x1 + 4k2p6yx1 + 2(k1L45 + k2p6x)

2 D3 = k2(p3y − p6y)(x1 − 1) + 2(k2(p3x − p6x) + k1(L12 − L45))x1

D4 = −2L01k1(L45 + L12)x1

2 D5 = 2k1L01L45(x1 − 1)

D6 = 2k1L01(L45 − L12)x1 (4–11)

2 k2(p3y(p6x + L3) − p3xp6y)x1 + 2L3(k1L12 + k2p3x)x1 E1 = −k2(p3y(L3 − p6x) + p3xp6y) 2 E2 = 2(1 + x1)(k1L12L45 + k2p3yp6y + k2p3xp6x)

2 k2(p3y(L3 − p6x) + p3xp6y)x1 + 2L3(k1L12 + k2p3x)x1 E3 = −k2(p3y(L3 + p6x) − p3xp6y)

E4 = −2k1L12L01L3x1

2 E5 = −2k1L12L01L45(1 + x1)

E6 = −2k1L12L01L3x1 (4–12)

2 F1 = 1 + x1

F2 = 0

2 F3 = 1 + x1

2 2 2 2 2 F4 = 2(L45 + L12)L3(1 − x1) − (L3 + L45 + 2L45L12 + L12)(1 + x1)

F5 = −8L3x1L45

2 2 2 2 2 F6 = (2L12 − 2L45)L3(1 − x1) − (L3 − 2L45L12 + L45 + L12)(1 + x1) . (4–13)

It is desired to form a system of equations

SOFLXOFL = 0 (4–14)

where SOFL, XOFL are respectively the square matrix(Sylvester matrix) and vector of ‘unknown coeﬃcients. They can be obtained by performing the operations listed in Table

22 Table 4-1. List of operations to obtain Sylvester Matrix for case 2 Equations # Equations # Unknowns Added Unknowns 2 {(4–5),(4–6)} 3 9 (x2, x2, 1) 2 (4–7)(x2, x2, 1) · d1 2 2 (x2, x2, 1) · d1 {(4–5),(4–6)}·d1 5 9 2 2 3 {(4–5),(4–6)}·d1 8 12 (x2, x2, 1) · d1 (4–7)·d1 3 {(4–5),(4–6)}·x2 16 16 x2 3 (4–7)·x2 x2 · d1 3 2 {(4–5),(4–6)}·d1x2 x2 · d1 2 3 3 {(4–5),(4–6)}·d1x2 x2 · d1 (4–7)·d1

4.1 The matrix SOFL and vector XOFL are written as    D6 0 0 0 0 0 D3 0 0 D5 0 0 D4 D1 D2 0       E6 0 0 0 0 0 E3 0 0 E5 0 0 E4 E1 E2 0       F6 0 0 F2 0 0 0 0 F1 F5 0 0 F4 0 0 F3         0 0 0 D2 0 0 D6 0 D1 0 0 0 0 D4 D5 D3       0 0 0 E2 0 0 E6 0 E1 0 0 0 0 E4 E5 E3       0 0 0 D5 0 0 0 D3 D4 0 D1 D2 0 0 0 D6         0 0 0 E5 0 0 0 E3 E4 0 E1 E2 0 0 0 E6       0 0 0 0 0 0 F6 F3 0 0 F1 F2 0 F4 F5 0  S =   OFL    0 D1 0 0 D4 0 0 0 0 D6 0 0 D5 D2 D3 0       0 E 0 0 E 0 0 0 0 E 0 0 E E E 0   1 4 6 5 2 3       0 0 F1 F3 F4 0 0 0 F2 F6 0 0 F5 0 0 0       0 D4 D1 D3 0 0 0 0 D2 0 0 0 0 D5 D6 0       0 E E E 0 0 0 0 E 0 0 0 0 E E 0   4 1 3 2 5 6       0 0 D4 D6 0 D1 0 0 D5 0 D2 D3 0 0 0 0       0 0 E4 E6 0 E1 0 0 E5 0 E2 E3 0 0 0 0   

0 F4 0 0 0 F1 0 0 0 0 F2 F3 0 F5 F6 0 (4–15)

23 £ 3 2 3 2 3 3 3 3 2 2 2 3 3 2 2 2¤T XOFL = 1, x2d1, d1x2, x2d1, x2, x2d1, d1, d1, d1x2, x2, x2d1, x2d1, x2, x2d1, x2d1, d1 (4–16)

The determinant of the Sylvester matrix is zero for common roots of Equations 4–5, 4–6,

4–7 and thus

fOFL(x1) =| SOFL |= 0 . (4–17)

This results in a 32 degree polynomial in x1. The solutions for x1 can be obtained by e solving this polynomial equation. To calculate corresponding values of x2, d1, let SOFL be the same as SOFL without its first column and last row, SÔFL be first column of SOFL b without its last element and XOFL be same as XOFL without its first row. Thus,     b e  SOFL(15×1) SOFL(15×15)   1(1×1)  SOFL(16×16) =   , XOFL(16×1) =   (4–18) b ×(1×1) ×(1×15) XOFL(15×1)

Thus, the ﬁrst 15 equations of Equation 4–14 may be written as

e b b SOFLXOFL = −SOFL (4–19)

e b b Each solution for x1 is substituted into SOFL, SOFL and XOFL is solved as

b e −1b XOFL = −SOFL SOFL . (4–20)

Values of d1 and x2 which correspond to a solution of x1 are the 6th and 9th elements of b vector XOFL. 4.2 Numerical Example

Values of the given parameters were selected to be the same as in the previous

example (Sec 3.2) with the exception of the free length of spring 1 which is now set to

L01 = 2.3 m. Equations 4–11, 4–12, 4–13 are evaluated and substituted into Equation

4–15. Equation 4–17 is obtained and solved to obtain values of x1. Corresponding values of x2, d1 are obtained from Equation 4–20. Solutions of Equation 4–17 were as follows

• Eight of the solutions of x1 were either ±i. Thus, Equation 4–17 may be divided 2 4 through by (1 + x1) resulting in a 24th degree polynomial.

24 Table 4-2. Twenty solutions for Case 2

x1 x2 d1 1 −2.880366 − 0.411872i −0.307013 − 1.456473i +26.02618 + 1.476993i 2 −2.880366 + 0.411872i −0.307013 + 1.456473i +26.02618 − 1.476993i 3 −1.833882 − 0.139340i −1.464914 − 0.949033i −20.35270 − 1.239320i 4 −1.833882 + 0.139340i −1.464914 + 0.949033i −20.35270 + 1.239320i 5 −1.674852 + 0.000000i +72.13693 + 0.000000i −19.69388 + 0.000000i 6 −1.492124 + 0.000000i −4.068392 + 0.000000i +19.64973 + 0.000000i 7 −0.944817 + 0.000000i +0.519148 + 0.000000i −7.534905 + 0.000000i 8 −0.928707 − 0.056278i +0.272806 − 0.408042i −3.794682 − 1.228946i 9 −0.928707 + 0.056278i +0.272806 + 0.408042i −3.794682 + 1.228946i 10 −0.766478 + 0.000000i +0.459257 + 0.000000i +5.926762 + 0.000000i 11 −0.172483 − 0.153043i +0.835170 + 0.288061i −0.516827 + 0.424489i 12 −0.172483 + 0.153043i +0.835170 − 0.288061i −0.516827 − 0.424489i 13 +0.169610 − 0.150949i −0.828087 + 0.306801i +0.450188 + 0.304645i 14 +0.169610 + 0.150949i −0.828087 − 0.306801i +0.450188 − 0.304645i 15 +0.711628 + 0.000000i −0.455891 + 0.000000i −5.417888 + 0.000000i 16 +0.843106 + 0.000000i +2.406830 + 0.000000i −16.46027 + 0.000000i 17 +0.869530 + 0.000000i −0.425974 + 0.000000i +6.719935 + 0.000000i 18 +1.002755 − 0.055558i −0.276958 − 0.470503i +1.780312 − 0.214965i 19 +1.002755 + 0.055558i −0.276958 + 0.470503i +1.780312 + 0.214965i 20 +1.091412 + 0.000000i +2.885304 + 0.000000i +18.04674 + 0.000000i

• Four solutions correspond to when points 2, 5 are coincident (i.e., d1 = 0). Value of x1 for d1 = 0 may be determined from system of equations Equations 4–5, 4–6, 4–5       2 D4 D5 D6 x2 0  E4 E5 E6   x2  =  0  (4–21) F F F 1 0 | 4 {z5 6 }

Sd1=0

fOF Ld1 (x1) =| Sd1=0 |= 0 (4–22) 2 2 2 2 2 2 2 ⇒ ((L3 − L45 + L12 + 2L3L12)x1 + L3 + L12 − L45 − 2L3L12) = 0 (4–23) 2 2 2 2 Remaining 24th degree polynomial can be divided by [(L3 − L45 + L12 + 2L3L12)x1 + 2 2 2 2 L3 + L12 − L45 − 2L3L12] resulting in a 20th degree polynomial.

• Remaining 20 solutions satisfy Equations 4–5, 4–6, 4–7. For this particular numerical example there are 8 real and 12 complex solution sets for x1, x2, d1.

A bifurcation diagram between the solution x1 and varying parameter L01 is shown in

Figure 4-1. It is interesting to observe that the four solutions for L01 = L02 = 0 case

bifurcate as the free length L01 is varied. At L01 = 2.3 m, the total number of solutions

25 Figure 4-1. Bifurcation diagram for solution of x1 and varying parameter L01

are eight (as numerically calculated), at some value of L01, the number of solutions increase to 9 or 10 also. Thus, it is possible to move from one value of x1 to another by slowly changing the free-length.

26 Figure 4-2. Eight real solutions for Case 2

27 CHAPTER 5 BOTH FREE LENGTHS ARE NON-ZERO

5.1 Equilibrium Analysis

Both non-zero free length implies

λ1 = L01/d1, λ2 = L02/d2 . (5–1)

It is important to observe that Equations 2–19, 2–20 are coupled to both Equations 2–22,

2–21. After applying the ‘tan-half angle concept’(Equation 2–24), the following equations are obtained.

2 2 2 (F1x2 + F2x2 + F3)d1 + (F4x2 + F5x2 + F6)d2 + (F7x2 + F8x2 + F9)d1d2 = 0 (5–2)

2 2 2 (G1x2 + G2x2 + G3)d1 + (G4x2 + G5x2 + G6)d2 + (G7x2 + G8x2 + G9)d1d2 = 0 (5–3)

2 2 2 (H1x2 + H2x2 + H3)d1 + (H4x2 + H5x2 + H6) = 0 (5–4)

2 2 2 (I1x2 + I2x2 + I3)d2 + (I4x2 + I5x2 + I6) = 0 (5–5) where

£ 2 ¤ F1 = L02k2 (p6y + p3y)(1 − x1) − 2(p6x + p3x)x1

¡ 2 ¢ F2 = 2L02k2 p6x(x1 − 1) − 2p6yx1

£ 2 ¤ F3 = L02k2 (p3y − p6y)(1 − x1) − 2(p3x − p6x)x1

F4 = −2(L45 + L12)L01k1x1

2 F5 = 2k1L01L45(x1 − 1)

F6 = 2(L45 − L12)L01k1x1

2 F7 = (p6y + p3y)k2(x1 − 1) + 2 [(p3x + p6x)k2 + (L45 + L12)k1] x1

2 F8 = 2(k2p6x + k1L45)(1 − x1) + 4k2p6yx1

2 F9 = (p3y − p6y)k2(x1 − 1) + 2 [(p3x − p6x)k2 + (L12 − L45)k1] x1 (5–6)

28 £ 2 2 ¤ G1 = k2L02 (1 + x1)(p3xp6y − p3yp6x) − 2L3p3xx1 + (1 − x1)L3p3y

£ 2 ¤ G2 = −2k2L02 (x1 + 1)(p3xp6x + p3yp6y)

£ 2 2 ¤ G3 = k2L02 (1 + x1)(p3yp6x − p3xp6y) − 2L3p3xx1 + (1 − x1)L3p3y

G4 = −2k1L12L01L3x1

2 G5 = −2(x1 + 1)k1L45L12L01

G6 = −2k1L12L01L3x1

£ 2 ¤ 2 G7 = 2k1x1L12 + (2p3xx1 − (1 − x1)p3y)k2 L3 + (1 + x1)(p3yp6x − p3xp6y)k2

2 2 G8 = 2(x1 + 1)k1L45L12 + 2(x1 + 1)(p3xp6x + p3yp6y)k2

£ 2 ¤ 2 G9 = 2k1x1L12 + (2p3xx1 − (1 − x1)p3y)k2 L3 + (1 + x1)(p3xp6y − p3yp6x)k2 (5–7)

2 H1 = (1 + x1)

H2 = 0

2 H3 = (1 + x1)

2 2 H4 = (L3 + L12 + L45) (x1 + 1) + 4(L12 + L45)L3

H5 = −8L3x1L45

2 2 H6 = (L3 + L12 − L45) (x1 + 1) + 4(L12 − L45)L3 (5–8)

2 I1 = (1 + x1)

I2 = 0

2 I3 = (1 + x1)

£ 2 2¤ 2 I4 = (L3 + p3x + p6x) + (p3y + p6y) (x1 + 1) + 4 [(p3y + p6x)x1 + p3x + p6x] L3

2 I5 = 4(p3yp6x − p3xp6y − L3p6y)(x1 + 1) + 8(p6y − p6xx1)L3

£ 2 2¤ 2 I6 = (L3 + p3x − p6x) + (p3y − p6y) (x1 + 1) + 4[(p3y − p6y)x1 + p3x − p6x]L3 (5–9).

It is desired to form a system of equations

SBFLXBFL = 0 (5–10)

29 where SBFL, XBFL are respectively the square matrix(Sylvester matrix) and vector of unknown coeﬃcients. To obtain minimum dimension of SBFL, Equations 5–2, 5–3 are

2 2 divided by d1d2, 5–4 is divided by d1, 5–5 is divided by d2. They are rewritten as

2 2 2 (F1x2 + F2x2 + F3)d2i + (F4x2 + F5x2 + F6)d1i + (F7x2 + F8x2 + F9) = 0 (5–11)

2 2 2 (G1x2 + G2x2 + G3)d2i + (G4x2 + G5x2 + G6)d1i + (G7x2 + G8x2 + G9) = 0 (5–12)

2 2 2 (H1x2 + H2x2 + H3) + (H4x2 + H5x2 + H6)d1i = 0 (5–13)

2 2 2 (I1x2 + I2x2 + I3) + (I4x2 + I5x2 + I6)d2i = 0 (5–14)

where d1i = 1/d1, d2i = 1/d2. SBFL, XBFL can be obtained by performing the operations listed in Table 5.1 The determinant of the Sylvester Matrix is zero for common roots of

Equations 5–11, 5–12, 5–13, 5–14 and thus

fBFL(x1) =| SBFL |= 0 . (5–15)

This results in a 104 degree polynomial in x1. The solutions for x1 can be obtained by e solving this polynomial equation. To calculate corresponding values of x2, d1, d2, let SBFL be the same as SBFL without its first column and last row, SˆBFL be first column of SBFL b without its last element and XBFL be same as XBFL without its first element(which is 1).     b e  SBFL(51×1) SBFL(51×51)   1(1×1)  SBFL(52×52) =   , XBFL(16×1) =   (5–16) b ×(1×1) ×(1×51) XBFL(51×1)

Thus, the ﬁrst 51 equations of Equation 5–10 may be written as

e b b SBFLXBFL = −SBFL (5–17)

e b b Each solution for x1 is substituted into SBFL, SBFL and XBFL is solved as

b e −1b ⇒ XBFL = −SBFL SBFL (5–18)

b d1, d2, x2 are elements of vector XBFL.

30 5.2 Numerical Example

Values of the given parameters were selected to be the same as in the previous example (Sec 3.2) with the exception of the free lengths of spring 1 and 2 which are now set to L01 = 5.1 m, L02 = 6.6309 m. Equations 5–6, 5–7, 5–8, 5–9 are calculated numerically. Next, the Sylvester Matrix (SBFL) was obtained by operations explained in

Section (5.1). Finally, Equation 5–15 is solved for x1. Observations for solutions of x1 were

• Circular points at inﬁnity - 26 of the solutions were equal to ±i. It must be the case 2 13 that the 104 degree polynomial can be divided by (1 + x1) . It is not surprising as 2 H1,H3,I1,I3 are equal to (1 + x1).

• Complex Solutions - 38 of the remaining 78 solutions were complex.

• Real, ‘extraneous Solutions - 16 of the solutions were real but did not satisfy Equations 5–2, 5–3, 5–4, 5–5.

• Real, ‘relevant Solutions - 24 of the solutions were real and satisﬁed Equations 5–2, 5–3, 5–4, 5–5. For two of these solutions (Case 3 and Case 6) - f1, f2 are zero.

31 (a) Case 1 (b) Case 2 (c) Case 3 - f1 = f2 = 0

(d) Case 4 (e) Case 5 (f) Case 6 - f1 = f2 = 0

(g) Case 7 (h) Case 8 (i) Case 9

(j) Case 10 (k) Case 11 (l) Case 12

Figure 5-1. Twenty-four real solutions for Case 3 (cases 1 to 12)

32 (a) Case 13 (b) Case 14 (c) Case 15

(d) Case 16 (e) Case 17 (f) Case 18

(g) Case 19 (h) Case 20 (i) Case 21

(j) Case 22 (k) Case 23 (l) Case 24

Figure 5-2. Twenty-four real solutions for Case 3 (cases 13 to 24)

33 Table 5-1. List of operations to obtain Sylvester Matrix for Case 3 Equations # Equations # Unknowns Added Unknowns 2 {(5–11),(5–12)} 4 15 (x2, x2, 1) 2 (5–13)(x2, x2, 1) · d1i 2 (5–14)(x2, x2, 1) · d2i 2 2 (x2, x2, 1) · d1i 2 2 (x2, x2, 1) · d2i 2 {(5–11),(5–12)}·d1i 8 18 (x2, x2, 1) · d1id2i {(5–11),(5–12)}·d2i 2 2 {(5–11),(5–12)}·d1id2i 12 24 (x2, x2, 1) · d1id2i 2 2 (5–13)·d2i (x2, x2, 1) · d1id2i (5–14)·d1i 2 2 3 {(5–11),(5–12)}·d1i 15 27 (x2, x2, 1) · d1i (5–13)·d1i 2 2 3 {(5–11),(5–12)}·d2i 18 30 (x2, x2, 1) · d2i (5–14)·d1i 2 2 3 {(5–11),(5–12)}·d1id2i 26 39 (x2, x2, 1) · d1id2i 2 2 3 {(5–11),(5–12)}·d1id2i (x2, x2, 1) · d1id2i 2 2 2 2 (5–13)·d2i (x2, x2, 1) · d1id2i 2 (5–14)·d1i (5–13)·d1id2i (5–14)·d1id2i 3 {(5–11),(5–12)}·x2 52 52 x2 3 (5–13)·x2 x2 · d1i 3 (5–14)·x2 x2 · d2i 3 2 {(5–11),(5–12)}·d2ix2 x2 · d1i 3 2 {(5–11),(5–12)}·d1id2ix2 x2 · d2i 3 (5–13)·d2ix2 x2 · d1id2i 3 2 (5–14)·d1ix2 x2 · d1id2i 2 3 2 {(5–11),(5–12)}·d1ix2 x2 · d1id2i 3 3 (5–13)·d1ix2 x2 · d1i 2 3 3 {(5–11),(5–12)}·d2ix2 x2 · d2i 3 3 (5–14)·d1ix2 x2 · d1id2i 2 3 3 {(5–11),(5–12)}·d1id2ix2 x2 · d1id2i 2 3 2 2 {(5–11),(5–12)}·d1id2ix2 x2 · d1id2i 2 (5–13)·d2ix2 2 (5–14)·d1ix2 (5–13)·d1id2ix2 (5–14)·d1id2ix2

34 Table 5-2. Twenty-four solutions for Case 3

x1 x2 d1 d2 1 +0.06791 +0.47666 −7.15173 +14.4690 2 +0.24086 −0.41749 +0.27141 +8.44826 3 +0.37697 −1.16995 +5.10000 +6.63085 4 +0.46257 −0.50356 +2.75137 +5.59366 5 +0.48104 +0.81960 +12.8823 +10.3232 6 +0.49871 −0.09772 +5.10000 +6.63085 7 +1.08109 −0.49303 −8.34619 +0.16126 8 +1.17125 −0.79791 +10.4030 +1.31517 9 +1.24786 −0.02641 +9.74132 +1.83447 10 +1.24970 −0.23039 −8.84363 +1.36263 11 +1.58864 −0.24201 −9.91765 +2.90077 12 +1.88396 −31.1760 +20.1231 −1.62330 13 +1.91068 +0.38270 −20.5096 −1.25716 14 +2.01111 +0.85549 +20.0955 −1.23264 15 +2.13470 +2.97004 −20.7903 −0.50454 16 +2.46232 −0.17160 +11.2958 +5.53768 17 −0.64549 +0.47906 +4.80076 +17.7501 18 −0.65507 +0.47724 +4.89600 −17.7508 19 −0.92267 +0.48824 −7.26801 −17.5157 20 −1.02457 −2.77350 +17.6764 −12.2530 21 −1.34644 +2.15516 +15.7880 +14.0808 22 −1.39506 −3.69685 −19.3426 −11.5960 23 −1.42815 −0.58529 +15.3028 +14.3383 24 −1.46480 +0.25896 −9.56783 +16.6180

35 CHAPTER 6 CONCLUSION The purpose of this thesis was to show the significant increase in complexity that results when springs with nonzero free lengths are incorporated in pre-stressed mechanisms. It has been shown that six equilibrium configurations exist for the case of a simple planar mechanism with two springs where both springs have zero free lengths. Twenty equilibrium configurations were found for the case where one of the springs had a nonzero free length. For the case where both springs had nonzero free lengths, seventy eight solutions sets were obtained once the circular points at infinity were disregarded. Sixteen of these seventy eight, did not satisfy the equation set which means that the presented elimination technique introduced extraneous roots. The remaining sixty two solutions satisfied the equations, but two solutions in the numerical example resulted in cases where the lines along the three legs did not intersect which is puzzling. Additional work needs to be done before this simple case is fully understood. The approach presented here does however bound the dimension of the solution. The goal is to extend this work to spatial devices in order to develop a thorough understanding of the nature of these pre-stressed mechanisms. The work can also be easily extended to study the human musculoskeletal system and microlevel study of cellular hardware using the cellular tensegrity theory.

36 APPENDIX A SHORT INTRODUCTION TO THEORY OF SCREWS Screw theory was developed by Sir Robert Stawell Ball in 1876, for application in kinematics and statics of mechanisms (rigid body mechanics). It is a way to express displacements, velocities, forces and torques in three dimensional space, combining both rotational and translational parts. The Theory of Screws is founded upon two celebrated theorems [23]. One relates to the displacement of a rigid body. The other relates to forces which act on a rigid body.

1. Reduction of the displacement of a rigid body to its simplest form. Fundamental theorem discovered by Chasles states Any given displacement of a rigid body can be eﬀected by a rotation about an axis combined with a translation parallel to that axis.

2. Reduction of of a system of forces applied to a rigid body to its simplest form. Fundamental theorem discovered by Poinsot states A force, and a couple in a plane perpendicular to the force, constitute an adequate representation of any system of forces applied to a rigid body. Pl¨ucker coordinates were introduced by Julius Pl¨ucker in the 19th century as a way to assign six homogenous coordinates to each line in projective 3-space. In Screw Theory, they are used to represent the coordinates of screws, twists and wrenches.

37 Deﬁning some basic terms used in Screw Theory

• Pitch of screw. Rectilinear distance through which the nut is translated parallel to axis of the screw, while the nut is rotated through the angular unit of circular measure. Pitch is thus a linear magnitude

• Screw. A straight line with which a deﬁnite linear magnitude termed the pitch is associated. In rigid body dynamics, velocities of a rigid body and the forces and torques acting upon it can be represented by the concept of a screw.

• Twist. A screw representing the velocity of a body. A body is said to receive a twist when it is rotated uniformly about the screw, while it is translated parallel to the screw, through a distance equal to the product of the pitch and circular measure of angle of rotation.

• Wrench. A screw representing forces and torques on a body. It denotes a force and couple in a plane perpendicular to the force. One way to conceptualize this is to consider someone who is fastening two wooden boards together with a metal screw. The person turns the screw (applies a torque), which then experiences a net force along its axis of rotation. A straight line can be deﬁned by two points. Assuming two points(1, 2) with point

vectors r1, r2 the equation of an arbitrary point lying on the line made by points 1, 2 can be written as

(r − r ) S = 2 1 (A–1) |r2 − r1|

(r − r1) × S = 0 (A–2)

⇒ r × S = r1 × S . (A–3)

Pl¨ucker coordinates of the line are {S; S0L} and they satisfy the following constraints

|S| = 1, S · S0L = 0 . (A–4)

It should be noted that dimensions of S and S0L are diﬀerent. Also, only four algebraic quantities are required to deﬁne a line due to the two constraints

|S| = 1, S · S0L = 0 . (A–5)

38 The representation of a screw is    S  $ =   . (A–6) S0L + hS

Five quantities are required to specify a screw, of these 4 are required to specify a line. The ﬁfth one in the pitch of the screw, h.

A twist is represented as θ · $1. A twist requires six algebraic quantities for its complete specification, of these five are required for complete specification of a screw. The sixth quantity, the amplitude of twist (θ) expresses the angle of rotation. The distance of translation is the product of amplitude of twist and pitch of the screw. If pitch is zero, the twist reduces to pure rotation around the screw $. If pitch is infinite, then finite twist is not possible except the amplitude be zero, in which case the twist reduces to pure translation parallel to the screw $.

A wrench is represented as f · $1. A wrench requires six algebraic quantities for its complete specification, of these five are required for complete specification of a screw. The sixth quantity, the intensity of wrench (f) expresses the magnitude of force. The moment of couple is the product of intensity of wrench and pitch of the screw. If pitch is zero, the wrench reduces to pure force along the screw $. If pitch is infinite the wrench reduces to couple in a plane perpendicular to the screw $.

39 APPENDIX B SYLVESTER MATRIX In mathematics, a Sylvester matrix (named after English Mathematician James

Joseph Sylvester) is a matrix associated to two polynomials that gives some information about those polynomials. If two polynomials have a common factor, then the determinant of the associated Sylvester Matrix is equal to zero. Given polynomials

n n−1 p(x) = anx + an−1x . . . a1x + a0 (B–1)

m m−1 q(x) = bmx + bm−1x . . . b1x + b0 (B–2)

of degrees n and m and roots αi, i = 1, 2, . . . n, βi, i = 1, 2, . . . m respectively. The resultant([24]) is deﬁned by

Yn Ym m n R(p, q) = an bm (αi − βj) (B–3) i=1 j=1

This is also given by the determinant of the corresponding Sylvester matrix. It can be

observed that for the resultant to be zero, the determinant of the Sylvester matrix should vanish. To construct the Sylvester matrix for the system p(x) = 0, q(x) = 0, equations of the form xkp(x) = 0 and xkq(x) = 0 may be added to the system. The enlarged system will have exactly the same solutions as the original system of two equations. Consider the

40 system of equations

p(x) = 0

xp(x) = 0 . . . . x(m−1)p(x) = 0, (B–4) q(x) = 0 . . . . x(n−1)q(x) = 0 .

The system may be written as a matrix equation       (n+m−1)  an an−1 ······ 0 0   x   0         ......   .   .   ......   .   .               0 0 ······ a a   x(m−1)   0   1 0          =   . (B–5)    (m−2)     bm bm−1 ······ 0 0   x   0         ......   .   .   ......   .   .        0 0 ······ b1 b0 1 0 | {z } Sp,q

The Sylvester matrix(Sp,q) associated with polynomials p(x) and q(x) is a square matrix of dimension (n + m) × (n + m). The determinant of the Sylvester matrix will vanish when p(x) and q(x) have a common root. The converse is also true. In order for there to be a common root for Eqns (B–1) and (B–2) , it is necessary that

det(Sp,q) = 0 . (B–6)

41 The concept of Sylvester Matrix can be extended to more than 2 equations. More importantly, it can be extended to more than one variable. Given a set of n polynomial

equations in m variables (p1(x1, ··· xm), ··· , pn(x1, ··· xm), it is possible to construct

Sylvester Matrix of p dimension by multiplying equations by combinations of x1, ·, xm e.g., x1x2 · p1(x1, ··· , xm). There is no deﬁnite algorithm to construct the Sylvester Matrix; the process solely depends on the nature of the polynomial equations. The Sylvester Matrix with the minimum dimension yields a non-zero determinant in the embedded variable.

Let S1(xi) is the minimum dimension Sylvester Matrix for the given set of n polynomial equations, then, for any Sylvester Matrix Sj(xi) with a greater dimension than that of

S1(xi), determinant of the Sylvester Matrix vanishes and does not give any information about the embedded variable. So, the Sylvester Matrix is unique by its dimension. More precisely, the determinant of the Sylvester Matrix is unique. To construct this minimum dimension Sylvester Matrix, it may be required to perform tricks on the set of given equations and in the process, change the nature of the polynomial. As the theory for Sylvester Matrix for multivariable, multi-polynomial-equation system is not developed, there may be introduction of ‘extraneous solutions’, that do not satisfy the set of given equations.

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44 BIOGRAPHICAL SKETCH Vishesh Vikas was born on the 31st of May, 1983 in New Delhi, India. He attended his high school at Delhi Public School R K Puram, Delhi. He recieved his Bachelor in

Technology in Mechanical Engineering from Indian Institute of Technology, Guwahati in May of 2005. After that, he worked at MAIA, INRIA Lorraine(LORIA), France. In 2007, he joined the Center for Intelligent Machines and Robotics(CIMAR) at the University of Florida, completing his Masters of Science degree in Mechanical Engineering in August of 2008. Upon completion of his MS, Vishesh will pursue PhD in Department of Aerospace and Mechanical Engineering at University of Florida.