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\title{2.  Kinematic foundations}
\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.  Kinematic foundations}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC}
\end{slide}

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% SLIDE 3
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\begin{slide}[R]{Kinematic foundations.}

We will focus on rigid motions in 
\begin{itemize}
\item[] the Euclidean plane ($\mathbb{E}^2$)
\item[] Euclidean three space ($\mathbb{E}^3$)
\item[] the sphere ($\mathbb{S}^2$)
\end{itemize}
Why the sphere?  Rigid motions of the sphere correspond to rotations
about a given point in $\mathbb{E}^3$.

\end{slide}
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% SLIDE 4
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\begin{slide}[R]{Kinematics foundations:  some definitions}

First, some general definitions.  Let $\mathbb{X}$ be the \emph{ambient space},
either $\mathbb{E}^2$, $\mathbb{E}^3$, or $\mathbb{S}^2$.

\begin{itemize}
\item
 A \emph{system} is a set of points in the space $\mathbb{X}$.

\item
A \emph{configuration} of a system is the location of every point in the
system.


\item
\emph{Configuration space}
is a metric space comprising all  
configurations of a given system.

(What kind of space is configuration space?  Devise a metric.)

(Note:  Every metric for cspace is sort of defective.)

\item
The \emph{degrees of freedom}
of a system is the dimension of the
configuration space.  (A less precise but roughly equivalent
definition: the minimum number of real numbers required to specify a
configuration.)

\end{itemize}



\end{slide}

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% SLIDE 5
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\begin{slide}[R]{Kinematics foundations:  systems, DOFs}

\begin{center}
%\tabular{lcc}
%\begin{unnumtable}
  \begin{tabular}{@{}lcc@{}}
%    \toprule
    \textbf{System} & \textbf{Configuration} & \textbf{DOFs} \\
 %   \colrule
    point in plane & $x$, $y$ & 2 \\
    point in space & $x$, $y$, $z$ & 3 \\
    rigid body in plane & $x$, $y$, $\theta$ & 3 \\
    rigid body in space & $x$, $y$, $z$, $\phi$, $\theta$, $\psi$ & 6
%    \botrule
  \end{tabular}
%\end{unnumtable}
%\endtabular
\end{center}

\end{slide}


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% SLIDE 6
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\begin{slide}[R]{Kinematics foundations:  rigid bodies, displacements}

Definitions:      
\begin{itemize}
\item[]
 A \emph{displacement}%
\index{displacements}
 is a change of configuration that does not change the
 distance between any pair of points, nor does it change the
 handedness of the system.


\item[]
A \emph{rigid body}%
\index{rigid body}
 is a system that is capable of displacements only.

\end{itemize}

\centerline{
\includegraphics[width=4in]{kinexs}}
\centerline{Transformations, rigid and otherwise.}


\end{slide}

%=======================================================================
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% SLIDE 7
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\begin{slide}[R]{Kinematics foundations:  moving and fixed planes}

We will consider displacements to apply to \emph{every} point
in the ambient space. 
E.g., displacements are described as motion of \emph{moving} plane relative
to \emph{fixed} plane.

\centerline{
\includegraphics[height=2in]{planes}}
\centerline{Moving and fixed planes.}


\end{slide}


%=======================================================================
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% SLIDE 8
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\begin{slide}[R]{Kinematics foundations:  rotations and translations}

 A \emph{rotation}%
\index{rotations}
 is a displacement that leaves at least one point fixed.


 A \emph{translation}%
\index{translation}
 is a displacement for which all points move equal distances 
 along parallel lines. 


\includegraphics[width=4in]{rotvsrot}

\end{slide}

%=======================================================================
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% SLIDE 9
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\begin{slide}[R]{Kinematics foundations:  digression for group theory}

A \emph{group} is a set of elements $X$ and a binary operator $\circ$
satisfying the following properties:
\begin{itemize}
\item[]\textbf{Closure:}  for all $x$ and $y$ in $X$, $x \circ y$ is in $X$.
\item[]\textbf{Associativity:}  for all $x$, $y$, and $z$ in $X$, $(x \circ y) \circ z$
is equal to $x \circ (y \circ z)$.
\item[]\textbf{Identity:}  there is some element, called $1$, such that for all $x$ in
$X$ $x \circ 1 = 1 \circ x = x$.
\item[]\textbf{Inverses:}  for all $x$ in $X$, there is some element called $x^{-1}$
such that $x \circ x^{-1} = x^{-1} \circ x = 1$.
\end{itemize}
(Did I remember them all?)

Some groups are commutative (Abelian) and some are not.  The integers with
addition are a commutative group.  Nonsingular $k$ by $k$ matrices with matrix
multiplication are a noncommutative group.

\end{slide}

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% SLIDE 10
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\begin{slide}[R]{Kinematics foundations:  Displacements as a group}
\begin{itemize}
\item[]
Every displacement $D$ can be described as
an operator on the ambient space $\mathbb{X}$, mapping every point $x$ to some
new point $D(x) = x^\prime$. 
\item[] The product of two displacements is the
composition of the corresponding operators, i.e.\ $(D_2 \circ D_1)
(\cdot) = D_2( D_1 (\cdot))$.  
\item[]The inverse of a displacement is just
the operator that maps every point back to its original position.  
\item[]The
identity is the null displacement, which maps every point to itself.
\end{itemize}
In other words:
\begin{quote}
The displacements, with functional composition, form a group.
\end{quote}
\end{slide}

%=======================================================================
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% SLIDE 11
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\begin{slide}[R]{Kinematics foundations:  $\mathbf{SE}(2)$, $\mathbf{SE}(3)$, and $\mathbf{SO}(3)$}
These groups of displacements have names:
\begin{itemize}
\item[]$\mathbf{SE}(2)$:  The special Euclidean group on the plane.
\item[]$\mathbf{SE}(3)$:  The special Euclidean group on $\mathbb{E}^3$.
\item[]$\mathbf{SO}(3)$:  The special orthogonal group.
\end{itemize}
Whence the names?
\begin{itemize}
\item[]\textbf{Special}:  they preserve handedness.  
\item[]\textbf{Orthogonal}:  referring to the connection with orthogonal matrices, which will be covered later.
\end{itemize}
\end{slide}

%=======================================================================
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% SLIDE 12
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\begin{slide}[R]{Kinematics foundations:  do displacements commute?}

Does $\mathbf{SO}(3)$ commute?  \textbf{NO!}  No, no, no.  (If you have found a commutative
way of representing spatial rotations, you are confused.)

\includegraphics[width=4in]{noncomm}

\end{slide}

%=======================================================================
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% SLIDE 13
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\begin{slide}[R]{Kinematics foundations:  do displacements commute?}

Does $\mathbf{SE}(3)$ commute?

Does $\mathbf{SE}(2)$ commute?

Does $\mathbf{SO}(2)$ commute?

\end{slide}

%=======================================================================
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% SLIDE 14
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\begin{slide}[R]{Time for a digression $\ldots$}

Next we look at $\mathbf{SE}(2)$, $\mathbf{SO}(3)$, and $\mathbf{SE}(3)$.

First, it helps if we contemplate the infinite $\ldots$
\end{slide}

%=======================================================================
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% SLIDE 15
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\begin{slide}[R]{The projective plane.}

The basic idea:
\begin{itemize}
\item[] Start with the Euclidean plane $\mathbb{E}^2$.
\item[] Add some points, the \textbf{ideal points} or the \textbf{points at infinity}.
\item[] Call the new structure the \textbf{projective plane}---$\mathbb{P}^2$.
\end{itemize}
You can do it formally by defining an ideal point for each set of parallel lines,
but we will employ a more concrete method using \textbf{homogeneous coordinates}.

\end{slide}

%=======================================================================
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% SLIDE 16
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\begin{slide}[R]{Homogeneous coordinates.}

Let the Cartesian coordinates of some point in $\mathbb{E}^2$ be
\[
        (\eta, \nu)
\]
Then we will say that
\[
        (x, y, w) \triangleq (w \eta, w \nu, w)
\]
are the \textbf{homogeneous coordinates} of the point, \textbf{provided}
\[
        w \neq 0
\]
To go from homogeneous to Cartesian:
\begin{equation} 
        \left(\begin{array}{c} x \\ y \\ w \\ \end{array}\right) \mapsto 
        \left(\begin{array}{c} x / w \\ y / w \\ \end{array}\right), w \neq 0
\end{equation} 


\end{slide}

%=======================================================================
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% SLIDE 17
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\begin{slide}[R]{Point in $\mathbb{E}^2$ versus line through origin of $\mathbb{E}^3$}


Scaling the homogeneous coordinates does \textbf{not} change the point!
\begin{equation}
        \left(\begin{array}{c} ax \\ ay \\ aw \\ \end{array}\right)     
        \mapsto 
        \left(\begin{array}{c} ax / aw \\ ay / aw \\ \end{array}\right)
        =
        \left(\begin{array}{c} x / w \\ y / w \\ \end{array}\right), a, w \neq 0
\end{equation}
So, homogeneous coordinates represent a point in $\mathbb{E}^2$ by a line through
the origin of $\mathbb{E}^3$.
\[
        \left(\begin{array}{c} x \\ y \end{array}\right)
        \leftrightarrow
        \left\{ \left(\left.\begin{array}{c} wx \\ wy \\ w \end{array}\right)\right| w \neq 0 \right\}
\]
\end{slide}
%=======================================================================
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% SLIDE 18
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\begin{slide}[R]{Central projection}

The Euclidean plane can be embedded as the $w = 1$ plane.

We can also embed a sphere of points satisfying $x^2 + y^2 + w^2 = 1$.

A line through the origin of $\mathbb{E}^3$
\begin{itemize}
\item[] intersects the sphere in \textbf{antipodal points}
\item[] intersects the $w=1$ plane at the appropriate point $(x/w, y/w)$.
\end{itemize}
These constructions are \textbf{central projection}, either to the sphere or to the plane.

\centerline{
\includegraphics[height=1.5in]{inftyproj}
}


\end{slide}

%=======================================================================
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% SLIDE 19
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\begin{slide}[R]{Ideal points}

\begin{itemize}
\item[]
The original idea:  extend $\mathbb{E}^2$ by adding some ideal points.

\item[]
Euclidean point:  line through origin of $\mathbb{E}^3$ intersecting $w=1$ plane.

\item[]
Ideal point:  line through origin of $\mathbb{E}^3$ parallel to $w=1$ plane.

\end{itemize}

With Cartesian coords, no place to put ideal points.  With homogeneous coordinates,
there's a big gaping hole!

\end{slide}
%=======================================================================
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% SLIDE 20
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\begin{slide}[R]{The projective plane}

So $\ldots$
\begin{itemize}
\item[]
define the projective plane $\mathbb{P}^2$ to be the set of lines through the origin of $\mathbb{E}^3$.

\item[]A line in $\mathbb{E}^2$ is represented by plane through origin of $\mathbb{E}^3$.

\item[]The ideal points form a line!  The \textbf{line at infinity}.  The equator of the embedded sphere.

\item[]``Parallel lines'' intersect at infinity.

\item[]Duality.  Two points determine a line.  Two lines determine a point.  Every axiom of the
projective plane has a dual axiom by switching ``line'' and ``point''.

\item[]Noneuclidean geometry!!!
\end{itemize}


\end{slide}

%=======================================================================
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% SLIDE 21
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\begin{slide}[R]{Next lecture:  plane kinematics}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC3}
\end{slide}



\end{document}
