cs223a-homework-2.tex

%insert copyright here
\documentclass{exam}
\usepackage{mathtools}
\begin{document}
\title{Stanford CS223A - Introduction to robotics \\
  Homework \#2}
\author{Arn-O}
\date{January 2014}
\maketitle

\begin{questions}

\question

\begin{parts}
\part

Quite a complex question that learns me the similarity transforms. The first really tricky point is that after a $T_{1}$ displacement, the frame transform is $T_{1}^{-1}$. So the transform between the frame $\{0\}$ and the frame $\{1\}$ can be written:

\begin{equation}
  P\{1\} = T_{1}^{-1}P\{0\}
\end{equation}

By definition, after the first displacement:

\begin{equation}
  P_{1}\{0\} = T_{1}P\{0\}
\end{equation}

And after the second displacement:

\begin{equation}
  P_{2}\{1\} = T_{2}P_{1}\{1\}
\end{equation}
 
And the result is:

\begin{equation}
  P_{2}\{0\} = T'_{2}P\{0\}
\end{equation}

The second displacement explained in the frame $\{0\}$ gives:

\begin{equation}
  T_{1}^{-1}P_{2}\{0\} = T_{2}T_{1}^{-1}P\{0\}
\end{equation}

And finally:

\begin{equation}
  T'_{2} = T_{1}T_{2}T_{1}^{-1}
\end{equation}

This could also be derivate from the similarity transform equation:

\begin{equation}
  B = T^{-1}AT
\end{equation}

Where $T$ is the frame transform from $\{A\}$ to $\{B\}$. In this case:

\begin{equation}
  T = T_{1}^{-1}
\end{equation}

\part

In this case, we use the similarity transform with $T_{3}$ after a $T'_{2}T_{1}$ displacement:

\begin{equation}
  T'_{3} = (T'_{2}T_{1})T_{3}(T'_{2}T_{1})^{-1}
\end{equation}

So with the result of the first part:

\begin{equation}
  T'_{3} = T_{1}T_{2}T_{3}T_{1}^{-1}T_{2}^{-1}
\end{equation}

\part

The same formula is applied again on this joint:

\begin{equation}
  T'_{4} = (T'_{3}T'_{2}T_{1})T_{4}(T'_{3}T'_{2}T_{1})^{-1}
\end{equation}

So if we develop this equation:

\begin{equation}
  T'_{4} = (T_{1}T_{2}T_{3}T_{1}^{-1}T_{2}^{-1}
            T_{1}T_{2}T_{1}^{-1}
            T_{1})
	    T_{4}
           (T_{1}T_{2}T_{3}T_{1}^{-1}T_{2}^{-1}
	    T_{1}T_{2}T_{1}^{-1}
	    T_{1})^{-1}
\end{equation}

After the simplification:

\begin{equation}
  T'_{4} = (T_{1}T_{2}T_{3})T_{4}(T_{3}^{-1}T_{2}^{-1}T_{1}^{-1})
\end{equation}

\part

From the fixed to the free end, the displacement is:

\begin{equation}
  T = T'_{4}T'_{3}T'_{2}T'_{1}
\end{equation}

Which gives:

\begin{equation}
  T = T_{1}T_{2}T_{3}T_{4}T_{3}^{-1}T_{2}^{-1}T_{1}^{-1}
      T_{1}T_{2}T_{3}T_{1}^{-1}T_{2}^{-1}
      T'_{2}T'_{1}
\end{equation}

\begin{equation}
  T = T_{1}T_{2}T_{3}T_{4}T_{1}^{-1}T_{2}^{-1}
      T_{1}T_{2}T_{1}^{-1}    
      T_{1}
\end{equation}

\begin{equation}
  T = T_{1}T_{2}T_{3}T_{4}
\end{equation}

And we come to the same result.

\end{parts}

\question

\begin{parts}
\part

I did not find any easy way to draw a schematic in LaTeX.

Just a couple of comments about the choice of the frames:

\begin{itemize}
  \item the frames $\{0\}$, $\{1\}$ and $\{2\}$ coincide
  \item the axis of the joint 1 and 2 intersect, so the $x_{1}$ axis is placed freely
  \item an extra parameter is required between the joint 2 and 3
\end{itemize}

\part

The Denavit-Hartenberg parameters are the following:

\begin{centering}

\begin{tabular}{|| c | c | c | c | c ||}
  \hline
  i & $a_{i-1}$ & $\alpha_{i-1}$ & $d_{i}$ & $\theta_{i}$ \\
  \hline
  1 & 0         & 0              & 0       & $\theta_{1}$ \\
  \hline
  2 & 0         & +90            & 0       & $\theta_{2}$ \\
  \hline
  3 & $l$       & +90            & $d_{3}$ & 0            \\
  \hline
\end{tabular}

\end{centering}

\part

Let's apply the forward kinematics formula from the DH parameters:

\begin{equation}
  \prescript{i-1}{i}T = \begin{bmatrix}
    c\theta_{i}              & -s\theta_{i}             & 0              & a_{i-1}             \\
    s\theta_{i}c\alpha_{i-1} & c\theta_{i}c\alpha_{i-1} & -s\alpha_{i-1} & -s\alpha_{i-1}d_{i} \\
   s\theta_{i}s\alpha_{i-1}  & c\theta_{i}s\alpha_{i-1} & c\alpha_{i-1}  & c\alpha_{i-1}d_{i}  \\
   0                         & 0                        & 0              & 1                   \\ 
			\end{bmatrix}
\end{equation}

This formula is applied to the several joints:

\begin{equation}
  \prescript{0}{1}T = \begin{bmatrix}
    c\theta_{1} & -s\theta_{1} & 0 & 0 \\
    s\theta_{1} & c\theta_{1}  & 0 & 0 \\
    0           & 0            & 1 & 0 \\
    0           & 0            & 0 & 1 \\ 
                      \end{bmatrix}
\end{equation}

\begin{equation}
  \prescript{1}{2}T = \begin{bmatrix}
    c\theta_{2} & -s\theta_{2} & 0  & 0 \\
    0           & 0            & -1 & 0 \\
    s\theta{2}  & c\theta{2}   & 0  & 0 \\
    0           & 0            & 0  & 1 \\ 
                      \end{bmatrix}
\end{equation}

\begin{equation}
  \prescript{2}{3}T = \begin{bmatrix}
    1 & 0 & 0  & l      \\
    0 & 0 & -1 & -d_{3} \\
    0 & 1 & 0  & 0      \\
    0 & 0 & 0  & 1      \\ 
                      \end{bmatrix}
\end{equation}

The matrix are multiplied to each other, from the first to the last:

\begin{equation}
  \prescript{0}{3}T = \prescript{0}{1}T\prescript{1}{2}T\prescript{2}{3}T
\end{equation}

\begin{equation}
  \prescript{0}{2}T = \begin{bmatrix}
    c_{12}     & -c_{1}s_{2} & s_{1}  & 0 \\
    s_{1}c_{2} & -s_{12}     & -c_{1} & 0 \\
    s_{2}      & c_{2}       & 0      & 0 \\
    0          & 0           & 0      & 1 \\ 
                      \end{bmatrix}
\end{equation}

\begin{equation}
  \prescript{0}{3}T = \begin{bmatrix}
    c_{12}     & s_{1}  & c_{1}s_{2} & lc_{12} + d_{3}c_{1}s_{2} \\
    s_{1}c_{2} & -c_{1} & s_{12}     & ls_{1}c_{2} + d_{3}s_{12} \\
    s_{2}      & 0      & -c_{2}     & ls_{2} - d_{3}c_{2}       \\
    0          & 0      & 0          & 1                         \\ 
                      \end{bmatrix}
\end{equation}

\end{parts}

\question
\begin{parts}

\part

Clearly not easy to draw the axis and the perpendiculars when the joints and the axis of the robot are not aligned. The schematic may become misleading at the end. According to me, the simple way to proceed is the use the definitions of the parameters.

Notably:

\begin{itemize}
  \item $\alpha_{i}$ is the angle between $z_{i}$ and $z_{i+1}$ along $x_{i}$
  \item $\theta_{i}$ is the angle between $x_{i-1}$ and $x_{i}$ along $z_{i}$
\end{itemize}

So if you manage to see the $x_{i}$ and $z_{i}$ axis, you can more easily see if a parameter is null or not.

They are several ways to define the D-H parameters. In my schematic, $d_{4} = 0$ and $d_{5}$ is not null.  

\part

With the configuration of my schematic, the D-H parameters are:

\begin{centering}

\begin{tabular}{|| c | c | c | c | c ||}
  \hline
  i & $a_{i-1}$ & $\alpha_{i-1}$ & $d_{i}$ & $\theta_{i}$ \\
  \hline
  1 & 0         & 0              & 0       & $\theta_{1}$ \\
  \hline
  2 & $a_{1}$   & $\alpha_{1}$   & $d_{2}$ & $\theta_{2}$ \\
  \hline
  3 & 0         & $\alpha_{2}$   & $d_{3}$ & $\theta_{3}$ \\
  \hline
  4 & 0         & $\alpha_{3}$   & 0       & $\theta_{4}$ \\
  \hline
  5 & $a_{5}$   & 0              & $d_{5}$ & $\theta_{5}$ \\
  \hline
\end{tabular}

The parameters of the schematic of the homework are variable. The other parameters are fixed.

\end{centering}

\end{parts}

\end{questions}

\end{document}