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\title{3.  Planar kinematics}
\subtitle{ Mechanics of Manipulation}
\author{Matt Mason}
\institution{Carnegie Mellon}
\email{matt.mason@cs.cmu.edu \\
http://www.cs.cmu.edu/\symbol{"7E}mason}
\slideCaption{Lecture 2. \hfill Mechanics of Manipulation }

\begin{document}

\maketitle

%=======================================================================
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% SLIDE 2
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\begin{slide}[R]{Lecture 2.  Planar kinematics.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC3}
\end{slide}

%=======================================================================
%
% SLIDE 3
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%

\begin{slide}[R]{Outline.}

\begin{itemize}
\item  One more general theorem:  Displacement $=$ translation $\circ$
       rotation.  
\item  Some fundamental theorems on planar motion.
\item  Main result:  every planar motion has a rotation center in
       projective plane. 
\item  Centrodes.
\end{itemize}

\end{slide}

%=======================================================================
%
% SLIDE 4
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%

\overlays{6}{
\begin{slide}[R]{Decomposition of displacements}

Theorem 2.2:
For any displacement $D$ of the Euclidean spaces $\mathbb{E}^2$ or
$\mathbb{E}^3$, and any point $O$, $D$ can be expressed as the
composition of a translation with a rotation about $O$. \\[6pt]

Proof: 
\fromSlide{2}{
\begin{itemize}
\item[]
Let $O^\prime$ be the image of $O$ under $D$\@.  
\item[]
Let $T$ be the translation taking $O$ to $O^\prime$. 
\fromSlide{3}{
\item[]
Consider the displacement $T^{-1} \circ D$.  Where does it map $O$?
\[      (T^{-1}\circ D) (O) = \onlySlide*{3}{?}\fromSlide{4}{O} \]}
\fromSlide{5}{%
\item[]
So $T^{-1}\circ D$ is a rotation; call it $R$\@. }
\fromSlide{6}{%
\item[]
So then $T \circ R = T \circ T^{-1} \circ D = D$ is the desired
decomposition.  QED} 
\end{itemize}}
%
\end{slide}}

%=======================================================================
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% SLIDE 5
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%

\overlays{6}{
\begin{slide}[R]{Decomposition of displacements}
Note:
\begin{itemize}
\item
Instead of $R = T^{-1}\circ D$, $D = T \circ R$,\\
we could have $S = D \circ T^{-1}$, and $D = S \circ T$.\\
So you can do rotation first or translation first $\ldots$
\fromSlide{2}{%
\item
$\ldots$ but rotation depends on order---$S \neq R$.}%
\fromSlide{3}{%
\item
Theorem 2.2 is basis for most common representation of displacements.}
\fromSlide{4}{%
\item
The decomposition is not unique:  it depends on the choice of $O$.}
\fromSlide{5}{%
\item
The proof extends to arbitrary $\mathbb{E}^n$.}
\fromSlide{6}{%
\item
Note how simple it is to prove using group theory!}
\end{itemize}
\end{slide}}

%=======================================================================
%
% SLIDE 6
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%

\begin{slide}[R]{Planar kinematics}

That is all we will do on ``general'' kinematics.  On to planar kinematics.  \\
What can we say about rigid motions of $\mathbb{E}^2$?

Theorem 2.3:  A planar displacement is completely determined by the
motion of any two points.\\[6pt] 

Proof:  Construct a coordinate frame $\ldots$



\end{slide}

%=======================================================================
%
% SLIDE 7
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%

\begin{slide}[R]{Planar kinematics:  every $D$ is an $R$ or a $T$}

Now for the big one:\\[6pt]
Theorem 2.4:  Every planar displacement is either a translation or a
rotation.\\[6pt] 

\begin{minipage}[t]{6cm}
Not a proof:
\begin{itemize}
\item[]
Pick two points $A$ and $B$.
\item[]
Let $A^\prime$ and $B^\prime$ be the images.
\item[]
Construct perpendicular bisectors.
\item[]
Intersection gives fixed point.
\item[]
Why?  Preserves distance from $A$ and from $B$, so $\ldots$
\item[]
Okay, not a proof, but a useful construction.
\end{itemize}
\end{minipage}
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{rotpole.eps}
\end{figure}
\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 8
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%

\begin{slide}[R]{Planar kinematics:  every $D$ is an $R$ or a $T$}

Theorem 2.4:  Every planar displacement is either a translation or a
rotation.\\[6pt] 

\begin{minipage}[t]{8cm}
Proof:
\begin{itemize}
\item[]
Pick any point $A$.  We can assume $A \neq A^\prime$.
 \item[]
Pick $B$ the
midpoint of the line segment $\overline{A\!A^\prime}$.  We can assume
$B^\prime$ is not on  $\overline{A\!A^\prime}$.
\item[]
Construct perp to $AB$ at $B$, and perp to  $A^\prime \! B^\prime$ at
$B^\prime$. They are not parallel.  Let $M$ be their intersection.
\item[]
Consider the rotation that maps $A$ to $A^\prime$ and $M$ to itself.
Where does it map $B$?  Preservation of distance gives two candidates,
and we can exclude one. 
\item[]
So our rotation maps $B$ to $B^\prime$.  It is the given
displacement.  QED.
\end{itemize}
\end{minipage}
\begin{minipage}[t]{2cm}
\begin{figure}[h!]
\includegraphics[width=2cm]{rotproof.eps}
\end{figure}
\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 9
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%

\overlays{4}{
\begin{slide}[R]{Planar kinematics.  Rotation centers}
\begin{minipage}[t]{6cm}
\begin{itemize}
\item[]
So, every planar displacement is a rotation or a translation.
\item[]
Consider again construction of rotation centers from
the motion of two points.  How does it fail when $\overline{A\!A^\prime}$
is parallel to $\overline{B\!B^\prime}$?
\fromSlide{2}{
\item[]
The perpendiculars are parallel.  There is no intersection, hence no
rotation center.} 
\fromSlide{3}{
\item[]
But, in the projective plane they \emph{do} intersect!!!}
\end{itemize}
\end{minipage}
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{rotpole.eps}
\end{figure}
\end{minipage}\\[18pt]
\fromSlide{4}{
\centerline{\emph{Every planar displacement is a rotation about a
point in the projective plane.}}} 
\end{slide}}

%=======================================================================
%
% SLIDE 10
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%

\begin{slide}[R]{Displacements, paths, trajectories.}
\begin{minipage}[t]{6cm}
\begin{description}
\item[Displacement]Discontinuous change of configuration. 
\item[Trajectory]Configuration a continuous function of
  time:  a continuous curve $q(t)$ in configuration space. 
\item[Path]  A curve $q(s)$ in configuration space parameterized
  perhaps by arc length.
\end{description}

For differentiable trajectory $q(t)$ or a path $q(s)$ we have
\emph{velocity} $dq/dt$ or \emph{differential change in configuration} $dq$. 

To construct rot'n center for diff'l displacement:
\end{minipage}
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{DispTrajPath}
\end{figure}
\begin{figure}[h!]
\includegraphics[width=4cm]{velpole.eps}
\end{figure}
\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 11
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%

\begin{slide}[R]{Planar kinematics.  ICs for 4bar linkages.}
\begin{minipage}[t]{6cm}
Terminology:
\begin{description}
\item[Rotation pole; rotation center] for displacements.
\item[Instantaneous center; IC; velocity center; velocity pole] for
  velocities or differential displacements.
\end{description}
Example:  Four bar linkages!
\begin{itemize}
\item[] Base link is fixed.
\item[] Two links that either translate or rotate w.r.t. base link.
\item[] The \emph{coupler link}, which can make all sorts of
  interesting motions.
\end{itemize}
Construct the ICs for two different four-bar linkages.
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[height=2.5cm]{bilat1}
\end{figure}
\begin{figure}[h!]
\includegraphics[height=2.5cm]{bilat2}
\end{figure}
\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 12
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%

\begin{slide}[R]{Planar kinematics.  Centrodes.}
\begin{minipage}[t]{6cm}
\begin{itemize}
\item[] Take an arbitrary continuous planar motion.  Generally the IC
  moves.
\item[] Plot the IC in the fixed plane.  That gives the \emph{fixed
    centrode}.
\item[] Plot the IC in the moving plane.  That gives the \emph{moving
    centrode}.
\item[] For any given time, the two curves must touch at the IC.  And
  the moving plane rotates about the IC.
\item[] I.e. the moving centrode rolls without slipping on the fixed
  centrode.
\item[] It even works for discontinuous motions.  If the centrodes are
  polygonal they are called \emph{central polygons} and you get a
  sequence of displacements.
\end{itemize}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{astroidcent}
\end{figure}
\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 13
%
%

\begin{slide}[R]{Problem set 1.}
\begin{minipage}[t]{6cm}
\begin{enumerate}
\item DOFs of a line in $\mathbb{E}^3$.
%\item Example of planar displacements that do not commute.
\item Centrodes (central polygons) for mobile robot.
\item Centrodes (central polygons) for fridge.
\item Centrode for Chebyshev's
  linkage.
  \emph{Show them my solution for Reuleaux's example.}
\end{enumerate}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[height=3cm]{floorbot}
\end{figure}
\begin{figure}[h!]
\includegraphics[height=2cm]{fridge}
\end{figure}
\end{minipage}
\end{slide}

%================================================================
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% SLIDE 14
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%

\begin{slide}[R]{Centrodes for Watt's linkage}
\begin{center}
\movie{0.57 \linewidth}{0.6 \linewidth}{../figs/WattsCentroden.mov}
\end{center}
\end{slide}

%=======================================================================
%
% SLIDE 15
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%

\begin{slide}[R]{Planar kinematics.  False ICs.}
\begin{minipage}[t]{6cm}
Review procedure for linkage IC construction
\begin{enumerate}
\item Reduce the constraints to point-velocity constraints.
\item Construct perpendiculars to allowed velocities at each point.
\item Intersection of perpendiculars are candidate ICs
\begin{itemize}
\item[] No intersection means no ICs.  It must be immobile.
\item[] But parallel lines intersect at infinity.
\end{itemize}
\end{enumerate}
But existence of intersection does not imply mobility!!!
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{bifalse}
\end{figure}
\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 15
%

\begin{slide}[R]{Next:  spherical, spatial kinematics; start on constraint.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC4}
\end{slide}



\end{document}
