Question 2 of 6: Cycloidal radial cam — motion program, profile & pressure angle

Question 2: Cycloidal radial cam — motion program, profile & 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. Constant cam speed $N=1800\ \text{rpm}$, so $\omega=1800\cdot2\pi/60=188.5\ \text{rad/s}$. Lift $h=25.4\ \text{mm}=0.0254\ \text{m}$. Rise interval $\beta_R=90^\circ=\pi/2$; dwell 90°–180°; fall interval $\beta_F=180^\circ=\pi$. Flat-faced follower.

Find. $s,v,a,j$ for rise and fall; $a_{\mathrm{max}}$ and $j_{\mathrm{max}}$ of the rise; base-circle radius and cam profile; pressure angles at 45° and 270°.

Approach. Write the cycloidal motion law on each segment, differentiate with respect to time via $\theta=\omega t$, evaluate the rise peaks, then size the base circle from the flat-faced curvature condition and generate the profile parametrically.

1. Cycloidal law — rise ($0\le\theta\le\beta_R$). With $\phi=\theta$, $$s=h\!\left[\frac{\phi}{\beta_R}-\frac{1}{2\pi}\sin\frac{2\pi\phi}{\beta_R}\right],\quad v=\frac{h\omega}{\beta_R}\!\left[1-\cos\frac{2\pi\phi}{\beta_R}\right],$$ $$a=\frac{2\pi h\omega^2}{\beta_R^{2}}\sin\frac{2\pi\phi}{\beta_R},\quad j=\frac{4\pi^{2}h\omega^{3}}{\beta_R^{3}}\cos\frac{2\pi\phi}{\beta_R}.$$
2. Cycloidal law — fall ($180^\circ\le\theta\le360^\circ$, $\psi=\theta-180^\circ$, interval $\beta_F$). $$s=h\!\left[1-\frac{\psi}{\beta_F}+\frac{1}{2\pi}\sin\frac{2\pi\psi}{\beta_F}\right],\quad v=-\frac{h\omega}{\beta_F}\!\left[1-\cos\frac{2\pi\psi}{\beta_F}\right],$$ $$a=-\frac{2\pi h\omega^{2}}{\beta_F^{2}}\sin\frac{2\pi\psi}{\beta_F},\quad j=-\frac{4\pi^{2}h\omega^{3}}{\beta_F^{3}}\cos\frac{2\pi\psi}{\beta_F}.$$ The dwell (90°–180°) has $s=h$, $v=a=j=0$.
3. Rise peaks. Cycloidal motion has $a_{\mathrm{max}}$ at $\phi=\beta_R/4$ and $j_{\mathrm{max}}$ at $\phi=0$: $$a_{\mathrm{max}}=\frac{2\pi h\omega^{2}}{\beta_R^{2}}=\frac{2\pi(0.0254)(188.5)^2}{(\pi/2)^2},\qquad j_{\mathrm{max}}=\frac{4\pi^{2}h\omega^{3}}{\beta_R^{3}}.$$ $a_{\mathrm{max}}=2298\ \text{m/s}^2\ (\approx 234g)$, $j_{\mathrm{max}}=1.73\times10^{6}\ \text{m/s}^3$; the peak rise velocity is $v_{\mathrm{max}}=2h\omega/\beta_R=6.10\ \text{m/s}$.
4. Base-circle design (flat-faced follower). The follower face must never lose contact, so the radius of curvature $\rho=R_0+s+s''$ must stay positive, where $s''=d^2s/d\theta^2$. The most negative value of $s+s''$ over the cycle is $-41.7\ \text{mm}$ (near $\phi=3\beta_R/4$ on the rise), hence $R_0>41.7\ \text{mm}$. Choose $R_0=45\ \text{mm}$.
5. Cam profile. For a radial flat-faced follower the profile is generated by $$x(\theta)=(R_0+s)\sin\theta+\frac{ds}{d\theta}\cos\theta,\qquad y(\theta)=(R_0+s)\cos\theta-\frac{ds}{d\theta}\sin\theta,$$ traced over $0\le\theta\le360^\circ$ (Figure 2.2). The required follower face width is set by the maximum offset of the contact point, $\max|ds/d\theta|=2h/\beta_R=32.3\ \text{mm}$ on the rise side.
6. Pressure angles. For a flat-faced follower the transmitted force is always normal to the flat face, i.e. always along the follower axis, so the pressure angle is identically zero: $\alpha(45^\circ)=\alpha(270^\circ)=0^\circ$, and indeed $\alpha=0^\circ$ at every cam angle.