A ship enters dry dock upright, with positive GM and a small trim by the stern. The gates close and the water is pumped out. At some point the stern touches the blocks while the bow is still afloat, and from that moment part of the ship’s weight is carried by the blocks instead of the water. That part is called the P force, and it costs stability. This chapter shows how big P gets, what it does to the GM, and how to check before docking that the ship can afford it. The working condition throughout: MV Ninja arrives for her docking at draughts 4.86 m forward and 5.42 m aft in salt water, with bunkers low, double bottoms empty and heavy spares staged on deck, giving a fluid KG of 11.72 m.
Four stages. First the ship floats over the blocks while pumping begins. Second, the stern touches: this starts the critical period, because the ship is losing stability and is not yet held by anything except one point of the keel. Third comes the critical instant, just as she is about to sit on the blocks along her whole length: P is at its largest while she is still unsecured, so the GM is at its lowest. Fourth, she lands, shores and side blocks go in, and only then does pumping resume. Every calculation in this chapter is aimed at the third stage.
While the stern rests on the blocks and the water keeps falling, the ship pivots about her centre of flotation and her trim reduces. By the critical instant the trim is gone: the keel is level with the blocks. So between first touch and the critical instant, the blocks have removed the whole trim the ship came in with. That takes a trimming moment, and the moment is P times its lever, the distance from the sternframe to F. Setting the two equal gives the working formula. A second, practical way to find P at any moment during the period is to watch the draughts: the mean draught falls by P divided by TPC, so the observed fall times the TPC gives the P at that moment.
MV Ninja enters dock at 4.86 m forward, 5.42 m aft, salt water. The blocks are level and her sternframe is at the after perpendicular. Find P at the critical instant (rows 5.00 m: 14798 t, TPC 32.28, MCTC 313.7, LCF 77.51, KM 12.356; 5.20 m: 15445 t, TPC 32.44, MCTC 318.1, LCF 77.24, KM 12.113).
Trim = 0.56 m by the stern. First pass at the AMD of 5.14 m gives LCF 77.32 m, so TMD = 5.42 − (0.56 × 77.32 ÷ 148) = 5.127 m.
Tables at 5.127 m (fraction 0.637, formed from the unrounded 5.1274 m and carried unrounded): displacement 15210 t, MCTC 316.5, TPC 32.38, LCF 77.34 m foap, KM 12.201 m.
P = 56 × 316.5 ÷ 77.34 = 229 t. The lever is 77.34 m because the sternframe is at the after perpendicular and F is 77.34 m forward of it.
Cross check by draughts: the mean draught will fall by 229 ÷ 32.38 = 7.08 cm by the critical instant, and 7.08 cm × 32.38 gives back 229 t. The water at the stern falls further, by the mean fall plus the stern’s share of the trim removed, 7.08 + 56 × 77.34 ÷ 148 = 36.3 cm, so the after draught at the critical instant is 5.42 − 0.363 = 5.057 m. On the day, watching the draught fall is the practical way to track P.
An upward force at the keel reduces stability the same way weight removed from the keel does: the centre of gravity effectively rises. The standard measure is a virtual loss of GM equal to P × KM ÷ W, where W is the displacement she entered with. One care is needed: the water level has fallen by the critical instant, the draught is smaller, and KM has changed. So KM is read again from the tables at the reduced displacement, W − P.
Continue: KG 11.72 m. Find the GM at the critical instant.
Step 1: P = 229 t, from Worked example 11.1.
Step 2: reduced displacement = 15210 − 229 = 14981 t. Tables by displacement (fraction 0.283): draught 5.057 m, KM at the critical instant = 12.287 m. Note it rose from 12.201 m: at smaller draughts this ship’s KM is larger.
Step 3: virtual loss = 229 × 12.287 ÷ 15210 = 0.185 m.
GM at the critical instant = 12.287 − 11.72 − 0.185 = 0.382 m. She entered with 0.481 m afloat, so the docking costs her 0.099 m, about a fifth of that. An alternative formula, P × KG ÷ (W − P), treats P as a weight discharged from the keel: GG1 = 229 × 11.72 ÷ 14981 = 0.179 m, giving a GM of 12.287 − 11.899 = 0.388 m which belongs to the reduced displacement 14981 t. The two GM figures differ because they refer to different displacements; the righting moments they describe are the same (15210 × 0.382 and 14981 × 0.388 are both 5812 t m unrounded). This book follows the P × KM ÷ W form on the examination sheet, whose GM goes with the entering displacement W.
Find the draughts forward and aft at the critical instant.
The mean draught falls by P ÷ TPC = 229 ÷ 32.38 = 7.1 cm. Taking that off the TMD, working in full figures, gives 5.057 m at the critical instant.
The trim is gone and the blocks are level, so this is the draught at both ends: forward 5.057 m, aft 5.057 m. It agrees with the draught the tables gave for the reduced displacement in Worked example 11.2, which is a free check.
Before docking, the question is asked the other way round: given the GM the ship must still have at the critical instant, how much trim can she afford? Work the same three numbers backwards: the loss she may spend, the largest P that loss allows, and the trim that produces that P.
The dock requires MV Ninja to keep at least 0.25 m of GM at the critical instant. Find her maximum trim, and say whether her present 0.56 m is acceptable.
GM afloat = 12.201 − 11.72 = 0.481 m. Loss she may spend = 0.481 − 0.250 = 0.231 m.
Largest P allowed = loss × W ÷ KM = 0.231 × 15210 ÷ 12.201 = 288 t.
Maximum trim = P × lever ÷ MCTC = 288 × 77.34 ÷ 316.5 = 70 cm = 0.70 m by the stern.
Her present trim is 0.56 m, inside the limit with 14 cm in hand: acceptable. This check uses the afloat KM throughout, which is slightly strict here because KM rises as the water falls; the margin is therefore a little better than it looks. If the trim had failed the test, the fixes are the usual ones: reduce the trim, or lower KG before docking.
After MV Ninja lands overall, suppose the water were pumped down a further 0.30 m before she was secured. What would P become, and what would it do to her stability?
Once she is on the blocks along her whole length, every centimetre of fall transfers weight from the water to the blocks across the whole keel: extra P = fall (cm) × TPC = 30 × 32.38 = 971 t.
Total P = 229 + 971 = 1200 t. Taking KM nominally at its critical instant value of 12.287 m, the virtual loss would be about 1200 × 12.287 ÷ 15210 = 0.969 m, twice her entire afloat GM, and the nominal GM would be 12.287 − 11.72 − 0.969 = −0.402 m. Unsecured, she would have no GM left after only a further 14.6 cm of fall, long before the dock was dry.
That is the whole reason for the drill: pumping stops at the critical instant, shores and side blocks go in, and only a secured ship has the water taken from under her. Once she is shored, the growing P is carried by the structure and the GM figure no longer governs.
The critical period runs from the stern touching the blocks to landing overall; the critical instant, at its end, is the danger point.
P at the critical instant = trim (cm) × MCTC ÷ distance of F from the blocks. On the day, P = observed fall of mean draught × TPC.
Virtual loss of GM = P × KM ÷ W, with KM read again at the reduced displacement W − P.
Maximum trim: spendable loss → largest P → largest trim, worked with the same three formulas backwards.
After landing overall, P grows by TPC for every centimetre of fall: the ship is secured before pumping resumes.