Radial cam with flat-faced follower — minimum-jerk motion design: 22-Mec-A2 Kinematics and Dynamics of Machines, May 2016, Q2

Question 2: Radial cam with flat-faced follower — minimum-jerk 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. Flat-faced follower; $\omega=100\ \text{rad/s}$ (constant); total lift $h=1.5\ \text{in}$; rise interval $\beta_r=75^\circ=1.3090\ \text{rad}$, dwell $[75^\circ,180^\circ]$, fall interval $\beta_f=180^\circ=\pi\ \text{rad}$.

Find. Motion program minimizing peak jerk; the $s,v,a,j$ equations and $j_{\mathrm{max}}$ for rise and fall; base-circle radius; pressure angles at the two stated cam angles.

Approach. Among the standard double-dwell programs that keep finite jerk (a fundamental requirement for a quiet cam), the cycloidal displacement has the lowest peak jerk coefficient ($4\pi^2\approx39.5$, versus 60 for 3-4-5 polynomial). It gives continuous acceleration (zero at both ends) — the classic “quiet” motion — so it is selected for both rise and fall.

Rise equations (let $x=\theta/\beta_r$, $0\le x\le1$):

$$s=h\!\left[x-\tfrac{1}{2\pi}\sin 2\pi x\right],\quad v=\tfrac{h\omega}{\beta_r}\!\left[1-\cos 2\pi x\right],$$ $$a=\tfrac{2\pi h\omega^{2}}{\beta_r^{2}}\sin 2\pi x,\quad j=\tfrac{4\pi^{2} h\omega^{3}}{\beta_r^{3}}\cos 2\pi x.$$

Fall equations (let $x=(\theta-180^\circ)/\beta_f$): identical form with $h\to$ falling, i.e. $s=h\!\left[1-x+\tfrac{1}{2\pi}\sin 2\pi x\right]$ and $v,a,j$ carrying the same magnitudes with $\beta_f$ in place of $\beta_r$ (opposite sign on $v$).

Figure 2.1 — Cycloidal rise profiles (normalized). Displacement $s$ is S-shaped; velocity $v$ starts and ends at zero; acceleration $a$ is a full sine (zero at both ends ⇒ no acceleration step); jerk $j$ is a cosine (finite, single-cycle) — the signature of quiet running.

Peak jerks. $j_{\mathrm{max}}=\dfrac{4\pi^{2}h\omega^{3}}{\beta^{3}}$. For the rise, $$j_{\mathrm{max,rise}}=\frac{4\pi^{2}(1.5)(100)^{3}}{(1.3090)^{3}}\approx \boxed{2.64\times10^{7}\ \text{in/s}^{3}}$$ and for the (gentler, longer) fall $$j_{\mathrm{max,fall}}=\frac{4\pi^{2}(1.5)(100)^{3}}{\pi^{3}}\approx 1.91\times10^{6}\ \text{in/s}^{3}.$$ Peak accelerations are $a_{\mathrm{max,rise}}=2\pi h\omega^2/\beta_r^2=5.50\times10^{4}\ \text{in/s}^2$ and $a_{\mathrm{max,fall}}=9.55\times10^{3}\ \text{in/s}^2$.
Base circle (flat-faced follower → cusp/radius-of-curvature limit, not pressure angle). The cam surface radius of curvature is $\rho=R_0+s+s''$, where $s''=\mathrm{d}^2s/\mathrm{d}\theta^2=(2\pi h/\beta^2)\sin2\pi x$. The most negative value of $s+s''$ over the cycle occurs on the fast rise: $\min(s+s'')=-4.14\ \text{in}$. Requiring $\rho>0$ everywhere gives $R_0>4.14\ \text{in}$; choose $$\boxed{R_0=5.0\ \text{in}}\quad(\rho_{\mathrm{min}}\approx0.86\ \text{in}>0,\ \text{no cusp}).$$ The flat face must overhang the contact by at least $\max|v/\omega|=2h/\beta_r=2.29\ \text{in}$ each side.
Pressure angles at $\theta=37.5^\circ$ and $127.5^\circ$. For a flat-faced follower the contact force is always normal to the face, i.e. parallel to the follower motion, so the pressure angle is identically zero at every cam angle: $$\phi(37.5^\circ)=\phi(127.5^\circ)=0^\circ.$$ Both are far below the 30° limit, so no pressure-angle modification is required. (Note $127.5^\circ$ falls in the dwell, where the follower is stationary and $\phi=0$ trivially.)

Quantity	Value
Selected program	Cycloidal (lowest peak jerk, quiet)
$j_{\mathrm{max}}$ rise / fall	2.64×10⁷ / 1.91×10⁶ in/s³
$a_{\mathrm{max}}$ rise / fall	5.50×10⁴ / 9.55×10³ in/s²
Base-circle radius $R_0$	5.0 in ($\rho_{\mathrm{min}}\approx0.86$ in)
Pressure angle @ 37.5° & 127.5°	0° (flat face) — satisfactory

Topic: Double-dwell cam-motion synthesis; cycloidal program; flat-faced-follower geometry.

Key relations: cycloidal $s=h[x-\sin2\pi x/2\pi]$; $j_{\mathrm{max}}=4\pi^2h\omega^3/\beta^3$; flat-face radius of curvature $\rho=R_0+s+s''$; flat-face pressure angle $\phi\equiv0$.

Why this works: “Quietness” is governed by jerk (rate of change of the inertia force). Any program with a jump in acceleration has infinite jerk; the cycloidal motion has $a=0$ at both boundaries and a smooth sinusoidal $a$, giving the smallest finite peak jerk of the common curves. For a flat-faced follower the normal always points along the follower axis, so the pressure angle is zero — the real limits are the radius of curvature (avoid a pointed/undercut cam, set by $R_0$) and the face width.

Common pitfall: Trying to “reduce” the flat-face pressure angle (it is already 0), or using the roller-follower pressure-angle formula. Also, forgetting that a shorter rise angle ($75^\circ$ here) drives the jerk up as $1/\beta^3$, making the rise — not the longer fall — the design-critical event.

Source: Norton, Design of Machinery §8 (cam design: SVAJ diagrams, cycloidal motion, flat-faced-follower geometry).

Question 2 of 6: Radial cam with flat-faced follower — minimum-jerk motion design

Question 2: Radial cam with flat-faced follower — minimum-jerk motion design (20 marks)