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\centerline{\bf ME 115(a):  Solution to Homework $\#4$}
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{\bf Problem 1:} (10 points)

The Euler-Angle representation of the moving body's orientation is
 \[ R = R_{\psi} R_{\phi} R_{\gamma} \]
where $R_{\psi}\ \in\ SO(3)$ represents rotation by angle $\psi$ about the body fixed $z$-axis,
$R_{\phi}\ \in\ SO(3)$ represents rotation by angle $\phi$ about the body fixed $y$-axis, and
$R_{\gamma}\ \in\ SO(3)$ represents rotation by angle $\gamma$ about the body fixed $z$-axis:
 \[ R_{\psi} =\begin{bmatrix} \cos\psi & -\sin\psi & 0\\ 
                \sin\psi & \cos\psi & 0\\
                0 & 0 & 1\end{bmatrix} \tab
    R_{\phi} =\begin{bmatrix} \cos\phi & 0 & \sin\phi\\ 
                0 & 1 & 0\\ 
                -\sin\phi & 0 & \cos\phi \end{bmatrix} \]

  \[ R_{\gamma}= \begin{bmatrix} \cos\gamma & -\sin\gamma & 0\\ 
                        \sin\gamma & \cos\gamma & 0 \\
                        0 & 0 & 1\end{bmatrix} \]
The spatial angular velocity of the rigid body is:
  \begin{eqnarray}
    \hat{\omega}^s &=& \dot{R}R^{-1} = 
       (\dot{R}_{\psi} R_{\phi} R_{\gamma} + R_{\psi} \dot{R}_{\phi} R_{\gamma}
      +  R_{\psi}R_{\phi}\dot{R}_{\gamma})R^{-1}_{\gamma} R^{-1}_{\phi}
	R^{-1}_{\psi} \nonumber \\
       &=& \dot{R}_{\psi} R^{-1}_{\psi} + R_{\psi}(\dot{R}_{\phi}R^{-1}_{\phi})
	R^{-1}_{\psi} + R_{\psi}R_{\phi}(\dot{R}_{\gamma}R^{-1}_{\gamma})
	R^{-1}_{\phi} R^{-1}_{\psi} \nonumber \\
      &=& \hat{\omega}^s_{\psi} + R_{\psi}\hat{\omega}^s_{\phi}R^{-1}_{\psi} + 
           R_{\psi}R_{\phi} \hat{\omega}^s_{\gamma} R^{-1}_{\phi}R^{-1}_{\psi}
  \end{eqnarray}
where $\omega^s_{\psi}=\dot{R}_{\psi}R^{-1}_{\psi}$, $\omega^s_{\phi} =
\dot{R}_{\phi}R^{-1}_{\phi}$, $\omega^s_{\gamma} =
\dot{R}_{\gamma}R^{-1}_{\gamma}$.  Converting to vector form:
   \begin{equation} \label{eq:angular}
    \vec{\omega}^s = (\hat{\omega}^s)^{\vee} =
      \vec{\omega}^s_{\psi} + R_{\psi}\ \vec{\omega}^s_{\phi} + R_{\psi}R_{\phi}\ 
      \vec{\omega}^s_{\gamma}
     \end{equation}
The angular velocities are simply:
   \[\vec{\omega}^s_{\psi} = (\dot{R}_{\psi}R^{-1}_{\psi})^{\vee}
                           = \begin{bmatrix} 0\\ 0\\ \dot{\psi}\end{bmatrix}
	 \tab\tab
     \vec{\omega}^s_{\phi} = (\dot{R}_{\phi}R^{-1}_{\phi})^{\vee}
                           =  \begin{bmatrix} 0\\ \dot{\phi}\\ 0\end{bmatrix}
          \tab\tab
    \vec{\omega}^s_{\gamma} = (\dot{R}_{\gamma}R^{-1}_{\gamma})^{\vee} \begin{bmatrix} 0\\ 0\\ \dot{\gamma}\end{bmatrix}
        \]

Substituting into Equation (\ref{eq:angular}) results in:
 \begin{equation}
    \vec{\omega}^s = \begin{bmatrix} 0\\ 0\\ \dot{\psi}\end{bmatrix} +
     \begin{bmatrix} -\dot{\phi}\sin\psi \\ 
                    \dot{\phi}\cos\psi \\ 0\end{bmatrix} +
     \begin{bmatrix}\dot{\gamma}\cos\psi \sin\phi \\
                    \dot{\gamma}\sin\psi \sin\phi \\
                    \dot{\gamma}\cos\phi \end{bmatrix}
   = \begin{bmatrix} -\dot{\phi}\sin\psi + \dot{\gamma}\cos\psi \sin\phi \\
        \dot{\phi}\cos\gamma + \dot{\gamma} \sin\psi \sin\phi \\
        \dot{\psi} + \dot{\gamma}\cos\phi \end{bmatrix}
   \end{equation}

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{\bf Problem 2:} (15 points) Problem 11(a,b,e) in MLS Chapter 2.

{\bf Part (a):} Recall that the matrix exponential of a twist, $\hat{\xi}$, is:
   \[ e^{\phi\hat{\xi}} = I + \frac{\phi}{1!} \hat{\xi} 
          + \frac{\phi^2}{2!}\hat{\xi}^2 + 
	              \frac{\phi^3}{3!}\hat{\xi}^3 + \cdots \]

First, let's consider the case of $\xi=(v,\omega)$, with $\omega=0$. If:
  \[ \hat{\xi} = \begin{bmatrix}0 & 0 & v_x \\
			      0 & 0 & v_y \\
			      0 & 0 & 0 \end{bmatrix} \]
then $\hat{\xi}^2=0$.  Thus
   \[ e^{\phi\hat{\xi}} = \begin{bmatrix}1 & 0 & \phi v_x \\ 
			      0 & 1 & \phi v_y \\ 
			      0 & 0 & 1 \end{bmatrix} = 
	\begin{bmatrix}I & \vec{v}\phi\\ \vec{0}^t & 1\end{bmatrix} \]
There are two approaches to 
to compute the exponential for the more general case in which $\omega\ne 0$.  One
could calculate the expansion in a brute force manner.  Alternatively, one
could first compute the exponential for a simpler matrix, and then transform
the exponential to obtain the desired result.  First, let's assume that 
$||\omega||=1$.  In this case, note that $\hat{\omega}^2 =
- I$, where $I$ is the $2\times 2$ identity matrix.  Let
  \begin{equation} \label{eq:twist}
     \hat{\xi}=\begin{bmatrix}0 & -\omega & v_x \\ 
			    \omega & 0 & v_y \\ 
			    0 & 0 & 0 \end{bmatrix} = 
	      \begin{bmatrix} \hat{\omega} & \vec{v} \\
			         \vec{0}^T & 0\end{bmatrix} \end{equation}
Be the general twist matrix that we want to exponentiate.  Let's transform
this matrix to a different coordinate system where the exponentiation will be
easier.  Let
 \[g = \begin{bmatrix} I & \hat{\omega}\vec{v} \cr \vec{0}^T & 1\end{bmatrix} \]
be the rigid body displacement between the coordinate system in which the
twist of Equation (\ref{eq:twist}) is originally defined, and a new 
coordinate system.  In the new coordinate system, the twist, $\hat{\xi}^{'}$, takes
the form::
  \[ \begin{array}{rl}
     \hat{\xi}^{'} &= g^{-1}\hat{\xi} g \\ 
	   &= \begin{bmatrix} I & -\hat{\omega}\vec{v} \\ 0 & 1\end{bmatrix}
	      \begin{bmatrix} \hat{\omega} & \vec{v} \\ 0 & 0\end{bmatrix}
      	      \begin{bmatrix} I & \hat{\omega}\vec{v} \\ 0 & 1\end{bmatrix} \\ 
          & = \begin{bmatrix} \hat{\omega} & (\hat{\omega}^2\vec{v} + \vec{v})\\
	                      0 & 0\end{bmatrix}
            = \begin{bmatrix} \hat{\omega} & 0 \\ 0 & 0\end{bmatrix}\end{array} \]
where we made use of the identity $\hat{\omega}^2 =-I$. That is, we have
chosen a coordinate system in which $\hat{\xi}^{'}$ corresponds to a pure
rotation.  Thus, the exponentiation of this twist leads to an algebraically 
simple formula: 
  \[ e^{\phi\hat{\xi}^{'}} = \begin{bmatrix} e^{\phi\hat{\omega}} & 0\\ 
			0 & 1\end{bmatrix}. \]
Using Eq. (2.35) on page 42 of the MLS text, we can transform this result back to the
original coordinates to obtain the exponential that we seek:
  \[ e^{\phi\hat{\xi}} = ge^{\phi\hat{\xi}^{'}}g^{-1} = 
    \begin{bmatrix} e^{\phi\hat{\omega}} & 
                 (I-e^{\phi\hat{\omega}})\hat{\omega}\vec{v}\phi\\
	           0 & 1 \end{bmatrix} \]
which is clearly an element of $SE(2)$.

{\bf Part(b):} It is easy to see from part (a) that the twist
$\xi=(v_x,v_y,0)^T$ maps directly to the planar translation $(v_x,v_y)$.  

The twist corresponding to pure rotation about a point $\vec{q}=(q_x,q_y)$ can
be thought of as the Ad-transformation of a twist, $\xi^{'}=(0,0,\omega)$,
which is pure rotation, by a transformation, $g$, which is pure translation by
$\vec{q}$:
  \begin{equation}\label{eq:foo}
     \xi = {\rm Ad}_{h}\xi^{'} = (h \hat{\xi}^{'} h^{-1})^{\vee}
  \end{equation}
where 
  \[ h=\begin{bmatrix} I & \vec{q}\\ 0 & 1\end{bmatrix} \tab
  {\rm and} \tab \hat{\xi}^{'} = \begin{bmatrix} \hat{\omega} & 0\\
	\vec{0}^T & 0\end{bmatrix}.\] 
Expanding Eq. (\ref{eq:foo}) gives:
  \[ \xi = (h \hat{\xi}^{'} h^{-1})^{\vee} = 
     {\begin{bmatrix} \hat{\omega} & -\hat{\omega}\vec{q}\\ \vec{0}^T & 0\end{bmatrix}}^{\vee} 
    = \begin{bmatrix} q_y\\ -q_x\\ 1\end{bmatrix} \]
assuming $\hat{\omega} = \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$.

{\bf Part (e):} Let $\hat{V}^b$ denote the planar body velocity:
   \[ \hat{V}^b = \begin{bmatrix} \hat{\omega}^b & \vec{v}^b\\
			        \vec{0}^T & 0\end{bmatrix} \]
where $\hat{\omega}^b\in so(2)$, $\vec{v}^b\in\real^2$. Then the planar
spatial velocity is:
  \[\begin{array}{rl} 
    \hat{V}^s &= Ad_g \hat{V}^b = g\hat{V}^b g^{-1} \\ 
      &= \begin{bmatrix} R & \vec{p}\\ \vec{0}^T & 1\end{bmatrix}
         \begin{bmatrix} \hat{\omega}^b & \vec{v}^b\\ \vec{0}^T & 0\end{bmatrix} 
         \begin{bmatrix} R^T & -R^T\vec{p}\\ \vec{0}^T & 0\end{bmatrix}\\ 
      &= \begin{bmatrix} R\hat{\omega}^bR^T & -R\hat{\omega}^b R^T\vec{p}
	       + R \vec{v}^b\\ \vec{0}^T & 0\end{bmatrix} \end{array} \]
Therefore:
  \[ \hat{\omega}^s = R \hat{\omega}^b R^T \tab 
     \vec{v}^s = R\vec{v}^b - R\hat{\omega}^b R^T \vec{p}
               = R\vec{v}^b - \hat{\omega}^s\vec{p} \]
The spatial angular velocity can be simplified as follows:
  \[ \hat{\omega}^s = R \hat{\omega}^b R^T = 
     \begin{bmatrix} r_{11} & r_{12}\\ r_{21} & r_{22}\end{bmatrix}
     \begin{bmatrix} 0 & -\omega\\ \omega & 0\end{bmatrix}
     \begin{bmatrix} r_{11} & r_{21}\\ r_{12} & r_{22}\end{bmatrix}
     = \omega \begin{bmatrix} 0 & -\det(R)\\ \det(R) & 0\end{bmatrix}
     = \omega \begin{bmatrix} 0 & -1\\ 1 & 0\end{bmatrix} = \hat{\omega}^b \]

Using this result:
  \begin{eqnarray}
    \vec{v}^s &=& R\vec{v}^b - \hat{\omega}^s \vec{p}
               = R\vec{v}^b + \omega^b \begin{bmatrix} p_y\\ -p_x\end{bmatrix} 
		 \end{eqnarray}
Therefore:
  \[ V^s = \begin{bmatrix} \vec{v}^s\\ \omega^s\end{bmatrix}
         = \begin{bmatrix} R & \begin{bmatrix} p_y \\ -p_x \end{bmatrix} \\
		         \vec{0}^T & 1\end{bmatrix} V^b \]


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{\bf Problem 3:} (10 points) Problem 13 in MLS Chapter 2.

Let $\xi_a$ take the form:
   \begin{equation}
   \xi_a = \begin{bmatrix} \vec{\rho}_a\times\vec{\omega}_a + h\vec{\omega}_a\\
            \vec{\omega}_a)\end{bmatrix} \ .\end{equation}

{\bf Part (a):} The configuration of $A$ relative to $B$ is given by 
$g_{ab}^{-1}$:
	\[ g_{ab}^{-1} = \begin{bmatrix} R_{ab}^T & -R_{ab}^T \vec{p}_{ab}\\
				\vec{0}^T & 1 \end{bmatrix} \]
Thus, the representation of $\vec{\rho}_a$ and $\vec{\omega}_a$ in $B$ is:
   \begin{eqnarray}
     \vec{\rho}_b &=& R_{ab}^T\vec{\rho}_a - R_{ab}^T\vec{p}_{ab}\nonumber \\
     \vec{\omega}_b &=& R_{ab}^T\vec{\omega}_a 
   \end{eqnarray}
Substituting these equations into the expression:
   \begin{eqnarray}
      \xi_a &=& \begin{bmatrix}
        (\vec{\rho}_b\times\vec{\omega}_b + h\vec{\omega}_b \\
        \vec{\omega}_b \end{bmatrix} \nonumber \\
       &=& \begin{bmatrix} 
           -R_{ab}^T\hat{\omega}_a\vec{\rho}_a - R_{ab}^T \hat{p}_{ab}
		\vec{\omega}_a + hR_{ab}^T\vec{\omega}_a \nonumber \\
          R_{ab}^T \vec{\omega}_a \end{bmatrix} \\
       &=& \begin{bmatrix} R_{ab}^T & -R_{ab}^T\hat{p}_{ab} \\ 0 & R_{ab}^T
	    \end{bmatrix} \begin{bmatrix}
        (\vec{\rho}_a\times\vec{\omega}_a + h\vec{\omega}_a \nonumber \\
        \vec{\omega}_a \end{bmatrix} \\
       &=& Ad_{g_{ab}^{-1}} \xi_z
   \end{eqnarray}

{\bf Part (b):} The screw is now transformed by a rigid motion $g=
\begin{bmatrix} R & \vec{p} \\ \vec{0}^T & 1\end{bmatrix}$.  In the new 
coordinates, the screw location is described by
  \begin{eqnarray}
    \vec{\rho}^{'} &=& \vec{p} + R\vec{\rho}_a \nonumber \\
     \vec{\omega}^{'} &=& R \vec{\omega}_a  \nonumber
  \end{eqnarray} 
Hence, 
  \begin{eqnarray}
      \xi^{'} &=& \begin{bmatrix} \rho^{'} \times \vec{\omega}^{'} + 
	h \vec{\omega}^{'} \\ \vec{\omega}^{'} \end{bmatrix}
             = \begin{bmatrix} (\vec{p}+R\vec{\rho}_a)\times (R\vec{\omega}_a)
			+ h R \vec{\omega}_a \\
	R \vec{\omega}_a \end{bmatrix} \nonumber \\
    &=& \begin{bmatrix} R & \hat p R \\ 0 & R\end{bmatrix}
        \begin{bmatrix} \vec{\rho}_a\times \vec{\omega}_a + h\vec{\omega}_a\\
	\vec{\omega}_a \end{bmatrix} \nonumber \\
    &=& Ad_g \xi
  \end{eqnarray}


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{\bf Problem 4:} (10 points) Problem 14 in MLS Chapter 2.

{\bf Part (a):} Let $g\in SE(3)$ denote a homogeneous transformation matrix and
$Ad_g$ the Adjoint transformation associated to displacement $g$:
  \[ g= \begin{bmatrix} R & \vec{p}\\ \vec{0}^T & 1\end{bmatrix} \tab
    Ad_g = \begin{bmatrix} R & \hat{p}R\\ 0 & R\end{bmatrix} \]
Then:
  \[ g^{-1}=\begin{bmatrix} R^T & -R^T\vec{p}\\ \vec{0}^T & 1\end{bmatrix} \tab
     Ad_{g^{-1}} = \begin{bmatrix} R^T & -\widehat{(R^T\vec{p})} R^T\\
       		                \vec{0}^T & R^T\end{bmatrix}
        = \begin{bmatrix} R^T & -R^T\hat{p}\\ 0 & R^T\end{bmatrix} \]
where we have made use of the identity $\widehat{(R^T\vec{p})} = R^T \hat{p} 
R$.  Let's now compute $Ad_g Ad_{g^{-1}}$:
  \[ Ad_g Ad_{g^{-1}} = 
     \begin{bmatrix} R & \hat{p}R\\ 0 & R\end{bmatrix}
     \begin{bmatrix} R^T & -R^T\hat{p}\\ 0 & R^T\end{bmatrix}
   = \begin{bmatrix} I & 0\\ 0 & I\end{bmatrix} \]
Hence, $Ad_{g^{-1}}$ must equal $(Ad_g)^{-1}$ since $Ad_g Ad_{g^{-1}} = I$.

{\bf Part (b):} If 
  \[   g_1=\begin{bmatrix} R_1 & \vec{p}_1\\ \vec{0}^T & 1\end{bmatrix} \tab
       g_2=\begin{bmatrix} R_2 & \vec{p}_2\\ \vec{0}^T & 1\end{bmatrix} \]
Then
  \[ g_1 g_2=\begin{bmatrix} R_1 R_2 & \vec{p}_1 + R_1 \vec{p}_2\\
			\vec{0}^T & 1\end{bmatrix} \]
Hence:
  \[ \begin{array}{rl}
    Ad_{g_1 g_2} &= 
     \begin{bmatrix} R_1 R_2 & (\vec{p}_1 + R_1 \vec{p}_2)^{\hat{}} R_1 R_2\\ 
	  0 & R_1 R_2 \end{bmatrix} \\
    &= \begin{bmatrix} R_1 R_2 & \hat{p}_1 R_1 R_2 +R_1\hat{p}_2 R_1^T R_1 R_2\\ 
		0 & R_1 R_2  \end{bmatrix} \\ 
    &= \begin{bmatrix} R_1 R_2 & \hat{p}_1 R_1 R_2 +R_1\hat{p}_2 R_2\\ 
		0 & R_1 R_2  \end{bmatrix} \\ 
    &= \begin{bmatrix} R_1 & \hat{p}_1 R_1\\  0 & R_1 \end{bmatrix}
       \begin{bmatrix} R_2 & \hat{p}_2 R_2\\  0 & R_2 \end{bmatrix} = Ad_{g_1}
	Ad_{g_2} \end{array}\]

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{\bf Problem 5:} (15 points) Problem 18(a,b,c,d) in MLS Chapter 2.

{\bf Part (a):} Let 
    \[ g_{ab}(t) =\begin{bmatrix} R_{ab}(t) & \vec{d}_{ab} \\ \vec{0}^T & 1\end{bmatrix} \]
denote the relative location of a moving body (with a reference frame ``$B$'' attached
to the moving body) with respect to a fixed observer in frame ``$A$.'' The body velocity
of the moving body is:
  \begin{equation}
    \vec{V}^b_{ab} (g_{ab}^{-1}(t)\dot{g}_{ab}(t))^{\vee} 
                  = \begin{bmatrix} \vec{v}^2_{ab} \\ \vec{\omega}^b_{ab} \end{bmatrix}
                  = \begin{bmatrix} R_{ab}^T \dot{\vec{d}}_{ab} \\
                      (R_{ab}^T \dot{R}_{ab})^{\vee}\end{bmatrix}
  \end{equation}

To show the desired result, 
  \[ \begin{bmatrix} R_{ab} & 0 \\ 0 & R_{ab} \end{bmatrix} \vec{V}^b_{ab}
     =    \begin{bmatrix} R_{ab} & 0\\ 0 & R_{ab} \end{bmatrix} 
         \begin{bmatrix} R_{ab}^T \dot{\vec{d}}_{ab} \\ \vec{\omega}^b_{ab} \end{bmatrix}
     = \begin{bmatrix} R_{ab}R_{ab}^T\dot{\vec{d}}_{ab} \\ R_{ab}\vec{\omega}^b_{ab}\end{bmatrix}
     = \begin{bmatrix} \dot{\vec{d}}_{ab} \\ \vec{\omega}^s_{ab} \end{bmatrix} =\vec{V}^h_{ab} \]
where we have used the fact that $\vec{\omega}_{ab}^s=R_{ab}\vec{\omega}_{ab}^b$.

{\bf Part (b):} There are many ways to solve this problem.  For example,
you could either start with Proposition 2.14 or Proposition 2.15 on
page 59 of MLS which relate the velocities of three frames, A, B, and C.  
Let's choose Prop. 2.15:
  \begin{equation}\label{eq:3frame}
  V_{ac}^b = Ad_{g_{bc}^{-1}} V_{ab}^b + V_{bc}^b \end{equation}
Using the fact that
  \[ V_{ac}^h = \begin{bmatrix} R_{ac} & 0\\ 0 & R_{ac}\end{bmatrix} V_{ac}^b\]
Eq. (\ref{eq:3frame}) can be written as:
  \begin{equation}\label{eq:3hybrid}\begin{array}{rl}
  V_{ac}^h &= \begin{bmatrix} R_{ac} & 0\\ 0 & R_{ac}\end{bmatrix}
	(Ad_{g_{bc}^{-1}} V_{ab}^b + V_{bc}^b)\\
  &= \begin{bmatrix} R_{ac} & 0\\ 0 & R_{ac}\end{bmatrix}
     \begin{bmatrix} R_{bc}^T & -R_{bc}^T\hat{p}_{bc} \\ 0 & R_{bc}^T
      \end{bmatrix}  	V_{ab}^b 
    + \begin{bmatrix} R_{ac} & 0\\ 0 & R_{ac}\end{bmatrix} V_{bc}^b\\
  &=\begin{bmatrix} R_{ab} & -R_{ab}\hat{p}_{bc}\\ 0 & R_{ab}\end{bmatrix}
     V_{ab}^b + \begin{bmatrix} R_{ab} & 0\\ 0& R_{ab}\end{bmatrix}
      \begin{bmatrix} R_{bc} & 0\\ 0& R_{bc}\end{bmatrix} V_{bc}^b\\ 
  &= \begin{bmatrix} I & -\widehat{(R_{ab}p_{bc})}\\ 0& I\end{bmatrix}
     \begin{bmatrix} R_{ab} & 0\\ 0& R_{ab}\end{bmatrix} V_{ab}^b + 
     Ad_{R_{ab}} V_{bc}^h\\
  & \quad \\ 
  &=Ad_{-R_{ab}p_{bc}} V_{ab}^h + Ad_{R_{ab}} V_{bc}^h   \end{array}
  \end{equation}


{\bf Part (c):} Let frames A and B be stationary ``spatial'' frames, and let Frame C be fixed to a
moving body.  Let $V_{bc}^h$ be the hybrid velocity of the body, as seen by an observer in the B
frame.  If we now want to express this velocity as seem by an observer in the A frame (i.e.,
changing the spatial frame), we need to calculate $V_{ac}^h$.  You can do this using the results of
part (b) of this problem, which derived the result:
  \begin{equation}\label{eq:3hybridb}
  V_{ac}^h =Ad_{-R_{ab}p_{bc}} V_{ab}^h + Ad_{R_{ab}} V_{bc}^h 
  \end{equation}
If you chose this approach, then since A and B are stationary, $V_{ab}^h = 0$.
Hence, Eq. (\ref{eq:3hybrid}) takes the form:
    \[ V_{ac}^h = Ad_{R_{ab}} V_{bc}^h \]
Hence, the hybrid velocity is dependent on the orientation of the
spatial frame, but not its position.

Alternatively, if you don't want to rely upon part (b), you can recall
that the expression for the hybrid velocity is:
   \[ V_{ac}^h = 
      \begin{bmatrix} \dot{\vec{p}}_{ac} \\ \vec{\omega}^s_{ac}\end{bmatrix} \]
Since $\vec{p}_{ac}=\vec{p}_{ab} + R_{ab}\vec{p}_{bc}$, and $\vec{p}_{ab}$ is
constant:
	\[ \dot{\vec{p}}_{ac} = R_{ab}\dot{\vec{p}}_{bc}. \]
Similarly, $\vec{\omega}_{ac}=R_{ab}\vec{\omega}_{bc}$.  Hence, $V_{ac}^h$
is dependent of $\vec{p}_{ab}$, but not $R_{ab}$.

{\bf Part (d):} Let A be a stationary spatial frame.  Let B and C be two
different frames attached to a moving body.  Let us assume that the velocity
of the rigid body is given by $V_{ab}^h$.  If we now switch the location of
the body fixed frame from position B to position C, the hybrid velocity of the
body is given by $V_{ac}^h$.  Since B and C are both fixed in the body, then
$V_{bc}^h=0$ in Eq. (\ref{eq:3hybrid}).  Hence Eq. (\ref{eq:3hybrid}) reduces
to:
   \[ V_{ac}^h = Ad_{-R_{ab}p_{bc}} V_{ab}^h\]
Hence, the hybrid velocity in only dependent on $p_{bc}$, the position of the
body frame, and not on $R_{bc}$, the orientation of the body fixed frame.
Alternatively, you could compute $V_{ac}^h$ in a ``brute force'' way:

  \[\begin{array}{rl}
    V_{ac}^h &= \begin{bmatrix} \dot{\vec{p}}_{ac} \\ 
                                \vec{\omega}_{ac}\end{bmatrix} \ 
     =\ \begin{bmatrix}{d\over{dt}} (\vec{p}_{ab} + R_{ab}\vec{p}_{bc}) \\
             (\dot{R}_{ac} R_{ac}^T)^{\vee}\end{bmatrix} \ 
     =\ \begin{bmatrix}  \dot{\vec{p}}_{ab} + 
		    \hat{\omega}_{ab}^s R_{ab}\vec{p}_{bc}) \\
             (\dot{R}_{ab}R_{bc} R_{bc}^T R_{ab}^T)^{\vee}\end{bmatrix} \\
     &= \begin{bmatrix} \dot{\vec{p}}_{ab} + 
		    \hat{\omega}_{ab}^s R_{ab}\vec{p}_{bc}) \\
             \vec{\omega}_{ab}^s\end{bmatrix}
      = Ad_{-R_{ab}p_{bc}} V_{ab}^h \end{array} \]

Thus, the result only depends upon $\vec{p}_{bc}$, and not $R_{bc}$.

%%%----------------------------------------------------------------------%%%
{\bf Problem 6:} (20 Points)  You were asked to calculate the velocity
of a moving rigid body assuming that you know the velocities of 3 non-collinear
points ($P$, $Q$, and $R$) in the body.

Let points $P$, $Q$, and $R$ be located at positions $\vec{r}_1$,
$\vec{r}_2$, and $\vec{r}_3$ with respect to a fixed observer.  Let
the velocities of each of these points be $\dot{\vec{r}}_1$,
$\dot{\vec{r}}_2$, and $\dot{\vec{r}}_3$.  Recall that a spatial
velocity has the form
   \[ \hat{V}^s = \begin{bmatrix}\hat{\omega}^s & \vec{v}^s \\ 
		\vec{0}^T & 0\end{bmatrix} \]
In general, let $\vec{p}$ denote the location of a particle in the moving
body, with respect to the observing frame.  Recall that the velocity of this
particle, $\dot{\vec{p}}$, as seen in the observing frame, is related to the
location of the particle (as seen in the stationary observing frame) by
$\dot{\vec{p}} = \hat{V}^s \vec{p}$---where we have used the same notation
for homogeneous coordinates.  Thus, for each of the particles the following
relationship holds:
    \begin{eqnarray}
      \dot{\vec{r}}_1 &=& \vec{\omega}^s \times \vec{r}_1 + \vec{v}^s 
		\label{eq:one} \\
        \dot{\vec{r}}_2 &=& \vec{\omega}^s \times \vec{r}_2 + \vec{v}^s 
		\label{eq:two} \\
        \dot{\vec{r}}_3 &=& \vec{\omega}^s \times \vec{r}_3 + \vec{v}^s 
		\label{eq:three} \end{eqnarray}
Hence, we have three simultaneous equations in the unknowns $\vec{\omega}^s$
and $\vec{v}^s$.  To solve these equations, we follow closely the technique
that was used in the handout entitled `` Computing the Screw Parameters of a
Rigid Body Displacement.''

{\bf Step 1:} subtract Equation \ref{eq:three} from Equations (\ref{eq:one})
and (\ref{eq:two}).  
  \begin{eqnarray}
      \dot{\vec{r}}_1 - \dot{\vec{r}}_3 &=& \vec{\omega}^s \times(\vec{r}_1
	  -\vec{r}_3) \label{eq:onethree}\\
      \dot{\vec{r}}_2 - \dot{\vec{r}}_3 &=& \vec{\omega}^s \times(\vec{r}_2
	  -\vec{r}_3) \label{eq:twothree}\end{eqnarray}

{\bf Step 2:} Take the cross product of the vector $(\dot{\vec{r}}_2 -
\dot{\vec{r}}_3)$ with Equation (\ref{eq:onethree}).  

  \begin{equation}\label{eq:temp}
    (\dot{\vec{r}}_2 - \dot{\vec{r}}_3) \times 
   (\dot{\vec{r}}_1 - \dot{\vec{r}}_3) = 
   (\dot{\vec{r}}_2 - \dot{\vec{r}}_3)\times (\vec{\omega}^s \times(\vec{r}_1
	  -\vec{r}_3)). \end{equation}

{\bf Step 3:} We now use two facts.  First, we use the triple vector identity 
$\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - 
(\vec{a}\cdot\vec{b})\vec{c}$.  Second, from Equation (\ref{eq:twothree})
we know that $\dot{\vec{r}}_2 - \dot{\vec{r}}_3$ is orthogonal to 
$\vec{\omega}$.  Hence, Equation (\ref{eq:temp}) reduces to 
  \[  (\dot{\vec{r}}_2 - \dot{\vec{r}}_3) \times  
    (\dot{\vec{r}}_1 - \dot{\vec{r}}_3) = \vec{\omega} 
   [(\dot{\vec{r}}_2 - \dot{\vec{r}}_3)\cdot (\vec{r}_1 -\vec{r}_3)] . \]
The quantity $\vec{\omega}$ is then easily found as 
   \[ \vec{\omega} = 
   [(\dot{\vec{r}}_2 - \dot{\vec{r}}_3)\cdot (\vec{r}_1 -\vec{r}_3)]^{-1}
 (\dot{\vec{r}}_2 - \dot{\vec{r}}_3) \times  
    (\dot{\vec{r}}_1 - \dot{\vec{r}}_3). \]

{\bf Step 4:} Any one of Equations (\ref{eq:one}), (\ref{eq:two}), or
(\ref{eq:three}) could be used to solve for $\vec{v}^s$.



\end{document}
