Question 7 of 10

Question 5 (20 marks: (a) 12, (b) 8)

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. Rise $h=40\,\text{mm}$ over $\beta_{\text{rise}}=0.015\,\text{s}$; dwell $0.005\,\text{s}$; fall $0.010\,\text{s}$; full cycle $T=0.030\,\text{s}$, so $\omega=2\pi/T=209.44\,\text{rad/s}$ ($\beta_{\text{rise}}=180^\circ$, dwell$=60^\circ$, $\beta_{\text{fall}}=120^\circ$ of cam rotation). In-line roller follower (no offset shown), spring-loaded (force-closed).

Find. (a) A rise displacement law satisfying the fundamental law of cam design with the smallest achievable peak follower acceleration, with its s-v-a-j curves; (b) a base-circle radius and the resulting maximum pressure angle.

(a) Rise motion law — chosen to minimise peak acceleration

Approach. The fundamental law of cam design requires the displacement, velocity and acceleration to be continuous through the whole cycle (so through the dwell–rise and rise–dwell boundaries too), ruling out simple harmonic motion (SHM has non-zero boundary acceleration $\Rightarrow$ an acceleration jump against the adjacent dwells). Among boundary-safe named curves, compare peak acceleration for the same $h,\beta$: cycloidal $a_{\max}=2\pi h/\beta^2$, 3-4-5 polynomial $a_{\max}=5.774\,h/\beta^2$, and the modified trapezoidal acceleration curve $a_{\max}\approx4.888\,h/\beta^2$ — closest to the unconstrained theoretical minimum ($4h/\beta^2$, achieved only by a discontinuous/illegal constant-acceleration curve) while keeping acceleration fully continuous. The modified trapezoid is therefore selected.

1. Construct the modified-trapezoidal acceleration shape over the rise (8 equal-time segments: sinusoidal ramp up to $+A$ over $[0,\beta/8]$, constant $+A$ over $[\beta/8,3\beta/8]$, sinusoidal ramp down to 0 by $\beta/2$, mirror-image negative half over $[\beta/2,\beta]$), then double-integrate numerically and scale so $s(\beta)=h$ exactly.
2. Read off the scaled peak values (with $h=0.04\,\text{m}$, $\beta=0.015\,\text{s}$): $$\boxed{a_{\max} = 869.0\ \text{m/s}^2}\qquad(\text{vs. }1117.0\ \text{m/s}^2\text{ for cycloidal — a 22\% reduction})$$ $$\boxed{v_{\max} = 5.333\ \text{m/s}}\ \Big(=\tfrac{2h}{\beta},\text{ same closed form as cycloidal}\Big)$$ Boundary check: $s(0){=}0$, $s(\beta){=}0.04000\,\text{m}$; $v(0){=}v(\beta){=}0$; $a(0){=}a(\beta){=}0$ — the fundamental law is satisfied exactly at both the dwell–rise and rise–dwell transitions.

(b) Base circle and pressure angle

Approach. For an in-line roller follower the pressure angle is $\varphi(\theta)=\tan^{-1}\!\big[(ds/d\theta)/(R_b+s)\big]$; scan the whole rise for several trial base-circle radii and pick the smallest $R_b$ that keeps $\varphi_{\max}$ comfortably under the usual $30^\circ$ design limit.

1. Scan candidate base circles (numerically, using $ds/d\theta=v(t)/\omega$ from the modified-trapezoid law):

   $R_b$ (mm)	30	40	50	60
   $\varphi_{\max}$	28.6°	23.6°	20.4°	18.1°

2. Select $R_b=50\,\text{mm}$ ($\ge$ 1.25× the 40 mm lift, a common proportioning rule), giving a comfortable margin below the 30° limit: $$\boxed{R_b = 50\ \text{mm}, \qquad \varphi_{\max} = 20.4^\circ\ \text{at }\theta\approx83^\circ\text{ into the rise}}$$
3. Lay the profile. With the base circle fixed, the pitch curve is $r(\theta)=R_b+s(\theta)$ (cam rotates clockwise, so $\theta$ is measured clockwise from the follower's start position); the physical cam surface is then the envelope of the roller (radius $r_f$, typically $r_f\le 0.4R_b\approx20\,\text{mm}$) rolled along that pitch curve. A check callout flags that no roller radius was specified in the source; $r_f=15\,\text{mm}$ is assumed as a typical proportion.