Space Jacobian
Modern Robotics Chapter 3
Space Jacobian
- If \(A, B \in \mathbb{R}^{n \times n}\) are both invertible then \((AB)^{-1} = B^{-1}A^{-1}\),
- If \(A \in \mathbb{R}^{n \times n}\) is constant and \(\theta(t)\) is a scalar function of t then \(d\left(e^{A \theta}\right) / d t=A e^{A \theta} \dot{\theta}=e^{A \theta} A \dot{\theta}\)
- \(\left(e^{A \theta}\right)^{-1}=e^{-A \theta}\)
n-link open chain whose forward kinematics is expressed in the following product of exponentials form: \[ T\left(\theta_{1}, \ldots, \theta_{n}\right)=e^{\left[\mathcal{S}_{1}\right] \theta_{1}} e^{\left[\mathcal{S}_{2}\right] \theta_{2}} \cdots e^{\left[\mathcal{S}_{n}\right] \theta_{n}} M \] The spatial twist \(\mathcal{V}_s\) is given by \(\left[\mathcal{V}_{s}\right]=\dot{T} T^{-1}\), where:
\begin{align*} \dot{T} &=\left(\frac{d}{d t} e^{\left[\mathcal{S}_{1}\right] \theta_{1}}\right) \cdots e^{\left[\mathcal{S}_{n}\right] \theta_{n}} M+e^{\left[\mathcal{S}_{1}\right] \theta_{1}}\left(\frac{d}{d t} e^{\left[\mathcal{S}_{2}\right] \theta_{2}}\right) \cdots e^{\left[\mathcal{S}_{n}\right] \theta_{n}} M+\cdots \\ &=\left[\mathcal{S}_{1}\right] \dot{\theta}_{1} e^{\left[\mathcal{S}_{1}\right] \theta_{1}} \cdots e^{\left[\mathcal{S}_{n}\right] \theta_{n}} M+e^{\left[\mathcal{S}_{1}\right] \theta_{1}}\left[\mathcal{S}_{2}\right] \dot{\theta}_{2} e^{\left[\mathcal{S}_{2}\right] \theta_{2}} \cdots e^{\left[\mathcal{S}_{n}\right] \theta_{n}} M+\cdots \end{align*}
Also, \[ T^{-1}=M^{-1} e^{-\left[\mathcal{S}_{n}\right] \theta_{n}} \cdots e^{-\left[\mathcal{S}_{1}\right] \theta_{1}} \] Calculating \(\dot{T}T^{-1}\), we obtain \[ \left[\mathcal{V}_{s}\right]=\left[\mathcal{S}_{1}\right] \dot{\theta}_{1}+e^{\left[\mathcal{S}_{1}\right] \theta_{1}}\left[\mathcal{S}_{2}\right] e^{-\left[\mathcal{S}_{1}\right] \theta_{1}} \dot{\theta}_{2}+e^{\left[\mathcal{S}_{1}\right] \theta_{1}} e^{\left[\mathcal{S}_{2}\right] \theta_{2}}\left[\mathcal{S}_{3}\right] e^{-\left[\mathcal{S}_{2}\right] \theta_{2}} e^{-\left[\mathcal{S}_{1}\right] \theta_{1}} \dot{\theta}_{3}+\cdots . \] The above can also be expressed in vector form by means of the Adjoint representation \[ \mathcal{V}_{s}=\underbrace{\mathcal{S}_{1}}_{J_{s 1}} \dot{\theta}_{1}+\underbrace{\operatorname{Ad}_{e^{\left[S_{1}\right] \theta_{1}}}\left(\mathcal{S}_{2}\right)}_{J_{s 2}} \dot{\theta}_{2}+\underbrace{\operatorname{Ad}_{e^{\left[S_{1}\right] \theta_{1}} e^{\left[S_{2}\right] \theta_{2}}}\left(\mathcal{S}_{3}\right)}_{J_{s 3}} \dot{\theta}_{3}+\cdots \] \(\mathcal{V}_s\) is a sum of n spatial twists of the form: \[ \mathcal{V}_{s}=J_{s 1}+J_{s 2}(\theta) \dot{\theta}_{1}+\cdots+J_{s n}(\theta) \dot{\theta}_{n} \] where each \(J_{s i}(\theta)=\left(\omega_{s i}(\theta), v_{s i}(\theta)\right)\) depends on the joint values \(\theta \in \mathbb{R}^n\) for \(i = 2, …, n\). In the matrix form,
\begin{align*} \mathcal{V}_{s} &=\left[\begin{array}{llll}J_{s 1} & J_{s 2}(\theta) & \cdots & J_{s n}(\theta)\end{array}\right]\left[\begin{array}{c}\dot{\theta}_{1} \\ \vdots \\ \dot{\theta}_{n}\end{array}\right] \\ &=J_{s}(\theta) \dot{\theta} \end{align*}
Formal definition
The space Jacobian \(J_s(\theta) \in \mathbb{R}^{6 \times n}\) relates the joint rate vector \(\dot{\theta} \in \mathbb{R}^n\) to the spatial twist \(\mathcal{V}_s\) via \[ \mathcal{V}_{s}=J_{s}(\theta) \dot{\theta} \] The ith column of \(J_s(\theta)\) is: \[ J_{s i}(\theta)=\operatorname{Ad}_{e^{\left[\mathcal{S}_{1}\right] \theta_{1} \ldots}} e^{\left[\mathcal{S}_{i-1}\right] \theta_{i-1}}\left(\mathcal{S}_{i}\right) \] for i = 2 ,…, n, with the first column \(J_{s1} = \mathcal{S}_{1}\)
Understand
The ith column is of the form \(Ad_{T_{i-1}}(\mathcal{S}_{i})\), where \(T_{i-1}=e^{\left[\mathcal{S}_{1}\right] \theta_{1}} \cdots e^{\left[\mathcal{S}_{i-1}\right] \theta_{i-1}}\); \(\mathcal{S}_i\) is the screw axis describing the ith joint axis in terms of the fixed frame with the robot in its zero position. \(\operatorname{Ad}_{T_{i-1}}\left(\mathcal{S}_{i}\right)\) is therefore the screw axis describing the ith joint axis after it undergoes the rigid body displacement \(T_{i-1}\). Physically this is the same as moving the first i-1 joints from their zero position to the current values \(\theta_{1}, \ldots, \theta_{i-1}\).