\documentclass[12pt]{article}
\usepackage{epsfig,amsmath}
%%%--------- Set up page size and formatting information ----------%%%
\def\newpage{\par\vfill\eject}	%new page macro
\def\tab{\ \ \ \ \ }		%tab in five spaces
\def\dtab{{\tab}{\tab}}         %double tab

\setlength{\textwidth}{6.5in}
\setlength{\textheight}{9.2in}
\setlength{\topmargin}{-0.31in}
\setlength{\oddsidemargin}{0.0in}
\parskip 2ex
\parindent=0pt
\begin{document}

%-----------------------------------------------------------------------------%
\centerline{\bf ME 115(b): Solution to Problem Set $\#4$}
\vskip 0.5 true in

%%%%----------------------------------------------------------------------%%%%
\vskip 0.2 true in
{\bf Problem $\# 1$:} Problem 2(b,d) of Chapter 5 in MLS text.\\
{\bf Part (b.)}\\
Yes, this grasp is force closure.\\

{\bf Part (d.)}\\
$B_1$, $B_2$ = $I$.

\[ G_1 = \begin{bmatrix}
I & S\begin{bmatrix} 0\\ -a\\ b\end{bmatrix}\\
0 & I \end{bmatrix} B_1 
\]

\[ G_2 = \begin{bmatrix}
I & S\begin{bmatrix} 0\\ a\\ b\end{bmatrix}\\
0 & I \end{bmatrix} B_1 
\]

The complete grasp matrix is
\[G = \begin{bmatrix} G_1 & G_2\end{bmatrix} \]
Note that each $G_i$ is full rank (since $B_i$ = $I$).

Force-closure $\longleftrightarrow$ G is full rank and there exists $F_N$ $\epsilon$ $\eta$(G) such that $F_N$ $\epsilon$ FC.  But G is full rank since each $G_i$ is full rank.  Furthermore, it has a null space of dimension 6 (nullity + rank  =12) and since FC = $\Re^6$ $\times$ $\Re^6$, \emph{every} $F_N$ $\epsilon$ $\eta$(G) is in FC.  Therefore, force closure dim($\eta$(G)) $= 12 - rank G = 6$.

The grasp is \emph{not} manipulable.  The grasping constraint is 
\[ \begin{bmatrix}
{B_1}^T  &  T_{{C_1}S}  & J_1 & 0\\
0 	&	{B_2}^T 	& T_{{C_2}S} 	& J_2\end{bmatrix} \dot\theta 
=
G^T \begin{pmatrix}V_0\\ \omega_0\end{pmatrix} = \begin{bmatrix} {G_1}^T\\ {G_2}^T\end{bmatrix} \begin{pmatrix} V_0\\ \omega_0\end{pmatrix} \]

Since ${G_1}^T$ is full rank, we must have $\Re$( ${B_1}^T$ $T_{{C_1}S}$  $J_1$) $\supseteq$ $\Re$(${G_1}^T$) = $\Re^6$.  But $J_1$ $\epsilon$ $Re^{6 \times 4}$ $\longrightarrow$ its range has dimension 4 away from singularities, therefore the grasp is not manipulable.
\bigskip

{\bf Problem $\# 2$:} Problem 3(b) of Chapter 5 in MLS text.\\
\[  R_{C_1} - \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0\end{bmatrix}\]
\[ \hat{P}_{C_1} = \begin{bmatrix} 0\\ -r \\ 0 \end{bmatrix} \]
\[ \hat{P}_{C_1} = \begin{bmatrix} 0 & 0 & -r\\ 0 & 0 & 0\\ r & 0 & 0\end{bmatrix} \]


\[  R_{C_2} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0\end{bmatrix}\]
\[ \hat{P}_{C_2} = \begin{bmatrix} 0\\ r \\ 0 \end{bmatrix} \]
\[ \hat{P}_{C_2} = \begin{bmatrix}  0 & 0 & r\\ 0 & 0 & 0 \\ -r & 0 & 0\end{bmatrix} \]

Plugging these into equations for $G_i$:
{\footnotesize 


\[ G_1 = \begin{bmatrix} R_{C_1} & 0\\ \hat{P}_{C_1}R_{C_1} & R_{C_1}\end{bmatrix}
\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0& 0& 0\\ 0& 0& 0& 0\\ 0 & 0& 0& 1\end{bmatrix} = \begin{bmatrix} 0&1& 0 & 0& 0& 0\\ 0&0&1&0&0&0\\ 1&0&0&0&0&0\\ -r&0&0&0&1&0\\ 0&0&0&0&0&1\\ 0&r&0&1&0&0\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0& 0& 0\\ 0& 0& 0& 0\\ 0 & 0& 0& 1\end{bmatrix} = \begin{bmatrix} 0 &1& 0& 0\\ 0&0&1&0\\ 1&0&0&0\\ -r&0&0&0\\ 0&0&0&1\\0&r&0&0\end{bmatrix} \]

\[ G_2 = \begin{bmatrix} R_{C_1} & 0\\ \hat{P}_{C_1}R_{C_1} & R_{C_1}\end{bmatrix}
\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0& 0& 0\\ 0& 0& 0& 0\\ 0 & 0& 0& 1\end{bmatrix} = \begin{bmatrix} 1&0& 0 & 0& 0& 0\\ 0&0&-1&0&0&0\\ 0&1&0&0&0&0\\ 0&r&0&1&0&0\\ 0&0&0&0&0&-1\\ -r&0&0&0&1&0\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0& 0& 0\\ 0& 0& 0& 0\\ 0 & 0& 0& 1\end{bmatrix}= \begin{bmatrix} 1 &0& 0& 0\\ 0&0&-1&0\\ 0&1&0&0\\ 0&r&0&0\\ 0&0&0&-1\\-r&0&0&0\end{bmatrix} \]
}

Plugging $G_i$s into the overall Grasp Map gives:

\[ G = \begin{bmatrix} G_1 & G_2 \end{bmatrix} = \begin{bmatrix}
0&1&0&0&1&0&0&0\\
0&0&1&0&0&0&-1&0\\
1&0&0&0&0&1&0&0\\
-r&0&0&0&0&r&0&0\\
0&0&0&1&0&0&0&-1\\
0&r&0&0&-r&0&0&0\end{bmatrix} \]

Now to write the friction cone conditions:

\begin{equation}
FC = FC_{C_1} \times FC_{C_2}
\end{equation}

\begin{equation}
FC_{C_1} = \{ f_c : \sqrt{ {(f'_{C_1})}^2 + {(f_{C_1})}^2}  \leq \mu {f^3}_{C_1}, \| {f^4}_{C_1}\| \leq 
\gamma {f^3}_{C_1} , {f^3}_{C_1} \geq 0 \}
\end{equation}

\begin{equation}
FC_{C_2} = \{ f_c : \sqrt{ {(f'_{C_2})}^2 + {(f_{C_2})}^2}  \leq \mu {f^3}_{C_2}, \| {f^4}_{C_2}\| \leq 
\gamma {f^3}_{C_2} , {f^3}_{C_2} \geq 0 \}
\end{equation}

The grasp map is full rank.
Additionally, the 'force closure by inspection' method shown in class shows that this  is a force closure grasp since  it is a soft-finger grasp  and a line between the two contact points can be drawn which is fully enclosed by both friction cones.


{\bf Problem $\# 3$:} Problem 4(a) of Chapter 5 in MLS text.\\

We can derive the wrench basis by looking at which directions are free to movement.
This contact model does not allow movement in the x,y, and z directions, or rotation along the x or z axis.  It does, however, allow rotation along the y axis.

Thus the wrench basis is:
\[ \begin{bmatrix}
1&0&0&0&0\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0\\
0&0&0&0&0\\
0&0&0&0&1
\end{bmatrix} \]

Friction cone:

\begin{equation}
\mu > 0
\end{equation}

\begin{equation}
\sqrt{ f_x + f_y} \leq \mu \cdot f_n
\end{equation}

\begin{equation}
f_n \geq 0
\end{equation}

\begin{equation}
|\tau_n| \leq \gamma_n f_n
\end{equation}

\begin{equation}
| \tau_y | \leq \gamma_y f_y
\end{equation}
{\bf Problem $\# 4$:} Problem 5(a) of Chapter 5 in MLS text.\\

There are many possible answers to this problem.  The easiest is to choose an equilateral triangle.  You can then show geometrically that there is no way that a line between two point contacts can be made such that it will pass through both friction cones (in the case where $\mu$ = $\tan$ 30.


{\bf Problem $\# 5$:} Problem 8(b) of Chapter 5 in MLS text.\\
We have an equilateral triangle and we wish to find all possible force-closure grasps for two contacts with friction and a coefficient of friction, $\mu$=$\tan$ 45.
The constraints that the line between contacts lie in the relevant friction cones become:

\begin{equation}
A:  (\eta_1 -\mu{\eta_1}^\bot) \otimes (p_2 - p_1) > 0
\end{equation}
\begin{equation}
B: (\eta_1 +\mu{\eta_1}^\bot) \otimes (p_2 - p_1) < 0
\end{equation}
\begin{equation}
C: (\eta_2 -\mu{\eta_2}^\bot) \otimes (p_1 - p_2) > 0
\end{equation}
\begin{equation}
D: (\eta_2 +\mu{\eta_2}^\bot) \otimes (p_2 - p_1) < 0
\end{equation}


\begin{equation}
\eta_1 = \begin{bmatrix} 0\\ 1 \end{bmatrix}
\end{equation}
\begin{equation}
{\eta_1}^\bot = \begin{bmatrix} -1\\ 0 \end{bmatrix}
\end{equation}
\begin{equation}
{\eta_2}= \begin{bmatrix}\cos 30 \\ -\sin 30\end{bmatrix}=\begin{bmatrix} \frac{\sqrt{3}}{2}\\ - \frac{1}{2}\end{bmatrix}
\end{equation}
\begin{equation}
{\eta_2}^\bot = \begin{bmatrix}\sin 30 \\ \cos 30\end{bmatrix}=\begin{bmatrix} \frac{1}{2}\\  \frac{\sqrt{3}}{2}\end{bmatrix}
\end{equation}
\begin{equation}
p_1= \begin{bmatrix} d_1\\ 0 \end{bmatrix}
\end{equation}
\begin{equation}
p_2=\begin{bmatrix} d_2 \sin 30\\ d_2 \cos 30\end{bmatrix} =\begin{bmatrix} \frac{1}{2} d_2\\ \frac{\sqrt{3}}{2} d_2\end{bmatrix}
\end{equation}

\begin{equation}
\mu = \tan 45 = 1
\end{equation}
Plugging in these values into A,B,C and D:

\begin{equation}
A:   d_1 + d_2 \frac{\sqrt{3}-1}{2}  > 0
\end{equation}
\begin{equation}
B: d_1 -d_2 \frac{\sqrt{3}-1}{2}  < 0
\end{equation}
\begin{equation}
C:  -(\frac{\sqrt{3}}{2}-\frac{1}{2})\frac{\sqrt{3}}{2}d_2 + (d_1 -\frac{d_2}{2})(\frac{1}{2}-\frac{\sqrt{3}}{2})> 0
\end{equation}
\begin{equation}
D: -(\frac{\sqrt{3}}{2}+\frac{1}{2})\frac{\sqrt{3}}{2} d_2 - (d_1-\frac{d_2}{2})(\frac{\sqrt{3}}{2}+\frac{1}{2})< 0
\end{equation}

{\bf Problem $\# 6$:} \\
{\bf Part $\# 1$:}
Yes, this grasp is force closure.  We can use the visual method we learned last week, we can see that a line connecting the two point contacts clearly runs through both friction cones.
\bigskip
{\bf Part $\# 2$:}
\[ R_1 = \begin{bmatrix} 0&1&0\\ 0&0&1\\ 1&0&0\end{bmatrix} \]
\[ P_1 = \begin{bmatrix} 0\\ -d\\0\end{bmatrix} \]
\[ R_1 = \begin{bmatrix} 1&0&0\\ 0&0&-1\\ 0&1&0\end{bmatrix} \]
\[ P_1 = \begin{bmatrix} 0\\ d\\0\end{bmatrix} \]

\[Wrench Basis = \begin{bmatrix} 
1&0&0\\
0&1&0\\
0&0&1\\
0&0&0\\
0&0&0\\
0&0&0\end{bmatrix} \]

\[ G_1 = \begin{bmatrix} R_1 & 0\\ \hat{P}_1R_1 & R_1 \end{bmatrix}
\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ 0&0&0\\ 0&0&0 \\ 0&0&0 \end{bmatrix}
= \begin{bmatrix} 0&1&0\\ 0&0&1\\ 1&0&0\\ -d&0&0\\ 0&0&0 \\ 0&d& 0 \end{bmatrix}
 \]

\[ G_2 = \begin{bmatrix} R_2 & 0\\ \hat{P}_2R_2 & R_2 \end{bmatrix}
\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ 0&0&0\\ 0&0&0 \\ 0&0&0 \end{bmatrix}
= \begin{bmatrix} 1&0&0\\ 0&0&-1\\ 0&1&0\\ 0&d&0\\ 0&0&0 \\ -d&0& 0 \end{bmatrix}
 \]

\[ G = \begin{bmatrix} G_1 & G_2 \end{bmatrix}= 
\begin{bmatrix} 0&1&0&1&0&0\\ 0&0&1&0&0&-1\\ 1&0&0&0&1&0\\ -d&0&0&0&d&0\\
0&0&0&0&0&0\\0&d&0&-d&0&0 \end{bmatrix}
\]



\end{document}