A single rotation to reach the final pose of a frame

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.

• 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. Commented Jul 9, 2023 at 17:57
• I understand @RollenS.D'Souza. Besides, in screw theory, the pitch accounts for the translational motion perpendicular to the plane of rotation I think Commented Jul 9, 2023 at 18:25
• Observing the transformation matrix, the rotation takes place around y axis, and the translation is within xz plane. Commented Jul 9, 2023 at 18:32

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.

• 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. Commented Jul 12, 2023 at 7:52
• @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 Commented Jul 12, 2023 at 13:13
• 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 $\omega^\top r = 0.$ Commented Jul 12, 2023 at 14:09