rotational kinema tics

7/29/2019 Rotational Kinema Tics

1/59

AOE 5204

Rotational Kinematics

Descr i pt i on of at t i t ude ki nemat i csusi ng r ef er ence f r ames, r ot at i on

mat r i ces, Eul er par amet er s, Eul erangl es, and quat er ni ons

Recall the fundamental dynamics equations

For both equations, we must relate momentum to

kinematics

~f = ddt

~p

~g = ddt

~h


2/59

AOE 5204

Translational vs Rotational Dynamics

Linear momentum

=mass velocity

d/dt (linear momentum)=

applied forces

d/dt (position)

=

linear momentum/mass

Angular momentum

=inertia angular velocity

d/dt (angular momentum)=

applied torques

d/dt (attitude)

=

angular momentum/inertia

~f = ddt

~p ~g =ddt

~h


3/59

AOE 5204

Translational Kinematics

Newtons second law can be written in

first-order state-vector form as

~r = ~p/m

~p = ~f

Here, ~p is the linear momentum, definedby the kinematics equation; i.e., ~p = m~r

Thus, the kinematics differential equationallows us to integrate the velocity tocompute the positionThe kinetics differential equation allows usto integrate the applied force to computethe linear momentumIn general, ~f

=~f

(~r

, ~p

)


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AOE 5204

Translational Kinematics (2)

Consider ~f = m~a expressed in an inertial frame:

mr1 = f1

mr = f

mr2 = f2

mr3 = f3

Equivalently

p1 = f1p = f p2 = f2

p3 = f3

r1 = p1/m

r = p/m r2 = p2/mr3 = p3/m

We need to develop rotational equations equivalent

to the translational kinematics equations


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AOE 5204

Reference Frame & Vectors

i1 i2

i3

~v

v1 v2

v3

Reference frame is

dextral triad oforthonormal unitvectors

Vector can be expressedas linear combination of

the unit vectors

Must be clear aboutwhich reference frame is

used

Fi n

i1, i2, i3

o

~v = v1i1 + v2i2 + v3i3


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AOE 5204

Orthonormal

Orthonormal means the base vectors are perpendicular(orthogonal) to each other, and have unit length(normalized)

This set of 6 (why not 9?) properties can be written in

shorthand as

We can stack the unit vectors in a column matrix

And then write the orthonormal property as

i1 i1 = 1 i1 i2 = 0 i1 i3 = 0

i2 i1 = 0 i2 i2 = 1 i2 i3 = 0i3 i1 = 0 i3 i2 = 0 i3 i3 = 1

ii ij =

1 if i = j0 if i 6= j

or ii ij = ij

nio = ni1 i2 i3oT

ni

on

i

oT

=

1 0 00 1 0

0 0 1

= 1


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AOE 5204

Dextral, or Right-handed

Right-handed means the ordering of the three

unit vectors follows the right-hand rule

Which can be written more succinctly as

Or even more succinctly as

i1 i1 = ~0 i1 i2 = i3 i1 i3 =

i2

i2 i1 = i3 i2 i2 = ~0 i2 i3 = i1i3 i1 = i2 i3 i2 = i1 i3 i3 = ~0

nion

ioT

=

~0 i3 i2i3 ~0 i1i2

i1 ~0

=

nio

ii ij = ijk ik

ijk =

1 for i,j,k an even permutation of 1,2,31 for i,j,k an odd permutation of 1,2,3

0 otherwise (i.e., if any repetitions occur)


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AOE 5204

Skew Symmetry

The [] notation defines a skew-symmetric 3 3

matrix whose 3 unique elements are the

components of the 3 1 matrix []

The same notation applies if the components ofthe 3 1 matrix [] are scalars instead of vectors

The skew-symmetry property is satisfied since

a =

a1a2

a3

a =

0 a3 a2

a3 0 a1a2 a1 0

(a)T = a

nio

=

~0 i3 i2i3 ~0 i1i2 i1 ~0


9/59

AOE 5204

Vectors

A vector is an abstract mathematical object with

two properties: direction and length

Vectors used in this course include, for example,

position, velocity, acceleration, force,

momentum, torque, angular velocity

Vectors can be expressed in anyreference frame

Keep in mind that the term state vector refers

to a different type of object -- specifically, a state

vector is generally a column matrix collecting all

the system states


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AOE 5204

Vectors Expressed in Reference Frames

i1i2

i3

~v

v1 v2

v3

~v = v1i1 + v2i2 + v3i3The scalars, v1, v2, and v3, arethe components of ~v expressedin Fi. These components arethe dot products of the vector ~vwith the three base vectors ofF

i.

Specifically,

v1 = ~vi1, v2 = ~vi2, v3 = ~vi3

Since thei vectors are unit vec-tors, these components may also

be written as

vj = v cosj, j = 1, 2, 3

where v = k~vk is the magnitudeor length of~v, and j is the angle

between ~v and ij for j = 1, 2, 3


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AOE 5204

Vectors Expressed in Reference Frames (2)

Frequently we collect the components of the vector into

a matrix

When we can easily identify the associated reference

frame, we use the simple notation above; however,when multiple reference frames are involved, we use a

subscript to make the connection clear. Examples:

v = v1

v2v3

vi denotes components in Fi

vo denotes components in Fo

vb denotes components in Fb

Note the absence of an overarrow


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AOE 5204

Further Vector Notation

Matrix multiplication arises frequently in dynamics and

control, and an interesting application involves the 3 1

matrix of a vectors components and the 3

1 matrixof a frames base vectors*

We frequently encounter problems of two types:

* Peter Hughes coined the term vectrix to denote this object

~v = [v1 v2 v3]

i1

i2

i3

= vTi

nio

= vTi

nio

= vTo {o} = vTb

nbo

=

Given vi and vb, determine the attitude ofFb with respect to Fi

Given the attitude of Fb with respect to Fi, and components vi,determine vb


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AOE 5204

Rotations from One Frame to Another

Suppose we know the components of vector ~v in Fb, denoted vb, andwe want to determine its components in Fi, denoted vi

In some sense this problem has three unknowns (the components of

vi); hence we expect to form a set of three equations and three unknowns

Specifically, we note that ~v = vTi

nio

= vTb

nbo

and seek a linear

transformation R such that nio = Rnbo With such a transformation, we can make a substition and have ~v =

vTi Rn

bo

= vTb

nbo

Since the base vectorsn

bo

are orthonormal, the coefficients on both

sides of the equation must be equal, so

v

T

i R = vT

b RT

vi = vb


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AOE 5204

Rotations (2)

Problem reminder: Knowing vb, determine vi

We have RTvi = vb

If we know R, then we just have to solve the linear system to determinevi

We know that R is a 3 3 matrix (i.e., R R33), and thatn

io

= Rn

bo

The latter can be expanded to

i1 = R11b1 + R12b2 + R13b3

i2 = R21b1 + R22b2 + R23b3

i3 = R31b1 + R32b2 + R33b3

What, for example, is R11?


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AOE 5204

Rotations (3)

Consider just one of the equations involve the components of R:

i1 = R11b1 + R12b2 + R13b3

Comparing this expression with the definition of the components of avector in a specific frame, we see that

R11 = i1 b1, R12 = i1 b2,

and in general, Rij = ii bj

Using direction cosines, R11 = cos11, R12 = cos12, and in general,Rij = cosij, where ij is the angle between ii and bj

Thus R is a matrix of direction cosines, and is frequently referred toas the DCM (direction cosine matrix)


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AOE 5204

Rotations (4)

Another way to describe R is to observe that its rows contain thecomponents of the base vectors of Fi expressed in Fb, and that itscolumns contain the components of the base vectors ofFb expressed inFi

These observations mean that the rows and columns are mutuallyorthogonal, and since the base vectors are unit vectors, the rows andcolumns are mutually orthonormal

Recalling earlier notation,n

ion

ioT

= 1, we can also write R as the

dot product ofn

io

withn

boT

, i.e.,

R =n

ion

boT

So, if we know the relative orientation of the two frames, we cancompute R and solve the required linear system to compute vi


17/59

AOE 5204

Rotations (5)

Assuming we have computed R, we just need to solve the requiredlinear system to compute vi

The linear system, RTvi = vb, is easily solved because of the previ-

ously observed fact that R is an orthonormal matrix

If we were to return to the beginning of this development and beginwith

RTnio = nbo instead of nio = Rnbowe would find that the same matrix R satisfies both equations, thusproving that R1 = RT, which is perhaps the most useful property oforthonormal matrices

Its application here leads to

vi = Rvb vi = Ribvb

We adopt the notation Rib

to denote the orthonormal matrix that trans-forms vectors from Fb to Fi


18/59

AOE 5204

Rotations (6)

Summary of Rotation Notation

Rotation matrix, orthonormal matrix, attitude matrix, and directioncosine matrix are synonymous

The inverse of the rotation matrix is simply its transpose

The orthonormal matrix that transforms vectors from Fb to Fi isdenoted Rib:

vi = Rib

vb and vb = Rbi

vi

The rotation matrix can be computed using

Rib = nio nboT

and Rbi = nbo nioT

Rib =

i1bi2b

i3b

= b1i b2i b3i


19/59

AOE 5204

Rotations (7)

More Properties of Rotation Matrices

Since R1 = RT, RTR = 1 (the identity matrix)

The determinant of R is unity: det R = 1

One of the eigenvalues of R is +1

Every rotation corresponds to a rotation about a single axis a through

an angle ; this fact implies that Ra = a, and hence a is the eigenvectorcorresponding to the unity eigenvalue

Rotations multiply; i.e., if Rab relates frames Fa and Fb, and Rbc

relates frames Fb and Fc, then

Rac = RabRbc


20/59

AOE 5204

Rotational Kinematics Representations

The rotation matrix represents the attitude

A rotation matrix has 9 numbers, but they are

not independent

There are 6 constraints on the 9 elements of a

rotation matrix (what are they?)

Rotation has 3 degrees of freedom There are many different sets of parameters that

can be used to represent or parameterize

rotations Euler angles, Euler parameters (aka quaternions),

Rodrigues parameters (aka Gibbs vectors),

Modified Rodrigues parameters,


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AOE 5204

Euler Angles

Leonhard Euler (1707-1783) reasoned that the

rotation from one frame

to another can bevisualized as a sequence

of three simple rotations

about base vectors

Each rotation is through

an angle (Euler angle)

about a specified axis

Consider the rotation from Fi toFb using three Euler angles, 1, 2,and 3

Thefi

rst rotation is about thei3axis, through angle 1

The resulting frame is denoted

Fi0 or ni0o The rotation matrix from Fi toFi0 is

Ri0i = R3(1)

Ri0i = R3(1) =

cos 1 sin 1 0 sin 1 cos 1 0

0 0 1

vi0 = R3(1)vi


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AOE 5204

Visualizing That 3 Rotation

Ri0i = R3(1) =

cos 1 sin 1 0 sin 1 cos 1 0

0 0 1

vi0 = R3(1)vi

i11

1

i2i0

2

i3 and

i0

3 are out of the page

i0

1


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AOE 5204

Euler Angles (2)

The second rotation is about thei0

3 axis, through angle 2

The resulting frame is denoted

Fi00 or ni00o The rotation matrix from Fi0 toFi00 is

Ri00i0

= R2(2)

The notation for the simple rota-tions is Ri(j), which denotes a ro-

tation about the ith axis. The sub-script on R defines which axis isused, and the subscript on de-fines which of the three angles in

the Euler sequence used

Ri00

i0

= R2(2)

=

cos 2 0 sin 20 1 0

sin 2 0 cos 2

vi00 = R2(2)vi0 = R2(2)R3(1)vi

Ri00i = R2(2)R3(1)

is the rotation matrix transformingvectors from Fi to Fi00

For an Ri rotation, the ith row andcolumn are always two zeros and

a one. The other two rows andcolumns have cos and sin in aneasily memorized pattern


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AOE 5204

Euler Angles (3)

The third rotation is about the i00

1

axis, through angle 3

The resulting frame is denoted Fb

orn

bo

The rotation matrix from Fi00 toFb is

Rbi00

= R1(3)

Rbi00

= R1(3)

=

1 0 00 cos 3 sin 30

sin 3 cos 3

Note that the cos terms are on thediagonal of the matrix, whereas thesin terms are on the off-diagonal

vb = R1(3)vi00

= R1(3)R2(2)R3(1)vi

Rbi = R1(3)R2(2)R3(1)

is the rotation matrix transformingvectors from Fi to Fb

3

3

i002

i00

3

b2

b3

i00

1 and b1 are out of the page


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AOE 5204

Euler Angles (4)

We have performed a 3-2-1 rotation from Fi to Fi00

Rbi = R1(3)R2(2)R3(1)

=

1 0 0

0 c3 s30 s3 c3

c2 0 s2

0 1 0s2 0 c2

c1 s1 0

s1 c1 00 0 1

=

c1c2 s1c2 s2

c3s1 + c1s2s3 c1c3 + s1s2s3 c2s3c1s2c3 + s1s3 s1s2c3

c1s3 c2c3

We can select arbitrary values of the three angles and compute a rotationmatrix. For example, let 1 = /3, 2 = /7, and 3 = /5:

R = 0.450484 0.780262 0.4338840.573114 0.625371 0.529576

0.684547 0.010099 0.728899

Conversely, given a rotation matrix, we can extract the Euler angles


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AOE 5204

Euler Angles (5)

To extract Euler angles from a given rotation matrix, equate elementsof the two matrices:

Rbi =

c1c2 s1c2 s2

c3s1 + c1s2s3 c1c3 + s1s2s3 c2s3c1s2c3 + s1s3 s1s2c3 c1s3 c2c3

=

0.450484 0.780262 0.4338840.573114 0.625371 0.529576

0.684547 0.010099 0.728899

Choose the easy one first:

R13 : s2 = 0.433884 2 = sin1 0.433884 = 0.448799 /7

R11 : c1c2 = 0.450484 1 = 1.047198

/3

Quadrant checks are generally necessary for the second and third calcu-lations.

Exercise: Select another Euler angle sequence and determine the valuesof the three s that give the numerical R above.


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AOE 5204

Euler Angles (6)

We just developed a 3-2-1 rotation, but there are other possibilities

There are 3 choices for the first rotation, 2 choices for the secondrotation, and 2 choices for the third rotation, so there are 322 = 12

possible Euler angle sequences

The Euler angle rotation sequences are

(1 2 3) (1 3 2) (1 2 1) (1 3 1)(2 3 1) (2 1 3) (2 3 2) (2 1 2)(3 1 2) (3 2 1) (3 1 3) (3 2 3)

The two left columns are sometimes called the asymmetric rotationsequences, and the two right columns are called the symmetricsequences

The roll-pitch-yaw sequence is an asymmetric sequence (which one?),whereas the -i- sequence (what is this?) is a symmetric sequence


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AOE 5204

Euler Angles (7)

The three simple rotation matricesare

R1() =

1 0 00 cos sin

0 sin cos

R2() = cos 0 sin

0 1 0sin 0 cos

R3() = cos sin 0 sin cos 0

0 0 1

(1,0,0) in 1st row and column

Cosines on diagonal, Sines on off-

diagonal, negative Sine on rowabove the 1st row

(0,1,0) in 2nd row and column

Cosines on diagonal, Sines on off-

diagonal, negative Sine on row

above the 2nd row

(0,0,1) in 3rd row and column

Cosines on diagonal, Sines on off-diagonal, negative Sine on row

above the 3rd row


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AOE 5204

Roll, Pitch and Yaw

Roll, pitch and yaw are Euler angles and are sometimes defined as a3-2-1 sequence and sometimes defined as a 1-2-3 sequence

Whats the difference?

The 3-2-1 sequence (we did earlier) leads to

Rbi =

c1c2 s1c2 s2c3s1 + c1s2s3 c1c3 + s1s2s3 c2s3

c1s2c3 + s1s3 s1s2c3 c1s3 c2c3

where 1 is the yaw angle, 2 is the pitch angle, and 3 is the roll angle

The 1-2-3 sequence leads to

Rbi =

c2c3 s1s2c3 + c1s3 s1s3 c1s2c3c2s3 c1c3 s122s3 s1c3 + c1s2s3

s2 s1c2 c1c2

where 1 is the roll angle, 2 is the pitch angle, and 3 is the yaw angle

AOE 5204


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AOE 5204

Roll, Pitch, Yaw (2)

Note that the two matrices are not the same

Rotations do not commute However, if we assume that the angles are small (appropriate for

many vehicle dynamics problems), then the approximations of the

two matrices are equal

3-2-1 Sequence

cos

1 and sin

Rbi

1 1 21 1 3

2 3 1

where 1 is the yaw angle, 2 is thepitch angle, and 3 is the roll angle

1-2-3 Sequence

cos

1 and sin

Rbi

1 3 23 1 1

2 1 1

where 1 is the roll angle, 2 is thepitch angle, and 3 is the yaw angle

Rbi

1

32

1

Rbi

1

Rbi

1

rollpitchyaw

AOE 5204


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AOE 5204

Perifocal Frame

Earth-centered, orbit-based,

inertial The P-axis is in periapsis

direction

The W-axis is perpendicularto orbital plane (direction of

orbit angular momentum

vector, )

The Q-axis is in the orbitalplane and finishes the triad

of unit vectors

In the perifocal frame, the

position and velocity vectorsboth have a zero component

in the W direction

P

P

Q

Q W

vrGG

AOE 5204


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AOE 5204

A One-Minute Course on Orbital Mechanics

i

I

J

K

n

Equatorialplane

Orbitalplane

Orbit is defined by 6 orbital elements(oes):

semimajor axis, a; eccentricity, e;

inclination, i; right ascension of ascending node, ;

argument of periapsis, ; and true anomaly,

AOE 5204


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AOE 5204

Inertial Frame to Perifocal Frame

R3() =

cos sin 0 sin cos 0

0 0 1

R1(i) =

1 0 00 cos i sin i

0 sin i cos i

R3() =

cos sin 0 sin cos 0

0 0 1

Rpi = R3()R1(i)R3() =

cc ci s s ci c s + c s si ss c ci c s ci c c s s si c

si s si c ci

Use a 3-1-3 sequence

Right ascension of the ascendingnode about K: R3(), rotates Ito n (the ascending node vector)

Inclination i about the node vec-tor: R1(i), rotates K to W (theorbit normal direction)

Argument of periapsis aboutthe orbit normal: R3(), rotates n

to P (the periapsis direction)

AOE 5204


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AOE 5204

Orbital Reference Frame

Same as roll-pitch-yawframe, for spacecraft

The o3 axis is in the nadirdirection

The o2 axis is in the negativeorbit normal direction

The o1 axis completes thetriad, and is in the velocityvector direction for circularorbits

In the orbital frame, positionand velocity both have zeroin the o2 direction, andposition has zero in the o1

direction as well You will find the rotation

from perifocal to orbitaleasier to visualize if youmake yourself two reference

frames

vG

rG

w

1o

2o

3o

AOE 5204


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AOE 5204

Perifocal Frame to Orbital Frame

Use a 2-3-2 sequence

270 about Q: R2(270), rotates

W to P (the nadir direction)

90 about the nadir vector:R3(90

), rotates o2 to W (thenegative orbit normal direction)

Negative true anomaly abouto2: R2(), rotates o3 to thetransverse direction

Rop = R2()R3(90)R2(270) =

s c 00 0 1c s 0

R2(270) =

0 0 10 1 01 0 0

R3(90) =

0 1 01 0 0

0 0 1

R2() =

cos 0 sin 0 1 0 sin 0 cos

AOE 5204


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AOE 5204

Inertial to Perifocal to Orbital

Rpi = R3()R1(i)R3() =

cc ci s s ci c s + c s si s

s c

ci c s ci c c

s s si c

si s si c ci

Rop = R2()R3(90)R2(270) =

s c 00 0 1

c

s 0

Roi = RopRpi

=

su c cu ci s su s + cu ci c cu sisi s si c ci

cu c + su ci s

cu s

su ci c

su si

where u = +

AOE 5204


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An Illustrative Example

In an inertial reference frame, an Earth-orbiting satellite has positionand velocity vectors:

~r = 6000~I + 10, 000 ~J + 5, 000 ~K [km]~v =

5~I

2 ~J + 1 ~K [km/s]

The orbital elements are (using elementary astrodynamics)

a = 12, 142 km, e = 0.22026, i = 23.986, = 46.469, = 321.80, = 113.98

We can use i, , and to compute Rpi as

Rpi =

0.95089 0.18060 0.25139

0.094620 0.94286 0.31946

0.29472 0.27998 0.91364

Then we can rotate position and velocity into Fp:

rp = [5156.3 11594 0.037610]T

vp = [5.3670 1.0932 2.7899 107

]T

AOE 5204


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An Illustrative Example (continued)We can use = 113.98 to compute Rop as

Rop =

0.91369 0.40642 00 0 1

0.40642 0.91369 0

We can then multiply RopRpi to get Roi, or we can use i, , andu = + to compute Roi:

Roi

= 0.83036 0.54821 0.0998580.29472 0.27998 0.91364

0.47291 0.78808 0.39406

Then we can rotate position and velocity into Fo:

ro = [0.66151 0.037610 12, 689]Tvo = [5.3481 2.7899 10

7 1.1824]T

AOE 5204


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Differential Equations of Kinematics

Given the velocity of a point and initialconditions for its position, we can compute itsposition as a function of time by integratingthe differential equation

We now need to develop the equivalentdifferential equations for the attitude whenthe angular velocity is known

Preview:

d/dt (attitude) = angular momentum/inertia

~r = ~v

= 0 sin 3/ cos 2 cos 3/ cos 2

0 cos 3 sin 31 sin 3 sin 2/ cos 2 cos 3 sin 2/ cos 2

123

= S

1

AOE 5204

E l A l d A l V l i


40/59

Euler Angles and Angular Velocity

One frame-to-frame at a time, just as we did for developing rotationmatrices

3-2-1 rotation from Fi to Fi0 to Fi00 to Fb

3-rotation from Fi to Fi0 about i3 i03 through 1

The angular velocity ofFi0 with respect to Fi is

~i0i = 1 i3 = 1 i

0

3

We can express ~i0i in any frame, but Fi and Fi0 are especially simple:

i

0

ii = [0 0 1]T

i0ii0 = [0 0 1]

T

Keep the notation in mind: i0ii is the angular velocity of Fi0 with

respect to Fi, expressed in Fi

AOE 5204

E l A l d A l V l i (2)


41/59

Euler Angles and Angular Velocity (2)

2-rotation from Fi0 to Fi00 about i0

2 i002 through 2

The angular velocity ofFi00 with respect to Fi0 is

~i00

i0

= 2 i02 = 2 i002

We can express ~i00i0 in any frame, but Fi0 and Fi00 are especially

simple:

i00i0

i0 = [0 2 0]T

i00i0

i00 = [0 2 0]T

Keep the notation in mind: i00i0

i0 is the angular velocity ofFi00 with

respect to Fi0 , expressed in Fi0

AOE 5204

E l A l d A l V l it (3)


42/59

Euler Angles and Angular Velocity (3)

1-rotation from Fi00 to Fb about i00

1 b1 through 3

The angular velocity ofFb with respect to Fi00 is

~bi00

= 3 i001 = 3b1

We can express ~bi00

in any frame, but Fi00 and Fb are especiallysimple:

bi00

i00 = [3 0 0]T

bi00

b = [3 0 0]T

Keep the notation in mind: bi00

b is the angular velocity of Fb with

respect to Fi00 , expressed in Fb

AOE 5204

Addi th A l V l iti


43/59

Adding the Angular Velocities

Angular velocities are vectors and add like vectors:

~bi = ~bi00

+ ~i00i0 + ~i

0i

We have expressed these three vectors in diff

erent frames; to add themtogether, we need to express all of them in the same frame

Typically, we want bib , so we need to rotate the 31 matrices we justdeveloped into Fb

We have

i0ii =

i0ii0 = [0 0 1]

T need Rbi0

i00i0

i0

=

i00i0

i00

= [02

0]

T

needRbi

00

bi00

i00 = bi00

b = [3 0 0]T need Rbb = 1

We previously developed all these rotation matrices, so we just needto apply them and add the results

AOE 5204

Complete the Operation


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Complete the Operation

What happens when 2 n/2, for odd n ?

What happens when the Euler angles and their rates are small?

Carry out the matrix multiplications and additions

bib =

bi00

b +i00i0

b + i0ib

and obtain

bib =

3 sin 2 1cos 3 2 + cos 2 sin 3 1 sin 3 2 + cos 2 cos 3 1

=

sin 2 0 1

cos 2 sin 3 cos 3 0cos 2 cos 3 sin 3 0

123

= S()

or

= S1() =

0 sin 3/ cos 2 cos 3/ cos 20 cos 3 sin 3

1 sin 3 sin 2/ cos 2 cos 3 sin 2/ cos 2

12

3

AOE 5204 Kinematic Singularity in the Differential


45/59

Kinematic Singularity in the Differential

Equation for Euler Angles

For this Euler angle set (3-1-2), the Euler rates go to

infinity when cos 2 0

The reason is that near 2 = /2 the first and third

rotations are indistinguishable

For the symmetric Euler angle sequences (3-1-3, 2-1-

2, 1-3-1, etc) the singularity occurs when 2 = 0 or

For the asymmetric Euler angle sequences (3-2-1, 2-3-1, 1-3-2, etc) the singularity occurs when 2 = /2 or

3/2

This kinematic singularity is a major disadvantage ofusing Euler angles for large-angle motion

There are attitude representations that do not have a

kinematic singularity, but 4 or more scalars are

required

AOE 5204

Linearizing the Kinematics Equation


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Linearizing the Kinematics Equation

Exercise: Repeat these steps for a 1-2-3 sequence and for a symmetric sequence.

bi

b

= S()

=

3 sin 2 1cos 3 2 + cos 2 sin 3 1 sin 3 2 + cos 2 cos 3 1

=

sin 2 0 1cos 2 sin 3 cos 3 0cos 2 cos 3 sin 3 0

123

If the Euler angles and rates are small, then sin i i and cos i 1:

bib

321

0 0 10 1 0

1 0 0

12

3

AOE 5204

Refresher: How To Invert a Matrix


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Refresher: How To Invert a MatrixSuppose you want to invert the n n matrix A, with elements Aij

The elements of the inverse are

A1ij =Cji|A|

where Cji is the cofactor, computed by multiplying the determinant ofthe (n 1) (n 1) minor matrix obtained by deleting the jth row andith column from A, by (1)i+j

This formulation is absolutely unsuitable for calculating matrix inversesin numerical work, especially with larger matrices, since it is computa-tionally expensive

One normally uses LU decomposition instead

Elementary row reduction is essentially LU decomposition

Note: In most cases, we do not need the inverse anyway; we need thesolution to a linear system

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Matrix Inversion Application


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Matrix Inversion ApplicationLet us invert the matrix

S() =

sin 2 0 1cos 2 sin 3 cos 3 0

cos 2 cos 3 sin 3 0

In row reduction, we augment the matrix with the identity matrix:

sin 2 0 1 1 0 0

cos 2 sin 3 cos 3 0 0 1 0cos 2 cos 3

sin 3 0 0 0 1

and apply simply row-reduction operations to transform the left 3 3block to the identity, leaving the inverse in the right 3 3 block

In this case, we must swap the first row with one of the other two rows,say the 3rd row, which amounts to a permutation by :

P =

0 0 10 1 0

1 0 0

Note that P

1 = P

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Matrix Inversion Application (2)


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Matrix Inversion Application (2)

cos 2 cos 3 sin 3 0 0 0 1cos 2 sin 3 cos 3 0 0 1 0 sin 2 0 1 1 0 0

Multiply row 2 by sin 3/ cos 3 and add to row 1.

Multiply row 1 by sin 3/ cos 3 and add to row 2.

c2 c3 + c2s3 tan 3 0 0 0 tan 3 1

0 c3 + s3 tan 3 0 0 1 tan 3s2 0 1 1 0 0

Divide row 1 by c2 c3 + c2s3 tan 3 and simplify

Multiply resulting row 1 by s2, add to row 3, and simplify

Divide row 2 by c3 + s3 tan 3 and simplify

1 0 0 0 s3/c2 c3/c20 1 0 0 c3

s3

0 0 1 1 s3 tan 2 c3 tan 2

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Eulers Theorem


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Euler s Theorem

00.5-0.5 00.5

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

The most general motion of a rigid body with a fixed point is a

rotation about a fixed axis.

The axis, denoted a, is called the eigenaxis or Euler axis

The angle of rotation, , is called the Euler angle or the principal Eulerangle

Here the black axes arethe base vectors of the in-ertial frame, and the red,blue, and yellow axes arethe base vectors of a bodyframe, rotated about

a = [2/2 2/2 2/2]T

through angle = /4

Why no subscript on a?

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Eulers Theorem (2)


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Euler s Theorem (2)Any rotation matrix can be expressed in terms of a and :

R = cos1 + (1 cos)aaT sina

Since a is an eigenvector of R with eigenvalue 1,

Ra = a

Check this result:

Ra = cos1 + (1 cos)aaT

sina a

= cos1a + (1 cos)aaTa sinaa= cosa + aaTa cosaaTa sinaa= cosa + a cosa sinaa (aTa = 1)

= cosa + a cosa (a

a = 0)= a

Also note that the trace of R is simply

traceR = 1 + 2 cos

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Extracting a and from R,


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g ,

and Integrating to Obtain a and

Given any rotation matrix we can compute the Euler axis a and angle using:

= cos1 1

2 (trace R 1)a =

1

2sin

RT R

What does one do about the sin

= 0 case?

One can show that the kinematics differential equations for a and are:

= aT

a = 12

a cot

2aa

So, this system of equations also has kinematic singularities, at = 0and = 2

AOE 5204

Quaternions (aka Euler Parameters)


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Quaternions (aka Euler Parameters)

Define another 4-parameter set of variables to represent the attitude:

q = a sin

2

q4 = cos

2

The 31 matrix q forms the Euler axis component of the quaternion, also

called the vector component. The scalar q4 is called the scalar component.Collectively, these four variables are known as a quaternion, or as theEuler parameters.

The notation q denotes the 4 1 matrix containing all four variables:

q =

qT q4

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Quaternions (2)


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Quaternions (2)

R(q) and q(R) :

R =

q24 qTq

1 + 2qqT 2q4q

q4 = 12

1 + trace R

q =1

4q4

R23 R32R31 R13R12 R21

Differential equations q :

q = 12

q

+ q41qT = Q(q)

Note that there is no kinematic singularity with these differential equa-tions

AOE 5204

Typical Problem Involving


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Angular Velocity and Attitude

Given initial conditions for the attitude

(in any form), and a time history ofangular velocity, compute R or any

other attitude representation as a

function of time Requires integration of one of the sets of

differential equations involving angular

velocity

AOE 5204

Spherical Pendulum Problem


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p

Illustration from Wolfram Research Mathworld

mathworld.wolfram.com/SphericalCoordinates.html

Use a rotation from Fi to Fr us-

ing two Euler angles, , and

The first (3) rotation is about the

i3 axis, through angle

The second (2) rotation is about

the e axis through

The three unit vectors havederivatives:

er = sine + e,

e = siner cose,e = er + cose

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Linearization for Small


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As with Euler angles, we are frequently interested in small attitude mo-tions.

If is small, then q = a sin(/2) a2

, and q4 1

Hence, for small ,

R =

q24 qTq

1 + 2qqT 2q4q

(1

0) 1 + 2(0)

2q

1 2q

Compare this expression with the previously developed R 1 forEuler angles. What is the equivalent expression for the (a,) representa-

tion?

Small rotations are commutative:

RcbRba

1 2q

2 1 2q

1 = 1 2q

2

2q1

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Other Attitude Representations


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p

We have seen direction cosines, Euler angles, Eulerangle/axis, and quaternions

Two other common representations Euler-Rodriguez parameters

Modified Rodriguez parameters

p = a tan

2

R = 1 +2

1 + pT

p

(pp

p)

p =1

2(ppT + 1 + p)

= a tan

4R =

1

1 + T

(1 (T)2)1 + 2T 2(1 (T)2)

=1

2

1 + T 1 +

T

21

AOE 5204

Typical Problem Involving


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Angular Velocity and Attitude

Given initial conditions for the attitude

(in any form), and a time history ofangular velocity, compute R or any

other attitude representation as a

function of time Requires integration of one of the sets of

differential equations involving angular

velocity

rotational kinema tics

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