4R-4P Serial-Parallel Hybrid Robot
==================================

Mathematics involved
--------------------

.. image:: ../examples/Jacobian/images/RRRRPPPP.png
   :alt: Alternative Text
   :width: 300
   :align: center

A figure of RRRRPPPP planar serial-parallel hybrid manipulator is shown
above. The corresponding adjacency matrix is given by

.. math:: \bf{M} = \left[\begin{matrix}L_1 & P & P & O & O & O & O\\A & L_2 & O & P & O & O & O\\A & O & L_3 & P & O & O & O\\O & O & O & L_4 & R & R & O\\O & O & O & A & L_5 & O & R\\O & O & O & O & O & L_6 & R\\O & O & O & O & O & O & L_7\end{matrix}\right]

Connecting paths:
~~~~~~~~~~~~~~~~~

All possible paths connecting the end-effector link from the base link,
are shown below.

.. math::


   \begin{matrix}
       \text{Path 1:} \;\;\; L_1-L_2-L_4-L_5-L_7 \\ 
       \text{Path 2:} \;\;\; L_1-L_3-L_4-L_5-L_7 \\ 
       \text{Path 3:} \;\;\; L_1-L_2-L_4-L_6-L_7 \\
       \text{Path 4:} \;\;\; L_1-L_3-L_4-L_6-L_7
   \end{matrix}

In order to check for possibility of redundant paths, the rank of the
connectivity matrix is considered.

.. math::


   \begin{matrix}
       \bf{v}^{(1)}=\bf{V}_{(1,2)}+\bf{V}_{(2,4)}+\bf{V}_{(4,5)}+\bf{V}_{(5,7)} \\ 
       \bf{v}^{(2)}=\bf{V}_{(1,3)}+\bf{V}_{(3,4)}+\bf{V}_{(4,5)}+\bf{V}_{(5,7)} \\ 
       \bf{v}^{(3)}=\bf{V}_{(1,2)}+\bf{V}_{(2,4)}+\bf{V}_{(4,6)}+\bf{V}_{(6,7)} \\ 
       \bf{v}^{(4)}=\bf{V}_{(1,3)}+\bf{V}_{(3,4)}+\bf{V}_{(4,6)}+\bf{V}_{(6,7)} \\ 
   \end{matrix}

.. math::


   \Rightarrow \begin{Bmatrix}
       \bf{v}^{(1)} \\
       \bf{v}^{(2)} \\
       \bf{v}^{(3)} \\
       \bf{v}^{(4)}
   \end{Bmatrix} = 
   \begin{bmatrix}
       1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\
       0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\
       1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 \\
       0 & 0 & 0 & 0 & 1 & 1 & 1 & 1
   \end{bmatrix}
   \begin{Bmatrix}
       \bf{V}_{(1,2)} \\
       \bf{V}_{(2,4)} \\
       \bf{V}_{(4,5)} \\
       \bf{V}_{(5,7)} \\
       \bf{V}_{(1,3)} \\
       \bf{V}_{(3,4)} \\
       \bf{V}_{(4,6)} \\
       \bf{V}_{(6,7)}
   \end{Bmatrix}

.. math::


   \Rightarrow \begin{Bmatrix}
       \bf{v}^{(k)}
   \end{Bmatrix} = 
   \begin{bmatrix}
       \bf{C}_{\bf{V}}
   \end{bmatrix}
   \begin{Bmatrix}
       \bf{V}_{(i,j)}
   \end{Bmatrix}

Even though there are 4 equations, the rank of the matrix
:math:`[\bf{C}_{\bf{V}}]` is 3. This shows that only three independent
connecting paths from base to end-effector exist and hence one of the
paths should be redundant. The set of independent connecting paths can
be found by performing the row-reduced echelon form or the row echelon
form of :math:`[\bf{C}_{\bf{V}}]^T`. The set of indices of pivot
columns indicates that the set of corresponding paths are independent.
By performing row-reduced echelon form on :math:`[\bf{C}_{\bf{V}}]^T`,
the list of pivot columns is :math:`(1,2,3)`, and hence the paths 1, 2
and 3 amount to a set of independent paths.

Now, given that the independent connecting paths are the first three
paths, the angular velocity connectivity matrix is considered as
follows.

.. math::


   \begin{matrix}
       \bf{\omega}^{(1)}=\bf{\Omega}_{(1,2)}+\bf{\Omega}_{(2,4)}+\bf{\Omega}_{(4,5)}+\bf{\Omega}_{(5,7)} \\ 
       \bf{\omega}^{(2)}=\bf{\Omega}_{(1,3)}+\bf{\Omega}_{(3,4)}+\bf{\Omega}_{(4,5)}+\bf{\Omega}_{(5,7)} \\ 
       \bf{\omega}^{(3)}=\bf{\Omega}_{(1,2)}+\bf{\Omega}_{(2,4)}+\bf{\Omega}_{(4,6)}+\bf{\Omega}_{(6,7)} 
   \end{matrix}

The :math:`\bf{\Omega}_{(i,j)}` terms corresponding to prismatic
joints, i.e., :math:`\bf{\Omega_{(1,2)}}`,
:math:`\bf{\Omega_{(1,3)}}`, :math:`\bf{\Omega_{(2,4)}}` and
:math:`\bf{\Omega_{(3,4)}}` are set to zero. The equations would then
become

.. math::


   \begin{matrix}
       \bf{\omega}^{(1)}=\bf{\Omega}_{(4,5)}+\bf{\Omega}_{(5,7)} \\ 
       \bf{\omega}^{(2)}=\bf{\Omega}_{(4,5)}+\bf{\Omega}_{(5,7)} \\ 
       \bf{\omega}^{(3)}=\bf{\Omega}_{(4,6)}+\bf{\Omega}_{(6,7)} 
   \end{matrix}

.. math::


   \Rightarrow \begin{Bmatrix}
       \bf{\omega}^{(1)} \\
       \bf{\omega}^{(2)} \\
       \bf{\omega}^{(3)} 
   \end{Bmatrix} = 
   \begin{bmatrix}
       1 & 1 & 0 & 0 \\
       1 & 1 & 0 & 0 \\
       0 & 0 & 1 & 1
   \end{bmatrix}
   \begin{Bmatrix}
       \bf{\Omega}_{(4,5)} \\
       \bf{\Omega}_{(5,7)} \\
       \bf{\Omega}_{(4,6)} \\
       \bf{\Omega}_{(6,7)}
   \end{Bmatrix}

.. math::


   \Rightarrow \begin{Bmatrix}
       \bf{\omega}^{(k)}
   \end{Bmatrix} = 
   \begin{bmatrix}
       \bf{C}_{\Omega}
   \end{bmatrix}
   \begin{Bmatrix}
       \bf{\Omega}_{(i,j)}
   \end{Bmatrix}

The rank of the matrix :math:`[\bf{C_{\Omega}}]` is 2, even though there
are three equations. Hence, only two independent equations exist. The
set of independent connecting paths can be found by performing
row-reduced echelon form or echelon form on :math:`[\bf{C_{\Omega}}]^T`.
The set of indices of pivot columns would indicate the set of
corresponding independent paths in the context of angular velocity. By
performing row-reduced echelon form on :math:`[\bf{C_{\Omega}}]^T`, the
list of pivoted columns is found to be (1,3), and hence the paths 1 and
3 amount to a set of independent paths in the context of angular
velocity.

Therefore, the independent linear velocities are :math:`\bf{v}^{(1)}`,
:math:`\bf{v}^{(2)}` and :math:`\bf{v}^{(3)}`, and the independent
angular velocities are :math:`\bf{\omega}^{(1)}` and
:math:`\bf{\omega}^{(3)}`.

.. math:: \bf{v}^{(1)}=\dot{d}_{(1,2)} \bf{\hat{n}_{(1,2)}} + \dot{d}_{(2,4)} \bf{\hat{n}_{(2,4)}} + \dot{\theta}_{(4,5)} \bf{\hat{k}} \times \left( \bf{a} - \bf{r}_{(4,5)} \right) + \dot{\theta}_{(5,7)} \bf{\hat{k}} \times \left( \bf{a} - \bf{r}_{(5,7)} \right)

.. math:: \bf{v}^{(2)}=\dot{d}_{(1,3)} \bf{\hat{n}}_{(1,3)} + \dot{d}_{(3,4)} \bf{\hat{n}}_{(3,4)} + \dot{\theta}_{(4,5)} \bf{\hat{k}} \times \left( \bf{a} - \bf{r}_{(4,5)} \right) + \dot{\theta}_{(5,7)} \bf{\hat{k}} \times \left( \bf{a} - \bf{r}_{(5,7)} \right)

.. math:: \bf{v}^{(3)}=\dot{d}_{(1,2)} \bf{\hat{n}}_{(1,2)} + \dot{d}_{(2,4)} \bf{\hat{n}}_{(2,4)} + \dot{\theta}_{(4,6)} \bf{\hat{k}} \times \left( \bf{a} - \bf{r}_{(4,6)} \right) + \dot{\theta}_{(6,7)} \bf{\hat{k}} \times \left( \bf{a} - \bf{r}_{(6,7)} \right)

.. math:: \bf{\omega}^{(1)} = \dot{\theta}_{(4,5)} \bf{\hat{k}} + \dot{\theta}_{(5,7)} \bf{\hat{k}}

.. math:: \bf{\omega}^{(3)} = \dot{\theta}_{(4,6)} \bf{\hat{k}} + \dot{\theta}_{(6,7)} \bf{\hat{k}}

.. math::


   \begin{Bmatrix}\bf{v}^{(1)} \\ \bf{\omega}^{(1)}\end{Bmatrix} = \begin{Bmatrix}\bf{v} \\ \bf{\omega}\end{Bmatrix} = \left[\begin{matrix}- a_{y} + r_{(4,5)y} & n_{(1,2)x} & 0\\a_{x} - r_{(4,5)x} & n_{(1,2)y} & 0\\1 & 0 & 0\end{matrix}\right]\begin{Bmatrix}\dot{\theta}_{(4,5)}\\\dot{d}_{(1,2)}\\\dot{d}_{(1,3)}\end{Bmatrix} + \left[\begin{matrix}0 & - a_{y} + r_{(5,7)y} & 0 & n_{(2,4)x} & 0\\0 & a_{x} - r_{(5,7)x} & 0 & n_{(2,4)y} & 0\\0 & 1 & 0 & 0 & 0\end{matrix}\right]\begin{Bmatrix}\dot{\theta}_{(4,6)}\\\dot{\theta}_{(5,7)}\\\dot{\theta}_{(6,7)}\\\dot{d}_{(2,4)}\\\dot{d}_{(3,4)}\end{Bmatrix}

Constraint equations:

.. math::


   \begin{Bmatrix}\bf{v}^{(2)}-\bf{v}^{(1)} \\ \bf{v}^{(3)}-\bf{v}^{(1)} \\ \bf{\omega}^{(3)}-\bf{\omega}^{(1)}\end{Bmatrix} = \bf{0}

.. math::


   \Rightarrow \left[\begin{matrix}0 & - n_{(1,2)x} & n_{(1,3)x}\\0 & - n_{(1,2)y} & n_{(1,3)y}\\a_{y} - r_{(4,5)y} & 0 & 0\\- a_{x} + r_{(4,5)x} & 0 & 0\\-1 & 0 & 0\end{matrix}\right]\begin{Bmatrix}\dot{\theta}_{(4,5)}\\\dot{d}_{(1,2)}\\\dot{d}_{(1,3)}\end{Bmatrix} + \left[\begin{matrix}0 & 0 & 0 & - n_{(2,4)x} & n_{(3,4)x}\\0 & 0 & 0 & - n_{(2,4)y} & n_{(3,4)y} \\- a_{y} + r_{(4,6)y} & a_{y} - r_{(5,7)y} & - a_{y} + r_{(6,7)y} & 0 & 0 \\a_{x} - r_{(4,6)x} & - a_{x} + r_{(5,7)x} & a_{x} - r_{(6,7)x} & 0 & 0 \\1 & -1 & 1 & 0 & 0 \end{matrix}\right]\begin{Bmatrix}\dot{\theta}_{(4,6)}\\\dot{\theta}_{(5,7)}\\\dot{\theta}_{(6,7)}\\\dot{d}_{(2,4)}\\\dot{d}_{(3,4)}\end{Bmatrix}=\begin{Bmatrix} 0 \\ 0 \\ 0\end{Bmatrix}

.. math::


   \bf{J_a} = \left[\begin{matrix}- a_{y} + r_{(4,5)y} & n_{(1,2)x} & 0\\a_{x} - r_{(4,5)x} & n_{(1,2)y} & 0\\1 & 0 & 0\end{matrix}\right]

.. math::


   \bf{J_p} = \left[\begin{matrix}0 & - a_{y} + r_{(5,7)y} & 0 & n_{(2,4)x} & 0\\0 & a_{x} - r_{(5,7)x} & 0 & n_{(2,4)y} & 0\\0 & 1 & 0 & 0 & 0\end{matrix}\right]

.. math::


   \bf{A_a} = \left[\begin{matrix}0 & - n_{(1,2)x} & n_{(1,3)x}\\0 & - n_{(1,2)y} & n_{(1,3)y}\\a_{y} - r_{(4,5)y} & 0 & 0\\- a_{x} + r_{(4,5)x} & 0 & 0\\-1 & 0 & 0\end{matrix}\right]

.. math:: \bf{A_p} = \left[\begin{matrix}0 & 0 & 0 & - n_{(2,4)x} & n_{(3,4)x} \\0 & 0 & 0 & - n_{(2,4)y} & n_{(3,4)y} \\- a_{y} + r_{(4,6)y} & a_{y} - r_{(5,7)y} & - a_{y} + r_{(6,7)y} & 0 & 0 \\a_{x} - r_{(4,6)x} & - a_{x} + r_{(5,7)x} & a_{x} - r_{(6,7)x} & 0 & 0 \\1 & -1 & 1 & 0 & 0 \end{matrix}\right]

.. math::


   \bf{\widetilde{J}} = \bf{J_a}-\bf{J_p}\bf{A^{-1}_p}\bf{A_a}