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\centerline{\bf ME 115(a):  Homework $\#6$ Solution}
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{\bf Problem 1:} (15 points) Find the Denavit-Hartenberg parameters for 
manipulators (ii) and (iv) in Figure 3.23 of the MLS text.

\begin{description}

\item{} {\bf Manipulator (ii):} The choice of the stationary frame is abitrary.  
Place its origin along joint axis 1, but not necessarily at the point
of coincidence of joint axes 1 and 2.  The tool frame origin is placed
in the middle of the ``U'', with its $x$-axis collinear with the mechanical
link axis, and with its $z$-axis parallel to joint axis 2.  In this case,
\[ 
   \begin{array}{cccc}
    a_0 = 0  & \alpha_0=0             & d_1\ne 0 & \theta_1=\ \rm{variable} \\
    a_1 = 0  & \alpha_1=\frac{\pi}{2} & d_2 = 0  & \theta_2=\ \rm{variable} \\
    a_2\ne 0 & \alpha_2=-\frac{\pi}{2}& d_3\ne 0 & \theta_3=\ \rm{variable} \\
    a_3\ne 0 & \alpha_3=\frac{\pi}{2} & d_4 = 0  & \theta_4=\ \rm{constant} 
   \end{array}
   \]     
where the value of $d_1$ will be determined by the location of the stationary
frame origin.

\item{}{\bf Manipulator (iv):} The choice of the stationary frame is abitrary.  For
simplicity, place the origin of the stationary frame at the point
where all three joints intersect.  Place the $z$-axis of the
stationary frame, $z_S$, collinear with the first joint axis.  Orient
the $x$-axis of the stationary frame to be orthogonal to both joint
axes 1 and 2 (pointing toward the right in Figure 3.23(i)).  Similary,
there are many choices for the tool frame.  Let's assume that the tool
frame is parallel with the link frame of link 3, (as determined using
the Denavit-Hartenberg procedure), but its origin lies at the tip of
the mechanism (in the ``U'' of Figure 3.23(iv)).  Then, the D-H
parameters are:

\[ 
   \begin{array}{cccc}
    a_0 = 0 & \alpha_0=0              & d_1=       0         & \theta_1=\ \rm{variable} \\
    a_1 = 0 & \alpha_1=-\frac{\pi}{2} & d_2 =      0         & \theta_2=\ \rm{variable} \\
    a_2 = 0 & \alpha_2= \frac{\pi}{2} & d_3 = \rm{variable}  & \theta_3=\ 0 \\
    a_3 = 0 & \alpha_3= 0             & d_4 = \rm{constant}  & \theta_4=\ \rm{constant} =0
   \end{array}
   \]     
where the constant $d_4$ depends upon the offset between the origin of
link frame 3 and the origin of the tool frame.
\end{description}


%%%----------------------------------------------------------------------%%%
{\bf Problem 2:} (20 points) Consider the simple manipulator (iii) associated 
with Prob.4 in Chapter 3 of the MLS text.  

\begin{itemize}
\item{} {\bf Part (a):} {\em Determine the Denavit-Hartenberg parameters of 
this manipulator}.

For simplicity, let us choose the z-axis of the stationary frame to be
collinear with the first joint axis.  The origin of the stationary
frame is located some distance below the point of intersection of the
first two axes.  Also, choose the tool frame origin to coincide with
the intersection point of the last three axes (the ``wrist'').  Also,
assume that the link 6 frame of the Denavit-Hartenberg approach is the
tool frame.
\[ \begin{array}{llll}
    a_0 = 0 & \alpha_0=0              & d_1\ne 0            & \theta_1=\ \rm{variable} \\
    a_1 = 0 & \alpha_1=\frac{\pi}{2}  & d_2 = 0             & \theta_2=\ \rm{variable} \\
    a_2 = 0 & \alpha_2=-\frac{\pi}{2} & d_3 =\rm{variable}  & \theta_3=0\ (\rm{constant}) \\
    a_3 = 0 & \alpha_3=\frac{\pi}{2}  & d_4 = 0             & \theta_4=\ \rm{variable} \\
    a_4 = 0 & \alpha_4=\frac{\pi}{2}  & d_5 = 0             & \theta_5=\ \rm{variable} \\
    a_5 = 0 & \alpha_5=\frac{\pi}{2}  & d_6 = 0             & \theta_6=\ \rm{variable} \\
    a_6 = 0 & \alpha_6=  0            & d_7 = 0             & \theta_7= 0 
   \end{array}  \]     

\item{} {\bf Part (b):} {\em Find the forward kinematics using the 
Denavit-Hartenberg approach}. To find the forward kinematics using the
Denavit-Hartenberg approach, one must use the formula
     \[ g_{ST}(\vec{\theta}) = g_{S1}(\theta_1)g_{12}(\theta_2)g_{23}(d_3)
                        g_{34}(\theta_4)g_{45}(\theta_5)g_{56}(\theta_6)g_{6T}.
      \]
where each $g_{i,i+1}$ is given by:
 \[ \begin{bmatrix} \cos\theta_{i+1} & -\sin\theta_{i+1} & 0 & a_i \\
   \sin\theta_{i+1}\cos\alpha_i &  \cos\theta_{i+1}\cos\alpha_i & -\sin\alpha_i 
                                & -d_{i+1}\sin\alpha_i\\
   \sin\theta_{i+1}\sin\alpha_i &  \cos\theta_{i+1}\sin\alpha_i & \cos\alpha_i 
                                & d_{i+1}\cos\alpha_i\\
      0 & 0 & 0 & 1    \end{bmatrix} \]
Plugging in the D-H parameters from above yields:
   \[\begin{array}{r}
     g_{ST} = \begin{bmatrix} \cos\theta_1 & -\sin\theta_1 & 0 & \\
                               \sin\theta_1 & \cos\theta_1 & 0 & 0 \\
			       0 & 0 & 1 & d_1 \\ 0 & 0 & 0 & 1\end{bmatrix}
      \begin{bmatrix} \cos\theta_2 & -\sin\theta_2 & 0 & 0 \\ 0 & 0 & -1 & 0\\
	-\sin\theta_2 & -\cos\theta_2 & 0 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}
      \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & d_3 \\ 0 & -1 & 0 & 0 \\
	0 & 0 & 0 & 1\end{bmatrix}  \\
      \begin{bmatrix} \cos\theta_4 & -\sin\theta_4 & 0 & 0\\ 0 & 0 & -1 & 0\\
	\sin\theta_4 & \cos\theta_4 & 0 & 0\\ 0 & 0 & 0 &1\end{bmatrix}
      \begin{bmatrix} \cos\theta_5 & -\sin\theta_5 & 0 & 0\\ 0 & 0 & -1 & 0\\
	\sin\theta_5 & \cos\theta_5 & 0 & 0\\ 0 & 0 & 0 &1\end{bmatrix}
      \begin{bmatrix} \cos\theta_6 & -\sin\theta_6 & 0 & 0\\ 0 & 0 & -1 & 0\\
	\sin\theta_6 & \cos\theta_6 & 0 & 0\\ 0 & 0 & 0 &1\end{bmatrix} g_{6T}
   \end{array}
  \]

\item{} {\bf Part (c):} {\em Find the forward kinematics using the 
Product-of-Exponentials Approach.}  This part of the problem was cancelled

\end{itemize}


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