# question about spatial velocity in the book <modern robotics>

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

• They 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 9 '19 at 6:18
• @hauptmech thanks. yes i m asking why $\dot p≠w_s×p$ instead. – E Takly Sep 9 '19 at 14:32
• 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 9 '19 at 18:30
• @Chuck ha, me too,really want to give up – E Takly Sep 9 '19 at 20:35

## 1 Answer

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: $$\dot{\mathbf{p}}=\mathbf{\omega} \times \mathbf{p}.$$

• The 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 9 '19 at 22:51
• legend, 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 9 '19 at 23:20
• @E Takly, you are right. – Ugo Pattacini Sep 10 '19 at 6:54