There is something I need to verify.

Say we have the following RPR robot manipulator.

enter image description here

The DH table

enter image description here

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

enter image description here

where the 3 by 3 inertia matrix is given by

enter image description here

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?


1 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.

  • $\begingroup$ I made some more research and you're right. Oddly enough the same book uses the same notation for both linear velocity Jacobian and the J matrix which relates the velocity of the COM to the derivatives of the joint variables. Thanks for pointing this out. $\endgroup$
    – csg
    Nov 16, 2017 at 22:11
  • $\begingroup$ No problem. Glad it helps. $\endgroup$ Nov 17, 2017 at 2:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.