Question 2 of 6: Radial cam — minimum-peak-acceleration design, flat-faced follower

Question 2: Radial cam — minimum-peak-acceleration design, flat-faced follower (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 $h = 12.7\ \text{mm}$; cam speed $\omega = 900\ \text{rpm} = 94.25\ \text{rad/s}$; rise interval $\beta_1 = 150^\circ = 2.618\ \text{rad}$, fall interval $\beta_2 = 180^\circ = 3.142\ \text{rad}$, with a 30° dwell between. Follower: flat-faced, radial (in-line).

Find. the motion program that minimizes peak acceleration; its $s,v,a,j$ for rise and fall; $a_{\mathrm{max}}$ and $j_{\mathrm{max}}$ on the rise; a base circle; and the pressure angles at 75° and 270°.

Approach. For a rise–dwell–fall (double-dwell) cam the acceleration must start and end at zero at every dwell edge; among such programs the modified-trapezoidal acceleration curve gives the lowest peak acceleration, so it is the correct choice for the stated objective. Build its $s,v,a,j$, scale to the lift, extract the peaks, size the base circle from the radius-of-curvature limit, and evaluate the flat-faced pressure angle.

Using the normalized cam angle $\xi = \theta/\beta \in [0,1]$, the modified-trapezoidal acceleration is a symmetric block with sinusoidal ramps (an eighth of the interval each) so that both $a$ and $j$ stay finite and $a=0$ at the ends. Writing the follower as $s = h\,S(\xi)$, its derivatives with respect to real time follow from $v = h\,\omega\,S'/\beta$, $a = h\,\omega^{2}S''/\beta^{2}$, $j = h\,\omega^{3}S'''/\beta^{3}$, where primes are $d/d\xi$. Integrating the standard shape gives the dimensionless peak factors

$C_a = 4.888,\qquad C_v = 2.000,\qquad C_j = 61.43,$

so that $a_{\mathrm{max}} = C_a\,h\,\omega^2/\beta^2$, $v_{\mathrm{max}} = C_v\,h\,\omega/\beta$, $j_{\mathrm{max}} = C_j\,h\,\omega^3/\beta^3$. (For comparison the constant-acceleration “parabolic” curve reaches only $C_a=4.0$ but with infinite jerk, and the popular cycloidal curve has $C_a=6.28$; the modified trapezoid is the lowest-peak-acceleration program with bounded jerk.)

1. Rise equations (0 ≤ θ ≤ 150°). With $\xi=\theta/\beta_1$, the follower is $s=h\,S(\xi)$ where $S(\xi)$ is the double integral of the normalized modified-trapezoid acceleration; $v,a,j$ scale as above. The acceleration block is $$a(\xi)=C_a\frac{h\omega^2}{\beta_1^2}\,b(\xi),\quad b(\xi)=\begin{cases}\sin(4\pi\xi)&0\le\xi<\tfrac18\\[2pt]1&\tfrac18\le\xi<\tfrac38\\[2pt]\sin\!\big(4\pi(\xi-\tfrac14)\big)&\tfrac38\le\xi<\tfrac12\\[2pt]\text{(mirror, negative)}&\tfrac12\le\xi\le1\end{cases}$$ with $v$ and $s$ obtained by successive integration subject to $s(0)=0,\ s(1)=h,\ v(0)=v(1)=0$.
2. Peak acceleration and jerk of the rise. $$a_{\mathrm{max}} = C_a\frac{h\,\omega^2}{\beta_1^2} = 4.888\times\frac{(0.0127)(94.25)^2}{(2.618)^2} = \boxed{80.5\ \text{m/s}^2}\ (\approx 8.2\,g)$$ $$j_{\mathrm{max}} = C_j\frac{h\,\omega^3}{\beta_1^3} = 61.43\times\frac{(0.0127)(94.25)^3}{(2.618)^3} = \boxed{3.64\times10^{4}\ \text{m/s}^3}$$ (peak follower velocity $v_{\mathrm{max}} = C_v h\omega/\beta_1 = 0.914\ \text{m/s}$).
3. Fall equations (180° ≤ θ ≤ 360°). Identical functional form with $\xi=(\theta-180^\circ)/\beta_2$ and the displacement inverted, $s = h\,[1-S(\xi)]$; the sign of $a$ reverses. Because $\beta_2=180^\circ>\beta_1$, the fall is gentler: $$a_{\mathrm{max,fall}} = C_a\frac{h\,\omega^2}{\beta_2^2} = 55.9\ \text{m/s}^2.$$ The rise therefore governs the peak acceleration of the cam.
4. Base circle (flat-faced follower). The follower contact radius is $\rho = R_0 + s + s''$ (with $s''=d^2s/d\theta^2$ in length units); a cusp is avoided when $\rho>0$ everywhere. Over the full cycle $\min(s+s'') \approx 0$, so any positive base radius satisfies the curvature limit; for a practical, robust cam choose $$\boxed{R_0 = 25\ \text{mm}}$$ (giving $\rho_{\min}\approx 25$ mm). The flat face must be wide enough for the contact offset $s'$: required half-width $= \max|s'| = C_v h/\beta_1 = 9.7\ \text{mm}$, so a face of $\gtrsim 22\ \text{mm}$ (with margin).
5. Pressure angles at 75° and 270°. For a flat-faced follower the contact normal is always perpendicular to the flat face — i.e. always parallel to the follower axis — so the pressure angle is identically zero at every cam angle: $$\phi(75^\circ)=0^\circ,\qquad \phi(270^\circ)=0^\circ.$$ Because the pressure angle can never exceed its limit, no pressure-angle iteration on the base circle is required; the base-circle size for a flat-faced follower is set instead by the radius-of-curvature (cusp) condition of step 4, not by pressure angle.