• Stage Turntable (PDF)

The Crane

• Step 1 and Figure

Step 2

The co-ordinate system of the joint has not been completely specified in the problem text, and some choices are free.
It is probably simplest if we align the bases when the angles are zero, i.e. that the boom moves in the \(xz\)-plane of of the basis of the joint. Then the boom is also parallel to the \(z\)-axis in the hand system.

With these assumptions the position of \(A\) in the joint basis is \[ R_y(\beta)\cdot(A^T+(0,0,b))\] where \(R_y(\beta)\) is a rotation by \(\beta\) around the \(y\)-axis.

Step 3

Similarly to above, a point \(C\) in the joint basis becomes \(C'\) in the base basis by the transformation \[ R_z(\alpha)\cdot(C^T+(0,0,a))\] where \(R_z(\alpha)\) is a rotation by \(\alpha\) around the \(z\)-axis.

To find the global co-ordinates of \(A\), we combine the two transformations and write \[ A = R_z(\alpha)\cdot( R_y(\beta)\cdot(A^T+(0,0,b)) + (0,0,a)) )\]