Chapter 7 answered the loading officer’s question: given the plan, what will the draughts be? The working ship asks it backwards far more often: the draughts are dictated, by a berth, a bar, a bridge or a charter party, and the plan must be found. Nothing new is needed: the same engine runs in reverse, with the demanded finish converted first into a displacement and an LCG, and then into tonnes and stowage positions. Every problem in this chapter starts from a condition this volume has already built.
Run forwards, the engine goes cargo plan → moments → Δ and LCG → draughts. Run in reverse, the demanded draughts give the TMD, the TMD gives the final displacement (and so the total cargo), and the demanded trim gives the final LCG (and so the moments the cargo must supply). The unknowns come out of one or two linear equations: the arithmetic of Chapter 7, read from the other end.
The tool is Chapter 6’s sharing rule, read from right to left: the demanded change at one end fixes the change of trim, the change of trim prices the moment, and the moment buys the shift.
MV Ninja completed Chapter 7 at draughts 8.99 m forward, 9.77 m aft (Δ 29719 t, TMD 9.391 m, TPC 35.22, MCTC 403.5 t m, LCF 71.91 m foap). Her berth has only 9.60 m of water alongside. Cargo can be shifted forward through 60 m. Find the weight to shift and the final draughts.
The after draught must come down 17 cm. A shift moves no weight across the rail, so the TMD stands still and the whole remedy is a rotation about F.
Change aft = CoT × LCF ÷ LBP, so CoT needed = 17 × 148 ÷ 71.91 = 34.99 cm by the head (call it 35.0 cm).
Moment = 34.99 × 403.5 = 14118 t m; weight = 14118 ÷ 60 = 235 t shifted 60 m forward.
Forward the ship takes the larger share: 34.99 × 76.09 ÷ 148 = 18.0 cm, so F = 8.99 + 0.18 = 9.17 m, A = 9.60 m. Check: the new trim of 0.43 m returns TMD = 9.60 − (0.43 × 71.91 ÷ 148) = 9.391 m: unmoved, as a shift promises.
The classic follow up: more cargo must come aboard, but the after draught, now sitting exactly on the berth’s limit, must not move. The trick is a cancellation: the parallel sinkage presses the stern down, so the load must also buy exactly enough head trim to lift the stern by the same amount. Setting the two equal fixes both the trimming moment and, with the weight known, the stowage position.
At draughts 9.17 m forward, 9.60 m aft, a further 400 t must be loaded with the after draught held at 9.60 m (TPC 35.22, MCTC 403.5, LCF 71.91 m foap). Where must it go?
Sinkage = 400 ÷ 35.22 = 11.36 cm: unopposed, the stern would settle to 9.71 m.
Required head trim: CoT × 71.91 ÷ 148 = 11.36, so CoT = 23.38 cm by the head, needing 23.38 × 403.5 = 9434 t m of head moment.
Arm forward of F = 9434 ÷ 400 = 23.58 m (carried unrounded; the arm is MCTC × LBP ÷ (TPC × LCF) whatever the weight), so the stowage is 71.91 + 23.58 = 95.49 m foap, call it 95.5 m. No hold is centred there: the booklet puts No. 2 hold’s centroid at 105.09 m foap and No. 3 hold’s at 80.82 m, so the parcel is shared, 242 t in No. 2 and 158 t in No. 3, which puts its centre at (242 × 105.09 + 158 × 80.82) ÷ 400 = 95.50 m foap; the residual centimetre is worth 400 × 0.01 ÷ 403.5, a hundredth of a centimetre of trim.
The forward draught takes everything: sinkage 11.36 plus the forward share 12.02, so F = 9.17 + 0.234 = 9.40 m, A = 9.60 m untouched.
MV Ninja floats at 8.10 m forward, 8.90 m aft (the Chapter 7 morning condition: Δ 26622 t, LCG 76.392 m foap). She is to complete loading on an even keel at 9.40 m, using No. 1 hold (lcg 128.41 m foap) and No. 5 hold (lcg 32.86 m foap, both from the booklet). Find the total cargo and the split (booklet row 9.40 m: Δ 29751 t, LCB 77.01 m).
The total. Even keel at 9.40 m means TMD 9.40 m exactly, and the booklet answers directly: final Δ = 29751 t, so cargo = 29751 − 26622 = 3129 t.
The split. Even keel demands final LCG = final LCB = 77.01 m foap, so the final moment must be 29751 × 77.01 = 2291125 t m; the cargo must supply 2291125 − 2033708 = 257417 t m.
| Item | Weight (t) | lcg (m foap) | Moment (t m) |
|---|---|---|---|
| Ship as floating | 26622 | 76.392 | 2033708 |
| No. 1 hold | x | 128.41 | 128.41x |
| No. 5 hold | 3129 − x | 32.86 | 102819 − 32.86x |
| Totals | 29751 | must equal 77.01 | 2136527 + 95.55x |
Let x go to No. 1: 128.41x + 32.86 × (3129 − x) = 257417, so 95.55x = 154598 and x = 1618 t in No. 1, leaving 1511 t for No. 5. Both weights positive and plausible for their holds: the algebra has produced a stowage plan, not just a number.
Verify the answer of Worked example 8.3 by running the forward engine of Chapter 7 on it.
Final moment = 2033708 + 1618 × 128.41 + 1511 × 32.86 = 2033708 + 207767 + 49651 = 2291126 t m; LCG = 2291126 ÷ 29751 = 77.010 m foap.
Against LCB 77.010 m the arm is a fraction of a millimetre, and the trim, 29751 × (LCB − LCG) ÷ 403.6, is under 0.01 cm: even keel at draught 9.40 m. The demand is reproduced.
This is the habit this volume has built since Chapter 2: every answer obtained one way is cheap to check the other way, and a reverse stowage plan that will not reproduce its own demand under the forward engine has no business on a signed loading plan.
Not every limit lies under the keel. A bridge on the river passage caps the ship from above, and the geometry inverts: now it is the high point that must be found, and trim decides which mast that is. The draught at any position x foap on a trimmed ship follows the straight waterline: d(x) = draught aft − trim × x ÷ LBP; the masthead’s height above water is its height above the keel minus that local draught. Stern trim lifts the bow, so the forward mast usually rides highest.
MV Ninja approaches a bridge whose underside is 20.90 m above the waterline; company policy requires 0.30 m of clearance. Draughts 8.99 m forward, 9.77 m aft (as Worked example 8.1’s start; TPC 35.22, MCTC 403.5, LCF 71.91). Foremast head 29.90 m above the keel at 108.0 m foap; aft masthead 29.60 m at 38.0 m foap. Can she pass, and if not, how much forepeak ballast (fore peak tank centroid 143.6 m foap, from the booklet) will let her?
Draught at the foremast: 9.77 − 0.78 × 108 ÷ 148 = 9.201 m, so the masthead stands 29.90 − 9.201 = 20.699 m above water: clearance 0.201 m. The aft masthead stands 20.030 m: clearance 0.870 m. The foremast governs, and 0.201 m fails the 0.30 m policy: she may not pass as she floats.
The masthead must come down another 9.9 cm, and forepeak ballast does two helpful things at once: parallel sinkage of 1 ÷ 35.22 = 0.0284 cm per tonne, and head trim worth (71.69 ÷ 403.5) × (36.09 ÷ 148) = 0.0433 cm per tonne at the foremast’s position: 0.0717 cm per tonne in all.
Ballast = 9.9 ÷ 0.0717 = 138 t, which leaves the clearance a fraction short (0.2998 m), so pump 140 t. Check run with 140 t: sinkage 3.98 cm, change of trim 140 × 71.69 ÷ 403.5 = 24.87 cm by the head; draughts become 9.158 m forward, 9.689 m aft, the foremast settles to 20.599 m above water, and the clearance is 0.301 m. The aft mast, deeper and further from the pivot’s favour, still clears by 0.85 m.
Run the engine backwards: demanded draughts give TMD, TMD gives Δ and the total; demanded trim gives LCG and the moments.
A shift buys attitude without touching the TMD: CoT = change aft × LBP ÷ LCF, then weight = moment ÷ distance.
To load with the after draught frozen, make the head trim share aft cancel the sinkage: the balance fixes the stowage position.
Two holds, one demand: the TMD fixes the total, one moment equation fixes the split, and both answers must come out positive.
Every reverse answer earns a forward check: if the loop does not close on the demand, the plan is wrong, not the booklet.