# 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 $$\mathbf{p}$$"

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?

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. Jul 9 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 Jul 9 at 18:25
• Observing the transformation matrix, the rotation takes place around y axis, and the translation is within xz plane. Jul 9 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. Jul 12 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 Jul 12 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 $\omega^\top r = 0.$ Jul 12 at 14:09