What could be the unit ( dimension) of the vectors spanning the Jacobian matrix?

Question

Let the general forward rate kinematics equation for a serial mechanism be given as follows:

$$ \begin{bmatrix} \boldsymbol{v} \\ \boldsymbol{\omega} \end{bmatrix} = \begin{bmatrix} \boldsymbol{J}_1 & \boldsymbol{J}_2 & \cdots &\boldsymbol{J}_n \\ \end{bmatrix} \begin{bmatrix} \dot{\boldsymbol{q}}_1 \\ \dot{\boldsymbol{q}}_1 \\ \vdots \\ \dot{\boldsymbol{q}}_n \end{bmatrix}$$

The Jacobian, $\boldsymbol{J}_i \in \mathbb{R}^{6}$,columns are formed by a direction and moment vectors, i.e., $\boldsymbol{s}_{i\parallel} $ and $(\boldsymbol{s}_{i\parallel} \times \boldsymbol{l}_i)$, respectively. I know $\boldsymbol{s}_{i\parallel}$ is a direction vector which is only dependent on the joint angles and $\boldsymbol{l}_i$ is a position vector to the location of $\boldsymbol{s}_{i\parallel}$. If $\dot{\boldsymbol{q}}_i$ has a unit of $rad/sec$, what would be the unit of $\boldsymbol{s}_{i\parallel} \times \boldsymbol{l}_i$ to get an approperiate dimension for $\boldsymbol{v}$ and $\boldsymbol{\omega}$?

Thank you in advance

Answer 1

The units of the linear velocity portions of your Jacobian (what you have denoted as $\boldsymbol{s}_{i \parallel} \times \boldsymbol{l}_{i}$) are whatever distance unit you have used to define $\boldsymbol{l}_{i}$ "divided by" radians. To some extent this may rely on the parameterization method of your robot's kinematic structure (i.e. DH parameters). Given that you are not mixing units when you calculate your Jacobian, it will be the same units as were used to calculate $\boldsymbol{l}_{i}$ - which physically represents the distance from the $i^\text{th}$ frame of your robot to the end-effector.

Assuming you use meters in your model parameterization, we have:

$$ \mathbf{J}(\mathbf{q}) = \frac{\partial f(\mathbf{q})}{\partial \mathbf{q}} $$

where $f(\cdot)$ is the forward kinematics map. So, since our forward kinematics have been defined in meters using an appropriate parameterization method, the linear velocity here is really not velocity as much as change in position of our end-effector with respect to change in joint angle. It should be noted that the units for our orientation space and individual joint angles are radians. Thus, our units for the linear Jacobian terms would simply be $\frac{\text{m}}{\text{rad}}$, and likewise the units of each $\boldsymbol{s}_{i \parallel}$ term are $\frac{\text{rad}}{\text{rad}}$. So, once multiplied by $\dot{\mathbf{q}}$, we have instantaneous linear velocity in terms of $\frac{\text{m}}{\text{sec}}$ and angular velocity in terms of $\frac{\text{rad}}{\text{sec}}$.