Anti-sway: the load is a pendulum

A load hanging on a rope is a pendulum. Accelerate the trolley and the load lags behind, then swings; stop it and the load swings on. A crane operator spends a lot of skill, and time, waiting for the load to settle. The drive can do better, because the swing is caused by the acceleration profile it commands. This lesson is course section 4.3.3 and Lab IV.

The pendulum

With a rope of length $L$, the load has the natural period

$$T_\text{sway} = 2\pi\sqrt{\frac{L}{g}} \tag{4.5}$$

For small angles, the sway angle $\theta$ under a trolley acceleration $a(t)$ obeys

$$\ddot\theta + \frac{g}{L}\,\theta = -\frac{a(t)}{L}$$

This is equation (4.1) of the Lab Works booklet. A steady acceleration $a_0$ tilts the rope back by about $a_0 / g$. A step in acceleration does not simply tilt it: it starts an oscillation about the new angle, and when the acceleration stops, the load keeps swinging with an amplitude set by when in the period the step ended.

FoundationStart here if this is new to you

Hold a bag by its handle and start walking briskly: the bag swings back, then forward, then back. Now start walking a little, pause for half a swing, then speed up again: if you time the second push right, the swing from the first push is cancelled by the second, and the bag travels with you, still. That is input shaping.

Two pulses, half a period apart

Split the acceleration into two equal half-steps, the second delayed by half a period, $T_\text{sway}/2$. The second half-step launches an oscillation exactly in antiphase with the first, and the two cancel. The trolley reaches the same speed, a little later.

Rope length L

4.0 m

Shaper tuned for L =

4.0 m

Retune the shaper to the rope length (measured L)

Step: 0.25 m/s² for 2 sTwo half-steps T/2 apart

• Sway, step profile (mrad)
• Sway, shaped profile (mrad)

Pendulum period T

4.01 s

Pulse spacing T/2 (tuned)

2.01 s

Residual sway

0.1 mrad · 0.0 cm

Reduction by shaping

100 %

The trolley stops accelerating at 4.01 s and then runs at 0.5 m/s.

θ″ + (g/L) θ = −a(t)/L ; T = 2π √(L/g)

Predict first

Lab IV: rope 4 m, trolley to 0.5 m/s. With the shaper tuned for 4 m, set the rope to 2 m without retuning. What happens to the residual sway?

It stays at zeroIt comes back, though smaller than with the plain stepIt becomes larger than with the plain step

What to read off the widget for Lab IV questions 3 to 5:

• With the plain step, the residual swing has the period $2\pi\sqrt{L/g}$: 4.01 s at 4 m.
• With the matched two-pulse profile, the residual sway falls by essentially 100 % in the linear model.
• Mismatch the rope length and part of the swing returns.

Beyond two pulses

Real anti-sway controllers go further: input shapers with more pulses that tolerate some error in $L$, jerk-limited profiles, or closed-loop feedback of the measured sway angle. All of them work on the same principle: the swing is excited by the acceleration the drive commands, so it can be removed in the controller.

ExplorerGo deeper: derivations and open questions

Why exactly half a period. Solve the pendulum equation for one acceleration step starting at $t = 0$, and write the residual oscillation as a phasor. Show that a second, equal step delayed by $\tau$ cancels it only if $\omega_n \tau = \pi$.

Robustness. A three-pulse shaper with amplitudes $\tfrac14, \tfrac12, \tfrac14$ spaced by $T/2$ tolerates a larger error in $L$. Build it in your Lab IV model and compare residual sway against $L$ for the two shapers. What does the extra robustness cost in time?