Three chapters have now leaned on interpolation without ever pausing over it: Chapter 7 entered the tables twice per problem, Chapter 8 entered them in reverse, and marks have quietly depended on fractions computed in the margin. This chapter slows the hand down once, so that everywhere else it can be fast: how the fence between the rows works, where its traps lie, and the two named ideas, the layer correction and the mean TPC rule, that the examination expects by name.
The hydrostatic booklet answers only at its own rows: 8.60 m, 8.80 m, 9.00 m and so on, every 20 cm. The ship almost never floats on a row. Interpolation is the standing agreement that over one panel of the fence every column may be treated as a straight line: false in principle, since hull forms curve, but false by millimetres over 20 cm, and the whole edifice of Chapters 7 and 8 stands on it.
The whole craft is one number: the fraction of the panel the ship has climbed, f = (value − lower row) ÷ (upper row − lower row). Compute it once, to three decimals, and run it through every column with the same rule: interpolated value = lower + f × (upper − lower). The bracket carries its own sign, which is the whole defence against the trap of Section 9.3.
MV Ninja floats on an even keel at 8.73 m. Produce her complete hydrostatic row.
f = (8.73 − 8.60) ÷ (8.80 − 8.60) = 0.13 ÷ 0.20 = 0.65: written once, used seven times.
| Column | at 8.60 m | at 8.80 m | gap | + 0.65 of gap | at 8.73 m |
|---|---|---|---|---|---|
| Δ (t) | 26941 | 27641 | +700 | +455 | 27396 |
| TPC | 34.92 | 34.98 | +0.06 | +0.04 | 34.96 |
| MCTC (t m) | 394.5 | 397.0 | +2.5 | +1.6 | 396.1 |
| KM (m) | 10.371 | 10.352 | −0.019 | −0.012 | 10.359 |
| KB (m) | 4.510 | 4.616 | +0.106 | +0.069 | 4.579 |
| LCB (m foap) | 77.51 | 77.38 | −0.13 | −0.08 | 77.43 |
| LCF (m foap) | 72.32 | 72.20 | −0.12 | −0.08 | 72.24 |
Every answer lands between its own two posts, which is the instant sanity check: an interpolated value outside its bracket is always an arithmetic slip, never a property of the ship.
Between 8.60 and 8.80 m, four of MV Ninja’s columns rise and three fall: KM, LCB and LCF all shrink as she sinks, KM because BM = I ÷ V falls faster than KB rises until KM reaches its minimum of 10.328 m at the 9.40 m row and turns up again, LCB and LCF because the hull fills out aft faster than forward as the waterline rises, so both centroids move aft at every step of the table. The formula is immune, because the bracket (upper − lower) comes out negative by itself; the sailor is not, because the tempting shortcut, take the gap off the table and add it, silently assumes every column climbs. The examiner knows which columns to set.
After loading, MV Ninja’s displacement is 28000 t. Find her TMD, MCTC, LCB and LCF (rows: 8.80 m: 27641 t, MCTC 397.0, LCB 77.38, LCF 72.20; 9.00 m: 28343 t, MCTC 399.3, LCB 77.26, LCF 72.10).
The known quantity is now in the displacement column, so the fraction is built in tonnes: f = (28000 − 27641) ÷ (28343 − 27641) = 359 ÷ 702 = 0.511.
TMD = 8.80 + 0.511 × 0.20 = 8.902 m; MCTC = 397.0 + 0.511 × 2.3 = 398.2 t m; LCB = 77.38 − 0.511 × 0.12 = 77.32 m; LCF = 72.20 − 0.511 × 0.10 = 72.15 m.
Nothing else changed: one fence, two gates. Enter by draught when the draught is known; enter by displacement when the weight is known, which after any big loading it is.
Chapter 6 caught the chief officer declaring the ship legal on an arithmetic mean of 9.600 m while the true mean stood at 9.615 m. That 1.5 cm slice between the two means has a name in the trade: the layer correction, and it has a closed formula, because both means live on the same straight waterline.
Derive the layer correction, evaluate it for the Chapter 6 condition (trim 1.04 m by the stern, LCF 71.84 m foap, TPC 35.28), and reconcile.
AMD is the waterline’s height at midships, 74.00 m foap; TMD is its height at F, 71.84 m foap. Between two points on a straight waterline the draught changes by trim × separation ÷ LBP, so:
Layer = TMD − AMD = Trim × (midships − LCF) ÷ LBP = 1.04 × 2.16 ÷ 148 = 0.0152 m: positive for stern trim with F abaft of midships, so the truth is deeper than the average.
In tonnes, carrying the unrounded layer of 1.518 cm: 1.518 × 35.28 = 53.5 t, in round figures 54 t: Chapter 6’s contraband. (The rounded 1.5 cm × 35.28 would give only 52.9 t; the 54 t needs the unrounded layer. The displacement column says the same: 30509 t at 9.615 m against 30456 t at 9.600 m, 53 t.) The draught survey form of Chapter 10 carries this same quantity as its printed “first trim correction”, and it will be met there again with its own paperwork.
TPC is itself a row value: it drifts as the ship sinks. For a small parcel the drift is invisible; pile on thousands of tonnes and holding the initial TPC quietly assumes the waterplane never grew. The course rule is exact on this: large weights go through the displacement column, and if TPC must serve, use the mean of the initial and final values. One worked example proves the whole footnote.
From TMD 8.509 m (Δ 26623 t, TPC 34.87), MV Ninja loads 2500 t. Find the new TMD three ways: initial TPC held; mean TPC; and by the displacement column (rows 9.20 m: 29046 t, TPC 35.16; 9.40 m: 29751 t, TPC 35.22).
Initial TPC held: sinkage = 2500 ÷ 34.87 = 71.7 cm, so TMD = 8.509 + 0.717 = 9.226 m.
Displacement column (the truth): final Δ = 26623 + 2500 = 29123 t; f = (29123 − 29046) ÷ 705 = 77 ÷ 705 = 0.109, so TMD = 9.20 + 0.109 × 0.20 = 9.222 m, where the final TPC reads 35.16 + 0.109 × 0.06 = 35.17.
Mean TPC: (34.87 + 35.17) ÷ 2 = 35.02; sinkage = 2500 ÷ 35.02 = 71.4 cm; TMD = 9.223 m.
Verdict: holding the initial TPC overshoots the truth by 0.4 cm, about 14 t of apparent cargo that does not exist; the mean closes to 0.1 cm. Over 2500 t the waterplane grew, and the displacement column is the only witness that watched it grow.
Speed drill, no new ideas: MV Ninja floats at 8.61 m forward, 9.23 m aft. Produce her displacement and LCG (rows as tabled above for 8.80 and 9.00 m).
Trim 0.62 m by the stern; AMD 8.92 m; first pass LCF at 8.92 m: 72.20 − 0.6 × 0.10 = 72.14 m; TMD = 9.23 − (0.62 × 72.14 ÷ 148) = 8.928 m.
f = (8.9278 − 8.80) ÷ 0.20 = 0.639, carrying the unrounded TMD (the rounded 8.928 m gives 0.640 and the same printed results): Δ = 27641 + 0.639 × 702 = 28090 t; MCTC = 397.0 + 0.639 × 2.3 = 398.5 t m; LCB = 77.38 − 0.639 × 0.12 = 77.303 m.
Separation of B and G = trim (cm) × MCTC ÷ Δ = 62 × 398.5 ÷ 28090 = 0.880 m; stern trim, so G is abaft B and LCG = 77.303 − 0.880 = 76.42 m foap. A second reading of LCF at 8.928 m gives 72.14 m again, so nothing changes; entered at the AMD of 8.92 m the table would have read 28062 t, 28 t short.
Every tool in this chapter fired once: TMD by the layer geometry, one fraction through four columns, two of them falling, and the row closed with Chapter 7’s formula. Under examination conditions this is three minutes, and all of it is checkable at a glance against the posts.
The booklet is a fence: bracket the ship with her two rows before touching a calculator.
One fraction, three decimals, every column: value = lower + f × (upper − lower), and the bracket carries the sign.
KM, LCB and LCF fall as she sinks: never add a gap by reflex, and check every answer lies between its posts.
Layer correction = trim × (midships − LCF) ÷ LBP: the named slice between the AMD and the TMD.
Large weights go by the displacement column; if TPC must serve, use the mean of initial and final.