Question 2 of 10

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(b) Eight-bar mechanism — transmission angles and Grashof classification

Given. Two four-bar loops share a common frame and are linked by a floating binary connecting rod (link 5): Loop I = {ground, link 2 (crank, driven by $\omega_2$), the ternary coupler link 4, link 3 (rocker, grounded)}; Loop II = {ground, link 7 (rocker, grounded), the ternary coupler link 6, link 8 (output rocker, grounded)}. Pin coordinates were digitised from the 1:2 scale drawing.

Find. The transmission angle at every coupler–rocker joint that carries motion from link 2 to link 8, and the Grashof type of each of the two four-bar loops.

Approach. $\omega_2$ enters Loop I at link 2, is carried by the rigid ternary link 4 to link 5, crosses to link 6 and drives Loop II's output rocker link 8. Motion transfer is efficient (low bearing/friction loss, no danger of locking) only where the transmission angle μ — the angle between the coupler and the driven rocker at their shared pin — stays clear of 0°/180°; the two joints that matter are therefore the coupler–rocker pin in each loop.

1. Digitise the four pivots of each loop: Loop I: $G_A=(216,318)$ (link 2's ground), $J_A=(188,275)$ (link2–link4), $J_B=(233,258)$ (link4–link3), $G_B=(271,310)$ (link 3's ground). Loop II: $G_C=(347,293)$ (link7's ground), $J_C=(403,260)$ (link7–link6), $J_D=(433,297)$ (link6–link8), $G_D=(420,340)$ (link 8's ground).
2. Loop I link lengths and Grashof sum: $$L_{\text{ground}}=|G_AG_B|=55.6,\quad L_2=|G_AJ_A|=51.3,\quad L_4=|J_AJ_B|=48.1,\quad L_3=|J_BG_B|=64.4$$ Sorted: $S=48.1$ (shortest, link 4), $L=64.4$ (longest, link 3), $P+Q=51.3+55.6=106.9$. $$S+L = 48.1+64.4 = 112.5 \;>\; P+Q = 106.9$$
3. Loop II link lengths and Grashof sum: $$L_{\text{ground}}=|G_CG_D|=86.8,\quad L_7=|G_CJ_C|=65.0,\quad L_6=|J_CJ_D|=47.6,\quad L_8=|J_DG_D|=44.9$$ Sorted: $S=44.9$ (link 8), $L=86.8$ (ground), $P+Q=65.0+47.6=112.6$. $$S+L = 44.9+86.8 = 131.7 \;>\; P+Q = 112.6$$
4. Apply the Grashof criterion $S+L$ vs $P+Q$ to both loops: in both cases $S+L>P+Q$, so $$\boxed{\text{Loop I and Loop II are both non-Grashof (Class III, "triple-rocker")}}$$ No link in either loop can complete a full revolution; link 2's $\omega_2$ arrow is valid only as the instantaneous angular velocity at the drawn position, not as a sustained full rotation.
5. Transmission angle, Loop I (at $J_B$, between the coupler link 4 direction $J_A\!\to\!J_B$ and the rocker link 3 direction $J_B\!\to\!G_B$): $$\vec u = J_B-J_A = (45,-17), \qquad \vec v = G_B-J_B=(38,52)$$ $$\mu_1=\cos^{-1}\!\left(\frac{\vec u\cdot\vec v}{|\vec u||\vec v|}\right) = \boxed{74.5^{\circ}}$$
6. Transmission angle, Loop II (at $J_D$, between coupler link 6 direction $J_C\!\to\!J_D$ and output rocker link 8 direction $J_D\!\to\!G_D$): $$\vec u = J_D-J_C=(30,37), \qquad \vec v = G_D-J_D=(-13,43)$$ $$\mu_2=\cos^{-1}\!\left(\frac{\vec u\cdot\vec v}{|\vec u||\vec v|}\right) = \boxed{55.9^{\circ}}$$ Both angles sit comfortably inside the recommended $45^\circ\text{–}135^\circ$ window, so at the drawn configuration motion transfers efficiently through both loops even though neither is a true crank-rocker.