18
The Dynamics of theClass 1 Shell
Tensegrity Structure
18.1 Introduction18.2 Tensegrity Definitions
A Typical Element • Rules of Closure for the Shell Class
18.3 Dynamics of a Two-Rod Element18.4 Choice of Independent Variables and
Coordinate Transformations18.5 Tendon Forces18.6 ConclusionAppendix 18.A Proof of Theorem 18.1Appendix 18.B Algebraic Inversion of the Q MatrixAppendix 18.C General Case for (n, m) = (i, 1)Appendix 18.D Example Case (n,m) = (3,1)Appendix 18.E Nodal Forces
Abstract
A tensegrity structure is a special truss structure in a stable equilibrium with selected membersdesignated for only tension loading, and the members in tension forming a continuous network ofcables separated by a set of compressive members. This chapter develops an explicit analyticalmodel of the nonlinear dynamics of a large class of tensegrity structures constructed of rigid rodsconnected by a continuous network of elastic cables. The kinematics are described by positionsand velocities of the ends of the rigid rods; hence, the use of angular velocities of each rod is avoided.
The model yields an analytical expression for accelerations of all rods, making the model efficientfor simulation, because the update and inversion of a nonlinear mass matrix are not required. Themodel is intended for shape control and design of deployable structures. Indeed, the explicitanalytical expressions are provided herein for the study of stable equilibria and controllability, butcontrol issues are not treated.
18.1 Introduction
The history of structural design can be divided into four eras classified by design objectives. In theprehistoric era, which produced such structures as Stonehenge, the objective was simply to opposegravity, to take static loads. The classical era, considered the dynamic
response and placed designconstraints on the eigenvectors as well as eigenvalues. In the modern era, design constraints couldbe so demanding that the dynamic response objectives require feedback control. In this era, the
Robert E. Skelton
University of California, San Diego
Jean-Paul Pinaud
University of California, San Diego
D. L. Mingori
University of California, Los Angeles
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control discipline followed the classical structure design, where the structure and control disciplineswere ingredients in a multidisciplinary system design, but no interdisciplinary tools were developedto integrate the design of the structure and the control. Hence, in this modern era, the dynamics ofthe structure and control were not cooperating to the fullest extent possible. The post-modern eraof structural systems is identified by attempts to unify the structure and control design for a commonobjective.
The ultimate performance capability of many new products and systems cannot be achieved untilmathematical tools exist that can extract the full measure of cooperation possible between thedynamics of all components (structural components, controls, sensors, actuators, etc.). This requiresnew research. Control theory describes how the design of one component (the controller) shouldbe influenced by the (given) dynamics of all other components. However, in systems design, wheremore than one component remains to be designed, there is inadequate theory to suggest how thedynamics of two or more components should influence each other at the design stage. In the future,controlled structures will not be conceived merely as multidisciplinary design steps, where a plate,beam, or shell is first designed, followed by the addition of control actuation. Rather, controlledstructures will be conceived as an interdisciplinary process in which both material architecture andfeedback information architecture will be jointly determined. New paradigms for material andstructure design might be found to help unify the disciplines. Such a search motivates this work.Preliminary work on the integration of structure and control design appears in Skelton
1,2
andGrigoriadis et al.
3
Bendsoe and others
4-7
optimize structures by beginning with a solid brick and deleting finiteelements until minimal mass or other objective functions are extremized. But, a very importantfactor in determining performance is the paradigm used for structure design. This chapter describesthe dynamics of a structural system composed of axially loaded compression members and tendonmembers that easily allow the unification of structure and control functions. Sensing and actuatingfunctions can sense or control the tension or the length of tension members. Under the assumptionthat the axial loads are much smaller than the buckling loads, we treat the rods as rigid bodies.Because all members experience only axial loads, the mathematical model is more accurate thanmodels of systems with members in bending. This unidirectional loading of members is a distinctadvantage of our paradigm, since it eliminates many nonlinearities that plague other controlledstructural concepts: hysteresis, friction, deadzones, and backlash.
It has been known since the middle of the 20th century that continua cannot explain the strengthof materials. While science can now observe at the nanoscale to witness the architecture of materialspreferred by nature, we cannot yet design or manufacture manmade materials that duplicate theincredible structural efficiencies of natural systems. Nature’s strongest fiber, the spider fiber,arranges simple nontoxic materials (amino acids) into a microstructure that contains a continuousnetwork of members in tension (amorphous strains) and a discontinuous set of members in com-pression (the
β
-pleated sheets in Figure 18.1).
8,9
This class of structure, with a continuous network of tension members and a discontinuousnetwork of compression members, will be called a Class 1 tensegrity structure. The importantlessons learned from the tensegrity structure of the spider fiber are that
1. Structural members never reverse their role. The compressive members never take tensionand, of course, tension members never take compression.
2. Compressive members do not touch (there are no joints in the structure).3. Tensile strength is largely determined by the local topology of tension and compressive
members.
Another example from nature, with important lessons for our new paradigms is the carbonnanotube often called the Fullerene (or Buckytube), which is a derivative of the Buckyball. Imagine
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a 1-atom thick sheet of a graphene, which has hexagonal holes due to the arrangements of materialat the atomic level (see Figure 18.2). Now imagine that the flat sheet is closed into a tube bychoosing an axis about which the sheet is closed to form a tube. A specific set of rules must definethis closure which takes the sheet to a tube, and the electrical and mechanical properties of theresulting tube depend on the rules of closure (axis of wrap, relative to the local hexagonal topol-ogy).
10
Smalley won the Nobel Prize in 1996 for these insights into the Fullerenes. The spider fiberand the Fullerene provide the motivation to construct manmade materials whose overall mechanical,thermal, and electrical properties can be predetermined by choosing the local topology and therules of closure which generate the three-dimensional structure from a given local topology. Bycombining these motivations from Fullerenes with the tensegrity architecture of the spider fiber,this chapter derives the static and dynamic models of a shell class of tensegrity structures. Futurepapers will exploit the control advantages of such structures. The existing literature on tensegritydeals mainly
11-23
with some elementary work on dynamics in Skelton and Sultan,
24
Skelton andHe,
25
and Murakami et al.
26
FIGURE 18.1
Nature’s strongest fiber: the Spider Fiber. (From Termonia, Y.,
Macromolecules
, 27, 7378–7381,1994. Reprinted with permission from the American Chemical Society.)
FIGURE 18.2
Buckytubes.
amorphouschain
β-pleated sheet
entanglement
hydrogen bond
yz
x6nm
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18.2 Tensegrity Definitions
Kenneth Snelson built the first tensegrity structure in 1948 (Figure 18.3) and Buckminster Fullercoined the word “tensegrity.” For 50 years tensegrity has existed as an art form with some archi-tectural appeal, but engineering use has been hampered by the lack of models for the dynamics.In fact, engineering use of tensegrity was doubted by the inventor himself. Kenneth Snelson in aletter to R. Motro said, “As I see it, this type of structure, at least in its purest form, is not likelyto prove highly efficient or utilitarian.” This statement might partially explain why no one botheredto develop math models to convert the art form into engineering practice. We seek to use scienceto prove the artist wrong, that his invention is indeed more valuable than the artistic scope that heascribed to it. Mathematical models are essential design tools to make engineered products. Thischapter provides a dynamical model of a class of tensegrity structures that is appropriate for spacestructures.
We derive the nonlinear equations of motion for space structures that can be deployed or heldto a precise shape by feedback control, although control is beyond the scope of this chapter. Forengineering purposes, more precise definitions of tensegrity are needed.
One can imagine a truss as a structure whose compressive members are all connected with balljoints so that no torques can be transmitted. Of course, tension members connected to compressivemembers do not transmit torques, so that our truss is composed of members experiencing nomoments. The following definitions are useful.
Definition 18.1
A given configuration of a structure is in a
stable equilibrium
if, in the absenceof external forces, an arbitrarily small initial deformation returns to the given configuration.
Definition 18.2
A tensegrity structure is a stable system of axially loaded members.
Definition 18.3
A stable structure is said to be a “Class 1” tensegrity structure if the membersin tension form a continuous network, and the members in compression form a discontinuous setof members.
FIGURE 18.3
Needle Tower of Kenneth Snelson, Class 1 tensegrity. Kröller Müller Museum, The Netherlands.(From Connelly, R. and Beck, A.,
American Scientist
, 86(2), 143, 1998. With permissions.)
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Definition 18.4
A stable structure is said to be a “Class 2” tense grity structure if the membersin tension form a continuous set of members, and there are at most tw o members in compressionconnected to each node.
Figure 18.4 illustrates Class 1 and Class 2 tensegrity structures.Consider the topology of structural members given in Figure 18.5, where thick lines indicate
rigid rods which tak e compressi ve loads and the thin lines represent tendons. This is a Class 1tense grity structure.
Definition 18.5
Let the topology of Figure 18.5 describe a three-dimensional structure by con-necting points A to A, B to B, C to C,…, I to I. This constitutes a “Class 1 tense grity shell” if thereexists a set of tensions in all tendons (
α
= 1
→
10,
β
= 1
→
n,
γ
= 1
→
m) such that thestructure is in a stable equilibrium.
FIGURE 18.4
Class 1 and Class 2 tense grity structures.
FIGURE 18.5
Topology of an (8,4) Class 1 tense grity shell.
1 2
tαβγ ,
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18.2.1 A Typical Element
The axial members in Figure 18.5 illustrate only the pattern of member connections and not theactual loaded configuration. The purpose of this section is two-fold: (i) to define a typical “element”which can be repeated to generate all elements, and (ii) to define rules of closure that will generatea “shell” type of structure.
Consider the members that make the typical
ij
element where
i
= 1, 2, …, n indexes the elementto the left, and
j
= 1, 2, …, m indexes the element up the page in Figure 18.5. We describe theaxial elements by vectors. That is, the vectors describing the
ij
element, are
t
1
ij
,
t
2
ij
, …
t
10
ij
and
r
1
ij
,
r
2
ij
, where, within the
ij
element,
t
α
ij
is a vector whose tail is fixed at the specified end of tendonnumber
α
, and the head of the vector is fixed at the other end of tendon number
α
as shown inFigure 18.6 where
α
= 1, 2, …, 10. The
ij
element has two compressive members we call “rods,”shaded in Figure 18.6. Within the
ij
element the vector
r
1
ij
lies along the rod r
1
ij
and the vector
r
2
ij
lies along the rod
r
2
ij
. The first goal of this chapter is to derive the equations of motion for thedynamics of the two rods in the
ij
element. The second goal is to write the dynamics for the entiresystem composed of
nm
elements. Figures 18.5 and 18.7 illustrate these closure rules for the case(
n,
m
) = (8,4) and (
n, m
) = (3,1).
Lemma 18.1
Consider the structure of Figure 18.5 with elements defined by Figure 18.6. A Class 2tensegrity shell is formed by adding constraints such that for all i =
1, 2, …
,
n, and for m > j >
1,
FIGURE 18.6
A typical
ij
element.
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(18.1)
This closes nodes n
2ij
and n
1(i+1)(j+1)
to a single node, and closes nodes n
3(i–1)j
and n
4i(j–1)
to a singlenode (with ball joints). The nodes are closed outside the rod, so that all tension elements are onthe exterior of the tensegrity structure and the rods are in the interior.
The point here is that a Class 2 shell can be obtained as a special case of the Class 1 shell, byimposing constraints (18.1). To create a tensegrity structure not all tendons in Figure 18.5 arenecessary. The following definition eliminates tendons
t
9
ij
and
t
10
ij
, (i
= 1
→
n, j = 1
→
m).
Definition 18.6
Consider the shell of Figures 18.4. and 18.5, which may be Class 1 or Class 2depending on whether constraints (18.1) are applied. In the absence of dotted tendons (labeled t
9
and t
10
), this is called a primal tensegrity shell. When all tendons t
9
, t
10
are present in Figure 18.5,it is called simply a Class 1 or Class 2 tensegrity shell.
The remainder of this chapter focuses on the general Class 1 shell of Figures 18.5 and 18.6.
18.2.2 Rules of Closure for the Shell Class
Each tendon exerts a positive force away from a node and
f
αβγ
is the force exerted by tendon
t
αβγ
and denotes the force vector acting on the node
n
α
ij
. All
f
α
ij
forces are postive in the directionof the arrows in Figure 18.6, where
w
α
ij
is the external applied force at node
n
α
ij
,
α
= 1, 2, 3, 4. Atthe base, the rules of closure, from Figures 18.5 and 18.6, are
t
9
i
1
= –
t
1
i
1
,
i
= 1, 2, …,
n
(18.2)
t
6
i
0
=
0
(18.3)
t
600
= –
t2n1 (18.4)
t901 = t9n1 = –t1n1 (18.5)
0 = t10(i–1)0 = t5i0 = t7i0 = t7(i–1)0, i = 1, 2, …, n. (18.6)
FIGURE 18.7 Class 1 shell: (n,m) = (3,1).
− + =+ =+ =+ =
t t
t t
t t
t t
1 4
2 3
5 6
7 8
ij ij
ij ij
ij ij
ij ij
0
0
0
0
,
,
,
.
fαij
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At the top, the closure rules are
t10im = –t7im (18.7)
t100m = –t70m = –t7nm (18.8)
t2i(m+1) = 0 (18.9)
0 = t1i(m+1) = t9i(m+1) = t3(i+1)(m+1)
= t1(i+1)(m+1) = t2(i+1)(m+1). (18.10)
At the closure of the circumference (where i = 1):
t90j = t9nj, t60(j–1) = t6n(j–1), t70(j–1) = t7n(j–1) (18.11)
t80j = t8nj, t70j = t7nj, t100(j–1) = t10n(j–1). (18.12)
From Figures 18.5 and 18.6, when j = 1, then
0 = f7i(j–1) = f7(i–1)(j–1) = f5i(j–1) = f10(i–1)(j–1), (18.13)
and for j = m where,
0 = f1i(m+1) = f9i(m+1) = f3(i+1)(m+1) = f1(i+1)(m+1). (18.14)
Nodes n11j, n3nj, n41j for j = 1, 2, …, m are involved in the longitudinal “zipper” that closes thestructure in circumference. The forces at these nodes are written explicitly to illustrate the closurerules.
In 18.4, rod dynamics will be expressed in terms of sums and differences of the nodal forces,so the forces acting on each node are presented in the following form, convenient for later use.The definitions of the matrices Bi are found in Appendix 18.E.
The forces acting on the nodes can be written in vector form:
f = Bd f d + Bo f o + Wow (18.15)
where
Wo = BlockDiag [ ,W1, W1, ],
f
f
f
f
f
f
f
f
f
f
w
w
w
=
=
=
=
11
21 1
MM
M M
m
d
d
d
md
o
o
mo
m
, , , ,
L L
B
B B
B B
B
B B
B B
B
B B
B
B
B
d o=
=
3 4
5 6
5
6 4
5 8
1 2
7
2
7
0 0
0 0
0 0
0 0
0
0
0 0
L
O O M
O O
M O O
K
L
O O M
M O O O
M O O
L L
,
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and
(18.16)
Now that we have an expression for the forces, let us write the dynamics.
18.3 Dynamics of a Two-Rod Element
Any discussion of rigid body dynamics should properly begin with some decision on how themotion of each body is to be described. A common way to describe rigid body orientation is touse three successive angular rotations to define the orientation of three mutually orthogonal axesfixed in the body. The measure numbers of the angular velocity of the body may then be expressedin terms of these angles and their time derivatives.
This approach must be reconsidered when the body of interest is idealized as a rod. The reasonis that the concept of “body fixed axes” becomes ambiguous. Two different sets of axes with acommon axis along the rod can be considered equally “body fixed” in the sense that all massparticles of the rod have zero velocity in both sets. This remains true even if relative rotation isallowed along the common axis. The angular velocity of the rod is also ill defined because thecomponent of angular velocity along the rod axis is arbitrary. For these reasons, we are motivatedto seek a kinematical description which avoids introducing “body-fixed” reference frames andangular velocity. This objective may be accomplished by describing the configuration of the systemin terms of vectors located only the end points of the rods. In this case, no angles are used.
We will use the following notational conventions. Lower case, bold-faced symbols with anunderline indicate vector quantities with magnitude and direction in three-dimensional space. Theseare the usual vector quantities we are familiar with from elementary dynamics. The same bold-faced symbols without an underline indicate a matrix whose elements are scalars. Sometimes wealso need to introduce matrices whose elements are vectors. These quantities are indicated with anupper case symbol that is both bold faced and underlined.
As an example of this notation, a position vector can be expressed as
In this expression, pi is a column matrix whose elements are the measure numbers of for the mutuallyorthogonal inertial unit vectors e1, e2, and e3. Similarly, we may represent a force vector as
Matrix notation will be used in most of the development to follow.
ff
f f
f
f
f
f
f
f
f
f
w
w
w
w
w
ijo
ijijd
ij
ij
ij
=
=
=
5
1
2
3
4
6
7
8
9
10
1
2
3
4
, ,
p e e e Epi
i
i
i
i
p
p
p
=
=[ ] .1 2 3
1
2
3
pi
fi
ˆ ˆ .f Efi i=
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We now consider a single rod as shown in Figure 18.8 with nodal forces and applied tothe ends of the rod.
The following theorem will be fundamental to our development.
Theorem 18.1 Given a rigid rod of constant mass m and constant length L, the governingequations may be described as:
(18.17)
where
The notation denotes the skew symmetric matrix formed from the elements of r:
and the square of this matrix is
The matrix elements r1, r2, r3, q1, q2, q3, etc. are to be interpreted as the measure numbers of thecorresponding vectors for an orthogonal set of inertially fixed unit vectors e1, e2, and e3. Thus,using the convention introduced earlier,
r = Er, = Eq, etc.
FIGURE 18.8 A single rigid rod.
f1 f 2
˙ ˜q Kq Hf+ =
q
p p
p p=
=
+−
1
2
1 2
2 1
ff f
f f H =
I
q
~
~=
+−
=
−
ˆ ˆ
ˆ ˆ , ,˙ ˙
.1 2
1 2
3
322 2
2 2 3
2
2m LL
T
0
0K
0 0
0 q q I
r~
r r = ~ =−
−−
0
0
0
3 2
3 1
2 1
1
2
3
r r
r r
r r
r
r
r
,
r~2 =− −
− −− −
r r r r r r
r r r r r r
r r r r r r
22
32
1 2 1 3
2 1 12
32
2 3
3 1 3 2 12
22
.
q
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The proof of Theorem 18.1 is given in Appendix 18.A. This theorem provides the basis of ourdynamic model for the shell class of tensegrity structures.
Now consider the dynamics of the two-rod element of the Class 1 tensegrity shell in Figure 18.5.Here, we assume the lengths of the rods are constant. From Theorem 18.1 and Appendix 18.A, themotion equations for the ij unit can be described as
(18.18)
(18.19)
where the mass of the rod αij is mαij and rαij = Lαij. As before, we refer everything to a commoninertial reference frame (E). Hence,
and the force vectors appear in the form
.
m
ij ij ij
ijij ij ij ij ij
ij ij ij ij
ij ij ij
ij
m
L
12 1 1 2
12 2 2 2 1
2 2 2 2
2 2 12
6
q f f
q q q f f
q q q q
q q
..
.. ^ ^
. . ..
= +
× = × −
+ =
=
^ ^
( ) ( )
. .
.
,
0
m
ij ij ij
m
ij ij ij ij ij
ij ij ij ij
ij ij ij
ij
ij
L
2
2
2
6
22
˙ ˆ ˆ
( ˙ ) (ˆ ˆ )
˙ . ˙ . ˙
.
,
q f f
q q q f f
q q q q
q q
3 3 4
4 4 4 4 3
4 4 4 4
4 4
= +
× = × −
+ =
=
0
q q q q1
11
12
13
2
21
22
23
3
31
32
33
4
41
42
43
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
q
q
q
q
q
q
q
q
q
q
q
q
∆ ∆ ∆ ∆
, , , ,
q q q q q1 1 2 3 4ij ijT
ijT
ijT
ijT
T∆ , , , ,[ ]
HI
q HI
q211
3
322
2
3
342
2 2
12
22
ijij L ij
ijij L ijm m
ij ij
=
=
0
0
0
0˜ , ˜ ,
HH
H f
f f
f f
f f
f f
ij
ij
ijij
ij ij
ij ij
ij ij
ij ij
=
+
−
+
−
1
2
1 2
1 2
3 4
3 4
0
0,
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
∆
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Using Theorem 18.1, the dynamics for the ij unit can be expressed as follows:
where
The shell system dynamics are given by
(18.20)
where f is defined in (18.15) and
18.4 Choice of Independent Variables and Coordinate Transformations
Tendon vectors tαβγ are needed to express the forces. Hence, the dynamical model will be completedby expressing the tendon forces, f, in terms of variables q. From Figures 18.6 and 18.9, it followsthat vectors and ij can be described by
(18.21)
(18.22)
To describe the geometry, we choose the independent vectors r1ij, r2ij, t5ij, for i = 1, 2, …, n, j =1, 2, …, m and ρρρρ11, t1ij, for i = 1, 2, …, n, j = 1, 2, …, m, and i < n when j = 1.
This section discusses the relationship between the q variables and the string and rod vectorstαβγ and rβij. From Figures 18.5 and 18.6, the position vectors from the origin of the reference frame,E, to the nodal points, p1ij, p2ij, p3ij, and p4ij, can be described as follows:
˙ ,q q H fij ij ij ij ij+ =ΩΩ
ΩΩ ΩΩ11
22 2 3
222
4 4 3ij
ij ijT
ijij
ij ijT
ijL L
=
=
− −
0 0
0
0 0
0˙ ˙ , ˙ ˙ ,q q I
q q I
ΩΩΩΩ
ΩΩij
ij
ij
=
1
2
0
0,
q q q q q q q= [ ]11 1 12 2 1T
nT T
nT
mT
nmT T
,..., , ,..., ,..., ,..., .
˙ ,q K q Hf+ =r
q q q q q q q
K
H = H H H H H H
= [ ]= [ ]
[ ]
11 1 12 2 1
11 1 12 2 1
11 1 12 2 1
TnT T
nT
mT
nmT T
r n n m nm
n n m nm
BlockDiag
BlockDiag
,..., , ,..., ,..., ,..., ,
,..., , ,..., ,..., ,..., ,
,..., , ,..., ,..., ,..., .
ΩΩ ΩΩ ΩΩ ΩΩ ΩΩ ΩΩ
ij
ρρ ρρij k
k
i
k
k
i
ik ik
k
j
ij
k
j
= + − + + −= =
−
=
−
=∑ ∑ ∑∑11 1 1
1
1 1
1
1
1 5
1
1
1
2
r t t t r
ˆ .ρρ ρρij ij ij ij ij = + r + t r 1 5 2−
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(18.23)
We define
(18.24)
Then,
(18.25)
FIGURE 18.9 Choice of independent variables.
p
p r
p
p r
1
2 1
3
4 2
ij ij
ij ij ij
ij ij
ij ij ij
=
= +
=
= +
ρρ
ρρ
ρρ
ρρ
ˆ
ˆ
q p p
q p p
q p p
q p p
1 2 1 1
2 2 1 1
3 4 3 1 2
4 4 3 2
2
2
ij ij ij ij ij
ij ij ij ij
ij ij ij ij ij
ij ij ij ij
= + = +
= − =
= + = +
= − =
∆∆
∆∆
∆∆
∆∆
ρρ
ρρ
r
r
r
r
ˆ
q
q
q
q
q
I I
I I
I I
I I
p
p
p
p
I I
I
I I
I
ij
ij ij
=
=−
−
=
∆∆
1
2
3
4
3 3
3 3
3 3
3 3
1
2
3
4
3 3
3
3 3
3
2
2
0 0
0 0
0 0
0 0
0 0
0 0 0
0 0
0 0 0
ij ij
ρρ
ρρr
r
1
2
ˆ
8596Ch18Frame Page 401 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
In shape control, we will later be interested in the p vector to describe all nodal points of thestructure. This relation is
p = Pq P = BlockDiag […,P1, …,P1,…] (18.26)
The equations of motion will be written in the q coordinates. Substitution of (18.21) and (18.22)into (18.24) yields the relationship between q and the independent variables t5, t1, r1, r2 as follows:
(18.27)
To put (18.27) in a matrix form, define the matrices:
and
P
I I
I I
I I
I I
I I
I I
I I
I I
1
3 3
3 3
3 3
3 3
13 3
3 3
3 3
3 3
12
=−
−
=
−
−
−
∆∆
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
q r t t t r
q r
q r t t t
1 11 1 1
1
1 1 1
2
5
1
1
1
1
1
2 1
3 11 1 1
1
1 1 1
2
5
11
1
2
2
ij k
k
i
k ik
k
j
ik
k
j
k
i
ij
ij ij
ij k
k
i
k ik
k
j
ik
k
j
k
i
= + − + +
−
=
= + − + +
= = =
−
=
−
= = ==
−
∑ ∑ ∑∑
∑ ∑ ∑∑
ρρ
ρρ
−
=
r
q r
2
4 2
ij
ij ij
l
r
r
t
t
ij
ij
ij
ij
ij
j m=
=
1
2
5
1
2 3for , ,..., ,
lr
r
t
l
t
r
r
t
( - )
11
11
111
211
511
1
1 1 1
1 1
2 1
5 1
2=
=
=
ρρ
, ,... , ,i
i
i
i
i
i nfor
l l l l l l l l= [ ]11 21 1 12 2 1
T TnT T
nT
mT
nmT T
, , , , , , , , , , ,K K K K
A
I I
I
I I I I
I
B
I I
I
I I I I
I
=−
=
−
− −
2
2 2 2
2
2 2 2
3 3
3
3 3 3 3
3
3 3
3
3 3 3 3
3
0 0
0 0 0
0 0 0
0 0
0 0 0
0 0 0
,
8596Ch18Frame Page 402 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Then (18.27) can be written simply
q = Ql, (18.28)
where the 12nm × 12nm matrix Q is composed of the 12 × 12 matrices A–H as follows:
(18.29)
n × n blocks of 12 × 12 matrices,
12n × 12n matrix,
Q22 = BlockDiag […,C, …,C],
Q32 = BlockDiag […,J, …,J],
C
I I
I
I I I
I
D
I I
I I=
−
−
=
3 3
3
3 3 3
3
3 3
3 3
2
2 2
2 2
2 2
0 0
0 0 0
0
0 0 0
0 0
0 0 0 0
0 0
0 0 0 0
,
E
I I
I I F
I I I
I I I=
−
−
=
2 2
2 2
2 2 2
2 2 2
3 3
3 3
3 3 3
3 3 3
0 0
0 0 0 0
0 0
0 0 0 0
0
0 0 0 0
0
0 0 0 0
,
J
I I
I I G
I I I
I I I=
=
−
0 0
0 0 0 0
0 0
0 0 0 0
0
0 0 0 0
0
0 0 0 0
2 2
2 2
2 2 2
2 2 2
3 3
3 3
3 3 3
3 3 3
,
Q
Q
Q Q
Q Q Q
Q Q Q Q
Q Q Q Q Q
=
11
21 22
21 32 22
21 32 32 22
21 32 32 32 22
0 0
0
L L
O M
O M
M M M M O O
,
Q
A
D B
D E B
D E E B
D E E E B
11
0
=
0 0L L L
O M
O M
O M
M M M O O
L
Q
F
D G
D E G
D E E
D E E E E G
21 =
0 0
0
L L L
O M
O M
O O M
M M M O O
8596Ch18Frame Page 403 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
where each Qij is 12n × 12n and there are m row blocks and m column blocks in Q. Appendix18.B provides an explicit expression for the inverse matrix Q, which will be needed later to expressthe tendon forces in terms of q.
Equation (18.28) provides the relationship between the selected generalized coordinates and anindependent set of the tendon and rod vectors forming l. All remaining tendon vectors may bewritten as a linear combination of l. This relation will now be established. The following equationsare written by inspection of Figures 18.5, 18.6, and 18.7 where
(18.30)
and for i = 1, 2, …, n, j = 1, 2, …,m we have
(18.31)
For j = 1 we replace t2ij with
For j = m we replace t6ij and t7ij with
where and i + n = i. Equation (18.31) has the matrix form,
t r1 1 1 1 1 11n n n= + −ρρ ρρ
t = + r
t
t t + r = r
t r
2 1 2 1
3 1
4 3 1 1 1
6 1 1 2
1ij ij i j i j
ij i j ij
ij ij ij ij ij i j
ij i j ij ij
j
j
ρρ ρρ
ρρ ρρ
ρρ ρρ
ρρ ρρ
− >
= −
= − + −
= − + <
− −
−
−
+ +
(ˆ ) , ( )
ˆ
ˆ
(ˆ ) , (
( ) ( )
( )
( )
( )( ) mm
j mij ij i j
ij ij ij ij ij ij ij
ij i j ij ij
ij i j i j ij
)
ˆ , ( )
ˆ
( )
ˆ ˆ .
( )( )
( )
( ) ( )
t
t r r t r
t r
t r
7 1 1
8 1 5 2
9 1 1
10 1 2 1
= − <
= + − = − − +
= − +
= + −
+ +
+
+ +
ρρ ρρ
ρρ ρρ
ρρ ρρ
ρρ ρρ
1
t2 1 1 1 1i i i= − +ρρ ρρ( ) .
t r r
t r
6 1 2 1 2
7 1 2 1
im i m i m im im
im im i m i m
= + − +
= − +
+ +
+ +
ˆ (ˆ )
ˆ (ˆ ).
( ) ( )
( ) ( )
ρρ ρρ
ρρ ρρ
ρρ ρρ ρρ ρρ0 j nj oj nj∆∆ ∆∆, ˆ ˆ ,
t
I I
r
ijd
ij
=
=
− −
∆∆
t
t
t
t
t
t
t
t
0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
2
3
4
6
7
8
9
10
3 3
1
ρρ
ρρrr
I
I
r
r2 1
3
3
1
2 1
+
−
− −i j i j( ) ( )
ˆ
0 0 0 0
0 0 0
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
ρρ
ρρ
8596Ch18Frame Page 404 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
(18.32)
+
−
− −
−− −
−
+
I
I
I I
I I
I
I I I
I I
I
r
r
3
3
3 3
3 3
3
3 3 3
3 3
3
1
2
0 0 0
0 0 0
0 0
0 0
0 0 0
0
0 0
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
ρρ
ρρ
ij
00 0 0
0 0 0 0
0 0 0
0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
I
I I
r
r
+I
I
3
3 3
1
2 1
3
3
−
+
ρρ
ρρ
( )i j
+ +
ρρ
ρρr
r
1
2 1 1
ˆ ,
( )( )i j
t
t
t
t
t
t
t
t
t
I
I
r
r
id
i
1
2
3
4
6
7
8
9
10 1
3
3
1∆∆
=
−
0 0 0 0
0 0 0
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
ρρ
ρρ
22 1 1
3
3
3
3 3
3
3 3 3
3 3
3
1
2
+
−
− −
−− −
−
−( )
ˆ
i
I
I
I I
I I
I
I I I
I I
I
r
r
0 0 0
0 0 0
0 0
0 0
0 0 0
0
0 0
0 0 0
3 ρρ
ρρ
+
−
+−
+
i
i
1
3
3
3 3
1
2 1 1
3
3
I
I
I I
r
r
I
I
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
0 0 0
ρρ
ρρ
( )00 0 0 0
0 0 0 0
0 0 0 0
+
ρρ
ρρr
r
1
2 1 2
ˆ ,
( )i
8596Ch18Frame Page 405 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Equation (18.25) yields
(18.33)
Hence, (18.32) and (18.33) yield
0 0
=
− −
t
t
t
t
t
t
t
t
t
I2
3
4
6
7
8
9
10
3
imd
im
II
r
r
I
I
3
1
2 1
3
3
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
+
−
−
ρρ
ρρ
( )i m
00
0 0 0
0 0 0
0 0
0 0
0 0 0
0
0 0
0 0 0
+
−
− −
−− −
−
−
ρρ
ρρr
r
I
I
I I
I I
I
I I I
I I
I
1
2 1
3
3
3
3 3
3
3 3
3 3
3
ˆ
( )i m
3
3
+− −
ρρ
ρρ
ρρ
ρρr
r
I I
I I
I
I I
r
r
1
2
3 3
3 3
3
3 3
1
2
ˆ ˆ
im
0 0 0 0
0 0 0 0
0 0 0 0
0 0
0 0
0 0 0 0
0 0 0
0 0
+( )
.
i m1
ρρ
ρρr
r
I I
I
I I
I
q1
2
12 3
12 3
3
12 3
12 3
3
ˆ
=
−
−
ij
ij
0 0
0 0 0
0 0
0 0 0
t
t
t
t
t
t
t
t
t
qijd
ij
i j=
=
∆∆
2
3
4
6
7
8
9
10
12
0 0 I I
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
3 3– –
( −− −+
1 1
12) ( )
–
–
0 0 0 0
0 0 I I
0 0 I I
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
3 3
3 3
q i j
∆∆
8596Ch18Frame Page 406 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
(18.34)
+ 12
–
–
– –
–
I I 0 0
I I 0 0
I I 0 0
0 0 I I
0 0 I I
I
3 3
3 3
3 3
3 3
3 3
33 3 3 3
3 3
3 3
3 3
3 3
I I I
I I 0 0
0 0 I I
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
I I 0 0
0 0 I I
–
– –
–
–
+
q ij
12
+
+
+ +
q
q
( )
( )( ),
–
–
i j
i j
1
1 1
12
0 0 0 0
0 0 0 0
0 0 0 0
I I 0 0
I I 0 0
0 0 0 0
0 0 0 0
0 0 0 0
3 3
3 3
t
t
t
t
t
t
t
t
t
qid
i
1
2
3
4
6
7
8
9
10 1
12
∆∆
=
−−
0 0 0 0
0 0 I I
0 0 I I
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
3 3
3 3
(( )i i− +
−−
− −−
−− −
−
+
−
1 1 1
12
12
I I 0 0
I I 0 0
I I 0 0
0 0 I I
0 0 I I
I I I I
I I 0 0
0 0 I I
I I 0 0
0 0 0 0
0 0 0 0
0
3 3
3 3
3 3
3 3
3 3
3 3 3 3
3 3
3 3
3 3
q
00 0 0
0 0 0 0
0 0 0 0
I I 0 0
0 0 I I
0 0 0 0
0 0 0 0
0 0 0 0
I I 0 0
I I 0 0
0 0 0 0
0 0 0 0
0 0 0 03 3
3 3
3 3
3 3
−
+−
−
+q( )i 1 1
12
+q( ) ,i 1 2
8596Ch18Frame Page 407 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Also, from (18.30) and (18.32)
(18.35)
With the obvious definitions of the 24 × 12 matrices E1, E2, E3, E4, Ê4, , E5, equations in(18.34) are written in the form, where q01 = qn1, q(n+1)j = qij,
(18.36)
=
− −
t
t
t
t
t
t
t
t
t
imd
im
2
3
4
6
7
8
9
10
12
∆∆
0 0 I I
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
3 3
+
−−
+
−−
− −q qi m i m( ) ( )1 1
12
12
0 0 0 0
0 0 I I
0 0 I I
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
I I 0 0
I I 0 0
I I 0 0
3 3
3 3
3 3
3 3
3 3
00 0 I I
0 0 I I
I I I I
I I 0 0
0 0 I I
0 0 0 0
0 0 0 0
0 0 0 0
0 0 I I
0 0 I I
0 0 0 0
I I 0 0
0 0 I I
3 3
3 3
3 3 3 3
3 3
3 3
3 3
3 3
3 3
3 3
− −−
−− −
−
+− −
−
q im
12
+q( ) .i m1
t Ir
r
I Ir
r
t I I q I I q
E q
1 1 31
2 11
3 31
2 1
1 112 3
12 3 11
12 3
12 3 1
6 11
n
n
n n
= −[ ]
+ [ ]
= −[ ] + [ ]=
, , ,
,
, , ˆ , ˆ
, , , , , ,
0 0 0 0 0
0 0 0 0
ρρ
ρρ
ρρ
ρρ
++
= [ ]
∈ ∈
= = [ ] ∈
× ×
×
E q
R
t R q R q R R
7 1
6 7
11
21
1
63 12
73 12
1 1 0 1 0 03 12
n
n
nn
,
, , , , , , ,
, .,
E 0 0 E
q
q
q
E E
0
LM
E4
t E q E q E q E q
t E q E q E q E q E q
t E q E q
ild
i i i i
ijd
i j i j ij i j i j
imd
i m i m
= + + +
= + + + +
= + +
− + +
− − + + +
− −
2 1 1 3 1 4 1 1 5 1 2
1 1 2 1 3 4 1 5 1 1
1 1 2 1
( ) ( ) ( )
( ) ( ) ( ) ( )( )
( ) ( )
ˆ ,
,
EE q E q3 4 1im i m+ +( ) .
8596Ch18Frame Page 408 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Now from (18.34) and (18.35), define
to get
(18.37)
and have the same structure as R11 except E4 is replaced by , and , respectively.Equation (18.37) will be needed to express the tendon forces in terms of q. Equations (18.28) and(18.37) yield the dependent vectors (t1n1, t2, t3, t4, t6, t7, t9, t10) in terms of the independent vectors(t5, t1, r1, r2). Therefore,
(18.38)
18.5 Tendon Forces
Let the tendon forces be described by
(18.39)
l t t t t t t t
t t t t
dn
dT dT dTndT dT
ndT
nmdT
T
ndT dT dT
ndT T
= [ ]= [ ]
1 1 11 21 1 12 2
1 1 1 2
, , , , , , ,
, , , , ,
K K K
K
l ld nm nm nm d nm= ∈ ∈ ∈+ ×Rq R R q R R( + ), , , ,( )24 3 12 12 24 3
R
R
R R
R R R
R R R
R R
R
R R
R R R R
R
E E E
E E E
=
∈ ∈
=
× ×
0
11 12
21 11 12
21 11 12
21 11
12
21 11
24 120
3 12
11
3 4 2
2 3 4
0 0
0
0
0 0
0
L L L
O M
O M
O M
M O O
M O O O
L L
L L
O
ˆ
, , ,ijn n n
MM
O M
M O
O O O
L
L L
O M
M O O O O M
M O O O
O O
L L
0
0
0
0
0 0 0
0 0
0
0
0 0 0
E E E
E E E
E
E 0 E E
R
E
E
E
E
R
2 3 4
2 3 4
4
4 2 3
12
5
5
5
5
=
,
i(( ) ),
, , ,
, , , , , , , ,
, , , .
i k i k i
i
k k
BlockDiag i
+ +
×
= > = >
= [ ] ∈ = →
= [ ] = −[ ]
= [ ]
0 0
0 0 0 0
0 0
if if
,
1 1
1 5
12
12
21 1 124 12
0 6 7 6 3 3
7 3 3
R
R E E E R
R E E E I I
E I I
(
L L
L
R11 R11 E4 E4
l RQld = .
ft
tα αα
αij ij
ij
ij
F=
8596Ch18Frame Page 409 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
For tensegrity structures with some slack strings, the magnitude of the force Fαij can be zero, fortaut strings Fαij > 0. Because tendons cannnot compress, Fαij cannot be negative. Hence, themagnitude of the force is
(18.40)
where
(18.41)
where is the rest length of tendon tαij before any control is applied, and the control is uαij,the change in the rest length. The control shortens or lengthens the tendon, so uαij can be positiveor negative, but . So uαij must obey the constraint (18.41), and
(18.42)
Note that for t1n1 and for α = 2, 3, 4, 6, 7, 8, 9, 10 the vectors tαij appear in the vector ld relatedto q from (4.7) by ld = Rq, and for α = 5, 1 the vectors tαij appear in the vector l related to q from(18.28), by l = Q–1 q. Let Pαij denote the selected row of R associated with tαij for αij = 1n1 andfor α = 2, 3, 4, 6, 7, 8, 9, 10. Let Pαij also denote the selected row of Q–1 when α = 5, 1. Then,
(18.43)
(18.44)
From (18.39) and (18.40),
fαij = – Kαij (q)q + bαij (q)uαij
where
(18.45)
(18.46)
Hence,
F k Lij ij ij ijα α α α= −( )t
kif L
k if Lij
ij ij
ij ij ij
αααα αα
αα αα αα
∆∆0
0
,
,
t
t
>
> ≤
L u Lij ij ijo
α α α− + ≥ 0
L ijoα > 0
L ijoα > 0
u Lij ijo
α α≤ > 0.
t qαα αα ααij ij ij
nm= ∈ ×R R, 3 12
t q qαα αα ααij
Tij
Tij
2= R R
K q q q Kαα αα αα αα αα αα ααij ij ij
o Tij
Tij ij ij
nmk L( ) ,= ( ) −
∈
− ×∆∆ R R R12 1 3 12R
b q q q q bαij ij
Tij
Tij ij ijk( ) ( ) , = ∈− ×∆∆
αα αα αα αα ααR R R R12 3 1
f
f
f
f
f
f
f
f
f
K
K
K
K
K
K
K
K
qijd
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
=
= −
2
3
4
6
7
8
9
10
2
3
4
6
7
8
9
10
8596Ch18Frame Page 410 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
or
(18.47)
and
or
(18.48)
Now substitute (18.47) and (18.48) into
Hence, in general,
or by defining
(18.49)
+
b
b
b
b
b
b
b
b
2
3
4
6
7
8
9
10
2
3
4
6
7
8
9
10
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
ij
u
u
u
u
u
u
u
u
f K q P uijd
ijd
ijd
ijd= − + ,
ff
f
K
Kq
bbij
o ij
ij
ij
ij
ij
ij
ij
ij
u
u=
= −
+
5
1
5
1
5
1
5
10
0
f K q P uijo
ijo
ijo
ijo= − + .
f
f
f
f
f
K
K
K
K
q
P
P
P
P
u
u
u
1
1 1
11
21
1
1 1
11
21
1
1 1
11
21
1
1 1
11
21
1
d
nd
d
nd
nd
d
nd
nd
d
nd
nd
d
nd
u
=
= −
+
M M O M
= − +K q P u1 1 1d d d
f
f
f
f
K
K
K
q
P
P
P
u
u
u
K q P2
12
22
2
12
22
2
12
22
2
12
22
2
2 2d
d
d
nd
d
d
nd
d
d
nd
d
d
nd
d=
= −
+
= − +M M O M 22 2
d du .
f K q P ujd
jd
jd
jd= − +
K
K
K
K
P
P
P
P
d
d
d
md
d
d
d
md
=
=
1
2
1
2
M M,
8596Ch18Frame Page 411 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
fd = –Kdq + Pdud.
Likewise, for forces (18.48),
(18.50)
fo = – Koq + Pouo.
Substituting (18.49) and (18.50) into (18.E.21) yields
f = –(BdKd + BoKo)q + BdPdud + BoPouo + Wow, (18.51)
which is written simply as
(18.52)
by defining,
f11o
f
f
f
f
K
K
K
q
P
P
P
u
u
u
1
11
21
1
11
21
1
11
21
1
11
21
1
o
o
o
no
o
o
no
o
o
no
o
o
no
=
= −
+
M M O M
f
f
f
f
K
K
K
q
P
P
P
u
u
u
jo
jo
jo
njo
jo
jo
njo
jo
jo
njo
jo
jo
njo
=
= −
+
1
2
1
2
1
2
1
2
M M O M
f K q P ujo
jo
jo
jo= − +
f Kq Bu W w= − + +˜ ˜ o ,
˜ ,K B K B K= +∆ d d o o
˜ , ,B B P B P=[ ]∆ d d o o
B P
B P B P
B P B P B P
B P B P
B P
B P
B P B P
d d
d d
d d d
d d
d
md
md
md
=
−
3 1 4 2
5 1 6 2 4 3
5 2 6 3
5 3
4
5 1 8
0 0
0
0
0 0
L L
O M
O O M
M O O O
M O O O
L L
,
8596Ch18Frame Page 412 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
(18.53)
(18.54)
In vector in (18.54), u1n1 appears twice (for notational convenience u1n1 appears in and in. From the rules of closure, t9i1 = – t1i1 and t7im = – t10im, i = 1, 2, …, n, but t1i1, t7im, t9i1, t10im all
appear in (18.54). Hence, the rules of closure leave only n(10m – 2) tendons in the structure, but(18.54) contains 10nm + 1 tendons. To eliminate the redundant variables in (18.54) define =Tu, where u is the independent set u , and is given by (18.54). We chooseto keep t7im in u and delete t10im by setting t10im = – t7im. We choose to keep t1i1 and delete t9i1 bysetting t9i1 = – t1i1, i = 1, 2, …, n. This requires new definitions of certain subvectors as follows in(18.57) and (18.58). The vector is now defined in (18.54). We have reduced the vector by2n + 1 scalars to u. The T matrix is formed by the following blocks,
B P
B P B P
B P B P
B P
B P
B P
o o
o o
o o
o
mo
mo
=
1 1 2 2
7 2 2 3
7 3
2
7
0 0
0
0
0 0
L
O M
M O O
M O O
L L
K B K B K
B K B K B K B K
B K B K B K B K B K
B K B K B K B K B K
B K B K B K B K= + =
+ + ++ + + ++ + + ++ + +d d o o
d o d o
d d d o o
d d d o o
d d d o
3 1 1 1 4 2 2 2
5 1 6 2 4 3 7 2 2 3
5 2 6 3 4 4 7 3 2 4
5 3 5 4 4 5 7 4 ++
+ + + ++ +
− − −
−
B K
B K B K B K B K B K
B K B K B K
2 5
5 2 6 1 4 7 1 2
5 1 6 7
o
md
md
md
mo
mo
md
md
mo
M
˜ ,
ˆ
ˆu
u
u
u
u
u
u
u
u
u
u
u
u
u
u
u
u
u
u
u
u
u
=
=
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
d
d
d
d
md
o
o
o
o
mo
d
d
d
d
md
o
o
o
o
mo
M
M
M
M
u u1d
u1o )
u∈ −n m( )10 2 u ∈ +R10 1nm
u u
8596Ch18Frame Page 413 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
(18.55)
where
(18.56)
There are n blocks labeled T1, n(m – 2) blocks labeled I8 (for m ≤ 2 no I8 blocks needed, seeappendix 18.D), n blocks labeled T2, nm blocks labeled I2 blocks, and n blocks labeled S.
The block becomes
(18.57)
T
T S
T S
I
I
T
T
I
I
I
I
=
∈ +( )× −( )( )
0 0 0 0 0L
O O
O
O
O
O
1
1
1
8
8
2
2
2
2
2
2
10 1 10 2R nm n m
T
I
1
6 6 1
1 6
1 6
8 70
1
=
∈×
×
×
×
0
0
0
R
TI
27 8 7
0 0 0 0 1 0 0=
−
∈ ×R
S
= −
∈×
×0
0
6 28 20 1
0R .
u1d
ˆ
ˆ
ˆ
ˆ
ˆ
, ˆ , , , , ...u
u
u
u
u
u 1
11
21
31
1
1
2 1
3 1
4 1
6 1
7 1
8 1
10 1
7 1 1 2 3d
d
d
d
nd
id
n
n
n
n
n
n
n
u
u
u
u
u
u
u
i n=
=
∈ =×∆∆
M
R ,, j = 1
8596Ch18Frame Page 414 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
The block becomes
(18.58)
The block is the block with the first element u1n1 removed, because it is included in From (18.17) and (18.52),
(18.59)
where,
Kp = H(q) (q),
B = H(q) (q)T,
D = H(q)Wo.
The nodal points of the structure are located by the vector p. Suppose that a selected set of nodalpoints are chosen as outputs of interest. Then
yp = Cp = CPq (18.60)
where P is defined by (18.26). The length of tendon vector tαij = is given from (18.44).Therefore, the output vector yl describing all tendon lengths, is
.
Another ouput of interest might be tension, so from (18.40) and (18.44)
.
The static equilibria can be studied from the equations
Kp(q)q = B(q)u + D(q)w, yp = CPq. (18.61)
umd
ˆ
ˆ
ˆ
ˆ
ˆ
, ˆ , , , , ... , .u
u
u
u
u
u md
md
md
md
nmd
imd
nm
nm
nm
nm
nm
nm
nm
u
u
u
u
u
u
u
i n j m=
=
∈ = =×∆∆
1
2
3
2
3
4
6
7
8
9
7 1 1 2 3
M
R
u1d u1
d uno1.
˙ ( ( ˙ ) ( )) ( ) ( ) ,q K q K q q B q u D q w+ + = +r p
K
B
ααij q
y yl ij=
M
Mαα ,
y q qαα ααααij
T Tijij
= ( ) 12
y yf ij ij ij ij ijF F k L=
= −M
Mαα αα αα αα αα, ( )
8596Ch18Frame Page 415 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Of course, one way to generate equilibria is by simulation from arbitrary initial conditions andrecord the steady-state value of q. The exhaustive definitive study of the stable equilibria is in aseparate paper.27
Damping strategies for controlled tensegrity structures are a subject of further research. Theexample case given in Appendix 18.D was coded in Matlab and simulated. Artificial critical dampingwas included in the simulation below. The simulation does not include external disturbances orcontrol inputs. All nodes of the structure were placed symmetrically around the surface of a cylinder,as seen in Figure 18.10. Spring constants and natural rest lengths were specified equally for alltendons in the structure. One would expect the structure to collapse in on itself with this giveninitial condition. A plot of steady-state equilibrium is given in Figure 18.11 and string lengths inFigure 18.12.
FIGURE 18.10 Initial conditions with nodal points on cylinder surface.
FIGURE 18.11 Steady-state equilibrium.
FIGURE 18.12 Tendon dynamics.
10 5 0 5 10
105051002468
101214161820
Initial Conditions Perspective View
10 8 6 4 2 0 2 4 6 8 1010
8
6
4
2
0
2
4
6
8
10 Initial Conditions Top View
10 5 0 5 10
105051002468
101214161820
Steady State Equilibrium Perspective View
10 8 6 4 2 0 2 4 6 8 10-10
-8
-6
-4
-2
0
2
4
6
8
10Steady State Equilibrium Top View
0 5 10 15 20 25 30 35 40 45 503
4
5
6
7
8
9
10
Ten
don
Leng
th
time
Tendon Dynamicst1, t7, t9, t10, (Diagonal Strings)
t4, t8, (Saddle Strings)
t5, t3, (Vertical Strings)
t2,t6, (Base, Top Strings)
8596Ch18Frame Page 416 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
18.6 Conclusion
This chapter developed the exact nonlinear equations for a Class 1 tensegrity shell, having nm rigidrods and n(10m – 2) tendons, subject to the assumption that the tendons are linear-elastic, and therods are rigid rods of constant length. The equations are described in terms of 6nm degrees offreedom, and the accelerations are given explicitly. Hence, no inversion of the mass matrix isrequired. For large systems this greatly improves the accuracy of simulations.
Tensegrity systems of four classes are characterized by these models. Class 2 includes rods thatare in contact at nodal points, with a ball joint, transmitting no torques. In Class 1 the rods do nottouch and a stable equilibrium must be achieved by pretension in the tendons. The primal shellclass contains the minimum number of tendons (8nm) for which stability is possible.
Tensegrity structures offer some potential advantages over classical structural systems composedof continua (such as columns, beams, plates, and shells). The overall structure can bend but allelements of the structure experience only axial loads, so no member bending. The absence ofbending in the members promises more precise models (and hopefully more precise control).Prestress allows members to be uni-directionally loaded, meaning that no member experiencesreversal in the direction of the load carried by the member. This eliminates a host of nonlinearproblems known to create difficulties in control (hysteresis, dead zones, and friction).
Acknowledgment
The authors recognize the valuable efforts of T. Yamashita in the first draft of this chapter.
8596Ch18Frame Page 417 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Appendix 18.A Proof of Theorem 18.1
Refer to Figure 18.8 and define
using the vectors and which locate the end points of the rod. The rod mass center is locatedby the vector,
(18.A.1)
Hence, the translation equation of motion for the mass center of the rod is
(18.A.2)
where a dot over a vector is a time derivative with respect to the inertial reference frame. A vector locating a mass element, dm, along the centerline of the rod is
(18.A.3)
and the velocity of the mass dm, v, is
(18.A.4)
The angular momentum for the rod about the mass center, hc, is
(18.A.5)
where the mass dm can be described using dx = L dρ as
(18.A.6)
Hence, (18.A.2) can be rewritten as follows:
(18.A.7)
where (18.A.1) and (18.A.3) yield
(18.A.8)
(18.A.4)–(18.A.8) yield
q p p q p p1 2 1 2 2 1
= + = −, ,
p1
p2
p qc
= 12 1
.
mm
c˙ ˙ (ˆ ˆ ),p q f f= = +
2 1 1 2
p
p p (p p ) q ( q= + − = + − ≤ ≤ =1 2 1 1 2
12
12
0 1ρ ρ ρ ρ) , , ,xL
v p q q= = + −˙ ˙ ( ) ˙ .12
121 2
ρ
h p p pc cm
dm= − − ×∫ ( ) ˙
dmmL
L d m d=
=( ) . ρ ρ
h p p p( c cm d= − ×∫ ( ) ˙ )
0
1
ρ
p p q− = −c
( ) .ρ 12 2
8596Ch18Frame Page 418 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
(18.A.9)
The applied torque about the mass center, is
Then, substituting hc and from (18.A.9) into Euler’s equations, we obtain
or
(18.A.10)
Hence, (18.A.2) and (18.A.10) yield the motion equations for the rod:
(18.A.11)
We have assumed that the rod length L is constant. Hence, the following constraints for hold:
Collecting (18.A.11) and the constraint equations we have
(18.A.12)
h q q q
q q q
q q
c m d
m d d
m
= − × + −
= × − + −
= × −
+ −
∫
∫∫
( ) ˙ ( ) ˙
˙ ( ) ˙ ( )
˙ ( )
ρ ρ ρ
ρ ρ ρ ρ
ρ ρ ρ
12
12
12
12
12
12
12
12
12
13
12
2 1 20
1
2 1 2
2
0
1
0
1
2
2
0
1
1
33
0
1
2
2 212
= ×
˙
˙
q
q qm
ττc,
ττc = × −12 2 2 1q f f(ˆ ˆ )
ττc
hc c=
˙ ( ˙ ˙ ˙ )
˙ (ˆ ˆ )
h q q q q
q q q f f
c
m
m
= × + ×
= × = × −
12
1212
2 2 2 2
2 2 2 2 1
m
m
2 1 1 2
6 2 2 2 2 1
˙ ˆ ˆ
( ˙ ) (ˆ ˆ )
q f f
q q q f f
= +× = × −
q2
q q
q q q q q q q q
q q q q q q
2 2
2
2 2 2 2 2 2 2 2
2 2 2 2 2 2
2 0
0
⋅ =
⋅ = ⋅ + ⋅ = ⋅ =
⋅ = ⋅ + ⋅ =
L
d
dt
d
dt
( ) ˙ ˙ ˙
( ˙ ) ˙ ˙ ˙
m
m
L
2 1 1 2
6 2 2 2 2 1
2 2 2 2
2 2
2
0
˙ ˆ ˆ
( ˙ ) (ˆ ˆ )
˙ ˙ ˙
q f f
q q q f f
q q q q
q q
= +× = × −
⋅ + ⋅ =⋅ =
8596Ch18Frame Page 419 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
We now develop the matrix version of (18.A.12). Recall that
Also note that ET⋅E = 3 × 3 identity. After some manipulation, (18.A.12) can be written as:
(18.A.13)
Introduce scaled force vectors by dividing the applied forces by m and mL
Then, (18.A.13) can be rewritten as
(18.A.14)
Solving for requires,
(18.A.15)
Lemma For any vector q, such that qTq = L2,
Proof:
Since the coefficient of in (18.A.15) has linearly independent columns by virtue of the Lemma,the unique solution for is
q E f Efqi i i i= =; ˆ ˆ
m
m
T T
T L
2 1 1 2
6 2 2 2 2 1
2 2 2 2
2 22
˙ ˆ ˆ
˜ ˙ ˜ (ˆ ˆ )
˙ ˙ ˙
.
q f f
q q q f f
q q q q
q q
= +
= −
= −
=
g f f g f f1 1 2 2 1 2 2
2 6= + = −∆ ∆(ˆ ˆ ) , (ˆ ˆ ) .
m mL
˙
˜ ˙ ˜ ( )
˙ ˙ ˙
.
q g
q q q g
q q q q
q q
1 1
2 2 2 22
2 2 2 2
2 22
== −= −=
L
L
T T
T
˙q2
˜˙
˙ ˙
˜q
q qq
g2
22
2 2
22
2T T L
=
−
−
0
0
˜ ˜q
q
q
qI
T
T
TL
= 2 3
0
0
0
0
0
0
3 2 1
3 1 2
2 1 3
3 2
3 1
2 1
1 2 3
23
q q q
q q q
q q q
q q
q q
q q
q q q
L
−−
−
−−
−
= I ∆.
˙q2˙q2
8596Ch18Frame Page 420 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
(18.A.16)
where the pseudo inverse is uniquely given by
It is easily verified that the existence condition for in (18.A.15) is satisfied since
Hence, (18.A.16) yields
(18.A.17)
Bringing the first equation of (18.A.14) together with (18.A.17) leads to
(18.A.18)
Recalling the definition of g1 and g2 we obtain
(18.A.19)
where we clarify
Equation (18.A.19) is identical to (18.17), so this completes the proof of Theorem 18.1.
Example 1
The generalized forces are now defined as
˙˜
˙ ˙
˜,q
q
q q qq
g22
2 2 2
22
2=
−
−
+
T T L0
0
˜ ˜ ˜ ˜ ˜q
q
q
q
q
q
q
q
q
q2
2
2
2
2
2
1
2
2
2 2
2T T
T
T T
T
T
T
L
=
=
+ −
−
˙q2
Iq
q
q
q q qq
g−
−
−
=
+˜ ˜
˙ ˙
˜,2
2
2
2 2 2
22
2 0T T T L0
0
˙˙ ˙
˜qq q
q q g22 2
2 2 22
2= − +T
L
˙
˙ ˜˙ ˙q
q I
q
q
I
q
g
gq q1
2 3
1
2
3
22
1
22 2
2
+
=
0 0
0
0
0T
L
˙
˙ ˜˙ ˙q
q I
q
q
I
q
f
fq q1
2 3
1
2
3
322
1
22 2
2 2
2
+
=
0 0
0
0
0T
L Lm
f
ff f
f f1
2
1 2
1 2
=
+−
ˆ ˆ
ˆ ˆ .
q p p1 1 211 21
12 22
11
12
= + =++
=
p p
p p
q
q,
q p p2 2 121 11
22 12
21
22
= − =−−
=
p p
p p
q
q,
8596Ch18Frame Page 421 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
From (18.A.16),
Example 2
Using the formulation developed in 18.A.3, 18.A.4, and 18.A.5 we derive the dynamics of a planartensegrity. The rules of closure become:
We define the independent vectors lo and ld:
The nodal forces are
FIGURE 18.A.1 A rigid bar of the length L and mass m.
g f f g f f1 1 211 21
12 222 2 2 1 2
21 11
22 12
2 2 6 6= + =++
= − =
−−
m m
f f
f f mL mL
f f
f f(ˆ ˆ ) , (ˆ ˆ )
˙
˙
˙
˙
˙ ˙
˙ ˙
q
q
q
q
q
q
q
q
qq q
Lq q
L
11
12
21
22
11
12
21
22
222
0 0 0 0
0 0 0 0
0 0 0
0 0 0
1 0 0 0
0 1 0 0
0 0212
222
2
212
222
2
= −
−
+−+
+
qq q
q q q21 22
21 22 2120 0
1
2
−
g
g
t t
t t
t t
t t
5 4
8 1
7 2
6 3
= −
=
= −
= −
lr
r
t
l
t
t
t
o d=
=
ρ
1
2
5
1
2
3
,
f
f f
f f
f f
f f
f f w f f w
f f w f f w
f f w f f w
f f w
=
+−+−
=
− + + − +− + − − +
+ + + − − ++ +
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) ( )
( ) ( )
( ) ( )
(
1 2
1 2
3 4
3 4
3 2 1 5 1 2
3 2 1 5 1 2
1 2 3 3 5 4
1 2 33 3 5 4) ( )− − − +
f f w
8596Ch18Frame Page 422 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
We can write
where
Or, with the obvious definitions for Bo, Bd, and W1, in matrix notation:
Bo fo + Bd fd + W1w. (18.A.20)
The nodal vectors are defined as follows:
and
We define
The relation between q and p can be written as follows:
f
I
I
I
I
f
I I I
I I I
I I I
I I I
f
I I
I I
I I
I I
w=−−
+
− −−
−
+−
−
2
2
2
2
2 2 2
2 2 2
2 2 2
2 2 2
2 2
2 2
2 2
2 2
o d
0 0
0 0
0 0
0 0
f f f
f
f
f
w
w
w
w
w
o d= [ ] =
=
5
1
2
3
1
2
3
4
, , .
f =
p
p r
p
p r
1
2 1
3
4 2
=
= +
=
= +
ρρ
ρρ
ρρ
ρρ
ˆ
ˆ
,
ˆ .ρρ ρρ= + + −r t r1 5 2
q p p r
q p p r
q p p r r t r
q p p r
1 2 1 1
2 2 1 1
3 4 3 2 1 5 2
4 4 3 2
2
2 2
= + = +
= − =
= + = + = + + −
= − =
∆∆
∆∆
∆∆
∆∆
ρρ
ρρ ρρˆ ( )
q
q
q
q
q
I I
I I
I I
I I
p
p
p
p
=
=−
−
1
2
3
4
2 2
2 2
2 2
2 2
1
2
3
4
0 0
0 0
0 0
0 0
,
8596Ch18Frame Page 423 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
and
(18.A.21)
We can now write the dependent variables ld in terms of independent variables lo. From (18.30)and (18.31):
t1 = ρρρρ + r1 –(ρρρρ + r1 + t5 – r2),
t2 = ρρρρ – (ρρρρ + r1 + t5 – r2).
By inspection of Figure 18.A.2, (18.30) and (18.31) reduce to:
t1 = r2 – t5
t2 = –r1 + r2 – t5
t3 = r1 + t5, (18.A.22)
or,
Equation (18.A.21) yields
FIGURE 18.A.2 A planar tensegrity.
q
q
q
q
q
I I
I
I I I I
I
r
r
t
Q=
=−
=
1
2
3
4
2 2
2
2 2 2 2
2
1
2
5
2
2 2 2
0 0
0 0 0
0 0 0
l
ρρ
o.
I
t
t
t
I I
I I I
I I
r
r
t
d =
=−
− −
1
2
3
2 2
2 2 2
2 2
1
2
5
0 0
0
0 0
ρρ
8596Ch18Frame Page 424 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Hence,
We can now write out the tendon forces as follows:
or
fd = –Kdq + Pdud
and
fo = [f5] = –[K5] q + [b5] [u5],
or
fo = – Koq + Pouo,
using the same definitions for K and b as found in (18.45) and (18.46), simply by removing the ijelement indices. Substitution into (18.A.20) yields:
With the matrices derived in this section, we can express the dynamics in the form of (18.20):
l
I I
I
I
I I I I
Q qo =
=
−
− −
=
= −
ρρr
r
t
0 0
0 0 0
0 0 0
q
q
q
q
1
2
5
2 2
2
2
2 2 2 2
1
2
3
4
112
2
2.
l
I I I I
I I I I q Rqd =−
− −−
=12
2 2 2 2
2 2 2 2
2 2 2 2I I I I
f
f
f
f
K
K
K
q
b
b
b
d
u
u
u
=
−
1
2
3
1
2
3
1
2
3
1
2
3
= + ,
f B K q P u B K q P u
B K B K q B P u B P u
= − + + − +
= − + + +
o o o o
o o o o o
( ) ( )
( ) ( ).
d d d d
d d d d d
˙ ( ) ,q K K q B Dw+ + = + r p u
Kr = 1,
K HKp = ˜ ,
B HB = ˜ ,
8596Ch18Frame Page 425 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
,
where
and
D HW H W = =o1 1
H H
I 0 0 0
0 q 0 0
0 0 I 0
0 0 0 q2
= =
1
2
322
322
2 12
22
mL
L
˜
˜
,
K
0 0 0 0
0 q q I 0 0
0 0 0 0
0 0 0 q q I
K B K B K
B B P B P
r
T
T
d d o o
d d o o
L
L
=
+
−
−
11
22 2 2
22
4 4 4
˙ ˙
˙ ˙
,
˜ ,
˜ [ , ],
∆∆
∆∆
˜ , ˜ ,
˙ ˙ ˙ ˙ , ˙ ˙ ˙ ˙ .
q q
q q q q
22 22
221 22
21 22 212 4
2 422
41 42
41 42 412
2 2 212
222
4 4 412
422
=−
−
=
−−
= + = +
q q q
q q q
q q q
q q q
q q q qT T
8596Ch18Frame Page 426 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Appendix 18.B Algebraic Inversion of the Q Matrix
This appendix will algebraically invert a 5 × 5 block Q matrix. Given Q in the form:
(18.B.1)
we define x and y matrices so that
Qx = y (18.B.2)
where
(18.B.3)
Solving (18.B.2) for x we obtain
x = Q–1y (18.B.4)
Substituting (18.B.1) and (18.B.3) into (18.B.2) and carrying out the matrix operations we obtain
(18.B.5)
Solving this system of equations for x will give us the desired Q–1 matrix. Solving each equationfor x we have
(18.B.6)
Elimination of x on the right side of (18.B.6) by substitution yields
Q
Q
Q Q
Q Q Q
Q Q Q Q
Q Q Q Q Q
=
11
21 22
21 32 22
21 32 32 22
21 32 32 32 22
0 0 0 0
0 0 0
0 0
0
x
x
x
x
x
x
y
y
y
y
y
y
=
=
1
2
3
4
5
1
2
3
4
5
,
Q x y
Q x Q x y
Q x Q x Q x y
Q x Q x Q x Q x y
Q x Q x Q x Q x Q x y
11 1 1
21 1 22 2 2
21 1 32 2 22 3 3
21 1 32 2 32 3 22 4 4
21 1 32 2 32 3 32 4 22 5 5
=+ =+ + =+ + =+ + + =
+
+
x Q y
x Q Q x y
x Q Q x Q x y
x Q Q x Q x Q x y
x Q Q x Q x Q x Q x
1 11 1
2 221
21 1 2
3 22 21 1 32 2 3
4 22 21 1 32 2 32 3 4
5 22 21 1 32 2 32 3 32 4
( +
( +
( +
(
1
1
1
1
== −= − −= − − −= − − − −
−
−
−
−
−
)
)
)
+ y5)
8596Ch18Frame Page 427 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
(18.B.7)
Or, in matrix form,
(18.B.8)
where we have used the notation with the following definitions
Using repeated patterns, the inverse may be computed by
(18.B.9)
x y
x y y
x y y y
x y y y y
x y y y y y
1 11 1
2 21 1 22 2
3 31 1 32 2 33 3
4 41 1 42 2 43 3 44 4
5 51 1 52 2 53 3 54 4 55 5
== += + += + + += + + + +
ΛΛΛΛ ΛΛΛΛ ΛΛ ΛΛΛΛ ΛΛ ΛΛ ΛΛΛΛ ΛΛ ΛΛ ΛΛ ΛΛ
Q−
1 =
ΛΛΛΛ ΛΛΛΛ ΛΛ ΛΛΛΛ ΛΛ ΛΛ ΛΛΛΛ ΛΛ ΛΛ ΛΛ ΛΛ
11
21 22
31 32 33
41 42 43 44
51 52 53 54 55
0 0 0 0
0 0 0
0 0
0
Λrow column,
ΛΛ
ΛΛ ΛΛ ΛΛ ΛΛ
ΛΛ
ΛΛ
11 111
21 221
21 111
221
21 11 22 21 11
31 221
32 221
21 111
221
21 111
41 221
32 221
32 221
21 111
221
32 221
21 1112
=
= − = − = −
= −
= − + −
−
− − −
− − − − −
− − − − − −
Q
Q Q Q Q Q Q
Q Q Q Q Q Q Q Q
Q Q Q Q Q Q Q Q Q Q Q Q Q–2222
121 11
1
51 221
32 221
32 221
21 111
221
32 221
21 111
221
32 221
32 221
32 221
21 111
221
21 111
22 33 44 55 221
32
3 3
− −
− − − − − − −
− − − − − − −
−
= − +
+ −
= = = =
=
Q Q
Q Q Q Q Q Q Q Q Q Q Q Q
Q Q Q Q Q Q Q Q Q Q Q Q
Q
ΛΛ
ΛΛ ΛΛ ΛΛ ΛΛ
ΛΛ ΛΛ4343 54 221
32 221= = − − −ΛΛ Q Q Q
ΛΛ ΛΛ
ΛΛ
42 53 221
32 221
32 221
221
32 221
52 221
32 221
32 221
32 221
221
32 221
32 221
221
32 2212
= = −
= − + −
− − − − −
− − − − − − − − −
Q Q Q Q Q Q Q Q
Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q
Q− =
1
11
21 22
21 32 222
21 32 32 223
212
31 32 32 224
213
312
31 32 32 22
ΛΛΛΛ ΛΛΘΘΛΛ ΛΛ ΛΛΘΘ ΛΛ ΘΘΛΛ ΛΛ ΛΛΘΘ ΛΛ ΘΘ ΛΛ ΘΘΛΛ ΛΛ ΛΛΘΘ ΛΛ ΘΘ ΛΛ ΘΘ ΛΛ ΘΘΛΛ ΛΛ ΛΛM M O O O O O
8596Ch18Frame Page 428 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
using
Only and powers of Θ need to be calculated to obtain for any (n, m).The only matrix inversion that needs to be computed to obtain is and , substantiallyreducing computer processing time for computer simulations.
ΘΘ ΛΛ
ΛΛ
ΛΛ ΛΛ ΛΛ
ΛΛ
ΛΛ ΛΛ ΛΛ
=
=
=
=
=
( )
.
I Q
Q
Q
Q
Q
−
−
−
−
−
22 32
11 111
21 22 21 11
22 221
32 22 32 22
ΛΛ ΛΛ ΛΛ ΛΛ11 22 21 32, , , , Q−1
Q−1 Q111− Q22
1−
8596Ch18Frame Page 429 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Appendix 18.C General Case for (n, m) = (i, 1)
In Appendix E the forces acting on each node are presented, making special exceptions for thecase when (j = m = 1). The exceptions arise for (j = m = 1) because one stage now contains bothclosure rules for the base and the top of the structure. In the following synthesis we use and
from (18.E.1) and and from (18.E.4). At the right, where (i = 1, j = 1):
At the center, where (1 < i < n, j = 1):
At the left end of the base in Figure 18.5, where (i = n, j = 1):
(18.C.1)
Using,
f111
f211 f311 f411
f
f f
f f
f f
f f
f f f f w f f f f w
f f f f w f f f
11
+
=
+−+−
=
− + + + + − − − +− + + + + − − −
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) ( )
( ) (
111 211
111 211
311 411
311 411
211 311 1 1 2 1 111 511 111 411 811 211
211 311 1 1 2 1 111 511 111 411
n n
n n −− +− − + + − − + +− − + − − − + +
f w
f f f f w f f f f w
f f f f w f f f f w
811 211
811 711 321 421 311 611 511 6 1 7 1 411
811 711 321 421 311 611 511 6 1 7 1 411
)
( ) ( )
( ) ( )
. +
+n n
n n
f
f f
f f
f f
f f
f f f f w f f f f
i
i i
i i
i i
i i
i i i i i i i i
1
1 1 2 1
1 1 2 1
3 1 4 1
3 1 4 1
1 1 1 2 1 1 2 1 3 1 1 1 1 1 4 1 5 1 8
=
+−+−
=
+ − + + + − − + −− −
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) (( ) ( ) ii i
i i i i i i i i i i
i i i i i i i
1 2 1
1 1 1 2 1 1 2 1 3 1 1 1 1 1 4 1 5 1 8 1 2 1
8 1 7 1 3 1 1 4 1 1 3 1 5 1 6 1
++ − + + − − − + − +
− − + + + − +− −
w
f f f f w f f f f w
f f f f w f f
)
( ) ( )
( ) (( ) ( )
( + ) ( + ) −− +− − + + − − + − +
− −
− −
f f w
f f f f w f f f f w6 1 7 1 1 4 1
8 1 7 1 3 1 1 4 1 1 3 1 5 1 6 1 6 1 7 1 1 4 1
( ) ( )
( + ) ( + ) ( ) ( )
+
+i i i
i i i i i i i i i i
)
( ) ( )
.
f
f f
f f
f f
f f
f f f f w f f f f
n
n n
n n
n n
n n
n n n n n n n n
1
1 1 2 1
1 1 2 1
3 1 4 1
3 1 4 1
1 1 1 2 1 1 3 1 2 1 1 1 1 1 4 1 5 1 8
=
+−+−
=
+ + − + + − − + −− −
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) (( ) ( ) nn n
n n n n n n n n n n
n n n n n n
1 2 1
1 1 1 2 1 1 3 1 2 1 1 1 1 1 4 1 5 1 8 1 2 1
7 1 8 1 311 411 3 1 6 1 1 5 1 6 1
++ + − + − − − + − +
− + − + + + − − + +− −
−
w
f f f f w f f f f w
f f f f w f f f f
)
( ) ( )
( ) (( ) ( )
( ) 77 1 1 4 1
7 1 8 1 311 411 3 1 6 1 1 5 1 6 1 7 1 1 4 1
( )
( ) ( )
n n
n n n n n n n n
−
− −
+− + − + + − − − + + +
w
f f f f w f f f f w
)
( ) ( )
.
8596Ch18Frame Page 430 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Or, in matrix notation, with the implied matrix definitions,
(18.C.2)
ff
ff
f
f
f
f
f
f
f
f
w
w
w
w
w
ijo
ijijd
ij
ij
ij
=
=
=
5
1
2
3
4
6
7
8
9
10
1
2
3
4
, , ,
f
I 0 0 0 0 0 0 0
I 0 0 0 0 0 0 0
0 0 0 I I 0 0 0
0 0 0 I I 0 0 0
f
I I I 0 0 I 0 0
I I I 0 0 I 0 0
0 0 0 I I I 0 0
0 0 0 I I I 0 0
3
3
3 3
3 3
3 3 3 3
3 3 3 3
3 3 3
3 3 3
11 1=−
−
+
− − −−
−− −
nd
f11d
+−−
+
−−−
+
+
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 I I 0 0 0 0 0
0 I I 0 0 0 0 0
f
I I
I I
I 0
I 0
f
I
I
0
0
f
I I 0 0
3 3
3 3
21
3 3
3 3
3
3
11
3
31 1
3 3
d on
II I 0 0
0 0 I I
0 0 I I
w3 3
3 3
3 3
11
−
−
.
f B f B f B f B f B f Ww11 1 1 11 11 21 21 11 11 1 1 1 1 11= + + + + +ˆ .nd
nd d d d d o o
n n
f
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
f
0 0 0 0
0 0 0 0
0 0 0 0 0
0 0
i id
1 1 1=
−−
+
− − −−
−−( )
I
I
I I
I I
I I I I
I I I I
I I I
3
3
3 3
3 3
3 3 3 3
3 3 3 3
3 3 3
00 0 0
f
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
f
0
0
0 0
0 0
f
− −
+−−
+
+( ) −(
I I I
I I
I I
I
I
3 3 3
3 3
3 3
3
3
id
id
i
1
1 1 1))
+
−−−
+−
−
1
1 1
o
io
i
I I
I I
I
I
I I
I I
I I
I I
3 3
3 3
3
3
3 3
3 3
3 3
3 3
0
0
f
0 0
0 0
0 0
0 0
w .
8596Ch18Frame Page 431 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Or, in matrix form,
(18.C.3)
Or, in matrix form,
(18.C.4)
Now assemble (18.C.2)–(18.C.4) into the form
(18.C.5)
where
Or, simply, (18.C.5) has the form (18.E.21), where
f B f B f B f B f B f Wwi
di
d did d
id o
io o
io
i1 01 1 1 11 1 21 1 1 01 1 1 11 1 1= + + + +− −ˆ .( ) ( + ) ( ) +
f
I 0 0 0 0 0 0 0
I 0 0 0 0 0 0 0
0 0 0 I I 0 0 0
0 0 0 I I 0 0 0
f
I I I 0 0 I 0 0
I I I 0 0 I 0 0
0 0 0 I I I 0 0
0
n nd
1
3
3
3 3
3 3
1 1
3 3 3 3
3 3 3 3
3 3 3
=−
−
+
− − −−
−−( )
00 0 I I I 0 0
f
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 I I 0 0 0 0 0
0 I I 0 0 0 0 0
f
0 I
0 I
0 0
0 0
f
− −
+−−
+
−( )
3 3 3
1
3 3
3 3
11
3
31 1
nd
dnoo
nlo
n n
+−−
+
−
+−
−
I 0
I 0
I 0
I 0
f
I
I
0
0
f
I I 0 0
I I 0 0
0 0 I I
0 0 I I
w
3
3
3
3
3
31 1
3 3
3 3
3 3
3 3
1 .
f B f B f B f B f B f B f Wwnd
nd d
nd d d o
no
no
no
n n n1 01 1 1 11 1 21 11 01 1 1 1 1 1 1 1 1 1= + + + + + +−( ) −( )ˆ .
f B f B f W w1 3 1 1 1 1 1= + +d o ,
f
f
f
f
ff
ff
f
f
f
w
w
w
w
j
j
j
nj
d nd j
o
jo
jo
njo
j
j
j
nj
=
=
=
=
1
21
1 1
1
1
2
1
2
M M M, , ,
B
B B B 0 0 B
0 B 0
0 0
0 B
B B 0 0 B B
B
B 0 0
B
0
B 03
1 1 11 21 1
01
21
1 1 21 01 11
1
11
01
11
=
=
nd d
nd
d
d
nd d d
o
o
o
L
O O O
M O O O O O M
O O O O
O O O O
L
L L
O O M
O O O M
M O O
ˆ
ˆ
ˆ
,
00 0 B BL 01 1o
no
,
W W W1 = [ ]BlockDiag K K, , , .
8596Ch18Frame Page 432 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
The next set of necessary exceptions that apply to the model (i,1) arises in the form of the Rmatrix that relates the dependent tendons set to the generalized coordinates . For any(i,1) case R takes the form following the same procedure as in (18.32) and (18.33).
(18.C.6)
where
The transformation matrix T that is applied to the control inputs takes the following form. Theonly exception to (18.55) is that there are no I8 blocks due to the fact that there are no stagesbetween the boundary conditions at the base and the top of the structure. The second set of I2
blocks is also not needed since m = 1. Hence, the appropriate T matrix for = Tu is
where S is defined by (18.56), and
f f f f f f w w W W = = 1 1 1 1 1, , , , ,d d o o o= = =
B B B Bd o= =3 1, .
(ld = Rq)
R =R
R 0
11˜ ,
˜
˜
˜
˜
˜
˜
˜
R
E E 0 E
E E E 0
0 E E E
E E E 0
0 E
E 0 0 E E
11
3 4 2
2 3 4
2 3 4
2 3 4
4
4 2 3
=
L L
O
O M
M O
O O O
L
˜ .E
I I
I I
I I
I I
I I
4
3 3
3 3
3 3
3 3
3 3
12
=
−
− −
−
0 0
0 0 0 0
0 0 0 0
0 0
0 0
0 0 0 0
0 0
0 0
u
T =
S
S
I
I
0 01
1
1
2
2
10 1 8
L
O O
O
T
T
′
′
∈ +( )×R n n,
8596Ch18Frame Page 433 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
There are n blocks, n S blocks, and n I2 blocks. The control inputs are now defined as
The block becomes
Appendix 18.D will explicitly show all matrix forms for the specific example (n,m) = (3,1).
′ =
−
∈ ×T
16
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
0 0 0 0 1 0
8R .
′T1
u
u
u = 1
1
d
o
.
u1d
u
u
u
u
u
u1
11
21
31
1
1
2 1
3 1
4 1
6 1
7 1
8 1
6 1 1 2 3 1′
′
′
′
′
′=
=
∈ = = =×d
d
d
d
nd
id
n
n
n
n
n
n
u
u
u
u
u
u
i n j m
M
L, , , , , ,R
8596Ch18Frame Page 434 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Appendix 18.D Example Case (n,m) = (3,1)
Given the equation for the dynamics of the shell class of tensegrity structures:
We explicitly write out the matrices that define the problem:
,
where Q11 is (36 × 36). Furthermore,
which is also a (36 × 36) matrix.
(36 × 36) = (36 × 36) [(36 × 75) ∗ (75 × 36) + (36 × 18) ∗ (18 × 36)]
In order to form , R is needed. In order to form Q–1 is needed. Therefore, we obtainR as follows:
˙ ( ( ˙ ) ( )) ( ) ( ) .q K q K q q B q u D q w+ + = +r p
q q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
Q l
A 0 0
D B 0
D E B
l
l
l
= =
=
= =
1
11
21
31
111
211
311
411
121
221
321
421
131
231
331
431
11 1
11
21
31
=
A 0 0
D B 0
D E B
r
r
r
r
r
r
r
r
r
r
r
ρρ11
111
211
511
111
121
221
521
121
131
321
531
K q
r ( ˙ ) =
ΩΩΩΩ
ΩΩ
11
21
31
0 0
0 0
0 0
K q HK H B K B K H B K B Kpd d o o d o( ) ˜ ( ) ( )= = + = +1 3 1 1 1
0
K
H
H
H
B B B B
B B B
B B B B
K
K
K
K
B
p
nd d
nd
d d d
nd d d
d
d
d
o
=
+
11
21
31
1 1 11 21 1
01 11 21
1 1 21 01 11
131
11
21
31
11
0 0
0 0
0 0
0
ˆ
ˆ
ˆ
00 0
0
0
B B
B B
K
K
K01 11
01 1
11
21
31
o o
ono
o
o
o
K11d K1
o,
8596Ch18Frame Page 435 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
where the matrix is dimension (75 × 36).
where the dimensions are (36 × 25) = (36 × 75) ∗ (75 × 25).
and the dimensions are (36 × 6) = (36 × 18) ∗ (18 × 6).
and the dimensions are (36 × 24) = (36 × 36) ∗ (36 × 31) ∗ (31 × 24).The control inputs uαij are defined as
l lt
tRq
R
Rq
t
t
t
t
E E
E E E
E E E
E E E
q
q
q
d dd
d
d
d
d
d
= =
=
=
1131
1
0
111
131
11
21
31
6 7
3 4 2
2 3 4
4 2 3
11
21
31
= ˜ [ ],
˜
˜
˜
0
,
B q HBT H B P B P T( ) ˜ [ , ]= = d d o o
B P B P
B B B B
B B B
B B B B
P
P
P
P
d d dn
d dnd
d d d
nd d d
nd
d
d
d
[ ] [ ]
= = 3 1
1 1 11 21 1
01 11 21
1 1 21 01 11
1 1
11
21
31
ˆ
ˆ
ˆ0
0 0 0
0 0 0
0 0 0
0 0 0
B P B P
B
B B
B B
P
P
P
o o o
o
o o
ono
o
o
o
[ ] [ ]
= = 1 1
11
01 11
01 1
11
21
31
0 0
0
0
0 0
0 0
0 0
B =
H
H
H
B P B P
T S
T S
T S
I
I
I
11
21
31
3 1 1 1
1
1
1
2
2
2
0 0 01
0 0
0 0
0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[ ]
′′
′
d o,
L L
uu
u
u
u
u
u
u
u
u
u
=
=
′
=
′′′
×d
o
d
o
d
d
d
o
o
o
1
1
11
21
31
11
21
31
24 1( )
8596Ch18Frame Page 436 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
where
The external forces applied to the nodes arise in the product Dw, where
with dimensions: (36 × 36) = (36 × 36) ∗ (36 × 36).
so
′ =
∈ = =×u id
n
n
n
n
n
n
u
u
u
u
u
u
i j1
2 1
3 1
4 1
6 1
7 1
8 1
6 1 1 2 3 1R , , , .
Bu∈ ×R( ) .36 1
D q HW H W
H
H
H
W
W
W
( ) = o = [ ]=
1
11
21
31
0 0
0 0
0 0
0 0
0 0
0 0
w w
w
w
w
= 1
11
21
31
36 1=
×( )
Dw ∈ ×R (36 1).
8596Ch18Frame Page 437 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Appendix 18.E Nodal Forces
At the base, right end of Figure 18.5, where (i = 1, j = 1):
At the center of the base, where (1 < i < n, j = 1):
At the left end of the base in Figure 18.5, where (i = n, j = 1):
(18.E.1)
At the second stage, where (1 ≤ i ≤ n, j = 2):
f
f f
f f
f + f
f f
f f f f w f f f f w
f f f f w f f f
11
111 211
111 211
311 411
311 411
211 311 1 1 2 1 111 511 111 411 811 211
211 311 1 1 2 1 111 511 111
=
+−
−
=
− + + + + + − − − +− + + + + − − −
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) ( )
( ) (n n
n n 411411 811 211
811 711 1011 321 42 1 311 611 511 212 112 411
811 711 1011 321 42 1 311 611 511 212 112 411
− +− + − + + + − + + +− + − + + − − + + +
f w
f f f f f w f f f f w
f f f f f w f f f f w
)
( ) ( )
( ) ( )
f
f f
f f
f + f
f f
f f f f w f f f f
i
i i
i i
i i
i i
i i i i i i i i
1
1 1 2 1
1 1 2 1
3 1 4 1
3 1 4 1
1 1 1 2 1 1 2 1 3 1 1 1 1 1 4 1 5 1
=
+−
−
=
+ − + + + − − + −− −
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) (( ) ( ) 88 1 2 1
1 1 1 2 1 1 2 1 3 1 1 1 1 1 4 1 5 1 8 1 2 1
8 1 7 1 10 1 3 1 1 4 1 1 3 1 5
i i
i i i i i i i i i i
i i i i i i i
++ − + + − − − − − +
− + − + + + −− −
+ +
w
f f f f w f f f f w
f f f f f w f
)
( ) ( )
( ) (( ) ( )
( ) ( ) 11 6 1 10 1 1 2 2 2 4 1
8 1 7 1 10 1 3 1 1 4 1 1 3 1 5 1 6 1 10 1 1 2 2 2 4 1
+ − + +− + − + + − − + − + +
−
+ + −
f f f f + w
f f f f f w f f f f f + wi i i i i
i i i i i i i i i i i i
( )
( ) ( ) ( )
)
( ) ( )
f
f f
f f
f + f
f f
f f f f w f f f f
n
n n
n n
n n
n n
n n n n n n n n
1
1 1 2 1
1 1 2 1
3 1 4 1
3 1 4 1
1 1 1 2 1 1 3 1 2 1 1 1 1 1 4 1 5 1
=
+−
−
=
+ + − + + − − + −− −
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) (( ) ( ) 88 1 2 1
1 1 1 2 1 1 3 1 2 1 1 1 1 1 4 1 5 1 8 1 2 1
7 1 8 1 10 1 311 411 3 1 10 1 1 5 1
n n
n n n n n n n n n n
n n n n n n
+
+ + − + − − − + − +
− + + − + + + − − +
− −
−
w
f f f f w f f f f w
f f f f f w f f
)
( ) ( )
( ) (
( ) ( )
( ) ff f f w
f f f f f w f f f f f w
6 1 1 2 2 2 4 1
7 1 8 1 10 1 311 411 3 1 10 1 1 5 1 6 1 1 2 2 2 4 1
n n n n
n n n n n n n n n n
+ + +
− + + − + + − − − + + + +
−
)
( ) ( )( )
8596Ch18Frame Page 438 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
(18.E.2)
At the typical stage (1 ≤ j < m, 1 ≤ i ≤ n):
f
f f
f f
f f
f f
f f f f f w f f f f f w
f f f f f
12
112 212
112 212
312 412
312 412
6 1 7 1 9 2 212 312 112 112 412 512 812 912 212
6 1 7 1 9 2 212
=
+−+−
=
− + − − + +( ) + − − + − + +( )− + − − +
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
n n n
n n n 312312 112 112 412 512 812 912 212
712 812 1012 322 422 312 512 612 10 2 113 213 412
712 812 1012 322 422 312 512 612 10 2 113 213 412
+( ) − − − + − + +( )− + + − + +( ) + − + − + + +( )− + + − + +( ) − − + − + + +
w f f f f f w
f f f f f w f f f f f w
f f f f f w f f f f f wn
n(( )
=
+−+−
=
− + − − +− − −
f
f f
f f
f f
f f
f f f f f
i
i i
i i
i i
i i
i i i i i
2
1 2 2 2
1 2 2 2
3 2 4 2
3 2 4 2
6 1 1 7 1 1 9 1 2 2 2 3
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) ( ) ( ) 22 1 2 1 2 4 2 5 2 8 2 9 2 2 2
6 1 1 7 1 1 9 1 2 2 2 3 2 1 2 1 2 4 2 5 2 8 2 9 2 2
+( ) + − − + − + +( )− + − − + +( ) − − − + − + +− − −
w f f f f f w
f f f f f w f f f f f wi i i i i i i
i i i i i i i i i i i i( ) ( ) ( ) 22
7 2 8 2 10 2 3 1 2 4 1 2 3 2 5 2 6 2 10 1 2 1 3 2 3 4 2
7 2 8 2 10 2 3 1 2 4 1 2
( )− + + − + +( ) + − + − + + +( )− + + − + +
+ + −
+ +
f f f f f w f f f f f w
f f f f f wi i i i i i i i i i i i
i i i i i
( ) ( ) ( )
( ) ( ) 33 2 5 2 6 2 10 1 2 1 3 2 3 4 2
2
1 2 2 2
1 2 2 2
3 2 4 2
3 2 4 2
i i i i i i i
n
n n
n n
n n
n n
( ) − − + − + + +( )
=
+−+−
−f f f f f w
f
f f
f f
f f
f f
( )
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
=
− + − − + +( ) + − − + − + +( )− + − − +
− − −
− − −
f f f f f w f f f f f w
f f f f6 1 1 7 1 1 9 1 2 2 2 3 2 1 2 1 2 4 2 5 2 8 2 9 2 2 2
6 1 1 7 1 1 9 1 2 2 2
( ) ( ) ( )
( ) ( ) ( )
n n n n n n n n n n n n
n n n n ff w f f f f f w
f f f f f w f f f f f w
f
3 2 1 2 1 2 4 2 5 2 8 2 9 2 2 2
7 2 8 2 10 2 312 412 3 2 5 2 6 2 10 1 2 1 3 2 3 4 2
7 2
n n n n n n n n
n n n n n n n n n n
n
+( ) − − − + − + +( )− + + − + +( ) + − + − + + +( )− +
−( )
ff f f f w f f f f f w 8 2 10 2 312 412 3 2 5 2 6 2 10 1 2 1 3 2 3 4 2n n n n n n n n n+ − + +( ) − − + − + + +( )
−( )
f
f f
f f
f f
f f
f f f f f w f f f f f w
1
11 21
11 21
31 41
31 41
6 1 7 1 9 21 31 11 11 41 51 81 91 21
j
j j
j j
j j
j j
n j n j nj j j j j j j j j
=
+−+−
=
− + − − + +( ) + − − + − + +− −
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) ( ) (( )− + − − + +( ) − − − + − + +( )
− + + − + +( ) + − + − +− −
+
f f f f f w f f f f f w
f f f f f w f f f f6 1 7 1 9 21 31 11 11 41 51 81 91 21
71 81 101 32 42 31 51 61 10 11 1
n j n j nj j j j j j j j j j
j j j j j j j j nj j
( ) ( )
( ) ++ +( )− + + − + +( ) − − + − + + +( )
+
+ +
f w
f f f f f w f f f f f w21 1 41
71 81 101 32 42 31 51 61 10 11 1 21 1 41
( )
( ) ( )
j j
j j j j j j j j nj j j j
8596Ch18Frame Page 439 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
(18.E.3)
(1 ≤ i ≤ n, j = m)
(18.E.4)
f
f f
f f
f f
f f
f f f f f w f f f
ij
ij ij
ij ij
ij ij
ij ij
i j i j i j ij ij ij ij ij ij
=
+−+−
=
− + − − + + + − − + −− − − − −
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) (( )( ) ( )( ) ( )
1 2
1 2
3 4
3 4
6 1 1 7 1 1 9 1 2 3 1 1 4 5 ff f w
f f f f f w f f f f f w
f f f f f
8 9 2
6 1 1 7 1 1 9 1 2 3 1 1 4 5 8 9 2
7 8 10 3 1 4 1
ij ij ij
i j i j i j ij ij ij ij ij ij ij ij ij
ij ij ij i j i
+ +− + − − + + − − − + − + +− + + − +
− − − − −
+ +
)
( ) ( )
(( )( ) ( )( ) ( )
( ) ( )) ( ) ( ) ( )
( ) ( ) ( ) ( )
) ( )
( ) (j ij ij ij i j i j i j ij
ij ij ij i j i j ij ij ij i j i j
+ + − + − + + +− + + − + + − − + − + +
− + +
+ + − +
w f f f f f w
f f f f f w f f f f f3 5 6 10 1 1 1 2 1 4
7 8 10 3 1 4 1 3 5 6 10 1 1 1 22 1 4
3 4
3 4
6 1 1 7 1 1 9 1
i j ij
nj
nj nj
nj nj
nj nj
nj nj
n j n j n j
( )
( )( ) ( )( ) ( )
)
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
(
+
− − − − −
+
=
+−+−
=
− + − −
w
f
f f
f f
f f
f f
f f f
1 2
1 2
ff f w f f f f f w
f f f f f w f f f f f w2 3 1 1 4 5 8 9 2
6 1 1 7 1 1 9 1 2 3 1 1 4 5 8 9 2
nj nj nj nj nj nj nj nj nj
n j n j n j nj nj nj nj nj nj nj nj nj
+ + + − − + − + +− + − − + + − − − + − + +−
− − − − −
) ( )
( ) ( )
(( )( ) ( )( ) ( )
ff f f f f w f f f f f w
f f f f f w f f f f(
7 8 10 31 41 3 5 6 10 1 1 1 2 1 4
7 8 10 31 41 3 5 6 10 1 1
nj nj nj j j nj nj nj n j n j n j nj
nj nj nj j j nj nj nj n j n
+ + − + + + − + − + + +− + + − + + − − + − +
− + +
−
) ( )
( ) (( ) ( ) ( )
) (( ) ( ) )j n j nj+ ++ +
1 2 1 4f w
f
f f
f f
f f
f f
f f f f f w f f f f f w
1m
m m
m m
m m
m m
n m n m m m nm nm m m m m m
=
+−+−
=
− + − + − + + − + − − + +− −
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( ) (( ) ( )
11 21
11 21
31 41
31 41
6 1 7 1 21 31 9 1 41 51 81 11 91 2nmnm
n m n m m m nm nm m m m m m nm
m m m m m m m nm nm m
)
( ) ( )
( ) ( )
(
( ) ( )− + − + − + − − + − − + +− + − + + + − + − + +−
− −f f f f f w f f f f f w
f f f f w f f f f w 6 1 7 1 21 31 9 1 41 51 81 11 91 2
71 81 32 42 31 51 61 6 7 41
ff f f f w f f f f w
f
f f
f f
f f
f f
f
71 81 32 42 31 51 61 6 7 41
1 2
1 2
3 4
3 4
m m m m m m m nm nm m
im
im im
im im
im im
im im
+ − + + − − + − + +
=
+−+−
=
−
) ( )
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
( 66 1 1 7 1 1 9 1 2 3 1 1 4 5 8 9 2
6 1 1 7 1 1 9 1 2 3
( )( ) ( )( ) ( )
( )( ) ( )( ) ( )
) ( )
(i m i m i m im im im im im im im im im
i m i m i m im im
− − − − −
− − − − −
+ − − + + + − − + − + +− + − − + +
f f f f w f f f f f w
f f f f f w11 1 4 5 8 9 2
7 8 3 1 4 1 3 5 6 1 6 7 1 4
7 8 3 1
im im m im im im im
im im i m i m im im i m im i m im
im im i
) ( )
( ) ( )
(( ) ( ) ( ) ( )
( )
− − − + − + +− + − + + + − − + + +− + −
+ + − −
+
f f f f f w
f f f f w f f f f w
f f f
i
mm i m im im i m im i m im
nm
nm nm
nm nm
nm nm
nm nm
+ + − − − + + +
=
+−+−
+ − −f w f f f f w
f
f f
f f
f f
f f
4 1 3 5 6 1 6 7 1 4
1 2
1 2
3 4
3 4
( ) ( ) ( )) ( )
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
=
− + − − + + + − − + − + +− + − −
− − − − −
− − − − −
( ) ( )
(( )( ) ( )( ) ( )
( )( ) ( )( ) ( )
f f f f f w f f f f f w
f f f f6 1 1 7 1 1 9 1 2 3 1 1 4 5 8 9 2
6 1 1 7 1 1 9 1 2
n m n m n m nm nm nm nm nm nm nm nm nm
n m n m n m nm ++ + − − − + − + +− + − + + + − + − + +− + − + +
− −
f w f f f f f w
f f f f w f f f f w
f f f f w
3 1 1 4 5 8 9 2
7 8 31 41 3 5 6 6 1 7 1 4
7 8 31 41 3
nm nm nm nm nm nm nm nm
nm nm m m nm nm nm n m n m nm
nm nm m m nm
) ( )
( ) ( )
(( ) ( )
)) ( )( ) ( )− − + − + +
− −f f f f w 5 6 6 1 7 1 4nm nm n m n m nm
8596Ch18Frame Page 440 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Or, in matrix notation, with the implied matrix definitions,
(18.E.5)
ff
ff
f
f
ff
f
ff
f
w
w
w
w
w
ijo
ijijd
ij
ij
ij
=
=
=
5
1
2
3
4
6
7
8
9
10
1
2
3
4
, ,
f f
I I I
I I I
I
I
I
I
I I I
I I
11
3
31
3 3 3
3 3 3
3
3
3
3
3 3 3
3 3
=
+
− −−
−
−
−−
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
0
0 0 0
0 0 0
0 0 0
0 0 0
0 n
d
00
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
I
f
f
3
11
3
3
123 3
3 3
+
−
+−−
d
d
+−−
−
+
−
+
+−
f
I
I
I
I
I
If
I
I
f
I
If
I I
I I
21
3
3
3
3
3
311
3
3
12
3
31 1
3 3
3 3
d
o on0
0
0
0
0
0
0
0
0
0
0 0
0 0
0 0 II I
I I
w3 3
3 3
11
0 0 −
.
f B f B f B f B f B f B f B f Ww11 1 1 11 11 12 12 21 21 11 11 12 12 1 1 1 1 11= + + + + + + +nd
nd d d d d d d o o o o
n n .
f
I
I
I
I
f
I I I
I I I
I
I
I
I
I I I
I I I
i id
1
3
3
3
3
1 1
3 3 3
3 3 3
3
3
3
3
3 3 3
3 3 3
=
+
− −−
−
−
−−
−
0 0 0
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
0 0 0
0
0
0 0 0
0 0 0
0 0 0
0 0 0
0
0
( )
fid1
+
−
+−−
+
+
0 0 0 0
0 0 0 0
0 0 0
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0
0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
0
I
I
fI I
I I
f
3
3
23 3
3 3
1 1 id
id( )
00
0
0
0
0
0
0
0
0
0
0
0
0 0
0 0
0 0
I
If
I
I
I
I
I
If
I
I
f
I I
I I
I I
3
31 1
3
3
3
3
3
31
3
3
2
3 3
3 3
3 3
+−−
−
+
−
+−
− ( )io
io
io
00 0 I I
w
3 3
1
−
i
8596Ch18Frame Page 441 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Or, in matrix form,
(18.E.6)
Or, in matrix form,
(18.E.7)
f B f B f B f B f B f
B f B f Ww
id
id d
id d
id d
id o
io
oio o
io
i
1 01 1 1 11 1 12 2 21 1 1 01 1 1
11 1 12 2 1
= + + + +
+ + +
− + −( ) ( ) ( )
.
f
I
I
I
I
f
I I I
I I I
I
I
I
I
I I I
I I I
n nd
1
3
3
3
3
1 1
3 3 3
3 3 3
3
3
3
3
3 3 3
3 3 3
=
+
− −−
−
−
−−
−
0 0 0
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
0 0 0
0
0
0 0 0
0 0 0
0 0 0
0 0 0
0
0
( )
fnd1
+
−
+−−
+
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
f f
0
0
0
0
I
I
3
3
nd d2
3 3
3 3
11
3
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
331 1
3
3
3
3
13
3
2
3
31 1
3 3
0
0
f
I
I
I
I
0
0
0
0
f
0
0
0
0
0
0
I
I
f
I
I
0
0
f
I I 0 0
+−−
+
−
+
−
+
−( )no
no
no
n
II I 0 0
0 0 I I
0 0 I I
w3 3
3 3
3 3
1
−
−
n .
f B f B f B f B f B f
B f B f B f Ww
nd
nd d
nd d
nd d d o
no
no
no o
no
n n n
1 01 1 1 11 1 12 2 21 11 01 1 1
1 1 12 2 1 1 1 1 1
= + + + +
+ + + +
− −( ) ( )
.
f
I I I
I I I
I
I
I I
I I
I I I
I I I
f12
3 3 3
3 3 3
3
3
3 3
3 3
3 3 3
3 3 3
123 3
3 3
=
− −−
−
−−
−−
+−−
0
0
0 0 0
0 0 0
0 0
0 0
0
0
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
d
+
−
+
−−
f
f
22
3
3
13
3
3
3
3
d
d
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
fnd1
8596Ch18Frame Page 442 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Or, in matrix form,
(18.E.8)
Or, in matrix form,
(18.E.9)
f
I
I
I
I
I
If
I
I
f+
−−
−
+−−
−
+
−
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 I 0
0 0 I 0
0 0 0 I
0 0 0 I
0
0
0
0
0
0
0
03
3
3
3
2
3
3
3
3
3
312
3
3
nd o
1313
3 3
3 3
3 3
3 3
12
o
I I
I I
I I
I I
w+−
−
0 0
0 0
0 0
0 0
.
f B f B f B f B f B f
B f B f Ww
12 12 12 21 22 12 13 1 1 2 2
11 12 12 13 12
= + + + +
+ + +
d d d d d dnd
nd
nd
nd
o o o o .
f f
I I I
I I I
I
I
I I
I I
I I I
I
i id
2
3
3
3
31 1
3 3 3
3 3 3
3
3
3 3
3 3
3 3 3
3
=
−−
+
− −−
−
−−
−−
−
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0
0
0 0 0
0 0 0
0 0
0 0
0 ( )
II I
f
f
3 3
2
3
3
3
3
3
3
3
0
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 I 0
0 0 I 0
0 0 0 I
0 0 0 I
+
−
+
−−
−
id
id
+−−
+−−
−
+
−
+
f
f
I
I
I
I
I
If
( )
( )
id
id
io
1 2
3 3
3 3
1 2
3
3
3
3
3
32
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
0
00
0
0
0
0
0
0 0
0 0
0 0
0 0
I
I
f
I I
I I
I I
I I
w3
3
3
3 3
3 3
3 3
3 3
2
−
+−
−
io
i .
f B f B f B f B f B f
B f B f Ww
i nd
id d
id d
id
nd
id d
id
oio o
io
i
2 1 1 1 12 2 12 3 2 1 2 21 1 2
11 2 12 3 2
= + + + +
+ + +
− − +( ) ( ) ( )
.
f f fnd
nd
23 3
3 3
12
3
3
3
31 1=
−−
+
−−
−
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
( )
8596Ch18Frame Page 443 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Or, in matrix form,
(18.E.10)
Or, in matrix form,
(18.E.11)
+
−−
−
+
− −−
−
−−
−−
−
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 I 0
0 0 I 0
0 0 0 I
0 0 0 I
0
0
0 0 0
0 0 0
0 0
0 0
0
3
3
3
3
1 2
3 3 3
3 3 3
3
3
3 3
3 3
3 3 3
3 3
f
I I I
I I I
I
I
I I
I I
I I I
I I
( )nd
00
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
0
0
0
0
0
I
f
f I
I
f
I
I
I
I
I
I
3
2
3
3
33
3
3
3
3
3
3
3
+
−
+
−
+−−
−
nd
nd
no 33
2
3 3
3 3
3 3
3 3
2
0
0
0 0
0 0
0 0
0 0
+−
−
f
I I
I I
I I
I I
w
no
n .
f B f B f B f B f B f B f
B f Ww
nd d
nd
nd
nd
nd d
nd d
nd o
no
ono
n
2 21 12 1 1 1 2 1 2 12 2 12 3 12 3
11 2 2
= + + + + +
+ +
− −( ) ( )
.
f f
I I I
I I I
I
I
I I
I I
I I I
I I
1
3
3
3
31
3 3 3
3 3 3
3
3
3 3
3 3
3 3 3
3
j n jd=
−−
+
− −−
−
−−
−−
−
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0
0
0 0 0
0 0 0
0 0
0 0
0( )
33 3
1
3
3
3
3
3 3
3 3
0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 I 0
0 0 I 0
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
I
f
f
+
−−
−
+−−
jd
njd
+
−
+−−
−
+
+
f
f
I
I
I
I
I
If
2
3
3
1 1
3
3
3
3
3
31
jd
jd
jo
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
0
0
0
0
0
0
0
( )
II
I
f
I I
I I
I I
I I
w3
3
1 1
3 3
3 3
3 3
3 3
1
−
+−
−
+( ) .jo
j
0 0
0 0
0 0
0 0
f B f B f B f B f B f
B f B f Ww
1 1 1 12 1 2 21 2 12 1 1
11 1 12 1 1 1
j nd
n jd d
jd
nd
njd d
jd d
jd
oj
o oj
oj
= + + + +
+ + +
− +
+
( ) ( )
( ) .
8596Ch18Frame Page 444 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Or, in matrix form,
(18.E.12)
f f
I I I
I I I
I
I
I I
I I
I I Iij i jd=
−−
+
− −−
−
−−
−−
− −
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0
0
0 0 0
0 0 0
0 0
0 0
0
3
3
3
31 1
3 3 3
3 3 3
3
3
3 3
3 3
3 3 3
( )( )
II I I
f
3 3 30
ijd
+
−−
−
+−−
− +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 I 0
0 0 I 0
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
3
3
3
3
13 3
3 3
1 f f( ) (i jd
i ))
( )
jd
i jd
ijo+
−
+−−
−
+
−
+
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
0
0
0
0
0
0
0
3
3
1
3
3
3
3
3
3
3
3
f
I
I
I
I
I
If
I
I
+−
−
+f
I I
I I
I I
I I
wi jo
ij( ) .1
3 3
3 3
3 3
3 3
0 0
0 0
0 0
0 0
f B f B f B f B f B f
B f B f Ww
ij nd
i jd d
ijd
nd
i jd d
i jd d
i jd
oijo o
i jo
ij
= + + + +
+ + +
− − − + +
+
1 1 1 12 2 1 21 1 12 1
11 12 1
( )( ) ( ) ( ) ( )
( ) .
f f
I I I
I I I
I
I
I I
I I
I I Inj n jd=
−−
+
− −−
−
−−
−−
− −
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0
0
0 0 0
0 0 0
0 0
0 0
0
3
3
3
31 1
3 3 3
3 3 3
3
3
3 3
3 3
3 3 3
( )( )
II I I
f
f
3 3 3
3
3
3
3
13 3
3 3
0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 I 0
0 0 I 0
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0
+
−−
−
+−−
−
njd
n jd ( ) 00 0
0 0 0 0
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
0
0
0
+
−
+−−
−
++
f
f
I
I
I
I
I
If
1
3
3
1
3
3
3
3
3
3
jd
n jd
njo
( ) 00
0
0
0
0 0
0 0
0 0
0 0
I
I
f
I I
I I
I I
I I
w
3
3
1
3 3
3 3
3 3
3 3
−
+−
−
+n jo
nj
( )
.
8596Ch18Frame Page 445 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Or, in matrix form,
(18.E.13)
Or, in matrix form,
(18.E.14)
Or, in matrix notation,
(18.E.15)
f B f B f B f B f B f
B f B f Ww
nj nd
n jd d
njd
nd
n jd d
jd d
n jd
onjo o
n jo
nj
= + + + +
+ + +
− − − +
+
1 1 1 12 2 1 21 1 12 1
11 12 1
( )( ) ( ) ( )
( ) .
f
I I I
I I I
I
I
I I
I I
I I
I I
f1
3 3 3
3 3 3
3
3
3 3
3 3
3 3
3 3
13 3
3 3
m md=
− −−
−
−−
−−
+−−
0
0
0 0 0
0 0 0
0 0
0 0
0 0
0 0
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
f2md
+−
−
+
−−
+−
−( )
0 0 0 0
0 0 0 0
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 03
3
3
3
3
3
3
31
3
f f
I
nmd
n md
II
I
I
I
If
I I
I I
I I
I I
W3
3
3
3
31
3 3
3 3
3 3
3 3
1−
−
+−
−
0
0
0 0
0 0
0 0
0 0
mo
m .
f B f B f B f B f B f Ww1 1 1 21 2 1 1 11 1 1m md
md d
md
nmd
nmd
nd
n md o
mo
m= + + + + +−( ) .
f fim i md=
−−
+−
−−
−
−( ) −( )
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 I
0 0 0 I
0 0 I 0
0 0 I 0
I 0 0 0
I 0 0 0
3
3
3
31 1
3
3
3
3
3
3
− −−
−
−−
−−
+−−
−( )f
+
I I I
I I I
I
I
I I
I I
I I
I I
f
i md
md
1
3 3 3
3 3 3
3
3
3 3
3 3
3 3
3 3
13 3
3 3
0
0
0 0 0
0 0 0
0 0
0 0
0 0
0 0
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
00 0 0 0
0 0 0 0
0
0
0 0
0 0
0 0
0 0
−−
−
+−
−
+( )f
+
I
I
I
I
I
If
I I
I I
I I
I I
w
i md
imo
im
1
3
3
3
3
3
3
3 3
3 3
3 3
3 3
.
f B f B f B f B f
+ B f Ww
im nd
i md
nmd
i md
md
imd d
i md
oimo
im
= + + +
+
−( ) −( ) −( ) +( )1 1 1 1 1 21 1
11 .
8596Ch18Frame Page 446 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
Or, in matrix notation,
(18.E.16)
Now assemble (18.E.5)–(18.E.16) into the form
(18.E.17)
where
f fnm n md=
−−
+−
−−
−
− −
0 0 0 I
0 0 0 I
0 0 0 0
0 0 0 0
I 0 0 0
I 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 I
0 0 0 I
0 0 I 0
0 0 I 0
I 0 0 0
I 0 0 0
3
3
3
31 1
3
3
3
3
3
3
( )( )
+
− −−
−
−−
−−
+−−
−f
I I I
I I I
I
I
I I
I I
I I
I I
f
( )n md
nmd
1
3 3 3
3 3 3
3
3
3 3
3 3
3 3
3 3
3 3
3 3
0
0
0 0 0
0 0 0
0 0
0 0
0 0
0 0
0 0 0 0
0 0 0 0
0 I I 0
0 I I 0
0 0 0 0
0 0 0 0
0 0 00 0
0 0 0 0
0
0
0 0
0 0
0 0
0 0
+−−
−
+−
−
f
I
I
I
I
I
If
I I
I I
I I
I I
w
1
3
3
3
3
3
3
3 3
3 3
3 3
3 3
md
nmo
nm .
f B f B f B B f
B f Ww
nm nd
n md
nmd
n md
md
nmd d
md
onmo
nm
= 1 1 1 1 1 1 21 1
11
( )( ) ( )
.
− − −+ + +
+ +
f
f B f B f B f B f W w1 3 1 4 2 1 1 2 2 1 1 = d d o o+ + + +
ff
ff
f
f
f
f
f
f
f
w
w
w
w
f
f
f1
1 1
1
1
2
1
2
1
2
1
2d nd j
d
jd
jd
njd
jo
jo
jo
njo
j
j
j
nj
j
j
j=
=
=
=
=, , , ,M M M M
ffnj
B
B B B 0 0 B
0 B 0
0 0
0 B
B B 0 0 B B
3
1 1 11 21 1
01
21
1 1 21 01 11
=
nd d
nd
d
d
nd d d
L
O O O
M O O O O O M
O O O O
O O O O
L
B B B
B B B
4 12 12
2 12 12
= … …
= … …
BlockDiag
BlockDiag
d d
o d
[ , , , ],
[ , , , ],
8596Ch18Frame Page 447 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
W1 = BlockDiag […, W, W, …].
Also, from (18.E.5)–(18.E.16)
(18.E.18)where
Also from (18.E.5)–(18.E.16)
(18.E.19)
(18.E.20)
Or, simply, the vector form of (18.E.17)–(18.E.19) is
(18.E.21)
where
B
B
B
B
B B
1
11
01
11
01 1
=
o
o
o
ono
0 0
0
0
0 0
L L
O O M
O O O M
M O O
L
,
f = B f B f B f B f B f W w 2 5 2 6 2 4 3 7 2 2 3 1 2d d d o o+ + + + + ,
B 0, B B B B5 5 7 11 11= [ ] = [ ], , , , ,BlockDiag o oL L
B
B
B
B
B
B B B
B
B
B B B
5
1
1
1
6
12 21 2
2
21
21 2 12
=
=
0 0 0
0
0
0
0 0 0
0 0
0
0
0
0
0 0
L
O O
O O O M
M O O O
L
L
O O O
O O O O M
M O O O O
O O O
L
nd
nd
nd
d dnd
nd
d
dnd d
,
.
f B f B f B f B f B f W w 3 5 2 6 3 4 4 7 3 2 4 1 3= + + + + +d d d o o ,
f = B f B f B f W w , m md
md
mo
m5 1 8 7 1( )− + + +
B
B B 0 0 B
B 0
0
0
0 B
B 0 0 B B
8
1 21
21
21 1
=
md d
nmd
nm
d
dnmd
md
L
O O O
O O O O M
M O O O O
O O O
f = B f B f W w d d o o o+ + ,
f
f
f
f
f
f
f
f
f
f
w
w
w
=1
1
21 1
MM
M M
m
d
d
d
md
o
o
mo
m
=
=
=
, , , ,
8596Ch18Frame Page 448 Wednesday, November 7, 2001 12:18 AM
© 2002 by CRC Press LLC
References
1. R. E. Skelton, Systems design, the absentee in system theory, American Automatic ControlConference, San Diego, CA, June 1999.
2. R. E. Skelton, A systems approach to integrate structure and control design, SPIE Conference onMathematics for Smart Structures, Newport Beach, CA, March 2000.
3. K. M. Grigoriadis, G. Zhu, and R. E. Skelton, Optimal redesign of linear systems, Journal ofDynamic Systems, Measurement and Control, 118(3), 598–605, September 1996.
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