Rotational Kinematics. Description of attitude kinematics using reference frames, rotation matrices, Euler parameters, Euler angles, and quaternions

Transcription

1 Rotational Kinematics Description of attitude kinematics using reference frames, rotation matrices, Euler parameters, Euler angles, and quaternions Recall the fundamental dynamics equations ~ f = d dt ~p ~g = d dt ~ h For both equations, we must relate momentum to kinematics

2 Translational vs Rotational Dynamics ~ f = d dt ~p Linear momentum = mass velocity ~g = d dt ~ h Angular momentum = inertia angular velocity d/dt (linear momentum) = applied forces d/dt (angular momentum) = applied torques d/dt (position) = linear momentum/mass d/dt (attitude) = angular momentum/inertia

3 Translational Kinematics Newton s second law can be written in first-order state-vector form as ~r = ~p/m ~p = ~ f Here, ~p is the linear momentum, defined by the kinematics equation; i.e., ~p = m ~r Thus, the kinematics differential equation allows us to integrate the velocity to compute the position The kinetics differential equation allows us to integrate the applied force to compute the linear momentum In general, ~ f = ~ f(~r, ~p)

4 Translational Kinematics (2) Consider ~ f = m~a expressed in an inertial frame: m r 1 = f 1 m r = f m r 2 = f 2 m r 3 = f 3 Equivalently ṗ 1 = f 1 ṗ = f ṗ 2 = f 2 ṗ 3 = f 3 ṙ 1 = p 1 /m ṙ = p/m ṙ 2 = p 2 /m ṙ 3 = p 3 /m We need to develop rotational equations equivalent to the translational kinematics equations

5 Reference Frame & Vectors ~v = v 1 î 1 + v 2 î 2 + v 3 î 3 v 1 v 3 î 3 ~v v2 î 1 î 2 Reference frame is dextral triad of orthonormal unit vectors o F i nî1, î2, î3 Vector can be expressed as linear combination of the unit vectors Must be clear about which reference frame is used

6 Orthonormal Orthonormal means the base vectors are perpendicular (orthogonal) to each other, and have unit length (normalized) î 1 î1 =1 î1 î2 =0 î1 î3 =0 î 2 î1 =0 î2 î2 =1 î2 î3 =0 î 3 î1 =0 î3 î2 =0 î3 î3 =1 This set of 6 (why not 9?) properties can be written in shorthand as î i îj = ½ 1 if i = j 0 if i 6= j We can stack the unit vectors in a column matrix nîo And then write the orthonormal property as nîo = nî1 î 2 î 3 o T o T nî = or = 1 î i îj = δ ij

7 Dextral, or Right-handed Right-handed means the ordering of the three unit vectors follows the right-hand rule î 1 î1 = ~0 î 2 î1 = î3 î 3 î1 = î2 î 1 î2 = î3 î 2 î2 = ~0 î 3 î2 = î1 Which can be written more succinctly as ε ijk = Or even more succinctly as nîo o T nî = î i îj = ε ijk î k ~0 î 3 î2 î3 ~0 î 1 î 2 î1 ~0 î 1 î3 = î2 î 2 î3 = î1 î 3 î3 = ~0 1 for i, j, k an even permutation of 1,2,3 1 for i, j, k an odd permutation of 1,2,3 0 otherwise (i.e., if any repetitions occur) = nî o

8 Skew Symmetry The[] notation defines a skew-symmetric 3 3 matrix whose 3 unique elements are the components of the 3 1 matrix [] o nî = The same notation applies if the components of the 3 1 matrix [] are scalars instead of vectors a = a 1 a 2 a 3 a = The skew-symmetry property is satisfied since (a ) T = a ~0 î3 î 2 î 3 ~0 î1 î2 î 1 ~0 0 a 3 a 2 a 3 0 a 1 a 2 a 1 0

9 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 any reference 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

10 Vectors Expressed in Reference Frames ~v = v 1 î 1 + v 2 î 2 + v 3 î 3 The scalars, v 1, v 2, and v 3, are v 3 î 3 the components of ~v expressed in F i. These components are the dot products of the vector ~v with the three base vectors of F i. Specifically, v 1 = ~v î1, v 2 = ~v î2, v 3 = ~v î3 ~v Since the î vectors are unit vectors, these components may also be written as v j = v cos α j, j =1, 2, 3 v 1 v2 î 1 î 2 where v = k~vk is the magnitude or length of ~v, andα j is the angle between ~v and îj for j =1, 2, 3

11 Vectors Expressed in Reference Frames (2) Frequently we collect the components of the vector into a matrix v = v 1 v 2 v 3 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 i denotes components in F i v o denotes components in F o v b denotes components in F b Note the absence of an overarrow

12 Further Vector Notation Matrix multiplication arises frequently in dynamics and control, and an interesting application involves the 3 1 matrix of a vector s components and the 3 1 matrix of a frame s base vectors* We frequently encounter problems of two types: î 1 nîo ~v =[v 1 v 2 v 3 ] î 2 = vt i î 3 nîo nˆbo = v T i = vo T {ô} = vb T = Given v i and v b, determine the attitude of F b with respect to F i Given the attitude of F b with respect to F i, and components v i, determine v b * Peter Hughes coined the term vectrix to denote this object

13 Rotations from One Frame to Another Suppose we know the components of vector ~v in F b,denotedv b,and we want to determine its components in F i, denoted v i In some sense this problem has three unknowns (the components of v i ); hence we expect to form a set of three equations and three unknowns nîo nˆbo Specifically, we note that ~v = vi T = vb T and seek a linear o o transformation R such that nî = R nˆb With such o a transformation, nˆbo we can make a substition and have ~v = vi nˆb T R = vb T nˆbo Since the base vectors are orthonormal, the coefficients on both sides of the equation must be equal, so v T i R = v T b R T v i = v b

14 Rotations (2) Problem reminder: Knowing v b, determine v i We have R T v i = v b If we know R, then we just have to solve the linear system to determine v i We know that R is a 3 3matrix(i.e., R R 3 3 ), and that nîo o = R nˆb The latter can be expanded to What, for example, is R 11? î 1 = R 11ˆb1 + R 12ˆb2 + R 13ˆb3 î 2 = R 21ˆb1 + R 22ˆb2 + R 23ˆb3 î 3 = R 31ˆb1 + R 32ˆb2 + R 33ˆb3

15 Rotations (3) Consider just one of the equations involve the components of R: î 1 = R 11ˆb1 + R 12ˆb2 + R 13ˆb3 Comparing this expression with the definition of the components of a vector in a specific frame,weseethat and in general, R 11 = î1 ˆb 1, R 12 = î1 ˆb 2, R ij = îi ˆb j Using direction cosines, R 11 =cosα 11, R 12 =cosα 12, and in general, R ij =cosα ij,whereα ij is the angle between îi and ˆb j Thus R is a matrix of direction cosines, and is frequently referred to as the DCM (direction cosine matrix)

16 Rotations (4) Another way to describe R is to observe that its rows contain the components of the base vectors of F i expressed in F b, and that its columns contain the components of the base vectors of F b expressed in F i These observations mean that the rows and columns are mutually orthogonal, and since the base vectors are unit vectors, the rows and columns are mutually orthonormal o nîo T Recalling earlier notation, nî = 1, we can also write R as the o nˆbo T dot product of nî with, i.e., o o T R = nî nˆb So, if we know the relative orientation of the two frames, we can compute R and solve the required linear system to compute v i

17 Rotations (5) Assuming we have computed R, wejustneedtosolvetherequired linear system to compute v i The linear system, R T v i = v b, is easily solved because of the previously observed fact that R is an orthonormal matrix If we were to return to the beginning of this development and begin with n o o nîo o R T î = nˆb instead of = R nˆb we would find that the same matrix R satisfies both equations, thus proving that R 1 = R T, which is perhaps the most useful property of orthonormal matrices Its application here leads to v i = Rv b v i = R ib v b We adopt the notation R ib to denote the orthonormal matrix that transforms vectors from F b to F i

18 Rotations (6) Summary of Rotation Notation Rotation matrix, orthonormal matrix, attitude matrix, and direction cosine matrix are synonymous The inverse of the rotation matrix is simply its transpose The orthonormal matrix that transforms vectors from F b to F i is denoted R ib : v i = R ib v b and v b = R bi v i The rotation matrix can be computed using o nˆbo T o o T R ib = nî and R bi = nˆb nî R ib = i 1b i 2b i 3b = b 1i b 2i b 3i

19 Rotations (7) More Properties of Rotation Matrices Since R 1 = R T, R T R = 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 Φ; this fact implies that Ra = a, and hence a is the eigenvector corresponding to the unity eigenvalue Rotations multiply; i.e., ifr ab relates frames F a and F b,andr bc relates frames F b and F c,then R ac = R ab R bc

20 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,

21 Euler Angles Leonhard Euler ( ) reasoned that the rotation from one frame to another can be visualized 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 F i to F b using three Euler angles, θ 1, θ 2, and θ 3 The first rotation is about the î3 axis, through angle θ 1 The resulting o frame is denoted F i 0 or nî0 The rotation matrix from F i to F i0 is R i0i = R 3 (θ 1 ) R i0i = R 3 (θ 1 )= cos θ 1 sin θ 1 0 sin θ 1 cos θ v i 0 = R 3 (θ 1 )v i

22 Visualizing That 3 Rotation R i0i = R 3 (θ 1 )= cos θ 1 sin θ 1 0 sin θ 1 cos θ v i 0 = R 3 (θ 1 )v i î 0 2 î 2 θ 1 î 0 1 θ 1 î 1 î3 and î0 3 are out of the page

23 The second rotation is about the î 0 3 axis, through angle θ 2 The nî00 resulting o frame is denoted F i 00 or Euler Angles (2) R i00 i 0 = R 2 (θ 2 ) = cos θ 2 0 sin θ sin θ 2 0 cosθ 2 The rotation matrix from F i 0 F i00 is R i00 i 0 = R 2 (θ 2 ) to v i 00 = R 2 (θ 2 )v i 0 = R 2 (θ 2 )R 3 (θ 1 )v i R i00i = R 2 (θ 2 )R 3 (θ 1 ) is the rotation matrix transforming vectors from F i to F i 00 The notation for the simple rotations is R i (θ j ), which denotes a rotation about the i th axis. The subscript on R defines which axis is used, and the subscript on θ defines which of the three angles in the Euler sequence used For an R i rotation, the i th row and column are always two zeros and a one. The other two rows and columns have cos θ and sin θ in an easily memorized pattern

24 Euler Angles (3) The third rotation is about the î00 1 axis, through angle θ 3 Theoresulting frame is denoted F b or nˆb The rotation matrix from F i 00 F b is R bi00 = R 1 (θ 3 ) ˆb 3 θ 3 î 00 3 ˆb 2 to R bi00 = R 1 (θ 3 ) = cosθ 3 sin θ 3 0 sin θ 3 cos θ 3 Note that the cos θ terms are on the diagonal of the matrix, whereas the sin θ terms are on the off-diagonal v b = R 1 (θ 3 )v i 00 = R 1 (θ 3 )R 2 (θ 2 )R 3 (θ 1 )v i R bi = R 1 (θ 3 )R 2 (θ 2 )R 3 (θ 1 ) is the rotation matrix transforming vectors from F i to F b θ 3 î 00 2 î00 1 and ˆb 1 are out of the page

25 Euler Angles (4) We have performed a rotation from F i to F i 00 R bi = R 1 (θ 3 )R 2 (θ 2 )R 3 (θ 1 ) = cθ 3 sθ 3 cθ 2 0 sθ sθ 3 cθ 3 sθ 2 0 cθ 2 = cθ 1 sθ 1 0 sθ 1 cθ cθ 1 cθ 2 sθ 1 cθ 2 sθ 2 cθ 3 sθ 1 +cθ 1 sθ 2 sθ 3 cθ 1 cθ 3 +sθ 1 sθ 2 sθ 3 cθ 2 sθ 3 cθ 1 sθ 2 cθ 3 +sθ 1 sθ 3 sθ 1 sθ 2 cθ 3 cθ 1 sθ 3 cθ 2 cθ 3 We can select arbitrary values of thethree angles and compute a rotation matrix. For example, let θ 1 = π/3, θ 2 = π/7, and θ 3 = π/5: R = Conversely, given a rotation matrix, we can extract the Euler angles

26 Euler Angles (5) To extract Euler angles from a given rotation matrix, equate elements of the two matrices: R bi = = Choose the easy one first: cθ 1 cθ 2 sθ 1 cθ 2 sθ 2 cθ 3 sθ 1 +cθ 1 sθ 2 sθ 3 cθ 1 cθ 3 +sθ 1 sθ 2 sθ 3 cθ 2 sθ 3 cθ 1 sθ 2 cθ 3 +sθ 1 sθ 3 sθ 1 sθ 2 cθ 3 cθ 1 sθ 3 cθ 2 cθ R 13 : sθ 2 = θ 2 =sin = π/7 R 11 :cθ 1 cθ 2 = θ 1 = π/3 Quadrant checks are generally necessary for the second and third calculations. Exercise: Select another Euler angle sequence and determine the values of the three θs that give the numerical R above.

27 Euler Angles (6) We just developed a rotation, but there are other possibilities There are 3 choices for the first rotation, 2 choices for the second rotation, and 2 choices for the third rotation, so there are 3 2 2=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 rotation sequences, and the two right columns are called the symmetric sequences The roll-pitch-yaw sequence is an asymmetric sequence (which one?), whereas the Ω-i-ω sequence (what is this?) is a symmetric sequence

28 Euler Angles (7) The three simple rotation matrices are R 1 (θ) = cosθ sin θ 0 sin θ cos θ (1,0,0) in 1 st row and column Cosines on diagonal, Sines on offdiagonal, negative Sine on row above the 1 st row R 2 (θ) = cos θ 0 sin θ sin θ 0 cosθ (0,1,0) in 2 nd row and column Cosines on diagonal, Sines on offdiagonal, negative Sine on row above the 2 nd row R 3 (θ) = cos θ sin θ 0 sin θ cos θ (0,0,1) in 3 rd row and column Cosines on diagonal, Sines on offdiagonal, negative Sine on row above the 3 rd row

29 Roll, Pitch and Yaw Roll, pitch and yaw are Euler angles and are sometimes defined as a sequence and sometimes defined as a sequence What s the difference? The sequence (we did earlier) leads to cθ 1 cθ 2 sθ 1 cθ 2 sθ 2 R bi = cθ 3 sθ 1 +cθ 1 sθ 2 sθ 3 cθ 1 cθ 3 +sθ 1 sθ 2 sθ 3 cθ 2 sθ 3 cθ 1 sθ 2 cθ 3 +sθ 1 sθ 3 sθ 1 sθ 2 cθ 3 cθ 1 sθ 3 cθ 2 cθ 3 where θ 1 is the yaw angle, θ 2 is the pitch angle, and θ 3 is the roll angle The sequence leads to cθ 2 cθ 3 sθ 1 sθ 2 cθ 3 +cθ 1 sθ 3 sθ 1 sθ 3 cθ 1 sθ 2 cθ 3 R bi = cθ 2 sθ 3 cθ 1 cθ 3 sθ 1 2θ 2 sθ 3 sθ 1 cθ 3 +cθ 1 sθ 2 sθ 3 sθ 2 sθ 1 cθ 2 cθ 1 cθ 2 where θ 1 is the roll angle, θ 2 is the pitch angle, and θ 3 is the yaw angle

30 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 Sequence Sequence cos θ 1andsinθ θ 1 θ 1 θ 2 R bi θ 1 1 θ 3 θ 2 θ 3 1 cos θ 1andsinθ θ 1 θ 3 θ 2 R bi θ 3 1 θ 1 θ 2 θ 1 1 where θ 1 is the yaw angle, θ 2 is the where θ 1 is the roll angle, θ 2 is the pitch angle, and θ 3 is the roll angle pitch angle, and θ 3 is the yaw angle θ 3 roll R bi 1 θ 2 R bi 1 pitch R bi 1 θ θ 1 yaw

31 Perifocal Frame Earth-centered, orbit-based, inertial The P-axis is in periapsis direction The W-axis is perpendicular to orbital plane (direction of orbit angular momentum vector, r v ) The Q-axis is in the orbital plane and finishes the triad of unit vectors In the perifocal frame, the position and velocity vectors both have a zero component in the W direction Qˆ Qˆ Ŵ Pˆ Pˆ

32 A One-Minute Course on Orbital Mechanics Equatorial plane ^ K ν ω ^ I Orbital plane Ω ^n ^ J i Orbit is defined by 6 orbital elements (oe s): semimajor axis, a; eccentricity, e; inclination, i; right ascension of ascending node, Ω; argument of periapsis, ω; and true anomaly, ν

33 Inertial Frame to Perifocal Frame Use a sequence Right ascension of the ascending node Ω about ˆK: R 3 (Ω), rotates Î to ˆn (the ascending node vector) R 3 (Ω) = cos Ω sin Ω 0 sin Ω cos Ω Inclination i about the node vector: R 1 (i), rotates ˆK to Ŵ (the orbit normal direction) R 1 (i) = cosi sin i 0 sin i cos i Argument of periapsis ω about the orbit normal: R 3 (ω), rotates ˆn to ˆP (the periapsis direction) R 3 (ω) = cos ω sin ω 0 sin ω cos ω R pi = R 3 (ω)r 1 (i)r 3 (Ω) = cωcω ci sω sω ci cω sω +cωsω si sω sω cω ci cω sω ci cω cω sω sω si cω si sω si cω ci

34 Orbital Reference Frame Same as roll-pitch-yaw frame, for spacecraft The o 3 axis is in the nadir direction The o 2 axis is in the negative orbit normal direction The o 1 axis completes the triad, and is in the velocity vector direction for circular orbits In the orbital frame, position and velocity both have zero in the o 2 direction, and position has zero in the o 1 direction as well You will find the rotation from perifocal to orbital easier to visualize if you make yourself two reference frames v r ô 1 ô 3 ô 2 ŵ

35 Perifocal Frame to Orbital Frame Use a sequence 270 about ˆQ: R 2 (270 ), rotates Ŵ to ˆP (the nadir direction) R 2 (270 ) = about the nadir vector: R 3 (90 ), rotates ô 2 to Ŵ (the negative orbit normal direction) R 3 (90 ) = Negative true anomaly ν about ô 2 : R 2 ( ν), rotates ô 3 to the transverse direction R 2 ( ν) = cos ν 0 sinν sin ν 0 cosν R op = R 2 ( ν)r 3 (90 )R 2 (270 )= sν cν cν sν 0

36 Inertial to Perifocal to Orbital R pi = R 3 (ω)r 1 (i)r 3 (Ω) = R op = R 2 ( ν)r 3 (90 )R 2 (270 )= cωcω ci sω sω ci cω sω +cωsω si sω sω cω ci cω sω ci cω cω sω sω si cω si sω si cω ci sν cν cν sν 0 R oi = R op R pi su cω cu ci sω su sω +cucicω cu si = si sω si cω ci cu cω +sucisω cu sω su ci cω su si where u = ω + ν

37 An Illustrative Example In an inertial reference frame, an Earth-orbiting satellite has position and 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= ,i=23.986, Ω =46.469, ω =321.80, ν = We can use i, Ω, andω to compute R pi as R pi = Then we can rotate position and velocity into F p : r p = [ ] T v p = [ ] T

38 An Illustrative Example (continued) We can use ν = to compute R op as R op = WecanthenmultiplyR op R pi to get R oi, or we can use i, Ω, and u = ω + ν to compute R oi : R oi = Then we can rotate position and velocity into F o : r o = [ , 689] T v o = [ ] T

39 Differential Equations of Kinematics Given the velocity of a point and initial conditions for its position, we can compute its position as a function of time by integrating the differential equation ~r = ~v We now need to develop the equivalent differential equations for the attitude when the angular velocity is known Preview: θ = 0 sin θ 3 / cos θ 2 cos θ 3 / cos θ 2 0 cosθ 3 sin θ 3 1 sinθ 3 sin θ 2 / cos θ 2 cos θ 3 sin θ 2 / cos θ 2 ω 1 ω 2 ω 3 = S 1 ω d/dt(attitude) = angular momentum/inertia

40 Euler Angles and Angular Velocity One frame-to-frame at a time, just as we did for developing rotation matrices rotation from F i to F i 0 to F i 00 to F b 3-rotation from F i to F i 0 about î3 î0 3 through θ 1 The angular velocity of F i 0 with respect to F i is ~ω i0i = θ 1 î 3 = θ 1 î 0 3 We can express ~ω i0i in any frame, but F i and F i 0 are especially simple: ω i0 i i = [0 0 θ 1 ] T ω i0 i i 0 = [0 0 θ 1 ] T Keep the notation in mind: ω i0 i i is the angular velocity of F i 0 with respect to F i,expressedinf i

41 Euler Angles and Angular Velocity (2) 2-rotation from F i 0 to F i 00 about î0 2 î00 2 through θ 2 The angular velocity of F i 00 with respect to F i 0 is ~ω i00 i 0 = θ 2 î 0 2 = θ 2 î 00 2 We can express ~ω i00 i 0 in any frame, but F i 0 and F i 00 are especially simple: ω i00 i 0 i 0 = [0 θ 2 0] T ω i00 i 0 i 00 = [0 θ 2 0] T Keep the notation in mind: ω i00 i 0 i 0 is the angular velocity of F i 00 with respect to F i 0, expressed in F i 0

42 Euler Angles and Angular Velocity (3) 1-rotation from F i 00 to F b about î00 1 ˆb 1 through θ 3 The angular velocity of F b with respect to F i 00 is ~ω bi00 = θ 3 î 00 1 = θ 3ˆb1 We can express ~ω bi00 in any frame, but F i 00 and F b are especially simple: ω bi00 i 00 = [ θ 3 0 0] T ω bi00 b = [ θ 3 0 0] T Keep the notation in mind: ω bi00 b respect to F i 00, expressedinf b is the angular velocity of F b with

43 Adding the Angular Velocities Angular velocities are vectors and add like vectors: ~ω bi = ~ω bi00 + ~ω i00 i 0 + ~ω i0 i We have expressed these three vectors in different frames; to add them together, we need to express all of them in the same frame Typically, we want ω bi b, so we need to rotate the 3 1 matriceswejust developed into F b We have ω i0 i i = ω i0 i i 0 = [0 0 θ 1 ] T need R bi0 ω i00 i 0 i 0 = ω i00 i 0 i 00 = [0 θ 2 0] T need R bi00 ω bi00 i 00 = ω bi00 b = [ θ 3 0 0] T need R bb = 1 We previously developed all these rotation matrices, so we just need to apply them and add the results

44 Complete the Operation Carry out the matrix multiplications and additions ω bi b = ωbi00 b + ω i00 i 0 b + ω i0 i b and obtain ω bi b = = θ 3 sin θ 2 θ1 cos θ 3 θ2 +cosθ 2 sin θ 3 θ1 sin θ 3 θ2 +cosθ 2 cos θ 3 θ1 sin θ cos θ 2 sin θ 3 cos θ 3 0 cos θ 2 cos θ 3 sin θ 3 0 θ 1 θ 2 θ 3 = S(θ) θ or θ = S 1 (θ)ω = What happens when θ 2 nπ/2, for odd n? 0 sin θ 3/ cos θ 2 cos θ 3 / cos θ 2 0 cosθ 3 sin θ 3 1 sin θ 3 sin θ 2 / cos θ 2 cos θ 3 sin θ 2 / cos θ 2 What happens when the Euler angles and their rates are small? ω 1 ω 2 ω 3

45 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 θ 2 0 The reason is that near θ 2 = π/2 the first and third rotations are indistinguishable For the symmetric Euler angle sequences (3-1-3, , 1-3-1, etc) the singularity occurs when θ 2 = 0 or π For the asymmetric Euler angle sequences (3-2-1, 2-3-1, 1-3-2, etc) the singularity occurs when θ 2 = π/2 or 3π/2 This kinematic singularity is a major disadvantage of using Euler angles for large-angle motion There are attitude representations that do not have a kinematic singularity, but 4 or more scalars are required

46 Linearizing the Kinematics Equation ω bi b = S(θ) θ θ 3 sin θ 2 θ1 = cos θ 3 θ2 +cosθ 2 sin θ 3 θ1 sin θ 3 θ2 +cosθ 2 cos θ 3 θ1 sin θ = cos θ 2 sin θ 3 cos θ 3 0 cos θ 2 cos θ 3 sin θ 3 0 θ 1 θ 2 θ 3 If the Euler angles and rates are small, then sin θ i θ i and cos θ i 1: ω bi b θ 3 θ 2 θ Exercise: Repeat these steps for a sequence and for a symmetric sequence. θ 1 θ 2 θ 3

47 Refresher: How To Invert a Matrix Suppose you want to invert the n n matrix A, with elements A ij Theelementsoftheinverseare A 1 ij = C ji A where C ji is the cofactor, computed by multiplying the determinant of the (n 1) (n 1) minor matrix obtained by deleting the j th row and i th column from A, by( 1) i+j This formulation is absolutely unsuitable for calculating matrix inverses in numerical work, especially with larger matrices, since it is computationally 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 the solution to a linear system

48 Matrix Inversion Application Let us invert the matrix S(θ) = sin θ cos θ 2 sin θ 3 cos θ 3 0 cos θ 2 cos θ 3 sin θ 3 0 In row reduction, we augment the matrix with the identity matrix: sin θ cos θ 2 sin θ 3 cos θ cos θ 2 cos θ 3 sin θ and apply simply row-reduction operations to transform the left 3 3 block to the identity, leaving the inverse in the right 3 3block In this case, we must swap the first row with one of the other two rows, say the 3 rd row, which amounts to a permutation by : P = Note that P 1 = P 1 0 0

49 Matrix Inversion Application (2) cos θ 2 cos θ 3 sin θ cos θ 2 sin θ 3 cos θ sin θ Multiply row 2 by sin θ 3 / cos θ 3 and add to row 1. Multiply row 1 by sin θ 3 / cos θ 3 and add to row 2. cθ 2 cθ 3 + cθ 2 sθ 3 tan θ tanθ cθ 3 + sθ 3 tan θ tan θ 3 sθ Divide row 1 by cθ 2 cθ 3 + cθ 2 sθ 3 tan θ 3 and simplify Multiply resulting row 1 by sθ 2, add to row 3, and simplify Divide row 2 by cθ 3 + sθ 3 tan θ 3 and simplify sθ 3/cθ 2 cθ 3 /cθ cθ 3 sθ sθ 3 tan θ 2 cθ 3 tan θ 2

50 Euler s Theorem The most general motion of a rigid body with a fixed point is a rotation about a fixed axis. The axis, denoted a, iscalledtheeigenaxis or Euler axis The angle of rotation, Φ, iscalledtheeuler angle or the principal Euler angle Here the black axes are thebasevectorsoftheinertial frame, and the red, blue, and yellow axes are thebasevectorsofabody frame, rotated about a =[ 2/2 2/2 2/2] T through angle Φ = π/ Why no subscript on a?

51 Euler s Theorem (2) Anyrotationmatrixcanbeexpressedintermsofa and Φ: R =cosφ1 +(1 cos Φ)aa T sin Φa Since a is an eigenvector of R with eigenvalue 1, Check this result: Ra = a Ra = cos Φ1 +(1 cos Φ)aa T sin Φa a = cosφ1a +(1 cos Φ)aa T a sin Φa a = cosφa + aa T a cos Φaa T a sin Φa a = cosφa + a cos Φa sin Φa a (a T a =1) = cosφa + a cos Φa (a a = 0) = a Also note that the trace of R is simply tracer =1+2cosΦ

52 Extracting a and Φ from R, and Integrating to Obtain a and Φ Given any rotation matrix we can compute the Euler axis a and angle Φ using: 1 Φ = cos 1 (trace R 1) 2 a = 1 2 sin Φ R T R What does one do about the sin Φ =0case? Onecanshowthatthekinematicsdifferential equations for a and Φ are: Φ = a T ω ȧ = 1 a cot Φ2 2 a a ω So, this system of equations also has kinematic singularities, at Φ = 0 and Φ =2π

53 Quaternions (aka Euler Parameters) Define another 4-parameter set of variables to represent the attitude: q = a sin Φ 2 q 4 = cos Φ 2 The 3 1 matrix q forms the Euler axis component of the quaternion, also called the vector component. The scalar q 4 is called the scalar component. Collectively, these four variables are known as a quaternion, orasthe Euler parameters. The notation q denotes the 4 1 matrix containing all four variables: q = q T q 4

54 Quaternions (2) R( q) and q(r) : R = q4 2 q T q 1 +2qq T 2q 4 q q 4 = ± traceR R 1 23 R 32 q = R 31 R 13 4q 4 R 12 R 21 Differential equations q : q = 1 2 q + q 4 1 q T ω = Q( q)ω Note that there is no kinematic singularity with these differential equations

55 Typical Problem Involving Angular Velocity and Attitude Given initial conditions for the attitude (in any form), and a time history of angular 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

56 Spherical Pendulum Problem Use a rotation from F i to F r using two Euler angles, θ, andφ The first (3) rotation is about the î 3 axis, through angle θ The second (2) rotation is about the ê θ ˆθ axis through φ The three unit vectors have derivatives: ê r =sinφ θê θ + φê φ, ê θ = sin φ θê r cos φ θê φ, ê φ = φêr +cosφ θê θ Illustration from Wolfram Research Mathworld mathworld.wolfram.com/sphericalcoordinates.html

57 Linearization for Small Φ As with Euler angles, we are frequently interested in small attitude motions. If Φ is small, thenq = a sin(φ/2) a Φ 2,andq 4 1 Hence, for small Φ, R = q 2 4 q T q 1 +2qq T 2q 4 q (1 0) 1 +2(0) 2q 1 2q Compare this expression with the previously developed R 1 θ for Euler angles. What is the equivalent expression for the (a, Φ) representation? Small rotations are commutative: R cb R ba 1 2q 2 1 2q 1 = 1 2q 2 2q 1

58 Other Attitude Representations We have seen direction cosines, Euler angles, Euler angle/axis, and quaternions Two other common representations Euler-Rodriguez parameters p = a tan Φ 2 R = p T p (p p p ) ṗ = 1 2 (ppt p )ω Modified Rodriguez parameters σ = a tan Φ 4 1 R = (1 (σ T σ) 2 )1 +2σσ T 2(1 (σ T σ) 2 )σ 1+σ T σ σ = 1 1 σ + σσ T 1+σT σ 1 2 2

59 Typical Problem Involving Angular Velocity and Attitude Given initial conditions for the attitude (in any form), and a time history of angular 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