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\title{21.  Pushing}
\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 21. \hfill Mechanics of Manipulation }

\begin{document}

\maketitle

%=======================================================================
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% SLIDE 2
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\begin{slide}[R]{Lecture 21.  Pushing.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC21.eps}
\end{slide}

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

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

\begin{itemize}
\item Finish the ``voting theorem''.
\begin{itemize}
\item We've proven that line of motion dictates rotation direction.
\item Prove that line of force dictates rotation direction.
\item Prove the voting theorem.
\end{itemize}
\item Application to stable pushing.
\end{itemize}

\end{slide}

%=======================================================================
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% SLIDE 4
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%
\begin{slide}[R]{Pushing}
Can we predict direction of rotation?
\begin{trivlist}
\item[]\defterm{Line of pushing} $l_P$ defined along vel of point in pusher.
\item[]\defterm{Line of motion} $l_M$ defined along vel of point in slider.
\item[]\defterm{Line of force} $l_F$ defined as usual.
\item[]Two edges of friction cone $l_L$ and $l_R$.
\end{trivlist}
\begin{figure}[h!]
\includegraphics[width=10cm]{pushrays}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\end{slide}

%=======================================================================
%
% SLIDE 5
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%
\begin{slide}[R]{Which way will it turn?}

  Easy to predict from $l_M$ or from $l_F$, but what you \emph{know}
  is $l_L$, $l_R$, and $l_P$.
\begin{trivlist}
\item[]Main result:  $l_L$, $l_R$, and $l_P$ vote on rotation direction.
\item[]First:  $l_M$ dictates rotation direction.
\item[]Second:  $l_F$ dictates rotation direction.
\end{trivlist}

\begin{figure}[h!]
\includegraphics[width=10cm]{pushrays}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\end{slide}

%=======================================================================
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% SLIDE 6
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\begin{slide}[R]{Line of motion dictates}

Theorem:  For quasistatic pushing of a rigid body in the plane, with uniform
coefficient of friction, the line of motion dictates the rotation
direction.

\begin{itemize}
\item[]Let $y$-axis be line of motion, let origin be contact point,
  let $x_{\mathrm{IC}}$ be IC coordinate, let $m\!_f(x_{\mathrm{IC}})$
  be frictional moment as function of IC.
\item[]Show $m\!_f(x_{\mathrm{IC}})$ is monotone decreasing.
\item[]Look at values at $0^+$, $0^-$, $\infty$, apply intermediate
  value theorem.
\end{itemize}

\begin{figure}[h!]
\includegraphics[width=7cm]{sl2prf}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\end{slide}

%=======================================================================
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% SLIDE 7
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\begin{slide}[R]{Line of force dictates $\ldots$}
Theorem:  For quasistatic pushing of a rigid body in the plane, with uniform
coefficient of friction, the line of force dictates the rotation
direction. \\
Proof:
\begin{itemize}
\item[]Choose origin at center of friction, construct limit surface.
\item[]Normals at $f_x$-$f_y$ plane are horizontal.
\item[]By convexity, normals in upper half point up, in lower half point down.
\end{itemize}

\end{slide}

%=======================================================================
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% SLIDE 8
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%
\begin{slide}[R]{Voting theorem}
\begin{minipage}[t]{7.5cm}
Theorem:  For quasistatic pushing of a planar rigid body with uniform
coefficient of friction, rotation direction is determined by
a vote $l_P$, $l_L$, and $l_R$.

\begin{itemize}
\item[]Construct voting tree.
\item[]If edges of friction agree, then so does line of force, and
  theorem follows.
\item[]Consider case where edges do not agree.
\begin{itemize}
\item[]$l_L$ votes $-$, $l_R$ votes $+$, and $l_P$ votes $-$.  The majority
is $-$.  
\item[]Assume positive rotation.  So $l_F$ and $l_M$ would vote $+$ by
  previous theorems.  If $l_M$ is right of $\vv r_0$ then it is right
  of $l_P$, so we have right sliding.  So $l_F = l_L$:  a contradiction.
\end{itemize}
\end{itemize}
\end{minipage}
\hfill
\begin{minipage}[t]{2.5cm}
\begin{figure}[h!]
\includegraphics[width=2.5cm]{thmsl4}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 9
%
%
\begin{slide}[R]{The voting theorem really works.}
\begin{trivlist}
\item[]Demo on overhead.
\item[]It tells you which way it turns but 
\begin{itemize}
\item[]\emph{not} how fast, and
\item[]\emph{not} about what IC.
\end{itemize}
\item[]Very useful when pushing with a translating edge.
\end{trivlist}

\end{slide}

%=======================================================================
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% SLIDE 10
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%
\begin{slide}[R]{Stable pushing}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]Sometimes we want to turn while pushing!
\item[]How can we achieve a stable push?
\item[]No slip of slider along pusher.
\item[]No rolling of slider on pusher.
\item[]Voting theorem by itself is not enough.  We need more
constraints on the IC.  
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{pushbox}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\begin{figure}[h!]
\includegraphics[width=4cm]{pushfail}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 11
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%
\begin{slide}[R]{Peshkin's bound}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]The voting theorem is a bound on IC's.  It tells you whether
the IC is in the positive plane, the negative plane, or the line at
infinity.  We need tighter bounds! 
\item[]Circumscribe slider support $R$ by a circle centered at center
of friction. 
\item[]Construct IC for every possible support dipod.
\item[]Conjecture:  resulting locus includes every possible support,
not just dipods. 
\item[]If we allow line of force to vary, locus sweeps out ``tip line''.
\item[]Note duality of tip line to contact point!
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{peshlocus}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\begin{figure}[h!]
\includegraphics[width=4cm]{tipline}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 12
%
%
\begin{slide}[R]{The ``bisector bound''}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]Construct line from contact to center of friction.
\item[]Construct perpendicular bisector.
\item[]IC is on c.o.f. side of perp bisector.
\item[]Proof never published.
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{bibound}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 13
%
%
\begin{slide}[R]{The vertical strip bound}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]Project support region $R$ onto pushing line of force.
\item[]IC must fall in inverse projection.
\item[]Proof:  Force balance impossible otherwise.
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[height=3cm]{vstrip}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\begin{figure}[h!]
\includegraphics[height=3cm]{vstriproof}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 14
%
%
\begin{slide}[R]{Not slipping off the pusher}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]Slipping of slider on pusher corresponds to left or right edge of FC.
\item[]No slipping:  interior of FC.
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{stab1}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\begin{figure}[h!]
\includegraphics[width=4cm]{stab2}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 15
%
%
\begin{slide}[R]{Not rolling off the pusher}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]Rolling corresponds to force through left or right corner of block.
\item[]Not rolling:  line of force between corners.
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{stab3}
%\includegraphics[height=3cm]{*file*}
\end{figure}\
\begin{figure}[h!]
\includegraphics[width=4cm]{stab4}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 16
%
%
\begin{slide}[R]{Combining constraints, planning a path}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]We eliminate all failure modes;
\item[]we can also incorporate nonholo constraints of the pusher;
\item[]and we plan a path using NHP.
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{stab5}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\begin{figure}[h!]
\includegraphics[width=4cm]{pushfinal}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}


%=======================================================================
%
% SLIDE last
%

\begin{slide}[R]{Next:  Parts orienting.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC22}
\end{slide}
\end{document}



\end{document}
