Before docking, the ship must be stable with an adequate GM, and the docking operation must not cause a list or loss of stability. As the water is pumped out, the keel-block reaction reduces the effective buoyancy, lowering KB and BM and hence GM; the ship must retain sufficient GM throughout so that it does not heel over on the blocks. Requirements:

• Adequate initial GM and even keel (or slight trim), with no list; ballast arranged to give even keel and to press up or empty tanks to avoid free surface.
• The docking weight, draft and trim must be such that the keel blocks contact evenly over the full keel length; excessive trim overloads the aft blocks.
• The ship must be central over the keel line and the dock level; the reaction builds gradually and the GM must remain positive at all stages.
• No weights moved during docking; the stability data (hydrostatics, docking plan) used to check the condition.

If the GM becomes too small or negative during docking, the vessel can heel and capsize on the blocks, so a stability margin is insisted on.

The half-ordinates of a waterplane at 15 m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0 m. There are 9 ordinates, so the waterplane length is 8 x 15 = 120 m.

(i) TPC.

Area A = 2 x (h/3)[y0 + y8 + 4(y1+y3+y5+y7) + 2(y2+y4+y6)]

= 2 x (15/3)[1 + 0 + 4(7+11+10.5+4) + 2(10.5+11+8)]

= 10[1 + 4x32.5 + 2x29.5] = 10[1 + 130 + 59] = 10 x 190 = 1900 m2.

TPC (salt) = A x 1.025/100 = 1900 x 1.025/100 = 19.48 t/cm.

(ii) Distance of the centre of flotation from midships.

First moment of area about the after perpendicular:

M = 2 x (h/3) x sum of (weighted y x distance). Using station distances 0,15,30,...,120 m:

M = 2 x (15/3)[1x0 + 7x15 + 10.5x30 + 11x45 + 11x60 + 10.5x75 + 8x90 + 4x105 + 0x120] with Simpson weights (1,4,2,4,2,4,2,4,1):

weighted sum = 1x0 + 4x(7x15) + 2x(10.5x30) + 4x(11x45) + 2x(11x60) + 4x(10.5x75) + 2x(8x90) + 4x(4x105) + 1x0

= 0 + 4x105 + 2x315 + 4x495 + 2x660 + 4x787.5 + 2x720 + 4x420 + 0

= 420 + 630 + 1980 + 1320 + 3150 + 1440 + 1680 = 10620.

M = 2 x 5 x 10620 = 106,200 m3.

Distance of centroid from AP = M/A = 106,200/1900 = 55.89 m. Midships is at 60 m from AP, so the centre of flotation is 60 - 55.89 = 4.11 m aft of midships.

(iii) Second moment of area about a transverse axis through the centre of flotation.

Second moment about AP: I_AP = 2 x (h/3) x sum of (weighted y x distance^2):

weighted sum = 1x0 + 4x(7x225) + 2x(10.5x900) + 4x(11x2025) + 2x(11x3600) + 4x(10.5x5625) + 2x(8x8100) + 4x(4x11025) + 1x0

= 0 + 4x1575 + 2x9450 + 4x22275 + 2x39600 + 4x59062.5 + 2x64800 + 4x44100 + 0

= 6300 + 18900 + 89100 + 79200 + 236250 + 129600 + 176400 = 735,750.

I_AP = 2 x 5 x 735,750 = 7,357,500 m4.

Transfer to the centre of flotation: I_CF = I_AP - A x (distance from AP to CF)^2 = 7,357,500 - 1900 x (55.89)^2 = 7,357,500 - 1900 x 3123.7 = 7,357,500 - 5,935,000 = 1,422,500 m4.

Answer: TPC = 19.48 t/cm; LCF = 4.11 m aft of midships; I about the transverse axis through CF = about 1.42 x 10^6 m4.