Five chapters of this volume have heeled MV Ninja sideways. This one tips her lengthways instead, and discovers that the longitudinal problem is the transverse one wearing different units: trim is a list measured in metres of draught difference, the centre of flotation plays the pivot, and MCTC is the exchange rate between moments and centimetres. One new habit matters more than any formula: every distance in a trimming calculation is measured from F, and F is not amidships.
Trim is the difference between the draughts at the forward and after perpendiculars: draughts of 9.08 m forward and 10.12 m aft make a trim of 1.04 m by the stern; equal draughts make even keel. Like a list, a trim is the ship’s answer to her weights: she trims until the longitudinal centre of buoyancy stands back under the longitudinal centre of gravity, and Chapter 7 will make that sentence into a calculation engine. This chapter builds its three parts: the pivot, the exchange rate, and the sharing rule.
A trimming ship rotates about the centroid of her waterplane: the centre of flotation, F, tabulated in the booklet as LCF, measured forward of the after perpendicular. MV Ninja’s waterplane at summer marks is fuller aft than forward, so F sits at 71.84 m foap: 2.16 m abaft of midships. Every trimming moment is a weight times its distance from F: a weight moved s metres contributes w × s directly, and a weight loaded or discharged contributes w times its distance from F. The exchange rate from moments to trim is the booklet’s MCTC column:
Because the ship rotates about F, the draught at F is the one draught that a change of trim cannot touch. It is therefore the draught that means anything: the true mean draught, the hydrostatic draught, the one to take into the tables. The load line mark itself is painted amidships; loading to the summer draught means bringing the draught at F to the summer draught, so that the ship carries her summer displacement and no more. The arithmetic mean of the end draughts equals it only when F is exactly amidships or the trim is zero: neither is usual.
MV Ninja is read at 9.08 m forward, 10.12 m aft. The chief officer notes the mean is 9.600 m, exactly the summer draught, and declares her at her marks. Check the claim.
Trim = 10.12 − 9.08 = 1.04 m by the stern; arithmetic mean draught = 9.600 m: to the eye, precisely legal.
TMD = 10.12 − (1.04 × 71.84 ÷ 148) = 10.12 − 0.505 = 9.615 m.
The draught at F is 1.5 cm more than the summer draught: at TPC 35.28 that is about 54 t more than the summer displacement, even though the mark amidships is exactly at the waterline. The arithmetic mean misleads because F is abaft of midships: with stern trim the waterline is deeper at F than at midships. The tables are entered at F, and loading to the summer draught means bringing the draught at F to 9.60 m.
MCTC, the moment to change trim one centimetre, is not an arbitrary tabulated number: it is the longitudinal GM at work. The longitudinal metacentre stands a ship’s length overhead, which is why vessels roll readily but barely pitch, and why trimming moments are priced in hundreds of tonne metres per centimetre.
The booklet’s summer row can be audited on the spot: MCTC 405.7 t m with Δ 30456 t and LBP 148 m gives GML = 405.7 × 14800 ÷ 30456 = 197.1 m: eighty eight times her transverse GM of 2.24 m. One row of the booklet proves the whole anatomy.
MV Ninja lies at even keel at her summer draught of 9.60 m. 250 t of cargo is shifted 60 m aft. Find the new draughts (MCTC 405.7 t m, LCF 71.84 m foap, LBP 148 m).
Trimming moment = 250 × 60 = 15000 t m by the stern; no weight crossed the rail, so there is no sinkage.
CoT = 15000 ÷ 405.7 = 37.0 cm by the stern.
Shares about F: aft = 37.0 × 71.84 ÷ 148 = 18.0 cm deeper; forward = 37.0 × 76.16 ÷ 148 = 19.0 cm shallower. The forward end moves further because it is further from the pivot. (The change of trim is rounded to 0.1 cm and the shares are taken from the rounded figure so that they add back to it; carried unrounded they are 17.9 and 19.0 cm, and the draughts are the same.)
New draughts: forward 9.60 − 0.190 = 9.41 m; aft 9.60 + 0.180 = 9.78 m. Check: the new trim, 9.78 − 9.41 = 0.37 m, is exactly the computed change of trim: the shares rebuilt the whole.
A loaded or discharged weight does two separate things, and the drill is to price them separately and add at the end: a parallel sinkage of w ÷ TPC centimetres as if the weight had gone aboard at F, then a change of trim of (w × distance from F) ÷ MCTC shared about F by the rule above.
353 t of stores are loaded exactly at the centre of flotation, 71.84 m foap, at summer marks. Find the change in the draughts.
Sinkage = 353 ÷ 35.28 = 10.0 cm; distance from F is zero, so the trimming moment is zero and the trim does not change: both draughts rise 10.0 cm together.
F is the only berth aboard where loading is purely vertical: everywhere else, the ship both sinks and swings.
Instead, 500 t is loaded 30.0 m abaft of F. Find the final draughts from even keel at 9.60 m.
Sinkage = 500 ÷ 35.28 = 14.2 cm, applied to both ends.
CoT = (500 × 30.0) ÷ 405.7 = 37.0 cm by the stern, shared as before: aft +18.0 cm, forward −19.0 cm.
Forward: 9.60 + 0.142 − 0.190 = 9.55 m. Aft: 9.60 + 0.142 + 0.180 = 9.92 m. Same 15000 t m as Worked example 6.2, so the same swing: a load s metres from F trims exactly like a shift of s metres, plus the sinkage.
The second mate argues that loading the 500 t “amidships, at 74.00 m foap” would add no trim at all. Test the claim.
Amidships is 74.00 − 71.84 = 2.16 m forward of F, so the load carries a trimming moment of 500 × 2.16 = 1080 t m by the head.
CoT = 1080 ÷ 405.7 = 2.7 cm by the head: small, but not nothing, and on a draught survey 2.7 cm is real money. The pivot is F, not the midship mark, and the booklet’s LCF column is where it lives.
LCF is a column, not a constant. Light, MV Ninja floats on a fine after body and F stands well forward of midships: 77.51 m foap at the 5.00 m ballast draught. Deep, the full stern takes its share of the waterplane and drags F abaft of midships: 71.84 m at summer marks. LCF changes with draught but not with density: fresh water moves the waterline, and F merely follows the shape. MCTC and TPC do change with density, because both contain the displacement: at 9.60 m in fresh water MCTC is 405.7 × 1.000 ÷ 1.025 = 395.8 t m and TPC is 35.28 ÷ 1.025 = 34.42 t/cm, so the same trimming moment changes the trim by more in fresh water. Within a single small problem the course convention holds LCF, TPC and MCTC constant; across large changes of draught they must be re read, which is Chapter 7’s business.
MV Ninja is loading at draughts 9.13 m forward, 9.71 m aft, and may not exceed her summer TMD of 9.60 m. How much more cargo may she load, and where should it go to keep the calculation honest?
Trim = 0.58 m by the stern; TMD = 9.71 − (0.58 × 71.84 ÷ 148) = 9.71 − 0.282 = 9.428 m.
Rise remaining to the marks = 9.600 − 9.428 = 0.172 m; carrying the unrounded TMD of 9.4285 m it is 17.15 cm.
Cargo = 17.15 × 35.28 = 605 t, loaded at F so the trim, and therefore the TMD arithmetic, is undisturbed. Loaded elsewhere it must be trimmed back or the change of trim allowed for: the deep end may not creep past its marks even though the TMD is lawful.
Trim is the list of the long axis: the draught difference at the perpendiculars, made by LCG and LCB out of line.
The ship trims about F, the centroid of the waterplane: all trimming distances are measured from the LCF column, not from midships.
The true mean draught is the draught at F: enter the tables with it, and trust no arithmetic mean while she is trimmed.
CoT = moment ÷ MCTC, and MCTC is the longitudinal GM at work: 197 m on MV Ninja, which is why she rolls but barely pitches.
A load is a sinkage plus a swing: w ÷ TPC at F, then (w × s) ÷ MCTC shared to the ends by their levers.