I think the equation(3.73) should be:
$\dot p-w_s$x$p=\dot p-\dot p=0$
have no idea how it can be wrong
and don't know the difference between infinitely large body and a body just include the fix-frame origin and body-frame origin
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1They are choosing to rewrite $\nu_s$ so you can see the idea of the infinitely large body in the equation. It's a valid way to write the equation, so not wrong. Are you asking why $\dot{p} \neq \omega_s \times p$ instead? – hauptmech Sep 09 '19 at 06:18
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@hauptmech thanks. yes i m asking why $\dot p≠w_s×p $ instead. – E Takly Sep 09 '19 at 14:32
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Welcome to Robotics, E Takly. Great first question! I'm looking forwards to an answer on this - questions like this, and the difficulty I had in trying to conceptually understand the physical meaning of spatial maths, are exactly why I wound up walking away from this approach to dynamics. I'm hoping a really good answer to this question might be enough to get something to click for me, too. – Chuck Sep 09 '19 at 18:30
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@Chuck ha, me too,really want to give up – E Takly Sep 09 '19 at 20:35
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A coordinate transformation of a point P from Frame 1 to Frame 0 is given by: $$ \mathbf{p}^0=\mathbf{o}^0_1+\mathbf{R}^0_1\mathbf{p}^1. $$ Differentiating with respect to time gives: $$ \dot{\mathbf{p}}^0=\dot{\mathbf{o}}^0_1+\mathbf{R}^0_1\dot{\mathbf{p}}^1+\dot{\mathbf{R}}^0_1\mathbf{p}^1. $$ Considering that $\dot{\mathbf{p}}^1=0$ as $\mathbf{p}^1$ is fixed in Frame 1, we come up with: $$ \dot{\mathbf{p}}^0=\dot{\mathbf{o}}^0_1+\mathbf{S}\left(\mathbf{\omega}^0_1\right)\mathbf{R}^0_1\mathbf{p}^1, $$ being $\mathbf{S}$ the skew matrix. This expression can be further reduced noting that $\mathbf{R}^0_1\mathbf{p}^1=\mathbf{r}^0_1$: $$ \dot{\mathbf{p}}^0=\dot{\mathbf{o}}^0_1+\mathbf{\omega}^0_1 \times \mathbf{r}^0_1, $$ which is eventually the known form of the velocity composition rule.
To come back to your statement, $\dot{\mathbf{p}}^0=\mathbf{\omega}^0_1 \times \mathbf{r}^0_1$ holds only when $\dot{\mathbf{o}}^0_1=0$, that is when there exists a pure rotational velocity without translational velocity.
This result can be obtained also by observing that a vector $\mathbf{p}$ can be expressed in terms of its magnitude $p$ and unit vector $\mathbf{\hat{p}}$ as: $$ \mathbf{p}=p \cdot \mathbf{\hat{p}}. $$ Thereby, the time derivative is: $$ \dot{\mathbf{p}}=\dot{p}\cdot\mathbf{\hat{p}}+p\cdot\dot{\mathbf{\hat{p}}}=\dot{p}\cdot\mathbf{\hat{p}}+\mathbf{\omega} \times \mathbf{p}. $$ Only in case there is a pure rotational velocity (i.e. $\dot{p}=0$), then you'll get: $$ \dot{\mathbf{p}}=\mathbf{\omega} \times \mathbf{p}. $$
Ugo Pattacini
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3The steps from $\dot{\mathbf{R}}^0_1\mathbf{p}^1$ to $\mathbf{S}\left(\mathbf{\omega}^0_1\right)\mathbf{R}^0_1\mathbf{p}^1$ are shown in the angular velocity wikipedia article.
Murray, Li, and Sastry remains my favorite explainer of robot maths and page 51-52 shows the steps as well.
– hauptmech Sep 09 '19 at 22:51 -
1legend, thank you. didn't notice the translation.so the equation(3.73) becomes: $\dot{\mathbf{p}}-\mathbf{\omega} \times \mathbf{p}=$$ \dot{p}\cdot\hat{p}+p\cdot\dot{\hat{p}} -\mathbf{\omega} \times \mathbf{p}=\dot{p}\cdot\hat{p}+\mathbf{\omega} \times \mathbf{p} - \mathbf{\omega} \times \mathbf{p} = \dot{p}\cdot\hat{p}. $ is it right? – E Takly Sep 09 '19 at 23:20
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