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In Modern Robotics v2, we have the following figures: enter image description here

enter image description here

I don't understand why in Eq. 3.73, we have the given expression for $v_s$. My thinking is:

The position $p$ is the position of the frame $b$ with respect to fixed frame $s$.

$\omega_s$ is supposed to be the angular velocity of $b$ wrt $s$. But how can $-p$ represent a "point on body [b] currently at the fixed-frame origin"? If $p$ is $p_{sb}$, then isn't $-p = p_{bs}$? Since if you sum $p_{sb}$ and $p_{bs}$ don't you get the $0$ vector, which means the linear velocity $v_s$ is 0?

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  • $\begingroup$ Please edit your question to explain the logic by which you encounter that sum in equation 3.73. $\endgroup$
    – Tully ♦
    Nov 27, 2022 at 22:14
  • $\begingroup$ There is not an equation on either of these pages where p is added to -p, so it's unclear what you are actually asking. The formulation of Equation 3.73 follows directly from the definitions of R, T, and p. $\endgroup$
    – Brandon J. DeHart
    Jan 11 at 19:20

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