Question 5 of 7: Orientations unattainable by a 3R mechanism with non-orthogonal twists (20 marks)

Question 5: Orientations unattainable by a 3R mechanism with non-orthogonal twists (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 3R wrist whose three axes meet at a fixed point, with link twists $\alpha_1 = \phi$ and $\alpha_2 = 180^\circ-\phi$, $\phi \neq 90^\circ$; all three joints turn freely through $360^\circ$ and self-collision is ignored.

Find. A geometric description of the orientations of link 3 that the mechanism cannot reach.

Approach. Separate the orientation into the direction of the link-3 axis and the spin about it; show that joints 1 and 3 supply the spin and azimuth freely, so the whole restriction falls on the polar angle that joint 2 controls, and bound that angle with the spherical law of cosines.

1. Decompose the orientation. Any orientation of link 3 can be described by the direction of its own axis $\hat{Z}_3$ (two parameters, a point on the unit sphere) plus a roll about that axis (one parameter). Joint 3 rotates link 3 about $\hat{Z}_3$, so the roll is always free; joint 1 rotates the whole chain about the fixed axis $\hat{Z}_1 = \hat{Z}_0$, so the azimuth of $\hat{Z}_3$ about $\hat{Z}_0$ is always free. Whatever cannot be reached must therefore be a restriction on the single remaining number: the polar angle $e$ between $\hat{Z}_0$ and $\hat{Z}_3$, which only joint 2 can change.
2. Set up the spherical triangle. Project the three joint axes onto a unit sphere centred on the common intersection point. The twists are the fixed arc lengths between neighbouring axes: $\hat{Z}_1$ to $\hat{Z}_2$ subtends $\alpha_1 = \phi$, and $\hat{Z}_2$ to $\hat{Z}_3$ subtends $\alpha_2 = 180^\circ - \phi$. Turning joint 2 by $\theta_2$ swings $\hat{Z}_3$ around $\hat{Z}_2$ on a cone, changing only the angle at the vertex $\hat{Z}_2$ of that triangle.
3. Apply the spherical law of cosines. For the triangle with sides $\alpha_1$ and $\alpha_2$ and included angle $\theta_2$, $$ \cos e = \cos\alpha_1\cos\alpha_2 + \sin\alpha_1\sin\alpha_2\cos\theta_2 = -\cos^{2}\phi + \sin^{2}\phi\,\cos\theta_2, $$ using $\cos(180^\circ-\phi) = -\cos\phi$ and $\sin(180^\circ-\phi) = \sin\phi$. Here $\theta_2$ is measured from the configuration in which the two twist planes coincide.
4. Bound the polar angle. As $\theta_2$ runs through a full turn, $\cos\theta_2$ covers $[-1,1]$, so $$ \cos e \in \left[-\cos^{2}\phi-\sin^{2}\phi,\ -\cos^{2}\phi+\sin^{2}\phi\right] = \left[-1,\ -\cos 2\phi\right], $$ and therefore $$ \boxed{\,|180^\circ-2\phi| \le e \le 180^\circ\,}. $$ The upper limit is always reached (the two twists then add to a straight angle and $\hat{Z}_3$ points opposite $\hat{Z}_0$); the lower limit is the obstruction.
5. State the unattainable set. The mechanism can never place the link-3 axis closer than $|180^\circ-2\phi|$ to the fixed axis $\hat{Z}_0$. Writing an orientation as a rotation matrix ${}^{0}_{3}R$ whose third column is $\hat{Z}_3$ expressed in $\{0\}$, the unattainable set is $$ \Big\{\,{}^{0}_{3}R \ :\ \angle\big(\hat{Z}_0,\ {}^{0}_{3}R\,\hat{Z}\big) \ \lt\ |180^\circ-2\phi| \,\Big\}, $$ that is, every orientation in which link 3 points into the polar cone of half-angle $|180^\circ-2\phi|$ around the base axis — including the case where link 3 is parallel to link 0's axis. Outside that cone every orientation is reachable, because joints 1 and 3 then supply the remaining two degrees of freedom.
6. Check the limiting cases and quantify the loss. Setting $\phi = 90^\circ$ gives a forbidden half-angle of $0^\circ$: the cone collapses, which is precisely the Z–Y–Z Euler mechanism of Figure 2(a) that reaches everything. For $\phi = 60^\circ$ the cone has half-angle $60^\circ$ and, since a spherical cap of half-angle $\gamma$ occupies a fraction $(1-\cos\gamma)/2$ of all directions, $\boxed{25\%}$ of axis directions are lost; at $\phi = 45^\circ$ the half-angle is $90^\circ$ and a full hemisphere — half of all directions — is unreachable. The further $\phi$ strays from $90^\circ$, the worse the wrist becomes.