Question 2 of 5

Question 2 (15 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.

Figure 2 models a one-armed handstand as a kinematic chain of seven links — Left Hand, Left Forearm, Left Arm, Trunk, Right Arm, Right Forearm, Right Hand — connected by six anatomical joints, with the right hand (G) fixed to the ground and the left hand (A) free.

[Figure not reproduced: Figure 2 (redrawn) — link-segment model of the arm-trunk-arm chain; each joint is classified J1 (1 DOF, hinge), J2 (2 DOF, saddle), or J3 (3 DOF, ball-and-socket) per Zatsiorsky. See the official exam paper.]

a) Chain sketch and joint DOF. Using the standard anatomical joint classification: the wrist is a saddle-type joint (flexion/extension + radial/ulnar deviation) = J2; the elbow is effectively a hinge (flexion/extension only, in this simplified model) = J1; the shoulder is a ball-and-socket joint (flexion/extension, ab/adduction, internal/external rotation) = J3. Reading the chain from the free left hand (A) to the fixed right hand (G): left wrist (A–B) J2, left elbow (B–C) J1, left shoulder (C–D, at the trunk) J3, right shoulder (D–E) J3, right elbow (E–F) J1, right wrist (F–G) J2. Links, in order: Left Hand (A), Left Forearm (B), Left Arm (C), Trunk (D), Right Arm (E), Right Forearm (F), Right Hand (G).

b) DOF of the entire chain (one hand fixed). This is an open (no-loop) serial chain anchored at G; for an open chain the system DOF equals the sum of the individual joint DOF — no loop-closure constraints reduce it, since every link's pose is fully and independently determined by the joint angles between it and the fixed base. $$\text{DOF} = \underbrace{2}_{\text{L wrist}}+\underbrace{1}_{\text{L elbow}}+\underbrace{3}_{\text{L shoulder}}+\underbrace{3}_{\text{R shoulder}}+\underbrace{1}_{\text{R elbow}}+\underbrace{2}_{\text{R wrist}} = \boxed{12\ \text{DOF}}$$

c) DOF with both hands on the ground. Fixing the left hand too closes a loop (ground → right arm chain → trunk → left arm chain → ground), so the general spatial mobility (Kutzbach–Grübler) equation is needed: $$M = 6(L-J-1)+\sum f_i$$ with $L=8$ links (7 moving + ground), $J=8$ joints (6 anatomical + 2 fixed hand–ground contacts, each $f=0$), and $\sum f_i = 12$ (the same anatomical total as part b): $$M = 6(8-8-1)+12 = -6+12 = \boxed{6\ \text{DOF}}$$ Physically: the six joint angles of one arm (say, the right, grounded chain) freely and uniquely set the trunk's full 6-DOF pose in space; the other six joint angles (left arm) are then over-determined by the requirement that the left hand land exactly on its fixed ground point, so they contribute no further independent freedom.

d) Classification. Both cases are a single sequential path with no branch point within the lettered chain, so both are serial. Case (b), with one free end, is open; case (c), with both ends fixed to the same ground reference, is closed.