1. ORIENTATION KINEMATICS: AXIS-ANGLE ROTATION
The final orientation of a body after a finite number of rotations is equivalent to a unique rotation about a unique axis.
Orientation Kinematics Problem: Refers to the determination of the rotation angle j of a rigid body with a fixed point O, about a globally fixed axis u (defining a unit vector u).
Let the body frame B(Oxyz) rotate by an angle j about a line indicated by vector u (represented by unit vector u). The unit vector is defined by its direction cosines u1, u2, u3. Thus, we have in the frame G(OXYZ):
u = u1 I + u2 J + u3 K with (u1)2 + (u2)2 + (u3)2 = 1.
To map the coordinates given in the local frame B(Oxyz) to the coordinates in the global frame G(OXYZ), use:
Gr = GRB Br
where GRB is the rotation matrix Ru,j around u by an angle j, (unit vector u is derived from u).

2. ORIENTATION KINEMATICS: AXIS-ANGLE ROTATION (Cont’d)
Matrix GRB = Ru,j maps frame B to global frame G is given by the following expression (known as Rodriguez or Euler-Lexell-Rodriguez formula:
GRB = Ru,j = I*cos(j) + uutvers(j) + v sin(j)
where: vers (j) = 1- cos(j) = 2sin2 (j/2), j being the rotation angle. This expression expands to:
GRB = Ru,j =
u1*u1*vers(j) + c j u2*u1*vers(j) – u3 s j u3*u1*vers(j) + u2 s j
u1*u2*vers(j) + u3 s j u2*u2*vers(j) + c j u3*u2*vers(j) – u1 s j
u1*u3*vers(j) – u2 s j u2*u3*vers(j) + u1 s j u3*u3*vers(j) + c j
Given the transformation matrix GRB, the axis u and the angle j, matrix v is defined as a function of GRB and also of the components of unit vector u as follows:
v = [GRB – (GRB)t ]/(2 sin j) =
cos(j) = [Tr(GRB ) – 1]/2
It is easy to verify that v is skew-symmetric, that is: vt = -v.
0 -u3 u2
u u1
-u2 u

3. Exercise # 15: ORIENTATION KINEMATICS
Find GRB when u = K (that is: u = [u1, u2, u3]t = [0 0 1]t).
Solution:
If the local frame B(Oxyz) rotates about the Z-axis, then u = K and the transformation matrix reduces to:
GRB =
0*vers(j) + c j *vers(j) – 1* s j *vers(j) + 0*s j
0*vers(j) + 1*s j 0*vers(j) + c j *vers(j) – 0*s j
0*vers(j) – 0 s j 0*vers(j) + 0*s j *vers(j) + c j
Since vers(j) = 1 – c(j)
GRB =
cos j - sin j
sin j cos j
which is the rotation matrix about the Z-axis of the global frame we have previously
encountered.

4. Exercise # 16: ORIENTATION KINEMATICS
Consider a fixed point at A and unit length of edges. Turn the cube 45 deg
about u = [1 1 1]t. Find the rotation angle j in radians, the unit vector u and the
rotation matrix Ru,j .
Solution:
j = p/4, (that is: sin(j) = cos(j) = 1/2 = ).
u = u/3 = [ , , ]t = t
GRB = Ru,j = I cos(j) + uut (1 - cos(j) ) + v sin(j) (vector v defined as before)
doing the calculations yields:
Ru,j = [ ; ; ]
1/3 1/3 1/3

5. Exercise # 16: ORIENTATION KINEMATICS (Cont’d)
Here we verify Ru,j using the Rodriguez formula:
Ru,j = I cos(j) + uut (1 - cos(j) ) + v sin(j)
I cos(j) =
* =
(1 - cos(j) ) = =
1/3
uut (1 - cos(j) ) =
1/3 1/3 1/3
*0.2929 =
v sin(j) =
* =
GRB = Ru,j =

6. Exercise # 17: ORIENTATION KINEMATICS
Consider a fixed point at A and unit length of edges. Turn the cube 45 deg
about v = [1 1 1]t. Find the global coordinates of its corners .
Solution:
We have the rotation matrix Ru,j :
Ru,j = I cos(j) + uut (1 - cos(j) ) + usin(j)
= [ ; ; ]
Local coordinates of the corners are Bri = with i = A, B, C, D, E, F, G, H.
B(1, 0, 0); C(1, 1, 0); D(0, 1, 0); E(0, 0, 1); F(1, 0, 1); G(1, 1, 1); H(0, 1, 1).
Using Gri = Ru,j Bri with i = A, B, C, D, E, F, G, H, we can verify that the coordinates of the corners after rotation are:
B(0.804, 0.505, -0.31); C(0.495, 1.31, 0.196); D(-0.31, 0.804, 0.505);
E(0.505, -0.31, 0.804); F(1.310, 0.196, 0.495); G(1,1,1); H(0.196, 0.495, 1.31).

7. Exercise # 18: ORIENTATION KINEMATICS
Consider a fixed point at A and unit length of edges. Turn the cube 45 deg
about v = [1 1 1]t. Determine points whose coordinates do not change and explain
the reason for the invariance.
Solution:
Point G coordinates do not change after rotation because it is on the rotation axis.
Points B, D, F, and G are in a symmetric plane indicated by u, so they will move on a circle. To verify this, we find the middle point of BH, given by 0.5(?rB + ?rH) or the middle point of FG 0.5(?rF + ?rG) and verify that they are on the u-axis with coordinates (0.5, 0.5, 0.5) for both ? = “B” or “G” ).

8. Exercise # 19: AXIS AND ANGLE OF A ROTATION MATRIX
A rigid body coordinate frame B, undergoes three Euler rotations (j, q, y) =
(30, 45, 60) degrees with respect to a global frame G. Find the rotation matrix
to transform the coordinates of B to G.
Solution:
We are looking for matrix GRB :
GRB = (BRG)t = (RZ,yRX,qRZ,j)t = (RZ,j)t ( RX,q)t ( RZ,y)t
= [ ; ; ]

9. Exercise # 20: AXIS AND ANGLE OF A ROTATION MATRIX
A rigid body coordinate frame B, undergoes three Euler rotations (j, q, y) =
(30, 45, 60) degrees with respect to a global frame G. Find the unique
angle-axis rotation matrix .
Solution:
We use the matrix GRB already found and use the following formulas:
j = acos[Tr(GRB) – 1)/2]
= acos( ) = 98 degrees
v = ((GRB) - (GRB)t )/2sin(j)
= [ ; ; ]
From this matrix we can extract u = [ ]t verifying that:
( )2 + ( )2 + ( )2 = 1

10. Exercise # 21: AXIS AND ANGLE OF A ROTATION MATRIX
A rigid body coordinate frame B, undergoes three Euler rotations (j, q, y) =
(30, 45, 60) degrees with respect to a global frame G. Verify the Rodriguez formula.
Solution:
Want to prove that
GRB = [ ; ; ]
can be obtained by using the expression:
GRB = Ru,j = I*cos(j) + uutvers(j) + v sin(j)
Where
u = [ ]t
j = 98 degrees
v = [ ]
[ ]
[ ]