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In section 5.1.1 of Modern Robotics by Park and Lynch (pg. 153 of 1st edition, pg. 177 of updated 1st edition), the following derivation is given for the Jacobian matrix.

Derivation

I don't understand this derivation. On the first line, we have an expression for $[\mathcal{V_s}]$, while in equation 5.6 have a very similar expression written for $\mathcal{V_s}$. The relationship between the two equations is unclear to me.

In "Definition 3.20" (pg. 85 of 1st edition, pg. 98 of updated 1st edition), there is some discussion of adjoint maps, and though I have reread it several times it does not clarify the issue for me. If anyone could provide more detailed steps that would be much appreciated.

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You are correct in noticing that Def. 3.20 is important in the move from $[\mathcal{V}_s]$ to $\mathcal{V}_s$ in Eq. 5.6. Specifically I think it's important to recognize 3 points:

  1. The adjoint map of a transform $T \in SE(3)$ can transform a twist represented in one frame into another frame. That is basically what is stated in Def. 3.20 when it says $\mathcal{V}^\prime = \left[Ad_T\right]\mathcal{V}$. This point is made even more explicitly just after Eq. 3.84 when it says:

More generally, for any two frames $\{c\}$ and $\{d\}$, a twist represented as $\mathcal{V}_c$ in $\{c\}$ is related to its representation $\mathcal{V}_d$ in $\{d\}$ by $\mathcal{V}_c = \left[Ad_{T_{cd}}\right]\mathcal{V}_d$

  1. At the end of Def. 3.20 they show an equivalent matrix expression for transforming a twist. So, the matrix form of $\mathcal{V}\in\mathbb{R}^6$ is $[\mathcal{V}]\in se(3)$ and the equivalent adjoint map expression is $$ \left[\mathcal{V}^\prime\right] = T[\mathcal{V}]T^{-1} $$

  2. The exponential coordinates for a homogeneous transformation $T$ are $\mathcal{S}\theta \in \mathbb{R}^6$ and they are related through the matrix exponential.

When moving from the first line to Eq. 5.6 all three of these facts are used. Let's just look at the second term in the top expression (and drop the $\dot{\theta}_2$); additionally, let's denote that term $\Gamma$ and use a $[\cdot]$ to denote that it is $\in se(3)$: $$ \left[\Gamma\right] = e^{[\mathcal{S}_1]\theta_1}\left[\mathcal{S}_2\right]e^{-[\mathcal{S}_1]\theta_1} \in se(3) $$

Looking at point (3) above we can recognize that $e^{[\mathcal{S}_1]\theta_1}$ and $e^{-[\mathcal{S}_1]\theta_1}$ are both $\in SE(3)$, and remember that $\left(e^{A\theta}\right)^{-1}=e^{-A\theta}$. So the expression above takes exactly the matrix form of the adjoint map from point (2) above. So we can move that expression back into the "vector" form of the adjoint map from point (1) above. $$ \Gamma = \left[Ad_{e^{[\mathcal{S}_1]\theta_1}}\right]\mathcal{S}_2 = Ad_{e^{[\mathcal{S}_1]\theta_1}}\left(\mathcal{S}_2\right) \in \mathbb{R}^6 $$ It can then easily be seen that $\Gamma = J_{s2}$ from Eq. 5.6. This same reasoning applies to all other $J_{s*}$ terms.

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  • $\begingroup$ Breaking the equations down term-by-term made this very clear, thank you. $\endgroup$ Commented Aug 24, 2023 at 3:08
  • $\begingroup$ Excellent. Glad I was able to help! $\endgroup$ Commented Aug 25, 2023 at 16:24

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