\documentclass[12pt]{article}

%%%--------- Set up page size and formatting information ----------%%%
\usepackage{epsfig}
%\usepackage{amsmath}
\def\newpage{\par\vfill\eject}  %new page macro
\def\tab{\ \ \ \ \ }            %tab in five spaces
\def\dtab{{\tab}{\tab}}         %double tab

\setlength{\textwidth}{6.5in}
\setlength{\textheight}{9.0in}
\setlength{\topmargin}{-0.51in}
\setlength{\oddsidemargin}{0.0in}
\parskip 2ex
\parindent=0pt

%\input /home/jwb/tex/reals.tex 

\input reals.tex
%%%----------------------------------------------------------------------%%%
\begin{document}
%\input /home/jwb/tex/reals.tex
\input reals.tex
\centerline{\bf ME 115(a):  Solution to Homework $\#1$}
\vskip 0.5 true in

%%%-------------------------------------------------------------------------%%%
{\bf Problem 1:}

{\bf Part b:} Recall that the location of the pole is fixed in both 
the moving and observer reference frames.  Hence, before displacement, the 
pole is located at some position $^B\vec{p}$ as seen by an observer in the 
fixed $B$ frame.  After displacement, the observer in the body fixed $C$ frame 
also sees the pole in his/her coordinates at point $^B\vec{p}$.  However, the 
moving body has displaced relative to the fixed observer by amount $D_{12}=
(\vec{d}_{12},R_{12})$.  But points in the observer and displaced 
reference frames are related by a coodinate transform.  Since the pole is at the 
same location in both the fixed and moving frames, it must be true that:
   \[ ^B\vec{p} = \vec{d}_{12} + R_{12}\ \  ^B\vec{p}. \]
This equation can be solved to find the pole location:
   \[ ^B\vec{p} = (I-R_{12})^{-1}\vec{d}_{12} \]
(Of course, you need to show the fact that $(I-R_{12})$ is invertible.  It will always be invertible, except when $R_{12}=I$.  In this case, the motion is a pure translation, and the pole is the ``pole at infinity.'')

%begin AMPT 1/18/06
{\bf Part a:} 
From the problem statement we know the relationship between coordinate frames $A$ and $B$.  
Thus:
\[^A\vec{p} = \vec{d}_{01} + R_{01}\\ ^B\vec{p}.\]
The problem statement asks for $^A\vec{p}$ as a function of $R_{01}$, $R_{12}$, $\vec{d}_{01}$, and $\vec{d}_{12}$.  This can be done by substituting in the statement for  $^B\vec{p}$ that was found above.


\[^A\vec{p} = \vec{d}_{01} + R_{01}\\  (I-R_{12})^{-1}\vec{d}_{12}   \]

{\bf Part c:}
Since reference frame $B$ is fixed to the rigid body,
 \[   ^C\vec{p} = ^B\vec{p}  \]
%end AMPT 1/18/06
{\bf Problem 2:}
To find the pole of the displacement: $D_1 = (x,y,\theta)= 
(1.0,2.0,30.0^o)$ and $D_2=(x,y,\theta)=(2.0, 2.0, 45.0^o)$,
substitute into the above results:



\[ ^B\vec{p} = (I-R_{12})^{-1}\vec{d}_{12} = 
     \left(    \left[
     \begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}
     \right]
      - 
      \left[
      \begin{array}{cc} \cos(45^o) & -\sin(45^o) \\
                            \sin(45^o) & \cos(45^o) \end{array}\right]
                            \right)
                            ^{-1}
       \left[
       \begin{array}{c} 2.0\\ 2.0 \end{array}
       \right]
        = \left[ \begin{array}{c} -1.4142\\ 3.4142 \end{array}  \right] \]


\[ ^A\vec{p} = \vec{d}_{01}+R_{01}\\^B\vec{p} = 
 \left[
     \begin{array}{c} 1.0  \\ 2.0 \end{array}
     \right]
      +
      \left[
      \begin{array}{cc} \cos(30^o) & -\sin(30^o) \\
                            \sin(30^o) & \cos(30^o) \end{array}\right]
  \left[
       \begin{array}{c} -1.4142\\ 3.4142 \end{array}
       \right]
        = \left[ \begin{array}{c} -1.9319\\ 4.2497 \end{array}  \right] \]





As we showed in Problem 1, $^B\vec{p} = ^C\vec{p}$.

         

%%%-------------------------------------------------------------------------%%%
\vskip 0.2 true in
{\bf Problem 3:} To show that a transformation is a pure rotation when 
viewed
in a reference frame at the pole,  select a new
reference frame, denoted by $D$, whose basis vectors are parallel to 
Frame B 
and whose origin lies at the pole of the displacement. 
Let $\vec{p}$ denote the location of the pole, as seen by an observer
in Frame B. The location of Frame $B$ relative to Frame $D$ is a pure
translation of amount $-^{1}\vec{p}$, and therefore, $D_{DB} = 
(-\vec{p},I)$.
The displacement of the body from the first position to the second 
position,
as now observed in Frame $D$, is obtained by a similarity transform
$D_{DB} D_{12} D_{DB}^{-1}$:
\begin{eqnarray}
   D_{DB} D_{12} D_{DB}^{-1} 
   &=& (-\vec{p},I)(\vec{d}_{12},R_{12})(-\vec{p},I)^{-1} \\ 
   &=& (-\vec{p},I)(\vec{d}_{12},R_{12})(+\vec{p},I) \\ 
   &=& (-\vec{p},I)((\vec{d}_{12}+R_{12}\vec{p}),R_{12}) \\ 
   &=& ((\vec{d}_{12}+(R_{12}-I)\vec{p}),R_{12}) 
\end{eqnarray}
Hence, if 
$\vec{p}=-(R_{12}-I)^{-1}\vec{d}_{12}=(I-R_{12})^{-1}\vec{d}_{12}$,
then $D_{DB} D_{12} D_{DB}^{-1} = (\vec{0},R_{12})$.  I.e., as viewed 
in
reference Frame $D$, the displacement is a pure rotation by amount 
$R_{12}$.

%%%-------------------------------------------------------------------------%%%
\vskip 0.2 true in
{\bf Problem 4:}
(problem 3(c) in chapter 2 of the MLS text). Let
   \[ R= \left[ \begin{array}{ccc} r_{11} & r_{12} & r_{13} \\
                         r_{21} & r_{22} & r_{23} \\
                         r_{31} & r_{32} & r_{33} \end{array}\right]
  = \left[ \begin{array}{ccc} \vec{r}_1 & \vec{r}_2 & \vec{r}_3 \end{array}.\right]  \]
Expanding $det(R)$ using cofactors, one finds:
   \begin{eqnarray*} det(R) &=& r_{11}(r_{22}r_{33}-r_{32}r_{23}) + 
               r_{21}(r_{32}r_{13}-r_{12}r_{33}) + 
               r_{31}(r_{12}r_{23}-r_{22}r_{13}) \\
     &=& \vec{r}_1^T \cdot (\vec{r}_2\times \vec{r}_3) \end{eqnarray*}

%%%-------------------------------------------------------------------------%%%
\vskip 0.2 true in
{\bf Problem 5:}


{\bf Part (a): } 
\begin{figure}
\begin{center}
\includegraphics[width=5 in]{Reflection.pdf}
\caption{Diagram of reflection process}
\label{fig:reflection}
\end{center}
\end{figure}

There are many ways that one can prove that 
reflections  preserve length.  Here is one approach (see Figure 
\ref{fig:reflection}).



Select any two non-identical points, $A$ and $B$, in a rigid body.  
After
reflection, those points become $A^{'}$ and $B^{'}$.  Form the right 
triangle
$ABD$, where the line $BD$ is chosen to be perpendicular to the line 
$AA^{'}$.
Similary, in the reflected body, form the right triangle 
$A^{'}B^{'}D^{'}$.
Simple geometric arguments show that since the distance $|BD|$ and
$|B^{'}D^{'}|$ are equal, and the distances $|AD|$ and $|A^{'}D^{'}|$ 
are
equal, then $|AB|=|A^{'}B^{'}|$.  Hence, the distance between $A$ and 
$B$ is
preserved under reflection.  Since $A$ and $B$ were chosen randomly, 
the
result will hold for any non-identical pair of points in the body.  
Thus,
distance is always preserved under reflection.

{\bf Part (b):} Generally, physically meaningful planar displacements 
are not
equivalent to a single reflection.  To see this, define three points 
$(A,B,C)$
in the body of Figure \ref{fig:reflection}.  Because the body is rigid, 
one
can think of points $(A,B,C)$ as forming a rigid triangle.  Consider 
the
triangle formed from the reflected points $(A^{'},B^{'},C^{'})$.  Note 
that
it is impossible  physically translate $(A,B,C)$ to 
$(A^{'},B^{'},C^{'})$. 
Finally, note that any rigid body planar displacement can generally be 
realized
as the result of two sequential reflections.
 
%%%-------------------------------------------------------------------------%%%
\vskip 0.2 true in
{\bf Problem 6:} You were to ``prove'' that a body undergoing spherical 
motion has three degrees of freedom.  

A body undergoing spherical motion has one fixed point.  Let the body 
consist
of $N$ particles. Let $P_1$ denote the particle lying at the fixed 
point.  A
point in 3-dimensional Euclidean space normally requires 3 independent
variables to fix its location.  However, since $P_1$ does not move, it
actually has $0$ degrees-of-freedom (DOF).  Now consider a particle 
$P_2$ in
the body.  Particle $P_2$ has 3 DOF as a particle.  However, it is 
constrained
to lie a fixed distance, $d_{12}$ from particle $P_1$ due to the fact 
that
$P_1$ and $P_2$ are part of the same rigid body.  The fixed distance
relationship imposes one constraint on $P_2$.  Next consider a point 
$P_3$,
which lie a fixed distance from $P_1$ and $P_2$. Therefore, there are 
two
constraints on its location.  Now, consider a particle $P_4$.  Since 
its must lie a fixed distance from $P_1$, $P_2$, and $P_3$, there are
three constraints on its motion.  Particles $P_5$, $\ldots$, $P_N$
similarly have 3 constraints.

The total number of degrees of freedom of the $N$ particles are: 
$3(N-1) + 0$
= $3N-3$.  The total number of constraints on these particles are: $1 + 
2 +
3(N-3)$ = $3N-6$.  Hence, the total net DOF of a body is the number of 
freedoms
of the particles minus the number of constraints that bind them into a 
rigid
body: $(3N-3)-(3N-6) = 3$.


\end{document} 