Question 4 of 7: Radial cam — minimum-peak-velocity rise

Question 4: Radial cam — minimum-peak-velocity rise (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. Rise $h=15$ cm over $\beta=90^\circ=\pi/2$ rad, then dwell, then fall; cam speed $\omega=100$ rad/s. In-line (radial) translating follower with a knife-edge tip, as Fig. (a) draws it. Objective for the rise: minimum peak velocity.

Find. The rise program, its $s(\theta)$ and $v(\theta)$, $v_{\max}$, a base circle and the pressure angle at $\theta=45^\circ$, and a neat cam-profile sketch.

Choice of motion program

Approach. Among the double-dwell smooth programs the peak velocity is $v_{\max}=C_v\,h\,\omega/\beta$; minimising $v_{\max}$ means choosing the smallest velocity coefficient $C_v$. From the candidate curves (Fig. b) the peak-velocity factors are:

The modified-sine program has the lowest $C_v=1.760$ of all the smooth (finite-jerk, double-dwell) curves offered, so it is the correct choice for minimum peak velocity. (Simple harmonic has a still lower $C_v=\pi/2$, but its infinite end-jerk violates the fundamental law of cam design and it is not among the listed double-dwell curves.)

Displacement and velocity equations (rise)

Let $x=\theta/\beta\in[0,1]$. The modified-sine acceleration is a sine wave whose central portion runs at one-third the end-portion frequency, with breakpoints at $x=\tfrac18$ and $x=\tfrac78$:

$$a(x)=C_a\frac{h\omega^2}{\beta^2}\;p(x),\qquad p(x)=\begin{cases}\sin(4\pi x), & 0\le x\le \tfrac18,\\[2pt] \cos\!\bigl(\tfrac{4\pi}{3}(x-\tfrac18)\bigr), & \tfrac18\le x\le \tfrac78,\\[2pt] -\cos\!\bigl(4\pi(x-\tfrac78)\bigr), & \tfrac78\le x\le 1.\end{cases}$$

Integrating once (velocity) and twice (displacement) with the dwell boundary conditions $s(0)=0,\ v(0)=0,\ s(\beta)=h,\ v(\beta)=0$ gives the standard closed form

$$v(\theta)=C_v\frac{h\omega}{\beta}\,V(x),\qquad s(\theta)=h\,S(x),$$

where $V(x)$ and $S(x)$ are the (normalised) integrals of $p(x)$; $S(x)$ climbs smoothly from 0 to 1 and $V(x)$ is a symmetric hump peaking at mid-rise. Evaluating the coefficients numerically confirms the textbook values $C_a=5.528$, $C_v=1.760$.

1. Maximum velocity. $$v_{\max}=C_v\frac{h\,\omega}{\beta}=1.760\cdot\frac{(0.15)(100)}{\pi/2}=\boxed{16.8\text{ m/s}}$$ occurring at $\theta=45^\circ$, where $s=h/2=7.5$ cm. (Peak acceleration $a_{\max}=C_a\,h\omega^2/\beta^2=3.36\times10^{3}$ m/s$^2$.)
2. Base circle and pressure angle at $\theta=45^\circ$. For an in-line (radial) translating follower the pressure angle is $$\phi=\arctan\!\frac{\mathrm ds/\mathrm d\theta}{R_b+s},\qquad \frac{\mathrm ds}{\mathrm d\theta}=\frac{v}{\omega}.$$ At mid-rise $\mathrm ds/\mathrm d\theta=v_{\max}/\omega=16.8/100=16.8$ cm/rad and $s=7.5$ cm — and since velocity peaks here, this is also the point of maximum pressure angle. Taking a first-trial base circle $R_b=15$ cm ($\approx h$): $$\phi=\arctan\frac{16.8}{15+7.5}=\arctan(0.747)=\boxed{36.8^\circ}.$$
3. Verdict and remedy. $36.8^\circ>30^\circ$, so the trial base circle is not satisfactory. The pressure angle is reduced by enlarging the base circle (or, equivalently, the prime circle) — increasing $R_b$ raises the denominator $R_b+s$. For example $R_b=22$ cm gives $\phi=\arctan\!\bigl(16.8/(22+7.5)\bigr)=29.7^\circ<30^\circ$. As instructed, no iteration is carried out; the design direction is simply "increase the base-circle radius (to about 22 cm)."