D(q) Inertia Matrix and the Jacobian Matrix

Question

There is something I need to verify.

Say we have the following RPR robot manipulator.

The DH table

yields 3 rotational matrices:

\begin{equation} R^0_1, R^0_2, R^0_3 \end{equation}

Using these rotational matrices, the linear velocity Jacobian matrices \begin{equation} Jv_1, Jv_2, Jv_3 \end{equation} and angular velocity Jacobian matrices \begin{equation} Jw_1, Jw_2, Jw_3 \end{equation} are derived.

Where each of these matrices are 3 by 1. So that combining the linear and angular velocity Jacobians yields the 6 by 3 Jacobian matrix of the manipulator:

\begin{equation} J = \begin{bmatrix} Jv_1 & Jv_2 & Jv_3 \\ Jw_1 & Jw_2 & Jw_3 \end{bmatrix} \end{equation}

The Euler Lagrange dynamics equation for a 3-DOF robot manipulator is

where the 3 by 3 inertia matrix is given by

where n is the number of DOF of the manipulator.

For the D(q) matrix to be 3 by 3, the linear and angular velocity Jacobian matrices must be 3 by 3 instead or 3 by 1.

Can you explain the mismatch of dimensions?

Am I supposed to augment the 3 by 1 matrices and obtain 3 by 3 matrices?

Answer 1

I think this is a matter of notations.

In the given formula for $D(q)$, the matrices $J_{vi}$ and $J_{\omega i}$ are not simply the direct extraction of columns of the Jacobian of the system.

$J_i$ is the matrix that relates $\dot{q}$ to the velocity (of the center of mass) of the link $i$. That is, if we write $v_1$ to denote the linear velocity of the center of mass of the first link, then $J_1$ will be such that

$$v_1 = J_1\dot{q}.$$

Since $\dot{q} \in \mathbf{R}^3$ and $v_1 \in \mathbf{R}^3$, the matrix $J_1$ is a $3 \times 3$ matrix.

Note also that since the velocity of link i is not affected by any joint $j > i$, the columns $j > i$ of the matrix $J_i$ will be zero.