Where every other calculation starts. Get displacement, the waterplane and Simpson's rules straight, and stability, strength and trim all become arithmetic you already know how to do.
A floating body pushes aside water until the weight of the water pushed aside equals its own weight. That single sentence is Archimedes' principle, and everything in this chapter is a consequence of it.
Two quantities come out of it, and it matters that you keep them separate. The volume of displacement, written ∇, is the volume of hull below the waterline — pure geometry. The displacement, written Δ, is the weight of the water that volume holds, which depends on how dense the water is.
So a ship moving from the sea into a river does not get heavier or lighter. Δ stays at whatever the ship weighs. What changes is ∇: fresh water is less dense, so a larger volume is needed to hold the same weight, and the ship settles deeper until it has one.
There is a second condition hiding behind flotation. The weight acts down through the centre of gravity G, and the buoyancy acts up through the centre of buoyancy B — the centroid of the underwater volume. For the ship to float without trimming, those two must be in the same vertical line. Load a weight forward and G moves forward; the ship trims until B has moved forward to sit under it again. That is the whole mechanism behind the trim calculation later in this chapter.
Δ is set by what the ship weighs. ∇ is set by the hull and the draught. Density is the exchange rate between them, and trim is the ship rearranging B to sit under G.
Form coefficients compare the actual hull with a box of the same length, breadth and draught. Each one answers a different question, and knowing which is which is worth easy marks.
The relation CP = CB/CM falls straight out of the definitions and is quoted often enough to be worth memorising rather than deriving under time pressure.
A useful sense check: CW is always larger than CB for a normal ship, because the waterplane is the fullest horizontal slice of the hull while the block coefficient is dragged down by the fined ends below the water.
The waterplane is only a flat shape, but almost every hydrostatic quantity you will be asked for comes from it.
Its area gives TPC. Sink the ship by one centimetre and you add a slab of underwater volume one centimetre thick with the area of the waterplane. The mass of water in that slab is the tonnes needed to sink the ship by that centimetre.
Its second moment of area gives BM. Heel the ship a little and the wedge of buoyancy that emerges on one side and immerses on the other shifts B sideways. How far it shifts depends on how much area the waterplane has far from the centreline — which is exactly what the second moment of area measures.
Because IT depends on breadth cubed, a beamier ship is dramatically stiffer — this is why beam, not depth, is the lever a designer pulls for stability.
Its centroid gives F, the centre of flotation. This is the point about which the ship trims. It is normally aft of amidships on a merchant hull, because the after waterplane is fuller than the fine forward end. Getting this wrong is the single commonest error in trim problems, and it is dealt with in section 5.
Area → TPC. Second moment → BM. Centroid → F. Three different properties of the same flat shape, answering three different questions.
You are given a table of offsets and asked for an area. Simpson's rules fit a smooth curve through the points and integrate it exactly — the first rule fits a parabola through every three points, the second fits a cubic through every four.
Count intervals, not ordinates. Seven ordinates is six intervals — even, so the first rule works. Six ordinates is five intervals, and neither rule applies without splitting the range or adding a half-ordinate near the end.
The same machinery gives moments, and this is where most of the marks are. Multiply each ordinate by its lever from the reference axis before summing, and you get the first moment; multiply by the lever squared and you get the second moment. Divide the first moment by the area and you have the centroid — that is how LCF is found.
One habit prevents most errors: lay the work out as a table with a column each for ordinate, multiplier, product, lever and moment. Examiners award method marks for a correct table even when the arithmetic goes astray at the end.
Loading a weight anywhere other than at F does two things at once, and the way to keep it straight is to do them one at a time.
Step one — put the weight at F. The ship sinks bodily, parallel to its original waterline, with no change of trim at all. How far it sinks is the weight divided by TPC.
Step two — slide the weight from F to where it really is. Shifting a weight that is already on board changes no displacement, so there is no further sinkage. What it produces is a trimming moment, w × d, where d is measured from F.
Then the trim has to be shared between the two ends. The ship pivots about F, so the end further from F moves more. If F is 3 m aft of amidships on a 120 m ship, it is 63 m from the forward perpendicular and 57 m from the after perpendicular, and the trim divides in that ratio.
Finally, combine: add the bodily sinkage to both draughts, then apply the trim change — one end up, the other down. A quick check at the end catches most slips. Did the mean draught increase for a weight added? Does the change of trim you calculated match the difference between the new draughts?
Sinkage first, at F. Then trim, about F. Never both at once, and never split the trim equally unless F really is amidships.
Everything here follows from one fact established in section 1: the ship's weight does not change when the water does.
Move from sea water to fresh, and the same 9000 t of ship must now be supported by water that is 2.5 % less dense. The volume of displacement therefore rises by about 2.5 %, and the ship sinks until it has found that extra volume.
The general result is the fresh water allowance — the sinkage in going from standard sea water to fresh water.
For dock water somewhere between the two, scale the allowance by how far the density has actually dropped:
This is the reason load line marks include a fresh water allowance above the summer line. A ship loading in a river to her fresh water mark will rise to exactly her summer marks by the time she reaches the open sea, having burned nothing and discharged nothing.
Three problems in the style GATE actually sets. Work each one on paper before reading the steps — recognition is not the same as being able to execute under time pressure.
A ship 120 m long has half-ordinates of the load waterplane, measured at equally spaced stations from the after perpendicular, of 0.5, 4.2, 6.8, 7.5, 7.1, 5.0 and 1.2 m. Find the waterplane area and the TPC in sea water.
L = 120 m, 7 half-ordinates ⇒ 6 equal intervals h = 120/6 = 20 m ρ = 1.025 t/m³
Find the waterplane area and the TPC in sea water
Seven ordinates means six intervals.
| Station | Half-ordinate y (m) | SM | Product |
|---|---|---|---|
| 0 | 0.5 | 1 | 0.5 |
| 1 | 4.2 | 4 | 16.8 |
| 2 | 6.8 | 2 | 13.6 |
| 3 | 7.5 | 4 | 30.0 |
| 4 | 7.1 | 2 | 14.2 |
| 5 | 5.0 | 4 | 20.0 |
| 6 | 1.2 | 1 | 1.2 |
| Σ | 96.3 |
An even number, so the first rule applies directly. Line up the Simpson multipliers 1, 4, 2, 4, 2, 4, 1 against the half-ordinates and sum the products.
Apply the rule.
This gives the area of one half of the waterplane, because the offsets given are half-breadths measured from the centreline.
Double it for the full waterplane.
TPC is the mass of water in a one-centimetre-thick slab of that waterplane.
One centimetre is 1/100 of a metre, so the volume is A_W/100 and the mass is that times the density.
AnswerA_W = 1284 m², TPC = 13.16 t/cm
The trap: offset tables give half-breadths almost every time. Forget the factor of two and every answer downstream — TPC, MCT, BM — is exactly half what it should be, and nothing in the arithmetic looks wrong.
A ship 120 m long floats at draughts of 6.40 m forward and 6.90 m aft. Displacement is 12 000 t, TPC is 18 t/cm and MCT1cm is 150 t·m/cm. The centre of flotation is 3 m aft of amidships. A weight of 90 t is loaded 40 m forward of amidships. Find the new draughts.
L = 120 m, d_F = 6.40 m, d_A = 6.90 m Δ = 12 000 t, TPC = 18 t/cm, MCT1cm = 150 t·m/cm LCF = 3 m aft of amidships, w = 90 t at 40 m forward of amidships
Find the new draughts
Split the problem in two.
Imagine the weight first placed at F: the ship sinks bodily with no trim at all. Then imagine it slid forward from F to its real position: that shift produces a pure trimming moment with no further sinkage.
Find the lever.
The weight is 40 m forward of amidships and F is 3 m aft of amidships, so the distance from F is the sum, not the difference.
Trimming moment and the resulting trim.
The weight is forward of F, so the ship trims by the head.
Split the trim between the two ends in proportion to their distance from F — not equally.
F is 3 m aft of amidships, so it is 63 m from the forward perpendicular and 57 m from the after perpendicular.
Add the bodily sinkage to both ends.
| Forward (m) | Aft (m) | |
|---|---|---|
| Initial draught | 6.400 | 6.900 |
| Bodily sinkage | +0.050 | +0.050 |
| Trim change | +0.135 | −0.123 |
| New draught | 6.585 | 6.827 |
Then apply the trim change: plus forward, minus aft.
Check the result before moving on.
The trim has changed from 0.50 m by the stern to 0.242 m by the stern — a change of about 0.258 m, which matches the 25.8 cm calculated. Displacement has increased by 90 t, and both draughts increased on average, which is right for a weight added.
Answerd_F = 6.59 m, d_A = 6.83 m
The trap: splitting the trim equally, 12.9 cm at each end. That is only correct when F is exactly amidships, which it almost never is. The examiner puts F off-centre precisely to see whether you noticed.
A ship of 9000 t displacement with a TPC of 15 t/cm moves from sea water into fresh water. How much does she sink, and why does the standard formula work?
Δ = 9000 t, TPC = 15 t/cm (sea water) ρ_sea = 1.025 t/m³, ρ_fresh = 1.000 t/m³
How much does she sink, and why does the standard formula work?
Start from what does not change.
The ship's weight is the same in both waters, so the buoyancy force is the same, so the mass of water displaced is the same 9000 t. What must change is the volume, because fresh water is less dense.
That extra volume has to come from the layer of hull between the old and new waterlines.
Its volume is the waterplane area times the sinkage.
Now the shortcut.
Doing that algebra in general gives the fresh water allowance directly — and because sea water is 1.025 and fresh is 1.000, the ratio always works out to the same constant.
Both routes agree.
Which is the point: the formula is not a magic number, it is this calculation done once and for all. For dock water of intermediate density, scale the allowance by how far the density has fallen.
AnswerThe ship sinks 150 mm
The trap: assuming displacement changes with density. It does not — the ship still weighs 9000 t. Only the volume of water needed to support that weight changes, and the draught follows the volume.
Δ = ρ∇Weight = density × volume; only ∇ changes with densityC_B = ∇/(L·B·T) C_M = A_M/(B·T)C_P = C_B/C_M, C_W = A_W/(L·B)TPC = A_W·ρ/100Tonnes to sink the ship 1 cmBM_T = I_T/∇ BM_L = I_L/∇I_T about the centreline, I_L about a transverse axis through FMCT1cm = Δ·GM_L/(100·L)GM_L ≈ BM_L in practiceSinkage = w/TPCBodily, for a weight placed at FTrim = w·d/MCT1cmd measured from F, split in the ratio of F to each perpendicularA = (h/3)(y₀ + 4y₁ + 2y₂ + … + yₙ)Simpson 1 — even number of intervalsA = (3h/8)(y₀ + 3y₁ + 3y₂ + 2y₃ + …)Simpson 2 — intervals in multiples of threeFWA (mm) = Δ/(4·TPC)DWA = FWA × (1025 − ρ_dock)/25