What's the difference between Spatial velocity and Body velocity?

Question

Let us suppose a rigid body $\{B\}$ is undergoing rotational and translational motion,and the spatial frame is $\{S\}$. In this case, we say that its configuration space is the Special Euclidean group, i.e.,$T:= SE(3)$

Spatial velocity is defined by $$\dot{T}T^{-1} =\left[ \begin{array} &\dot{R} &\dot{p} \\0 &0 \end{array} \right] \left[\begin{array}{cc} {R}^T & -R^T{p} \\0 &1 \end{array}\right] = \left[\begin{array}{cc} \dot{R}{R}^T & \dot{p}-\dot{R}R^T{p} \\0 &0 \end{array}\right]= \left[\begin{array}{cc} [\omega_s] & v_s \\0 &0 \end{array}\right]$$

The corresponding body velocity is $$T^{-1}\dot{T} =\left[\begin{array}{cc} {R}^T & -R^T{p} \\0 &1 \end{array}\right]\left[ \begin{array} &\dot{R} &\dot{p} \\0 &0 \end{array} \right] = \left[\begin{array}{cc} {R}^T\dot{R} & R^T\dot{p} \\0 &0 \end{array}\right]= \left[\begin{array}{cc} [\omega_b] & v_b \\0 &0 \end{array}\right]$$

From first formula, $v_s=\dot{p}-\dot{R}R^T{p},$ it is not the linear velocity of the rigid body frame expressed in the spatial frame(indeed,quantity would be $\dot{p}$).

I cann't understand the physical interpretation of $v_s$.

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Reference:Murray, Richard M. A mathematical introduction to robotic manipulation. CRC press, 2017.

and related question What's the difference between Spatial angular velocity and Body angular velocity?

Answer 1

"Spatial velocity" v.s. "body velocity": It's rather confusing to think of "body velocity" as some mathematical magic of swapping $\dot{T}$ and $T^{-1}$ in the definition. Its correct definition is "spatial velocity as observed in a frame that coincides with the current body frame".

Consider a different fixed spatial frame $S'$. Then $T_{S'S}$ is constant and $T_{S'B}(t) = T_{S'S}T_{SB}(t) \ \forall t$. The spatial velocity observed in $S'$ is $$\dot{T}_{S'B}(t)T_{S'B}(t)^{-1} = T_{S'S}(\dot{T}_{SB}(t) T_{SB}(t)^{-1}) T_{S'S}^{-1}$$ Now imagine that $S'$ coindicdes with $B$ at time $t$, i.e. $T_{S'S} = T_{BS}(t) = T_{SB}(t)^{-1}$. Then the above becomes $$T_{SB}(t)^{-1}\dot{T}_{SB}(t), \mbox{ or simply } T^{-1}\dot{T}$$

So "spatial velocity" and "body velocity" are derived from the same concept.

Spatial velocity is indepedent of the choice of the body frame. Imagine another frame $B'$ that moves together with the body, so $T_{BB'}$ is constant. The spatial velocity for $B'$ is $$\dot{T}_{SB'}T_{SB'}^{-1} = \dot{T}_{SB}(T_{BB'}T_{BB'}^{-1})T_{SB}^{-1} = \dot{T}_{SB}T_{SB}^{-1}$$ This is great, because the spatial velocity should describe the motion of the entire body, without relying on which body frame is chosen.

Physical meaning of $v_s$: You probably wonder why $v_s \not= \dot{p}$. It's because if the body rotates, $\dot{p}$ depends on which body frame you choose. Therefore $v_s$ is defined from only the frame $S$ itself.

In fact, imagine a point $c$ that (1) moves together with the body, (2) coincides with the origin of $S$ at time $t$. Then its linear velocity observed in $S$ at time $t$ is $\dot{p} - w_s \times p = \dot{p} - \dot{R}R^Tp = v_s$.

Answer 2

I think $v_s = \dot{p} - \omega_s\times p,$ is the combination of velocity $\dot{p}$ caused by translation and velocity $\omega\times{(-p)}$ caused by rotation,so it is the instantaneous velocity at spatial frame exressed in spatial frame.