adcs notes part1

8/9/2019 ADCS Notes Part1

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Course Notes

Spacecraft Attitude Dynamics and Control

Politecnico di Milano

Prof. Franco Bernelli Zazzera

Part 1:Attitude dynamics and kinematics


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Spacecraft Attitude Dynamics and Control Course notes

1

Note for the reader

These short notes are in support of the course Attitude dynamics and control, they are not

intended to replace any textbook. Interested readers are encouraged to consult also printed

textbooks and archival papers.

Acknowledgments

I must really thank Ivan Ferrario and Mauro Massari, former students in my class, whose patient

work gave the first impulse to the preparation of the current notes.


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Fundamental quantities

Angular momentum of a rigid body

Let us start by defining the physical quantities that are fundamental and recurrent in the rotationalmotion of rigid spacecraft. These quantities are the angular momentum and the kinetic energy. The

angular momentum of a rigid body B is defined referring to a reference frame attached to the

rigid body and moving with it, with origin O:

Considering an infinitesimal point mass, it is defined as:

dmvrhd o =

Therefore, for the whole rigid body B:

dmvrhBo =

The velocity is expressed as:

iBo vrvv ++=

where vo is the velocity of the origin of the reference system attached to the rigid body, is theangular velocity of the rigid body and v iBis the relative velocity between any point of the body and

the origin of the reference system. For a rigid body, this relative velocity is always zero.Evaluating the integral yields:

( ) ( ) +=+= BBoB oo dmrrdmrvdmrvrh

We must now evaluate the vector cross products:

z

v

B r

O

y

x


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( )

( )

+

+

+

=

=

=

=

yzyxzx

xyxyzz

xzzxyy

rr

xy

zx

yz

r

z

y

x

r

y

2

zx

2

z

x

2

yz

2

y

z

2

xy

2

x

yx

xz

zy

z

y

x

The first term of the integral can be expressed as a function of the first moment of mass, that is a

measure of the distance between the origin of the reference system and the center of mass, while

integrating the terms of the second vector term, since angular velocity is constant at any point of the

rigid body, we can form the so called inertia matrix and write the angular momentum as:

+= ISvh ooo

where Sois the first moment of mass and I is the inertia matrix that is expressed as:

( )( )

( )

+===

=+==

==+=

=

=

B

22zzBzyBzx

ByzB

22yyByx

BxzBxyB

22xx

Bo

dmyxIdmzyIdmzxI

dmyzIdmzxIdmyxI

dmxzIdmxyIdmzyI

I

dmrS

Inertia matrix

The inertia matrix is symmetric and the terms along the diagonal are called inertia moments, while

the off-diagonal terms are called inertia products. Inertia moments are always positive, while there

is no general rule for the inertia products. The inertia matrix, by its definition, must respect precise

constraints among its elements. It is possible to define 6 triangular inequalities that relate the inertia

moments, which represent constraints on the admissible values of inertia moments. These

constraints are on the sum of two inertia moments and on the difference between two inertia

moments, for example:

( ) ( )

( ) ( ) zz2222

yyxx

zz22222

yyxx

IdmyxdmxyII

Idmyxdmz2yxII

=+=

=+++=+

The remaining inequalities are obtained by simple rotation of the labels x, y and z.

There exist also 3 inequalities that relate the inertia products:


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( ) ( )

zyxx

xx

zy

222

xx

I2I

0I0A

I2Admzy2dmzydmzyI

+=+=+=

Again, remaining inequalities are obtained by simple rotation of the labels x, y and z. Overall, 9

constraints relate the elements of the inertia matrix.

Rotation kinetic energy of a rigid body

Kinetic energy is defined as:

( ) ( ) ( ) ( ) ( )( ) ++=++== B oooB ooB dmrrrv2vvdmrvrvdmvvT2

The first term represents the kinetic energy due to the linear motion of the rigid body, not of interest

in this case since we study only rotational motion. Therefore, the kinetic energy due to the angular

motion is expressed as:

( ) ( ) ( )( ) ( ) ( ) +=+= BooB orot dmrrSv2dmrrrv2T2

Expanding the term to be integrated we have:

( ) ( ) xy2xz2yz2xyzxyzrr xyxzzy22

y

22

x

22

x

22

z

22

z

22

y +++++=

That, once integrated, leads to:

( ) ( ) xyxyxzxzzyyz2

zzz

2

yyy

2

xxxB I2I2I2IIIdmrr +++++=

We can then write the kinetic energy using matrix notation, as:

+= ISv2T2 oo

By choosing the origin O coincident with the center of mass of the system, Sovanishes and the

angular momentum and kinetic energy become:

=

=

IT2

Ih

Principal axes and principal inertia moments

We can now try to find a further simplification of the expressions of the angular momentum and

kinetic energy, trying to simplify the inertia matrix. In order to do so, we must change the reference

system, keeping the origin coincident with the center of mass but such that the kinetic energy is a


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purely quadratic expression in the components of , thus reducing the inertia matrix to diagonalform.

Any vector can be transformed from one reference system to another by multiplication by the

direction cosine matrix. This matrix contains the projections on the second reference frame of the

axis defining the first reference frame.

= a'

a represents the direction cosines matrix, that is orthogonal (aT=a

-1). Applying the multiplication

rule to the expression of the kinetic energy leads to:

( ) ( )'aI'aT2 TT =

and, in matrix notation1:

''I''aIa'T2TTT ==

where I is the inertia matrix in the new reference frame (a I aT).

Coming back to the problem of finding a rotation capable of transforming the inertia matrix intodiagonal form, we can write this problem as:

''',, zyx

T IIIdiagaIa =

This is the standard eigenvalue problem for matrix I. Once the eigenvalue problem is solved, the

rotation matrix a is composed by the eigenvectors associated to the eigenvalues of I, that are Ix, Iy

and Iz. Each row of rotation matrix a is one eigenvector. Eigenvectors correspond to the direction

cosines of the new reference frame, called principal inertia frame, with respect to the starting

reference frame. The principal inertia frame is therefore reference frame with origin in the center of

mass of the rigid body and such that all inertia products are equal to zero. Calling Ix, Iy, Iz, the

principal inertia moments, the angular momentum and kinetic energy become:

2

zz

2

yy

2

xx

zzyyxx

IIIIT2

IIIIh

++==

++==

1Matrix notation for vector dot product (inner product or scalar product) consists in the following:

2

T

121 vvvv =

z

z

y

y

x

x


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It is pointed out that a plane of symmetry (in terms of mass distribution) includes 2 principal inertia

axes, and then the third is uniquely defined as the axis orthogonal to the plane and passing through

the center of mass. It is also pointed out that, normally, it is easy to evaluate the inertia matrix in a

geometrically meaningful reference. Should this not be a principal axis frame, it can be calculated

by solving the eigenvalue problem as described.

The principal inertia matrix includes always the maximum possible and the least possible inertiamoment, no other reference can show along the diagonal greater or smaller inertia moments.

Inertia ellipsoid, kinetic energy ellipsoid, angular momentum ellipsoid

Assuming that the angular velocity has a constant direction, we can write:

=

where is the unit vector defining the direction of . The kinetic energy can be written in the form:

( ) 2zxxzzyyzyxxy2

zzz

2

yyy

2

xxx IIII2IIIIT2 =+++++==

The expression of Ican now be evaluated as:

+

+

+

+

+

=

zxxz

zy

yz

yxxy

2

zzz

2

y

yy

2

xxx III2IIII

with:

zz

y

y

x

x

lk

lj

l

i

=

=

=

=

=

=

defined as the direction cosines of unit vector . In a principal inertia reference system we canwrite:

2

zz

2

yy

2

xx lIlIlII ++=

that can be transformed into:

1I1

Il

I1

Il

I1

Il

z

2

z

y

2

y

x

2

x=++

This is an ellipsoid, called inertia ellipsoid, in which the coordinates of each point represent the

components of the generic direction divided by the square root of the inertia moment associated


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to that direction, while the semi-axes are inversely proportional to the square root of each principal

inertia moment.

Semi-axes: zyx I/1,I/1,I/1

Point coordinates Il,Il,Il zyx

Representing this ellipsoid in a Cartesian space, in a principal inertia frame, it is possible to

calculate the inertia moment associated to any direction. Tracing any direction, the intersection with

the ellipsoid occurs at a distance from the origin equal to I1 , where is the unit vector of the

given direction. In the same way, the intersection of the ellipsoid with the coordinate axes allows to

calculate the principal inertia moments.

Now we can write the expression of the kinetic energy as a function of the generic inertia moment

in the direction :

2

zz

2

yy

2

xx

2IIIIT2 ++==

This can be written also as:

1IT2IT2IT2 z

2

z

y

2

y

x

2

x =

+

+

that represents another ellipsoid, called kinetic energy ellipsoid, in a space defined by angular

velocities. This ellipsoid represents all possible angular velocities compatible with the given kinetic

energy T. The intersection of the ellipsoid with any direction is at distance IT2 from the origin.

A similar representation holds for the angular momentum:

2

z

2

z

2

y

2

y

2

x

2

x

2IIIhhh ++==


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that can be written as:

1IhIhIh

2

z

2

2

z

2

y

2

2

y

2

x

2

2

x =

+

+

Finally, we have a third ellipsoid, called angular momentum ellipsoid, which represents the

compatibility of the angular velocity with the given level of angular momentum.


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Geometrical/graphical description of torque-free motion

Polhode and herpolhode

The inertia ellipsoid has no intersection with the kinetic energy ellipsoid, unless the two arecoincident, while the angular momentum ellipsoid must intersect the kinetic energy ellipsoid at least

in one point in order to represent a real motion. In fact, if no energy dissipation exists and no torque

is applied, both kinetic energy and angular momentum must be constant. So, given T and h, angular

motion must evolve in such a way that angular velocity vector identifies the intersection of the

kinetic energy ellipsoid with the angular momentum ellipsoid. In fact, angular velocity must be

compatible with the energy level and with the angular momentum defined by the initial conditions

of motion. The intersection of the kinetic energy ellipsoid with the angular momentum ellipsoid

generates two lines, called polhodes.

So far we have looked at the problem in a body frame, not fixed in inertial space. In an inertial

frame, we will observe vector h as a fixed vector, if no torque is applied. Then, if no dissipation is

present, we have:

tcosh

tcoshT

=

==

We can observe that the projection of over h is constant:

tcoshT

hh

h ===

Now, considering a plane orthogonal to h, at a distance h from the center of mass, the terminalpoint of vector lies always on this plane due to the conservation of the projection of over h.This plane is called invariant plane. All this is a consequence of conservation of kinetic energy T

with time:

-2

-1

0

1

2

-2

-1

0

1

2-1.5

-1

-0.5

0

0.5

1

1.5

x

Poloide

y

z


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( ) ( )

hd0hd

0hd

0hdhddt

1h

dt

d

=

=

=+=

The line traced by the angular velocity on the invariant plane is in general not closed, it is calledherpolhode. The herpolhode is in any case a curve limited in space. In the body axes reference

frame the derivative of the angular velocity vector is always tangent to the kinetic energy ellipsoid,

while in the inertial space the kinetic energy ellipsoid rolls on the invariant plane as the angular

velocity draws the herpolhode.

From the mathematical point of view, the intersection of the two ellipsoids is evaluated as:

1

IhIhIh

1

IT2

IT2

IT2

2

z

2

2

z

2

y

2

2

y

2

x

2

2

x

z

2

z

y

2

y

x

2

x

=

+

+

=

+

+

Equating the two left hand terms, equal to 1, we find the polhode:

0T2

1

h

II

T2

1

h

II

T2

1

h

II

2

zz

2

z2

y

y

2

y2

xx

2

x =

+

+

In order to obtain a real solution, the three terms into brackets must show differences in sign.

Multiplying each term by h2we have three terms like:

T2

hI

2

Assuming Ix> Iy> Izwe obtain the conditions for a sign change:

T2

hI

2

x> andT2

hI

2

z<

It is not necessary to impose conditions on Iysince the change in sign is already guaranteed by the

previous conditions. The final constraint is then:

z

2

x IT2

hI >>

that guarantees the possibility of having a real rotational motion.

Given the result so far obtained, it is difficult to make any consideration on the polhode as a line in

space, while it is interesting to analyze its projections onto the three coordinate planes. We have:


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T2III

1T2

I

T2

I

T2

I

2

zz

2

yy

2

xx

2

zz

2

yy2

xx

=++

=

+

+

Looking at the projection onto the plane (x-y), we reduce to two dimensions by eliminating the

coordinate z:

2

yy

2

xx

2

zz IIT2I =

Substitution into the polhode equation gives:

( ) ( )

( ) ( )1

TI2h

III

TI2h

III

0h

hTI2

h

III

h

III

01h

IT2

h

I

h

II

h

I

h

II

0T2

1

h

IT2

h

I

h

II

h

I

T2

1

T2

1

h

II

z

2

zyy2

y

z

2

zxx2

x

2

2

z

2

zyy2

y2

zxx2

x

2

z

2

z

2

y2

yy2

z

2

x2

xx

2

z

2

z

2

y2

yy2

z

2

x2

xx

=

+

=

+

+

=+

+

=

+

+

+

that represents a planar conic curve. Operating in the same way, we have the projections onto the

(y-z) plane:

( ) ( )1

TI2h

III

TI2h

III

x

2

xzz2

z

x

2

xyy2

y =

+

and onto the (x-z) plane:

( ) ( )1

TI2h

III

TI2h

III

y

2

yzz2

z

y

2

yxx2

x =

+

Analyzing the planar conic curves and the signs of all coefficients, we have that:

?TI2h

0TI2h

0TI2h

y

2x

2

z

2


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Therefore we know that the projections on planes (x-y) and (y-z) are ellipses, since coefficients are

of the same sign, while projection on plane (x-z) is a hyperbola since the signs of the coefficients

are different.

The fact that, looking at the polhode in the (x-z) plane, i.e. looking at it from the y axis, we see a

hyperbola can be considered as a sort of instability. In fact, if the angular velocity is slightly shifted

from the x-axis or from the z-axis it will remain confined in a region close to the axis, while a slight

shift from the y-axis would cause a dramatic departure from the original condition. All this can be

seen by looking at the kinetic energy ellipsoid and at the trajectories that are drawn on it by the

angular velocity vector. It is possible to divide the ellipsoid into areas corresponding to different

initial conditions:

The inequalities hold for a given ellipsoid where the energy level has been fixed.

z h2

= 2TIz

2TIx< h2


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Attitude parameters

It is always necessary to be able to switch from a reference frame to another reference frame, such

as the case of representing the principal inertia frame (body frame) as seen by an inertial frame or

by an Earth-centered reference frame or even by a local-vertical-local-horizontal frame.

To define the orientation of one frame with respect to another, three parameters are the minimal setrequired. Often redundant parameters are used, i.e. more than three, either in order to improve the

physical insight into the transformation or to simplify some numerical analysis.

Direction cosines

Each axis of the reference frame that we target is defined in the current reference frame by the three

components of its unit direction vector. Assembling the components of the three axes we obtain the

direction cosines matrix:

=

321

321

321

www

vvv

uuu

A

in which each row represents one axis. This matrix allows switching from the representation of a

vector in one reference to another reference. As example, let consider a planar rotation:

3 3

Y X

Satellite

3 Z

Earth

Sun 2 2

Earth

2 1

11

Y y

X

X is the satellite local vertical direction x

Y is the satellite velocity direction z Satellite

Z is the third direction, orthogonal to X and Y

Z

x,y,z are the satellites principal inertia axes


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We can now see how to treat a sequence of two rotations:

The generic rotation of vector a from the reference 123 into reference uvw is expressed as:

123uvw aAa =

In the same way, the second rotation from reference uvw to reference uvw is expressed as:

uvw'w'v'u a'Aa =

Then, the overall rotation from reference 123 to reference uvw becomes:

123uvw123'w'v'u aA'Aa'Aa"Aa ===

Concluding, in order to compose a sequence of two rotations it is sufficient to build the overall

direction cosine matrix as the product of the two individual rotations (direction cosine matrix),

taken in the reversed order compared to the sequence of rotations.

A'A"A =

Euler axis and angle

From matrix algebra it is known that real, orthogonal matrices have one unit eigenvalue, to whichwe associate the eigenvector e:

eeA = (the eigenvalue is 1)

Therefore, vector e does not change due to the rotation represented by matrix A. This is possible

only if the rotation occurs around axis e. This axis is called Euler axis, and the rotation amplitude

around this axis is called Euler angle. We can now try to relate the direction cosines matrix A with

3 w

v

w u

v

u

2

123 Inertial (Sun)

uvw Geocentric

uvw Principal inertia

1


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vector e. If we consider elementary rotations around each coordinate axis 1, 2 and 3, we can notice

that:

( )

( )

( )

=

=

=

=

=

=

0

0

1

e

cossin0

sincos0

001

A

0

1

0

e

cos0sin

010

sin-0cos

A

1

0

0

e

100

0cossin

0sincos

A

1

2

3

In each case, the trace of matrix A is the same and equals:

( ) += cos21Atr

Then it is possible to infer the relation between Euler angle and rotation matrix A:

( )( )1Atr2

1cos =

Given matrix A it is possible to evaluate e and with simple operations.

Matrix A has the following form in relation to components of the Euler axis e and angle :

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

++

++

++

=

cos1ecossinecos1eesinecos1ee

sinecos1eecos1ecossinecos1ee

sinecos1eesinecos1eecos1ecos

A2

3132231

132

2

2321

23132121

or:

( ) [ ]+= esineecos1cosIA T where:

[ ]

=

0ee

e0e

ee0

e

12

13

23

represents cross product matrix operator.

If we know the attitude parameters expressed as Euler axis and angle, it is therefore straightforward

to evaluate the direction cosine matrix A. The inverse operation is also possible, given the rotation

matrix A evaluate Euler axis and angle, in fact:


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( )( )

( )

( )

( )

=

=

=

=

sin2

AAe

sin2

AA

e

sin2

AAe

1Atr2

1cos

21123

1331

2

32231

1

It is remarked that the direct transformation, from Euler axis/angle to rotation matrix A, is always

possible, while the inverse transformation, from rotation matrix A to Euler axis/angle, is not always

defined. In fact, when sinbecomes zero the Euler axis is undetermined. This occurs when = n ,that physically means that the axis is not uniquely determined, i.e., it is possible to reach the final

configuration through at least two rotations of same amplitude but different axis. When the inverse

transformation is not determined, the condition is known as a singularity in the attitude

parameterization.

It is also remarked that the inverse trigonometric function arcos returns two possible values of the

Euler angle ( ) that in turn provide two different axes ( e ). This is actually different from the

singularity condition, since we are always looking at the same axis, once from the positive and once

from the negative direction, and to each condition we associate the same rotation angle with

opposite sign. This is physically the same condition.

Euler axis/angle parameters are useful since they are only 4 parameters (the constraint condition is

associated to the fact that e is a unit vector), but unfortunately it is not possible to infer a relation for

sequence of rotations.

Quaternion

A quaternion is a vector of four parameters, linked to the Euler axis/angle parameters through the

relation:

=

=

=

=

2cosq

2sineq

2sineq

2sineq

4

33

22

11

It can be noticed that the quaternion is normalized:

1qqqq2

4

2

3

2

2

2

1 =+++


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In some cases it is convenient to divide the quaternion into a vector part q and a scalar part q4:

4

3

2

1

q,

q

q

q

q

=

The direct transformation, given the quaternion evaluate the direction cosine matrix A, leads to:

( ) ( )

( ) ( )

( ) ( )

+++

+++

++

=2

4

2

3

2

2

2

141324231

4132

2

4

2

3

2

2

2

14321

42314321

2

4

2

3

2

2

2

1

qqqqqqqq2qqqq2

qqqq2qqqqqqqq2

qqqq2qqqq2qqqq

A

or:

[ ]+= q2qqq2IqqqA 4TT2

4

where:

[ ]

=

0qq

q0q

qq0

q

12

13

23

The inverse transformation, from matrix A to quaternion, is represented as:

( )

( )

( )

( )

+++=

=

=

=

2

1

3322114

2112

4

3

1331

4

2

3223

4

1

AAA12

1q

AAq4

1q

AAq4

1q

AAq4

1q

The sign ambiguity has the same explanation given for the Euler axis/angle case. We are always

looking at the same axis, once from the positive and once from the negative direction, and to each

condition we associate the same rotation angle with opposite sign. This is physically the same

condition. Quaternions have no singular condition even in the inverse transformation. In fact,

should q4 become zero it is always possible to evaluate one of the other components of the

quaternion, which will be surely different from zero due to the normalization constraint, and then

evaluate the remaining three components. The inverse transformation has then the following three

further sets of alternative equations:


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( )

( )

( )322321

24

31132

1

23

211221

22

33221121

AA4q

1q

AA

4q

1q

AA4q

1q

AAA1q

=

+=

+=

+=

( )

( )

( )133132

34

32233

2

33

211232

31

33221132

AA4q

1q

AA

4q

1q

AA4q

1q

AAA1q

=

+=

+=

+=

( )

( )

( )211243

44

32234

3

42

311343

41

33221143

AA4q

1q

AA

4q

1q

AA4q

1q

AAA1q

=

+=

+=

+=

It is even possible to select the set of equations to adopt on the basis of the maximization of the

square root term, for a reduced truncation error in the following operations. Unfortunately,

quaternions have no physical meaning and therefore their use is not intuitive.

Adopting quaternions, it is possible to express a sequence of two consecutive rotations, combining

quaternion components in the reversed order of rotations. Given a first rotation represented by

quaternion q, a second rotation represented by quaternion q, the overall rotation is represented by

quaternion q given by:

q

qqqq

qqqq

qqqq

qqqq

"q

4321

3412

2143

1234

=

Gibbs vector

Gibbs vector is a minimal attitude parameterization, since it includes only three parameters. It can

also be seen as a non-normalized quaternion, and can be inferred from Euler axis/angle:

==

==

==

2tane

q

qg

2tane

q

qg

2tane

q

qg

3

4

33

2

4

22

1

4

11

Also in this case it is possible to represent the direct transformation from Gibbs vector to rotation

matrix A:

( ) ( )

( ) ( )

( ) ( )

++

++

++

+++=

2

3

2

2

2

1132231

132

2

3

2

2

2

1321

231321

2

3

2

2

2

1

2

3

2

2

2

1ggg1ggg2ggg2

ggg2ggg1ggg2

ggg2ggg2ggg1

ggg1

1A

or:


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) [ ]( )2

T2

g1

g2gg2Ig1A

+

+=

where:

[ ]

=

0gg

g0g

gg0

g

12

13

23

The inverse transformation is given by:

+++

=

+++

=

+++

=

332211

21123

332211

13312

332211

32231

AAA1

AAg

AAA1

AAg

AAA1

AAg

The inverse transformation becomes singular when = (2n + 1) .

Two consecutive rotations (first rotation expressed by vector g, second rotation represented by

vector g) can be merged into a single Gibbs vector g with the following rule:

'gg1

g'g'gg"g

+=

Also this parameterization has a poor physical meaning and furthermore it has singularity

configuration in the inverse transformation, therefore part of the advantages of having only threeparameters are lost.

Euler angles

The attitude can be represented by means of three Euler angles, which are a minimal representation

and have a clear physical interpretation. The concept is based on the fact that it is always possible to

make two orthogonal frames overlap by appropriate rotation of one of them three times around its

reference axes.

Let us make an example to overlap systems 1,2,3 and u,v,w.

We start by rotating by angle around axis 3, that brings the triad in the new configuration 1,2,3.Now rotate by angle around axis 1, that brings the triad in the configuration 1,2,3.Finally, rotate by angle around axis 3 to superimpose the triad to the reference u,v,w.


23/82


22

3=3

2

2

1 1

3 3

2

2

1=1

3=w v

2

1 u

The direction cosine matrix can be written by adopting the rule of consecutive rotations. In fact, it is

possible to combine the elementary rotation matrices since each rotation is around one of the

coordinate axes:

( ) ( ) ( ) ( )= 313313 AAA,,A

where:

( )

( )

( )

=

=

=

100

0cossin

0sincos

A

cossin0

sincos0

001

A

100

0cossin

0sincos

A

3

1

3

The labels 1,2,3 of the rotation matrices do not indicate in this case the axis of rotation, but ratherthe rotation type, i.e., the family of axes around which the triad is rotating. The overall rotation

matrix is labeled according to the rotation sequence and the rotation angles associated to each

individual rotation:

( ) ,,A313

In this case it is fundamental to remind that matrix products must be made in the reversed order as

compared to the rotations. It is also possible to superimpose the two triads by a different set of

rotation axes, and in this case the values of the rotation angles will also change radically.

This is true even if only the last rotation is changed, all angles will change their magnitude. Matrix

A313is then written as:

+

+

=

cossincossinsin

sincoscoscoscossinsincossincoscossin

sinsincoscossinsincoscossinsincoscos

A313

It can be noticed that the inverse transform, from matrix A to Euler angles, in this case takes the

form:


24/82


23

( )

=

=

=

23

131

32

311

33

1

A

Atan

A

Atan

Acos

In this case we notice a singularity when sin tends to zero. This means that the first and lastrotations are actually around the same physical direction in space, therefore it is not possible to

distinguish them individually, but rather only the sum or difference of the two.

With Euler angles we have a different rotation matrix A for each combination of rotation axes. Two

consecutive rotations cannot be around the same axis, they would become one single rotation,

therefore the possible sequences are 12, out of which 6 have all indexes different and 6 are

characterized by the same first and third index:

312,213,123,321,231,132 all different indexes

313,323,212,232,131,121 first and third index equal

The different sets of Euler angles are singular for different values of the second rotation angle ,depending on the fact that the indexes are all different or the first and third index coincide. When all

indexes are different, the singularity condition is = (2n+1)/2,while for equal first and third indexthe singularity occurs when = n.Here follow the symbolic expressions of the direction cosine matrix for the 12 possible sequences,

assuming that the first rotation is always represented by angle , the second by angle and the thirdby angle .

++

++

=

coscossincossin

sincoscossinsincoscossinsinsincossin

sinsincossincos-cossinsinsincoscoscos

A123

+

+

=

coscossinsinsinsincoscossinsincossin

sincoscoscossin

cossinsinsincossinsincossincoscoscos

A132

++

++=

coscossinsinsin-cossinsincoscossinsin

cossinsinsincoscoscossinsincossincos

sincos-sincoscos

A231

++

++

=

coscossin-cossin

sincoscossinsincoscossinsincoscossin

sincossinsincos-cossinsinsinsincoscos

A213


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24

+

+

=

coscossincoscossinsinsinsincoscossin

sincoscoscossin-

cossin-sincossinsincossinsinsincoscos

A312

++

++=

coscossinsincoscossinsincoscossinsin

cossinsinsinsincoscossincossinsincos

sin-sincoscoscos

A321

+

+

=

coscoscossinsincossincoscossinsincos

coscossinsincoscossinsincoscossinsin

sincossinsincos

A121

++=

cossinsincoscoscoscossinsincos-sinsin

cossincoscossincoscoscossinsinsincos-

sinsinsincoscos

A131

++

=

coscoscossinsinsincos-cossincoscossinsincoscossinsin

coscossinsincos-sinsincossinsincoscos

A212

+

+

=

cossinsincoscossinsincoscossinsincos

sinsincossincos

cossincoscossinsincoscoscoscossinsin

A232

+

+

=

cossincossinsin

sincoscoscoscossinsincossincoscossin

sinsincoscossinsincoscossinsincoscos

A313

++

=

cossinsinsincos

sinsincossinsincoscoscoscossinsincos-

sincoscossincoscossincoscoscossinsin

A323

If angles are small, i.e., up to 10- 15, we can assume cos x =1, sin x = x, x*x =0 (with x in

radians), therefore the dependence on the order of rotations is lost and we can write:

( ) ( ) ( ) ....,,A,,A

1

1

1

,,A 213321312 ===

=

It is reminded that in any case the correspondence between axis and angle must be kept. Matrix Acan also be written as:

[ ]= anglesIA

If the first and last indexes are coincident, due to loss of importance of the order of rotations, we are

always in the condition that the first and last rotations are almost around the same physical direction

in space, so it is like having only two rotations:


26/82


25

( )

+

=

10

1

01

,,A313

For small angles, quaternion can be simplified and directly connected to Euler angles when the

three indexes are different. For the sequence 312 we can write:

=

=

=

=

1q

2

1q

2

1q

2

1q

4

3

2

1

Attitude parameterization adopting Euler angles has a clear physical meaning but is numericallytime consuming. Moreover, the rule of consecutive rotations, although existing, is really hard to

implement due to its complexity, therefore it will not be illustrated.


27/82


28/82


27

Then, knowing we can integrate dA/dt in order to update matrix A. Since integration is donenumerically, after some steps matrix A will no longer be orthogonal. It is therefore necessary to

orthogonalize matrix A after a certain number of integration steps.

This operation is long and complex, if done exactly, so that approximate solutions to the

orthogonalization are normally used. In this way, approximate solutions can be evaluated with a

higher frequency (at every integration step) since the numerical process is fast enough. It can bedemonstrated that, starting from an initial guess A0(t) of the direction cosines matrix, not orthogonal

due to numerical errors, the recursive formula

Ak+1(t)= Ak(t)*3/2- Ak(t)*AkT(t)*Ak(t)/2

converges rapidly, with increasing k, to the exact value of A. In a first order approximation it is

possible to adopt a single step iteration

A(t)= A0(t)*3/2- A0(t)*A0T(t)*A0(t)/2

Quaternions

The procedure adopted to derive the kinematic relation for direction cosines can now be repeated to

analyze the quaternion kinematics. We can assume that quaternion at time t + t is given by thecomposition of quaternion at time t and a quaternion representing the rotation in the interval t:

( ) ( )

=

=

=

=

=+

2cosq

2sineq

2sineq

2sineq

withtq

qqqq

qqqq

qqqq

qqqq

ttq

4

w3

v2

u1

4321

3412

2143

1234

Then:

( ) ( )tq2

sin

0eee

e0ee

ee0e

eee0

2cosIttq

wvu

wuv

vuw

uvw

+

=+

For short intervals t we can approximate:

2

t

22sin1

2cost

=

=

=

=

and evaluate eu, ev, ew, as a function of = e and therefore write:


29/82


28

( ) ( )

=

+=+

0

0

0

0

:where

tqt2

1Ittq

wvu

wuv

vuw

uvw

Taking the limit for t0:

( ) ( )( )tq

2

1

t

tqttqlim

dt

dq

0t=

+=

Therefore, also in this case, knowing the angular velocity we can integrate the derivative of thequaternion to propagate attitude. As for the direction cosines, there is a problem of normalization,

but since q is a vector its normalization is trivial.

Gibbs vector

It is not possible to evaluate the derivative of the Euler axis/angle parameterization, due to the lack

of rule for consecutive rotations. For the Gibbs vector this can instead be evaluated, with the usual

procedure already adopted for quaternion and direction cosines:

( ) ( ) ( )( ) 'gtg1tg'g'gtgttg

+=+

where:

t2

1

2tane'g

tsmall= ==

The derivative is then:

( ) ( )( ) ( )[ ]tgtgtg2

1

dt

dg+=

Euler angles

Euler angles have no convenient rule for combining two consecutive rotations, however it is

possible to calculate the derivatives of the angles for each sequence of rotations. Let consider the

sequence 313 (,,):


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29

If we have variation of angle alone, then rotation will be about axis 3, so that we can write:

3= &

If we have variation of alone, then rotation will be about axis 1, so that:

'1= &

Again, if we have variation of angle alone, rotation will be about axis w, so that:

w= &

The three contributions are independent one from the other, therefore by simple combination of the

three contributions we obtain:

w'13 &&& ++=

3=3

w=3 v

2

2

2

& & &

1 1 u

Taking the projections of on axes u,v,w :

++==

+=++==

+=++==

www'1w3w

v'1v3vwv'1v3v

u'1u3uwu'1u3u

w

v

u

&&&

&&&&&

&&&&&

Inner products 3 u, 3 v, 3 w represent the third column of matrix A313, while inner products 1u, 1v,

1w represent the first column of matrix A313if we consider = 0, in fact the first column representaxis 1, that is coincident with axis 1 in case = 0. Finally, w w = 1 since w is a unit vector.

We then have:

+=

=

+=

&&

&&

&&

cos

sincossin

cossinsin

w

v

u

from which the kinematic relations are obtained as:


31/82


32/82


31

( )

( )

++=

=

+=

cos

sinsincos

sincos

cos

sincos

321sequence

vwu

wv

vw

&

&

&

( )

( )

+=

=

+=

sin

coscossin

sincos

sin

cossin

121sequence

wvu

wv

wv

&

&

&

( )

( )

+=

+=

=

sin

cossincos

sincos

sin

sincos

131sequence

wvu

vw

wv

&

&

&

( )

( )

+=

+=

=

sin

cossincos

sincos

sin

sincos

212sequence

uwv

wu

uw

&

&

&

( )

( )

+=

=

+=

sin

cos

cossin

sincos

sin

cossin

232sequence

uwv

uw

uw

&

&

&

( )

( )

+=

=

+=

sin

coscossin

sincos

sin

cossin

313sequence

vuw

vu

vu

&

&

&

( )

( )

+=

+=

=

sincossincos

sincos

sin

sincos

323sequence

vuw

uv

vu

&

&

&

Euler angles are integrable, this meaning that:


33/82


32

+=

+=

+=

t

00

t

00

t

00

dt

dt

dt

&

&

&

This is not true for angular velocities, that when integrated do not provide angles, since angular

velocity is not a fixed vector. To verify this statement, we recall the integrability condition:

x

B

y

ABdyAdxdz

=

+=

+=+=+= dBdAdcosdsinsindtcossinsin uu&&

=

=

0B

cossinA

not integrable

The same integrability condition verified for Euler angles:

dtBdtAdtsin

cosdt

sin

sind

sin

cos

sin

sinvuvuvu +=

+

=

+

=&

( ) ( ) ( ) ( )

( )

( )

( ) ( )

sincoscossin

sin

1

dt

A

:therefore

sindt

sin

coscos

dt

dtsininsin

dt

cos

sin

1

dt

sin

sin

dt

A

222

v

vv

vv

vv

2

vv

=

=

=

=

=

=

=

&

&

In similar way we can evaluate( )dt

B

u

and check that it has the same expression, this

demonstrates integrability of Euler angles.


34/82


33

Euler equations

Euler equations are the analytic form adopted to describe the attitude motion of a rigid body. We

start by writing the fundamental equation:

Mdthd =

In body axes the angular momentum vector is:

++

++

++

=

zzzyzyxzx

zyzyyyxyx

zxzyxyxxx

III

III

III

h

Writing the fundamental equation in body axes we have:

Mhh =+&

and expanding the expression of the angular momentum we obtain a system of first order

differential equations, nonlinear and strongly coupled:

=+++++

=+++++

=+++++

z2xxyxzyzyxyy

2yxyzyxzyxxxzzzyzyxzx

y2zxzyzxyzxxx

2xxzzxzzyxyzzyzyyyxyx

x2yyzyzzzyxxz

2zyzzyyyzxxyzxzyxyxxx

MIIIIIIIII

MIIIIIIIII

MIIIIIIIII

&&&

&&&

&&&

If we refer the dynamics to the principal inertia axes, we can simplify the system dynamics sinceinertia products will vanish:

( )( )

( )

=+

=+

=+

zxyxyzz

yzxzxyy

xyzyzxx

MIII

MIII

MIII

&

&

&

In these equations, if the coupling term vanishes then the rotation around one axis will be driven

only by the torque around the same axis.

Rotational dynamics of a satellite is then described by joining Euler equations and an attitudeparameterization.

M Euler Attitude ,,

? equations Parameters

0 0,0,0


35/82


34

On a satellite we have torques influencing directly the dynamics (Euler equations) and induce

changes in the components of angular velocity. Adopting one attitude parameterization we can,

knowing the angular velocity, calculate the attitude parameters that indicate the orientation of the

satellite in space and that allow to evaluate the position-dependent torques. In Euler equations it is

therefore required to assign initial conditions for angular velocity, while in attitude parameterization

we must assign the initial satellite attitude.

If the reference system is not centered in the system center of mass, in deriving the rotationaldynamics equations we must also consider the linear motion of the origin of the reference frame.

With reference to the following figure, assuming O is the center of mass and A the origin of the

reference system, assuming that we define with Mo and Fo the torque and force referred to the

system center of mass, we have

OOOO

amF;Mdt

hd==

AAOA rrmhh &+= ; OAOA FrMM += ; AOA raa &&=

that is

AAAO rrmhh &= ; OAAO FrMM = ; AAO raa &&+=

Substituting these expressions in the equation of the angular momentum derivative, we have

( )AAAdtd

OAAO

O rrmhFrMdt

hdM &==

and, since OO amF =

( ) ( )AAAdtd

AAAA rrmhramrM &&& =+

For a rigid body we have 0rA=& , 0rA=&& , therefore AO hh = , AO aa = and

AAO

AAA

A aSdt

hdamr

dt

hdM +=+=

where AS stands for the static moment of the rigid body with respect to point A. If the reference

axes are parallel to the principal inertia axes, then we can write

x

y

z

AO

RA

RO

MO

rA

FO


36/82


35

( ) ( )( ) ( )

( ) ( )

=++

=++

=++

zxyyxxyxyzz

yzxxzzxzxyy

xyzzyyzyzxx

MaSaSIII

MaSaSIII

MaSaSIII

&

&

&

It is pointed out that the same result would have been obtained by considering the equation of the

derivative of the angular momentum vector, expressed in the general form += ISvh ooo ,

with no additional assumptions on the reference frame (fixed or with origin in the system center of

mass).

Even in the simplest case, assuming as reference system the principal inertia system, Euler

equations are difficult to integrate and therefore numerical integrations is adopted. In some

simplified cases we can obtain an analytical solution of the equations, to be used to verify the

numerical integration results.

Torque-free motion of a simple spin satellite

Assume the satellite is axial-symmetric. Having two equal principal inertia moments, one equation

is decoupled from the other two.

Assume that Ix= Iy:

( )

( )

=

=+

=+

zzz

yxzzxyy

xyzyzxx

MI

MIII

MIII

&

&

&

Assume that no external torque is applied, then we can study the torque-free motion knowing the

initial conditions:

( )

( )

=

=+

=+

0I

0III

0III

zz

xzzxyy

yzyzxx

&

&

&

The last equation leads immediately to:

z0z const ==

Furthermore, since no torque is applied, angular momentum and kinetic energy are constant.

Combining the first two equations, the first multiplied by xand the second multiplied by y, weobtain:

( )

( ) const

0dt

d

0

0II

2y

2x

2y

2x

yyxx

yyxxxx

=+

=+

=+

=+

&&

&&


37/82


36

The magnitude of is then constant. The same result would have been obtained by considering theconstant kinetic energy. To have the magnitude constant, both the x and y components of the

angular velocity must be constant or the projection of the angular velocity on the x-y plane is on a

circle. In fact we know that the angular momentum is constant, therefore:

( ) zxyzzyxx hhkIjiIhconsth

+=++=

=

hz is constant, while hxy is parallel to xy (component on x-y plane) but not constant. Thereforevectors h and describe cones with axis coincident with axis z if we look at the motion from abody axes reference. If we look at the motion from an inertial reference, h will be fixed and andthe unit vector z describe cones with axes coincident with vector h. Furthermore, and h areconstantly on a plane intersecting axis z. In case the only non-zero component of the angular

velocity is the z component, then motion is referred to as simple spin satellite motion.

We can now solve the first two equations. Define:

( )

x

zxz

I

II =

as a characteristic constant of the rigid body, the equations become:

=

=+

0

0

xy

yx

&

&

Deriving the first and substituting the value of y& from the second equation, we have:

0x2

x =+&&

z

hz

z h

y

xy

hxy

x


38/82


37

This is the typical equation of an harmonic system that has the following solution:

( ) ( )tsinbtcosax += ( ) ( )tcosbtsinax +=&

with initial conditions:

( )( )

( )

( ) y0y

y0y

x0x

x0x

0

0

0

0

&&

&&

=

=

=

=

We then have:

( ) ( )tsin

tcos x0x0x

&+=

From the second equation we obtain:

x

y

&

= so that:

( ) ( )

( ) ( )tcostsin

tsintcos

y0x0y

y0x0x

+=

=

In order to demonstrate that xydescribes a circle, we can write it as a complex number:

( ) ( )[ ] ( ) ( )[ ]tcositsintsinitcosi y0x0yxxy +=+=

or:

( ) ( )[ ] ( ) ( )[ ]

( ) ( ) ( )[ ]tsinitcosi

tsinitcositsinitcos

0y0xxy

0y0xxy

++=

+++=

that is a harmonic motion on a plane, with magnitude given by the initial values of and constantangular velocity equal to .

At this point we must analyze the case in which Izis the maximum inertia moment and the case in

which Izis the minimum inertia moment:

Now we can look at the angles that h and form with the z axis:

Izmaximum > 0 Izminimum < 0


39/82


38

=

==

=

tanI

I

I

I

h

htan

tan

z

x

zz

xyx

z

xy

z

xy

If Iz is the maximum inertia moment, then > 0 and < , while if Iz is the minimum inertiamoment, < 0and > .

So far we have seen the motion in the body axes frame, and analyzed the motion of the angular

momentum vector around the z body axis.

In reality vector h is fixed in an inertial frame, so if we allow one of the inertial frame axis to be

coincident with h we will see the body axis rotating around it, along with the angular velocity

vector.

The z axis draws a cone with axis coincident with h, with aperture , while the angular velocity

draws a cone with axis coincident with z, with aperture . Moreover we recall that draws also acone with axis coincident with h. We then have two cones, one fixed (around h) called spacecone, and one that revolves on it (around z), called body cone. The two cones revolve one

outside the other if is negative, while they revolve one inside the other if is positive.

We can now apply one attitude representation to the axial symmetric satellite:

3 h 3 h

z z

2 2

< 0 > 0

1 1

z

hz

z h

y

xyhxy

x


40/82


39

( )

x

zxz

I

II =

Adopting the Euler angles with sequence 313, we can represent the inertial reference with the third

axis coincident with h.

+=

=

=

&&

&

&

cos

cossin

sinsin

z

y

x

We notice that is constant due to the satellite symmetry, and axis z draws a cone of amplitude around axis h.

Taking time derivatives, we have:

+=

=

+=

&&&&&

&&&&&

&&&&&

cos

sinsincossin

sincossinsin

z

y

x

Replacing in the Euler equations we represent the dynamics as a function of ,,and recallingthat:

0

0

yyxx

z

=+

=

&&

&

from the second equation we obtain:

( ) 0sincossinsincossinsincossincossinsinsinsin

2222

22222222

==+

=++

&&&&&&

&&&&&&&&&&

Since must be constant the following must hold:tcos0 == &&&&

From the first equation we now have:

tcos0 == &&&

h = 3z

y

2

1 xx


41/82


40

We now have the following expressions:

=

+=

+=

0

00

00

t

t

&

&

We must now calculate the initial values of the derivatives of e , and to do so we make use ofthe first two Euler equations:

0sinsinsinsincossin

0cossinsincossinsin

=

=++

&&&&&

&&&&&

Since the second time derivative of is zero we have:

+=

=

cos

z&

&

=

=

cosI

II

II

z

x

z

z

x

zx

&

&

In our case Iz< Ixand therefore the time derivative of is positive.

Solution of Euler equations in the phase plane ( ),&

Reconsider the Euler equations:

( )( )

( )

=+

=+

=+

0III

0III

0III

xyxyzz

zxzxyy

yzyzxx

&

&

&

If we assume that the external torques are zero, then angular momentum and kinetic energy will be

constant:

consth

const2T

2 =

=

We now look for a method to decouple and solve the three equations, making use of the

conservation laws just mentioned. Starting from the first equation, taking its time derivative, we

have:

0IIIII yzyzyzyzxx =++ &&&&

From the second and third equations we can compute the time derivative of the angular velocity

components, and substitute it in the above equation:


42/82


41

( ) ( ) 0I

IIII

I

IIIII x

2

z

y

xzyzx

2

y

z

yx

yzxx =

+

+&&

Now we evaluate:

2

yzy

2

xzx

2

y

2

y

2

x

2

xz

2

IIIIIITI2h += so that:

( ) ( )

( ) ( )

y

xzx

2

xz

2

zy

2

y

xzx

2

xz

2

yzy

2

y

I

IIITI2hII

IIITI2hIII

=

=

Substituting in the previous equation, we get to:

( ) ( )( ) ( ) 0I

IIII

I

II

I

II

I

hTI2

I

III x

2

z

y

xzyz

3

x

y

zx

z

yx

x

y

2

z

z

yx

xx =

+

+

+&&

Now evaluate:2

xyx

2

zzy

2

x

2

x

2

z

2

zy

2IIIIIITI2h +=

so that:

( ) ( )

( ) ( )

z

xxy

2

xy

2

yz

2

z

xxy

2

xy

2

zyz

2

z

I

IIITI2hII

IIITI2hIII

+=

+=

Substituting in the previous equation, we get to:

( ) ( )( ) ( ) ( )( )0

I

II

I

II

I

TI2h

I

II

I

II

I

II

I

hTI2

I

III

3

x

z

xy

y

xzx

z

y

2

y

xz3

x

y

zx

z

yx

x

y

2

z

z

yx

xx =

+

+

+

+&&

or:

( )( ) ( )( ) ( )( )0

III

IIIII2

III

TI2hIITI2hII3

x

zyx

xxyxz

x

zyx

y

2

xzz

2

xy

x =

+

++&&

The other two equations are obtained by following the same procedure, simply with a permutation

of all the operation indexes:

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )0

III

IIIII2

III

TI2hIITI2hII

0III

IIIII2

III

TI2hIITI2hII

3

z

zyx

zzyzx

z

zyx

x

2

zyy

2

zx

z

3

y

zyx

yyxyz

y

zyx

z

2

yxx

2

yz

y

=

+

++

=

+

++

&&

&&


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42

The three equations all have the same structure:

0QP3 =++&&

Integrating the equation we obtain:

KQ2

1P

422 =++&

that, for the x equation, becomes:

x

2

xxx

2

x

2

x

x

4

xx

2

xx

2

x

KQ2

1P

KQ2

1P

=

++

=++

&

&

The last expression can be considered as a conic section in the phase plane ( ),& . We can evaluatethe sign of each coefficient.

Assuming that Iz> Iy> Ix, we have:

z

2

x IT2

hI 0 Pz ?

Qx> 0 Qy< 0 Qz> 0

Due to the structure of the coefficients, if the angular velocity component is large enough the Q

term will predominate over the P term, while if the angular velocity component is small enough the

P term will predominate over the Q term. Then, in the y phase plane if the angular velocity is large

the conic section will be a hyperbola, if the angular velocity is small the conic section will be an

ellipse. In the x and z phase planes the conic sections are ellipses for large velocities, for smallvelocities the type of conic section is undefined. This is not really important, since in any case, even

for divergent phase plane trace (hyperbola) it is clear that as the angular velocity grows the trace

will change into an ellipse. Notice that in the y phase plane (intermediate inertia phase plane),

should the velocity grow, we would have a diverging motion. This is not possible due to the

conservation of kinetic energy, therefore the conclusion is that in the y phase plane the solution

must be such that the angular velocity is small enough to prevent the trace from being a hyperbola.

x xPx> 0 Px< 0

Qx> 0 Qx> 0

x x

y

y


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43

Attitude stability

At this stage we can discuss attitude stability with time, starting by the analysis of Euler equations

in case of torque-free motion:

( )( )

( )

=+

=+

=+

0III

0III

0III

xyxyzz

zxzxyy

yzyzxx

&

&

&

There is a distinction between torque-free stability and forced-motion stability. The latter depends

on the type of torque applied, rarely known in a suitable analytical form, therefore we will mostly

concentrate on torque-free stability. To analyze stability the first step consists in evaluating an

equilibrium configuration. This is obtained by imposing zero time derivatives of the system states

and then solving the corresponding system of equations:

( )( )

( )

=

=

=

0II

0II

0II

xyxy

zxzx

yzyz

The solution is found for one single angular velocity component different from zero. This condition

is known as simple spin satellite. Assume, without loss of generality, that the non-zero component

is the z component:

0

0

yx

z

==

Attitude stability of simple spin satellites

To analyze the stability of this condition, we must first write the linearized equations of motion,

considering small perturbations around the equilibrium position:

( )( )( ) yyyy

xxxx

zzz

=+=

=+=

+=

Angular velocity components are now expressed as the sum of the nominal value plus the small

perturbation. In the rest of the analysis, should the assumption of small perturbations hold this will

mean that the condition is stable. Rewriting the equations, the product of two perturbations become

negligible, so that the linearized equations are:


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44

( )

( )

=

=+

=+

0I

0III

0III

zz

zxzxyy

yzyzxx

&

&

&

The derivative are taken only for the perturbation component of the angular velocity, since thenominal components are constant by definition. The system can now be integrated:

constz=

The perturbation of the z component is therefore constant. For the two remaining equations we can

write:

( )( ) 2z

yx

xzyz2

x

2

x

II

IIII

0

=

=+&&

The solution is then of the general form:ti

x ea =

The condition that guarantees a stable motion, harmonic, is 2> 0, while for 2< 0 the solution isexponentially diverging, so unstable.

The condition for stability is verified when:

xzyzxzyz2

IIandIIorIIandII0 >

that means that Izmust be either the maximum or the minimum inertia moment.

Stability of simple spin satellites with energy dissipation

Now we look at how the two previous stability conditions are changed due to presence of energy

dissipation, that is:

tcosh

0T

=


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45

maxzminz IITT >

Therefore, in case of energy loss, the satellite will tend to the motion condition that has the

minimum energy, rotating around the maximum inertia axis.

The only stable condition, with the usual assumptions on the inertia moments, is then:

xyz III >>

We can now analyze how the satellite reorients itself from a generic motion condition to the stable

one. For simplicity we take the case of axial symmetric satellite.

h

Izz

Inn

We take as reference a rotating system, whose plane n-z includes the three vectors ,z and h. Thisreference is also principal inertia, and vector h has only 2 components, along the two axes n and z:

2

nn

2

zz

2

n

2

n

2

z

2

z

2

IIT2

IIh

+=

+=

The energy loss can be represented as:

0I2I2T2 nnnzzz


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46

00

00

0

zz

zz

zz

>


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49/82


48

( )

( )

=++

=++

0KnK1n

0Kn1Kn

yy

2

xyy

xx

2

yxx

&&&

&&&

where the coefficients Kxand Kyare non-dimensional and always in the range between 1 and +1:

y

xz

y

x

yz

x

I

IIK

I

IIK

=

=

To evaluate the stability we can take the Laplace transform of the system, and analyze the

characteristic polynomial:

( ) ( )( ) ( )

( )

( )

=

+

+

0

0

s

s

nKsK1sn

1KsnnKs

y

x

2

y

2

y

x

2

x

2

The roots of the characteristic polynomial must have non positive real part in order to have a stable

system:

( )[ ]

( )( ) ( )( )

( )( )( )

( )

( ) 0KKnKK1sns

0KKnKKKK1KKsns

0KKnK1K1nKnKnss

01KK1nsKnsKns

0sAdet

yx

4

yx

224

yx

4

yxyxyx

224

yx

4

xy

2

y

2

x

224

xy

22

y

22

x

22

=+++

=+++++

=++++

=++

=

For stability, s2cannot be real or complex, since in this case its square root will correspond to two

complex numbers with phase difference equal to , so one will have positive real part. The onlystable condition will then be s

2

=>

Since n2 is positive, it is possible to solve the characteristic equation for the variable n

2s

2, so the

stability conditions are written as:

( ) ( )

( ) ( ) 0KKKK1KK1

0KK10KK4KK1

yx

2

yxyx

2

yxyx

2

yx

>>+

>>+

The first condition is trivial, so the second is the real stability condition, that is transformed into:

yz

xz

II

II

>

> or

yz

xz

II

II

>

we have two inequalities whose solution depends on Irand r :

( )

( )

>+

>+

0III

0III

rrzxz

rrzyz

For given rrz I,, we can evaluate the conditions on Ix, Iy, Iz that guarantee stability. This is

normally not the best way to look at the problem, since in general we want to analyze stability for


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51

given z , Ix, Iy, Iz, evaluating the conditions on Ir and r . So, normally we evaluate the

characteristics of the rotor that guarantee the stability of the overall system. If 0z = then the

system is stable for any r , while if 0r = we have the same result of the simple spin satellite.

The most interesting case is the general case, for which stability holds if:

( )

( )


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52

rsnnnrrrzzz

nnnrrrzzz

2

nn

2

rr

2

zz

TTIIIT

I2I2I2T2

IIIT2

&&&&&&

&&&&

+=++=

++=

++=

We can evaluate an equivalent angular velocity as:

( )

( )

( ) ( )errrezzzrs

rrzzennn

n

rrzze

IITT

0h2hsinceIII

I

II

+=+

=+=

+=

&&&&

&&&&

It is an arbitrary distribution of the energy loss, with the first term relative to the satellite and the

second relative to the rotor:

zzzs IT = && misalignment of h

rrrr IT = && viscous energy loss due to the relative velocitywith:

( )

( )err

ezz

=

=

We therefore have:

+

=

r

r

z

s

ennn

TTI

&&

&

Now, evaluate angle :

h

hsin n=

Taking the time derivative:

+=

+=

=

+

=

+

=

=

+

=

+

=

=

+

===

=

r

r

z

s

r

r

z

s2

en

2r

r

z

sen

r

r

z

sen

n

r

r

z

sen

nn

r

r

z

sen

n

r

r

z

se

n

nnnnn

nn

T

TA

T

T

2sinh

2I

2sinh

T

T2I

hsincosh

T

TI

hcosh

T

TI

Icosh

T

TI

cosh

T

T

cosh

I

cosh

I

h

Icos

&&&&

&&&&

&&&&

&&

&&&

&&

since:

20


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53

therefore ( ) 0sin > and where A > 0.

For stability we should have 0

+

&&

Assuming the contribution of the nominal value of h is given mostly by the presence of the rotor,

that is:

zzrr II >>

we then have:

n

rrrr

n

rr

n

rrzz

n

rre

II

I

I

I

I

I

I

=

=

=

In the case Ir>> In(case only theoretical, since this would imply a rotor bigger than the satellite):

>

&&

In general, the rotor energy loss can be tuned with some viscous damping mechanism. Notice that

now we have a condition that is no longer purely geometrical. In fact, in the case 0r=& , then the

rotor will show no energy loss and rT& would be zero, so that even with energy loss in the satellite

the system would be unstable. Therefore the rotor should generate some energy loss to ensurestability.

Stability diagrams

We can now take a look at a few diagrams, known as canonical representations, used to evaluate

attitude stability of satellites.


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54

Let consider a plane including the three points on the axes at unit distance from the origin,

represented by the equation:

1CBA =++

or, defining:

zyx

z

zyx

y

zyx

x

III

IC;

III

IB;

III

IA

++=

++=

++=

1I

I

I

I

I

I zyx =++

Connecting the three mid points on the sides of the triangle, we form a second triangle (a,b,c) that

includes all the possible combinations of inertia moments.

In fact, the following holds:

- In point a: Ix = 0 and Iy= Iz- In point b: Iy = 0 and Ix= Iz- In point c: Iz = 0 and Iy= Ix

Along the bisectors two inertia moments are equal:

- Along the bisector from a: Iz= Iy- Along the bisector from b: Iz= Ix- Along the bisector from c: Ix= Iy

At the intersection of the three bisectors the three inertia moments are all equal.

C

1 b a

b a

1 B

1 c c

A


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55

Iy= 0 Ix= 0

Iz>Ix>Iy Iz>Iy>Ix

Stability area of simple

spin satellites with energy loss

Ix>Iz>Iy Iy>Iz>Ix

+ Stability area of simple spin

satellites with no energy loss Ix>Iy>Iz Iy>Ix>Iz

Iz= 0

Alternatively, through a simple change of variables, it is possible to represent in the same way the

plane as a function of Kx, Ky, Kz, in fact:

z

xy

z

y

xzy

x

yzx

I

IIK

I

IIK

IIIK

=

=

=

It is then possible to reduce the representation to the Kx Ky plane. In this case, typically the

coefficients are renamed as Kp, Kr, and Ky, where the correspondence with the previously defined

Kx, Ky, and Kzcoefficients is:

Kx= Ky Yaw

Ky= Kr Roll

Kz= Kp Pitch

Ky1 Kp= 0

Kp0

-1 Izmax 1

Kr

Izmin

-1


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56

It is possible to see that:

xryyyx

xryzxzry

yyxzyzyx

IKIIKI

IKIIIIKI

IKIIIIKI

+=+

+==

+==

Therefore:

( ) ( )1KI

I1K r

x

y

y =

If Iy > Ixwe then have Kp> 0, (Kr - 1) < 0, (Ky - 1) < 0, so that ( ) yrry KK1K1K >+=

>+=

z

r

y

rr

z

r

x

ry

I

IK

I

IK

>

>

If the satellite angular velocity and the rotor angular velocity have the same sign, the stability region

is increased as shown in the figure.


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57

Simple spin

Dual spin

If the angular velocity of satellite and rotor have opposite signs, then the stability region is reduced.


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Disturbing torques

Now we can analyze the disturbing torques that act on the satellite:

M Dynamics Kinematics

Start with the torque due to gravity gradient:

Gravity gradient torque

The gravity field is not uniform, therefore there could be a torque acting on the satellite. This is

mainly true for large satellites, and even if the resulting torque is small the effect can be

considerable due to the long time of action.

Z=PitchY=Roll X=Yaw

r

dm

f

R

We can evaluate the torque generated by the elementary force f, due to the gravity acting on the

elementary mass dm:

( )

( ) ++

=

++

=

B

3

t

3

t

dmrRrR

GmrM

rRrR

dmGmrdM

r is much smaller than R, therefore it can be considered as a perturbation:


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( )

( )

=

=

+

++=++=+

2

3

22

53

2

3

23

2

3

22

232

3223

R

rR31Rr

R

rR21

2

31R0ratexpansionseriesthetaking

R

rR21R

R

rR2

R

r1RrR2rRrR

Then:

( ) +

=

B

23

t dmrRR

rR31r

R

mGM

If the reference system has origin in the center of mass:

( ) ( ) ( ) 0SRdmRrrrdmRrr 0==+=+ so that:

( ) ( )( ) +=+

=

B

5

t

B

23

t dmrR3rRrRmGdmrR

RrR3r

RmGM

Finally, we get to:

( )( ) =B

5

t dmRrRrR

mG3M

Taking as reference system the LVLH frame (x-yaw, y-roll, z-pitch), we have:

iRR

kzjyixr

=

++=

So that:

( ) ( ) ( ) ( )kIjIR

mG3dmkxyjxz

R

mG3dmkyjzx

R

mG3RdmkyjzRx

R

mG3M xyxz3

t

B

3

t

B

3

t

B

5

t +====

We therefore have a torque with components in roll and pitch, and in case one principal axis is

aligned with the yaw axis then the torque will vanish, since in this case Ixz e Ixy are zero.

Furthermore the torque is inversely proportional to the third power of the distance from the

attracting body (Earth).

In order to evaluate the effect of the torque on the satellite dynamics, this has to be evaluated in the

principal axis frame and inserted in the Euler equations.Taking as reference the principal inertia axes, the expression of r does not change but:

)kcjcicRR 321 ++=

where c1, c2, c3are the direction cosines of the radial direction in the principal axes. Then:


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60

( )

++=B

12

31

23

3213

t dm

ycxc

xczc

zcyc

zcycxcR

mG3M

Evaluating all the terms under the integral sign, considering that the reference is principal inertia,

we have:

( )( )( )

( )( )

( )

=

= 21xy

31zx

32yz

3

t

B

21

22

31

22

32

22

3

t

ccII

ccII

ccII

R

mG3dm

ccyx

ccxz

cczy

R

mG3M

Therefore if one of the principal axes is aligned with the radial direction the torque is zero because

only one of the direction cosines is non-zero.

Stability of simple spin satellites subject to gravity gradient torque

In this case the Euler equations become:

( ) ( )

( ) ( )

( ) ( )

=+

=+

=+

12xy3

txyxyzz

31zx3

tzxzxyy

23yz3

tyzyzxx

ccIIR

Gm3III

ccIIR

Gm3III

ccIIR

Gm3III

&

&

&

This torque is generated by the same force field that causes the orbital motion, so it is actingcontinuously, and it depends on the attitude of the satellite (the orientation of the radial direction in

the principal inertia frame). We can introduce a set of three small Euler angles to define the relative

orientation of the principal axes and the LVLH frame:

=

+

=

0

0

1

1

1

1

c

c

c

n1

1

1

xy

xz

yz

3

2

1

z

y

x

xy

xz

yz

z

y

x

&

&

&

In this way we can write the system dynamics as a function of the Euler angles and their

derivatives, including the gravity gradient torque that now becomes an integral part of the system

dynamics. Assuming that n is equal to one rotation per orbit, then:


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3

t

3

2

22

22

R

Gm

R

K

R

K

R

1

R

Vn ==

==

Therefore, in case of circular orbit, the attitude dynamics including gravity gradient becomes:

( ) ( )( ) ( )

( ) ( )

=+

=+

=+

12xy

2

xyxyzz

31zx

2

zxzxyy

23yz2

yzyzxx

ccIIn3III

ccIIn3III

ccIIn3III

&

&

&

where n is the nominal angular velocity along the orbit.

In this set of equations we have a twofold coupling between angular velocities and angles. Attitude

is represented by the direction cosines c1, c2, c3.

Assuming the principal inertia axes are sufficiently close to the corresponding axes of the LVLH

frame, we can write:

Y

Z X

=

+

=

0

0

1

1

1

1

c

c

c

n11

1

xy

xz

yz

3

2

1

z

y

x

xy

xz

yz

z

y

x

&

&

&

We can now substitute angular velocities and direction cosines with the corresponding terms

depending on angles and their derivatives. Neglecting the terms that include two angles or

derivatives (small terms) we get to the linearized form:

( ) ( )( ) ( ) ( )

( )

=

=+++

=++

zxy

2

zz

yzx

2

y

2

xzxzyxyy

x2

yzyxyzxx

IIn3I

IIn3nIInIIII

0nIInIIII

&&

&&&

&&&

That can be simplified into:


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62

( ) ( )

( ) ( )

( )

=+

=+++

=++

0IIn3I

0IIn4nIIII

0nIInIIII

zxy

2

zz

yxz

2

xzyxyy

x

2

yzyxyzxx

&&

&&&

&&&

The third equation is decoupled from the first two. We can now look at the stability of the system.

Starting from the z axis, the stability condition is:

Iy> Ix therefore Kp> 0 => Kr> Ky

KyPitch instability

Kr

Consider now the first two equations. Divide the first by Ixand the second by Iy:

( )

( )

=++

=++

0nK4nK1

0nKn1K

y

2

rxry

x

2

yyyx

&&&

&&&

The characteristic equation is:

( )( ) ( ) ( )[ ]( )

( ) 0KKn4KKK31sns

0KKn4KKKK1K4Ksns0K1ns1KnsKn4sKns

yr

2

yrr

224

yr

2

yryrry

224

ryr

22

y

22

=++++

=+++++=++

The conditions for stability are:

( )0KK

K16KKK3K1

04cb

02

4cbb

0c

0cbxx

yr

yr2

yrr

2

2

2

>

>++

>

=++

That is:


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63

>

>++

0KK

KK4KKK31

yr

yryrr

The presence of gravity gradient therefore reduces the stability region. In detail, the first condition

can be analyzed by evaluating a few solutions for particular values of coefficient Ky:

K4KK31KK

K4K211K

3

1K0K

2

ry

rry

ry

=++=

=+=

==

1 Ky

7 8

1

-1 6 6 1 Kr5 2

4 3

-1

The unmarked regions represent the stable regions. In fact, in region (1) we have:

Iz> Iy> Ix Ip> Ir> Iy always

In region (2) we have:

Iy> Ix> Iz Ir> Iy> Ip limited by the additional curve

In the remaining regions we have some kind of instability. In regions (3) and (4) we have yaw and

roll instability, in regions (5) and (7) yaw, roll and pitch instability while in regions (6) and (8) only

pitch instability. Furthermore in region (8) we have Ip> Iy> Irwhile in region (7) Iy> Ir> Ip.

Now assume we have an orbiting satellite with three different inertia moments as in the following

figure:


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64

Im IM

II

Im= I minimumIM= I maximum

II= I intermediate

To have stability identified by the region (1), the satellite must be oriented in orbit as follows:

IM II Im

To have stability identified by the region (2) instead orientation must be:

Im

IM

II

In order to stabilize a generic configuration, we can have one mass positioned at an appropriate

distance from the main body

adcs notes part1

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