Radial cam design — minimum-velocity fall and pressure angle: 22-Mec-A2 Kinematics and Dynamics of Machines, May 2018, Q3

Question 3: Radial cam design — minimum-velocity fall and pressure angle (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. Total lift $h=40\ \text{mm}=0.040\ \text{m}$; fall interval $\beta_f=200^\circ=3.491\ \text{rad}$; rise interval $\beta_r=120^\circ=2.094\ \text{rad}$; cam speed $N=1800\ \text{rpm}\Rightarrow \omega=1800\cdot 2\pi/60=188.5\ \text{rad/s}$.

Find. (i) the minimum-peak-velocity fall program with its $s$-$v$-$a$ peaks; (ii) a base circle and the maximum pressure angles.

Approach. The fundamental law requires the $s$-curve to be continuous in $s,v,a$ (finite jerk) across the dwell boundaries. Among the smooth double-dwell programs the one with the lowest peak velocity coefficient is the modified-sine curve, so it is chosen for the fall. Then size the prime circle so the peak pressure angle stays under $30^\circ$.

Select the fall program. Objective = minimise $v_{\max}$. The smooth double-dwell curves have velocity coefficients $C_v$: parabolic $2.00$ (but infinite jerk — violates the law), simple harmonic $1.571$ (but finite acceleration step at the ends — violates the law), cycloidal $2.000$, and modified sine $C_v=1.760$, which is the smallest $C_v$ among curves that keep acceleration continuous. Hence the fall uses the modified-sine displacement.
Peak velocity of the fall. $$v_{\max}=C_v\,\frac{h\,\omega}{\beta_f}=1.760\cdot\frac{0.040\,(188.5)}{3.491}=\boxed{3.80\ \text{m/s}}.$$
Peak acceleration of the fall. Modified sine has $C_a=5.528$: $$a_{\max}=C_a\,\frac{h\,\omega^2}{\beta_f^{\,2}}=5.528\cdot\frac{0.040\,(188.5)^2}{3.491^2}=\boxed{6.45\times10^{2}\ \text{m/s}^2}.$$ The $s$-$v$-$a$ curves for the fall are sketched below (rest-to-rest, smooth, finite peaks).

Modified-sine fall ($180^\circ\!\to\!360^\circ$): displacement $s$ ($40\!\to\!0$ mm), velocity $v$ (peak $3.80$ m/s), acceleration $a$ (peak $645$ m/s$^2$). All continuous — the fundamental law is satisfied.

Choose a base (prime) circle and check the pressure angle. For a translating roller follower the pressure angle is $\phi=\tan^{-1}\!\big[(ds/d\theta)/(R_p+s)\big]$, with $R_p$ the prime-circle radius. Take $R_p=50\ \text{mm}$. The steeper interval is the short $120^\circ$ rise; using a cycloidal rise ($\,ds/d\theta$ peaks at $2h/\beta_r=38.2\ \text{mm/rad}$ near mid-rise, $s=20$ mm): $$\phi_{\text{rise,max}}=\tan^{-1}\!\frac{38.2}{50+20}\approx\boxed{29.5^\circ}\;(<30^\circ,\ \text{OK}).$$ For the longer $200^\circ$ fall ($ds/d\theta$ peak $=C_v h/\beta_f=20.2\ \text{mm/rad}$): $$\phi_{\text{fall,max}}=\tan^{-1}\!\frac{20.2}{50+s^\ast}\approx\boxed{16.7^\circ}\;(\ll30^\circ,\ \text{OK}).$$

(ii) verdict. With $R_p=50\ \text{mm}$ both peak pressure angles are below the $30^\circ$ limit, so the base circle is acceptable. If a case exceeded $30^\circ$, the standard fixes (stated, not carried out) are: increase the base/prime circle radius (largest single lever on $\phi$), and/or add a suitable follower offset in the direction that reduces the rise-side angle, and/or lengthen the steep interval $\beta$. Increasing $R_p$ is the first move.

Quantity	Value
Fall program (min $v_{\max}$)	Modified sine ($C_v=1.760,\ C_a=5.528$)
Fall peak velocity $v_{\max}$	$3.80\ \text{m/s}$
Fall peak acceleration $a_{\max}$	$645\ \text{m/s}^2$
Prime circle chosen $R_p$	$50\ \text{mm}$
Max pressure angle (rise / fall)	$29.5^\circ$ / $16.7^\circ$ — both $<30^\circ$

Topic: Double-dwell cam program selection under an optimisation objective, and pressure-angle sizing of a radial roller cam.

Key relations: $v_{\max}=C_v h\omega/\beta$, $a_{\max}=C_a h\omega^2/\beta^2$; modified sine $C_v=1.760,\ C_a=5.528$; pressure angle $\phi=\tan^{-1}[(ds/d\theta)/(R_p+s)]$.

Why this works: the fundamental law bars any discontinuity in $s,v,a$ (an $a$-jump means infinite jerk and impact). Among law-abiding curves the objective picks the coefficient to minimise: min-velocity → modified sine; min-acceleration → modified trapezoid; min-jerk → cycloidal. Pressure angle falls as the prime circle grows because the denominator $R_p+s$ increases while the slope $ds/d\theta$ is fixed by the motion program.

Common pitfall: choosing parabolic for “lowest velocity” — its jerk is infinite, so it violates the fundamental law even though $C_v$ looks attractive. Also checking $\phi$ only at maximum slope: the true max can shift because $s$ (hence $R_p+s$) varies.

Question 3 of 7: Radial cam design — minimum-velocity fall and pressure angle

Question 3: Radial cam design — minimum-velocity fall and pressure angle (20 marks)