So I thought I understood well enough what a Jacobian was (in the context of an $n$-DOF robot) -- a function that takes a vector of n joint positions and returns an $n \times n$ matrix that can be multiplied with a vector of $n$ joint velocities to return a velocity vector for the end effector.

I'm using ROS and MoveIt, so I actually already have a function to calculate the Jacobian for my robot from the URDF.

However, I'm reading lecture notes from the 2005 MIT Intro to Robotics course, and in one (mission-critical, it seems) portion of chapter 7 (between pages 11 and 12), he refers to "$3 \times n$ Jacobian matrices relating the centroid linear velocity and the angular velocity of the $i^\text{th}$ link to joint" as $J^L$ and $J^A$.

He introduces Jacobians in Chapter 5, and indeed I looked through all the rest of the course material, and I don't think he ever explains what these matrices are or how to compute them.

Could someone enlighten me as to what he's talking about?


1 Answer 1


Short answer

Robot Dynamics and Control by Spong et al. (especially Chapter 5) can definitely help you on this matter.

Long answer

First of all, you are partially correct about a Jacobian. It is indeed a function of joint values (say $q \in \mathbf{R}^n$). However, as a Jacobian maps a joint velocity to an end-effector velocity, its dimension is not generally $n \times n$, but rather $m \times n$, where $n$ is the number of DOFs your robot has and $m$ is the dimension of the velocity vector you want to map a joint velocity vector to.

Let's consider the following equation $$\dot{x} = J(q)\dot{q},$$ where $\dot{x}$ is an end-effector velocity vector. The vector $\dot{x}$ is generally a 6-vector that is (generally) a vertical stack of linear and angular velocity, i.e. $$\dot{x} = \begin{bmatrix}v\\ \omega\end{bmatrix}.$$

Looking closely, one can see that the first three rows of $J(q)$ are mapping the joint values $q$ to the linear end-effector velocity $v \in \mathbf{R}^{3}$. Similarly, the last three rows are mapping $\dot{q}$ to the angular end-effector velocity $\omega \in \mathbf{R}^{3}$. Therefore, one can actually writes $$J = \begin{bmatrix}J^L\\J^A\end{bmatrix},$$ where $J^L$ and $J^A$ are such that $v = J^L \dot{q}$ and $\omega = J^A \dot{q}.$ Computing $J^L$ and $J^A$ can then be done accordingly by analyzing the relationship between the joint velocity and the end-effector velocity.

For detailed analysis (and also how $J^L$ and $J^A$ can be computed), you can consult Spong's book I mentioned above.

  • $\begingroup$ Wow, thank you so much both for the formatting and for the answer! $\endgroup$
    – River Tam
    Commented Jan 7, 2018 at 22:12

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