A compartment is bilged when it is holed and floods freely to the sea. The water inside is not cargo and it is not ballast: it comes and goes with the damage, and its level always matches the sea outside. Because of that, the examination treats bilging by the lost buoyancy method: the ship’s weight never changes, but the flooded space stops supporting her. Examination bilging problems use box shaped hulls, so this chapter works on a box with MV Ninja’s length and beam: 148 m × 24.2 m, depth 13.5 m, floating at 4.000 m even keel in salt water, KG 12.80 m. Displacement is 148 × 24.2 × 4.0 × 1.025 = 14684.6 t, the volume of displacement is 14326.4 m³, and both stay at those numbers through every problem below: the lost buoyancy method never changes the weight of the ship, so it never changes the volume she must displace, and KG is unchanged too. The KG is deliberately high, close to the deck, so that the loss of stability shows clearly; a box this size at so light a draught would otherwise have a GM of many metres.
The flooded compartment no longer holds the ship up. She must recover exactly that much buoyancy, and the only place it can come from is the intact part of the hull sinking deeper. So the lost volume equals the sinkage times the intact waterplane area, and every bilging problem starts by turning that sentence around.
A compartment symmetrical about midships and about the centreline sinks the ship bodily: no trim, no list. That does not make it harmless. The waterplane has lost a slice of its length, BM is built on the waterplane, and KB has risen with the draught. The GM has to be recomputed, and it usually comes out smaller.
An empty midship compartment 16 m long, full beam, is bilged. Find the new draught and the new GM.
Sinkage = (16 × 24.2 × 4.0) ÷ ((148 − 16) × 24.2) = 0.485 m, so the new draught is 4.485 m.
Before: KB = 2.000 m; BM = LB³ ÷ 12V = 148 × 24.2³ ÷ 12 ÷ 14326.4 = 12.201 m; KM = 14.201 m; GM = 14.201 − 12.80 = 1.401 m.
After: KB = half the new draught = 2.242 m. BM uses the intact waterplane but the original volume: BM = 132 × 24.2³ ÷ 12 ÷ 14326.4 = 10.882 m; KM = 13.124 m; GM = 0.324 m.
More than three quarters of the GM is gone with the ship still upright and level. Midship damage looks calm and is not.
A hold full of cargo floods far less than an empty one, because the cargo already fills most of the space. The fraction the water can actually use is the permeability. The cargo’s own density sets the solid part; the gaps between packages, the broken stowage, set the rest. The neatest way to use it is the effective length: multiply the compartment length by the permeability and run the ordinary sinkage formula.
The same 16 m compartment is instead full of cargo of relative density 0.90 and stowage factor 1.48 m³/t. Find the new draught and the new GM if it is bilged.
Solid stowage factor = 1 ÷ 0.90 = 1.111 m³/t. Broken stowage = 1.48 − 1.111 = 0.369 m³/t. Permeability = 0.369 ÷ 1.48 = 0.2492 = 24.9%.
Effective length = 16 × 0.2492 = 3.99 m.
Sinkage = (3.99 × 4.0) ÷ (148 − 3.99) = 0.111 m; new draught 4.111 m.
For the GM the effective length is removed from the waterplane too: new KB = 4.111 ÷ 2 = 2.055 m; BM = (148 − 3.99) × 24.2³ ÷ 12 ÷ 14326.4 = 11.872 m; GM = 2.055 + 11.872 − 12.80 = 1.127 m, against 0.324 m for the same compartment empty.
Empty, this hold sank her 49 cm and left 0.324 m of GM; loaded, 11 cm and 1.127 m. Cargo is not protection, but it is time.
Bilge a compartment at the bow and the lost buoyancy sits as far from the centre of flotation as it can get. The sinkage is modest; the trimming moment is not. The method is the Chapter 7 engine with two changes: F is now the centroid of the intact waterplane, and the MCTC belongs to that intact waterplane too. For the box: F sits at the middle of the intact length, and MCTC = W × BML ÷ 100L with BML = B(L − l)³ ÷ 12 ÷ V, where L is still the full length between perpendiculars and V the original volume. Strictly MCTC = W × GML ÷ 100L; the examination convention takes GML ≈ BML, because BML is hundreds of metres and KG − KB only a few.
An empty forward end compartment 12 m long, full beam, is bilged. Find the final draughts.
Sinkage = (12 × 24.2 × 4.0) ÷ (136 × 24.2) = 0.353 m. Lost buoyancy = 12 × 24.2 × 4.0 × 1.025 = 1190.6 t.
The intact waterplane runs from the stern to 136 m foap, so its centroid F is at 68 m foap. The compartment centroid is at 142 m foap: lever = 74 m. Trimming moment = 1190.6 × 74 = 88104 t m.
MCTC of the intact waterplane: BML = 24.2 × 136³ ÷ 12 ÷ 14326.4 = 354.09 m; MCTC = 14684.6 × 354.09 ÷ (100 × 148) = 351.3 t m per cm. Change of trim = 88104 ÷ 351.3 = 250.8 cm by the head.
Shares about F at 68 m foap: forward (80 ÷ 148) × 250.8 = +135.6 cm; aft (68 ÷ 148) × 250.8 = −115.2 cm.
FINAL: forward 4.000 + 0.353 + 1.356 = 5.709 m; aft 4.000 + 0.353 − 1.152 = 3.201 m. A third of a metre of sinkage and two and a half metres of trim: with end damage, the trim is the story. (Had GML been used exactly, 2.176 + 354.09 − 12.80 = 343.5 m, the MCTC would be 340.8 and the trim 258.5 cm, 8 cm more; the examination answer is the BML one.)
Bilge a compartment on one side and the same logic runs across the ship instead of along it. The intact waterplane is now lopsided, so its centroid F, and with it the axis the ship rolls about, shifts away from the damage; the centre of buoyancy shifts the same distance. G has not moved, so the ship lists towards the damaged side until B comes back under G. The working needs one new tool: the moment of inertia of the lopsided waterplane about its new axis, found by taking it about the damaged side first and transferring with the parallel axis rule. Five steps, always in the same order: sinkage, shift, inertia, new GM, list.
An empty amidships side compartment, 14 m long and 8 m wide, lies against the port shell inboard to a longitudinal bulkhead. It is bilged. Find the list.
Step 1, sinkage: a = 14 × 8 = 112 m²; intact area = 3581.6 − 112 = 3469.6 m²; sinkage = 112 × 4.0 ÷ 3469.6 = 0.129 m; new draught 4.129 m; new KB = 2.065 m.
Step 2, the shift: the compartment centroid is 12.1 − 4.0 = 8.1 m from the centreline. Shift of F and of B = 112 × 8.1 ÷ 3469.6 = 0.2615 m to starboard.
Step 3, inertia about the new axis: about the port shell, I = 148 × 24.2³ ÷ 3 − 14 × 8³ ÷ 3 = 699176 − 2389 = 696787 m⁴. The new axis lies 12.1 + 0.2615 = 12.3615 m from the port shell, so I = 696787 − 3469.6 × 12.3615² = 166611 m⁴.
Step 4, new GM: BM = 166611 ÷ 14326.4 = 11.630 m; KM = 2.065 + 11.630 = 13.694 m; GM = 13.694 − 12.80 = 0.894 m.
Step 5, the list: tan of the list = 0.2615 ÷ 0.894 = 0.2924, so the list is 16.3 degrees to port, towards the damage. At sea this would be a serious situation: with a list of 16 degrees and a hole in the side, any opening on the low side that is not closed is a route for progressive flooding. The list formula treats GM as constant, which at 16 degrees is an approximation; the true angle would come from the damaged GZ curve.
A watertight flat caps the flooding. What is lost depends on where the flat sits and on which side of it the hole is: bilged below a flat that is under the waterline, only the space below the flat floods and the whole waterplane keeps working; bilged above it, the lost volume runs from the flat to the waterline and the waterplane loses its slice as usual; and with the flat above the waterline, check whether the sinkage takes the ship past the flat before finishing the sum.
A midship double bottom tank, 16 m long, full beam and 1.4 m deep, is bilged below its watertight tank top. Find the new draught and the new GM.
Lost volume = 16 × 24.2 × 1.4 = 542.08 m³. The tank top holds, so the waterplane is untouched: the intact area is the full 3581.6 m².
Sinkage = 542.08 ÷ 3581.6 = 0.151 m; new draught 4.151 m.
No trim, no list, and BM unchanged at 12.201 m because the whole waterplane still works over the same volume. KB has moved, though: the intact volume is the whole box to 4.151 m (14868.5 m³, centroid 2.0757 m) less the tank (542.08 m³ at 0.7 m), so new KB = (14868.5 × 2.0757 − 542.08 × 0.7) ÷ 14326.4 = 2.128 m; KM = 2.128 + 12.201 = 14.329 m; GM = 14.329 − 12.80 = 1.529 m, a rise of 0.128 m. Buoyancy lost low in the ship is regained at the waterline, so B rises and, with BM unchanged, M rises with it.
The flat has turned major damage into fifteen quiet centimetres and a small gain in stability, which is exactly what it is there for. The same 16 m of hull bilged above that tank top would lose the waterplane slice and leave a GM of only 0.184 m.
Bilging is lost buoyancy at constant displacement, so displacement, volume of displacement and KG are unchanged: sinkage = lost volume ÷ intact waterplane area.
A midship compartment sinks her level but rebuilds the GM: new KB from the new draught, new BM from the intact waterplane over the original volume.
Permeability = broken stowage ÷ stowage factor; use it as an effective length and run the ordinary formula.
End damage: F and MCTC belong to the intact waterplane; the trimming moment is lost buoyancy times its lever to the new F.
Side damage: shift = a × s ÷ intact area; inertia about the new axis by the parallel axis rule; tan of the list = shift ÷ new GM.