Radial cam with flat-faced follower — minimum-acceleration program: 22-Mec-A2 Kinematics and Dynamics of Machines, May 2017, Q2

Question 2: Radial cam with flat-faced follower — minimum-acceleration program (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 translating follower; $\omega=1500$ rpm $=157.08\ \text{rad/s}$ (constant); total lift $h=2\ \text{in}$; rise angle $\beta_r=90^\circ=\pi/2$, top dwell $180^\circ$, fall angle $\beta_f=90^\circ=\pi/2$. Objective: minimise peak follower acceleration (bounded jerk).

Find. the motion program and its $s,v,a,j$ equations; $j_{\mathrm{max}}$ for rise and fall; a base circle; the pressure angles at $45^\circ$ and $315^\circ$.

Choice of program

Among the standard double-dwell (rise–dwell–fall) programs with finite jerk, the modified-trapezoidal acceleration curve has the lowest peak acceleration ($C_a=4.888$). Simple harmonic ($C_a=4.93$) and cycloidal ($C_a=2\pi=6.28$) are higher; pure trapezoidal has a slightly lower $C_a=4.0$ but infinite jerk (it violates the fundamental cam law and the problem explicitly asks for a finite $j_{\mathrm{max}}$). Hence the minimum-acceleration answer is the modified trapezoid: a sine-ramped trapezoid built from eight segments — sinusoidal jerk in the corner ramps, constant acceleration on the plateaus, and a smooth zero-crossing at mid-event. Its dimensionless factors (rise; identical form for the fall) are

$$C_a=4.888,\qquad C_v=2.000,\qquad C_j=61.43 = 4\pi C_a .$$

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

Writing the normalized displacement, velocity, acceleration and jerk as $s=h\,S(x)$, $v=\dfrac{h\omega}{\beta_r}V(x)$, $a=\dfrac{h\omega^2}{\beta_r^2}A(x)$, $j=\dfrac{h\omega^3}{\beta_r^3}J(x)$, the modified-trapezoid shape functions are (first quarter $0\le x\le\tfrac18$ shown; the remaining seven segments follow by the sine-ramp / constant / mirror construction):

$$A(x)=C_a\sin\!\Big(4\pi x\Big),\quad V(x)=\frac{C_a}{4\pi}\big[1-\cos(4\pi x)\big],\quad S(x)=\frac{C_a}{4\pi}\Big[x-\frac{\sin(4\pi x)}{4\pi}\Big],\quad J(x)=4\pi C_a\cos(4\pi x).$$

with $A$ held at $+C_a$ over $\tfrac18\le x\le\tfrac38$, sine-ramped back to $0$ at $x=\tfrac12$, then the negative mirror over $\tfrac12\le x\le1$. The peaks are $A_{\mathrm{max}}=C_a$, $V_{\mathrm{max}}=C_v$, $|J|_{\mathrm{max}}=C_j$ (at $x=0,\,\tfrac12,\,1$). The rise $s,v,a,j$ diagrams:

Rise motion program (modified-trapezoidal acceleration). $s$ rises smoothly $0\to h$; $v$ is a single hump; $a$ is the sine-cornered trapezoid (equal positive/negative plateaus, zero at ends and mid); $j$ is finite everywhere, peaking at the start, middle and end of the rise.

Peak acceleration and jerk

With $h=2\ \text{in}$, $\omega=157.08\ \text{rad/s}$, $\beta_r=\pi/2=1.5708\ \text{rad}$:

$$a_{\mathrm{max}}=C_a\frac{h\omega^2}{\beta_r^2}=4.888\cdot\frac{2(157.08)^2}{(1.5708)^2}=9.78\times10^{4}\ \text{in/s}^2\;(\approx 253\,g),$$

$$j_{\mathrm{max}}=C_j\frac{h\omega^3}{\beta_r^3}=61.43\cdot\frac{2(157.08)^3}{(1.5708)^3}=\boxed{1.23\times10^{8}\ \text{in/s}^3}.$$

The fall uses the same modified-trapezoid over $\beta_f=90^\circ=\pi/2$ (mirror in $s$), so because $\beta_f=\beta_r$ its peak jerk is identical:

$$j_{\mathrm{max,fall}}=C_j\frac{h\omega^3}{\beta_f^3}=1.23\times10^{8}\ \text{in/s}^3.$$

(Had the fall used a different angle, $j_{\mathrm{max}}$ would scale as $1/\beta^3$.)

Base circle and cam profile

For a flat-faced follower the cam contour must stay convex: the radius of curvature $\rho=R_b+s+s''(\theta)$ must be positive everywhere, where $s''=\mathrm{d}^2s/\mathrm{d}\theta^2=a/\omega^2$. Evaluating $s+s''$ over the rise and fall, its minimum is $-2.51\ \text{in}$, so

$$R_b > -\min(s+s'') = 2.51\ \text{in}\;\Rightarrow\; \text{choose } \boxed{R_b=3.0\ \text{in}}\;(\rho_{\min}\approx0.5\ \text{in}>0).$$

The follower face must also be wide enough to keep contact: the contact point slides $\ell=\mathrm{d}s/\mathrm{d}\theta=v/\omega$ off-centre, whose maximum is $V_{\max}h/\beta_r = 2.55\ \text{in}$, so a face of at least $\pm2.6\ \text{in}$ about the stem is required.

Cam profile (schematic) for $R_b=3.0$ in: a $90^\circ$ rise, $180^\circ$ top dwell and $90^\circ$ fall, with the flat-faced follower resting on the contour. The contour is convex everywhere ($\rho>0$).

Pressure angles at $\theta=45^\circ$ and $315^\circ$

This is the design point of a flat-faced follower: the contact normal is always perpendicular to the (flat) follower face, i.e. it always lies along the direction of follower travel. The pressure angle is therefore zero at every cam angle:

$$\varphi(45^\circ)=\varphi(315^\circ)=\boxed{0^\circ}.$$

Both are far below the $30^\circ$ limit, so no modification is required. (A flat-faced follower is limited by radius of curvature / undercutting, handled through the base-circle choice above, not by pressure angle. Were this a roller follower with $\varphi>30^\circ$, the standard fix is to increase the base circle — or reduce the lift or lengthen the event angle — which lowers $\mathrm{d}s/\mathrm{d}\theta$ relative to $R_b+s$; the question asks us to state, not iterate, this remedy.)

Quantity	Value
Program (minimises peak acceleration, finite jerk)	Modified trapezoid ($C_a{=}4.888,\,C_v{=}2.0,\,C_j{=}61.43$)
Rise peak acceleration	$9.78\times10^{4}\ \text{in/s}^2$ ($\approx253\,g$)
Rise peak jerk $j_{\mathrm{max}}$ ($\beta_r=90^\circ$)	$1.23\times10^{8}\ \text{in/s}^3$
Fall peak jerk $j_{\mathrm{max}}$ ($\beta_f=90^\circ$)	$1.23\times10^{8}\ \text{in/s}^3$ (same)
Base circle	$R_b=3.0\ \text{in}$ ($\rho_{\min}>0$)
Pressure angle at $45^\circ$ and $315^\circ$	$0^\circ$ (flat-faced) — no modification needed

Topic: Double-dwell cam synthesis; the modified-trapezoidal acceleration program; flat-faced-follower geometry.

Key relations: $a_{\mathrm{max}}=C_a h\omega^2/\beta^2$, $j_{\mathrm{max}}=C_j h\omega^3/\beta^3$; radius of curvature (flat-faced) $\rho=R_b+s+s''$; pressure angle of a flat-faced translating follower $\varphi\equiv0$.

Why this works: Minimising peak acceleration minimises peak inertia force ($F=ma$) and contact stress, but jerk must stay finite to avoid vibration and noise (the fundamental law of cam design). The modified trapezoid is the trapezoid’s low $C_a$ softened by sine ramps that bound the jerk — the best compromise, so it is the standard “minimum-acceleration” answer. A flat face makes the transmission angle ideal ($\varphi=0$) but shifts the design limit to curvature: too small a base circle would make $\rho<0$ and the cam would undercut.

Common pitfall: choosing pure trapezoidal for “lowest acceleration” and ignoring its infinite jerk; or trying to reduce a flat-faced follower’s pressure angle (it is already zero) instead of checking the radius of curvature. Also forgetting the follower face must span the full $\mathrm{d}s/\mathrm{d}\theta$ offset.