Question 3 of 7: Radial cam with flat-faced follower

Question 3: Radial cam with 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. Flat-faced translating follower; $\omega=1500$ rpm $=157.08$ rad/s (const); rise $h=2$ in over $\beta=90^\circ=\pi/2$; dwell; fall $h=2$ in over $\beta=90^\circ$. Objective: minimum peak acceleration with finite jerk.

Find. Motion program & its $s,v,a,j$ equations; $a_{\mathrm{max}}$, $v_{\mathrm{max}}$, $j_{\mathrm{max}}$ (rise and fall); base circle; pressure angles at $45^\circ$ and $315^\circ$.

Approach. Among finite-jerk double-dwell programs the modified-trapezoidal acceleration curve has the smallest peak acceleration, so it is the correct choice for “minimize maximum acceleration.” Use its standard coefficients, then note that a flat-faced follower always has zero pressure angle, so the design constraint is convexity (base-circle), not pressure angle.

1. Program choice. The modified-trapezoid builds the acceleration from three sine-ramp blends (rise-hold-fall of $a$) so that jerk stays finite while the acceleration plateau is as low as possible. Its peak factors are $$C_a=4.888,\quad C_v=2.000,\quad C_j=61.43$$ with $a_{\mathrm{max}}=C_a\dfrac{h\omega^2}{\beta^2},\ v_{\mathrm{max}}=C_v\dfrac{h\omega}{\beta},\ j_{\mathrm{max}}=C_j\dfrac{h\omega^3}{\beta^3}$. In normalized form $x=\theta/\beta\in[0,1]$ the acceleration is $a(x)=A\,p(x)$ with $p$ = $\sin(4\pi x)$ on $[0,\tfrac18]$, $1$ on $[\tfrac18,\tfrac38]$, $\cos\!\big(4\pi(x-\tfrac38)\big)$ on $[\tfrac38,\tfrac58]$, $-1$ on $[\tfrac58,\tfrac78]$, $-\cos\!\big(4\pi(x-\tfrac78)\big)$ on $[\tfrac78,1]$; $v$ and $s$ are its successive integrals scaled so $s(1)=h$. The fall uses the mirror ($s_{\text{fall}}=h-s_{\text{rise}}$).
2. Peak acceleration (rise = fall, same $\beta$). $$a_{\mathrm{max}}=4.888\frac{(2)(157.08)^2}{(1.5708)^2}=9.78\times10^{4}\ \text{in/s}^2=\boxed{2483\ \text{m/s}^2}\ (\approx 253g)$$
3. Peak velocity. $v_{\mathrm{max}}=2.000\dfrac{(2)(157.08)}{1.5708}=400\ \text{in/s}=\boxed{10.16\ \text{m/s}}$ (sets the flat-follower face width, $\pm v_{\mathrm{max}}/\omega=\pm 2.55$ in).
4. Peak jerk. Because $\beta_{\text{rise}}=\beta_{\text{fall}}=90^\circ$, the rise and fall jerks are equal in magnitude: $$j_{\mathrm{max}}=61.43\frac{(2)(157.08)^3}{(1.5708)^3}=1.23\times10^{8}\ \text{in/s}^3=\boxed{3.12\times10^{6}\ \text{m/s}^3}$$
5. Base circle (convexity). For a flat follower the radius of curvature is $\rho=R_b+s+s''$ (with $s''=d^2s/d\theta^2$); a cusp/undercut occurs if $\rho\le0$. The most negative $s+s''$ over the cycle is $-2.51$ in, so choose $$R_b\ge 2.51\ \text{in}\ \Rightarrow\ \boxed{R_b=3.0\ \text{in}}\ (\rho_{\min}\approx0.5\ \text{in}\gt 0)$$ The resulting contour is sketched below.
6. Pressure angles. For a flat-faced follower the contact normal is always parallel to the follower axis, so the pressure angle is identically zero at every cam angle: $$\phi(45^\circ)=\phi(315^\circ)=\boxed{0^\circ}\ (\lt 30^\circ,\ \text{satisfied automatically})$$ Hence no pressure-angle modification is needed. (If contact stress or a cusp were the concern, the remedy would be a larger base circle, as sized above — not a follower offset.)