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The following question is from Kinematic Analysis of Robot Manipulators by Carl D. Crane, III, Joseph Duffy:

"The transformation that relates the A and B coordinate systems is given as That is the rotation of frame B with respect to frame A.

$${}^{A}_{B}T= \left[ \begin{matrix} 0.866025 & 0 & 0.5 & 0.26795 \\ 0 & 1 & 0 & 0 \\ -0.5 & 0 & 0.866025 & 1\\ 0&0&0&1 \\ \end{matrix}\right] $$

Coordinate system B can be obtained from coordinate system A by initially aligning it with A and then rotating coordinate system B about an axis $\mathbf{m}$ by an angle $\gamma$ where the rotation axis passes through a point $\mathbf{p}$. Determine $\mathbf{m}$, $\gamma$ and $\mathbf{p}$"[1]

My first question is about the question itself. As far as I know a final pose of a body in 3D space can be achieved by a rotation about an axis and a translation about that same axis (The screw theory). If a body in 3D space has only experienced rotation and if we know the rotation matrix, then we can use equivalent angle-axis formulation to find the unit vector parallel to the rotation axis and the angle of rotation. In 2D a single rotation suffices to attain a final pose of a planar body, no matter how many translations and rotations that planar body has performed. Checking the transformation matrix ${}^{A}_{B}T$, there is a translational difference between the origins of frames A and B which means some translation occured during the transformation. How is it possible that the frame B can be obtained by a single rotation about an axis as the question suggests?

My second question is how do you solve the question?

[1]Carl D. Crane III, Joseph Duffy (1998) Kinematic Analysis of Robot Manipulators.Cambridge University Press.

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    $\begingroup$ The axes are initially aligned, and then $B$ is rotated about a rotation axis [that] passes through a point $p$ (not through the origin!). Such a rotation moves B away from A both translationally and rotationally. $\endgroup$ Commented Jul 9, 2023 at 17:57
  • $\begingroup$ I understand @RollenS.D'Souza. Besides, in screw theory, the pitch accounts for the translational motion perpendicular to the plane of rotation I think $\endgroup$
    – Ali Kıral
    Commented Jul 9, 2023 at 18:25
  • $\begingroup$ Observing the transformation matrix, the rotation takes place around y axis, and the translation is within xz plane. $\endgroup$
    – Ali Kıral
    Commented Jul 9, 2023 at 18:32

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Let $T_B^A = (R_B^A, {r}_A^B)$ where $R_B^A$ is the rotation and ${r}_A^B$ is the translation. The critical observation is that ${p}$ must be invariant under the transformation since we are rotating about an axis through ${p}.$ Invariance of ${p}$ demands,

$$R_B^A\, {p} + {r}^B_A = {p},$$

which you can solve under appropriate conditions: when the translation is orthogonal to the axis of rotation. Once you've determined ${p},$ everything else falls in place. Take any vector ${v}_B$ and compute,

$$R_B^A\, {v}_B + {r}^B_A = {v}_A.$$

Subtract ${p}$ from both sides and use the characteristic equation defining ${p}$ to find,

$$R_B^A \left( {v}_B - {p} \right) = {v}_A - {p}.$$

From this point onwards $m$ and $\gamma$ are computable directly from $R_B^A$ as we can explicitly see that rotating about point $p$ is done using $R_B^A.$ Section 2.8.2 of your text should cover this.

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  • $\begingroup$ Rp+r=p, then p=(I-R)^(-1)r. (I-R) is singular thus not invertible despite the fact that the translation is orthogonal to the axis of rotation, as rotation is about the y axis and the plane of translation is the xz plane. $\endgroup$
    – Ali Kıral
    Commented Jul 12, 2023 at 7:52
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    $\begingroup$ @AliKıral I never talked about invertibility. Of course it isn't invertible since 1 is an eigenvalue of R. It is solvable however since r is in the image of I-R $\endgroup$ Commented Jul 12, 2023 at 13:13
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    $\begingroup$ If $\omega$ is the axis of rotation then observe that $\omega^\top R p + \omega^\top r = \omega^\top p$ implies that $\omega^\top p + \omega^\top r = \omega^\top p$. Deduce that $\omega^\top r = 0.$ $\endgroup$ Commented Jul 12, 2023 at 14:09

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