The counterweight that knows the load

The counterweight that follows the jib removed the larger of the two loads that move on a luffing crane. A 50 m jib weighing 300 kN changes its own moment about the mast by 6,591 kN·m as it luffs from 15° to 85°, more than twice what a 60 kN payload at the tip ever adds, and a counterweight linked to the luff cancels that change at every angle. What the linkage cannot cancel is the payload, because the payload leaves the ground in about a second and a linkage moves only when the jib does. The best a linked counterweight can do is wait at the half-load position, so that the step at pick-up runs from $-Qx/2$ to $+Qx/2$. At 48.3 m that is ±1,449 kN·m, a floor no linkage gets under.

That essay ended on a crane that knows what it is lifting. A load cell in the hoist reads the rope’s force continuously. It cannot read the payload before the pick, but it reads the rope taking up the payload, and the rope takes it up over a time the hoist decides. A counterweight on its own motor, positioned from that reading, is not linked to the jib or limited to half of anything. It could, in principle, balance the payload exactly.

Whether it should is a different question, and the answer depends on something the step at pick-up was standing in front of.

A step that is really a ramp #

The payload does not leave the ground at the instant the hoist starts. The rope is slack, then taut, and then its force climbs as the hoist winds in and the rope, the jib and the mast stretch, until the force equals the payload’s weight and the load lifts off. How fast it climbs is the hoist’s creep speed times the stiffness of everything between the drum and the hook. A normal hoist at normal speed takes up a few tonnes in a second or two. The same hoist on a slow creep takes ten or twenty.

To the mast the take-up is a ramp of moment, the rope’s force times the radius, and a load cell sees the ramp as it happens. A counterweight that moves while the ramp climbs has time to arrive, provided the ramp is slow enough and its drive is fast enough. The linked counterweight had no such time because it never knew the ramp was happening.

The linked crane starts at −1,449 kN·m — the counterweight holding half the payload’s moment for a payload that is not yet there — and rises with the rope’s force. At lift-off it reaches $+0.5\,Qx$, and then a little more, because a load leaving the ground surges: the hoisting factor $\varphi_2$, typically between 1.1 and 1.3 for a crane at a moderate hoisting speed, multiplies the payload for a moment as the rope and the jib spring back from their stretch. With $\varphi_2 = 1.2$ the linked crane’s mast reaches $0.7\,Qx = 2{,}028$ kN·m.

The driven counterweight starts where the rope does, at nothing, and follows the rope’s force half a second behind the cell. With a 20 s take-up it keeps pace: by lift-off it holds its chosen share of the payload and the mast carries the rest plus the surge. With a 3 s take-up it is still travelling when the payload leaves the ground, and the mast takes nearly all of the payload’s moment and the surge on top — 3,101 kN·m, more than the linked crane ever sees. A load cell turns the step into a ramp, and the ramp is only as slow as the crane agrees to pick.

How slow is slow enough #

The drive’s speed is a moment rate: the counterweight’s weight times the speed it travels along its track. A counterweight of 1,500 kN moving at a tenth of a metre a second changes its moment by 150 kN·m every second.

The peak falls quickly once the take-up is slow enough for the drive. At 150 kN·m/s a 2 s take-up leaves the mast 1.12 times the payload’s moment, worse than the linked crane; 10 s leaves 0.87; beyond that the curve is flat at 0.86, half a second of lag above the floor. A slower drive needs a slower pick, and a drive of 50 kN·m/s needs half a minute. A faster one, 400 kN·m/s, is at the floor with a 5 s take-up.

The comparison that decides whether this is practical is with the drive the crane already needs. A counterweight that replaces the luffing linkage has to follow the jib’s own moment as it luffs, and the jib’s moment changes fastest with the jib nearly upright: 173 kN·m/s for a luff from 15° to 85° in a minute, with the payload share included. A drive fast enough to replace the linkage is fast enough for a pick of about ten seconds. The load cell costs no extra motor. It costs the operator patience at the moment of take-up, and a slow take-up is something a crane’s control system can enforce.

The case the step was hiding #

With the step turned into a ramp, the driven counterweight can be made to balance any share of the payload it likes, and the obvious share is all of it. A fully balanced crane carries its payload through the whole lift with nothing in the mast except the surge at lift-off, $(\varphi_2 - 1)Qx$: 0.2 times the payload’s moment.

That arithmetic is right, and it leaves out a load case every crane standard requires. A payload can be let go. A sling parts, a grab opens, a magnet drops its load, a load snagged on a neighbouring structure tears free. When it goes, the rope’s force drops to nothing in a fraction of a second and the counterweight is still wherever it was. A crane whose counterweight was balancing the payload is suddenly a crane with that much unbalanced moment pulling backwards. And the drop is not static. The jib and the mast were stretched forward by the payload, and released at once they spring back past their new equilibrium. Crane standards cover the spring-back with a rebound factor: $\beta_3$, 0.5 for a release no faster than a grab’s, 1.0 for one as sudden as a magnet’s.

So for a counterweight that holds a share $\beta$ of the payload, the mast’s moment per unit of the payload’s moment $Qx$ is, at lift-off and when let go,

$$m_{\text{lift}} = \varphi_2 - \beta, \qquad m_{\text{release}} = -(\beta + \beta_3).$$

The first falls as the counterweight takes more of the payload, and the second grows. Balancing the payload is a hedge between lifting it and losing it.

The two lines cross at

$$\beta^* = \frac{\varphi_2 - \beta_3}{2}, \qquad m^* = \frac{\varphi_2 + \beta_3}{2},$$

for a release of the whole payload. With $\varphi_2 = 1.2$ and $\beta_3 = 0.5$ the best share is 0.35 and the mast is designed for 0.85 of the payload’s moment. The linked crane’s half sits on the release line at 1.00. The fully balanced crane sits at the far end of it, at 1.50, which is the worst of all three: half as much again as the crane that knew nothing about its payload.

The half was a hedge all along #

The formula has a case worth reading off. With no surge, $\varphi_2 = 1$, and no rebound, $\beta_3 = 0$, the best share is exactly one half. That is the linked crane’s position, and it means the half was never only about the step at pick-up. It was already the balance between picking the payload up and losing it, for a crane that does both statically. The step at pick-up and the step at release are the same step taken in opposite directions, and a counterweight that cannot tell them apart sits midway between them.

The load cell changes what the counterweight knows during the lift. It does not change the release, because a release is as sudden as the pick-up used to be, and nothing a load cell reads warns of it. So the knowledge buys exactly the difference between the two steps’ dynamics. A pick can be made slow, so its surge is modest and the counterweight can follow it. A release cannot be made slow, so its rebound is whatever the failure makes it. The best share moves down from a half by the amount the release is worse than the pick.

With no rebound the surge at lift-off is the worse of the two, and the hedge balances more than half — 0.60 — to give 0.60 against the linked crane’s 0.70. With a rebound of 0.5 it balances 0.35. With a rebound of 1, as sudden as a magnet’s, it balances a tenth of the payload and carries 1.10, against 1.50 for the linked crane and 2.00 for the fully balanced one. The worse the release, the less of the payload a counterweight should hold. In the limit of a severe rebound, the best counterweight for the payload is no counterweight, and the load cell’s knowledge is used only to stay out of the way.

The fully balanced crane’s line is the highest everywhere. That is the result least expected of a machine that knows its load: balancing the payload exactly is never the best use of knowing it, because the release is always at least as bad as no balance at all.

The lift, stage by stage #

Each option puts its worst moment in a different place, and a lift taken in stages shows where.

The linked crane spreads its moment evenly. It waits at −0.5 with the rope slack, the counterweight holding half a payload that is not there, rises to 0.7 at lift-off and carries 0.5, and is thrown back to −1.0 when the load goes. The fully balanced crane is the quietest in normal working: nothing waiting, 0.2 at lift-off, nothing carried. Its whole design is the one stage nobody plans for, at −1.5. The hedged crane does what a hedge does. Nothing waiting, its largest forward moment at lift-off and its largest backward moment at release, both 0.85, and neither larger than the other.

That symmetry is the reason it is best. A structure designed for the larger of two moments in opposite directions uses its strength best when the two are equal. The mast is a cantilever whose base must carry the moment either way, and whose foundation, where weight is the only thing resisting it, must keep its resultant inside the footprint either way, whether it tips or slides. Any share other than the hedge makes one direction govern and leaves strength unused in the other.

What the same mast will lift #

A mast is designed for a moment, and the moment it can carry decides what the crane may lift at each radius: its load chart, read off the mast’s response to a payload placed at each radius in turn as an influence line is read for a bridge.

For a mast designed to carry the linked crane’s moment with 60 kN at 48.3 m, the hedged driven crane may lift 70.6 kN at the tip, and proportionally more at every radius. That is 18% from the same steel, bought with a motor, a load cell and a slow take-up. It is real and it is modest, and the reason it is modest is the release.

The dashed line is what the load cell would be worth if no payload could ever be let go: a fully balanced crane leaving the mast only the lift-off surge, 0.2 of the payload’s moment, and a mast that would allow five times the payload. That line runs off the top of the figure and is not a load chart. Long before the mast carried 300 kN at 48 m, the jib, the hoist rope and the counterweight’s own reach would have run out. But it measures what the release case costs. Of everything a load cell might buy, the possibility of losing the payload takes back nearly all of it.

Where the counterweight’s own limits come in #

A driven counterweight that hedges at 0.35 of the payload needs a smaller range of travel than a linked one at half, and it carries the jib’s whole change of moment as the linked one did. The total the drive must supply is the jib’s change, 6,591 kN·m, plus the payload share at the longest radius, $0.35 \times 2{,}898 = 1{,}014$. That is a little less than the linked counterweight’s 7,909, so the same counterweight and track serve.

What a driven counterweight adds is a new thing to fail. A linked counterweight cannot be in the wrong place, because its position is the jib’s angle. A driven one can, if its motor stops, its brake slips, its load cell drifts or its controller is wrong about the radius. Its failure cases are a counterweight stranded far out with the jib raised, which is the crane left for the night with the linkage taken away, and a counterweight running out while a payload is let go, which is the release made worse. A driven crane has to be checked with its counterweight at each end of its travel and every payload state, and the 18% gain at the tip is worth only what those checks leave of it.

By hand, at the tip #

The payload’s moment at the tip is $60 \times 48.3 = 2{,}898$ kN·m. The linked crane, holding half, carries at lift-off $(1.2 - 0.5) \times 2{,}898 = 2{,}028$ kN·m forwards, and at release $(0.5 + 0.5) \times 2{,}898 = 2{,}898$ backwards. The release governs, and the mast is designed for 2,898.

The hedge balances $(1.2 - 0.5)/2 = 0.35$ of the payload. At lift-off it carries $(1.2 - 0.35) \times 2{,}898 = 2{,}463$ kN·m, and at release $(0.35 + 0.5) \times 2{,}898 = 2{,}463$ backwards: the two equal, 0.85 of the payload’s moment. A mast of 2,898 kN·m then allows $2{,}898/(0.85 \times 48.3) = 70.6$ kN at the tip.

The fully balanced crane carries $0.2 \times 2{,}898 = 580$ kN·m at lift-off and $1.5 \times 2{,}898 = 4{,}347$ backwards when the load goes. That is half as much again as the linked crane, from a counterweight that did everything it was asked.

A crane reduced to its mast #

The model is the mast’s moment and nothing else, and its simplifications are these.

The factors are code values, not a dynamic analysis. The hoisting factor and the rebound factor stand for a vibration of the jib, the rope and the mast that a real crane would have to be analysed for, and a load applied or removed suddenly does not overshoot by one factor everywhere in a structure: different parts of the crane see different multiples of their change of force. The mast’s base moment is the part the factors are calibrated for.

The payload is let go whole. A partial release — a two-legged sling losing one leg, a load whose part tears free — releases a share of the payload, and the best share for the counterweight rises towards the static half as the released share falls. The solver here takes the released share as a parameter. Every figure lets all of the payload go, because that is the case crane standards require when a load can be lost at all.

The take-up is a straight ramp. A real rope’s force climbs as the rope straightens, not linearly, and a payload resting partly on something else can come off in a jerk. A slow take-up is a slow average, and the drive’s lag is measured against it.

Slewing, wind and the counterweight’s own inertia are left out. The first two add to the mast’s moment in ways no counterweight addresses. The third is the mass that helps by being late in reverse: a heavy counterweight started and stopped quickly by a motor puts its own inertia into the mast, and a drive that has to move 150 kN·m/s of moment starts and stops a large mass.

What the load cell knows and what it cannot #

The linked counterweight’s limit was a step. The step’s size was the payload’s moment, and its sharpness was the second it took a payload to leave the ground. The load cell softens the first step by turning it into a ramp, but a load that can be removed suddenly has a second step, as sharp as the first used to be, that no instrument sees coming. A balance chosen against the first step alone was always going to be wrong about the second.

That is a general shape, not a crane’s. A structure that adapts to a load it can measure is designed by the moment the measurement stops being true. A counterweight is a moment and not a weight, and the moment it should be is the one that leaves the structure least exposed, either way, to the load arriving and the load going. Knowing the load narrows the first exposure and widens the second. The best answer sits where the two are equal, and it moves with how badly the load can go.

Still open: whether the release can be made slow #

The hedge moves down because a release is fast and a pick is slow. A crane that could slow its own release would move it back up. A hoist drum with a brake that pays out against a falling rope force, or a damped spring at the jib’s head that lets the hook rise a little before the jib feels the loss, would turn the release into a ramp as the load cell turned the pick into one. With a rebound factor driven towards nothing by a device on the rope, the best share climbs back past a half towards the full payload. Whether a device of that kind can be made reliable enough to be relied on in the case it exists for — a release that is by definition the crane’s failure — and whether the mast saved is worth more than the device and its own failures, is the question of whether the second step can be softened the way the first was, or only hedged.