Question 11.23: For the yaw–pitch–roll sequence ϕyaw = 50°, θpitch = 90°, an...
For the yaw–pitch–roll sequence \phi_{\text {yaw }} = 50°, \theta_{\text {pitch }} = 90°, and \psi_{\text {roll }} = 120°, calculate
(a) the quaternion and
(b) the rotation angle and the axis of rotation.
(a) Substituting the given angles into Eq. (11.119) yields the direction cosine matrix
[ Q ]_{X x}=\left[\begin{array}{ccc}0 & 0 & -1 \\0.93969 & 0.34202 & 0 \\0.34202 & -0.93969 & 0\end{array}\right] (a)
[Q]_{X x}=\left[\begin{array}{ccc}\cos \phi \cos \theta & \sin \phi \cos \theta & -\sin \theta \\\cos \phi \sin \theta \sin \psi-\sin \phi \cos \psi & \sin \phi \sin \theta \sin \psi+\cos \phi \cos \psi & \cos \theta \sin \psi \\\cos \phi \sin \theta \cos \psi+\sin \phi \sin \psi & \sin \phi \sin \theta \cos \psi-\cos \phi \sin \psi & \cos \theta \cos \psi\end{array}\right] (11.119)
Substituting the components of [Q]_{X x} into Eq. (11.159), we get
[ K ]=\left[\begin{array}{rrrr}-0.11401 & 0.31323 & -0.21933 & 0.31323 \\0.31323 & 0.11401 & -0.31323 & 0.44734 \\-0.21933 & -0.31323 & -0.11401 & -0.31323 \\0.31323 & 0.44734 & -0.31323 & 0.11401\end{array}\right]
[ K ]=\frac{1}{3}\left[\begin{array}{cccc}Q_{11}-Q_{22}-Q_{33} & Q_{21}+Q_{12} & Q_{31}+Q_{13} & Q_{23}-Q_{32} \\Q_{21}+Q_{12} & -Q_{11}+Q_{22}-Q_{33} & Q_{32}+Q_{23} & Q_{31}-Q_{13} \\Q_{31}+Q_{13} & Q_{32}+Q_{23} & -Q_{11}-Q_{22}+Q_{33} & Q_{12}-Q_{21} \\Q_{23}-Q_{32} & Q_{31}-Q_{13} & Q_{12}-Q_{21} & Q_{11}+Q_{22}+Q_{33}\end{array}\right] (11.159)
The following is a MATLAB script that implements the power method described in Algorithm 11.2.
The quaternion is the eigenvector associated with \lambda_{\max }, so that
\widehat{ q }=\left\{\begin{array}{r}0.40558 \\0.57923 \\-0.40558 \\\hdashline 0.57923\end{array}\right\}
Observe that \|\widehat{ q }\|=1 . \widehat{ q } must be a unit quaternion.
(b) From Eq. (11.146), we find that the principal angle is
\theta=2 \cos ^{-1}\left(q_4\right)=2 \cos ^{-1}(0.57923)=54.604^{\circ}
\widehat{ p } \otimes \widehat{ q }=\left\{\frac{p_4 q +q_4 p + p \times q }{p_4 q_4- p \cdot q }\right\} (11.146)
and the Euler axis is
\hat{ u }=\frac{0.40558 \hat{ I }+0.57923 \hat{ J }-0.40558 \hat{ K }}{\sin \left(54.604^{\circ} / 2\right)}=0.4975 \hat{ I }+0.71056 \hat{ J }-0.49754 \hat{ K }
K = [-0.11401 0.31323 -0.21933 0.31323
0.31323 0.11401 -0.31323 0.44734
-0.21933 -0.31323 -0.11401 -0.31323
0.31323 0.44734 -0.31323 0.11401];
v0 = [1 1 1 1]0; %Initial estimate of the eigenvector.
v0 = v0/norm(v0); %Normalize it.
lamda_new = v00*K*v0; %Rayleigh quotient (norm(v0) = 1)
% estimate of the eigenvalue.
lamda_old = 10*lamda_new; %Just to begin the iteration.
no_iterations = 0; %Count the number of iterations.
tolerance = 1.e-10;
while abs((lamda_new - lamda_old)/lamda_old) > tolerance
no_iterations = no_iterations + 1;
lamda_old = lamda_new;
v = v0;
vnew = K*v/norm(K*v);
lamda_new = vnew0*K*vnew; %Rayleigh quotient (norm(vnew) = 1).
v0 = vnew;
end
no_iterations = no_iterations
disp(‘ ‘)
lamda_max = lamda_new
eigenvector = vnew
The output of this program to the Command Window is as follows:
no_iterations =
12
lamda_max =
1
eigenvector =
0.40558
0.57923
-0.40558
0.57923