Cams and Followers

Cam

A cam is a rotating or oscillating machine element that gives a required, often irregular, motion to another element called the follower by direct contact.

Classification of cams

• Disc (plate) cam: profile cut on a disc, follower moves radially or at an angle to the axis.
• Cylindrical (drum) cam: groove cut on a cylinder, follower moves parallel to the axis.
• Translating (wedge) cam: cam reciprocates and moves the follower.
• Face cam / grooved cam: follower held by a groove, so no spring needed (positive drive).

Classification of followers

Nomenclature

          follower
             |
        trace point *
             |  .  pitch curve
         ____|___
       /   (prime) \
      |   ( base    |   cam profile
       \  circle   /
         ---------

• Base circle: smallest circle drawn from the cam centre, tangent to the cam profile.
• Trace point: reference point on the follower used to generate the pitch curve (knife edge, or roller centre).
• Pitch curve: path traced by the trace point relative to the cam. For a knife-edge follower, the pitch curve is the cam profile.
• Prime circle: smallest circle from the cam centre tangent to the pitch curve; radius = base circle radius + roller radius.
• Pressure angle ($\alpha$): angle between the follower's direction of motion and the common normal at the point of contact. It should not exceed about $30^\circ$ for translating followers to avoid jamming.
• Lift or stroke ($h$): maximum displacement of the follower from its lowest position.
• Angle of ascent, dwell, descent, and action: cam angles during rise, rest at top or bottom, and fall.

Derivation

In SHM the follower moves like the projection of a point moving on a circle of diameter equal to the lift $h$. Let the cam turn through $\theta$ during the rise angle $\beta$. The point on the auxiliary circle turns through $\pi\theta/\beta$ while the cam turns $\theta$.

$$s = \frac{h}{2}\left(1 - \cos\frac{\pi\theta}{\beta}\right)$$

With $\theta = \omega t$, differentiating with respect to time:

$$v = \frac{ds}{dt} = \frac{\pi h\omega}{2\beta}\sin\frac{\pi\theta}{\beta}$$ $$a = \frac{dv}{dt} = \frac{\pi^2 h\omega^2}{2\beta^2}\cos\frac{\pi\theta}{\beta}$$

Maximum velocity is at mid-rise ($\theta = \beta/2$) and maximum acceleration is at the start ($\theta = 0$).

Numerical

Data: $h = 0.040$ m, $\beta = 120^\circ = 2.0944$ rad, $\omega = \dfrac{2\pi \times 300}{60} = 31.416$ rad/s, $\theta = 45^\circ$.

$$\frac{\pi\theta}{\beta} = \frac{\pi \times 45}{120} = 1.1781\ \text{rad} = 67.5^\circ$$

At $\theta = 45^\circ$:

$$\begin{aligned} s &= 0.020(1 - \cos 67.5^\circ) = 0.020(1 - 0.3827) = 12.35\ \text{mm} \\ v &= \frac{\pi(0.040)(31.416)}{2(2.0944)}\sin 67.5^\circ = 0.871\ \text{m/s} \\ a &= \frac{\pi^2(0.040)(987.0)}{2(2.0944)^2}\cos 67.5^\circ = 17.00\ \text{m/s}^2 \end{aligned}$$

Maximum values:

$$v_{max} = \frac{\pi h\omega}{2\beta} = 0.942\ \text{m/s}, \qquad a_{max} = \frac{\pi^2 h\omega^2}{2\beta^2} = 44.41\ \text{m/s}^2$$

Answer: at $45^\circ$: $s = 12.35$ mm, $v = 0.871$ m/s, $a = 17.00$ m/s$^2$; $v_{max} = 0.942$ m/s, $a_{max} = 44.41$ m/s$^2$.

Displacement data

Rise (UARM, $h = 30$ mm, $\beta = 90^\circ$). The first half (0 to $45^\circ$) is uniform acceleration, $s = 2h(\theta/\beta)^2$; the second half is uniform retardation, $s = h - 2h(1-\theta/\beta)^2$.

Return (SHM, $\beta = 120^\circ$). $s = \dfrac{h}{2}\left(1 + \cos\dfrac{\pi\phi}{120^\circ}\right)$ measured from the top, where $\phi$ is the angle from the start of return.

Remaining angles: dwell at top $45^\circ$, dwell at bottom $105^\circ$. Check: $90 + 45 + 120 + 105 = 360^\circ$.

Drawing procedure

1. Choose a scale (full size). Mark the cam centre $O$.
2. Draw the base circle of radius 35 mm and the prime circle of radius $35 + 10 = 45$ mm (the roller centre is the trace point).
3. Draw the offset circle of radius 10 mm about $O$. The follower line is tangent to this circle, 10 mm to the right of the vertical through $O$. Mark the lowest trace point $A_0$ where this line meets the prime circle.
4. Draw the displacement diagram: a base line of length for $360^\circ$ divided into rise 90, dwell 45, return 120, dwell 105. Plot the curve using the table above (parabolic for rise, cosine curve for return).
5. Because the cam rotates clockwise, use inversion: keep the cam fixed and rotate the follower anticlockwise. Starting from $A_0$, divide the prime circle into the same angles (15 degrees steps for rise, 30 degrees for return) in the anticlockwise direction.
6. Through each division point on the prime circle, draw a tangent to the offset circle (follower position lines). Along each tangent, measure from the prime circle the displacement from the table, outward. This gives the roller-centre points $A_1, A_2, \dots$.
7. Join the points with a smooth curve: this is the pitch curve.
8. With each point as centre and radius 10 mm (roller radius), draw circles. The inner envelope touching all the circles is the cam profile.
9. Mark the dwell portions as arcs of the prime circle (pitch) and the base circle (profile).

Check

The pressure angle is greatest near mid-rise; measure it between the follower line and the normal at contact and confirm it is below about $30^\circ$. Offset on the correct side lowers the pressure angle during rise.

Theory

For a radial flat-faced follower, the contact point lies at a distance $x = \dfrac{ds}{d\theta}$ from the follower axis, and the radius of curvature of the cam profile is

$$\rho = r_b + s + \frac{d^2 s}{d\theta^2}$$

where $r_b$ is the base circle radius. If $\rho$ becomes zero or negative, the profile has a cusp (undercutting).

Data

$h = 20$ mm, $\beta = 90^\circ = \pi/2$ rad, $r_b = 30$ mm, $\omega = \dfrac{2\pi \times 600}{60} = 62.83$ rad/s.

SHM during rise:

$$s = \frac{h}{2}\left(1 - \cos\frac{\pi\theta}{\beta}\right),\quad \frac{ds}{d\theta} = \frac{\pi h}{2\beta}\sin\frac{\pi\theta}{\beta},\quad \frac{d^2s}{d\theta^2} = \frac{\pi^2 h}{2\beta^2}\cos\frac{\pi\theta}{\beta}$$

(a) Maximum velocity and acceleration

$$\left(\frac{ds}{d\theta}\right)_{max} = \frac{\pi h}{2\beta} = \frac{\pi(20)}{2(\pi/2)} = 20.0\ \text{mm/rad}$$ $$v_{max} = \omega\left(\frac{ds}{d\theta}\right)_{max} = 62.83 \times 0.0200 = 1.257\ \text{m/s}$$ $$a_{max} = \omega^2\left(\frac{d^2s}{d\theta^2}\right)_{max} = \omega^2\,\frac{\pi^2 h}{2\beta^2} = (3948)(0.040) = 157.9\ \text{m/s}^2$$

where $\dfrac{\pi^2 h}{2\beta^2} = \dfrac{\pi^2 (20)}{2(\pi/2)^2} = 40$ mm.

(b) Minimum radius of curvature

Put $x = \pi\theta/\beta$ (from 0 to $\pi$):

$$\rho = 30 + 10(1 - \cos x) + 40\cos x = 40 + 30\cos x\ \text{mm}$$

The minimum is at the end of the rise.

$$\rho_{min} = 40 - 30 = 10.0\ \text{mm}$$

It is positive, so the profile is not undercut.

(c) Face width

The contact point is at distance $x = ds/d\theta$ from the follower axis. During the rise it moves from 0 to a maximum of $20.0$ mm, always on the same side of the axis. So the contact travel is $20.0$ mm. Adding 5 mm clearance at each end:

$$\text{face width} = 5 + 20.0 + 5 = 30\ \text{mm}$$

Answer: (a) $v_{max} = 1.257$ m/s, $a_{max} = 157.9$ m/s$^2$; (b) $\rho_{min} = 10.0$ mm; (c) face width = 30 mm.