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Let $p$ is the position vector from the origin of frame {s} (i.e. inertial frame) to the origin of the body frame {b}. Take a look at the following picture, the vectors $\omega_b,v_b$ represent the angular and linear velocities of frame{b} attached to the moving robot expressed in the body frame. The vector $\omega_s$ is the angular velocity of frame {b} expressed in the inertial frame {s}. Surprisingly, the vector $v_s$ is not the linear velocity of the body frame's origin expressed in the inertial frame {s} (i.e. $\dot{p} \neq v_s$). The actual formula is $$ \dot{p} = v_s + \dot{R}R^T p $$ The notation confuses me. In the book I'm reading, it is

the physical meaning of $v_s$ can now be inferred: imagining the moving body to be infinitely large, $v_s$ is the instantaneous velocity of the point on this body currently at the fixed-frame origin, expressed in.

Could anyone provide different explanation what does exactly this vector mean? Why do we need it if it is not the linear velocity of the origin of frame {b} expressed in the inertial frame {s}?

enter image description here

Reference: Modern Robotics Mechanics, Planning, and Control

CroCo
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1 Answers1

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Anything to do with rotations is often difficult and usually counter-intuitive.
I have used the symbols that you have quoted but simplified the diagram a little.

enter image description here

The top image is the initial condition with frame{s} at position $K$ and frame{b} at position $L$.
The displacement from $K$ to $L$ is $\vec p$.
I have introduced another set of coordinate axes attached to the "extended" robot and positioned at $K$ and called the unit vectors $\hat X_{\rm b}$ and $\hat X_{\rm b}$.
A key idea is that the two sets of axes attached to the robot do not move relative to one another as the robot is rigid.

Now let the robot move with both a rotation and a translation relative to frame{s} to a new position in a time $\Delta t$ with the two sets of axes attached to the robot moving to positions $K'$ and $L'$ as shown in the lower figure.

The displacement from position $L$ to position $L'$ is $\Delta \vec p$ and the displacement from position $K$ to position $K'$ is $\Delta \vec q$.

Relative to frame{s}, the frame attached the left hand side of the robot has undergone a displacement of $\Delta \vec q$ in a time of $\Delta t$ so $\vec v_{\rm s} = \dfrac{\Delta \vec q}{\Delta t}$ and this must also be the velocity of the robot frame at position $L'$ as the robot is rigid.
From the diagram it can be seen that $\vec v_{\rm s} \ne \dfrac{\Delta \vec p}{\Delta t}$.

Farcher
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  • Thanks for the answer. So using $s$ in the notation $v_s$ may not be a good approach, right? Also, why we need $v_s$? since we are interested in $\dot{p}$ to apply it to the Newton's law to derive the motion equation. – CroCo Dec 24 '21 at 10:42
  • @CroCo The equation for $\dot p$, which you belated published, shows that it can be decomposed into a translation $v_{\rm s}$ and a rotation $\dot{R}R^T p$. – Farcher Dec 24 '21 at 11:04