Ship Stability, Theory and Practice • Volume One: Foundations of Ship Stability

Chapter 6 — The Centre of Gravity

Loading, Discharging and Shifting Weights

Learning objectives

By the end of this chapter you will be able to:

  1. define the centre of gravity G and the height KG measured from the keel;
  2. state the three rules: G moves towards a loaded weight, away from a discharged weight, and parallel to a shifted weight;
  3. calculate the shift of G from GGH/V = (w × s) ÷ (∆ ± w) for loading and discharging, and GGH/V = (w × s) ÷ ∆ for shifting;
  4. find the final KG for any number of weights by moments about the keel, KG = Σ Vertical Moments ÷ ∆;
  5. apply the sign rules: loaded and raised weights give positive moments, discharged and lowered weights negative;
  6. explain why a suspended weight acts at the head of the crane or derrick from the instant of lift off;
  7. combine the final KG with the tabulated KM to find the metacentric height GM = KM − KG.

The first five chapters answered one question: how deep does the ship float? From here the series turns to a sharper one: will she float upright, and will she return when disturbed? Everything in that story is a contest between two points, and this chapter introduces the one the officer controls. The centre of gravity G is the point through which the ship's entire weight acts, and every tonne loaded, discharged or moved shifts it. The officer cannot choose the weather or the hull form; the position of G is the one great lever in their hands.

6.1 The centre of gravity and KG

However the cargo, fuel, water and structure are scattered through the ship, their combined weight acts as a single force, ∆, pulling vertically downwards through one point: the centre of gravity, G. Its height above the keel is KG, and the keel point K is the reference for every vertical position on board, just as the perpendiculars of Chapter 3 anchor the horizontal ones. Facing G across the ship is the centre of buoyancy B of Chapter 1, through which the water pushes up. Where these two forces stand relative to one another decides whether the ship sits upright, lists, or worse, and that contest is the business of Chapter 7. This chapter's task is the prerequisite: knowing exactly where G is at every stage of loading.

Two forces, two points: G and B G the whole weight of the ship, ∆, acts downwards through G B the buoyancy force acts upwards through B, the centre of the underwater volume K KG G is fixed by how the ship is loaded; B is fixed by the shape of the underwater volume. Chapter 7 brings the two together. This chapter is about where G is, and how the officer moves it.
Figure 6.1   The whole weight of the ship acts downwards through G; the buoyancy acts upwards through B. This chapter is about where G is.

6.2 The three rules

Every problem in this chapter obeys three short rules, worth committing to memory before any formula:

The three rules of G

The distances obey one family of formulas from the MCA sheet, where w is the weight, s the distance between G and the weight (or the distance the weight is moved), and the subscripts H and V simply say whether the shift of G being calculated is horizontal or vertical:

GGH/V = (w × s) ÷ ∆      GGH/V = (w × s) ÷ (∆ ± w) MCA formula sheet, September 2020 — the first for a shifted weight, the second for loading (+) or discharging (−)
The three rules of the centre of gravity LOADING w G G moves straight towards the loaded weight GG₁ = (w × s) ÷ (∆ + w) DISCHARGING w G G moves straight away from the discharged weight GG₁ = (w × s) ÷ (∆ − w) SHIFTING w G G moves parallel to the shift of the weight GG₁ = (w × s) ÷ ∆ s is the distance between G and the weight (loading and discharging), or the distance the weight is moved (shifting). MCA formula sheet, September 2020.
Figure 6.2   The three rules with their formulas. Note the denominators: the final displacement for loading and discharging, the unchanged displacement for a shift.

6.3 One weight at a time

Worked example 6.1

MV Ninja floats at a displacement of 28343 t with KG 7.60 m. She loads 500 t of deck cargo on the hatch covers at Kg 14.60 m (the moulded depth is 13.50 m, so a stow on the hatch covers has its centre a little above the deck). Find her new KG (a) by the GG₁ formula and (b) by moments about the keel.

(a) s = 14.60 − 7.60 = 7.00 m above G

GGV = (w × s) ÷ (∆ + w) = (500 × 7.00) ÷ 28843 = 3500 ÷ 28843 = 0.121 m upwards

new KG = 7.600 + 0.121 = 7.721 m

(b) KG = Σ Vertical Moments ÷ ∆ = (28343 × 7.60 + 500 × 14.60) ÷ 28843 = 222706.8 ÷ 28843 = 7.721 m

Two methods, one answer. The weight went on above G, so G climbed towards it, exactly as rule one promises.

Worked example 6.1 in a picture: 500 t of deck cargo w = 500 t at Kg 14.60 m G G₁ G climbs towards w s = 14.60 − 7.60 = 7.00 m above G GG₁ = (w × s) ÷ (∆ + w) = (500 × 7.00) ÷ 28843 = 0.121 m upwards The weight sits above G, so G climbs towards it along the centreline. Had the cargo gone into the double bottom, G would have moved down instead.
Figure 6.3   Worked example 6.1 drawn out. The added weight sits 7.00 m above G, and G moves towards it.
Worked example 6.2

MV Ninja lies at her summer displacement of 30456 t with KG 7.50 m. She discharges 705 t of salt water ballast from the No.2 double bottom tanks, port and starboard, whose centre of gravity the booklet gives as Kg 1.12 m. Find her new KG.

s = 7.50 − 1.12 = 6.38 m below G

GGV = (w × s) ÷ (∆ − w) = (705 × 6.38) ÷ 29751 = 4497.9 ÷ 29751 = 0.151 m

The weight left from below G, so G moves directly away from it: upwards. New KG = 7.500 + 0.151 = 7.651 m (check by moments: 228420 − 789.6 = 227630.4 t m, and 227630.4 ÷ 29751 = 7.651 m). The two No.2 double bottom tanks hold 2 × 471.3 = 942.6 m³, or 966 t of salt water, so 705 t is within what they can give.

Pumping out low ballast raises the centre of gravity, one of the quiet traps of ship operation. Notice too that the new displacement, 29751 t, is exactly the winter figure from Chapter 5: the 705 t of Chapter 4 at work again.

Worked example 6.3

At a displacement of 29751 t, 400 t of cargo is shifted from the bottom of a hold, Kg 3.20 m, to the hatch covers, Kg 14.60 m. Find the rise of G.

s = 14.60 − 3.20 = 11.40 m, the distance the weight itself moves

GGV = (w × s) ÷ ∆ = (400 × 11.40) ÷ 29751 = 4560 ÷ 29751 = 0.153 m upwards

The displacement is unchanged, so the denominator is simply ∆: nothing has come aboard or left, the weight has only moved, and G moves parallel to it.

6.4 Many weights: moments about the keel

A real loading involves dozens of weights, and chasing G through them one GG₁ at a time would be miserable. The professional method takes moments about the keel: every weight is multiplied by its Kg, the moments are summed with their signs, and one division delivers the final KG:

KG = Σ Vertical Moments ÷ ∆ MCA formula sheet, September 2020

The sign rules for vertical moments

Many weights, one method: moments about the keel K, the keel: the common reference every Kg is measured from here Item Weight Kg Moment about K Light ship 4950 t 8.86 m 43857 t m No.1 hold grain 4503 t 7.94 m 35754 t m No.2 hold grain 5308 t 7.70 m 40872 t m No.3 hold grain 5038 t 7.72 m 38893 t m No.4 hold grain 5308 t 7.70 m 40872 t m No.5 hold grain 4640 t 8.02 m 37213 t m Heavy fuel oil 509 t 12.65 m 6439 t m Diesel oil 35 t 11.45 m 401 t m Fresh water 165 t 11.86 m 1957 t m Totals 30456 t 246258 t m KG = Σ Vertical Moments ÷ ∆ = 246258 ÷ 30456 = 8.086 m (8.09 m) MCA formula sheet, September 2020; figures from Worked example 6.4 Loaded weights carry positive moments; discharged weights carry negative ones. The keel never moves, which is what makes it the ideal referee.
Figure 6.4   The moments table of Worked example 6.4 as a picture. The keel is the common reference; every Kg in the booklet is measured from it.
Worked example 6.4

MV Ninja loads for sea. Light ship 4950 t at KG 8.86 m (booklet). All five holds are filled with bulk grain at a stowage factor of 1.30 m³/t, the hold masses and centres coming from the booklet's capacity table; bunkers and fresh water are as shown. Find the final KG and, using the booklet KM of 10.330 m at the summer draught, the sailing GM.

ItemWeight (t)Kg (m)Moment about K (t m)
Light ship49508.8643857
No.1 hold grain45037.9435754
No.2 hold grain53087.7040872
No.3 hold grain50387.7238893
No.4 hold grain53087.7040872
No.5 hold grain46408.0237213
Heavy fuel oil (topside tanks)50912.656439
Diesel oil3511.45401
Fresh water16511.861957
Totals30456246258

KG = Σ Vertical Moments ÷ ∆ = 246258 ÷ 30456 = 8.086 m, carried in later chapters as 8.09 m

GM = KM − KG = 10.330 − 8.086 = 2.244 m, carried as 2.24 m

Three checks worth making every time. The total weight is exactly the summer displacement, 30456 t, so she sails at her marks at a draught of 9.600 m, which is why the KM of that draught is the right one. The final KG of 8.09 m is well below the booklet's maximum permissible KG at this displacement, 9.642 m by interpolation between 9.753 m at 30000 t and 9.631 m at 30500 t, so she complies with a margin of 1.55 m. And the answer sits between the smallest and largest Kg in the table, as any weighted average must. Each liquid has been taken as a solid weight at the centre of its tank; the free surface correction for these seven slack tanks, 0.027 m, is added in Chapter 9 to give a fluid KG of 8.113 m.

Worked example 6.5

On passage, MV Ninja burns 400 t of heavy fuel from the topside tanks, Kg 12.65 m. Using the loaded condition of Worked example 6.4 (∆ 30456 t, KG 8.086 m), find the new KG (a) by the GG₁ formula and (b) by adjusting the moments table.

(a) s = 12.65 − 8.086 = 4.564 m above G; the weight leaves from above G, so G moves away, downwards:

GGV = (w × s) ÷ (∆ − w) = (400 × 4.564) ÷ 30056 = 1825.6 ÷ 30056 = 0.061 m down; new KG = 8.086 − 0.061 = 8.025 m

(b) new moments = 246258 − (400 × 12.65) = 246258 − 5060 = 241198 t m; KG = 241198 ÷ 30056 = 8.025 m

The methods agree to the millimetre because the working carried KG as 8.086 m; started from the rounded 8.09 m, route (a) would give 8.029 m, four millimetres from route (b) through rounding alone. The physics is worth noticing: burning fuel from high tanks lowers G, but burning it from double bottom tanks would raise G, just as the discharge in Worked example 6.2 did. Where the consumables live decides which way the ship's stability drifts on passage.

6.5 The suspended weight

One situation catches people out on ships and in examinations alike. The moment a crane or derrick takes the weight of a load, before the load has physically risen a centimetre, the load behaves as if it were placed at the head of the crane. The wire can only pull along its own length, so the weight's effect on the ship acts from the point of suspension. G jumps accordingly, at lift off, all at once.

The suspended weight: g leaps to the crane head at lift off 60 t crane head, Kg 21.50 m the weight acts HERE from the instant it leaves the floor hold floor, Kg 3.00 m: where the weight was resting effective rise s = 18.50 m GG₁ = (w × s) ÷ ∆ = (60 × 18.50) ÷ 26000 = 0.043 m rise in G before the load moves an inch A weight on a crane or derrick behaves as if placed at the head of it. Heavy lifts are planned around this jump in G, not around the lift itself.
Figure 6.5   From the instant the load leaves the floor, its weight acts at the crane head. G rises before the load visibly moves.
Worked example 6.6

MV Ninja, displacement 26000 t and KG 7.30 m, uses her crane to lift 60 t from the bottom of a hold, Kg 3.00 m. The crane head is at Kg 21.50 m. Find the KG (a) while the load hangs from the crane and (b) after it is landed on the hatch covers at Kg 14.00 m.

(a) At lift off the weight acts at the crane head, an effective shift of s = 21.50 − 3.00 = 18.50 m:

GGV = (w × s) ÷ ∆ = (60 × 18.50) ÷ 26000 = 0.043 m; KG = 7.300 + 0.043 = 7.343 m

(b) Landed on the hatch covers the weight acts at its stowed position, a net shift from the hold of s = 14.00 − 3.00 = 11.00 m:

GGV = (60 × 11.00) ÷ 26000 = 0.025 m; KG = 7.300 + 0.025 = 7.325 m

The worst moment of the whole operation is while the load hangs: KG is higher during the lift than after it. Heavy lift plans are checked against condition (a), not condition (b).

6.6 Sideways, and onwards to Chapter 7

Everything above worked in the vertical, but nothing in the three rules or the moments method cares which way is up. Load a weight off the centreline and G moves towards it, horizontally; take moments about the centreline instead of the keel and the same arithmetic delivers the athwartships position of G:

GGH = Σ Horizontal Moments ÷ ∆ MCA formula sheet, September 2020
The same rules work sideways: G leaves the centreline centreline port side w G G₁ GGᴴ s metres off the centreline GGᴴ = Σ Horizontal Moments ÷ ∆ MCA formula sheet, September 2020: moments now taken about the centreline instead of the keel Nothing new is needed: the same three rules and the same moments method, turned through ninety degrees. A ship whose G is off the centreline will not float upright. How far she leans is the list, and tan(List) = GGᴴ ÷ GM is where Chapter 7 begins.
Figure 6.6   The same rules turned through ninety degrees. A G off the centreline means the ship will not float upright.

A ship whose G stands off the centreline cannot float upright: she lists until the buoyancy shifts across to stand beneath G again. For small angles the list is given by tan(List) = GGH ÷ GM, and GM, the metacentric height, is the difference between the KM of the hydrostatic table and the KG this chapter taught you to find:

Why KG matters: the ladder to GM K G at KG 8.09 m M at KM 10.330 m KG: set by the loading, found by moments GM = KM − KG = 10.330 − 8.09 = 2.24 m KM comes from the hydrostatic table (10.330 m at the summer draught); KG comes from this chapter’s moments table. Their difference, the metacentric height, is the master figure of Chapter 7.
Figure 6.7   The ladder from K to G to M. KG comes from the moments table; KM from the hydrostatic table; their difference is the GM of Chapter 7.

Interactive: the GG₁ simulator

Choose the operation and watch G move on the section. The simulator applies the correct MCA formula, with the correct denominator, and states the direction.

GG₁ = – New KG = – m
K G faded dot: G before green dot: G after (movement exaggerated)

Interactive: the moments table laboratory

Build your own loading. Each row takes a weight (negative to discharge) and its Kg; the table keeps the running totals and delivers the final KG and, with your KM, the GM. It opens preloaded with Worked example 6.4 (the laboratory keeps each moment unrounded, so its total reads 246257 t m where the printed table, with each row rounded, gives 246258 t m; the KG is 8.086 m either way).

Σ weights = – t Σ moments = – t m Final KG = – m GM = KM − KG = – m

Chapter summary

Self test questions

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