Question 3 of 7: Cam–follower motion design

Question 3: Cam–follower motion design (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. Cam speed $\omega=1500\ \text{rpm}=157.1$ rad/s (CW); centre distance $24$ in; lift $h=15^\circ=0.2618$ rad; rise interval $\beta_R=75^\circ$, fall interval $\beta_F=105^\circ$, bottom dwell $180^\circ$.

Find. a base-circle choice, the fall motion law, and the S-V-A-J (position–velocity–acceleration–jerk) diagrams.

Approach. Enforce the fundamental law (continuous position, velocity and acceleration across every dwell/segment boundary), select the double-dwell law with the smallest peak-velocity coefficient for the fall, and size the base circle from the dwell geometry with a pressure-angle limit.

1. Fundamental law and boundary conditions. The follower must have continuous $\phi,\dot\phi,\ddot\phi$ everywhere (finite jerk). The bottom dwell forces $\dot\phi=\ddot\phi=0$ at $\theta=0$ (start of rise) and at $\theta=180^\circ$ (end of fall); the rise–fall cusp at $\theta=75^\circ$ is a velocity zero shared by both segments.
2. Choose the fall law — minimize the maximum velocity. Among the smooth double-dwell laws, the peak-velocity coefficients are: parabolic $2.0$ (but infinite jerk — violates the law), cycloidal $2.0$, 3-4-5 polynomial $1.875$, and modified sine $C_v=1.760$ — the lowest. Hence the fall uses the modified-sine law: $$\dot\phi_{\mathrm{max}}=C_v\,\frac{h}{\beta_F}\,\omega=1.760\,\frac{0.2618}{1.833}(157.1)=\boxed{39.5\ \text{rad/s}}.$$ Its peak acceleration ($C_a=5.528$) is $\ddot\phi_{\mathrm{max}}=C_a\,h\,\omega^2/\beta_F^2\approx1.06\times10^4\ \text{rad/s}^2$, and its jerk stays finite, so the fundamental law holds.
3. Rise segment. The rise (no special objective) uses a smooth double-dwell law — here cycloidal — giving $\dot\phi_{\mathrm{max}}=2.0\,(h/\beta_R)\,\omega=55.3$ rad/s (larger than the fall because the rise interval $75^\circ$ is shorter than the fall $105^\circ$).
4. (i) Base circle. During the dwell ($\theta=180^\circ\!\to\!360^\circ$) the follower rests at $\phi=30^\circ$ and the roller rides a circular arc — the base circle. Size its radius $R_b$ so the maximum pressure angle over the rise/fall stays within the good-practice limit $\le 30^\circ$ for an oscillating follower: with $\tan\alpha=(d\phi/d\theta)/(\text{effective radius})$, the peak $d\phi/d\theta=C_v h/\beta$ occurs on the fall. Taking the follower arm $\approx$ half the $24$-in centre distance and requiring $\alpha_{\mathrm{max}}\le30^\circ$ gives a base-circle radius on the order of $R_b\approx 8$–$10$ in (measured to the roller-centre/prime circle); a larger $R_b$ lowers the pressure angle further at the cost of cam size. The dwell arc is then the $\phi=30^\circ$ reference from which rise and fall are laid out.
5. (ii) Diagrams. The four stacked curves above show $\phi$ (30→45→30°, then flat), $\dot\phi$ (zero at $0,75,180^\circ$), $\ddot\phi$ (zero at the dwell boundaries, continuous through $75^\circ$) and $\dot{\ddot{\phi}}$ (finite everywhere) — the signature of a fundamental-law-compliant program with a modified-sine fall.