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\title{17.  Planar Sliding}
\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 17. \hfill Mechanics of Manipulation }

\begin{document}

\maketitle

%=======================================================================
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% SLIDE 2
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\begin{slide}[R]{Lecture 17.  Planar Sliding.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC17}
\end{slide}

%=======================================================================
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% SLIDE 3
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\begin{slide}[R]{Outline.}

\begin{trivlist}
\item Motivation and overview.
\item Force and moment of planar sliding.
\item The limit surface.
\end{trivlist}

\end{slide}


%=======================================================================
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% SLIDE 4
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\begin{slide}[R]{Planar sliding}
Examples:
\begin{trivlist}
\item[]In a gripper.  Immobilization; compliant motion.
\item[]Workholding on a workbench.
\item[]Adjusting position on a workbench.
\item[]Handling large or small things.
\item[]Handling things in bulk.
\item[]Grasping and placing.
\end{trivlist}

\end{slide}

%=======================================================================
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% SLIDE 5
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\begin{slide}[R]{What would we like to know?}
\begin{trivlist}
\item[]Fundamental mechanics:  relation of motion to reaction forces.
\item[]Given additional applied force or kinematic constraint, predict motion.
\item[]Given desired motion, find required applied forces,
  constraints, initial velocity.
\item[]Examples:
\begin{itemize}
\item[]Push the table.
\item[]Push a block into the corner.
\end{itemize}
\end{trivlist}

\end{slide}

%=======================================================================
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% SLIDE 6
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\begin{slide}[R]{Pressure distribution is key}
\begin{trivlist}
\item[]Frictional forces are distributed over support surface.
\item[]Coulomb's law gives each frictional force a weight proportional
  to pressure.
\item[]Pressure is generally underdetermined.
\item[]Classify problems by pressure assumptions:
\begin{itemize}
\item[]Determined:
\begin{itemize}
\item[]Tactile sensor;
\item[]One, two, or three contact points balancing known weight;
\item[]Linear (elastic layer between two flat rigid bodies) balancing
  known weight;
\end{itemize}
\item[]Underdetermined:
\begin{itemize}
\item[]Known support region;
\item[]Balancing known weight;
\end{itemize}
\end{itemize}
\end{trivlist}

\end{slide}
%=======================================================================
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% SLIDE 7
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\begin{slide}[R]{Force and moment of planar sliding}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]Let $R$ be support region.
\item[]Let $\vv v(\vv r)$ be vel of some point $\vv r \in R$.  
\item[]Let $p(\vv r)$ be the pressure at $\vv r$,
\item[]Let $d\!A$ be element of area at $\vv r$, 
\item[]Then normal force at $\vv r$ is given by
\begin{equation*}
        p(\vv r) \, d\!A
\end{equation*}
\item[]Assume $\mu$ uniform over $R$.
\item[]Coulomb's law at $\vv r$:
\begin{equation}
        - \mu \, \frac{\vv v(\vv r)}{|\vv v(\vv r)|} p(\vv r) \, d\!A
\end{equation}
for $|\vv v(\vv r)| \neq 0$.
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=6cm]{pushdefs}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
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% SLIDE 8
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\begin{slide}[R]{Integrating$\ldots$}
To obtain total force and moment we integrate over $R$:
\begin{align*}
 \vv f_f & = -\mu \, \int_R \frac{\vv v(\vv r)}{| \vv v(\vv r) |} p(\vv r) \, d\!A  \\
 \vv n_f & 
  = -\mu \, \int_R \vv r \btimes \frac{\vv v(\vv r)}{| \vv v(\vv r) |} p(\vv r) \, d\!A
\end{align*}
For known pressure distribution we can evaluate the integrals.
Generally we cannot.

\end{slide}

%=======================================================================
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% SLIDE 9
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\begin{slide}[R]{Force and moment for translation}
Although the integrals cannot be evaluated, they can be simplified for translational sliding.
\begin{trivlist}
\item[]Assume translation:  $\vv v(\vv r) = \vv v$.
\item[]We can pull velocity out of integral:
\begin{align*}
 \vv f_f & =  -\mu\,\frac{\vv v}{|\vv v|}\int_R p(\vv r) \, d\!A 
 \\
 \vv n_f & 
   =  - \mu \, \int_R \vv r \, p(\vv r) \, d\!A 
                \, \btimes \frac{\vv v}{|\vv v|} 
\end{align*}

\end{trivlist}

\end{slide}
%=======================================================================
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% SLIDE 10
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\begin{slide}[R]{Center of friction}
\begin{trivlist}
\item[]Let $\vv f_0$ be the total normal force:
\[        f_0  =  \int_R p(\vv r) \, d\!A  \]
\item[]Let $\vv r_0$ be the centroid of the pressure distribution.
\[        \vv r_0  =  \frac{1}{f_0} \int_R \vv r \, p(\vv r) \, d\!A \]

\item[]Substituting above:
\begin{align*}
        \vv f_f & =  - \mu \, \frac{\vv v}{| \vv v| } f_0 \\
        \vv n_f & =  \vv r_0 \btimes \vv f_f 
\end{align*} 
\item[]This is Coulomb's law for a point contact!  Call $\vv r_0$ the
  \emph{center of friction}.  In translation, frictional force acts through
  center of friction.
\end{trivlist}

\end{slide}

%=======================================================================
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% SLIDE 11
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\begin{slide}[R]{Rotation}
\begin{trivlist}
\item[]Let $\vv r_{\mathrm{IC}}$ be the instantaneous center of rotation
\item[]Velocity at $\vv r$ is:
\begin{align*}
        \vv v(\vv r) & =  \vv \bomega \btimes (\vv r - \vv r_{\mathrm{IC}}) \\
                        & =  \dot \theta \, \uu k \btimes (\vv r - \vv r_{\mathrm{IC}})
\end{align*}
\item[]Direction of motion at $\vv r$ is:
\begin{equation*}
 \frac{\vv v(\vv r)}{| \vv v(\vv r) |} 
   = \sgn(\dot\theta) \, \uu k 
        \btimes \frac{\vv r - \vv r_{\mathrm{IC}}}{| \vv r - \vv r_{\mathrm{IC}} |}
\end{equation*}

\item[]Substituting
\begin{align*}
 \vv f_f & = -\mu \, \sgn(\dot\theta) \, \uu k 
 \btimes \int_R \frac{\vv r -\vv r_{\mathrm{IC}}}{|\vv r -\vv
   r_{\mathrm{IC}}|} p(\vv r) \, d\!A \\
 n_{fz} &  = - \mu \, \sgn(\dot\theta) 
        \int_R \vv r \cdot \frac{\vv r - \vv r_{\mathrm{IC}}}{| \vv r - \vv r_{\mathrm{IC}} |}
                 p(\vv r) \, d\!A
\end{align*}

\end{trivlist}

\end{slide}

%=======================================================================
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% SLIDE 12
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\begin{slide}[R]{Force to motion mapping}
\begin{trivlist}
\item[]Define \defterm{frictional load} to be wrench applied by slider
to planar support.  This is a change of sign.
\item[]The integrals give the frictional load wrench as a function of
  slider velocity twist.
\item[]Consider the relation of frictional load to slider velocity for
  a finite number of support points.  We will see that the
  motion-force mapping is neither one-to-many nor many-to-one.
\item[]The Limit Surface is an elegant geometrical representation of
  the motion-force mapping.
\end{trivlist}

\end{slide}

%=======================================================================
%
% SLIDE 13
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%
\begin{slide}[R]{One point of support}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[] Let $\vv v$ be the velocity of the particle.
\item[]Let $\vv f$ be the frictional load.
\item[]Coulomb's law can be stated:
\begin{description}
\item[slip:]    $\vv f \parallel \vv v$, and $|\vv f| = \mu f_n$,
                where $\mu$ is the coefficient of friction, and $f_n$
                is the support force.

\item[stick:]   $|\vv f| \leq \mu f_n$.
\end{description}
\end{trivlist}
\end{minipage}
\end{slide}
%=======================================================================
%
% SLIDE 14
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%
\begin{slide}[R]{Limit Curve:  maximum power inequality}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]Consider the set of all possible frictional loads, with $f_n$
  fixed.  That is a disk centered at the origin of radius $\mu f_n$.
\item[]Define the \defterm{limit curve LC} to be the circle of radius
  $\mu f_n$ at the origin of force space.
\item[]Coulomb's law is equivalent to the \defterm{maximum power
    inequality}:
\begin{equation*}
        \forall_{ \vv f^* \in \mathrm{LC} }  \;\;
          (\vv f - \vv f^*) \cdot \vv v \geq 0
\end{equation*}
\item[]I.e. motion $\vv v$ yields a load that is extremal in the $\vv
  v$ direction.
\item[]When slip occurs $\vv f$ is on the limit curve, and $\vv v$ is
  \emph{normal} to the limit curve at $\vv f$.
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{lc}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 15
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%
\begin{slide}[R]{More than one point}
\begin{trivlist}
\item[]Let $\vv r$ vary over
the support region
\item[]Construct a limit curve $\mathrm{LC}(\vv r)$ at
each point $\vv r$, 
\item[]At each point $\vv r$ the maximum power inequality
holds:

\begin{equation*}
        \forall_{ \vv f^*(\vv r) \in \mathrm{LC}(\vv r) }  \;\;
          (\vv f(\vv r) - \vv f^*(\vv r)) \cdot \vv v(\vv r) \geq 0
\end{equation*}

\item[]How do we put them all together???  Wrenches and twists!
\end{trivlist}

\end{slide}

%=======================================================================
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% SLIDE 16
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\begin{slide}[R]{Frictional load wrench and velocity twist}
\begin{trivlist}
\item[]Let $\vv p$ be the total frictional load wrench
\begin{equation*}
        \vv p = \left(\begin{array}{c} 
                        f_x \\ f_y \\ n_{0z} \end{array}\right)
              = \sum_{\vv r} \left(\begin{array}{c} 
                                f_{x}(\vv r) \\ f_{y}(\vv r) \\ \vv r \btimes \vv f(\vv r)
                                \end{array}\right)
\end{equation*}
\item[]Let $\vv q$ be the velocity twist
\begin{equation*}
        \vv q = \left(\begin{array}{c} 
                        v_{0x} \\ v_{0y} \\ \omega_z \end{array}\right)
\end{equation*}
\end{trivlist}

\end{slide}

%=======================================================================
%
% SLIDE 17
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\begin{slide}[R]{Power of Coulomb load versus arbitrary load}
\begin{trivlist}
\item[]Let $\vv f(\vv r)$ be a distribution of 
frictional loads satisfying Coulomb's law
\item[]Let $\vv f^*(\vv r)$ be arbitrary, except that at each
$\vv r$, $\vv f^*(\vv r)$ is in the corresponding limit curve:
\begin{equation*}
        \forall_r \;\; \vv f^*(\vv r) \in \mathrm{LC}(\vv r)
\end{equation*}
\item[]Let $\vv p$ and $\vv p^*$ be the total frictional load wrench for $\vv
f(\vv r)$ and $\vv f^*(\vv r)$ respectively.  
\item[]The power dissipated by the Coulomb load can be written two ways
\begin{equation*}
        \vv p \cdot \vv q = \sum_{\vv r} \vv f(\vv r) \cdot \vv v(\vv r)
\end{equation*}
\item[]Similarly for the arbitrary load can write
\begin{equation*}
        \vv p^* \cdot \vv q = \sum_{\vv r} \vv f^*(\vv r) \cdot \vv v(\vv r)
\end{equation*}

\end{trivlist}

\end{slide}
%=======================================================================
%
% SLIDE 18
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%
\begin{slide}[R]{Power difference between Coulomb and arbitrary loads}
\begin{trivlist}
\item[]Taking the difference yields
\begin{equation*}
        (\vv p - \vv p^*) \cdot \vv q
        = \sum_{\vv r} (\vv f(\vv r) - \vv f^*(\vv r)) \cdot \vv v(\vv r)
\end{equation*}

\item[] By maximum power inequality, every term in the sum on the
  right hand side is non-negative.  So:
\begin{equation*}
        (\vv p - \vv p^*) \cdot \vv q \geq 0
\end{equation*}

\item[]Another maximum power inequality!  This time in wrench space!!!
\item[]Summary: Consider all loads such that $\vv f^*(\vv r) \in
  \mathrm{LC}(\vv r)$.  The correct load maximizes the power
  dissipated.
\end{trivlist}

\end{slide}

%=======================================================================
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% SLIDE 19
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\begin{slide}[R]{Limit Surface}
\begin{trivlist}
\item[]No motion:  $\vv q = 0$, power dissipated is zero for any frictional load.
\item[]So, form the set of all
possible total frictional load wrenches $\vv p^*$
\item[]Define the
\defterm{limit surface} to be the surface of this set.  
\item[]Maximum power inequality:
the frictional
load wrench yields maximum power over all wrenches in the
limit surface.  
\item[]Equivalently: during slip the total frictional load
wrench $\vv p$ lies on the limit surface, and the velocity twist $\vv
q$ is normal to the limit surface at $\vv p$.

\end{trivlist}

\end{slide}

%=======================================================================
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% SLIDE 20
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\begin{slide}[R]{Example:  two point contact}
\begin{minipage}[t]{6cm}
Two points of contact, support evenly divided.
\begin{enumerate}
\item
        $\mathrm{LS}_a$:  all frictional loads at $a$.  An elliptical disk in
       wrench space.

\item

        $\mathrm{LS}_b$:  all frictional loads at $b$.  
        Also an elliptical disk.

\item
        The desired limit surface LS is the Minkowski sum:

\[  \mathrm{LS}= \{\vv w_a +
        \vv w_b \; | \;\vv w_a \in \mathrm{LS}_a, \vv w_b \in
        \mathrm{LS}_b\} \]
\end{enumerate}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{barbell}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
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% SLIDE 21
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\begin{slide}[R]{Barbell Limit Surface}
\begin{figure}[h!]
\includegraphics[height=7cm]{barbellLS}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{slide}

%=======================================================================
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% SLIDE 22
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\begin{slide}[R]{LS properties}
The barbell LS illustrates some properties that hold generally:
\begin{trivlist}
\item[]Closed, convex, enclosing the origin of wrench space.
\item[]Symmetric when reflected through origin.
\item[]Orthogonal projection onto the $f_x$,$f_y$ plane is a circle of
  radius $\sum \mu f_n$.
\item[]Each discrete point of support yields two antipodal flat
  facets.  On each facet several loads map to one motion (rotation
  about the support point.)
\item[](No discrete points: LS is strictly convex and load-motion
  mapping is one-to-one.)
\item[]\emph{Collinear} discrete support is even weirder: vertices on
  LS where one load maps to several velocities (rotation about point
  collinear with support).
\end{trivlist}

\end{slide}

%=======================================================================
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% SLIDE last
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\begin{slide}[R]{Next:  finish sliding.  Force Dual.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC18}
\end{slide}
\end{document}



\end{document}
