Chapter 6 built the parts: the pivot at F, MCTC the exchange rate, the TMD, the sharing rule. This chapter bolts them into the engine that answers the question every loading plan ends with: after all this cargo, what will her draughts be? The method is the longitudinal twin of the KG moments table of Volume One: weights times levers, this time measured from the after perpendicular, meeting the booklet twice, once on the way in and once on the way out.
A floating ship carries her weight down through G and her buoyancy up through B. If LCG and LCB stand on the same vertical, she floats at rest; if not, the pair form a couple that rotates her about F until the shifting underwater volume brings B back beneath G. The trim she takes up to achieve this is priced by the booklet:
The bracket LCB − LCG is the trimming lever and its sign gives the direction: positive when G is abaft B, and then the trim is by the stern; negative when G is forward of B, and then the trim is by the head. The deep end is always the end G is closer to. Trim by the stern is positive throughout, so a trim of −25 cm is 25 cm by the head.
Every full trim problem in the examination room runs the same seven steps, and the discipline is worth more than any single formula: the booklet is consulted twice, at the floating condition and again at the final displacement, because Δ, LCB, LCF and MCTC all drift as she sinks.
MV Ninja lies at draughts 8.10 m forward, 8.90 m aft. Find her displacement and LCG (booklet rows: 8.40 m: Δ 26242 t, MCTC 391.9, LCB 77.65, LCF 72.46, TPC 34.80; 8.60 m: Δ 26941 t, MCTC 394.5, LCB 77.51, LCF 72.32, TPC 34.92; LBP 148 m).
Trim = 8.90 − 8.10 = 0.80 m by the stern. First pass LCF at the 8.50 m mean: 72.39 m; TMD = 8.90 − (0.80 × 72.39 ÷ 148) = 8.509 m.
Interpolate at 8.509 m (8.5087 m before rounding), fraction 0.544 up the interval: Δ 26622 t, MCTC 393.3 t m, LCB 77.574 m, LCF 72.384 m, TPC 34.87.
LCB − LCG = trim (cm) × MCTC ÷ Δ = 80 × 393.3 ÷ 26622 = 1.182 m.
Stern trim means LCG stands abaft of LCB: LCG = 77.574 − 1.182 = 76.392 m foap (76.39 m). Two draught readings and the booklet have given both the displacement and the position of the centre of gravity.
From the condition of Worked example 7.1 the following is worked: load 2000 t in No. 2 hold (lcg 105.09 m foap) and 1500 t in No. 4 hold (56.55 m); discharge 800 t from No. 3 hold (80.82 m); take 300 t of heavy fuel oil, RD 0.950, into the No. 2 HFO tanks port and starboard (32.91 m). Find the final displacement and LCG. (Compartment positions are the booklet's.)
Capacity check on the bunkers first: 300 ÷ 0.950 = 315.8 m³ against 2 × 174.8 = 349.6 m³ in the pair of tanks, so the oil fits.
| Item | Weight (t) | lcg (m foap) | Moment (t m) |
|---|---|---|---|
| Ship as floating | 26622 | 76.392 | 2033708 |
| Load No. 2 hold | +2000 | 105.09 | +210180 |
| Load No. 4 hold | +1500 | 56.55 | +84825 |
| Discharge No. 3 hold | −800 | 80.82 | −64656 |
| Bunkers, No. 2 HFO tanks P and S | +300 | 32.91 | +9873 |
| Totals | 29622 | 76.765 | 2273930 |
Final LCG = 2273930 ÷ 29622 = 76.765 m foap. The table is the KG table of Volume One rotated ninety degrees: same discipline, levers now measured from the after perpendicular, discharges subtracting weight and moment alike.
Complete the problem: find MV Ninja’s final draughts (booklet rows: 9.20 m: Δ 29046 t, MCTC 401.4, LCB 77.13, LCF 72.00; 9.40 m: Δ 29751 t, MCTC 403.6, LCB 77.01, LCF 71.91).
Enter with Δ = 29622 t: fraction (29622 − 29046) ÷ 705 = 0.817 up the interval, so TMD 9.363 m, MCTC 403.2 t m, LCB 77.032 m, LCF 71.926 m: every constant differs from the value read in Worked example 7.1, which is exactly why the booklet is consulted twice.
Final trim = Δ × (LCB − LCG) ÷ MCTC = 29622 × (77.032 − 76.765) ÷ 403.2 = 29622 × 0.267 ÷ 403.2 = 19.6 cm by the stern (positive: G abaft B).
Draught aft = TMD + trim × LCF ÷ LBP = 9.363 + 0.196 × 71.926 ÷ 148 = 9.363 + 0.095 = 9.458 m (9.46 m).
Draught forward = 9.458 − 0.196 = 9.262 m (9.26 m). Check: the pair give trim 0.196 m and TMD 9.458 − (0.196 × 71.926 ÷ 148) = 9.363 m: the draughts return the TMD they were built from. Had the initial constants (LCB 77.574, MCTC 393.3) been kept, the lever would have read 0.809 m and the trim 60.9 cm, three times the true figure.
A final 500 t parcel remains. The master wants MV Ninja to sail on an even keel. Where must it be stowed (booklet row 9.60 m: Δ 30456 t, MCTC 405.7, LCB 76.90 m)?
Even keel demands final LCG = final LCB. At Δ = 29622 + 500 = 30122 t, fraction (30122 − 29751) ÷ 705 = 0.526 between the 9.40 and 9.60 rows: LCB = 76.952 m foap, MCTC 404.7.
Taking moments: required lcg = (76.952 × 30122 − 2273930) ÷ 500 = (2317948 − 2273930) ÷ 500 = 44018 ÷ 500 = 88.0 m foap: 7.2 m forward of the centre of No. 3 hold (80.82 m), so in the forward part of No. 3.
A check on the size of the task: the LCG has to move only 0.187 m forward (from 76.765 to 76.952 m), and 500 t does that from a lever of about 11 m. Stowed right at the forward perpendicular the same parcel would give 74.1 cm of trim by the head, and at the after perpendicular 108.7 cm by the stern; nothing beyond those limits is available from 500 t. Had the answer fallen outside the hull, the demand could not be met with 500 t, and the options would be a heavier parcel nearer amidships or a ballast transfer: the same moments, written differently.
From the Worked example 7.1 condition, a single 300 t parcel is loaded into No. 5 hold, lcg 32.86 m foap. Find the final draughts twice: once by the small tool of Chapter 6 (sinkage plus moment about F, constants held), and once by the full engine (constants re read). Compare.
Route A. Sinkage = 300 ÷ 34.87 = 8.6 cm. Arm from F = 72.384 − 32.86 = 39.52 m abaft; CoT = 300 × 39.52 ÷ 393.3 = 30.1 cm by the stern, shared 14.7 cm aft, 15.4 cm forward. Final: F 8.10 + 0.086 − 0.154 = 8.032 m; A 8.90 + 0.086 + 0.147 = 9.133 m.
Route B. New Δ = 26922 t, new LCG = (2033708 + 300 × 32.86) ÷ 26922 = 2043566 ÷ 26922 = 75.907 m; re entering the tables at fraction 0.973: TMD 8.595 m, MCTC 394.4, LCB 77.514 m, LCF 72.324 m; lever 1.607 m and the trim formula gives 26922 × 1.607 ÷ 394.4 = 109.7 cm by the stern. Final: A 8.595 + 1.097 × 72.324 ÷ 148 = 9.131 m, F 9.131 − 1.097 = 8.034 m.
Agreement: 2 mm at each end, less than a draught mark can be read to. For one small parcel the Chapter 6 tool is quick and adequate; for a full cargo plan only the engine, with its second visit to the booklet, gives the correct answer.
Trim is caused by a couple: Trim (cm) = Δ × (LCB − LCG) ÷ MCTC, positive and by the stern when G is abaft B.
The floating ship gives her LCG: TMD into the tables, then LCG = LCB − trim × MCTC ÷ Δ for stern trim.
The grand table is the KG table turned lengthways: weights times levers foap, discharges negative, one division at the end.
Visit the booklet twice: Δ, LCB, LCF and MCTC all drift with draught, and the final row is the one that prices the answer.
Finish at F: draught aft = TMD + trim × LCF ÷ LBP, forward = aft − trim, and the answer must loop back to its own TMD.