Learning objectives
By the end of this chapter you will be able to:
- explain why a suspended weight acts at the point of suspension from the instant the wire takes the strain;
- calculate the virtual rise of G and the equal loss of GM when a weight is lifted;
- handle a lift from the quay as a weight added at the head of the gear;
- find the list while a load hangs outboard, using the GM of the lifted condition;
- work a complete lift stage by stage, and identify the worst stage on which the lift is judged;
- size the heaviest permissible lift against a list or GM limit;
- plan counter ballast for a heavy lift, and price its free surface honestly.
Twice already this book has met the strangest rule in ship stability. In Chapter 6 a 60 t lift sent KG upwards before the load had visibly moved; in Chapter 7 the same load, swung outboard, listed the ship with a GM that had already shrunk. Both times the treatment was a single worked example and a promise. This chapter is the promise kept: the suspended weight in full, and with it the safe use of MV Ninja's own cranes, the everyday heavy machinery of a geared bulker's trade.
11.1 The point of suspension
A wire rope can pull only along its own length. Hang a load from a crane and let it swing where it will: whatever the load does, the force it exerts on the ship is the pull in the wire, and that pull passes always through the point of suspension, the head of the crane or derrick. So far as the ship's stability is concerned, the load might as well be bolted to the head. This is true from the instant the wire takes the strain, while the load still rests apparently innocent on the deck or the quay, and it remains true until the moment the load lands and the wire falls slack.
11.2 The virtual rise of G, and the loss of GM
For a weight already on board, the lift is simply Chapter 6's shift formula applied all at once: the weight w moves, in effect, from its stowage to the head, a vertical distance d, the moment the wire goes taut.
And since KM has not moved, every millimetre that G rises is a millimetre of GM lost: the loss of GM equals the virtual rise of G. The word virtual earns its keep: nothing physical has climbed to the head, yet the ship behaves in every respect as if it had, and behaves so until the wire is slacked.
MV Ninja, displacement 26000 t, KG 7.300 m, booklet KM 10.406 m, lifts 60 t with her crane from the bottom of a hold, Kg 3.00 m. The crane head is at Kg 21.50 m. Find the loss of GM the instant the wire takes the strain, and the GM during the lift.
d = 21.50 − 3.00 = 18.50 m
GGV = (w × d) ÷ ∆ = (60 × 18.50) ÷ 26000 = 0.043 m, so KG = 7.300 + 0.043 = 7.343 m, exactly as Worked example 6.6 found.
GM before = 10.406 − 7.300 = 3.106 m; GM during = 10.406 − 7.343 = 3.063 m: a loss of 0.043 m, equal to the rise.
Modest here, because the ship is stiff and the load light. Scale w up towards a crane's full working load on a tender ship, and this same line of arithmetic is the difference between a routine lift and a dangerous one.
11.3 Outboard: the list during the lift
A lift is rarely straight up and down: the whole purpose of the gear is reach. The moment the load swings outboard of the centreline, the suspended weight pulls G sideways exactly as Chapter 7 taught, and the ship takes a list. Both effects of the suspension now act together, and both against her: G has risen, so GM is down; and GGH grows with every metre of outreach.
MV Ninja, displacement 26000 t, KG 7.300 m, is to load a 120 t transformer from the quay on her port side using her heavy derrick, head at Kg 21.50 m. At the moment the derrick takes the whole weight off the quay, the load hangs 15.0 m outboard of the centreline, about 3 m beyond the ship's side (her breadth is 24.20 m). All tanks are pressed full or empty. The booklet KM at 26120 t is 10.400 m. Find the KG, GM and list at that moment.
Lifting from the quay, the weight comes on to the ship the instant it leaves the ground, acting at the head:
new ∆ = 26000 + 120 = 26120 t
KG = (26000 × 7.300 + 120 × 21.50) ÷ 26120 = (189800 + 2580) ÷ 26120 = 192380 ÷ 26120 = 7.365 m
GM = 10.400 − 7.365 = 3.035 m
GGH = (120 × 15.0) ÷ 26120 = 1800 ÷ 26120 = 0.0689 m; tan(List) = 0.0689 ÷ 3.035 = 0.0227, so List = 1.3° to port, towards the quay
Note the bookkeeping: a lift from ashore is a loading, done with the moments table at the increased displacement, with the weight's Kg taken as the head. Compare Worked example 11.1, where the 60 t was already on board and the lift was a shift. Same principle, different table entries.
11.4 The whole lift, stage by stage
A complete heavy lift is four different ships in the space of ten minutes, and the officer of the watch should be able to write down all four before the wire is bent on. The transformer of Worked example 11.2 tells the whole story.
Continue Worked example 11.2: the transformer is swung inboard until it hangs plumb over No.3 hold, then landed on the tank top at Kg 2.50 m. Find the ship's KG and GM at each remaining stage, and confirm the landed condition by two routes.
Plumbed over the hold: the weight still acts at the head, so KG remains 7.365 m and GM 3.035 m; but the outreach is gone, GGH = 0, and she stands upright again. The virtual rise outlives the list.
Landed, route A (shift from head to tank top): GGV = 120 × (21.50 − 2.50) ÷ 26120 = 0.087 m down; KG = 7.365 − 0.087 = 7.278 m
Landed, route B (moments from scratch): KG = (26000 × 7.300 + 120 × 2.50) ÷ 26120 = 190100 ÷ 26120 = 7.278 m
GM = 10.400 − 7.278 = 3.122 m. Two routes, one answer; and she finishes stiffer than she began, because 120 t now lies deep in the ship. The dangerous stage was never the destination: it was the journey through the head.
Port regulations limit MV Ninja's list during cargo work to 2°. With the ship at 26000 t, KG 7.300 m, head at Kg 21.50 m and full outreach 15.0 m, what is the heaviest single lift she may take from the quay? (Booklet KM by interpolation at each trial displacement.)
The condition to satisfy is tan(2°) = 0.03492 = GGH ÷ GM with every quantity depending on w, so the equation is solved by trial. Try w = 180 t:
∆ = 26180 t; KG = (189800 + 180 × 21.50) ÷ 26180 = 193670 ÷ 26180 = 7.398 m
KM at 26180 t (interpolated) = 10.398 m, so GM = 10.398 − 7.398 = 3.000 m
GGH = (180 × 15.0) ÷ 26180 = 0.1031 m; tan(List) = 0.1031 ÷ 3.000 = 0.0344, List = 1.97°: just inside the limit.
Try w = 185 t: ∆ = 26185 t; KG = 193777.5 ÷ 26185 = 7.400 m; KM = 10.398 m; GM = 2.997 m; GGH = 2775 ÷ 26185 = 0.1060 m; tan(List) = 0.0354, List = 2.03°: just outside.
Interpolating between the two trials, maximum lift ≈ 183 t (a direct check at 183 t gives KG 7.399 m, GM 2.998 m, GGH 0.1048 m, tan(List) = 0.0350, List = 2.0°). Notice why the answer is not simply proportional: every added tonne raises G and thins GM even as it adds listing moment, so the list climbs faster than the load; had GM stayed at its unlifted 3.106 m, the same 183 t would have listed her only 1.9°. Figure 11.5 draws the curve.
11.5 Preparing the ship: counter ballast, and its price
The professional lift begins hours before the wire is bent on. The GM of the lifted condition is calculated and checked against the booklet; slack tanks are pressed up or stripped so the free surfaces of Chapter 9 do not gnaw at a GM already thinned by the lift; and, for a heavy lift at reach, ballast is often run to the off side beforehand, so the ballast moment stands ready to meet the lifting moment and the ship works nearly upright through the worst stage. A transfer between a pair of double bottom tanks at the same height leaves KG untouched, but it usually leaves both tanks slack, and their free surface moments, i × density of the liquid for each tank from the booklet tank tables, must be priced before pumping.
Before the 120 t transformer lift of Worked example 11.2 (moment at full outreach 120 × 15.0 = 1800 t m to port), ballast is to be run from No.3 D.B. (P) to No.3 D.B. (S). From the booklet the centres of the two tanks are 7.56 m either side of the centreline, both at Kg 1.12 m, and the free surface inertia of each is i = 2669 m4; the ballast is salt water. How much should be transferred to hold her upright at the worst stage, and what does the transfer cost if it leaves both tanks slack?
The transfer distance is 2 × 7.56 = 15.12 m. Counter moment required = 1800 t m, so w = 1800 ÷ 15.12 = 119 t to starboard (119 × 15.12 = 1799 t m against the lift's 1800 t m). Both tanks are at the same Kg, so KG is unchanged; but before the strain is taken the ballast alone lists her 1.4° to starboard (GGH = 1799 ÷ 26000 = 0.0692 m against a fluid GM of 2.895 m).
The price: two slack tanks. The booklet lists the free surface inertia i, and FSM = i × ρ = 2669 × 1.025 = 2736 t m per tank of salt water; ΣFSM = 2 × 2736 = 5471 t m; FSC = 5471 ÷ 26120 = 0.209 m; the fluid GM at the worst stage falls from 3.035 to 2.826 m. Without the counter ballast, but with the two tanks slack, the same stage would list her 1.4° to port (tan(List) = 0.0689 ÷ 2.826 = 0.0244) against the 1.3° of Worked example 11.2 with solid tanks.
Ample, on this stiff ship in this condition. But the lesson generalises: the cure has a cost, and on a tender ship the free surface bill can exceed the listing problem it was meant to solve. Price both sides before pumping, with the booklet tank data (FSM = i × density of the liquid), exactly as Chapter 9 taught; and when the lift is over, press one tank full and strip the other.
A small general cargo ship, displacement 8000 t, breadth 18.0 m, KG 7.20 m, KM 7.95 m, is to lift 45 t from the quay with her derrick: head at Kg 17.0 m, the load 11.0 m outboard of the centreline at lift off. Find her list at the moment of lift off, and comment.
∆ = 8045 t; KG = (8000 × 7.20 + 45 × 17.0) ÷ 8045 = 58365 ÷ 8045 = 7.255 m
GM = 7.95 − 7.255 = 0.695 m (taking KM as unchanged for so small a weight)
GGH = (45 × 11.0) ÷ 8045 = 0.0615 m; tan(List) = 0.0615 ÷ 0.695 = 0.0885, so List = 5.1°
A smaller sideways shift of G than the transformer gave MV Ninja (0.0615 m against 0.0689 m) gives this ship a list four times greater, five degrees against 1.3°, because her GM in the lifted condition is under a quarter of the bulker's. The gear does not know or care how stiff the ship is; the officer must. The GM check comes first, always, and a lift that a big ship shrugs off can put a small one's deck edge towards the water.
11.6 Onwards
Every calculation in this chapter leaned on one number read from the booklet: KM at the working displacement. Chapter 12 asks where that column really comes from, and answers with a picture, the metacentric diagram, in which KB and KM are drawn as curves against draught and the ship's initial stability can be read straight from her geometry.
Interactive: conduct the lift yourself
Set the load, take the strain, and swing it outboard. G climbs the ladder the instant the wire goes taut; the ship settles to her calculated list as you swing; the verdict panel judges the lift. The drawn heel is exaggerated four times.
Interactive: the four stage calculator
The whole of Section 11.4 as a live table, preloaded with the transformer lift. Change any figure and watch all four stages follow. The lift is from the quay; the load lands at the hold Kg.
| Stage | ∆ (t) | KG (m) | GM (m) | List |
|---|---|---|---|---|
| 1. alongside, before | – | – | – | upright |
| 2. at the head, over the quay | – | – | – | – |
| 3. plumbed over the hold | – | – | – | upright |
| 4. landed in the hold | – | – | – | upright |
Chapter summary
- A suspended weight acts at the point of suspension, from the instant the wire takes the strain until it falls slack.
- For a weight already aboard, GGV = (w × d) ÷ ∆ with d from stowage to head; the loss of GM equals the rise of G.
- A lift from the quay is a loading: the weight enters the moments table at the head's Kg, at the increased displacement.
- Swung outboard, GGH = (w × a) ÷ ∆ and tan(List) = GGH ÷ GM, with the GM of the lifted condition.
- The lift is judged at its worst stage: load at the head, at full outreach; never on the comfortable stages either side.
- Counter ballast is sized by moments, w × d against the lift's moment, and its free surface is priced before pumping: FSM = i × density of the liquid, with i from the booklet tank tables.
Self test questions
Work each question with pencil and paper first. Your score appears in the bar below.