On the Use of Quaternions and Euler1

Rodrigues Symmetric Parameters with2

Moments and Moment Potentials3

Nur Adila Faruk Senan a, Oliver M. O’Reilly a,∗4

aDepartment of Mechanical Engineering, University of California at Berkeley,5

Berkeley, CA 94720, U. S. A.6

Abstract7

This paper contains a comprehensive discussion of the use of Euler-Rodrigues sym-8

metric (or Euler symmetric) parameters to parameterize a potential energy function.9

By exploiting the equivalence of these parameters to unit quaternions, several rep-10

resentations for moments derivable from moment potentials are established. These11

representations are applied to a system of two rigid bodies connected by an elastic el-12

ement, and the issue of identification of the potential energy function using moment13

measurements. It is also shown how the representations can be used to prescribe14

constraint moments, and how they illuminate the stiffness matrices associated with15

certain moment potentials.16

Key words: Rigid body dynamics, rotations, stiffness matrices, Euler parameters,17

quaternions, moment potentials, potential energy.18

1 Introduction19

In rigid body dynamics, representations for the conservative moments asso-20

ciated with a moment potential have been the subject of several works. De-21

pending on the parameterization of the rotation, distinct representations are22

obtained. For Euler angles, a representation featuring a dual basis is discussed23

in [1,2]. Prior to these works, Antman [3] had presented a representation when24

∗ Corresponding author: Address: 6137 Etcheverry Hall, Department of MechanicalEngineering, University of California at Berkeley, Berkeley, CA 94720, U. S. A.

Email address: oreilly@berkeley.edu (Oliver M. O’Reilly).URL: http://www.me.berkeley.edu/faculty/oreilly/index.html (Oliver

M. O’Reilly).

Preprint submitted to Elsevier 27 October 2008

the rotation is parameterized by an axis and angle of rotation and Simmonds25

[4] established a representation featuring the Rodrigues vector. When the rigid26

body motion is represented as a screw motion, expressions for the associated27

conservative moments were recently established in Howard et al. [5] and Zefran28

and Kumar [6].29

Of particular interest in this paper is the situation where Euler-Rodrigues sym-30

metric (or Euler symmetric) parameters are used. In this case, representations31

for the components (relative to a body fixed basis) of a conservative moment32

associated with a moment potential are available (see, e.g., [7–9]). However,33

derivations of the representations and discussions of the precise role played34

by the Euler parameter constraint are absent from the literature. These issues35

are important when attempting to extend the representation to the case of a36

system of two coupled rigid bodies (cf. Figure 1). Systems of this type arise37

in many areas of mechanics and, in particular, in a vertebral motion segment38

in the spine [10–12]. This segment features two vertebral discs and their as-39

sociated interverterbral disc and the associated moment potential models the40

elastic behavior of the intervertebral disc. Accurately estimating the moment41

potential from moment measurements for such systems is also difficult and,42

in this respect, the Euler-Rodrigues symmetric parameter representation has43

several distinct advantages over traditional methods which predominantly fea-44

ture Euler angles. First, an elegant direct formulation exists for estimating the45

Euler-Rodrigues symmetric parameters associated with a series of measure-46

ments of a rotation (see, e.g., [13–15]). Second, as shown later in the present47

paper, the estimation of a moment potential which is a quadratic function of48

the Euler-Rodrigues symmetric parameters can be easily achieved.49

a1

b1

x1

x2

F1

M1

F2

M2

X1

X2

B2

E

B1

O

Fig. 1. Schematic of two rigid bodies B1 and B2 which are connected by an elasticelement E.

With the help of results from the theory of generalized inverses [16], new proofs50

2

of several representations for the conservative moment M associated with a51

moment potential U are established. In addition, the role played by the Euler52

parameter constraint in calculating these representations is elaborated upon.53

Specifically, the equivalence of calculating M by imposing this constraint after54

the derivatives of U have been computed is shown. The key to this equiva-55

lence lies in the exploitation of a motion which was originally discussed by56

Gauss [17]. The representations for M are shown to be compatible with repre-57

sentations for the generalized forces featuring in Lagrange’s equations for the58

rotational motion of a rigid body that are discussed in the literature (e.g., in59

[9,18–21]). To further illustrate the representations, various examples of mo-60

ment potentials and their associated conservative moments are discussed. In61

addition, it is shown how the representations for M naturally lead to repre-62

sentations for constraint moments, and how they can be used to characterize63

the conservative moments present in two rigid bodies connected by an elastic64

element. The final contribution of the present paper is a demonstration of65

how the potential energy function can be estimated from measurements of the66

moment and Euler-Rodrigues symmetric parameters.67

For additional background information on rotations and their parameteriza-68

tions, the reader is referred to the Shuster’s authorative review [22].69

Notation70

In the present paper, a tensor notation for the rotation is employed following71

[23,24]. This notation allows one to clearly track a variety of frames of refer-72

ence, and enables succinct derivations of multiple representations of various73

vectors. A rapid summary is provided here. All vectors in Euclidean three-74

space E3 are denoted by bold-faced letters. The set {E1,E2,E3} denotes a75

fixed right-handed orthonormal basis for E3.76

The tensor product a⊗ b of any two vectors is defined as77

(a ⊗ b) c = (b · c) a. (1.1)78

for any vector c. A tensor can be considered as a linear operator which trans-79

forms a vector in E3 to another vector in E

3. For example, a ⊗ b transforms80

c to the vector (b · c) a.81

The linear operator ǫa is defined as82

(ǫa)b = −a × b. (1.2)83

for all vectors a and b. Given a skew-symmetric tensor V = −VT , where84

3

the subscript T denotes transpose, its axial vector v is defined as the unique85

vector which satisfies the identity86

v × a = Va, (1.3)87

for all vectors a. The primary example of an axial vector in this paper is the88

angular velocity vector ω which is associated with the skew-symmetric tensor89

RRT where R is a rotation tensor.90

2 Background on Euler-Rodrigues Symmetric Parameters and Quater-91

nions92

The use of four Euler-Rodrigues symmetric (or Euler symmetric) parameters93

to parameterize a rotation (or proper-orthogonal transformation) dates to Eu-94

ler [25] in 1771 and Rodrigues [26] in 1840. We denote these parameters by95

the pair (β0, β) where β is a scalar and β is a vector. In 1843, Hamilton made96

his discovery of quaternion multiplication and shortly afterwards Cayley [27]97

published results showing how quaternions could be used to parameterize a98

rotation. 199

Defining a quaternion as the pair (q0,q) where q0 is a scalar and q is a vector,100

then the following correspondences between a quaternion and a set of Euler-101

Rodrigues symmetric parameters exist: β0 = q0√

qand the vector β = q

√q

where102

q = q2

0+ q · q. (2.1)103

When q = 1, the quaternion (q0,q) is known as a unit quaternion. Unit quater-104

nions satisfy the Euler parameter constraint:105

q2

0+ q · q = 1. (2.2)106

Thus, a unit quaternion can be used to define a set of Euler-Rodrigues sym-107

metric parameters and vice versa. In the four-dimensional space parameterized108

by the components of a quaternion, the set of all unit quaternions define a109

unit sphere which is known as the 3-sphere S3.110

1 For additional background on the relationships between the works of Cayley,Gauss, Hamilton, and Rodrigues, the reader is referred to Altmann [28,29].

4

2.1 The Rotation Tensor and its Components111

The parameters q0√

qand q

√q

can be used to define a rotation about an axis r112

through an angle φ using the identifications113

β0 =q0√q

= cos

(

φ

2

)

, β =q√q

= sin

(

φ

2

)

r. (2.3)114

The resulting representation of the rotation tensor R is115

R = R (q0,q) =1

q

[(

q2

0− q · q

)

I + 2q ⊗ q − 2q0 (εq)]

. (2.4)116

Suppose that R transforms Ek to a vector ek: REk = ek where k = 1, 2, 3. It117

can be shown that {e1, e2, e3} is a right-handed orthonormal basis for E3 and118

that R =∑

3

i=1ei ⊗ Ei.119

As Rr = r, is easy to show that120

rk = r · ek = r ·Ek, βk = β · ek = β · Ek, qk = q · ek = q · Ek.(2.5)121

This common feature of the vectors r, β, and q is the source of several inter-122

esting identities pertaining to their time derivatives.123

Examining the components Rik = ek · Ei of the tensor R one finds that124

R =

R11 R12 R13

R21 R22 R23

R31 R32 R33

=

(

2q2

0− q

q

)

1 0 0

0 1 0

0 0 1

+2

q

q2

1q1q2 q1q3

q1q2 q2

2q2q3

q1q3 q2q3 q2

3

+2q0

qQ, (2.6)

where125

Q = Q (q) =

0 −q3 q2

q3 0 −q1

−q2 q1 0

. (2.7)

5

Examining (2.6) in closer detail, it becomes evident that126

qRik =[

q0 q1 q2 q3

]

Fik

q0

q1

q2

q3

, (i = 1, 2, 3, k = 1, 2, 3) . (2.8)127

Each of the nine 4× 4 matrices Fik are symmetric and proper-orthogonal. For128

example,129

F11 =

1 0 0 0

0 1 0 0

0 0 −1 0

0 0 0 −1

, F31 =

0 0 −1 0

0 0 0 1

−1 0 0 0

0 1 0 0

. (2.9)130

The matrices Fik will play a role in constructing stiffness matrices later on.131

2.2 Angular Velocity Vectors132

The angular velocity vector ω associated with the rotation tensor R is defined133

as the axial vector of the skew-symmetric tensor RRT . Differentiating the134

identity ek = REk with respect to time t, it can be shown that135

ek = ω × ek. (2.10)136

Given any vector a, the corotational derivative of a =∑

3

i=1aiei is defined as137

oa=

3∑

i=1

aiei. (2.11)138

Thus,139

a =oa +ω × a. (2.12)140

For vectors, such as the axis of rotation r, the parameter vector β, and the141

quaternion vector q, which have the same components in the fixed {E1,E2,E3}142

6

and corotational {e1, e2, e3} bases, the time derivative and corotational time143

derivatives are simply related:144

r = RT or, β = RT

o

β, q = RToq . (2.13)145

To verify the first of these results, one notes that r =∑

3

k=1rkEk =

∑

3

i=1riei.146

Computingor=

∑

3

i=1riei and then using the identity RTei = Ei confirms147

(2.13)1. The proofs of the other two results are identical.148

It can be shown that ω has the representation 2149

ω =2

q(q0q − q0q + q × q) . (2.14)150

With some lengthy but straightforward manipulations with the help of (2.13)151

and (2.14) it can be shown that152

q0q + q × q = q0

oq −q× o

q . (2.15)153

Consequently, ω has the equivalent representation154

ω =2

q

(

q0

oq −q0q − q× o

q)

. (2.16)155

Taking the Ei and ek components of ω leads to the results 3156

ω · E1

ω · E2

ω · E3

= A

q0

q1

q2

q3

,

ω · e1

ω · e2

ω · e3

= C

q0

q1

q2

q3

, (2.17)157

2 See Sections 6.5 and 6.9 of [24].3 By way of background, the components of the column vector (2.17)2 are equivalentto Equation (312) in Shuster [22], and these are also the representation for thecomponents ωi that are used in [18,21]. Similarily, the components of the columnvector (2.17)1 are identical to the components of the vector ωI presented in Equation(322) in Shuster [22] (cf. [30]).

7

where158

A =2

q

−q1 q0 −q3 q2

−q2 q3 q0 −q1

−q3 −q2 q1 q0

, C =2

q

−q1 q0 q3 −q2

−q2 −q3 q0 q1

−q3 q2 −q1 q0

. (2.18)159

The matrices A and C have generalized inverses:160

A− =q

4AT C− =

q

4CT . (2.19)161

That is, AA− = I and CC− = I, where I is the 3 × 3 identity matrix.162

For completeness, it is noted that a different angular velocity ω0 can be defined163

as the axial vector of RT R. It can be shown that164

ω0 = RT ω. (2.20)165

Using the identities Rq = q, (2.13)3, and RT (a× b) =(

RTa× RTb)

, and166

the representation (2.16), it follows that ω0 has the representation167

ω0 =2

q(q0q − q0q − q × q) . (2.21)168

As discussed in [31], this representation is convenient to use when computing169

Lagrange’s equations of motion.170

e1

e2

e3

e1

e2e3

Sphere of radius qRSphere of radius R

mutation

Fig. 2. Schematic of a mutation of a sphere of radius R. As a result of the mutation,the sphere is rotated into a sphere of radius qR. The parameters of the rotation andthe expansion are prescribed by the quaternion (q0,q), and q = q2

0+ q · q.

8

2.3 Mutations and Rotations171

If one considers a rigid body with a fixed point, then the motion of the rigid172

body is a rotation about the fixed point. In this case, the motion can be173

parameterized using a set of Euler-Rodrigues symmetric parameters or their174

corresponding unit quaternion. The deformation gradient for such a rigid con-175

tinuum is simply given by F = R (q0,q).176

A more general situation arises when a motion of a continuum is parameter-177

ized by a quaternion whose magnitude is not subject to the Euler parameter178

constraint (2.2). Such a motion was discussed by Gauss [17] and he referred179

to it as a “mutation of space.” 4 During this motion, the continuum is free to180

rigidly rotate and can also experience uniform expansions and contractions.181

The deformation gradient of the continuum is F = qR (q0,q) (see Figure 2).182

As noted by O’Reilly and Varadi [31] this motion can be visualized using a183

toy known as the Hoberman sphere.184

3 Derivatives of Functions of Quaternions185

Of particular interest in this paper are the derivative of scalar-valued functions186

of a quaternion and the derivatives of these functions when their domain is187

restricted to the set of unit quaternions. For functions of the former type, one188

can use the correspondence between quaternions and rotations to define two189

equivalent representations:190

V = V (q0, q1, q2, q3) = V (R (q0,q) , q) . (3.1)191

The restriction of V to the set of unit quaternions is denoted by a subscript c:192

Vc = Vc (q0, q1, q2, q3) = Vc (R (q0,q)) . (3.2)193

In this section, representations for combinations of the derivatives of V and Vc194

which arise when expressing V and Vc using the angular velocity vector ω are195

established. The resulting representations ((3.9), (3.10), (3.16), and (3.17))196

are useful in computing constraint moments and conservative moments.197

4 It is known [28,29,32] that this posthumously published work by Gauss can beconsidered to contain Hamilton’s formula for quaternion multiplication and theRodrigues’ formula for the composition of two rotations.

9

3.1 The Unconstrained Case198

The first case of interest arises where the Euler parameter constraint is not199

imposed. Here, the function V can represent the potential energy of a body200

which is undergoing a “mutation.”201

Assuming that the components of the quaternion are functions of time, one202

computes that203

V = tr

(

∂V

∂RRT

(

RRT)T)

+∂V

∂qq, (3.3)204

where tr denotes the trace operator and205

∂V

∂R=

3∑

i=1

3∑

k=1

∂V

∂Rik

Ei ⊗ Ek. (3.4)206

As RRT is a skew-symmetric tensor, the first term in this equation can be207

replaced by a vector product: 5208

tr

(

∂V

∂RRT

(

RRT)

)

= v · ω. (3.5)209

Here, the vectors ω and v are axial vectors. That is,210

ω × a=(

RRT)

a,

v × a=

∂V

∂RRT −R

(

∂V

∂R

)T

a, (3.6)

for any vector a. Thus,211

V = v · ω +∂V

∂qq. (3.7)212

Computing the derivative of V , equating it to the derivative to V , and using213

5 Further details on calculations of this type can be found in Section 6.10 of [24].

10

a representation of the form (2.16), it can be shown that (3.7) implies that214

∂V∂q0

∂V∂q1

∂V∂q2

∂V∂q3

= CT

v · e1

v · e2

v · e3

+

2∂V∂q

q0

2∂V∂q

q1

2∂V∂q

q2

2∂V∂q

q3

. (3.8)215

Since the columns of CT are orthogonal to a quaternion, the term containing216

the vector v can be eliminating by multiplying each of the individual equa-217

tions in (3.8) by q0, q1, q2, and q3, respectively, and then adding the resulting218

expressions. This computation leads to the conclusion that219

∂V

∂q=

1

2q

(

∂V

∂q0

q0 +∂V

∂q1

q1 +∂V

∂q2

q2 +∂V

∂q3

q3

)

. (3.9)220

To solve for v, it should be recalled that the matrix CT has the generalized221

inverse q

4C Using this generalized inverse and the representation (3.9) for ∂V

∂q,222

one can solve (3.8) for the components of v. To do this, one appeals to a result223

from Rao and Mitra [16]: The general solution to the equation Bx = b is x =224

B−b+(I − B−B) y where y is an arbitrary vector and B has a generalized inverse225

B−. As q

4CCT = I for the case at hand, the solution for the ei components of226

v is easily obtained. A similar procedure using (2.14) instead of (2.16) yields227

the Ei components of v. Thus,228

v =1

2

(

−∂V

∂q0

q + q0

∂V

∂q+ q × ∂V

∂q

)

=1

2

(

−∂V

∂q0

q + q0

∂v

∂q− q × ∂v

∂q

)

, (3.10)

where229

∂V

∂q=

3∑

i=1

∂V

∂qi

Ei,∂v

∂q=

3∑

i=1

∂V

∂qi

ei. (3.11)230

The first of (3.10) readily provides the components v ·Ek while (3.10)2 can be231

used to easily compute v ·ek. It should be emphasized that the representations232

(3.9) and (3.10) are valid when the components q0 and q are not subject to a233

constraint.234

11

q0 q1

q2

q3

S3

(

∂Vc

∂q0

∂Vc

∂q1

∂Vc

∂q2

∂Vc

∂q3

)

(

vc · ∂ω∂q0

vc · ∂ω∂q1

vc · ∂ω∂q2

vc · ∂ω∂q3

)

2λ(

q0 q1 q2 q3

)

P

Fig. 3. Schematic of the 3-sphere S3. At a point P on this sphere, the derivative ofa function Vc of the quaternion parameters can be divided into a component normalto this surface and a component tangent to the surface (cf. equation (3.15)).

3.2 The Unit Quaternion Case235

For the case where the components of the quaternion are restricted so as to236

satisfy q = 1, then the procedure leading to (3.10) needs modification. For237

this case, one examines the expression for Vc and compares it to (3.7) to find238

that239

∂Vc

∂q0

q0 +3∑

k=1

∂Vc

∂qk

qk = vc · ω. (3.12)240

Here, vc is the axial vector associated with the derivative of Vc with respect241

to R:242

vc × a=

∂Vc

∂RRT − R

(

∂Vc

∂R

)T

a, (3.13)

for any vector a. Further, q0 and q in (3.12) are subject to the constraint243

q0q0 + q · q = 0, (3.14)244

which is obtained by differentiating the Euler parameter constraint q = 1.245

12

One now seeks the solution vc to (3.12) for all (q0, q) which satisfy (3.14).246

With the help of the representation (2.14) for ω, it is easy to argue that247

∂Vc

∂q0

∂Vc

∂q1

∂Vc

∂q2

∂Vc

∂q3

= CT

vc · e1

vc · e2

vc · e3

+

2λq0

2λq1

2λq2

2λq3

, (3.15)248

where λ is arbitrary. A graphical schematic of this solution is shown in Figure249

3. Here, the vector associated with λ is normal to the 3-sphere S3, while the250

vector associated with vc · ω is tangent to S3.251

Following the steps which lead from (3.8) to (3.10), one finds that multiplica-252

tion of (3.15) by q

4C leads to the elimination of λ. As a result, the representa-253

tions for vc are254

vc =1

2

(

−∂Vc

∂q0

q + q0

∂Vc

∂q+ q × ∂Vc

∂q

)

=1

2

(

−∂Vc

∂q0

q + q0

∂vc

∂q− q × ∂vc

∂q

)

. (3.16)

In these representations, Vc is evaluated on the set of unit quaternions. That255

is, q0 and q are equivalent to a set of Euler-Rodrigues symmetric parameters.256

Multiplying each of the individual equations in (3.15) by an Euler-Rodrigues257

symmetric parameter, adding the resulting four equations, and invoking the258

Euler parameter constraint allows one to solve for λ:259

λ =1

2

(

∂Vc

∂q0

q0 +∂Vc

∂q· q)

. (3.17)260

It is interesting to note that if Vc is a quadratic function of q0, q1, q2, q3, then261

λ = Vc. The representation (3.16)1 for vc can also be inferred from Equation262

(6.42)3 on page 197 of [24].263

3.3 Imposing the Constraint264

The representations for v and vc have distinct similarities. It is useful to show265

how vc can be computing by imposing the Euler parameter constraint on the266

representation for v. This justifies existing computations of the derivatives of267

13

functions of the Euler-Rodrigues symmetric parameters where the parameter268

constraint is initially ignored, and imposed only after the derivatives have been269

computed.270

To proceed, one presumes that the Euler parameter constraint can be satisfied271

by setting q = 1 and imposing either one of the following conditions:272

q0 =√

1 − q · q, or q0 = −√

1 − q · q. (3.18)273

That is,274

q0 = q0 (q) . (3.19)275

In addition, one defines276

Vc = Vc (q) = V (q0 (q) ,q) . (3.20)277

Application of the chain rule demonstrates that278

∂Vc

∂q0

= 0,

∂Vc

∂qk

=∂q0

∂qk

∂V

∂q0

+∂V

∂qk

=−qk

q0

∂V

∂q0

+∂V

∂qk

. (3.21)

With the help of these expressions for the partial derivatives of Vc, it follows279

from (3.10) and (3.16) that280

−∂Vc

∂q0

q + q0

∂Vc

∂q+ q × ∂Vc

∂q=

(

−∂V

∂q0

q + q0

∂V

∂q+ q × ∂V

∂q

)

q=1

. (3.22)281

Consequently,282

vc = (v)q=1. (3.23)283

and one can use the unconstrained function V to compute vc by first calcu-284

lating v and then imposing the Euler parameter constraint.285

14

4 A Potential Energy286

Consider a potential energy function U which depends on all of the quaternion287

components: q0, q1, q2, and q3. This function has the representations288

U = U (q0,q) = U (q,R (q0,q)) . (4.1)289

To compute the conservative moment M and pressure p associated with U , it290

is assumed that the mechanical power of M and p are equal to the negative291

of the time-rate of change of U : 6292

U = −M · ω − pq. (4.2)293

Next, one expresses the derivative of U using the representations (3.9) and294

(3.10):295

U = u · ω +∂U

∂qq, (4.3)296

where297

u=1

2

(

−∂U

∂q0

q + q0

∂U

∂q+ q × ∂U

∂q

)

,

∂U

∂q=

1

2q

(

∂U

∂q0

q0 +∂U

∂q· q)

. (4.4)

Consequently, (4.2) is equivalent to298

(M + u) · ω +

(

p +∂U

∂q

)

q = 0. (4.5)299

Assuming that M and p are independent of q and ω, then (4.5) holds for all300

motions if, and only if,301

M=−1

2

(

−∂U

∂q0

q + q0

∂U

∂q+ q × ∂U

∂q

)

6 A physical interpretation of the pressure p can be found in [31]. To do this, theseauthors used the theory of a pseudo-rigid body (or Cosserat point or homogeneouslydeformable continuum).

15

=−1

2

(

−∂U

∂q0

q + q0

∂u

∂q− q × ∂u

∂q

)

,

p =− 1

2q

(

∂U

∂q0

q0 +∂U

∂q· q)

. (4.6)

These are the final desired representations for M and p.302

5 A Moment Potential303

The case of a moment potential Uc can be considered by restricting the do-304

mains of the functions U (q0,q) and U (q,R (q0,q)) to the set of unit quater-305

nions. In this case, one seeks to determine the conservative moment M which306

satisfies the identity307

Uc = −M · ω, (5.1)308

for all q0 and q where q = 1.309

To find a prescription for M, the developments in Section 4 are paralleled and310

and representations of the form (3.16) are appealed to. This enables one to311

establish that312

(M + uc) · ω = 0, (5.2)313

where314

uc =1

2

(

−∂Uc

∂q0

q + q0

∂Uc

∂q+ q × ∂Uc

∂q

)

. (5.3)315

Solutions M to (5.2) are now sought for all unit quaternions.316

Paralleling the arguments used in Section 3.2 the solution to (5.2) is expressible317

in the form318

−CT

M · e1

M · e2

M · e3

−

2λq0

2λq1

2λq2

2λq3

=

∂Uc

∂q0

∂Uc

∂q1

∂Uc

∂q2

∂Uc

∂q3

, (5.4)319

where λ is arbitrary.320

16

It is straightforward to solve for λ and M from (5.4), and the final form of the321

results are quoted:322

M=−1

2

(

−∂Uc

∂q0

q + q0

∂Uc

∂q+ q × ∂Uc

∂q

)

=−1

2

(

−∂Uc

∂q0

q + q0

∂uc

∂q− q × ∂uc

∂q

)

,

λ=−1

2

(

q0

∂Uc

∂q0

+ q · ∂Uc

∂q

)

=−1

2

(

q0

∂Uc

∂q0

+ q · ∂uc

∂q

)

. (5.5)

The representation (5.5)1 for M is identical to Equation (6.42)3 on page 197323

of [24]. In the literature (e.g., [7,8,18,21]), the representation (5.5)2 for M is324

commonly used. It is interesting to note that if Uc is a quadratic function of325

q0, q1, q2, q3, then λ = −Uc.326

6 Applications of the Representations327

6.1 Constraints328

Suppose that a constraint on the rotation of a rigid body is represented us-329

ing the Euler-Rodrigues symmetric parameters. Then a natural question to330

ask is what is a prescription for the constraint moment associated with this331

constraint? In the context of Euler-Rodrigues symmetric parameters, a dis-332

cussion of this topic can be found in Nikravesh et al. [19]. Here, their work is333

supplemented by placing the prescription in the context of the present work.334

Consider a constraint of the form335

Φc (q0, q1, q2, q3, t) = 0. (6.1)336

The subscript c denotes the fact that the Euler parameter constraint has been337

imposed. This constraint implies that338

∂Φc

∂q0

q0 +∂Φc

∂q1

q1 +∂Φc

∂q2

q2 +∂Φc

∂q3

q3 +∂Φc

∂t= 0. (6.2)339

With some algebraic manipulation, this equation can be expressed in the com-340

17

pact form341

g · ω + h = 0, (6.3)342

where343

g =1

2

(

−∂Φc

∂q0

q + q0

∂Φc

∂q+ q × ∂Φc

∂q

)

, h =∂Φc

∂t. (6.4)344

The expression for g is obtained by paralleling the derivation of (3.16) in345

Section 3.346

A prescription for the constraint moment is347

Mc = µg, (6.5)348

where µ is a function of time that is determined from the equations of motion.349

This prescription is known as the Lagrange prescription of constraint forces350

and moments (cf. [2,24]), and is identical to that which would be obtained351

were the Lagrange-D’Alembert principle invoked. 7352

As an example consider the case where the axis of rotation of a rigid body is353

constrained to lie in the e1 − e3 plane. Thus,354

Φc = q2. (6.6)355

The computation of g is straightforward and the corresponding constraint356

moment is obtained from (6.5):357

Mc =µ

2(q0E2) =

µ

2(q0e2) . (6.7)358

Physically, this moment lies in the plane perpendicular to the axis of rotation359

and ensures that the axis remains in the e1 − e3 plane.360

6.2 Generalized Forces361

In the development of the equations of motion for rigid bodies whose rotation is362

parameterized using Euler-Rodrigues symmetric parameters, Lagrange’s equa-363

7 That is, the virtual work performed by Mc in any motion compatible with theconstraint is zero.

18

tions for the rotational motion are364

d

dt

(

∂T

∂qA

)

− ∂T

∂qA

= MR · ∂ω

∂qA

+ µqA, (A = 0, 1, 2, 3) . (6.8)365

Here, T = 1

2ω · (Jω) is the rotational kinetic energy of the rigid body, J is366

the inertia tensor of the rigid body relative to its center of mass, and MR is367

the resultant moment relative to the center of mass of the rigid body. The368

Lagrange multiplier µ in (6.8) ensures that the Euler parameter constraint is369

enforced. It can be shown (see, e.g., [19,31]) that µ = −2T .370

With the assistance of (2.17), one finds that371

MR · ∂ω∂q0

MR · ∂ω∂q1

MR · ∂ω∂q2

MR · ∂ω∂q3

= AT

MR · E1

MR · E2

MR · E3

= CT

MR · e1

MR · e2

MR · e3

. (6.9)372

If MR = M is conservative, then one can use (5.4) to reduce the expressions373

for the generalized forces to374

M · ∂ω∂q0

M · ∂ω∂q1

M · ∂ω∂q2

M · ∂ω∂q3

= −

2λq0

2λq1

2λq2

2λq3

−

∂Uc

∂q0

∂Uc

∂q1

∂Uc

∂q2

∂Uc

∂q3

. (6.10)375

When these expressions are inserted into (6.8), the multiplier λ can be sub-376

sumed into µ.377

6.3 A System of Two Rigid Bodies378

As a further application of the representations for the conservative moment,379

consider two rigid bodies which are connected by an elastic element (see Figure380

1). The elastic element could be a series of elastic springs connecting the two381

bodies or an elastic continuum. As discussed in [11,12], an example of such382

a system is a vertebral motion segment consisting of two vertebra and an in-383

tervertebral disc. Existing treatments of the conservative forces and moments384

in this system have exclusively featured parameterizations of the relative ro-385

tation between the vertebrae using Euler angles. The developments in this386

19

section can easily be used to develop a treatment based on Euler-Rodrigues387

symmetric parameters.388

To elaborate, one rigid body is denoted by B1 and it is assumed that a right-389

handed orthonormal body fixed basis {a1, a2, a3} is attached to this body. The390

corresponding basis affixed to the second rigid body, which is denoted by B2,391

is {b1,b2,b3}. Material points X1 on B1 and X2 on B2 are also selected. The392

position vectors of these points relative to a fixed point O are denoted by x1393

and x2, respectively.394

A relative rotation tensor Q =∑

3

i=1bi ⊗ ai can be defined. This tensor has395

an associated angular velocity vector ω which is the angular velocity of B2396

relative to B1. That is,397

ω = −1

2ǫ[

QQT]

= ω2 − ω1. (6.11)398

where ωα is the angular velocity vector of Bα (α = 1, 2). 8399

It is assumed that the elastic element provides restoring forces and moments on400

each of the bodies. To prescribe these forces and moments, the developments in401

O’Reilly and Srinivasa’s treatment [2] are followed and the following potential402

energy function is defined:403

U = U (y,Q) = Uc (y,Q (q0,q)) , (6.12)404

where y = x2−x1 is the displacement of X2 relative to X1. Here, it is assumed405

that Q is parameterized by a set of Euler-Rodrigues symmetric parameters.406

The combined power of the conservative forces and moments is then equated407

to the time rate of change of −Uc:408

Uc = −F1 · x1 − F2 · x2 −M2 · ω2 −M1 · ω1. (6.13)409

Here, F1 is the resultant force on B1, M1 is the resultant moment relative to410

X1 on B1, F2 is the resultant force on B2, and M2 is the resultant moment411

relative to X2 on B2.412

With some alterations, the earlier developments for the representations of413

a conservative moment can now be invoked. The major changes are as fol-414

lows: In (5.5), the basis {E1,E2,E3} is replaced by {a1, a2, a3}, and the basis415

{e1, e2, e3} is replaced by {b1,b2,b3}. Following [2], one solves (6.13) for the416

8 For further details on the derivation of the relative angular velocity vector see[33] or Section 6.7 of [24].

20

conservative forces and moments. With the help of the representations (5.5)1,2,417

it can be shown that418

F2 = −F1 =−∂Uc

∂y,

M1 =−1

2

(

−∂Uc

∂q0

q + q0

∂Uc

∂q+ q × ∂Uc

∂q

)

,

M2 = −M1 =1

2

(

−∂Uc

∂q0

q + q0

∂uc

∂q− q × ∂uc

∂q

)

, (6.14)

where419

q =3∑

k=1

qkak =3∑

k=1

qkbk,∂uc

∂q=

3∑

k=1

∂Uc

∂qk

bk,∂Uc

∂q=

3∑

k=1

∂Uc

∂qk

ak. (6.15)420

Notice that the representation one uses for M1 is easy to express in terms of the421

{a1, a2, a3} basis, while that for M2 is easy to express in the {b1,b2,b3} basis.422

If load cells are mounted on the bodies, this feature of the representations423

facilitates the measurement of the conservative forces and moments.424

7 Examples of Potential Energy Functions425

Several examples of potential energy functions featuring rotations parameter-426

ized by Euler-Rodrigues symmetric parameters are present in the literature.427

For example, Arribas et al. [34] discuss Mac Cullagh’s gravitational potential428

energy for a body of mass m orbiting a fixed point O of mass M . This function429

is quartic in the Euler-Rodrigues symmetric parameters and yields a conserva-430

tive moment which is a quadratic function of the Euler-Rodrigues symmetric431

parameters. Another case of interest lies in a potential energy function which432

is a quadratic function of the Euler-Rodrigues symmetric parameters. In this433

section, the simplest such function is first considered and then the problem of434

estimating this function for a more general case is discussed. The resulting es-435

timation procedure is far easier to implement than that associated with Euler436

angles.437

7.1 A Quadratic Potential Energy Function438

Consider the simple case of the rigid body shown in Figure 4. The body is439

pin jointed to the fixed point 0 and thus performs a fixed axis rotation about440

21

O

X

e2

E3

E1

E2

g

Fig. 4. A rigid body that is free to rotate about the fixed point O. The basis{e1, e2, e3} corotates with the rigid body.

E3 = e3. The gravitational potential energy of the rigid body is441

Ug = mgx ·E2, (7.1)442

where x = L0e2 is the position of the center of mass X of the body, with443

L0 being a constant. Since E3 is the axis of rotation, the Euler-Rodrigues444

symmetric parameters are445

β0 = cos

(

θ

2

)

, β = sin

(

θ

2

)

E3. (7.2)446

Noting that ei ·Ek = Rki, one finds (with the help of (2.8)) that Ug = mgL0e2 ·447

E2 can be easily expressed as a function of the quaternion parameters:448

Ug =mgL0

q

[

q0 q1 q2 q3

]

F22

q0

q1

q2

q3

, F22 =

1 0 0 0

0 −1 0 0

0 0 1 0

0 0 0 −1

. (7.3)449

With the help of (5.5)1,2, one finds that450

M = 2mgL0q0q3E3 = mgL0 sin(θ)E3 = mgL0 sin(θ)e3, (7.4)451

as expected.452

22

More generally, a quadratic function of the quaternion components has the453

representation454

U =1

2

[

q0 q1 q2 q3

]

K

q0

q1

q2

q3

, K =

k00 k01 k02 k03

k01 k11 k12 k13

k02 k12 k22 k23

k03 k13 k23 k33

. (7.5)455

With the help of (5.5)1,2, representations for the moment components can be456

found. For the particular case where the stiffness matrix K in (7.5) is diagonal,457

these expressions reduce to458

M ·E1

M ·E2

M ·E3

=−1

2

(k11 − k00) q1q0 − (k22 − k33) q2q3

(k22 − k00) q2q0 − (k33 − k11) q1q3

(k33 − k00) q3q0 − (k11 − k22) q1q2

,

M · e1

M · e2

M · e3

=−1

2

(k11 − k00) q1q0 + (k22 − k33) q2q3

(k22 − k00) q2q0 + (k33 − k11) q1q3

(k33 − k00) q3q0 + (k11 − k22) q1q2

. (7.6)

The arbitrary function λ can be determined to be459

λ = −(

Uc −1

2k00

)

= −U. (7.7)460

For the isotropic case k = k11 = k22 = k33, the expression for the moment461

simplifies to462

M=−1

2(k − k00) q0q

=−1

4(k − k00) sin(φ)r. (7.8)

Linearizing this expression, one obtains the representation for a moment −Kφ463

about an axis r where the torsional stiffness K = 1

4(k − k00). Thus, the ex-464

pression for U in this case represents the moment potential for a torsional465

spring.466

These simple examples illustrate the interesting feature that, regardless of the467

rotational complexity of the rigid body motion, the stiffness matrix associated468

23

with any potential featuring linear combinations of ek · Ei always assumes469

an extremely simple form. This is because the dot products ek · Ei can each470

be represented by matrices Fik (cf. (2.8)) that are permutations of the 4 × 4471

identity matrix with at most two of the components possibly equal to −1.472

As noted earlier, the 4 × 4 matrices in the expressions for ei · Ek are proper473

orthogonal and symmetric. This structure is inherited by the resulting stiffness474

matrices. In particular, one cannot expect the stiffness matrix associated with475

U to be positive definite.476

7.2 Estimating a Quadratic Moment Potential477

Given a conservative moment, one can, in essence, go backwards and compute478

the associated potential energy function. Such a calculation has applications479

in numerous areas of mechanics.480

The discussion presented here is restricted to the case where U is a quadratic481

function of the Euler-Rodrigues symmetric parameters. Thus, U is given by482

(7.5). Imposing the Euler parameter constraint as in Section 3.3, Uc can be483

computed:484

Uc =1

2

[

q1 q2 q3

]

k11 − k00 k12 k13

k21 k22 − k00 k23

k13 k23 k33 − k00

q1

q2

q3

(7.9)

+ q0 (q)[

k01 k02 k03

]

q1

q2

q3

+1

2k00. (7.10)

The derivatives of Uc with respect to q1, q2, and q3 have the representations485

∂Uc

∂q1

∂Uc

∂q2

∂Uc

∂q3

= SK, (7.11)

where S and K are486

24

S =S (q) =

q1 0 0 q2 q3 0 q0 (q) 0 0

0 q2 0 q1 0 q3 0 q0 (q) 0

0 0 q3 0 q1 q2 0 0 q0 (q)

,

KT =[

k11 − k00 k22 − k00 k33 − k00 k12 k31 k23 k01 k02 k03

]

. (7.12)

Thus, with the help of (5.5)1,2, one can write487

M ·E1

M ·E2

M ·E3

= TEK,

M · e1

M · e2

M · e3

= TeK, (7.13)

where488

TE =TE (q) = −1

2(q0 (q) I + Q (q)) S,

Te =Te (q) = −1

2(q0 (q) I − Q (q)) S. (7.14)

Either one of (7.13) provides a system of three equations for nine unknowns:489

k11 − k00, k22 − k00, . . . , k31. If at least three moment readings are available,490

then one can, in principle, solve the nine resulting equations for the nine491

unknown stiffnesses, provided that the three quaternions associated with the492

three moment readings are linearly independent of each other.493

For example, suppose that a set of three measurements are made. Labeling the494

measurements with subscripts A, B and C, one can use (7.13)1 to compute495

that496

MA · E1

...

MB · E2

...

MC · E3

=

TE(qA)

TE(qB)

TE(qC)

K = TEK. (7.15)

Inverting the 9x9 matrix TE then yields the stiffnesses.497

In the case where the stiffness matrix is purely diagonal (as in the example in498

Section 7.1), then it is easy to show that a single measurement of M suffices499

25

to compute Uc provide the axis of rotation associated with M has non-zero500

components along each Ei and that the angle of rotation is neither 0,±π2, nor501

π.502

Acknowledgment503

The work of the authors was partially supported by the National Science504

Foundation under Grant No. 0726675.505

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