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Take a look at the below picture, I would like to derive the Jacobian matrix without differentiation. In this Modern Robotics book, the screw theory is used. I've derived the forward kinematic using PoE formula which stated here:

$$ \begin{align} M &=\begin{bmatrix} 1&0&0& L_1+L_2\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} \\ \mathcal{S}_2 &= [0,0,1,L_1,0,0]^T \\ \mathcal{S}_1 &= [0,0,1,0,0,0]^T \\ e^{[\mathcal{S}_2]\theta_2} &= \begin{bmatrix} c_{\theta_2} &-s_{\theta_2} &0&-L_1(c_{\theta_2}-1) \\s_{\theta_2} &c_{\theta_2} &0&-L_1s_{\theta_2}\\0&0&1&0\\0&0&0&1 \end{bmatrix} \\ e^{[\mathcal{S}_1]\theta_1} &= \begin{bmatrix} c_{\theta_1} &-s_{\theta_1} &0&0 \\s_{\theta_1} &c_{\theta_1} &0&0\\0&0&1&0\\0&0&0&1 \end{bmatrix} \\ T_2^0 &= e^{[\mathcal{S}_1]\theta_1} e^{[\mathcal{S}_2]\theta_2}M \\ &= \begin{bmatrix} c_{\theta_1+\theta_2}&-s_{\theta_1+\theta_2}&0&L_2c_{\theta_1+\theta_2}+L_1c_{\theta_1} \\ s_{\theta_1+\theta_2}&c_{\theta_1+\theta_2}&0&L_2s_{\theta_1+\theta_2}+L_1s_{\theta_1} \\0&0&1&0\\0&0&0&1\end{bmatrix} \end{align} $$ In the book, the authors differentiate the last column of $T_2^0$ to obtain this: $$ \begin{align} \dot{x} &= -L_1\dot{\theta}_1 s_{\theta_1} - L_2(\dot{\theta}_1+\dot{\theta}_2)s_{\theta_1+\theta_2} \\ \dot{y} &= L_1\dot{\theta}_1 c_{\theta_1} + L_2(\dot{\theta}_1+\dot{\theta}_2)c_{\theta_1+\theta_2} \end{align} $$ Or more compactly as $$ \begin{align} \begin{bmatrix} \dot{x}\\\dot{y}\end{bmatrix} &= \begin{bmatrix} (-L_1s_{\theta_1}-L_2s_{\theta_1+\theta_2})&(-L_2s_{\theta_1+\theta_2})\\(L_1c_{\theta_1}+L_2c_{\theta_1+\theta_2})&(L_2c_{\theta_1+\theta_2})\end{bmatrix} \begin{bmatrix} \dot{\theta}_1\\\dot{\theta}_2\end{bmatrix} \\ \dot{\mathbb{x}} &= [J_1(\theta) \ J_2(\theta)]\dot{\theta} \\ \dot{\mathbb{x}} &= J(\theta) \dot{\theta} \end{align} $$ where $J(\theta)$ the Jacobian matrix. The authors claim the Jacobian matrix can be obtained without differentiation using this formula:

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I couldn't reach to the same result with the above formula. For the first column, $J_{1s}=\mathcal{S}_1$ which is not the same result obtained by differentiation. Any suggestions?

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2 Answers 2

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It seems there are two types of Jacobian matrix which are the geometric and analytic ones. They are not the same but related. The one I've provided in my question is the geometric Jacobian expressed in the spatial frame. The same one expressed in the body frame is

$$ \begin{align} \mathcal{B}_1 &= [0,0,1,0,L_1+L_2,0]^T \\ \mathcal{B}_2 &= [0,0,1,0,L_2,0]^T \\ \mathcal{V}_b &= \text{Ad}_{e^{-[\mathcal{B}_2]\theta_2}}(\mathcal{B}_1) \dot{\theta}_1 + \mathcal{B}_2\dot{\theta}_2 \\ e^{-[\mathcal{B}_2]\theta_2} &= \begin{bmatrix} c_{\theta_2} &s_{\theta_2} &0&L_2(c_{\theta_2}-1) \\-s_{\theta_2} &c_{\theta_2} &0&-L_2s_{\theta_2}\\0&0&1&0\\0&0&0&1 \end{bmatrix} \\ % R_{e^{-[\mathcal{B}_2]\theta_2}} &= \begin{bmatrix} c_{\theta_2} &-s_{\theta_2} &0 \\s_{\theta_2} &c_{\theta_2} &0\\0&0&1 \end{bmatrix} \\ % p_{e^{-[\mathcal{B}_2]\theta_2}} &= [L_2(c_{\theta_2}-1), \ L_2s_{\theta_2} \ ,0]^{T}\\ % % \text{Ad}_{e^{-[\mathcal{B}_2]\theta_2}} &= \begin{bmatrix} R_{e^{-[\mathcal{B}_2]\theta_2}}& 0\\ [p_{e^{-[\mathcal{B}_2]\theta_2}}]R_{e^{-[\mathcal{B}_2]\theta_2}} & R_{e^{-[\mathcal{B}_2]\theta_2}} \end{bmatrix}\\ %% J_{b_1} &= \text{Ad}_{e^{-[\mathcal{B}_2]\theta_2}}(\mathcal{B}_1) \\ J_{b_2} &= \mathcal{B}_2 \\ J_{b}(\theta) &=[ J_{b_1}, \ J_{b_2}] \end{align} $$ Converting the geometric Jacobian to the analytic one is straightforward. $$ \begin{align} J_v(\theta) &= J_b(\theta)_{(4:6,:)} \\ &= \begin{bmatrix} L_1s_{\theta_2} & 0 \\ L_2+L_1c_{\theta_2}& L_2\\0&0\end{bmatrix} \\ R_{sb} &= \begin{bmatrix} c_{(\theta_1+\theta_2)}&-s_{(\theta_1+\theta_2)}&0\\s_{(\theta_1+\theta_2)} &c_{(\theta_1+\theta_2)}&0\\ 0&0&1\end{bmatrix} \\ J_a(\theta) &= R_{sb} J_v(\theta) \end{align} $$ where $J_a(\theta)$ is the analytic Jacobian matrix.

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For this case shouldn't the "space Jacobian" $J_s$ be $6 \times 2$ while the "Jacobian" $J$ is $2 \times 2$?

It seems like the "space Jacobian" operates on the joint velocities to deliver the end-effector space twist, while the "Jacobian" operates on the joint velocities to deliver the end-effector velocity in the space-fixed Cartesian basis.

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    $\begingroup$ I strongly suspect this is the answer. I tried reading Featherstone's spatial math papers and a lot of it was very difficult for me to follow because the symbols and terminology were borrowed/adapted from "conventional" calculus and used for spatial math. I found myself frequently confused because the results I was getting didn't match what I was expecting or what the book stated because I didn't use the "spatial" operator. $\endgroup$
    – Chuck
    Dec 30, 2021 at 15:22
  • $\begingroup$ CroCo and I are working through Park and Lynch's book. It is easier to understand than Featherstone and is more pedagogical. I find the two complement each other well though. If I am struggling on a concept in one, I will go to the other book and see what kind of insights the other author offers. $\endgroup$
    – HelpMe
    Dec 31, 2021 at 17:50
  • $\begingroup$ @Chuck indeed this is not an answer. I've solved the problem and I will post it sooner. It seems there are two types of Jacobian matrix but they are related. $\endgroup$
    – CroCo
    Dec 31, 2021 at 23:45

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