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\title{26.  Dynamics and frictional contact}
\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 26. \hfill Mechanics of Manipulation }

\begin{document}

\maketitle

%=======================================================================
%
% SLIDE 2
%

\begin{slide}[R]{Lecture 26.  Dynamics and frictional contact.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC26}
\end{slide}

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

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

\begin{trivlist}
\item Review of frictional contact (static and quasistatic)
\item Planar dynamic single contact
\item Inconsistency and indeterminacy
\item Graphical methods (acceleration center)
\item Multiple contact
\end{trivlist}

\end{slide}

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

\begin{slide}{Review of frictional contact}
\begin{trivlist}
\item Static problems
\begin{itemize}
  \item Friction cones in wrench space
  \item Polyhedral convex cones in 2D and 3D
  \item Graphical methods in 2D
  \item Applications to grasp and fixture design
\end{itemize}
\item Quasistatic problems
\begin{itemize}
  \item Planar sliding
  \item Maximum power inequality
  \item The Limit Surface
  \item Application to stable pushing and parts orienting
\end{itemize}
\item Today:  planar dynamic.  Applications to all the above.
\end{trivlist}
\end{slide}

%=========
%
% Slide 5
%

\begin{slide}[R]{Recall sliding rod}
\begin{minipage}[t]{7cm}
\begin{trivlist}
\item
From Lecture 16:

\centerline{
\begin{tabular}{@{}cccccl@{}}
 \\ \hline
  $\dot p_{cn}$ & $\ddot p_{cn}$ & $\dot p_{ct}$ & $\ddot p_{ct}$ & & 
\\ \hline
 $<0$ &      &       &          &                        & impact \\    
 $>0$ &      &       &          &                        & separation \\
 $=0$ & $<0$ &       &          &                        & impact \\
 $=0$ & $>0$ &       &          &                        & separation \\
 $=0$ & $=0$ & $<0$  &          & $f_t = \mu f_n$        & left sliding \\
 $=0$ & $=0$ & $>0$  &          & $f_t = - \mu f_n$      & right sliding \\
 $=0$ & $=0$ & $=0$  & $<0$     & $f_t = \mu f_n$        & left sliding \\
 $=0$ & $=0$ & $=0$  & $>0$     & $f_t = -\mu f_n$       & right sliding \\
 $=0$ & $=0$ & $=0$  & $=0$     & $|f_t| \leq |\mu f_n|$ & fixed
\\ \hline
\end{tabular}}
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{3cm}
\begin{figure}[h!]
\includegraphics[width=3cm]{rod}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

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

\begin{slide}[R]{Basic approach:  case analysis of contact modes}
\begin{trivlist}
\item Look at table of previous slide.  Force is not expressed as
  function of state!  Contact mode is function of acceleration, not
  just position and velocity.  Solving for acceleration leads to
  circularity?
  
\item But if contact mode is given, friction is not so weird---no
  circularity.  So $\ldots$
\begin{enumerate}
\item
        Enumerate the contact modes.

\item
        Solve the mechanics problem associated with each contact mode
        to obtain velocities and accelerations at each contact point.

\item
        Discard the solution if the contact velocities and
        accelerations are not consistent with the hypothesized contact
        mode.

\end{enumerate}
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 7
%
%
\begin{slide}[R]{Inconsistency and nondetermism}
\begin{trivlist}
\item Solve a Newtonian mechanics problem for each contact mode?  How
  strange!  Presumably only one of them gives a solution consistent
  with the assumed contact mode.  Wrong!
\begin{itemize}
\item[]  Sometimes two or more modes are valid.  Nondeterminism.
\item[]  Sometimes there are \emph{no} valid solutions.  Inconsistency.
\end{itemize}
(Except, when there are no valid solutions, maybe treating it as
impact will save the day.)
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 8
%
%
\begin{slide}[R]{Example:  the sliding rod}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item To write equations of motion using contact point $\vv p_c$, start
  with kinematics:
\begin{align*}
 \vv p_c & =  \vv p_0 - l \left(\begin{array}{c} 
                                \cos\theta \\ \sin\theta \end{array}\right) \\
 \dot {\vv p}_c & =  \dot{\vv p}_0 
                        - l \left(\begin{array}{c} 
                                -\sin\theta \\ \cos\theta \end{array}\right) \\
 \ddot {\vv p}_c & = \ddot{\vv p}_0
                        + l \left(\begin{array}{c} 
                                \cos\theta \\ \sin\theta \end{array}\right) 
                            \dot\theta^2
                        - l \left(\begin{array}{c} 
                                -\sin\theta \\ \cos\theta 
                                \end{array}\right) \ddot\theta
\end{align*}
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=6cm]{roddefs}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

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

\begin{slide}[R]{Sliding rod equation of motion}
\begin{trivlist}
\item
\begin{align*}
 \vv f_n + \vv f_t + \vv f_a & =  m \ddot{\vv p}_0 \\
 (\vv p_c - \vv p_0) \btimes (\vv f_n + \vv f_t) + n_a & =  I \ddot\theta 
\end{align*}
Solve for $\ddot{\vv p}_0$ and $\ddot\theta$ and substitute:
\begin{multline*}
\label{eq:star}
 \ddot{\vv p}_c  =  \frac{1}{m} ( \vv f_n + \vv f_t + \vv f_a) 
 + l \dot\theta^2 \left(\begin{array}{c} 
     \cos\theta \\ \sin\theta 
   \end{array}\right)\\
 - \frac{l}{I} \left(\begin{array}{c} 
     - \sin\theta \\ \cos\theta
   \end{array}\right)
 \left[(\vv p_c - \vv p_0) 
   \btimes (\vv f_n + \vv f_t ) + n_a \right] 
\end{multline*}

\item We use this equation and choose different parameters to
  illustrate inconsistency and nondeterminism.
\end{trivlist}
\end{slide}

%=========
%
% Slide 10

\begin{slide}[R]{Inconsistency}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.8\linewidth]{ingeom}
\end{center}
\end{figure}
\end{slide}

%==========
%
% Slide 11

\begin{slide}[R]{Inconsistency (the math)}
\begin{trivlist}
\item We give the object a leftward velocity on a horizontal support
  with a gravitational field:
\begin{align*}
 \dot {\vv p}_0 & =  \left(\begin{array}{c} -1 \\ 0\\ \end{array}\right) \\
 \dot\theta & =  0 \\
 \vv f_a & =  \left(\begin{array}{c} 0 \\ -mg \\ \end{array}\right) \\
 n_a & =  0 
\end{align*}

\item
Left sliding?  Maybe, we'll check it on next slide.

\item
Separation?  Nope---with no contact force rod would accelerate downward.
\end{trivlist}
\end{slide}

%==========
%
% Slide 12

\begin{slide}[R]{Inconsistency example:  try left sliding}
\begin{trivlist}
\item
\begin{equation*}
        \vv f_t = \mu \vv f_n
\end{equation*}
or
\begin{equation*}
 \vv f_t + \vv f_n = f_n \left(\begin{array}{c} \mu \\ 1 \\ \end{array}\right)
\end{equation*}
The moment is 
\begin{align*}
 \lefteqn{ (\vv p_c - \vv p_0) \btimes (\vv f_n + \vv f_t) }\\
 & =  -l \left(\begin{array}{c}\cos\theta \\ \sin\theta \end{array}\right)
        \btimes f_n \left(\begin{array}{c} \mu \\ 1 \\ \end{array}\right) \\
 & =  l f_n ( \mu \sin\theta - \cos\theta) \\
 & =  l f_n     \frac{1}{\cos\alpha } ( -\cos(\alpha+\theta)) 
\end{align*}
where $\alpha = \tan^{-1}\mu$.
\end{trivlist}
\end{slide}

%==========
%
% Slide 13

\begin{slide}[R]{Left sliding analysis continued}
\begin{trivlist}
\item
Substituting into equation of motion
\begin{equation*}
 \ddot{\vv p}_c  =  \frac{1}{m} f_n \left(\begin{array}{c}
                                                \mu \\ 1 \\ 
                                                \end{array}\right)
                        + \left(\begin{array}{c}
                                0 \\ -g \\ 
                                \end{array}\right) + \vv 0
         + \frac{l^2 f_n}{I} \left(\begin{array}{c}
                                        -\sin\theta \\ cos\theta \\ 
                                        \end{array}\right)
                                \frac{\cos(\alpha+\theta)}{\cos\alpha} \nonumber
\end{equation*}
Now the left sliding mode specifies that $\ddot p_{cn} = 0$:
\begin{equation}
 \frac{f_n}{m} - g 
 + \frac{l^2 f_n}{I} \frac{\cos\theta}{\cos\alpha}\cos(\alpha+\theta) 
 = 0
\end{equation}
or
\begin{equation}
 f_n (\frac{1}{m} + 
      \frac{l^2}{I}\frac{\cos\theta}{\cos\alpha}\cos(\alpha+\theta)) 
 = g
\end{equation}
\end{trivlist}
\end{slide}

%=============
%
% Slide 14
%

\begin{slide}[R]{Inconsistency, left sliding mode, continued}
\begin{trivlist}
\item Now $g$ is positive, and $f_n$ is positive, so all values of
  $m$, $l$, $I$, $\alpha$, and $\theta$ must satisfy
\begin{equation*}
 (\frac{1}{m} 
  + \frac{l^2}{I}\frac{\cos\theta}{\cos\alpha}\cos(\alpha+\theta)) > 0
\end{equation*}

\item
It turns out that for some $\theta$ this condition is violated by
large values of $\mu$, or small values of $I$\@.  For example, if $m =
I = 1$, $l = 4$, $\alpha = 30$, and $\theta = 75$, then
the equation is violated.  There is no contact mode that
works.
\end{trivlist}
\end{slide}

%==========
%
% Slide 15
%

\begin{slide}[R]{Frictional indeterminacy example}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item
Change gravity, start with motionless rod:
\begin{align*}
        \vv f_a &= \left(\begin{array}{c}-mg\\0\\ \end{array}\right) \\
        n_a  &= 0 \\
        \dot{\vv p}_0 &= \vv 0 \\
        \dot\theta & =  0 
\end{align*}
Three modes to worry about.  Separation won't work: $\vv f_t + \vv f_n
= \vv 0$, hence $\ddot p_{cn} = 0$, i.e.\ there is no normal
acceleration.  That leaves \emph{left-sliding}, and \emph{fixed}.
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[height=6cm]{ambrod}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%===============
%
% Slide 16
%

\begin{slide}[R]{Indeterminacy:  left sliding}
\begin{trivlist}
\item
For left sliding, there is a solution---if $\vv f_t = \mu \vv f_n = \vv 0$, 
then we get $\ddot p_{ct} = -g$, i.e.\ the rod falls straight down, skimming 
the surface but with zero contact force. 
\end{trivlist}
\end{slide}

%==============
%
% Slide 17
%

\begin{slide}[R]{Indeterminacy:  fixed}
\begin{trivlist}
\item
 Let $s = {f_t}/{f_n}$.  (Coulomb's law:  $-\mu \leq s \leq \mu$.)
\begin{equation*}
        \vv f_t + \vv f_n = f_n \left(\begin{array}{c}s\\1\\ \end{array}\right)
\end{equation*}
Substituting into equation of motion with $\ddot {\vv p}_c = \vv 0$,
\begin{multline*}
 \vv 0 = \frac{1}{m}
           \left( f_n \left(\begin{array}{c}s\\1\\ \end{array}\right)
                + \left(\begin{array}{c}-mg\\0\\ \end{array}\right)\right)
   + \frac{l}{I} \left(\begin{array}{c}-\sin\theta\\ \cos\theta\\ 
                                \end{array}\right)
        l \left(\begin{array}{c}\cos\theta \\ \sin\theta \\ \end{array}\right)
        \btimes f_n \left(\begin{array}{c} s\\ 1\\ \end{array}\right)
                                                \nonumber 
\end{multline*}
Given the physical parameters $m$, $I$, $g$, $l$, and the initial
angle $\theta$, this yields two equations in $F_n$ and $s$.  $\mu$
might have to be large, but solutions exist.
\end{trivlist}
\end{slide}

%=================
% Slide 18
%

\begin{slide}[R]{Graphical methods for planar dynamics}
\begin{trivlist}
\item
Force dual transform works for planar rigid body dynamics problems!
\begin{itemize}
\item Assume body is initially at rest, accelerating with twist $\vv a$ due to
total applied wrench $\vv f$.
\item Let origin be the center of mass of planar body.
\item Let unit of length be the radius of gyration.
\item Then Newton's second law gives:
\begin{equation*} 
        \vv a = \left(\begin{array}{c} 
                                a_x \\ a_y \\ \alpha_z 
                              \end{array}\right) 
                  = \frac{1}{m} \left(\begin{array}{c} 
                                        f_x \\ f_y \\ n_z 
                                  \end{array}\right) \label{eq:N2WT}
\end{equation*} 
\end{itemize}
Recall force dual is projection of wrench $\vv f$.  Acceleration
center is projection of twist $\vv a$.  They're the same!
\end{trivlist}
\end{slide}

%==================
%
% Slide 19
%

\begin{slide}[R]{Graphical method for frictional contact}
\begin{trivlist}
\item
  Let applied force $\vv f$ be sum of contact wrench $\vv c$ and some
  additional applied wrench $\vv w$ to be determined.
\begin{equation*}
        \vv a = \frac{1}{m} \left(\vv w + \vv c\right) 
\label{eq:N2contact}
\end{equation*}
where $\vv c$ is the positive weighted sum of those contact normals or
friction cone edges consistent with the contact mode
\begin{equation*}
        \vv c \in \pos(\{\vv c_i\}).
\end{equation*}

\item Define a relation: wrench $\vv w$ and twist $\vv a$ are related
  iff the first equation holds for some choice of contact forces for the
  assumed contact mode.
\end{trivlist}
\end{slide}

%==========
% 
% Slide 20
%

\begin{slide}[R]{Extend relation to rays}
\begin{trivlist}
\item The graphical methods work with \emph{rays} in wrench and twist
  space, so extend the relation as follows: the pair $\vv w, \vv a$
  are related if and only if
\begin{equation*}
        \exists_{s \geq 0, s_i \geq 0} \; \vv w + \sum s_i \vv c_i = s m \vv a
\end{equation*}
which is equivalent to
\begin{equation*}
        \vv w \in \pos(\{\vv a\} \cup \{- \vv c_i\}).  \label{eq:graphDyn}
\end{equation*}
\end{trivlist}

\end{slide}

%-========
% 
% Slide 21
% 

\begin{slide}[R]{Force dual for dynamic contact problem}
\begin{trivlist}
\item For a given contact mode,
\begin{itemize}
\item Let $\{-\vv c_i\}$ be the negated contact normals or friction
  cone edges consistent with the contact mode;
\item Let $\vv a$ be any acceleration twist consistent with the contact mode;
\item Let $\vv a^\prime$ and $-\vv c_i^\prime$ be the corresponding
  force duals;
\item Then the possible applied wrenches must have force duals satisfying:
\begin{equation*}
        \vv w^\prime \in \conv(\{\vv a^\prime\} \cup \{- \vv c_i^\prime\}) 
\end{equation*}
\end{itemize}
\end{trivlist}
\end{slide}

%========
%
% Slide 22
%

\begin{slide}[R]{Multiple contact problems!!!}
\begin{trivlist}
\item Here's the most amazing part: the method for works for multiple
  contacts.
  
\item First we need a way of representing contact modes.  For the
  $i$th contact we will abbreviate the mode $m_i$ as follows:
\begin{tabular}{ll}
p & kinematically infeasible (``penetration'') \\
s & separation \\
l & left sliding \\
r & right sliding \\
f & fixed 
\end{tabular}

\item
Then the contact mode of the body can be written as a string
\[
        m_1 m_2 \ldots m_n
\]
\end{trivlist}
\end{slide}

%=======
%
% Slide 23
%

\begin{slide}[R]{How many contact modes for $n$ contacts?}
\begin{trivlist}
\item
$5^n$?

\item
Use variation on Reuleaux.
\end{trivlist}
\end{slide}

%========
%
% Slide 24
%

\begin{slide}[R]{The method}
\begin{trivlist}
\item
Then, for each contact mode:
\begin{enumerate}
\item identify the set of acceleration centers;
        
      \item identify the set of possible contact forces $\vv c_i$,
        negate them, and map them to their force duals;
        
      \item form the convex hull of the acceleration centers and
        negated contact force duals, to obtain the set of applied
        wrenches represented as force duals.

\end{enumerate}
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 25
%
%
\begin{slide}[R]{Analysis of mode rsrs}
\begin{figure}[h!]
\includegraphics[height=6cm]{wrmode}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\end{slide}

%================
%
% Slide 26
%
%
\begin{slide}[R]{Contact modes for wrench in tray}
\begin{figure}[h!]
\includegraphics[height=6cm]{wrrsrs}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\end{slide}



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

\begin{slide}[R]{Next:  Impact.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC27}
\end{slide}
\end{document}



\end{document}