2 attitude and kinematics of coordinate framesrenaissance.ucsd.edu/courses/mae207/tewari_chap02.pdf2...

2

Attitude and Kinematics of Coordinate Frames

2.1 Aims and Objectives

• To present the kinematic modeling useful in all flight dynamicderivations and analyses.

• To offer a rigorous vector analysis of rotational kinematics.• To derive the basic identities of coordinate transformations in a manner

that can be easily utilized for both translational and rotational equationsof motion in subsequent chapters.

• To introduce several alternative kinematic representations (Euler angles,Euler axis/principal angle, rotation matrix, quaternion, Rodrigues andmodified Rodrigues parameters) and their evolution in time.

2.2 Basic Definitions and Vector Operations

The basic entity in dynamics is a vector . We shall denote vectors by bold-face symbols, and draw them as arrows. A vector has both magnitude anddirection and is represented by magnitudes along any three mutually perpen-dicular axes, called a coordinate frame (or, a reference frame). Each axis of acoordinate frame is represented by a unit vector , defined as a vector of unitmagnitude. Let us consider a vector, A, written in terms of its magnitudesalong a coordinate frame consisting of a triad formed by the unit vectors,i, j,k, as follows:

A = Axi + Ayj + Azk , (2.1)

or, simply as

A =

⎧⎨⎩

Ax

Ay

Az

⎫⎬⎭ , (2.2)

where it is understood that the components , Ax, Ay , Az multiply i, j,k, respec-tively. A component is the magnitude of the vector projected on to a particular

10 2 Attitude and Kinematics of Coordinate Frames

axis. As we shall see below, it is possible to transfer the components of a vec-tor to another coordinate frame by using a coordinate transformation. Two(or more) vectors can be added, or subtracted by adding, or subtracting theirrespective components referred to the same coordinate frame. There are twodistinct ways in which vectors can be multiplied: the scalar product , and thevector product . As the names suggest, the scalar product of two vectors isa scalar (a quantity with only magnitude), whereas a vector product of twovectors is a vector. The scalar product (also called dot product) of two vectors,A and B, is defined by

A · B .= AB cos θ , (2.3)

where A, B are the respective magnitudes, and θ is the angle between the twovectors. From this definition, it is clear that the dot product of any two vectorswith the same direction is the product of their respective magnitudes, whiletwo mutually perpendicular (orthogonal) vectors have a zero dot product.Therefore, it follows that

A · i = Ax ;A · j = Ay ;A · k = Az , (2.4)

and we can write the dot product of two vectors, A = Axi + Ayj + Azk andB = Bxi + Byj + Bzk, as

A ·B = AxBx + AyBy + AzBz = {Ax Ay Az}⎧⎨⎩

Bx

By

Bz

⎫⎬⎭ = AT B . (2.5)

In Eq. (2.5) we have denoted the transpose of a vector with a superscriptT, which is obtained by arranging the components in a row [rather than thecolumn form of Eq. (2.2)]. Hence, the scalar product is the sum of products ofthe respective components of the two vectors. For a refresher on vectors andmatrices, please refer to Kreyszig [4].

The vector product (also called the cross product), A × B, is defined asfollows:(a) The magnitude is given by | A × B |= AB sin θ, where θ is the anglebetween the two vectors.(b) The direction of A×B is normal to the plane formed by the two vectors,and is given by the right-hand rule, i.e., when the curled fingers of the righthand point from A to B, the thumb points in the direction of A×B (Fig. 2.1).

From this definition, it is clear that the vector product of any two vectorswith the same (or opposite) direction is zero, while two mutually perpendicularvectors have a vector product with magnitude equal to the product of theirrespective magnitudes. Furthermore, it follows that A × B = −B × A. Acoordinate frame i, j,k is said to be right-handed if its triad of unit vectorsis arranged such that i × j = k. Therefore, with reference to a right-handedframe, we can write the vector product of two vectors, A = Axi + Ayj + Azkand B = Bxi + Byj + Bzk, as follows:

2.2 Basic Definitions and Vector Operations 11

A B

A B

Fig. 2.1. Vector product of two vectors, A and B.

A × B = (AyBz − AzBy)i + (AzBx − AxBz)j + (AxBy − AyBx)k . (2.6)

By introducing the short-hand notation of the determinant of a square matrix,we can write Eq. (2.6) as

A × B =

∣∣∣∣∣∣i j k

Ax Ay Az

Bx By Bz

∣∣∣∣∣∣ . (2.7)

The foregoing definitions allow us to derive the following identities involvingthe vector product:

A × (B × C) = B(A · C) − C(A · B) . (2.8)(A × B) × C = B(A · C) − A(B · C) . (2.9)

(A× B) · (C× D) = (A · C)(B ·D) − (A · D)(B ·C) . (2.10)(A × B) × (C× D) = (A · (B× D))C − (A · (B × C))D . (2.11)

By using MATLAB, one can easily compute scalar and vector products ofvectors by using the in-built commands dot and cross, respectively.

Example 2.1. Let us find scalar and vector products of A = −2i + 5j− k andB = i + 3k with the following MATLAB commands:

>>A=[-2;5;-1];B=[1;0;3];>>s=dot(A,B) %scalar products = -5

>>C=cross(A,B) %vector product C=AxBC = 15

5-5

>>D=cross(B,A) %vector product D=BxAD = -15

-55


>>E=cross(A,cross(A,B)) % vector triple product E=Ax(AxB)E = -20

-25-85

You may verify these results with hand calculations of Eqs. (2.5) and (2.7).

A(t+ t)A(t)

A

Fig. 2.2. A changing vector.

The time derivative of a vector, A, is defined by the following:

dAdt

.= limΔt→0

A(t + Δt) − A(t)Δt

= limΔt→0

ΔAΔt

, (2.12)

where ΔA denotes the total change caused by changes in both magnitudeand direction, as depicted in Fig. 2.2. It can be seen from Fig. 2.2 that anisosceles triangle is formed by extending the vector A(t) until its magnitudebecomes equal to A(t + Δt). In the limit Δt → 0, we have Δθ → 0, and thusα → π

2 . Hence, the two dashed lines in Fig. 2.2 represent the extension (orcontraction) and rotation of A, respectively, and we have

limΔt→0

ΔA = [A(t + Δt) − A(t)]A(t)A(t)

+ ω × A(t)Δt , (2.13)

where ω is the angular velocity of A, directed into the plane of Fig. 2.2 withmagnitude given by

ω =dθ

dt

.= limΔt→0

Δθ

Δt. (2.14)

Upon substitution of Eq. (2.13) into Eq. (2.12), we have

dAdt

=dA

dt

AA

+ ω × A , (2.15)

2.3 Coordinate Systems and Rotation Matrix 13

where the first term on the right-hand side represents time derivative due toa change in the magnitude, and the second, that due to rotation. An observerrotating with the same angular velocity as A would notice only the first term,while a stationary observer would notice both the terms.

2.3 Coordinate Systems and Rotation Matrix

As noted above, a vector can be expressed in a variety of coordinate frames,each of which is represented by a triad of unit vectors. We will confine ourdiscussion to right-handed coordinate frames. Any number of different coordi-nate frames can be derived by rotating the orthogonal axes about the origin,O. Sometimes, it may be necessary to use a coordinate frame with a differentorigin, O′. Hence, the general transformation from one coordinate frame toanother consists of a translation of the origin, and a rotation of the axes aboutthe new origin. However, the translation of the origin is handled quite easilyby specifying the displacement vector, R from O to O′ (Fig. 2.3), and its timederivatives, such that the displacement, velocity, and acceleration, r,v, a, re-spectively, in the original frame are related to their counterparts r′,v′, a′, inthe translated frame by

r = R + r′ ,

v =dRdt

+ v′ , (2.16)

a =d2Rdt2

+ a′ .

Since the translation of coordinate frames is trivial, we shall focus instead on

Y

Z

Y

OX

R

Z

O

rr

X

Fig. 2.3. Translation of a coordinate frame.


XX

YO

ZZ

Y

e

Fig. 2.4. Rotation of a coordinate frame.

coordinate transformations involving rotation. Consider a frame (OXYZ ) withaxes OX, OY, and OZ, denoted by unit vectors i, j,k, respectively. The frameis rotated about the origin, O, to produce a new frame, (OX ′Y ′Z ′) denoted byi′, j′,k′ (Fig. 2.4). Now, consider a vector, A, alternately expressed in termsof its components in the original and rotated frames, as follows:

A = Axi + Ayj + Azk = A′xi

′ + A′yj

′ + A′zk

′ , (2.17)

which can be written in a matrix form as

A = (i j k)

⎧⎨⎩

Ax

Ay

Az

⎫⎬⎭ = (i′ j′ k′)

⎧⎨⎩

A′x

A′y

A′z

⎫⎬⎭ . (2.18)

In order to find the relationship between the two sets of components, we takescalar products A · i′, A · j′, and A · k′, resulting in⎧⎨

⎩A′

x

A′y

A′z

⎫⎬⎭ =

⎛⎝ i′ · i i′ · j i′ · k

j′ · i j′ · j j′ · kk′ · i k′ · j k′ · k

⎞⎠⎧⎨⎩

Ax

Ay

Az

⎫⎬⎭ . (2.19)

We can also take scalar products of A with i, j, and k, to derive the following:⎧⎨⎩

Ax

Ay

Az

⎫⎬⎭ =

⎛⎝ i · i′ i · j′ i · k′

j · i′ j · j′ j · k′

k · i′ k · j′ k · k′

⎞⎠⎧⎨⎩

A′x

A′y

A′z

⎫⎬⎭ . (2.20)

We rewrite Eq. (2.19) as ⎧⎨⎩

A′x

A′y

A′z

⎫⎬⎭ = C

⎧⎨⎩

Ax

Ay

Az

⎫⎬⎭ , (2.21)

2.3 Coordinate Systems and Rotation Matrix 15

where C is the following matrix consisting of the cosines of angles (Eq. (2.3))between the axes of the two coordinate frames and is thus called the directioncosine matrix, or the rotation matrix :

C.=

⎛⎝ i′ · i i′ · j i′ · k

j′ · i j′ · j j′ · kk′ · i k′ · j k′ · k

⎞⎠ . (2.22)

The coordinate transformation, Eqs. (2.19), can also be expressed as follows:⎧⎨⎩

i′

j′

k′

⎫⎬⎭ = C

⎧⎨⎩

ijk

⎫⎬⎭ . (2.23)

From Eqs. (2.19) and (2.20), it is clear that the rotation matrix has the fol-lowing property:

CT C = CCT = I , (2.24)

from which it follows that C−1 = CT . A matrix with this property is said tobe orthogonal , since the vectors formed out of the columns of the matrix areorthogonal. From Eq. (2.24), we can also deduce the fact that the determinant,| C |= ±1. A rotation for which | C |= 1 is called a proper rotation. It is easy tosee from the definition of the rotation matrix, Eq. (2.21), that two successiverotations of a coordinate frame can be represented simply by multiplying therotation matrices of individual rotations as follows:

C′′ = C′C , (2.25)

where the orientation C′′ is obtained by first undergoing a rotation C, followedby a rotation C′. Other properties of the rotation matrix are discussed in thefollowing section.

Example 2.2. Find the rotation matrix that produces the right-handed co-ordinate frame i′, j′,k′ given by i′ = 0.1i + 0.2j +

√1 − 0.01 − 0.04k, and

j′ = −0.1i− 0.9726095077j+√

1 − 0.01 − 0.97260950772k.We check the orthogonality of the given vectors, calculate the third axes

of the right-handed frame, and then find C according to Eq. (2.22) as follows:>> iprime=[0.1;0.2;sqrt(1-0.01-0.04)],jprime=[-0.1;-0.9726095077;

sqrt(1-0.01-0.9726095077^2)]iprime = 0.10000000000000 jprime = -0.10000000000000

0.20000000000000 -0.972609507700000.97467943448090 0.20983504362133

>>dot(iprime,jprime)ans = 1.111077896354118e-010

>> kprime=cross(iprime,jprime) %k’=i’xj’kprime = 0.98994949366004

-0.11845144781022-0.07726095077000

>>i=[1;0;0];j=[0;1;0];k=[0;0;1];


>> C=[dot(iprime,i),dot(iprime,j),dot(iprime,k);...%rotation matrixdot(jprime,i),dot(jprime,j),dot(jprime,k);...dot(kprime,i),dot(kprime,j),dot(kprime,k)]

C = 0.10000000000000 0.20000000000000 0.97467943448090-0.10000000000000 -0.97260950770000 0.209835043621330.98994949366004 -0.11845144781022 -0.07726095077000

>> det(C) %determinant of Cans = 1.0000

Hence, the rotation matrix consists of the axes of the rotated frame (ex-pressed in the unit vectors of the original axes) as its columns. You may verifythat the rotation considered in this example is proper.

2.4 Euler Axis and Principal Angle

From the discussion given above, it is clear that a rotation of a coordinateframe can be represented by a rotation matrix, C. A rotation can also bedescribed by specifying the axis of rotation, as well as the angle by whichthe frame has been rotated. This common experience is formalized in Euler’stheorem, which states that the relative orientation of any pair of coordinateframes is uniquely determined by a rotation by angle, Φ, about a fixed axisthrough the common origin, called the Euler axis. This unique rotation istermed the principal angle. A graphical depiction of the principal rotation isshown in Fig. 2.4, where a counter-clockwise rotation is considered positive bythe right-hand rule. Euler’s theorem thus provides an alternative descriptionof rotation using unit vector, e, representing the direction of Euler axis, andthe principal rotation angle, Φ. Before using the new representation, we mustknow how these two quantities can be derived. An insight into the Euleraxis can be obtained by analyzing the eigenvalues and eigenvectors [4] of therotation matrix. Let c be an eigenvector associated with the eigenvalue, λ ofC:

Cc = λc . (2.26)

By premultiplying Eq. (2.26) by the Hermitian conjugate [4] of each side, wehave

(Cc)H(Cc) = λλcHc , (2.27)

or, since C is real and satisfies Eq. (2.24),

(λλ − 1)cHc = 0 , (2.28)

which implies thatλλ = 1 , (2.29)

because c is nonzero. Equation (2.29) states the fact that all eigenvalues of Chave unit magnitudes. Now, C, being a (3 × 3) matrix, has three eigenvalues.Since complex eigenvalues occur in conjugate pairs, it follows that one of theeigenvalues of C must be real, for which Eq. (2.26) becomes

2.4 Euler Axis and Principal Angle 17

Cc1 = c1 , (2.30)

and which implies that the eigenvector, c1, associated with λ = 1 is unchangedby the rotation. It also follows from Eqs. (2.24) and (2.30) that c1

T c1 = 1,i.e., c1 is a unit vector. Therefore, it is clear that Euler axis—being invariantunder coordinate frame rotation—is represented by e .= c1.

The other two eigenvalues of C, being complex conjugates with a unit mag-nitude, can be written as λ2,3 = e±iβ .= cosβ ± i sinβ. Thus, from Eq. (2.26),we have

Cc2,3 = e±iβc2,3 . (2.31)

From the complex plane representation of a vector (also due to Euler), thefactor eiβ multiplying a vector, implies a rotation by angle β. Thus, the eigen-vectors, c2,3 (complex conjugates), undergo a rotation by angle β when thecoordinate frame is rotated about the axis c1. A consequence of C being or-thogonal [Eq. (2.24)] is that its eigenvectors are mutually perpendicular. Sincec2,3 are perpendicular to c1, their rotation must be equal to the angle of co-ordinate frame rotation. Therefore, Φ

.= β. A simple method of obtaining theprincipal rotation angle is through the trace of C [4]:

traceC = λ1 + λ2 + λ3 = 1 + eiΦ + e−iΦ = 1 + 2 cosΦ , (2.32)

or,

cosΦ =12(traceC − 1) . (2.33)

There are two values of Φ, differing only in sign, that satisfy Eq. (2.33), eachhaving Euler axis, e, in opposite directions. This does not cause any ambiguity,because a rotation by Φ about e is the same as a rotation by −Φ about −e.

Example 2.3. Let us find the principal rotation angle and Euler axis for thetransformation given in Example 2.2, with the use of the built-in MATLABprogram eig.m, that enables eigenvalue analysis of a square matrix, as follows:

>> [c,D]=eig(C) %eigenvectors, c, and diagonal matrix of eigenvalues, D

c = -0.7378 -0.0179 - 0.4770i -0.0179 + 0.4770i-0.0343 0.7067 0.7067-0.6742 -0.0164 + 0.5220i -0.0164 - 0.5220i

D = 1.0000 0 00 -0.9749 + 0.2225i 00 0 -0.9749 - 0.2225i

>> phi=acos(0.5*(trace(C)-1)) %rotation angle (rad.)phi = 2.9172

>> phi=acos(real(D(2,2))) %confirm rotation angle (rad.)phi = 2.9172

Thus, the rotation angle is Φ = ±2.9172 rad. (±167.1447◦), and Euleraxis is given by the eigenvector of C corresponding to the real eigenvalue,e = ±(−0.7378i−0.0343j−0.6742k). It is also clear that the complex conjugateeigenvectors satisfy Eq. (2.31).


The rotation matrix can be derived from the Euler axis/principal angleusing Euler’s formula (Exercise 2.5).

The simplest coordinate transformations are rotations about the axes of acoordinate frame, called elementary rotations. A positive rotation of (OXYZ )about OX by angle Φ is represented by the rotation matrix

C1.=

⎛⎝1 0 0

0 cosΦ sin Φ0 − sinΦ cosΦ

⎞⎠ , (2.34)

whereas the rotation matrix for a positive rotation about OY by the sameangle is

C2.=

⎛⎝ cosΦ 0 − sinΦ

0 1 0sin Φ 0 cosΦ

⎞⎠ . (2.35)

Similarly, a positive rotation about OZ by Φ is given by

C3.=

⎛⎝ cosΦ sinΦ 0

− sinΦ cosΦ 00 0 1

⎞⎠ . (2.36)

A more complicated coordinate transformation can be derived by multiplesingle-axis rotations in a given sequence, using the elementary rotation ma-trices, as demonstrated below.

Example 2.4. Find the rotation matrix for a coordinate transformation ob-tained by a 20◦ rotation about OZ, followed by a −65◦ rotation about OX ′.

We begin by representing the first rotation as follows:⎧⎨⎩

i′

j′

k′

⎫⎬⎭ = C3(20◦)

⎧⎨⎩

ijk

⎫⎬⎭ . (2.37)

The next rotation is by an angle −65◦ about i′. Thus, we write the finalorientation as ⎧⎨

⎩i′′

j′′

k′′

⎫⎬⎭ = C1(−65◦)

⎧⎨⎩

i′

j′

k′

⎫⎬⎭ , (2.38)

or, by substituting Eq. (2.37) into Eq. (2.38), we have⎧⎨⎩

i′′

j′′

k′′

⎫⎬⎭ = C1(−65◦)C3(20◦)

⎧⎨⎩

ijk

⎫⎬⎭ . (2.39)

2.5 Euler Angles 19

Therefore, the required rotation matrix is the following:

C = C1(−65◦)C3(20◦)

=

⎛⎝ cos(20◦) sin(20◦) 0

− sin(20◦) cos(−65◦) cos(20◦) cos(−65◦) sin(−65◦)sin(20◦) sin(−65◦) − cos(20◦) sin(−65◦) cos(−65◦)

⎞⎠

=

⎛⎝ 0.9397 0.3420 0

−0.1445 0.3971 −0.9063−0.3100 0.8517 0.4226

⎞⎠ .

The Euler axis and principal angle for this rotation are obtained in the samemanner as in Example 2.3 to be e = ±(0.9501i + 0.1675j − 0.2630k) andΦ = ±67.6836◦, respectively.

In terms of the elements (i, j) of C—denoted by cij—Eqs. (2.30) and (2.33)lead to the following explicit expressions for the components of Euler axis, e:

e1 =c23 − c32

2 sinΦ,

e2 =c31 − c13

2 sinΦ, (2.40)

e3 =c12 − c21

2 sinΦ.

It is clear from Eqs. (2.40) that Euler axis is defined only for nonzero rotations.

2.5 Euler Angles

We are now in a position to consider the orientation, or attitude, of a coor-dinate frame, relative to another frame. Such a description is quite useful inrepresenting the attitude of a flight vehicles (or any other object), to which acoordinate frame is rigidly attached, in reference to a second coordinate frame.Two previously discussed attitude representations are via the rotation matrixand the Euler-axis/principal angle combination. From the foregoing discus-sion, it is clear that a general orientation can also be obtained by using suc-cessive rotations about the axes of the reference frame. The largest number ofsuch rotations needed to uniquely specify a given orientation, called rotationaldegrees of freedom, is three. Hence, we can employ three angles, each about aparticular cordinate axis, to describe a given orientation. Such a representationof the attitude by three angles is called an Euler angle representation, and theconcerned angles are known as Euler angles. The sequence of axial rotationsis of utmost importance in the Euler angle representation. You can convinceyourself about this fact by rotating a book by three elementary rotations of90◦, each about a different axis, and then carrying out the rotations about the


YO

YZ

Z

Z,Z

XX ,X

Y ,Y

X

Fig. 2.5. The Euler angle orientation, (ψ)3, (θ)2, (φ)1.

same axes in a different sequence. The final attitude of the book will be differ-ent in each set of rotations. This important property of elementary rotationscan also be seen by simply reversing the sequence of rotations in Example 2.4.

We can specify the Euler angles and the axes of sequential rotations us-ing notation such as (ψ)3, (θ)2, (φ)1, which denotes a rotation of (OXYZ ) byangle ψ about OZ, resulting in the intermediate orientation, (OX ′Y ′Z ′), fol-lowed by a rotation by angle θ about OY ′, resulting in (OX ′′Y ′′Z ′′), andthen a final rotation by angle φ about OX ′′, to produce the new orientation,(OX ′′′Y ′′′Z ′′′). This Euler angle orientation, which is a description of the at-titude of an aircraft relative to a local horizon using the Euler angles, yaw(ψ), pitch (θ), and roll (φ), is depicted in Fig. 2.5. The rotation matrix forthe orientation (ψ)3, (θ)2, (φ)1, is the following:

C = C1(φ)C2(θ)C3(ψ) = (2.41)⎛⎝ cos θ cosψ cos θ sin ψ − sin θ

(sin φ sin θ cosψ − cosφ sin ψ) (sinφ sin θ sin ψ + cosφ cosψ) sinφ cos θ(cosφ sin θ cosψ + sin φ sin ψ) (cosφ sin θ sinψ − sin φ cosψ) cosφ cos θ

⎞⎠ .

However, it is not necessary that all the three coordinate axes shouldbe involved in describing a particular orientation. For example, astronomersand physicists traditionally employ the classical Euler angles, (ω)3, (i)1, (Ω)3,to represent the orientation of an orbital plane in reference to a celestialcoordinate frame (Chapter 5). A set of Euler angles that begins and endswith the same axis—such as (ω)3, (i)1, (Ω)3—is said to be symmetric. Thesymmetric and asymmetric sets of Euler angles are qualitatively different,and it is generally much easier to handle the former.

Example 2.5. Find the rotation matrix, Euler axis, and principal angle for theorientation (α)1, (β)2, (α)3, where α = −45◦, β = 45◦.

2.5 Euler Angles 21

We shall employ MATLAB for this calculation, by first evaluating the ele-mentary rotation matrices, and then multiplying them in the proper sequenceto produce the final orientation:

>> dtr=pi/180;alfa=-45*dtr;beta=45*dtr;>> C1=[1 0 0;0 cos(alfa) sin(alfa);0 -sin(alfa) cos(alfa)]C1 = 1.0000 0 0

0 0.7071 -0.70710 0.7071 0.7071

>> C2=[cos(beta) 0 -sin(beta);0 1 0;sin(beta) 0 cos(beta)]C2 = 0.7071 0 -0.7071

0 1.0000 00.7071 0 0.7071

>> C3=[cos(alfa) sin(alfa) 0;-sin(alfa) cos(alfa) 0;0 0 1]C3 = 0.7071 -0.7071 0

0.7071 0.7071 00 0 1.0000

>> C=C3*C2*C1 %rotation matrix for the final orientationC = 0.5000 -0.8536 0.1464

0.5000 0.1464 -0.85360.7071 0.5000 0.5000

>> [c,D]=eig(C) %eigenvectors and eigenvalues of Cc = 0.6786 0.1405+0.5000i 0.1405-0.5000i

-0.2811 0.6786 0.67860.6786 0.1405-0.5000i 0.1405+0.5000i

D = 1.0000 0 00 0.0732+0.9973i 00 0 0.0732-0.9973i

>> Phi=acos(0.5*(trace(C)-1))/dtr %principal angle (deg.)Phi = 85.8009

Hence, the rotation matrix for the given orientation is

C =

⎛⎝ 0.5 −0.8536 0.1464

0.5 0.1464 −0.85360.7071 0.5 0.5

⎞⎠ ,

and the Euler axis and principal angle are e = ±(0.6786i−0.2811j+0.6786k),and Φ = ±85.8009◦, respectively.

In order to specify the attitude by Euler angles, we must be able to deter-mine them uniquely from the rotation matrix. It is clear from Eq. (2.41) thatthe Euler angles for the representation, (ψ)3, (θ)2, (φ)1, can be determinedaccording to the following inverse transformation:

φ = tan−1 c23

c33,

θ = − sin−1 c13, (2.42)

ψ = tan−1 c12

c11,

where cij represents the element (i, j) of C. Of course, neither c11 nor c33 mustvanish; otherwise the angles φ and ψ cannot be determined.


Example 2.6. Let us find the Euler angles for the attitude given in Example2.2, using the representation (ψ)3, (θ)2, (φ)1:

φ = tan−1 c23

c33= tan−1 0.20983504362133

−0.07726095077= −69.7864◦ or 110.2136◦,

θ = sin−1 c−13 = sin−1(−0.9746794344809) = −77.079◦ or − 102.921◦,

ψ = tan−1 c12

c11= tan−1 0.2

0.1= 63.4349◦ or − 116.5651◦ .

You may verify the accuracy of the calculated Euler angles by formingthe rotation matrix according to Eq. (2.41) and comparing with the result ofExample 2.2.

It is clear from Example 2.6 that the Euler angles are not unique. Fur-thermore, there are certain orientations for which the Euler angles cannot bedetermined at all from the rotation matrix, C. In such a case, the Euler anglerepresentation is said to be singular , and becomes useless. An example of asingular orientation is (ψ)3, (±90◦)2, (φ)1, for which c11 = c12 = c23 = c33 = 0,and the angles φ and ψ become indeterminate. For most aircraft, usage of theEuler angle representation (ψ)3, (θ)2, (φ)1 does not cause a problem, because(ψ)3, (±90◦)2, (φ)1 is rarely encountered. However, the same cannot be saidof a fighter aircraft, a missile, or a spacecraft, where the vertical attitude is apossibility. Of course, in such a case one can switch to a different set of Eulerangles (e.g., (ψ)3, (θ)1, (φ)3), for which a particular singularity is avoided, butthe new representation would have singularity at some other orientation. Thus,a single Euler angle representation cannot be utilized where an arbitrary ori-entation is possible. This deficiency in attitude representation by three angles(or any three parameters) leads one to search for nonsingular representa-tions that must necessarily involve more than three parameters. Two obviouschoices of nonsingular attitude representations are the set of nine elements ofthe rotation matrix, and the set of four orientation parameters arising out ofthe Euler axis/principal angle combination. Equally obvious is the fact thatany four (or more) orientation parameters are not mutually independent [theelements of the rotation matrix obey Eq. (2.24), and the components of theEuler axis must produce a unit vector]. The Euler angles—like any other threeparameter set—have the advantage of being mutually independent, and thusform a minimal set for attitude representation. However, their use is limitedto those applications where the principal rotation is restricted to nonsingularorientations.

Example 2.7. Find the rotation matrix and the principal rotation for the at-titude (ψ)3, (90◦)2, (φ)1, which is singular in terms of Euler angles.

By substituting θ = 90◦ into Eq. (2.41), we get the following rotationmatrix:

C =

⎛⎝ 0 0 −1

sin(φ − ψ) cos(φ − ψ) 0cos(φ − ψ) − sin(φ − ψ) 0

⎞⎠ ,

2.6 Euler Symmetric Parameters (Quaternion) 23

We next determine the Euler axis, e = e1i + e2j + e3k, according toEq. (2.30): ⎛

⎝ 0 0 −1sin(φ − ψ) cos(φ − ψ) 0cos(φ − ψ) − sin(φ − ψ) 0

⎞⎠⎧⎨⎩

e1

e2

e3

⎫⎬⎭ =

⎧⎨⎩

e1

e2

e3

⎫⎬⎭ ,

which, along with e12 + e2

2 + e32 = 1, yields

e1 = ± 1 − cos(φ − ψ)√3 + cos(φ − ψ)2 − 4 cos(φ − ψ)

,

e2 =sin(φ − ψ)

1 − cos(φ − ψ)e1,

e3 = −e1 .

The appropriate sign of e1 is obtained from that of the principal angle, Φ,which is derived from Eq. (2.33) as follows:

Φ = cos−1 12(traceC − 1) = cos−1 1

2{cos(φ − ψ)}

orcos

Φ

2=

1√2

cosφ − ψ

2.

Thus, there is no difficulty in determining the rotation matrix and principalrotation for this case for which the Euler angle representation is singular.

2.6 Euler Symmetric Parameters (Quaternion)

Since Euler axis/principal angle representation is free from singularities, avery useful representation can be derived from it, called Euler symmetricparameters , or the quaternion. A quaternion is a special set composed of fourmutually dependent scalar parameters, q1, q2, q3, q4, such that the first threeform a vector, called the vector part,

q .=

⎧⎨⎩

q1

q2

q3

⎫⎬⎭ , (2.43)

and the fourth, q4, represents the scalar part . The quaternion for attituderepresentation can be derived from the Euler axis, e, and principal rotationangle, Φ, as follows:

qi.= ei sin

Φ

2(i = 1, 2, 3),

q4.= cos

Φ

2. (2.44)


It is clear from Eq. (2.44) that q1, q2, q3, q4, must satisfy the constraint equa-tion

q21 + q2

2 + q23 + q2

4 = 1 . (2.45)

This constraint implies that the quaternion yields only three independent,scalar parameters, as in the principal angle/Euler axis, or the Euler angleattitude representations. Since the four elements of the quaternion satisfy theconstraint equation, Eq. (2.45), it can be said that attitude orientations varyalong the surface of a four-dimensional unit sphere without any singularity.This fact is also evident from the principal angle, Φ, and the elements ofthe unit vector, e, representing the Euler axis. The chief advantage of thequaternion over the principal angle/Euler axis combination (which is alsoa four-parameter, nonsingular representation) lies in that the former doesnot require computationally intensive trigonometric function evaluations whenderived from the rotation matrix. The rotation matrix, C, can be writtenin terms of the quaternion by substituting the definitions of Eq. (2.44) intoEuler’s formula (to be derived in Exercise 2.5), leading to

C = (q24 − qT q)I + 2qqT − 2q4S(q) , (2.46)

where S(q) is the following skew-symmetric matrix function formed out of theelements of vector q:

S(q) =

⎛⎝ 0 −q3 q2

q3 0 −q1

−q2 q1 0

⎞⎠ . (2.47)

We can write Eq. (2.46) in terms of the individual quaternion elements asfollows:

C =

⎛⎝ q2

1 − q22 − q2

3 + q24 2(q1q2 + q3q4) 2(q1q3 − q2q4)

2(q1q2 − q3q4) −q21 + q2

2 − q23 + q2

4 2(q2q3 + q1q4)2(q1q3 + q2q4) 2(q2q3 − q1q4) −q2

1 − q22 + q2

3 + q24

⎞⎠ , (2.48)

which yields the following expressions for calculating the quaternion elementsfrom the elements of the rotation matrix, cij :

q1 =c23 − c32

4q4,

q2 =c31 − c13

4q4, (2.49)

q3 =c12 − c21

4q4,

whereq4 = ±1

2√

1 + c11 + c22 + c33 = ±12

√1 + traceC . (2.50)

Note that two signs are possible in deriving the quaternion from C. How-ever, just as in the case of principal angle/Euler axis derivation from the

2.6 Euler Symmetric Parameters (Quaternion) 25

rotation matrix [Eq. (2.33)] this does not cause any ambiguity, because a ro-tation by Φ about e is the same as a rotation by −Φ about −e. Thus, thereis no loss of generality in taking the positive sign in Eq. (2.50). Of course, thederivation given above is valid only if q4 �= 0. If q4 is close to zero, one canemploy an alternative derivation, such as the following:

q2 =c12 + c21

4q1,

q3 =c31 + c13

4q1, (2.51)

q4 =c23 − c32

4q1,

whereq1 = ±1

2√

1 + c11 − c22 − c33 . (2.52)

Similarly, the two remaining alternative derivations of the quaternion fromthe rotation matrix involve division by q2 and q3, respectively. Among allthe four possible derivations, the greatest numerical accuracy is obtained forthe one that has the largest denominator term, which implies the largestargument in the square root. There are efficient algorithms that employ sucha procedure for the computation of the quaternion, such as the one encodedin the MATLAB file, quaternion.m, given in Table 2.1.

The quaternion is a compact, nonsingular representation of attitude thatresults in algebraic (rather than trigonometric) expressions for the elementsof the rotation matrix. Another benefit in using the quaternion over otherattitude representations is in its easy combination for successive rotations.When the orientation q′′, q′′4 is obtained by first undergoing a rotation q, q4

followed by a rotation q′, q′4, we can substitute Eq. (2.48) on both sides of therelationship

C(q′′, q′′4 ) = C(q′, q′4)C(q, q4) , (2.53)

in order to obtain the following simple product, called the composition rule:⎧⎪⎪⎨⎪⎪⎩

q′′1q′′2q′′3q′′4

⎫⎪⎪⎬⎪⎪⎭ =

⎛⎜⎜⎝

q′4 q′3 −q′2 q′1−q′3 q′4 q′1 q′2q′2 −q′1 q′4 q′3−q′1 −q′2 −q′3 q′4

⎞⎟⎟⎠⎧⎪⎪⎨⎪⎪⎩

q1

q2

q3

q4

⎫⎪⎪⎬⎪⎪⎭ . (2.54)

The composition rule of Eq. (2.54) is the defining property of the quater-nion [1], [2]. Any set (q, q4) that satisfies this rule is called a quaternion.1 The1 The quaternion predates vectors and matrices. It was discovered by Hamilton

(1805–1865), who spent many years deriving the rather abstract quaternion alge-bra. However, the relationships for symmetric Euler parameters (quaternion forrotation) that are employed in attitude representation can be derived withoututilizing the quaternion algebra. The symmetric Euler parameters form a specialquaternion that obeys the constraint given by Eq. (2.45).


Table 2.1. M-file quaternion.m for Deriving the Quaternion from Rotation Matrix

function q=quaternion(C)%(c) 2006 Ashish TewariT=trace(C);qsq=[1+2*C(1,1)-T;1+2*C(2,2)-T;1+2*C(3,3)-T;1+T]/4;[x,i]=max(qsq);if i==4

q(4)=sqrt(x);q(1)=(C(2,3)-C(3,2))/(4*q(4));q(2)=(C(3,1)-C(1,3))/(4*q(4));q(3)=(C(1,2)-C(2,1))/(4*q(4));

endif i==3

q(3)=sqrt(x);q(1)=(C(1,3)+C(3,1))/(4*q(3));q(2)=(C(3,2)+C(2,3))/(4*q(3));q(4)=(C(1,2)-C(2,1))/(4*q(3));

endif i==2


endif i==1


end

efficiency of the quaternion relationship, Eq. (2.54), for the combined rotationis evident from the fact that it involves only 16 multiplications, whereas therotation matrix representation of the same combination [Eq. (2.53)] requires27. A MATLAB code called quatrot.m, which implements Eq. (2.54), is givenin Table 2.2. Note that the code considers the quaternion as a row vector.

Table 2.2. M-file quatrot.m for Combining Two Rotations in Terms of Quaternion

function qpp=quatrot(q,qp)% Program for combining two successive rotations, given by quaternions% ’q’ and ’qp’, to produce the final orientation given by the quaternion% ’qpp’. All the quaternions are stored here as row vectors.% (c) 2006 Ashish Tewariqpp=q*[qp(4) qp(3) -qp(2) qp(1);

-qp(3) qp(4) qp(1) qp(2);qp(2) -qp(1) qp(4) qp(3);-qp(1) -qp(2) -qp(3) qp(4)]’;

Example 2.8. Derive the quaternion representation for the orientation givenin Example 2.5, where we had utilized Eq. (2.53) to obtain the compositerotation matrix formed out of three successive, elementary rotations.

2.7 Rodrigues Parameters (Gibbs Vector) 27

We first derive the quaternion for each elementary rotation, and then em-ploy the relationship of Eq. (2.54) twice to produce the quaternion for the finalrepresentation. In order to do so, we use the following MATLAB statementsto call quaternion.m (Table 2.1), and quatrot.m (Table 2.2). It is presumedthat the three elementary rotation matrices have been calculated and storedin the MATLAB workspace (Example 2.5).

>> q1=quaternion(C1) %quaternion for the first elementary rotation

q1 = -0.3827 0 0 0.9239

>> q2=quaternion(C2) %quaternion for the second elementary rotation

q2 = 0 0.3827 0 0.9239

>> q3=quaternion(C3) %quaternion for the third elementary rotation

q3 = 0 0 -0.3827 0.9239

>> qp1=quatrot(q1,q2) %quaternion for intermediate rotation

qp1 = -0.3536 0.3536 -0.1464 0.8536

>> qp2=quatrot(qp1,q3) %quaternion for final rotation

qp2 = -0.4619 0.1913 -0.4619 0.7325

>> C=C3*C2*C1 % check: final rotation matrix

C = 0.5000 -0.8536 0.14640.5000 0.1464 -0.85360.7071 0.5000 0.5000

>> q=quaternion(C) %confirm quaternion for final rotation

q = -0.4619 0.1913 -0.4619 0.7325

Hence, the final orientation is represented by q1 = −0.4619, q2 = 0.1913,q3 = −0.4619, and q4 = 0.7325, which is compatible with the final rotationmatrix. The accuracy of the calculated quaternion can also be confirmed bychecking that it yields the correct principal rotation as follows:

Φ = 2 cos−1 q4 = ±85.8009◦

ande =

qsinΦ

2

= ±(0.6786i− 0.2811j + 0.6786k) .

Another set of four parameters related to the quaternion is the complexCayley–Klein matrix [3], which is sometimes used in attitude representations.However, we shall not consider its derivation here.

2.7 Rodrigues Parameters (Gibbs Vector)

A set of three attitude parameters, ρ = (ρ1, ρ2, ρ3)T , called Rodrigues parame-ters, or the Gibbs vector, can be directly derived from the quaternion (q, q4)as follows:


ρ.=

qq4

, (2.55)

which, when substituted into Eq. (2.44), yields

ρ = e tanΦ

2. (2.56)

The composition rule for Rodrigues parameters can be derived from that forthe quaternion [Eq. (2.54)] to be the following:

ρ′′ =ρ + ρ′ − ρ′ × ρ

1 − ρ · ρ′ , (2.57)

where ρ′′ represents the final orientation obtained by combining ρ and ρ′. Byusing Euler’s formula (Exercise 2.5), one can derive the following expressionfor the rotation matrix in terms of Rodrigues parameters:

C = (I − S(ρ))(I + S(ρ))−1 , (2.58)

where S(ρ) is the following skew-symmetric matrix formed out of the elementsof ρ:

S(ρ) =

⎛⎝ 0 −ρ3 ρ2

ρ3 0 −ρ1

−ρ2 ρ1 0

⎞⎠ . (2.59)

Clearly, Rodrigues parameters can be derived from the rotation matrix ele-ments, cij , as follows:

ρ1 =c23 − c32

1 + traceC,

ρ2 =c31 − c13

1 + traceC, (2.60)

ρ3 =c12 − c21

1 + traceC.

It is evident from Eq. (2.56) [as well as from Eq. (2.60)] that Rodrigues para-meters have a singularity at Φ = nπ (n = 1, 3, 5, . . .), hence their use is limitedto principal rotations of Φ < 180◦. The three-parameter set is thus similar tothe Euler angles in being incapable of representing an arbitrary orientation.

2.8 Modified Rodrigues Parameters

In order to extend the applicability of Rodrigues parameters for principal ro-tations greater than 180◦, a modified three-parameter set is defined as follows:

p .=q

1 + q4, (2.61)

2.8 Modified Rodrigues Parameters 29

which, when substituted into Eq. (2.44), yields

p = e tanΦ

4. (2.62)

Clearly, the new set p = (p1, p2, p3)T —called the modified Rodrigues para-meters—is nonsingular for principal rotations of Φ < 360◦. However, thereis a singularity at Φ = 360◦. Most nonspinning aerospace vehicles have ori-entations with Φ < 360◦ and are thus represented by the set of modifiedRodrigues parameters without singularity. Since p is a minimal representa-tion, it is advantageous over the four-parameter quaternion in reducing thenumber of kinematic equations to be solved for the attitude.

The rotation matrix can be expressed in terms of the modified Rodriguesparameters with the use of Eq. (2.46) as follows:

C = I +4(pT p− 1)(1 + pT p)2

S(p) +8

(1 + pT p)2S2(p) , (2.63)

where S(p) is the following skew-symmetric matrix formed out of the elementsof p:

S(p) =

⎛⎝ 0 −p3 p2

p3 0 −p1

−p2 p1 0

⎞⎠ . (2.64)

By substituting the relationship between the Rodrigues and modified Ro-drigues parameters,

ρ =2p

1 − pT p, (2.65)

into Eq. (2.57), we can derive the following rather complicated compositionrule for the modified Rodrigues parameters:

p′′ =(1 − pT p)p′ + (1 − p′T p′)p − 2p′ × p

1 + (pT p)(p′T p′) − 2p · p′ . (2.66)

Example 2.9. Derive the modified Rodrigues parameters for the orientationgiven in Example 2.5, for which the quaternion was calculated in Example 2.8.

We begin by applying Eq. (2.61) to the calculated quaternion:

p =1

1 + 0.7325

⎧⎨⎩

−0.46190.1913−0.4619

⎫⎬⎭ =

⎧⎨⎩

−0.26660.1104−0.2666

⎫⎬⎭ ,

leading to the skew-symmetric matrix [Eq. (2.64)],

S(p) =

⎛⎝ 0 0.2666 0.1104

−0.2666 0 0.2666−0.1104 −0.2666 0

⎞⎠ ,


which corresponds to the following rotation matrix according to Eq. (2.63):

C =

⎛⎝ 0.5 −0.8536 0.1464

0.5 0.1464 −0.85360.7071 0.5 0.5

⎞⎠ .

This is the same result as that of Example 2.5.

2.9 Attitude Kinematics

We are now in a position to address the evolution of a coordinate frame’sattitude with time. We can adopt any one of the various attitude representa-tions considered above to describe the changing attitude. Let us first take therotation matrix representation. Consider a rotating coordinate frame, (oxyz),with unit vectors, i, j,k, representing the axes ox, oy, oz, respectively, whosechanging attitude relative to an arbitrary, fixed frame is of interest. When theattitude is changing with time, the rotation matrix representing the orienta-tion of the rotating frame relative to a fixed frame is a function of time, C(t).In order to find this function, consider an infinitesimal principal rotation, ΔΦ,in a small time interval, Δt, measured after a given time, t. The rotation issmall enough for us to approximate cosΔΦ ≈ 1, sinΔΦ ≈ ΔΦ. Furthermore,we assume that that Δt is so small that the axis of rotation, e(t), remains es-sentially unchanged. These assumptions allow us to write the rotation matrixrepresenting attitude change in the time interval Δt with the use of Euler’sformula (Exercise 2.5) as follows:

C(Δt) ≈ I − ΔΦS(e) , (2.67)

or,C(Δt) ≈ I − S(ΔΦe) , (2.68)

where

S(ΔΦe) =

⎛⎝ 0 −ΔΦez ΔΦey

ΔΦez 0 −ΔΦex

−ΔΦey ΔΦex 0

⎞⎠ . (2.69)

We chose to resolve the instantaneous axis of rotation in the rotating frame,because C(Δt) describes the frame’s orientation at time t + Δt, relative to itsown previous orientation at time t, rather than that relative to the fixed frame.Thus, e(t) has its components, ex, ey, ez, resolved along the instantaneousaxes of the rotating frame, i, j,k, at time t. Let the rotation matrices C(t) andC(t + Δt) denote the attitudes of the rotating frame relative to a fixed frameat times t and t + Δt, respectively. The rotation matrix, C(Δt), describesthe evolution of the attitude in the time interval, Δt, caused by a principalrotation of ΔΦ about e. Thus, C(Δt) denotes the rotation required to produceC(t+Δt) from C(t). Now, these rotation matrices must obey the composition

2.9 Attitude Kinematics 31

rule for successive rotations, Eq. (2.25), which implies that the final orientationat time t + Δt is related to the initial attitude at time t by

C(t + Δt) = C(Δt)C(t) . (2.70)

By substituting Eq. (2.68) into Eq. (2.70), we can write the time derivativeof the rotation matrix as follows:

dC

dt

.= limΔt→0

C(t + Δt) − C(t)Δt

= −S(ω)C(t) , (2.71)

whereω(t) .= lim

Δt→0

ΔΦeΔt

(2.72)

is called the angular velocity vector of the coordinate frame. It is important toemphasize that the angular velocity, ω(t), has its components, ωx, ωy, ωz, attime t resolved along the axes of the rotating coordinate frame (rather thanthe fixed frame). The skew-symmetric matrix of angular velocity componentsis thus the following:

S(ω) =

⎛⎝ 0 −ωz ωy

ωz 0 −ωx

−ωy ωx 0

⎞⎠ . (2.73)

Equation (2.71) can also be derived by simply differentiating with time bothsides of the following defining equation of the rotation matrix:⎧⎨

⎩ijk

⎫⎬⎭ = C

⎧⎨⎩

IJK

⎫⎬⎭ , (2.74)

Oy

x

x(t)

z(t+ t)

x(t+ t)

z

z(t)( )t

y(t+ t)

y(t)

Fig. 2.6. A coordinate frame rotating with angular velocity, ω(t).


resulting in ⎧⎨⎩

ω × iω × jω × k

⎫⎬⎭ =

dC

dt

⎧⎨⎩

IJK

⎫⎬⎭ . (2.75)

By applying the result of Exercise 2.1, and substituting (2.74) into Eq. (2.75),we have

−S(ω)

⎧⎨⎩

ijk

⎫⎬⎭ =

dC

dtCT

⎧⎨⎩

ijk

⎫⎬⎭ , (2.76)

from which Eq. (2.71) follows through the orthogonality property, Eq. (2.24).Recall that the angular velocity, ω(t), is resolved along the axes of the ro-

tating coordinate frame (oxyz), as shown in Fig. 2.6. Such a choice of axes isvery convenient when depicting a rotating rigid body (Chapter 13). In such acase, the axes of the rotating coordinate frame are fixed to the rigid body andcan be used to represent the attitude and angular velocity of the body relativeto an inertial (or space-fixed) frame. On rare occasions, it is desired to alter-natively express the angular velocity with components along the axes of aninertial frame. In such a case the angular velocity is termed space-referenced,or inertial angular velocity, and is related to the body referenced angularvelocity by ωI = CT (t)ω(t) according to the coordinate transformation ofEq. (2.21).

The differential equation for the rotation matrix is Eq. (2.71), which, witha given function ω(t) and initial condition, C(0), should be solved for C(t)in order to describe the evolution of the attitude. The solution, C(t), mustsatisfy the orthogonality condition Eq. (2.24); by differentiating both sides ofwhich, we have

ddt

CCT =ddt

CT C = 0 . (2.77)

This implies that CCT and CT C are equal and constant matrices. Thus, if C(t)satisfies Eq. (2.24) at some initial time t = 0, it does so at all other times.One can easily verify (Exercise 2.13) that the matrix differential equation,Eq. (2.71), satisfies Eq. (2.77).

Example 2.10. Consider a coordinate frame, i, j,k, rotating with an angularvelocity ω(t) = 0.1i − 0.5j − k rad/s. If the original attitude of the frame isgiven by

C =

⎛⎝ 0.1399200225 −0.9857942023 0.0929095147

−0.9432515656 −0.1612425105 −0.29030559210.3011625330 −0.0470174803 −0.9524129804

⎞⎠ ,

determine the attitude of the frame after one second.Note that since the given angular velocity components resolved in the

rotating frame are constant, the matrix differential equation, Eq. (2.71), isnow a set of linear algebraic equations with constant coefficients, S(ω), and


can be integrated easily using the evolution rule of Eq. (2.70). It is to be notedthat Eq. (2.70) is valid only for an infinitesimal time interval, Δt. Since thegiven time of 1 s is not small enough given the assigned angular speed, we mustbreak it into several smaller intervals of equal length, and apply Eq. (2.70) toeach interval, beginning with the given attitude at t = 0. Such an approach iscalled time-marching and is frequently applied to the solution of linear, vector(or matrix) differential equations.

We begin the solution by writing a MATLAB code called rotevolve.m (Ta-ble 2.3), which obtains the rotation matrix by the time-marching approach ofEq. (2.70). During each time step, the rotation angle is given by ΔΦ =| ω | Δt,while the axis of rotation is approximated by e = ω

|ω| . Needless to say, theseapproximations are accurate only if a reasonably small Δt is employed. Thesize of a time step should be much smaller than the period of rotation andis generally dictated by the desired tolerance in satisfying the orthogonal-ity condition Eq. (2.24). Since the computation error accumulates over thesteps, a reasonably small time step must be used for a given time interval.In rotevolve.m, the time step is obtained by 106Δt = 2π

|ω| . In order to invokerotevolve.m for the present example, the following MATLAB statements areused:

>> C= [0.1399200225 -0.9857942023 0.0929095147;-0.9432515656 -0.1612425105 -0.2903055921;0.3011625330 -0.0470174803 -0.9524129804];% rotation matrix (t=0)

>>C*C’ %test for orthogonality of given rotation matrix

ans = 1.0000 0.0000 0.00000.0000 1.0000 0.00000.0000 0.0000 1.0000

>> Cf=rotevolve(C,[0.1;-0.5;-1],1) %call rotevolve for final rot. matrix

Cf = 0.94711618664267 -0.315392238061307 -0.0592086317903142-0.313796220025831 -0.871634028765343 -0.3765523943592750.0671544766324447 0.375218475325714 -0.924501393941303

Therefore, the attitude at t = 1 s is given by the rotation matrix

C =

⎛⎝ 0.94711618664267 −0.315392238061307 −0.0592086317903142

−0.313796220025831 −0.871634028765343 −0.3765523943592750.0671544766324447 0.375218475325714 −0.924501393941303

⎞⎠ .

We can verify the orthogonality of this matrix, within a tolerance of 10−5.However, since the time-step size is only Δt = 5.5975× 10−6 s, the computa-tion requires 178,651 steps, which is rather inefficient.

The angular velocity, ω(t), is generally time-varying and is obtained from theattitude dynamics (Chapter 13) (rather than kinematics). As seen above, evenwhen ω(t) has constant elements, a large number of inefficient matrix productsmust be computed for obtaining the attitude using the rotation matrix. Whenthe angular velocity, ω(t), does not have constant elements, it becomes rathercumbersome to solve the differential equation, Eq. (2.71), with a time-varyingcoefficient matrix S(ω), for the nine elements of C(t). Instead, it is much easier


Table 2.3. M-file rotevolve.m for the Time Evolution of the Rotation Matrix

function c=rotevolve(c0,w,T)%function for evolving the rotation matrix with a constant%body-referenced angular velocity%c0=rotation matrix at t=0%w=angular velocity vector (3x1) (rad/s)%T=final time (s)%(c)2006 Ashish TewariS=[0 -w(3,1) w(2,1);w(3,1) 0 -w(1,1);-w(2,1) w(1,1) 0];dt=2*pi/(10^6*norm(w))cdt=eye(3)-S*dt;t=dt;c=cdt*c0;while t<=T

c=cdt*c;t=t+dt;

end

to employ other attitude representations, such as Euler angles, quaternion, orthe modified Rodrigues parameters.

Y

X

O

Z,z

z ,z

y

xx , x

y

y

Fig. 2.7. The Euler angle orientation, (ψ)3, (θ)1, (φ)3.

Let us consider the Euler angle representation, (ψ)3, (θ)1, (φ)3, which isoften employed for spacecraft (as well as orbital plane) attitude. Figure 2.7depicts the three successive, elementary rotations required in this representa-tion from a stationary frame, (OXYZ), with axes along I,J,K, respectively, tothe rotating frame (Oxyz) with axes i, j,k, respectively. The rotation matrixrelating the final orientation of (Oxyz) to (OXYZ) can be obtained by

C = C3(φ)C1(θ)C3(ψ) = (2.78)⎛⎝(cos ψ cosφ − sin ψ sin φ cos θ) (sin ψ cosφ + cosψ sin φ cos θ) sin φ sin θ−(cosψ sin φ + sin ψ cosφ cos θ) (− sinψ sin φ + cosψ cosφ cos θ) cosφ sin θ

sinψ sin θ − cosψ sin θ cos θ

⎞⎠.


In terms of the local angular rates, φ̇, θ̇, ψ̇, the angular velocity, ω(t), canbe expressed as follows:

ω(t) = φ̇k + θ̇i′ + ψ̇K , (2.79)

wherei′ = i cosφ − j sinφ (2.80)

andK = i sinφ sin θ + j cosφ sin θ + k cos θ . (2.81)

Upon substituting Eqs. (2.80) and (2.81) into Eq. (2.79), we have

ω(t) =

⎧⎨⎩

ωx

ωy

ωz

⎫⎬⎭ =

⎧⎨⎩

ψ̇ sin φ sin θ + θ̇ cosφ

ψ̇ cosφ sin θ − θ̇ sin φ

ψ̇ cos θ + φ̇

⎫⎬⎭ , (2.82)

or, ⎧⎨⎩

ψ̇

θ̇

φ̇

⎫⎬⎭ =

1sin θ

⎛⎝ sin φ cosφ 0

cosφ sin θ − sin φ sin θ 0− sinφ cos θ − cosφ cos θ sin θ

⎞⎠⎧⎨⎩

ωx

ωy

ωz

⎫⎬⎭ . (2.83)

Equation (2.83) is the required kinematic relationship between the Euler an-gles, ψ, θ, φ, and the angular velocity in the form of three, coupled, nonlinearfirst-order ordinary differential equations. Note that this relationship has sin-gularities at θ = nπ (n = 0, 1, 3, . . .), which are the inherent singularities ofthe Euler angle representation, (ψ)3, (θ)1, (φ)3. Hence, this representation islimited to rotations of 0 < θ < π, which are applicable to spacecraft attitudesand the orientation of orbital planes. A different Euler angle representation,(ψ)3, (θ)2, (φ)1 (Fig. 2.5), is commonly employed in aircraft applications andhas singularities at θ = nπ

2 (n = 1, 3, . . .), which limits its application to−π

2 < θ < π2 . The kinematic relationship for (ψ)3, (θ)2, (φ)1 will be discussed

in Chapter 4.

Example 2.11. Solve the problem given in Example 2.10 using the Euler anglerepresentation, (ψ)3, (θ)1, (φ)3.

We begin by deriving the (ψ)3, (θ)1, (φ)3 Euler angles for the initial at-titude prescribed in Example 2.10. By using the third column of the givenrotation matrix and Eq. (2.78), we have

θ = cos−1 c33 = cos−1(−0.9524129804) = 162.2532◦,

sin φ =c13

sin θ=

0.09290951470.3048106211

= 0.3048106211,

cosφ =c23

sin θ=

−0.29030559210.3048106211

= −0.9524129804 ,

which yield φ = θ = 162.2532◦. The third column of the given rotation matrixand Eq. (2.78) produce


sin ψ =c31

sin θ=

0.3011625330.3048106211

= 0.98803162409,

cosφ =c32

sin θ=

−0.04701748030.3048106211

= 0.1542514499 ,

or, ψ = 81.1266◦. Now, we write a code called euler313evolve.m, tabulated inTable 2.4, which provides the kinematic equations for integration accordingto the intrinsic Runge–Kutta, nonlinear differential equation solver (AppendixA) of MATLAB (ode45.m). The program is invoked as follows:

>> dtr=pi/180; psi=81.1266*dtr; theta=162.2532*dtr; phi=theta;% Euler angles>> [t,x]=ode45(@euler313evolve,[0 1],[psi theta phi]’); %Runge-Kutta solver>> plot(t,x/dtr) %plot of Euler angles’ time history

Fig. 2.8. The time history of the Euler angles, (ψ)3, (θ)1, (φ)3 with ω =(0.1,−0.5,−1)T .

The resulting time history of the Euler angles, obtained in only 41 timesteps, is shown in Fig. 2.8. Note that θ does not cross the singularity atθ = 180◦. This is merely due to the specific value of the constant angu-lar velocity and cannot be ensured in a general case where angular velocitycomponents could be time-varying. The final values obtained at t = 1 s areψ = 169.8530◦, θ = 157.5933◦, and φ = 188.9358◦. We can easily verify that asubstitution of these angles into Eq. (2.78) results in the same final rotationmatrix as that derived in Example 2.10.

There is no qualitative modification required in the above implementedRunge–Kutta approach (Appendix A) for solving the nonlinear differentialequations when the angular velocity components are varying with time. Thus,the Euler angle representation for attitude kinematics is much more compact


Table 2.4. M-file euler313evolve.m for the Time Evolution of the 3-1-3 Euler Angles

function xdot=euler313evolve(t,x)%x(1)=psi, x(2)=theta, x(3)=phi% (c) 2006 Ashish Tewariw=[0.1,-0.5,-1];% angular velocity in rad/sxdot(1,1)=(sin(x(3))*w(1)+cos(x(3))*w(2))/sin(x(2));xdot(2,1)=cos(x(3))*w(1)-sin(x(3))*w(2);xdot(3,1)=w(3)-(sin(x(3))*cos(x(2))*w(1)+cos(x(3))*cos(x(2))*w(2))/sin(x(2));

than the rotation matrix. However, it suffers from singularities and cannot beapplied when principal rotations are larger than θ = 180◦. For a compact kine-matical calculation that has a larger range of validity, we must look toward theattitude representations that are based upon the Euler-axis/principal anglecombination, such as the quaternion and the modified Rodrigues parameters.

The quaternion has an advantage in being a compact (albeit nonminimal)as well as nonsingular representation for attitude kinematics. Thus, modernflight dynamic applications generally employ the quaternion. The kinematicalequations of the rotating frame, (oxyz ), in terms of the quaternion can beobtained from its composition rule, Eq. (2.54), as follows. As in the foregoingdiscussion, we consider an infinitesimal principal rotation, ΔΦ, in a small timeinterval, Δt, such that the axis of rotation, e(t) = exi + eyj + ezk, remainsessentially unchanged. In Eq. (2.54), we substitute the orientation at t + Δt,{q′′, q′′4 } = {q(t+Δt), q4(t+Δt)}, which is obtained from the initial attitude,q(t), q4(t), by undergoing a rotation [q′, q′4] = [q(Δt), q4(Δt)]. Thus, from thedefinition of quaternion we have

q′1 = ex sinΔΦ

2,

q′2 = ey sinΔΦ

2, (2.84)

q′3 = ez sinΔΦ

2,

q′4 = cosΔΦ

2.

By substituting these into Eq. (2.54), approximating cosΔΦ ≈ 1, sinΔΦ ≈ΔΦ, and utilizing Eq. (2.72), we can write the following equation for the timeevolution of the quaternion:

{q(t + Δt), q4(t + Δt)}T ≈ [I +12ΩΔt]{q(t), q4(t)}T , (2.85)

where Ω is the following skew-symmetric matrix of the angular velocity com-ponents:

Ω =

⎛⎜⎜⎝

0 ωz −ωy ωx

−ωz 0 ωx ωy

ωy −ωx 0 ωz

−ωx −ωy −ωz 0

⎞⎟⎟⎠ . (2.86)


Therefore, the time derivative of the quaternion is the following:

d{q, q4}T

dt

.= limΔt→0

{q(t + Δt), q4(t + Δt)}T − {q(t), q4(t)}T

Δt

=12Ω{q(t), q4(t)}T . (2.87)

The linear, algebraic form of the matrix differential equation, Eq. (2.87), isan obvious advantage of the quaternion representation. Unlike the Euler an-gles, it is not required to evaluate trigonometric functions in the process ofsolving the kinematic equations. The matrix Ω is either constant, or time-varying, depending upon whether the components of the angular velocity arechanging with time. In either case, we can adopt a numerical scheme (Appen-dix A) to integrate this linear, ordinary differential equation for the quater-nion, q(t), q4(t). We can break the final time into smaller intervals and utilizeEq. (2.87) in each interval to produce a time history of the changing quater-nion. The splitting of time into smaller intervals is necessary when the angularvelocity components (Ω) are time-varying, wherein Eq. (2.87) is employed witha different Ω in each time interval. The size of the time interval in such a casewould be determined by the rate of change of angular velocity components.Such a quasi-steady approximation of the time-varying coefficient matrix iscommonly employed.

Example 2.12. Solve the problem given in Examples 2.10 and 2.11 using thequaternion.

Since Ω is a constant matrix here, we can write the solution to the result-ing linear, time-invariant state equations involving the quaternion using thematrix exponential (Chapter 14) as follows:

{q(t), q4(t)}T = e12Ωt{q(0), q4(0)}T , (2.88)

where q(0), q4(0) represents the initial attitude at t = 0. The matrix exponen-tial, eA, of a square matrix, A, can be computed using an efficient algorithm(Chapter 14), such as that implemented in the intrinsic MATLAB functionexpm.m, which we will employ here. Equation (2.88) is implemented in theprogram, quatevolve.m, which is tabulated in Table 2.5. The computation ofthe final quaternion is carried out by the following MATLAB statements,assuming the initial rotation matrix, C, is already stored in the workspace:

>> q0=quaternion(C) %quaternion at t=0 using "quaternion.m"

q0 = 0.750595682341698 -0.64250495082845 0.131253102359508 -0.0810316784134013

>> qf=quatevolve(q0’,[0.1;-0.5;-1],1)’ % final quaternion

qf = 0.967373385090801 -0.162601824240507 0.002053412854426 -0.194281146902704

>> Cf=rotquat(qf’) %rotation matrix

Cf = 0.94711286 -0.31539123 -0.0592081039-0.31379548 -0.87163097 -0.37655259880.067153772 0.375217044 -0.9245012389


The final rotation matrix is computed above from the final quaternionaccording to Eq. (2.46), which is implemented in the M-file rotquat.m (Ta-ble 2.6). Note that the final attitude agrees with that calculated in Example2.10 up to the fifth decimal place, which was the tolerance specified in Ex-ample 2.10. However, since only one matrix multiplication is required (apartfrom the steps used in calculating the matrix exponential) in the specified timeinterval (compared to 178,651 in Example 2.10), this computation is more ef-ficient and suffers from a smaller cumulative truncation error. Of course, thequaternion computation does not have any singularities and thus can be car-ried out for any angular velocity. The same cannot be said for the Euler anglecomputation in Example 2.11, where a singularity was avoided only by thegiven value of the angular velocity and the initial attitude.

Table 2.5. M-file quatevolve.m for the Time Evolution of the Quaternion

function q=quatevolve(q0,w,T)%function for evolving the quaternion with a constant%body-referenced angular velocity%q0=quaternion at t=0 (4x1) vector%w=angular velocity vector (3x1) (rad/s)%T=final time (s)%(c) 2006 Ashish TewariS=[0 w(3,1) -w(2,1) w(1,1);-w(3,1) 0 w(1,1) w(2,1);

w(2,1) -w(1,1) 0 w(3,1);-w(1,1) -w(2,1) -w(3,1) 0];q=expm(0.5*S*T)*q0;

Table 2.6. M-file rotquat for the Computation of Rotation Matrix from the Quater-nion

function C=rotquat(q)% rotation matrix from the quaternion% (c) 2006 Ashish TewariS=[0 -q(3,1) q(2,1);q(3,1) 0 -q(1,1);-q(2,1) q(1,1) 0];C=(q(4,1)^2-q(1:3,1)’*q(1:3,1))*eye(3)+2*q(1:3,1)*q(1:3,1)’-2*q(4,1)*S;

Finally, we consider the kinematical representation possible through themodified Rodrigues parameters, p = (p1, p2, p3)T . Being a minimal repre-sentation based upon the quaternion, p, reduces the number of kinematic,first-order differential equations required by the latter. However, its singu-larity at Φ = 360◦ limits its application to principal rotations of Φ < 360◦,which gives a larger range of validity than that of the other minimal represen-tations (Euler angles and Rodrigues parameters). By substituting Eq. (2.61)into Eq. (2.87), we have

dpdt

=12

[S(p) + ppT +

(1 − pT p

2

)I

]ω , (2.89)

where S(p) is given by Eq. (2.64). Being a nonlinear, ordinary differentialequation, Eq. (2.89) can be integrated in time using a Runge–Kutta algorithm(Appendix A) to produce the instantaneous value of p(t), given an initialcondition, p(0), and the time-varying angular velocity, ω(t).


Example 2.13. Solve the problem given in Examples 2.10–2.12 using the modi-fied Rodrigues parameters.

We begin by writing a code called mrpevolve.m, tabulated in Table 2.7,which provides the scalar kinematic equations, Eq. (2.89), for integration ac-cording to the intrinsic Runge–Kutta nonlinear differential equation solver(Appendix A) of MATLAB (ode45.m). The final rotation matrix is calculatedfrom the final modified Rodrigues vector through Eq. (2.63), which is imple-mented in the program rotmrp.m (Table 2.8). The computations are given bythe following statements, beginning with the initial quaternion from Exam-ple 2.12:

>> p0=q0(1:3)/(q0(4)+1) %MRP at t=0 from initial quaternion (Example 2.12)

p0 = 0.816780801590412 -0.699158976143123 0.142826579846517

>> [t,p]=ode45(@mrpevolve,[0 1],p0’); %Runge-Kutta solver>> pf=p(size(t,1),:) % final MRP vector at t=1 second

pf = 1.2006339199379 -0.201809630219251 0.0025485478218541

>> Cf=rotmrp(pf’) %final rotation matrix using "rotmrp.m"

Cf = 0.947112860753656 -0.315391232427005 -0.0592081033575685-0.313795474675653 -0.871630965841634 -0.3765525985291590.0671537718084939 0.375217043714908 -0.924501239067713

>> plot(t,p) %plot of MRP time history>> for i=1:size(t,1);phi(i)=4*atan(norm(p(i,:)))*180/pi;end>> plot(t,phi) %plot of principal angle time history

The final rotation matrix agrees with that computed in Example 2.12 up tonine decimal places. The resulting plots for the time evolution of the modifiedRodrigues parameters, and the principal angle, Φ = 4 tan−1 | p | are shownin Figs. 2.9 and 2.10, respectively. As in the case of Euler angles (Example2.12), the Runge–Kutta iteration is carried out with 41 time steps.

Table 2.7. M-file mrpevolve.m for the Time Derivatives of the MRP

function pdot=mrpevolve(t,p)% program for calculating the time derivatives of the modified Rodrigues% parameters, with a constant angular velocity, w (3x1)% (c) 2006 Ashish Tewariw=[0.1;-0.5;-1]; % angular velocity in rad/sS=[0 -p(3,1) p(2,1);p(3,1) 0 -p(1,1);-p(2,1) p(1,1) 0];G=0.5*(eye(3)+S+p*p’-0.5*(1+p’*p)*eye(3));pdot=G*w;

Table 2.8. M-file rotmrp.m for the Rotation Matrix from the MRP

function C=rotmrp(p)% rotation matrix from the modified Rodrigues parameters% (c) 2006 Ashish TewariS=[0 -p(3,1) p(2,1);p(3,1) 0 -p(1,1);-p(2,1) p(1,1) 0];C=eye(3)+4*(p’*p-1)*S/(1+p’*p)^2+8*S*S/(1+p’*p)^2;


Fig. 2.9. The time history of the modified Rodrigues vector, p, with ω =(0.1,−0.5,−1)T .

Fig. 2.10. The time history of the principal angle, Φ = 4 tan−1 | p |.


2.10 Summary

The coordinates of a vector, A, can be resolved in various frames of reference,each of which consists of a right-handed triad formed by the unit vectorsi, j,k. The transformation from a frame of reference to another (also calledthe attitude of a reference frame relative to another) involves an orthogo-nal rotation matrix, C, the elements of which can be represented by variousalternative kinematic parameters, such as Euler angles, Euler axis/principalrotation, Euler symmetric parameters (quaternion), Rodrigues and modifiedRodrigues parameters, etc. Of these, the three-parameter sets are inherentlysingular, whereas the four-parameter sets—such as the quaternion—are non-singular, but also nonminimal. The attitude kinematics is governed by a setof nonlinear ordinary differential equations in terms of either the rotation ma-trix or any of the alternative attitude parameters, which must be integrated intime in order to specify the instantaneous attitude of a reference frame withrespect to another. When the angular velocity of a reference frame relative toanother is constant, the attitude kinematics is described by a linear system ofdifferential differential equations.

Exercises

2.1. Show that the vector product can be written as follows:

a × b = S(a)b ,

where S(a) is the following skew-symmetric matrix formed out of the elementsof a:

S(a) =

⎛⎝ 0 −a3 a2

a3 0 −a1

−a2 a1 0

⎞⎠ .

A skew-symmetric matrix has the property S(a)T = −S(a).

2.2. Find the Euler angles from the elements of the rotation matrix for therepresentation (ψ)3, (θ)1, (φ)3. What are the points of singularity for this rep-resentation?

2.3. Derive the general expressions for the Euler axis, e, and the principalangle, Φ, in terms of a symmetric Euler angle representation. Can such simpleexpressions be obtained for an asymmetric set?

2.4. The singularity of an Euler angle representation prevents the Euler an-gles from describing infinitesimal rotations about singular orientations. Usingthe representation (ψ)3, (θ)1, (φ)3, show that if an infinitesimal rotation isperformed about OY from the initial orientation of φ = θ = ψ = 0, the Eulerangles change instantaneously to finite values. [In a gyroscope (Chapter 14),

2.10 Summary 43

this peculiarity of Euler angles manifests itself as the phenomenon of gimballock , in which a jamming of the gimbals occurs at a singular attitude, sincethey are physically incapable of rotating instantaneously by finite angles.]

2.5. For a general orientation, show that the rotation matrix can be expressedin terms of the Euler axis and principal angle as follows:

C = I cosΦ + (1 − cosΦ)eeT − sin ΦS(e) ,

where S(e) is the following skew-symmetric matrix (Exercise 2.1) formed outof the elements of e:

S(e) =

⎛⎝ 0 −e3 e2

e3 0 −e1

−e2 e1 0

⎞⎠ .

This relationship between the rotation matrix and principal rotation an-gle/Euler axis is called Euler’s formula.

2.6. Derive Eq. (2.46) using the result of Exercise 2.5 and the definition ofthe quaternion given in Eq. (2.44). Note the similarity in the form of E andQ.

2.7. Obtain the expressions for the quaternion elements in terms of the ele-ments of the rotation matrix and employing (a) q2, and (b) q3, respectively,in the denominator.

2.8. Derive the composition rule for the quaternion, Eq. (2.54).

2.9. Derive the composition rule for the Rodrigues parameters, Eq. (2.57).

2.10. Show that the derivation of the Rodrigues parameters from the rotationmatrix, Eq. (2.60), is true.

2.11. Derive the composition rule for the modified Rodrigues parameters,Eq. (2.66).

2.12. Derive an expression for the modified Rodrigues vector, p, in terms ofthe elements of the rotation matrix, cij .

2.13. By differentiating CCT (or CT C) with time according to the chain rule,and substituting into the result the differential equation for the rotation ma-trix, Eq. (2.71), show that Eq. (2.77) is satisfied.

2.14. Using the space-referenced angular velocity, ωI = CT ω, derive an ex-pression for the time derivative of the rotation matrix.

2.15. Derive the kinematic relationship between the time derivatives of theEuler angles, (ψ)3, (θ)2, (φ)1, and the body-referenced angular velocity com-ponents.


2.16. Write a MATLAB program for calculating the time history of quater-nion, given a constant body-referenced angular velocity. Use this program toplot the variation of q(t), q4(t) for Example 2.12.

2.17. Write a MATLAB program for calculating the time history of the modi-fied Rodrigues parameters, given a varying body-referenced angular velocity,ω = (−0.01, 0.5 sin3t,−0.5 cos3t)T . Use this program to compute and plotthe variation of p(t), given p(0) = (0.1,−0.2, 0.3)T . Also, plot the variationof the principal angle with time.

2 attitude and kinematics of coordinate framesrenaissance.ucsd.edu/courses/mae207/tewari_chap02.pdf2...

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