2 Rotation Kinematics

Transcription

1 X P x P r P 2 Rotation Kinematics Consider a rigid body with a fixed point. Rotation about the fixed point is the only possible motion of the body. We represent the rigid body by a body coordinate frame B, that rotates in another coordinate frame, as is shown in Figure 2.1. We develop a rotation calculus based on transformation matrices to determine the orientation of B in, and relate the coordinates of a body point P in both frames. Z B z Z P z P y P y Y P X x Y FIURE 2.1. A rotated body frame B in a fixed global frame, aboutafixed point at O. 2.1 Rotation About lobal Cartesian Axes Consider a rigid body B with a local coordinate frame Oxyz that is originally coincident with a global coordinate frame OXYZ.Point O of the body B is fixed to the ground and is the origin of both coordinate frames. If the rigid body B rotates α degrees about the Z-axis of the global coordinate frame, then coordinates of any point P of the rigid body in the local and global coordinate frames are related by the following equation r = Q Z,α B r (2.1) R.N. Jazar, Theory of Applied Robotics, 2nd ed., DOI / _2, Springer Science+Business Media, LLC 2010

2 34 2. Rotation Kinematics where, r = X Y Z B r= x y z (2.2) and Q Z,α = cos α sin α 0 sin α cos α (2.3) Similarly, rotation β degrees about the Y -axis, and γ degrees about the X-axis of the global frame relate the local and global coordinates of point P by the following equations r = Q Y,β B r (2.4) r = Q X,γ B r (2.5) where, Q Y,β = cos β 0 sinβ sin β 0 cosβ (2.6) Q X,γ = cosγ sin γ. (2.7) 0 sinγ cos γ Proof. Let (î, ĵ, ˆk) and (Î,Ĵ, ˆK) be the unit vectors along the coordinate axes of Oxyz and OXY Z respectively. The rigid body has a space fixed point O, that is the common origin of Oxyz and OXY Z. Figure 2.2 illustrates the top view of the system. The initial position of a point P is indicated by P 1. The position vector r 1 of P 1 can be expressed in body and global coordinate frames by B r 1 = x 1 î + y 1 ĵ + z 1ˆk (2.8) r 1 = X 1 Î + Y 1 Ĵ + Z 1 ˆK (2.9) x 1 = X 1 y 1 = Y 1 z 1 = Z 1 (2.10) where B r 1 refers to the position vector r 1 expressed in the body coordinate frame B, and r 1 refers to the position vector r 1 expressed in the global coordinate frame. If the rigid body undergoes a rotation α about the Z-axis then, the local frame Oxyz, pointp, and the position vector r will be seen in a

3 2. Rotation Kinematics 35 Y y B y 1 =Y 1 P 1 r 1 x 1 =X 1 x X FIURE 2.2. Position vector of point P when local and global frames are coincident. second position, as shown in Figure 2.3. Now the position vector r 2 of P 2 is expressed in both coordinate frames by B r 2 = x 2 î + y 2 ĵ + z 2ˆk (2.11) r 2 = X 2 Î + Y 2 Ĵ + Z 2 ˆK. (2.12) Using the definition of the inner product and Equation (2.11) we may write X 2 = Î r 2 = Î x 2î + Î y 2ĵ + Î z 2ˆk (2.13) Y 2 = Ĵ r 2 = Ĵ x 2î + Ĵ y 2ĵ + Ĵ z 2ˆk (2.14) Z 2 = ˆK r 2 = ˆK x 2 î + ˆK y 2 ĵ + ˆK z 2ˆk (2.15) or equivalently X 2 Y 2 Z 2 = Î î Î ĵ Î ˆk Ĵ î Ĵ ĵ Ĵ ˆk ˆK î ˆK ĵ ˆK ˆk x 2 y 2. (2.16) z 2 The elements of the Z-rotation matrix, Q Z,α, are called the direction cosines of B r 2 with respect to OXY Z. Figure 2.4 shows the top view of the initial and final configurations of r in both coordinate systems Oxyz and OXY Z. Analyzing Figure 2.4 indicates that Î î =cosα, Î ĵ = sin α, Î ˆk =0 Ĵ î =sinα, Ĵ ĵ =cosα, Ĵ ˆk =0 (2.17) ˆK î =0, ˆK ĵ =0, ˆK ˆk =1.

4 36 2. Rotation Kinematics Y B y Y 2 P 2 y 2 = y 1 r 2 x x 2 = x 1 α X 2 X FIURE 2.3. Position vector of point P when local frames are rotated about the Z-axis. Combining Equations (2.16) and (2.17) shows that we can find the components of r 2 by multiplying the Z-rotation matrix Q Z,α and the vector B r 2. X 2 Y 2 Z 2 = cos α sin α 0 sin α cos α It can also be shown in the following short notation x 2 y 2 z 2. (2.18) r 2 = Q Z,α B r 2 (2.19) where Q Z,α = cos α sin α 0 sin α cos α (2.20) Equation (2.19) says that the vector r at the second position in the global coordinate frame is equal to Q Z times the position vector in the local coordinate frame. Hence, we are able to find the global coordinates of a point of a rigid body after rotation about the Z-axis, if we have its local coordinates. Similarly, rotation β about the Y -axis and rotation γ about the X-axis are described by the Y -rotation matrix Q Y,β and the X-rotation matrix Q X,γ respectively. Q Y,β = cos β 0 sinβ sin β 0 cosβ (2.21)

5 2. Rotation Kinematics 37 B Y y Y 2 P 2 y 2 = y 1 Y 1 r 2 r 1 P 1 x x 2 = x 1 α X 2 X 1 X FIURE 2.4. Position vectors of point P before and after the rotation of the local frame about the Z-axis of the global frame. Q X,γ = cosγ sin γ (2.22) 0 sinγ cos γ The rotation matrices Q Z,α, Q Y,β,andQ X,γ are called basic global rotation matrices. We usually refer to the first, second and third rotations about the axes of the global coordinate frame by α, β, and γ respectively. Example 3 Successive rotation about global axes. The final position of the corner P (5, 30, 10) oftheslabshowninfigure 2.5 after 30 deg rotation about the Z-axis, followed by 30 deg about the X- axis, and then 90 deg about the Y -axis can be found by first multiplying Q Z,30 by [5, 30, 10] T to get the new global position after first rotation X 2 Y 2 = Z 2 cos 30 sin 30 0 sin 30 cos = (2.23) and then multiplying Q X,30 and [ 10.68, 28.48, 10.0] T to get the position of P after the second rotation X 3 Y 3 Z 3 = cos 30 sin 30 0 sin30 cos = (2.24) and finally multiplying Q Y,90 and [ 10.68, 19.66, 22.9] T to get the final position of P after the third rotation. The slab and the point P in first, second,

6 38 2. Rotation Kinematics Z X 30 P Y FIURE 2.5. Corner P of the slab at first position. third, and fourth positions are shown in Figure 2.6. X 4 cos 90 0 sin 90 Y 4 = = Z 4 sin 90 0 cos (2.25) Example 4 Time dependent global rotation. Consider a rigid body B that is continuously turning about the Y -axis of at a rate of 0.3rad/ s. The rotation transformation matrix of the body is: Q B = cos 0.3t 0 sin0.3t sin 0.3t 0 cos 0.3t (2.26) Any point of B willmoveonacirclewithradiusr = X 2 + Z 2 parallel to (X, Z)-plane. X Y = Z = cos 0.3t 0 sin0.3t sin 0.3t 0 cos 0.3t x cos 0.3t + z sin 0.3t y z cos 0.3t x sin 0.3t x y z (2.27) X 2 + Z 2 = (x cos 0.3t + z sin 0.3t) 2 +(z cos 0.3t x sin 0.3t) 2 = x 2 + z 2 = R 2 (2.28)

7 2. Rotation Kinematics 39 Z P 3 P 4 P 1 P 2 X Y FIURE 2.6. Corner P and the slab at first, second, third, and final positions. Consider a point P at B r = T.Aftert =1s, the point will be seen at: X cos sin0.3 Y = = (2.29) Z sin cos and after t =2s,at: X cos 0.6 sin Y = sin 0.6 cos Z = (2.30) 0 We can find the global velocity of the body point P by taking a time derivative of r P = Q B Y,β r P (2.31) cos 0.3t 0 sin0.3t Q Y,β = (2.32) sin 0.3t 0 cos 0.3t Therefore, the global expression of its velocity vector is: z cos 0.3t x sin 0.3t v P = Q B Y,β r P =0.3 0 (2.33) x cos 0.3t z sin 0.3t

8 40 2. Rotation Kinematics z 2 Z P 1 P 2 y2 X B x 2 Y FIURE 2.7. Positions of point P in Example 5 before and after rotation. Example 5 lobal rotation, local position. If a point P is moved to r 2 =[4, 3, 2] T after a 60 deg rotation about the Z-axis, its position in the local coordinate is: B r 2 = Q 1 Z,60 r 2 (2.34) x 1 2 cos 60 sin 60 0 y 2 = sin 60 cos = z The local coordinate frame was coincident with the global coordinate frame before rotation, thus the global coordinates of P before rotation was also r 1 = [4.60, 1.95, 2.0] T. Positions of P before and after rotation are showninfigure Successive Rotation About lobal Cartesian Axes The final global position of a point P in a rigid body B with position vector r, after a sequence of rotations Q 1, Q 2, Q 3,..., Q n about the global axes can be found by r = Q B B r (2.35) where, Q B = Q n Q 3 Q 2 Q 1 (2.36)

9 2. Rotation Kinematics 41 and, r and B r indicate the position vector r in the global and local coordinate frames. Q B is called the global rotation matrix. It maps the local coordinates to their corresponding global coordinates. Since matrix multiplications do not commute the sequence of performing rotations is important. A rotation matrix is orthogonal; i.e., its transpose Q T is equal to its inverse Q 1. Q T = Q 1 (2.37) Rotation about global coordinate axes is conceptually simple because the axes of rotations are fixed in space. Assume we have the coordinates of every point of a rigid body in the global frame that is equal to the local coordinates initially. The rigid body rotates about a global axis, then the proper global rotation matrix gives us the new global coordinate of the points. When we find the coordinates of points of the rigid body after the first rotation, our situation before the second rotation is similar to what we had before the first rotation. Example 6 Successive global rotation matrix. The global rotation matrix after a rotation Q Z,α followed by Q Y,β and then Q X,γ is: Q B = Q X,γ Q Y,β Q Z,α (2.38) cαcβ cβsα sβ = cγsα + cαsβsγ cαcγ sαsβsγ cβsγ sαsγ cαcγsβ cαsγ + cγsαsβ cβcγ Example 7 Successive global rotations, global position. The end point P of the arm shown in Figure 2.8 is located at: X Y 1 = l cos θ = 1 cos 75 = (2.39) Z 1 l sin θ 1sin The rotation matrix to find the new position of the end point after 29 deg rotation about the X-axis, followed by 30 deg about the Z-axis, and again 132 deg about the X-axis is Q B = Q X,132 Q Z,30 Q X, 29 = and its new position is at: X Y 2 = Z (2.40) = (2.41)

10 42 2. Rotation Kinematics Y 1.0 P 75 Z X FIURE 2.8. The arm of Example 7. Example 8 Twelve independent triple global rotations. Consider a rigid body in the final orientation after a series of rotation about global axes. We may transform its body coordinate frame B from the coincident position with a global frame to any final orientation by only three rotations about the global axes provided that no two consequence rotations are about the same axis. In general, there are 12 different independent combinations of triple rotations about the global axes. They are: 1 Q X,γ Q Y,β Q Z,α 2 Q Y,γ Q Z,β Q X,α 3 Q Z,γ Q X,β Q Y,α 4 Q Z,γ Q Y,β Q X,α 5 Q Y,γ Q X,β Q Z,α 6 Q X,γ Q Z,β Q Y,α 7 Q X,γ Q Y,β Q X,α 8 Q Y,γ Q Z,β Q Y,α 9 Q Z,γ Q X,β Q Z,α 10 Q X,γ Q Z,β Q X,α 11 Q Y,γ Q X,β Q Y,α 12 Q Z,γ Q Y,β Q Z,α (2.42) The expanded form of the 12 global axes triple rotations are presented in Appendix A. Example 9 Order of rotation, and order of matrix multiplication. Changing the order of global rotation matrices is equivalent to changing the order of rotations. The position of a point P of a rigid body B is located at B r P = T. Its global position after rotation 30 deg about X-

11 2. Rotation Kinematics 43 axis and then 45 deg about Y -axis is at r P = Q 1 Y,45 Q B X,30 r P (2.43) = = and if we change the order of rotations then its position would be at: r P = Q 2 X,30 Q B Y,45 r P (2.44) = = These two final positions of P are d = r P 1 r P 2 =4.456 apart. Example 10 F Repeated rotation about global axes. If we turn a body frame B about X-axis γ rad, where, α = 2π n n N (2.45) then we need to repeat the rotation n times to turn the body back to its original configuration. We can check it by multiplying Q X,α by itself until we achieve an identity matrix. So, any body point of B will be mapped to the same point in global frame. To show this, we may find that Q X,α to the power m as: m Q m X,α = m 0 cosγ sin γ = 0 cos 2π sin 2π 0 sinγ cos γ n n 0 sin 2π cos 2π n n = 0 cosm 2π sin m 2π n n (2.46) 0 sinm 2π cos m 2π n n If m = n, then we have an identity matrix Q n X,α = 0 cosn 2π sin n 2π n n 0 sinn 2π cos n 2π = (2.47) n n Repeated rotation about any other global axis provides the same result.

12 44 2. Rotation Kinematics Let us now rotate B about two global axes repeatedly, such as turning α about Z-axis followed by a rotation γ about X-axis, such that α = 2π n 1 γ = 2π n 2 {n 1,n 2 } N. (2.48) We may guess that repeating the rotations n = n 1 n 2 times will turn B back to its original configuration. [Q X,γ Q Z,α ] n1 n2 =[I] (2.49) As an example consider α = 2π 3 and γ = 2π 4.Weneed13 times combined rotation to achieve the original configuration Q B = Q X,γ Q Z,α = (2.50) Q 13 B = I (2.51) We may turn B back to its original configurationbylowernumberof combined rotations if n 1 and n 2 have a common divisor. For example if n 1 = n 2 =4, we only need to apply the combined rotation three times. In a general case, determination of the required number n to repeat a general combined rotation Q B to turn back to the original orientation is an unsolved question. my Q B = Q Xi,α j i =1, 2, 3 (2.52) j=1 α j = 2π n j m, n j N (2.53) Q n B = [I] n =? (2.54) 2.3 lobal Roll-Pitch-Yaw Angles The rotation about the X-axis of the global coordinate frame is called a roll, the rotation about the Y -axis of the global coordinate frame is called a pitch, and the rotation about the Z-axis of the global coordinate frame is called a yaw. The global roll-pitch-yaw rotation matrix is: Q B = Q Z,γ Q Y,β Q X,α (2.55) cβcγ cαsγ + cγsαsβ sαsγ + cαcγsβ = cβsγ cαcγ + sαsβsγ cγsα + cαsβsγ sβ cβsα cαcβ

13 2. Rotation Kinematics 45 Z Z Z X Y X Y X Y Roll Pitch Yaw FIURE 2.9. lobal roll, pitch, and yaw rotations. Figures 2.9 illustrates 45 deg roll, pitch, and yaw rotations about the axes of a global coordinate frame. iven the roll, pitch, and yaw angles, we can compute the overall rotation matrix using Equation (2.55). Also we are able to compute the equivalent roll, pitch, and yaw angles when a rotation matrix is given. Suppose that r ij indicates the element of row i and column j of the roll-pitch-yaw rotation matrix (2.55), then the roll angle is µ α =tan 1 r32 (2.56) r 33 and the pitch angle is β = sin 1 (r 31 ) (2.57) and the yaw angle is provided that cos β 6= 0. µ γ =tan 1 r21 r 11 (2.58) Example 11 Determination of roll-pitch-yaw angles. Let us determine the required roll-pitch-yaw angles to make the x-axis of the body coordinate B parallel to u, while y-axis remains in (X, Y )-plane. Because x-axis must be along u, we have u = Î +2Ĵ +3ˆK (2.59) î = u u = 1 14 Î Ĵ ˆK (2.60) and because y-axisisin(x, Y )-plane, we have ĵ = ³Î ĵ Î + ³Ĵ ĵ Ĵ =cosθ Î +sinθĵ. (2.61)

14 46 2. Rotation Kinematics The axes î and ĵ must be orthogonal, therefore, 1/ 14 2/ 14 3/ cos θ sin θ 14 0 = 0 (2.62) θ = deg. (2.63) We may find ˆk by a cross product. ˆk = î 1/ 14 ĵ = 2/ 14 3/ = Hence, the transformation matrix Q B is: Î î Î ĵ Î ˆk 1/ Q B = Ĵ î Ĵ ĵ Ĵ ˆk = 2/ ˆK î ˆK ĵ ˆK ˆk 3/ (2.64) (2.65) Now it is possible to determine the required roll-pitch-yaw angles to move the body coordinate frame B from the coincidence orientation with to the final orientation. µ µ α = tan 1 r32 0 =tan 1 =0 (2.66) r 33 ³ β = sin 1 (r 31 )= sin 1 3/ rad (2.67) µ Ã γ = tan 1 r21 =tan 1 2/! 14 r 11 1/ rad (2.68) Rotation About Local Cartesian Axes Consider a rigid body B with a space fixed point at O. Thelocalbody coordinate frame B(Oxyz) is coincident with a global coordinate frame (OXY Z), where the origin of both frames are on the fixed point O. Ifthe body undergoes a rotation ϕ about the z-axis of its local coordinate frame, as can be seen in the top view shown in Figure 2.10, then coordinates of any point of the rigid body in local and global coordinate frames are related by the following equation B r = A z,ϕ r. (2.69) The vectors r and B r are the position vectors of the point in global and local frames respectively r = X Y Z T B r = x y z T (2.70) (2.71)

15 2. Rotation Kinematics 47 and A z,ϕ is the z-rotation matrix. cos ϕ sin ϕ 0 A z,ϕ = sin ϕ cos ϕ 0 (2.72) Similarly, rotation θ about the y-axis and rotation ψ about the x-axis are described by the y-rotation matrix A y,θ and the x-rotation matrix A x,ψ respectively. A y,θ = A x,ψ = cos θ 0 sin θ sin θ 0 cosθ cosψ sin ψ 0 sin ψ cos ψ (2.73) (2.74) Proof. Vector r indicates the position of a point P of the rigid body B where it is initially at P 1. Using the unit vectors (î, ĵ, ˆk) along the axes of local coordinate frame B(Oxyz), and(î,ĵ, ˆK) along the axes of global coordinate frame (OXY Z), the initial and final position vectors r 1 and r 2 in both coordinate frames can be expressed by B r 1 = x 1 î + y 1 ĵ + z 1ˆk (2.75) r 1 = X 1 Î + Y 1 Ĵ + Z 1 ˆK (2.76) B r 2 = x 2 î + y 2 ĵ + z 2ˆk (2.77) r 2 = X 2 Î + Y 2 Ĵ + Z 2 ˆK. (2.78) The vectors B r 1 and B r 2 are the initial and final positions of the vector r expressed in body coordinate frame B(Oxyz), and r 1 and r 2 are the initial and final positions of the vector r expressed in the global coordinate frame (OXY Z). The components of B r 2 canbefoundifwehavethecomponentsof r 2. Using Equation (2.78) and the definition of the inner product, we may write or equivalently x 2 y 2 z 2 x 2 = î r 2 =î X 2 Î +î Y 2 Ĵ +î Z 2 ˆK (2.79) y 2 = ĵ r 2 =ĵ X 2 Î +ĵ Y 2 Ĵ +ĵ Z 2 ˆK (2.80) z 2 = ˆk r 2 = ˆk X 2 Î + ˆk Y 2 Ĵ + ˆk Z 2 ˆK (2.81) = î Î î Ĵ î ˆK ĵ Î ĵ Ĵ ĵ ˆK ˆk Î ˆk Ĵ ˆk ˆK X 2 Y 2. (2.82) Z 2

16 48 2. Rotation Kinematics B Y y Y 2 P 2 y 2 = Y 1 Y 1 r 2 r 1 P 1 x X 2 x 2 = X 1 X 1 ϕ X FIURE Position vectors of point P before and after rotation of the local frame about the z-axis of the local frame. The elements of the z-rotation matrix A z,ϕ are the direction cosines of r 2 with respect to Oxyz. So, the elements of the matrix in Equation (2.82) are: î Î =cosϕ, î Ĵ =sinϕ, î ˆK =0 ĵ Î = sin ϕ, ĵ Ĵ =cosϕ, ĵ ˆK =0 (2.83) ˆk Î =0, ˆk Ĵ =0, ˆk ˆK =1 Combining Equations (2.82) and (2.83), we can find the components of B r 2 by multiplying z-rotation matrix A z,ϕ and vector r 2. x 2 cos ϕ sin ϕ 0 y 2 = sin ϕ cos ϕ 0 X 2 Y 2 (2.84) z Z 2 It can also be shown in the following short form where, A z,ϕ = B r 2 = A z,ϕ r 2 (2.85) cos ϕ sin ϕ 0 sin ϕ cos ϕ (2.86) Equation (2.85) says that after rotation about the z-axis of the local coordinate frame, the position vector in the local frame is equal to A z,ϕ times the position vector in the global frame. Hence, after rotation about the z-axis,weareabletofind the coordinates of any point of a rigid body in local coordinate frame, if we have its coordinates in the global frame. Similarly, rotation θ about the y-axis and rotation ψ about the x-axis are described by the y-rotation matrix A y,θ and the x-rotation matrix A x,ψ respectively.

17 2. Rotation Kinematics 49 A y,θ = cos θ 0 sin θ sin θ 0 cosθ (2.87) A x,ψ = cosψ sin ψ (2.88) 0 sin ψ cos ψ We indicate the first, second, and third rotations about the local axes by ϕ, θ, andψ respectively. Example 12 Local rotation, local position. If a local coordinate frame Oxyz has been rotated 60 deg about the z- axis and a point P in the global coordinate frame OXY Z is at (4, 3, 2), its coordinates in the local coordinate frame Oxyz are: x y = z cos 60 sin 60 0 sin 60 cos = (2.89) Example 13 Local rotation, global position. If a local coordinate frame Oxyz has been rotated 60 deg about the z-axis and a point P in the local coordinate frame Oxyz is at (4, 3, 2), itsposition in the global coordinate frame OXY Z is at: X Y = Z cos 60 sin 60 0 sin 60 cos T = (2.90) 2.0 Example 14 Successive local rotation, global position. The arm shown in Figure 2.11 has two actuators. The first actuator rotates the arm 90 deg about y-axis and then the second actuator rotates the arm 90 deg about x-axis. If the end point P is at B r P = T (2.91) then its position in the global coordinate frame is at: r 2 = [A x,90 A y, 90 ] 1 B r P = A 1 y, 90 A 1 x,90 B r P = A T y, 90 A T x,90 B r P = (2.92) 9.5

18 50 2. Rotation Kinematics P B Z z X x y Y FIURE Arm of Example Successive Rotation About Local Cartesian Axes The final global position of a point P in a rigid body B with position vector r, aftersomerotationsa 1, A 2, A 3,..., A n about the local axes, can be found by B r = B A r (2.93) where, B A = A n A 3 A 2 A 1. (2.94) B A is called the local rotation matrix and it maps the global coordinates to their corresponding local coordinates. Rotation about the local coordinate axis is conceptually interesting because in a sequence of rotations, each rotation is about one of the axes of the local coordinate frame, which has been moved to its new global position during the last rotation. Assume that we have the coordinates of every point of a rigid body in a global coordinate frame. The rigid body and its local coordinate frame rotate about a local axis, then the proper local rotation matrix relates the new global coordinates of the points to the corresponding local coordinates. If we introduce an intermediate space-fixed frame coincident with thenewpositionofthebodycoordinateframe,wemaygivetherigidbody a second rotation about a local coordinate axis. Now another proper local rotation matrix relates the coordinates in the intermediate fixed frame to thecorrespondinglocalcoordinates.hence,thefinal global coordinates of the points must first be transformed to the intermediate fixed frame and second transformed to the original global axes.

19 2. Rotation Kinematics 51 Example 15 Successive local rotation, local position. A local coordinate frame B(Oxyz) that initially is coincident with a global coordinate frame (OXYZ) undergoes a rotation ϕ =30degabout the z-axis, then θ =30degabout the x-axis, and then ψ =30degabout the y-axis. The local coordinates of a point P located at X =5, Y =30, Z =10 can be found by x y z T T = Ay,ψ A x,θ A z,ϕ Thelocal rotation matrix is B A = A y,30 A x,30 A z,30 = and coordinates of P in the local frame is: x y = z = (2.95) (2.96) Example 16 Successive local rotation. The rotation matrix for a body point P (x, y, z) after rotation A z,ϕ followed by A x,θ and A y,ψ is: B A = A y,ψ A x,θ A z,ϕ cϕcψ sθsϕsψ cψsϕ + cϕsθsψ cθsψ = cθsϕ cθcϕ sθ (2.97) cϕsψ + sθcψsϕ sϕsψ cϕsθcψ cθcψ Example 17 Twelve independent triple local rotations. Euler proved that: Any two independent orthogonal coordinate frames with a common origin can be related by a sequence of three rotations about the local coordinate axes, where no two successive rotations may be about the same axis. In general, there are 12 different independent combinations of triple rotation about local axes. They are: 1 A x,ψ A y,θ A z,ϕ 2 A y,ψ A z,θ A x,ϕ 3 A z,ψ A x,θ A y,ϕ 4 A z,ψ A y,θ A x,ϕ 5 A y,ψ A x,θ A z,ϕ 6 A x,ψ A z,θ A y,ϕ (2.98) 7 A x,ψ A y,θ A x,ϕ 8 A y,ψ A z,θ A y,ϕ 9 A z,ψ A x,θ A z,ϕ 10 A x,ψ A z,θ A x,ϕ 11 A y,ψ A x,θ A y,ϕ 12 A z,ψ A y,θ A z,ϕ The expanded form of the 12 local axes triple rotation are presented in Appendix B.

20 52 2. Rotation Kinematics Z z' B' X ϕ x' y' Y 2.6 Euler Angles FIURE First Euler angle. The rotation about the Z-axis of the global coordinate is called precession, the rotation about the x-axis of the local coordinate is called nutation, and the rotation about the z-axis of the local coordinate is called spin. The precession-nutation-spin rotation angles are also called Euler angles. Euler angles rotation matrix has many application in rigid body kinematics. To find the Euler angles rotation matrix to go from the global frame (OXY Z) to the final body frame B(Oxyz), we employ a body frame B 0 (Ox 0 y 0 z 0 ) as shown in Figure 2.12 that before the first rotation coincides with the global frame. Let there be at first a rotation ϕ about the z 0 -axis. Because Z-axis and z 0 -axis are coincident, we have: B 0 r = B0 A r (2.99) cos ϕ sin ϕ 0 B 0 A = A z,ϕ = sin ϕ cos ϕ 0 (2.100) Next we consider the B 0 (Ox 0 y 0 z 0 ) frame as a new fixed global frame and introduce a new body frame B 00 (Ox 00 y 00 z 00 ). Before the second rotation, the two frames coincide. Then, we execute a θ rotation about x 00 -axisasshown in Figure The transformation between B 0 (Ox 0 y 0 z 0 ) and B 00 (Ox 00 y 00 z 00 ) is: B 00 r = B00 A B 0 B0 r (2.101) B 00 A B 0 = A x,θ = cosθ sin θ (2.102) 0 sin θ cos θ Finally we consider the B 00 (Ox 00 y 00 z 00 ) frame as a new fixed global frame and consider the final body frame B(Oxyz) to coincide with B 00 before the

21 2. Rotation Kinematics 53 z' B'' z" θ y" B' y' x' x" FIURE Second Euler angle. third rotation. We now execute a ψ rotation about the z 00 -axis as shown in Figure The transformation between B 00 (Ox 00 y 00 z 00 ) and B(Oxyz) is: B r = B A B 00 B00 r (2.103) cos ψ sin ψ 0 B A B 00 = A z,ψ = sin ψ cos ψ 0 (2.104) By the rule of composition of rotations, the transformation from (OXY Z) to B(Oxyz) is B r = B A r (2.105) where, B A = A z,ψ A x,θ A z,ϕ cϕcψ cθsϕsψ cψsϕ + cθcϕsψ sθsψ = cϕsψ cθcψsϕ sϕsψ + cθcϕcψ sθcψ (2.106) sθsϕ cϕsθ cθ and therefore, Q B = B A 1 = B A T =[A z,ψ A x,θ A z,ϕ ] T cϕcψ cθsϕsψ cϕsψ cθcψsϕ sθsϕ = cψsϕ + cθcϕsψ sϕsψ + cθcϕcψ cϕsθ. (2.107) sθsψ sθcψ cθ iven the angles of precession ϕ, nutation θ, and spin ψ, we can compute the overall rotation matrix using Equation (2.106). Also we are able

22 54 2. Rotation Kinematics z z" B y ψ B'' y" x x" FIURE Third Euler angle. to compute the equivalent precession, nutation, and spin angles when a rotation matrix is given. If r ij indicates the element of row i and column j of the precessionnutation-spin rotation matrix (2.106), then, θ =cos 1 (r 33 ) (2.108) µ ϕ = tan 1 r31 (2.109) r 32 µ ψ =tan 1 r13 (2.110) r 23 provided that sin θ 6= 0. Example 18 Euler angle rotation matrix. The Euler or precession-nutation-spin rotation matrix for ϕ = deg, θ =41.41 deg, andψ = 40.7deg would be found by substituting ϕ, θ, and ψ in Equation (2.106). B A = A z, 40.7 A x,41.41 A z,79.15 (2.111) = Example 19 Euler angles of a local rotation matrix. The local rotation matrix after rotation 30 deg about the z-axis, then rotation 30 deg about the x-axis, and then 30 deg about the y-axis is B A = A y,30 A x,30 A z,30 (2.112) =

23 2. Rotation Kinematics 55 and therefore, the local coordinates of a sample point at X =5, Y =30, Z =10are: x y = = (2.113) z The Euler angles of the corresponding precession-nutation-spin rotation matrix are: θ = cos 1 (0.75) = deg µ 0.65 ϕ = tan 1 = deg µ 0.43 ψ = tan 1 = 40.7deg (2.114) 0.50 Hence, A y,30 A x,30 A z,30 = A z,ψ A x,θ A z,ϕ when ϕ =79.15 deg, θ =41.41 deg, and ψ = 40.7deg. In other words, the rigid body attached to the local frame moves to the final configuration by undergoing either three consecutive rotations ϕ =79.15 deg, θ =41.41 deg, andψ = 40.7deg about z, x, and z axes respectively, or three consecutive rotations 30 deg, 30 deg, and 30 deg about z, x, andy axes. Example 20 Relative rotation matrix of two bodies. Consider a rigid body B 1 with an orientation matrix B 1 A made by Euler angles ϕ =30deg, θ = 45 deg, ψ =60deg, and another rigid body B 2 having ϕ =10deg, θ =25deg, ψ = 15 deg, with respect to the global frame. To find the relative rotation matrix B1 A B2 to map the coordinates of second body frame B 2 to the first body frame B 1,weneedtofind the individual rotation matrices first. B 1 A = A z,60 A x, 45 A z,30 (2.115) = B 2 A = A z,10 A x,25 A z, 15 (2.116) = The desired rotation matrix B 1 A B2 may be found by B 1 A B2 = B1 A A B2 (2.117)

24 56 2. Rotation Kinematics which is equal to: B 1 A B2 = B1 B A 2 A T (2.118) = Example 21 Euler angles rotation matrix for small angles. The Euler rotation matrix B A = A z,ψ A x,θ A z,ϕ for very small Euler angles ϕ, θ, and ψ is approximated by B A = 1 γ 0 γ 1 θ (2.119) 0 θ 1 where, γ = ϕ + ψ. (2.120) Therefore, in case of small angles of rotation, the angles ϕ and ψ are indistinguishable. Example 22 Small second Euler angle. If θ 0 then the Euler rotation matrix B A = A z,ψ A x,θ A z,ϕ approaches to B A = = cϕcψ sϕsψ cψsϕ + cϕsψ 0 cϕsψ cψsϕ sϕsψ + cϕcψ cos (ϕ + ψ) sin(ϕ + ψ) 0 sin (ϕ + ψ) cos(ϕ + ψ) (2.121) and therefore, the angles ϕ and ψ are indistinguishable even if the value of ϕ and ψ are finite. Hence, the Euler set of angles in rotation matrix (2.106) is not unique when θ =0. Example 23 Euler angles application in motion of rigid bodies. The zxz Euler angles are good parameters to describe the configuration of a rigid body with a fixed point. The Euler angles to show the configuration of a top are shown in Figure 2.15 as an example. Example 24 F Angular velocity vector in terms of Euler frequencies. A Eulerian local frame E (o, ê ϕ, ê θ, ê ψ ) can be introduced by defining unit vectors ê ϕ, ê θ,andê ψ as shown in Figure Although the Eulerian frame is not necessarily orthogonal, it is very useful in rigid body kinematic analysis.

25 2. Rotation Kinematics 57 B z θ Z ê ϕ ϕ ê ψ ψ y X ϕ ψ x Y ê θ FIURE Application of Euler angles in describing the configuration of a top. The angular velocity vector ω B of the body frame B(Oxyz) with respect to the global frame (OXY Z) can be written in Euler angles frame E as thesumofthreeeulerangleratevectors. E ω B = ϕê ϕ + θê θ + ψê ψ (2.122) where, the rate of Euler angles, ϕ, θ, and ψ are called Euler frequencies. To find ω B inbodyframewemustexpresstheunitvectorsê ϕ, ê θ, and ê ψ shown in Figure 2.16, in the body frame. The unit vector ê ϕ = T = ˆK is in the global frame and can be transformed to the body frame after three rotations. Bê ϕ = B A ˆK = Az,ψ A x,θ A z,ϕ ˆK = sin θ sin ψ sin θ cos ψ cos θ (2.123) The unit vector ê θ = T = î 0 is in the intermediate frame Ox 0 y 0 z 0 andneedstogettworotationsa x,θ and A z,ψ to be transformed to the body frame. Bê θ = B A Ox 0 y 0 z 0 î0 = A z,ψ A x,θ î 0 = cos ψ sin ψ (2.124) 0 The unit vector ê ψ is already in the body frame, ê ψ = T = ˆk.

26 58 2. Rotation Kinematics B Z z ê ϕ ê ψ θ y x X ϕ ψ Y ê θ FIURE Euler angles frame ê ϕ, ê θ, ê ψ. Therefore, ω B is expressed in body coordinate frame as sin θ sin ψ B ω B = ϕ sin θ cos ψ + θ cos ψ sin ψ + ψ cos θ 0 ³ = ϕ sin θ sin ψ + θ cos ψ î ³ ϕ sin θ cos ψ θ sin ψ ĵ ³ ψ ϕ cos θ + ˆk (2.125) and therefore, components of ω B in body frame Oxyz are related to the Euler angle frame Oϕθψ by the following relationship: B ω B = B E A E ω B (2.126) sin θ sin ψ cos ψ 0 = sin θ cos ψ sin ψ 0 (2.127) cos θ 0 1 ω x ω y ω z Then, ω B can be expressed in the global frame using an inverse transformation of Euler rotation matrix (2.106) ϕ sin θ sin ψ + θ cos ψ ω B = B A 1 B ω B = B A 1 ϕ sin θ cos ψ θ sin ψ ϕ cos θ + ψ = ³ θ cos ϕ + ψ sin θ sin ϕ Î + ³ θ sin ϕ ψ cos ϕ sin θ Ĵ ³ + ϕ + ψ cos θ ˆK (2.128) and hence, components of ω B in global coordinate frame OXY Z are related to the Euler angle coordinate frame Oϕθψ by the following relation- ϕ θ ψ

27 2. Rotation Kinematics 59 ship. ω B = E Q E ω B (2.129) 0 cosϕ sin θ sin ϕ = 0 sinϕ cos ϕ sin θ (2.130) 1 0 cosθ ω X ω Y ω Z Example 25 F Euler frequencies based on a Cartesian angular velocity vector. The vector B ω B, that indicates the angular velocity of a rigid body B with respect to the global frame writteninframeb, is related to the Euler frequencies by B ω B = B E A E ω B (2.131) B ω B = ω x sin θ sin ψ cos ψ 0 ϕ ω y = sin θ cos ψ sin ψ 0 θ. (2.132) ω z cos θ 0 1 ψ The matrix of coefficients is not an orthogonal matrix because, B A T E 6= B A 1 E (2.133) sin θ sin ψ sin θ cos ψ cos θ B A T E = cos ψ sin ψ 0 (2.134) sin ψ cos ψ 0 B A 1 1 E = sin θ cos ψ sin θ sin ψ 0. (2.135) sin θ cos θ sin ψ cos θ cos ψ 1 It is because the Euler angles coordinate frame Oϕθψ is not an orthogonal frame. For the same reason, the matrix of coefficients that relates the Euler frequencies and the components of ω B ω B = E Q E ω B (2.136) ω B = ω X 0 cosϕ sin θ sin ϕ ϕ ω Y = 0 sinϕ cos ϕ sin θ θ (2.137) ω Z 1 0 cosθ ψ is not an orthogonal matrix. Therefore, the Euler frequencies based on local and global decomposition of the angular velocity vector ω B must solely be found by the inverse of coefficient matrices E ω B = B A 1 B E ω B (2.138) ϕ sin ψ cos ψ 0 θ 1 = sin θ cos ψ sin θ sin ψ 0 ω x ω y (2.139) sin θ ψ cos θ sin ψ cos θ cos ψ 1 ω z ϕ θ ψ

28 60 2. Rotation Kinematics and E ω B = Q 1 E ω B (2.140) ϕ cos θ sin ϕ cos θ cos ϕ 1 θ 1 = sin θ cos ϕ sin θ sin ϕ 0 ω X ω Y. (2.141) sin θ ψ sin ϕ cos ϕ 0 ω Z Using (2.138) and (2.140), it can be verified that the transformation matrix B A = B A E Q 1 E would be the same as Euler transformation matrix (2.106). The angular velocity vector can thus be expressed as: ω B = î ĵ ˆk ω x ω y = Î Ĵ ˆK ω X ω Y ω z ω Z = ϕ ˆK êθ ˆk θ (2.142) ψ Example 26 F Integrability of the angular velocity components. The integrability condition for an arbitrary total differential of f = f (x, y) df = f 1 dx + f 2 dy = f f dx + dy (2.143) x y is: f 1 y = f 2 (2.144) x The angular velocity components ω x, ω y,andω z along the body coordinate axes x, y, andz can not be integrated to obtain the associated angles because ω x dt =sinθsin ψdϕ+cosψdθ (2.145) and (sin θ sin ψ) θ 6= cos ψ ϕ. (2.146) However, the integrability condition (2.144) is satisfied by the Euler frequencies. From (2.139), we have: dϕ = sin ψ sin θ (ω x dt)+ cos ψ sin θ (ω y dt) (2.147) dθ = cosψ (ω x dt) sin ψ (ω y dt) (2.148) dψ = cos θ sin ψ sin θ (ω x dt)+ cos θ cos ψ sin θ For example, the second equation indicates that cos ψ = θ (ω x dt) sin ψ = (ω y dt)+ (ω z dt) sin θ θ (ω y dt) (2.149) (2.150)

29 and therefore, (cos ψ) ψ = sin ψ (ω y dt) (ω y dt) ( sin ψ) ψ = cos ψ (ω x dt) (ω x dt) It can be checked that dϕ and dψ are also integrable. 2. Rotation Kinematics 61 = sin ψ cos θ cos ψ sin θ = sin ψ cos θ cos ψ sin θ (2.151) (2.152) Example 27 F Cardan angles and frequencies. The system of Euler angles is singular at θ =0,andasaconsequence, ϕ and ψ become coplanar and indistinguishable. From 12 angle systems of Appendix B, each with certain names, characteristics, advantages, and disadvantages, the rotations about three different axes such as B A = A z,ψ A y,θ A x,ϕ are called Cardan or Bryant angles. The Cardan angle system is not singular at θ =0, and has some application in mechatronics and attitude analysis of satellites in a central force field. B A = cθcψ cϕsψ + sθcψsϕ sϕsψ cϕsθcψ cθsψ cϕcψ sθsϕsψ cψsϕ + cϕsθsψ sθ cθsϕ cθcϕ (2.153) The angular velocity ω of a rigid body can either be expressed in terms of the components along the axes of B(Oxyz), or in terms of the Cardan frequencies along the axes of the non-orthogonal Cardan frame. The angular velocity in terms of Cardan frequencies is therefore, ω x ω y ω B = ϕa z,ψ A y,θ ω z ϕ θ ψ = = θa z,ψ cos θ cos ψ sin ψ 0 cos θ sin ψ cos ψ 0 sin θ 0 1 cos ψ cos θ + ψ sin ψ cos θ 0 sin ψ cos ψ 0 tan θ cos ψ tan θ sin ψ ϕ θ ψ (2.154) (2.155) ω x ω y ω z. (2.156) In case of small Cardan angles, we have B A = 1 ψ θ ψ 1 ϕ (2.157) θ ϕ 1 and ω x ω y = 1 ψ 0 ϕ ψ 1 0 θ. (2.158) ω z θ 0 1 ψ

30 62 2. Rotation Kinematics z yaw θ B x ϕ roll ψ pitch y FIURE Local roll-pitch-yaw angles. 2.7 Local Roll-Pitch-Yaw Angles Rotation about the x-axis of the local frame is called roll or bank, rotation about y-axis of the local frame is called pitch or attitude, and rotation about the z-axis of the local frame is called yaw, spin, or heading. The local roll-pitch-yaw angles are shown in Figure The local roll-pitch-yaw rotation matrix is: B A = A z,ψ A y,θ A x,ϕ cθcψ cϕsψ + sθcψsϕ sϕsψ cϕsθcψ = cθsψ cϕcψ sθsϕsψ cψsϕ + cϕsθsψ (2.159) sθ cθsϕ cθcϕ Note the difference between roll-pitch-yaw and Euler angles, although we show both utilizing ϕ, θ, and ψ. Example 28 F Angular velocity and local roll-pitch-yaw rate. Using the roll-pitch-yaw frequencies, the angular velocity of a body B with respect to the global reference frame is ω B = ω x î + ω y ĵ + ω zˆk = ϕê ϕ + θê θ + ψê ψ. (2.160) Relationships between the components of ω B in body frame and roll-pitchyaw components are found when the local roll unit vector ê ϕ and pitch unit vector ê θ are transformed to the body frame. The roll unit vector ê ϕ = T transforms to the body frame after rotation θ and then

31 2. Rotation Kinematics 63 rotation ψ. Bê ϕ = A z,ψ A y,θ 1 0 = 0 cos θ cos ψ cos θ sin ψ sin θ (2.161) The pitch unit vector ê θ = T transforms to the body frame after rotation ψ. Bê θ = A z,ψ 0 1 = sin ψ cos ψ (2.162) 0 0 The yaw unit vector ê ψ = T is already along the local z-axis. Hence, ω B canbeexpressedinbodyframeoxyz as B ω B = ω x cos θ cos ψ ω y = ϕ cos θ sin ψ + θ sin ψ cos ψ + ψ 0 0 ω z sin θ 0 1 cos θ cos ψ sin ψ 0 ϕ = cos θ sin ψ cos ψ 0 θ (2.163) sin θ 0 1 ψ and therefore, ω B in global frame OXY Z in terms of local roll-pitch-yaw frequencies is: ω B = ω X ω Y = B A 1 ω x θ sin ψ + ϕ cos θ cos ψ ω y = B A 1 θ cos ψ ϕ cos θ sin ψ ω Z ω z ψ + ϕ sin θ = = ϕ + ψ sin θ θ cos ϕ ψ cos θ sin ϕ θ sin ϕ + ψ cos θ cos ϕ 1 0 sinθ 0 cosϕ cos θ sin ϕ 0 sinϕ cos θ cos ϕ ϕ θ ψ (2.164) 2.8 Local Axes Versus lobal Axes Rotation The global rotation matrix Q B is equal to the inverse of the local rotation matrix B A and vice versa, where Q B = B A 1, B A = Q 1 B (2.165) Q B = A 1 1 A 1 2 A 1 3 A 1 n (2.166) B A = Q 1 1 Q 1 2 Q 1 3 Q 1 n. (2.167)

32 64 2. Rotation Kinematics Also, premultiplication of the global rotation matrix is equal to postmultiplication of the local rotation matrix. Proof. Consider a sequence of global rotations and their resultant global rotation matrix Q B to transform a position vector B r to r. r = Q B B r (2.168) The global position vector r can also be transformed to B r usingalocal rotation matrix B A. B r = B A r (2.169) Combining Equations (2.168) and (2.169) leads to r = Q B B A r (2.170) B r = B A Q B B r (2.171) and hence, Q B B A = B A Q B = I. (2.172) Therefore, the global and local rotation matrices are the inverse of each other. Q B = B A 1 Q 1 B = B A (2.173) Assume that Q B = Q n Q 3 Q 2 Q 1 and B A = A n A 3 A 2 A 1 then, Q B = B A 1 B A = Q 1 B and Equation (2.172) becomes = A 1 1 A 1 2 A 1 3 A 1 n (2.174) = Q 1 1 Q 1 2 Q 1 3 Q 1 n (2.175) Q n Q 2 Q 1 A n A 2 A 1 = A n A 2 A 1 Q n Q 2 Q 1 = I (2.176) and therefore, or Q n Q 3 Q 2 Q 1 = A 1 1 A 1 2 A 1 3 A 1 n A n A 3 A 2 A 1 = Q 1 1 Q 1 2 Q 1 3 Q 1 n (2.177) Q 1 1 Q 1 2 Q 1 3 Q 1 n Q n Q 3 Q 2 Q 1 = I (2.178) A 1 1 A 1 2 A 1 3 A 1 n A n A 3 A 2 A 1 = I. (2.179) Hence, the effect of in order rotations about the global coordinate axes is equivalent to the effect of the same rotations about the local coordinate axes performed in the reverse order.

33 2. Rotation Kinematics 65 Example 29 lobal position and postmultiplication of rotation matrix. The local position of a point P after rotation is at B r = T. If the local rotation matrix to transform r to B r is given as B A z,ϕ = cos ϕ sin ϕ 0 sin ϕ cos ϕ = cos 30 sin 30 0 sin 30 cos (2.180) then we may find the global position vector r by postmultiplication B A z,ϕ by the local position vector B r T, r T = B r T B A z,ϕ = cos 30 sin 30 0 sin 30 cos = (2.181) instead of premultiplication of B A 1 z,ϕ by B r. r = B A 1 z,ϕ B r cos 30 sin 30 0 = sin 30 cos = (2.182) 2.9 eneral Transformation Consider a general situation in which two coordinate frames, (OXY Z) and B(Oxyz) with a common origin O, are employed to express the components of a vector r. Thereisalwaysatransformation matrix R B to map the components of r from the reference frame B(Oxyz) to the other reference frame (OXY Z). r = R B B r (2.183) In addition, the inverse map, B r = R 1 B r, can be done by B R B r = B R r (2.184) where, and R B = B R =1 (2.185) B R = R 1 B = R T B. (2.186)

34 66 2. Rotation Kinematics Proof. Decomposition of the unit vectors of (OXY Z) along the axes of B(Oxyz) Î = (Î î)î +(Î ĵ)ĵ +(Î ˆk)ˆk (2.187) Ĵ = (Ĵ î)î +(Ĵ ĵ)ĵ +(Ĵ ˆk)ˆk (2.188) ˆK = ( ˆK î)î +(ˆK ĵ)ĵ +(ˆK ˆk)ˆk (2.189) introduces the transformation matrix R B to map the local frame to the global frame Î Ĵ ˆK = Î î Î ĵ Î ˆk Ĵ î Ĵ ĵ Ĵ ˆk ˆK î ˆK ĵ ˆK ˆk î ĵ ˆk = R B î ĵ ˆk (2.190) where, R B = = Î î Î ĵ Î ˆk Ĵ î Ĵ ĵ Ĵ ˆk ˆK î ˆK ĵ ˆK ˆk cos(î,î) cos(î,ĵ) cos(î,ˆk) cos(ĵ,î) cos(ĵ,ĵ) cos(ĵ,ˆk) cos( ˆK,î) cos( ˆK,ĵ) cos( ˆK,ˆk). (2.191) Each column of R B is decomposition of a unit vector of the local frame B(Oxyz) in the global frame (OXY Z). R B = î ĵ ˆk = ˆr V1 ˆr V2 ˆr V3 (2.192) Similarly, each row of R B is decomposition of a unit vector of the global frame (OXY Z) in the local frame B(Oxyz). R B = B Î T B Ĵ T B ˆKT = ˆr H 1 ˆr H2 (2.193) ˆr H3 The elements of R B are direction cosines of the axes of (OXY Z) in frame B(Oxyz). This set of nine direction cosines then completely specifies the orientation of the frame B(Oxyz) in the frame (OXY Z), andcan be used to map the coordinates of any point (x, y, z) to its corresponding coordinates (X, Y, Z).

35 2. Rotation Kinematics 67 Alternatively, using the method of unit vector decomposition to develop the matrix B R leads to: B r = B R r = R 1 B r (2.194) î Î î Ĵ î ˆK B R = ĵ Î ĵ Ĵ ĵ ˆK ˆk Î ˆk Ĵ ˆk ˆK cos(î, Î) cos(î, Ĵ) cos(î, ˆK) = cos(ĵ, Î) cos(ĵ, Ĵ) cos(ĵ, ˆK) (2.195) cos(ˆk, Î) cos(ˆk, Ĵ) cos(ˆk, ˆK) and shows that the inverse of a transformation matrix is equal to the transpose of the transformation matrix. R 1 B = R T B (2.196) A matrix with condition (2.196) is called orthogonal. Orthogonality of R comes from this fact that it maps an orthogonal coordinate frame to another orthogonal coordinate frame. The transformation matrix R has only three independent elements. The constraint equations among the elements of R will be found by applying the orthogonality condition (2.196). r 11 r 12 r 13 r 21 r 22 r 23 r 31 r 32 r 33 R B R B T = [I] (2.197) = (2.198) r 11 r 21 r 31 r 12 r 22 r 32 r 13 r 23 r 33 Therefore, the dot product of any two different rows of R B is zero, and the dot product of any row of R B with the same row is one. r r r13 2 = 1 r r r23 2 = 1 r r r33 2 = 1 r 11 r 21 + r 12 r 22 + r 13 r 23 = 0 r 11 r 31 + r 12 r 32 + r 13 r 33 = 0 r 21 r 31 + r 22 r 32 + r 23 r 33 = 0 (2.199) These relations are also true for columns of R B, and evidently for rows and columns of B R. The orthogonality condition can be summarized in the following equation: ˆr Hi ˆr Hj = ˆr T H i ˆr Hj = 3X r ij r ik = δ jk (j, k =1, 2, 3) (2.200) i=1

36 68 2. Rotation Kinematics where r ij is the element of row i and column j of the transformation matrix R, andδ jk is the Kronecker s delta. δ jk =1if j = k, and δ jk =0if j 6= k (2.201) Equation (2.200) gives six independent relations satisfied by nine direction cosines. It follows that there are only three independent direction cosines. The independent elements of the matrix R cannot obviously be in the same row or column, or any diagonal. The determinant of a transformation matrix is equal to one, R B =1 (2.202) because of Equation (2.197), and noting that R B R B T = R B RB T = R B R B = R B 2 =1. (2.203) Using linear algebra and row vectors ˆr H1,ˆr H2, and ˆr H3 of R B,weknow that R B = ˆr T H 1 (ˆr H2 ˆr H3 ) (2.204) and because the coordinate system is right handed, we have ˆr H2 ˆr H3 = ˆr H1 so R B = ˆr T H1 ˆr H1 =1. Example 30 Elements of transformation matrix. The position vector r of a point P may be expressed in terms of its components with respect to either (OXY Z) or B (Oxyz) frames. Body and a global coordinate frames are shown in Figure If r = 100Î 50Ĵ ˆK, and we are looking for components of r in the Oxyz frame, then we have to find the proper rotation matrix B R first. The row elements of B R are the direction cosines of the Oxyz axes in the OXY Z coordinate frame. The x-axis lies in the XZ plane at 40 deg from the X-axis, and the angle between y and Y is 60 deg. Therefore, î Î î Ĵ î ˆK cos 40 0 sin 40 B R = ĵ Î ĵ Ĵ ĵ ˆK = ĵ Î cos 60 ĵ ˆK ˆk Î ˆk Ĵ ˆk ˆK ˆk Î ˆk Ĵ ˆk ˆK = ĵ Î 0.5 ĵ ˆK (2.205) ˆk Î ˆk Ĵ ˆk ˆK and by using B R R B = B R B R T = I r r r 21 r r 32 r 31 r 32 r r 23 r 33 = (2.206)

37 2. Rotation Kinematics 69 Z z x 40 B X 60 y Y FIURE Body and global coordinate frames of Example 30. we obtain a set of equations to find the missing elements r r 23 = r r 33 = 0 r r = 1 r 21 r r 32 + r 23 r 33 = 0 r r r33 2 = 1 (2.207) Solving these equations provides the following transformation matrix: B R = (2.208) and then we can find the components of B r B r = B R r = = (2.209) Example 31 lobal position, using B r and B R. The position vector r of a point P may be described in either (OXY Z) or B (Oxyz) frames. If B r = 100î 50ĵ + 150ˆk, and the following B R is

38 70 2. Rotation Kinematics the transformation matrix to map r to B r B r = B R r = r (2.210) then the components of r in (OXY Z) would be r = R B B r = B R T B r = = (2.211) Example 32 Two points transformation matrix. The global position vector of two points, P 1 and P 2,ofarigidbodyB are: r P1 = r P2 = (2.212) The origin of the body B (Oxyz) is fixed on the origin of (OXY Z), and the points P 1 and P 2 arelyingonthelocalx-axis and y-axis respectively. To find R B,weusethelocalunitvectors î and ĵ to obtain ˆk î = ĵ = r P1 r P1 = r P2 r P2 = (2.213) (2.214) ˆk = î ĵ =ĩ ĵ = = (2.215) where ĩ is the skew-symmetric matrix corresponding to î, and ĩ ĵ is an alternative for î ĵ.

39 2. Rotation Kinematics 71 Hence, the transformation matrix using the coordinates of two points r P1 and r P2 would be R B = î ĵ ˆk = (2.216) Example 33 Length invariant of a position vector. Describing a vector in different frames utilizing rotation matrices does not affect the length and direction properties of the vector. Therefore, length of a vector is an invariant The length invariant property can be shown by r = r = B r. (2.217) r 2 = r T r = R B B r T R B B r = B r T RB T R B B r = B r TB r. (2.218) Example 34 Skew symmetric matrices for î, ĵ, andˆk. The definition of skew symmetric matrix ã correspondingtoavectora is defined by ã = 0 a 3 a 2 a 3 0 a 1. (2.219) a 2 a 1 0 Hence, ĩ = (2.220) j = (2.221) k = (2.222) Example 35 Inverse of Euler angles rotation matrix. Precession-nutation-spin or Euler angle rotation matrix (2.106) B R = A z,ψ A x,θ A z,ϕ cϕcψ cθsϕsψ cψsϕ + cθcϕsψ sθsψ = cϕsψ cθcψsϕ sϕsψ + cθcϕcψ sθcψ (2.223) sθsϕ cϕsθ cθ

40 72 2. Rotation Kinematics must be inverted to be a transformation matrix to map body coordinates to global coordinates. R B = B R 1 = AT z,ϕa T x,θa T z,ψ cϕcψ cθsϕsψ cϕsψ cθcψsϕ sθsϕ = cψsϕ + cθcϕsψ sϕsψ + cθcϕcψ cϕsθ (2.224) sθsψ sθcψ cθ The transformation matrix (2.223) is called a local Euler rotation matrix, and (2.224) is called a global Euler rotation matrix. Example 36 F roup property of transformations. AsetS together with a binary operation defined on elements of S is called a group (S, ) if it satisfies the following four axioms. 1. Closure: Ifs 1,s 2 S, thens 1 s 2 S. 2. Identity: There exists an identity element s 0 such that s 0 s = s s 0 = s for s S. 3. Inverse: Foreachs S, there exists a unique inverse s 1 S such that s 1 s = s s 1 = s Associativity: Ifs 1,s 2,s 3 S, then(s 1 s 2 ) s 3 = s 1 (s 2 s 3 ). Three dimensional coordinate transformations make a group if we define the set of rotation matrices by S = R R 3 3 : RR T = R T R = I, R =1 ª. (2.225) Therefore, the elements of the set S are transformation matrices R i, the binary operator is matrix multiplication, the identity matrix is I, and the inverse of element R is R 1 = R T. S is also a continuous group because 5. The binary matrix multiplication is a continuous operation, and 6. The inverse of any element in S is a continuous function of that element. Therefore, S is a differentiable manifold. A group that is a differentiable manifold is called a Lie group. Example 37 F Transformation with determinant 1. An orthogonal matrix with determinant +1 corresponds to a rotation as described in Equation (2.202). In contrast, an orthogonal matrix with determinant 1 describes a reflection. Moreover it transforms a right-handed coordinate system into a left-handed, and vice versa. This transformation does not correspond to any possible physical action on rigid bodies.

41 2. Rotation Kinematics 73 Example 38 Alternative proof for transformation matrix. Starting with an identity î î ĵ ˆk ĵ =1 (2.226) ˆk we may write Î Ĵ ˆK = Î Ĵ ˆK î ĵ ˆk î ĵ ˆk. (2.227) Since matrix multiplication can be performed in any order we find Î Î î Î ĵ Î ˆk î î Ĵ = Ĵ î Ĵ ĵ Ĵ ˆk ĵ = R B ĵ (2.228) ˆK ˆK î ˆK ĵ ˆK ˆk ˆk ˆk where, R B = Î Ĵ ˆK î ĵ ˆk. (2.229) Following the same method we can show that î B R = ĵ Î Ĵ ˆK. (2.230) ˆk 2.10 Active and Passive Transformation Rotation of a local frame when the position vector r of a point P is fixed in global frame and does not rotate with the local frame, is called passive transformation. Alternatively, rotation of a local frame when the position vector B r of a point P is fixed in the local frame and rotates with the local frame, is called active transformation. Surprisingly, the passive and active transformations are mathematically equivalent. In other words, the rotation matrix for a rotated frame and rotated vector (active transformation) is the same as the rotation matrix for a rotated frame and fixed vector (passive transformation). Proof. Consider a rotated local frame B(Oxyz) with respect to a fixed global frame (OXY Z), as shown in Figure P is a fixed point in the global frame, and so is its global position vector r. Position vector of P can be decomposed in either a local or global coordinate frame, denoted by B r and r respectively.