Every calculation so far has assumed a whole hull. This chapter opens a compartment to the sea and asks whether the ship can live with it. The surprise, for a ship like MV Ninja, is how little the answer has to do with stability.
5.1 What actually changes
When a compartment is opened to the sea, three things change at once. The ship loses buoyancy, so she sinks deeper until the rest of the hull makes up the shortfall. She loses it at a particular place, so unless that place is at the centre of flotation she also trims, and unless it is on the centreline she also lists. And if the opened space reaches up through the waterline, she loses waterplane area, which reduces both BM and MCTC. The first effect is the obvious one. The third is usually what damages the stability. The second is what sinks ships.
Notice the distinction hidden in the third. A double bottom tank whose top is well below the waterline loses no waterplane area at all when bilged, because the water inside reaches the tank top and stops. A cargo hold open from the tank top to the deck loses the whole of its waterplane. Two compartments of similar volume can have entirely different effects on GM.
5.2 Permeability
A flooded compartment does not fill completely. The permeability is the proportion the sea can actually get into, and it is applied to the compartment volume, to its waterplane area and to its moment of inertia alike.
| space | permeability | regime |
|---|---|---|
| cargo, coal or stores | 0.60 | older SOLAS passenger ship rules (floodable length) |
| machinery spaces | 0.85 | older passenger ship rules and SOLAS II-1 probabilistic rules |
| passenger and crew spaces, accommodation, voids | 0.95 | both |
| spaces appropriated to stores | 0.60 | SOLAS II-1 probabilistic rules |
| dry cargo spaces | 0.70 at the deepest subdivision draught, more at lighter draughts | SOLAS II-1 probabilistic rules |
| spaces intended for liquids | 0 or 0.95, whichever is the worse | SOLAS II-1 probabilistic rules |
| loaded cargo hold of a bulk carrier | 0.90 | SOLAS chapter XII |
| empty cargo hold of a bulk carrier | 0.95 | SOLAS chapter XII |
The last line repays attention. A space intended for liquids is taken at whichever of nought and 0.95 gives the worse result, because a full tank floods not at all while an empty one floods completely, and the ship must survive either.
5.3 Two accounts of the same accident
Lost buoyancy treats the flooded space as no longer part of the ship. The displacement and the KG are unchanged; the new draught is the one at which the intact hull, less the flooded space, again displaces the original weight, and KB, BM and KM are all worked out afresh. There is no free surface correction, because the flooded space is not a tank in the ship; it is part of the sea.
Added weight treats the floodwater as a weight loaded aboard. The displacement rises, the KG moves towards the height of the floodwater, and the intact hydrostatics are used unaltered. A free surface correction must be applied, because by this accounting the floodwater is a slack tank.
Both give the same draught, the same trim and the same behaviour. They do not give the same GM, and they should not be expected to. GM is a lever measured against a displacement, and the two methods do not use the same displacement. What they agree on, exactly, is the product:
displacement × GM is the same by either method
5.4 Worked example 5.1: No.3 hold bilged
MV Ninja is in her summer departure condition, 30456 t on even keel at 9.600 m, solid KG 8.09 m, GM 2.240 m (the free surface of the intact tanks is left aside so that the damage effects stand alone; with it the figures are KG 8.113 m and GM 2.217 m, and every residual GM below would be 0.023 m less). Her side is opened in way of No.3 hold, which in this chapter is taken as a box 24.20 m long and 23.8 m broad between bulkheads at 68.70 and 92.90 m from the after perpendicular, with an effective flat floor 2.20 m above the keel; permeability 0.60.
By lost buoyancy the new draught is 10.421 m, a sinkage of 0.821 m, with 2841.2 m³ of buoyancy lost at Kg 6.311 m. The waterplane falls from 3468.1 to 3122.5 m². KB becomes 5.395, BM 4.820, KM 10.215, and with KG unchanged the residual GM is 2.125 m. By added weight, 2912.3 t of sea water comes aboard, the displacement rises to 33368.3 t, the KG falls to 7.9347 m, a free surface correction of 0.5011 m applies (i × 1.025, the floodwater being sea water), and the GM is 1.9399 m. Same draught, different number, same moment: 30456 × 2.1254 and 33368.3 × 1.9399 are both 64731 tonne metres.
The trim is only 72.4 centimetres by the head because No.3 hold lies close to the centre of flotation: the new centre of flotation is at 70.599 m, the damaged MCTC 410.5 t m/cm, the draughts 10.800 m forward and 10.076 m aft. The margin line, 13.424 m above the keel, is clear by 2.62 m. She is in no difficulty at all.
5.5 Worked example 5.2: No.1 hold bilged
Nothing changes except the position of the damage. No.1 hold is smaller than No.3, so less buoyancy is lost and every stability figure is very slightly better. The result is far worse.
| No.3 hold bilged | No.1 hold bilged | |
|---|---|---|
| volume of buoyancy lost, m3 | 2841.2 | 2416.5 |
| mean draught, m | 10.421 | 10.299 |
| residual GM, m | 2.125 | 2.198 |
| shift of the centre of flotation, m aft | 1.017 | 5.351 |
| lever from the new centre of flotation, m | 10.201 | 62.111 |
| MCTC, intact at that draught, t m/cm | 413.9 | 412.7 |
| MCTC, damaged, t m/cm | 410.5 | 339.0 |
| trim by the head, cm | 72 | 454 |
| draught forward, m | 10.800 | 12.804 |
| draught aft, m | 10.076 | 8.266 |
| clearance to the margin line, m | 2.62 | 0.62 |
The lever from the new centre of flotation is 62.1 metres instead of 10.2, and the damaged MCTC has fallen from 412.7 to 339.0 t m/cm because the lost waterplane is a long way from the centre of flotation and takes a great deal of longitudinal inertia with it (the parallel axis transfer of the intact waterplane to its new centroid, 5.35 m further aft, is part of that sum). The trim is 454 centimetres by the head. The bow goes to 12.804 m, within 0.62 m of the margin line, and the stern rises to 8.266 m, lifting a good part of the propeller and rudder out of the water.
The residual GM is 2.198 m, marginally better than in worked example 5.1, and of no help. This is the shape of many of the bulk carrier losses of the 1980s and 1990s: the ship does not capsize, she goes down by the head.
Where the hand method starts to strain
A trim of 4.54 metres is three per cent of the length between perpendiculars. The classical calculation assumes a single value of MCTC, a fixed centre of flotation and a waterplane that keeps its shape, and none of those assumptions holds well at a trim of that size. What the hand calculation does give, in ten minutes and without a computer, is the thing the officer needs to know at once: which end is going down, and roughly how far.
5.6 Worked example 5.3: damage off the centreline
No.2 double bottom water ballast tank on the starboard side (Appendix A: 471.3 m³, Kg 1.12 m, 105.39 m from the after perpendicular, 7.53 m off the centreline) is bilged, empty, permeability 0.95. Because the top of the tank is below the waterline no waterplane is lost: she sinks bodily 0.130 m to 9.730 m, where the full waterplane of 3446 m² serves the original volume. There is no loss of BM and no free surface in the added weight account either, the tank being pressed full to its top. The GM actually rises, from 2.240 to 2.382 m, because 447.7 m³ of buoyancy has been lost at Kg 1.12 m and replaced at the waterline, so B and M both rise. What she gets instead is a heel of 2.7 degrees towards the damage, well inside the fifteen allowed, and a trim of 38 cm by the head.
Volume is not what matters. Where the volume sits, relative to the waterplane and to the centre of flotation, is what matters.
5.7 Any one hold
SOLAS chapter XII requires bulk carriers of 150 metres in length and upwards, carrying solid bulk cargoes of density 1000 kg/m³ and above, to withstand the flooding of any one cargo hold when loaded to the summer load line, at a permeability of 0.90 for a loaded hold and 0.95 for an empty one. The length is the Load Line length, which for MV Ninja lies between her 148 metres between perpendiculars and her 152 metres overall, so she sits at the edge of that band, which makes her a useful ship to run the test on. The animation and the table use the classic 0.60.
| hold | lost m3 | draught m | trim cm | forward m | aft m | GM m | margin clear m (deeper end) | verdict at 0.60 |
|---|---|---|---|---|---|---|---|---|
| No.1 | 2416.5 | 10.299 | 454 | 12.804 | 8.266 | 2.198 | 0.62 | survives |
| No.2 | 2853.8 | 10.425 | 284 | 11.961 | 9.123 | 2.125 | 1.46 | survives |
| No.3 | 2841.2 | 10.421 | 72 | 10.800 | 10.076 | 2.125 | 2.62 | survives |
| No.4 | 2788.7 | 10.406 | -115 | 9.823 | 10.977 | 2.127 | 2.45 | survives |
| No.5 | 2631.3 | 10.361 | -307 | 8.857 | 11.931 | 2.193 | 1.49 | survives |
At 0.60 she keeps the margin line clear in every hold, but the margin narrows sharply towards the ends: 2.62 metres in hand with No.3 flooded, 1.49 with No.5 (at the stern, which is now the deeper end), 0.62 with No.1. The residual GM meanwhile varies only between 2.125 and 2.198 metres across all five cases. At the chapter XII permeability of 0.90 the same calculation puts the bow 2.1 m under the margin line with No.1 flooded and 0.2 m under with No.2, and the stern 0.2 m under with No.5. The position of the damage decides the outcome, the permeability decides the margin, and the residual stability never enters the argument.
5.8 The criteria, and what a hand method can settle
After damage the ship must satisfy all of the following in her final equilibrium condition. The list is the one commonly taught; the figures were not all in one instrument (the margin line, the factor of subdivision and the 0.05 m residual GM are from the older SOLAS passenger ship rules, the 15 or 17 degrees, 0.10 m and 20 degrees from the Load Line and cargo ship criteria), and the current text of the instrument that applies to a ship governs.
| criterion | requirement | settled by hand? |
|---|---|---|
| the margin line | not immersed at any point | yes |
| angle of heel from unsymmetrical flooding | not more than 15 degrees, or 17 where no part of the deck is immersed | yes |
| residual metacentric height | not less than 0.05 m by lost buoyancy (passenger ship rules); positive (Load Line rules) | yes |
| maximum residual righting lever | not less than 0.10 m | yes, up to the angle at which the deck edge immerses |
| range of the residual curve | not less than 20 degrees | no, it needs damaged cross curves |
| openings | no opening through which progressive flooding may occur to be immersed | yes, from the arrangement |
By hand the damaged righting lever is estimated from the wall sided formula using the damaged GM and BM, and it is reliable only as far as the angle at which the deck edge enters the water. For worked example 5.1 that angle is 14.3 degrees at the mean draught of 10.421 m, and the residual lever there is already 0.563 metres, more than five times the tenth of a metre required. The range criterion cannot be settled this way and this chapter does not pretend otherwise.
5.9 Floodable length, the margin line and the factor of subdivision
The margin line is drawn along the ship’s side 76 millimetres below the upper surface of the bulkhead deck at side. For MV Ninja that puts it at 13.424 m above the keel. The floodable length at any point is the greatest length of compartment, centred there, which may be opened to the sea without immersing the margin line. The permissible length is the floodable length multiplied by a factor of subdivision, and it is the permissible length that governs where the bulkheads may go.
Worked at No.1 hold, loaded, permeability 0.60, her floodable length is 25.34 metres. No.1 hold is 22.40 metres long, so she has 2.94 metres in hand and no more. A factor of subdivision of 0.60 would give a permissible length of 15.20 metres, and at the chapter XII permeability of 0.90 the floodable length falls to 16.91 m, shorter than the hold. That comparison is why bulk carriers are not built like passenger ships, and why the regulatory answer was SOLAS chapter XII rather than more bulkheads.
Chapter 5 in seven lines
- Damage loses buoyancy, loses it at a place, and if the space reaches through the waterline loses waterplane area. The third hurts GM, the second sinks ships.
- Permeability is the proportion of a space the sea can get into. Spaces for liquids are taken at nought or 0.95, whichever is worse.
- Lost buoyancy keeps the displacement and the KG. Added weight keeps the hull form, and needs a free surface correction.
- The two methods give different GM and the same displacement times GM: 64731 t m for No.3 hold bilged. The identity is exact.
- Worked example 5.2: No.1 hold flooded gave a residual GM of 2.198 m, better than the amidships case, and put the bow within 0.62 m of the margin line.
- A tank whose top is below the waterline loses no waterplane area, so it costs no BM.
- Her floodable length at No.1 hold is 25.34 m against a hold 22.40 m long. A factor of subdivision of 0.60 would allow 15.20 m.
Test yourself
Questions
- Define permeability, and state the values customarily used for machinery spaces, for dry cargo spaces and for spaces intended for liquids. Explain the rule for the last of these.
- Set out the lost buoyancy and added weight methods side by side, stating for each what is held constant and what has to be recalculated.
- Explain why the two methods give different values of GM for the same damage, and state the quantity on which they must agree.
- MV Ninja is bilged in No.3 hold in the condition of worked example 5.1. Starting from the lost volume at the intact waterline, 2557.4 m3, and the area lost, 345.6 m2, verify the first pass sinkage of 0.826 m, and explain why the exact answer, 0.821 m, needs a second pass for a ship when it did not for a box.
- A double bottom tank whose top lies well below the waterline is bilged. Explain why the waterplane area is unchanged, and state the consequences for BM and for the free surface correction in each of the two methods.
- Explain why flooding No.1 hold is far more dangerous than flooding No.3 hold, even though less buoyancy is lost, and identify the two quantities in the calculation that account for the difference.
- Define the margin line and state its position. Explain why it is the criterion that decides the outcome for a bulk carrier while the residual GM criterion almost never does.
- Define floodable length and permissible length, and explain the part played by the factor of subdivision.
- MV Ninja’s floodable length at No.1 hold is 25.34 m at permeability 0.60 and the hold is 22.40 m long. Comment on this result, state what it becomes at the chapter XII permeability of 0.90, and explain what SOLAS chapter XII requires of larger bulk carriers instead of further subdivision.
- A candidate calculates a damaged condition by hand and obtains a trim of six metres. State what he should conclude, and what he should do next.
- In the summer departure condition No.5 hold (24.10 m long, 22.3 m broad, plan area 537.4 m2, centre 32.85 m from the after perpendicular, floor 2.20 m) is bilged, permeability 0.60. Find the damaged mean draught, the residual GM and the draughts at the perpendiculars. (Answer: 10.361 m, 2.193 m, 8.857 m forward and 11.931 m aft, the trim being 307 cm by the stern.)
- Both No.2 double bottom tanks, port and starboard, empty, are bilged at permeability 0.95 in the summer departure condition. Find the sinkage and the residual GM, and explain why there is no heel. (Answer: 0.259 m, 2.526 m.)
- For the damaged condition of worked example 5.2 (mean draught 10.299 m, GM 2.198 m, BM 4.946 m), find the angle at which the deck edge immerses at the mean draught and the residual righting lever there by the wall sided formula, and say why the answer is less useful than it looks. (Answer: 14.8 degrees, 0.607 m; at the forward perpendicular the freeboard is only 0.70 m.)
Looking ahead
This chapter asked a deterministic question: open this compartment, and show that she survives it. The answer was a yes or a no, and the yes depended entirely on which compartment was chosen. That is both the strength of the method and its weakness. Chapter 6 replaces the question with a different one. Over the whole range of damages that could actually happen to a ship of this form, weighted by how likely each one is, what fraction does she survive? The answer is a number between nought and one, the attained subdivision index A, which must reach a required index R. The hull is the same and the flooding is the same. Only the question has changed.