Chapter 9: The Free Surface Effect | Ship Stability, Volume One

Learning objectives

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

explain why a slack tank behaves as a virtual rise of the ship's centre of gravity;
calculate the free surface moment from I = (l × b³) ÷ 12 and FSM = I × ρ_T;
apply the free surface correction FSC = FSM ÷ ∆ to obtain the fluid KG and fluid GM;
state and use the b³ law, and show that subdivision by n bulkhead spaces divides the moment by n²;
read the tabulated free surface inertias i (m⁴) from the stability booklet, convert each to a moment with the density of the liquid in the tank, and sum them over the slack tanks;
use the fluid GM, never the solid one, in list and stability calculations;
state what the effect does and does not depend on, and how it is ended.

Everything so far has treated the ship's contents as solid: cargo that stays where it is stowed. But a ship's tanks are full of liquid, and liquid obeys its own rule: its surface stays level whatever the ship does. Heel the ship a degree and every slack tank quietly pours its weight downhill; bring her back and the weight runs home again. The result is one of the most important, least visible ideas in the subject: the ship behaves exactly as if her G were higher than the moments table says. This chapter measures that virtual rise, straight from three lines on the MCA sheet and the free surface inertia column of MV Ninja's own tank tables, each entry multiplied by the density of the liquid in that tank.

9.1 The mechanism: a wandering weight

Picture a double bottom tank half full of ballast in a heeled ship. The liquid surface stays parallel to the sea, so a wedge of water has crossed to the low side: the liquid's own centre of gravity g has shifted towards the heel, exactly like the cargo shift of Chapter 6, except that this one happens by itself, at every angle, in both directions. The shifting weight subtracts righting lever at every heel, and the loss turns out to be identical to what a genuine rise of the ship's G would cost. It is therefore treated as one: a virtual rise of G, called the free surface correction:

FSC = FSM ÷ ∆ MCA formula sheet, September 2020 — FSM is the free surface moment of the slack tank

Figure 9.1 The liquid surface stays level, its g wanders with every heel, and the ship behaves as if G had climbed by FSC.

The two GMs

Solid GM = KM − KG from the moments table: the value if nothing on board could move.
Fluid GM = solid GM − FSC (equivalently, fluid KG = solid KG + FSC): the value the ship actually exhibits.
Every stability and list calculation must use the fluid GM. Examiners and the sea are equally unforgiving on this point.

9.2 Calculating the moment

The free surface moment of a rectangular tank comes from the second moment of area of its liquid surface about the tank's own centreline, times the density of the liquid in the tank:

I = (l × b³) ÷ 12 FSM = I × ρ_T MCA formula sheet, September 2020 — ρ_T is the density of the liquid in the tank, in t/m³

Worked example 9.1

A double bottom tank on MV Ninja is 20.0 m long and 10.0 m wide, partly filled with salt water ballast (ρ_T = 1.025 t/m³). At the summer displacement of 30456 t with solid GM 2.24 m, find the free surface moment, the correction, and the fluid GM.

I = (l × b³) ÷ 12 = (20.0 × 10.0³) ÷ 12 = 1666.7 m⁴

FSM = I × ρ_T = 1666.7 × 1.025 = 1708 t m

FSC = FSM ÷ ∆ = 1708 ÷ 30456 = 0.056 m

fluid GM = 2.24 − 0.056 = 2.18 m

One slack tank has quietly done what loading about 300 t of cargo onto the weather deck would have done to GM, and it will keep doing it until the tank is pressed up or emptied.

Worked example 9.2

The same tank is now built with a watertight centreline bulkhead, making two tanks each 20.0 m by 5.0 m, both slack with salt water. Find the total free surface moment and compare.

Each tank: I = (20.0 × 5.0³) ÷ 12 = 208.3 m⁴; FSM = 208.3 × 1.025 = 213.5 t m

Both tanks together: 2 × 213.5 = 427 t m, against 1708 t m undivided: exactly one quarter.

One bulkhead, n = 2, divides the moment by n² = 4. The liquid, its weight and its surface area are completely unchanged; only the breadth of each surface has halved, and b³ does the rest.

Worked example 9.3

The undivided tank of Worked example 9.1 instead holds heavy fuel oil of density 0.950 t/m³. Find the free surface moment and correction.

FSM = I × ρ_T = 1666.7 × 0.950 = 1583 t m; FSC = 1583 ÷ 30456 = 0.052 m

Note which density appears: ρ_T, the liquid in the tank, not the water the ship floats in. A tank of light oil steals a little less GM than the same tank of sea water. The booklet's tank tables list i, the inertia of the surface in m⁴, not a moment: the officer multiplies it by the density of whatever the tank holds on the day.

9.3 The b³ law and the shipbuilder's answer

Look hard at I = (l × b³) ÷ 12. Length enters once; breadth enters cubed. The shape of the liquid surface matters enormously more than its size, and the athwartships direction is the one that counts, because that is the direction the liquid runs when the ship heels.

Figure 9.3 Three tanks of identical surface area and identical contents. Only the breadth differs, and the moments differ ninefold.

The cubed law hands the shipbuilder a cheap and powerful cure. Divide a tank with longitudinal bulkheads into n equal spaces: each surface keeps the full length but only b ÷ n of breadth, so each carries the original moment divided by n³, and the n of them together give the original divided by n². One bulkhead quarters the moment; two cut it to a ninth.

Figure 9.4 Subdivision at work on the tank of Figure 9.3: 1476 to 369 to 164 t m, the moment falling as n².

9.4 The booklet does the arithmetic: tabulated inertias

Real tanks are not rectangles, so the naval architect computes the true second moment of the liquid surface for every tank and prints it in the stability booklet's tank tables: MV Ninja's carry a column headed "i, free surface (m⁴)" for every space, from the fore peak to the fresh water tanks, with the instruction that FSM = i × RD of the liquid in the tank. The column is the inertia in m⁴, not a moment in tonne metres. In service the calculation collapses to a sum: identify every slack tank, multiply its tabulated i by the density of its contents (salt water 1.025, heavy fuel oil 0.950, diesel oil 0.850, fresh water 1.000), add up the moments, divide once by the displacement.

Figure 9.5 Worked example 9.4 as a picture: the Chapter 6 departure condition with the booklet inertia of every slack tank multiplied by the density of its contents.

Worked example 9.4

MV Ninja sails in the Chapter 6 departure condition: ∆ 30456 t, solid KG 8.086 m (8.09 m to two places), KM 10.330 m. Her slack tanks, with the booklet's tabulated inertias i, are: No.1 H.F.O. (P) and (S) 147 m⁴ each and No.2 H.F.O. (P) and (S) 139 m⁴ each, holding heavy fuel oil of density 0.950; the Diesel Oil tank 28 m⁴, diesel oil 0.850; Fresh Water (P) 145 m⁴ and Fresh Water (S) 103 m⁴, fresh water 1.000. Find the fluid KG and fluid GM.

FSM = i × ρ_T for each tank: 147 × 0.950 = 139.65; 147 × 0.950 = 139.65; 139 × 0.950 = 132.05; 139 × 0.950 = 132.05; 28 × 0.850 = 23.80; 145 × 1.000 = 145.00; 103 × 1.000 = 103.00 t m

ΣFSM = 139.65 + 139.65 + 132.05 + 132.05 + 23.80 + 145.00 + 103.00 = 815.2 t m

FSC = ΣFSM ÷ ∆ = 815.2 ÷ 30456 = 0.027 m

fluid KG = 8.086 + 0.027 = 8.113 m; fluid GM = 10.330 − 8.113 = 2.217 m

A modest theft this time, 0.027 m, because the slack tanks are narrow wing and service tanks. Note the professional habit: the fluid KG, not just the fluid GM, is what gets checked against the booklet's maximum KG table, and 8.113 m still clears the 9.64 m limit at this displacement by more than a metre and a half. Had the inertias been used as if they were moments in tonne metres the sum would have been 848 t m and the correction 0.028 m: a small difference here, but an error of principle.

9.5 The fluid GM at work

Worked example 9.5

On passage at her winter displacement of 29751 t, solid KG 7.95 m (KM 10.328 m), MV Ninja carries out a ballast exchange with the No.2 and No.3 double bottom pairs slack. The booklet's tabulated inertias are 2782 m⁴ for each No.2 tank and 2669 m⁴ for each No.3 tank, and the ballast is salt water (1.025). During the exchange, 300 t of ballast stands transferred 12.0 m across the ship. Find the list, and the list that would have been predicted from the solid GM.

solid GM = 10.328 − 7.95 = 2.378 m

Σi = 2782 + 2782 + 2669 + 2669 = 10902 m⁴; ΣFSM = 10902 × 1.025 = 11175 t m; FSC = 11175 ÷ 29751 = 0.376 m

fluid GM = 2.378 − 0.376 = 2.002 m, say 2.00 m

GG_H = (300 × 12.0) ÷ 29751 = 0.121 m

tan(List) = 0.121 ÷ 2.002 = 0.0604, so List = 3.5°

The solid GM would have promised tan(List) = 0.121 ÷ 2.378 = 0.0509, only 2.9°. Half a degree of surprise on a calm day; the difference between prediction and reality grows exactly as fast as the slack surfaces do, and this is why ballast exchanges are sequenced one pair at a time.

Figure 9.2 Worked example 9.5 on the ladder: the booklet inertias, multiplied by the density of the ballast, lift G virtually by 0.38 m, and the fluid GM is what the ship actually shows.

Worked example 9.6

MV Ninja's tank tables list No.3 cargo hold as a salt water ballast space for heavy weather, with a free surface inertia of 27317 m⁴. To compare its free surface with that of the double bottom tanks, suppose the hold were slack in the condition of Worked example 9.5 (solid GM 2.378 m, ∆ 29751 t), and find the fluid GM, taking the displacement and KM as unchanged for the purpose of the comparison.

FSM = 27317 × 1.025 = 28000 t m; FSC = 28000 ÷ 29751 = 0.941 m

fluid GM = 2.378 − 0.941 = 1.44 m

One slack space, and almost a metre of metacentric height, nearly forty percent of the ship's stiffness, is gone: the hold's surface spans nearly the full breadth of the ship, its inertia is almost ten times that of the widest double bottom tank (2782 m⁴), and b³ is merciless. This is why hold ballast is carried pressed full or not at all, and why the filling and emptying stages are the anxious hours of the operation.

9.6 What matters, what does not, and how it ends

The formula asks only for the shape of the liquid surface and the density of the liquid. That has three consequences that catch people out. The depth of liquid scarcely matters: a few centimetres slopping across a wide tank top costs the same moment as a half full tank, so long as the surface spans the full breadth. The height of the tank in the ship does not matter at all: a slack topside tank and a slack double bottom of the same shape steal the same GM, because the correction is about the surface's freedom, not its position. And the effect ends only when the surface itself disappears: a pressed up tank has no surface to shift, an empty tank has nothing to shift, and everything in between pays in full.

Figure 9.6 The correction depends on the surface, not the quantity or the position. The only cures are full, empty, or subdivided.

The working discipline

Keep the number of slack tanks to the minimum the day's work allows; press up or strip the rest.
Take the inertia i of every slack tank from the booklet, multiply by the density of its contents, sum the moments, and divide by ∆ once.
Carry the correction as a fluid KG in the loading calculation, and check that against the maximum KG table.
Sequence ballast work one tank or one pair at a time, so the slack surfaces never accumulate.

One appointment remains for Volume One. Every KG in this book has stood on the light ship figure in the booklet, and that figure is not calculated but measured, once, with the ship newly built: a known weight is moved a known distance, the whisper of a list is read from a plumb line, and tan(List) = GG_H ÷ GM is run backwards to surrender the GM, and with it the light KG.

Figure 9.7 The inclining experiment: the list formula run backwards, the subject of the next chapter.

Interactive: the free surface calculator

Shape a tank, choose its liquid, add bulkheads, and watch the moment obey b³ and n² in real time.

length l (m) breadth b (m) b = 10.0 m ρ_T (t/m³)

bulkhead spaces n n = 1 ∆ (t) solid GM (m)

I total = – m⁴ FSM = – t m FSC = – m fluid GM = – m

Interactive: the fluid list laboratory

The Worked example 9.5 scenario, adjustable: sum the slack FSMs, shift some ballast, and compare the honest fluid list with the optimistic solid one.

∆ (t) solid GM (m) ΣFSM (t m) ΣFSM = 11175 t m

w shifted (t) across s (m)

FSC = – m fluid GM = – m solid list = – fluid list = –

Chapter summary

A slack tank's liquid runs downhill at every heel: the ship behaves as if G had risen by FSC = FSM ÷ ∆.
For a rectangular surface, I = (l × b³) ÷ 12 and FSM = I × ρ_T, with the tank liquid's own density.
Breadth is cubed: subdivision into n spaces divides the moment by n².
In service the booklet's i column is multiplied by the density of each slack tank's contents, the moments are summed and the sum is divided by ∆ once.
Fluid GM = solid GM − FSC; fluid KG = solid KG + FSC; the fluid figures are the ones the ship obeys.
Depth of liquid and height of tank scarcely matter; the effect ends only at pressed full, stripped empty, or subdivided.

Self test questions

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