**A ship of 19200 M.T. displacement, 235m long has KM 10.2m, KG 9.4m, TPC 32 MCTC 254tm is floating at draft forward of 9.42m and aft 10.88m in water of RD 1.005. The centre of floatation is 5m abaft amidships, centre of buo yancy is 3m forward of amidships. Vessel is drydocked in water of density 1.020. Calculate her virtual GM on taking blocks all over.**

**Solution –**

**Solution –**

**Given W = 19200, LBP = 235m, KM = 10.2m, KG = 9.4m, TPC = 32, MCTC = 254tm, draft fwd = 9.42m, draft aft = 10.88m in RD 1.005**

**In this Question first change of trim has to be considered due to change of density.**

** We have BB _{1 }=**

**(change in v/w vol.) ⨯ (Distance between LCB + LCF) /Final volume****We can calculate, v/w vol. in 1.005**

** = 19200/1.005**

** = 19104.477m ^{2 }**

**v/w vol. in 1.020**

** = 19200/1.020**

** = 18823.529m ^{3}**

**BB _{1 }= 280.95 ⨯ 8/18823.529**

**= 0.119m****Since COF is abaft midship vessel will him by stern**

**Change in trim = (W⨯BB _{1}/ MCTC)_{ }**

**= 19200 ⨯ 0.119/252.76 ⨯ 100****MCTC in 1.025 = 254**

** MCTC in 1.020 = 252.76**

** Trim caused ( T _{C}) = 0.090m**

**Change in trim aft = (0.090 ⨯11.5)/235 = 0.043m**

** Change in trim fwd = (0.090 – 0.043) = 0.047m**

**Also,**

**FWA = (W40 ⨯ TPC)**

** = 19200 /40 ⨯ 32**

** = 15cm,**

**DWA = 0.015 ⨯ 15/0.025**

** = 9cm**

| Fwd | Aft |

Drafts | 9.42m | 10.88m |

DWA | -0.09m | -0.09m |

| 9.33m | 10.79m |

| -0.047m | +0.043m |

| 9.283m | 10.833 |

**Therefore, Trim now = 1.55m**

** Now P = (trim ⨯ MCTC)/d**

** = 155 ⨯ 252.76/112.5**

** = 348.25**

**MCTC in 1.020 = 252.76**

**We know that –**

** GG1 = P ⨯ KM/W**

** = 0.185m,**

**Old GM = (KM – KG)**

** = (10.2 – 9.4)m**

** = 0.8m**

**Virtual GM = (0.8 – 0.185) m**

** = 0.615m**

**Virtual GM = (0.8 – 0.185) m**

**= 0.615m**

**A vessel of displacement 6220t, length 144m, CF 2m abaft midships, KM 6.9m, MCTC = 150, TPC = 17t, KG 6.2m at a draft of F 4.10m, A 6.20m, is floating in a dry dock were the depth of water is 7.0m above the top of the blocks. Calculate her virtual GM (i) when the level is lowered by 1.5m, (ii) when the water level is lowered a further 1 m.**

**Solution –**

**Solution –**

**Draft fwd = 4.10m, draft aft = 6.20m,**

** Trim = 2.10m**

**Corn to aft draft = 2.10 ⨯ 70/144**

** = 1.021m**

**Hydrostatic draft = (6.20 – 1.021)m**

** = 5.179m**

**When level is lowered by 1.5m the water level = 5.5m above. It means stern has already taken to the block while forward end is still afloat.**

**Change in draft aft = (5.5 – 6.20)m = 0.70m**

** 70 = P/17 +( P⨯70/150) ⨯70/144**

** P = 245.03 tonnes**

**GG _{1} = P ⨯ KM/W**

**= 245.03 ⨯ 6.9/6220**

**= 0.272m****Virtual GM = (Old GM – GG _{1})**

**= (0.7 – 0.272)m**

**= 0.428m****When level is further lowered by 1.0m water level = 4.5m. Since water level is lower than hydrostatic draft ship has taken over block overall.**

**Then P = (TPC ⨯ mean rise)**

** = 0.679 ⨯ 100 ⨯ 17**

** P = 1154.3 tonnes.**

**GG1 = (P⨯KM)/W**

** = ( 1154.3 ⨯ 6.9 )/ 6220**

** = 1.280m**

**Therefore, New Virtual GM = (0.7 – 1.280)m **

** = -0.580m **

**Therefore, New Virtual GM = (0.7 – 1.280)m**

**= -0.580m**

**A vessel is to be dry docked in a damaged condition. She is floating at F 12.60m and A 9.00m. it is estimated that she will take the blocks 15m aft of the forward perpendicular. Estimated data: KG 8.9m, KM 9.4m, TPC 21.3 t/cm, MCTC 290 tm/cm, AF 76m, LBP 165m displacement 19500 tonnes.**

**Is it safe to dry dock the vessel in this condition? Support your answer with calculation.****Estimate the drafts at the perpendiculars on taking to the blocks fore and aft.**

**Solution –**

**Solution –**

**Trim = 3.60m, d = 74m**

**P = (trim ⨯ MCTC)/d**

** = (360 ⨯ 290)/74**

** = 1410.8 tonnes**

**GG _{1 }= (P ⨯ KM)/W**

**= (1410.8 ⨯ 9.4)/19500**

**= 0.680m****Old GM = 0.5m,**

** GG _{1 }= 0.680m**

**New GM = -0.18m**

**Since GM is –ve it is not safe to dry dock vessel in this condition. **

**New GM = -0.18m**

**Since GM is –ve it is not safe to dry dock vessel in this condition.**

**Trim caused = Tc = (1410.8 ⨯ 74/290 ⨯ 100)**

**= 3.600m**

**Change in draft aft = 3.60 ⨯ 76/165**

** = 1.658m**

**Change in draft fwd = 1.942m**

** Mean rise = (1410.8/21.3 ⨯ 100)**

** = 0.662m**

| Fwd | Aft |

Draft | 12.60m | 9.00m |

| -0.662m | -0.662m |

| 11.938m | 8.338m |

| -1.942m | +1.658m |

| 9.996m | 9.996m |

Method 2 =

** Corn to aft draft = (3.6 ⨯ 76)/165**

** = 1.658m**

**Hydrostatic draft = 10.658m,**

** Mean rise = 0.662m**

**Final draft fwd and aft = (10.658 – 0.662) **

** = 9.996m**

**Final draft fwd and aft = (10.658 – 0.662)**

**= 9.996m**

**A vessel, displacement 9013t, KG 8.0m, TPC 21.89, MCTC 162.7 tm, AF 72.013m, FSM 900 tm, enters a SW drydock drawing 3.4m fwd and 5.8m Aft. Find the virtual GM when the level of water has fallen one metre after the stem has taken to the blocks, the KM then being 9.151m. LBP = 145 m.**

**Solution –**

**Solution –**

**Clearly stern will take block once level of water reaches 5.8m. i.e equal to aft draft present level = 4.8m**

**Corn to aft draft = Trim ⨯ LCF/LBP**

** = 2.4 ⨯ 72.013/145**

** = 1.192m**

**Hydrostatic draft = 4.608 m**

** Since present level of water is greater than hd ship has not taken block overall.**

**Reduction in aft draft = 100cm**

** Or, 100 = P/21.89 + (P ⨯ 72.013/162.7) ⨯ 72.013/145**

** Or, P _{1} = 376.64 tonnes**

**GG _{1} = P ⨯ KM/W**

**= 0.382m****Old GM = 1.151m, FSC = 900/(2013 – 376.64)**

** = 0.104m,**

**Fluid GM = 1.047m**

** New GM = (1.047 – 0.382)m **

** = 0.665m**

**Fluid GM = 1.047m**

**New GM = (1.047 – 0.382)m**

**= 0.665m**

**A vessel of length 140m, displacement 5210t, KG 5.8m, KM 6.1m, MCTC 130tm, TPC 14t, CF 2m abaft midlength is afloat in a drydock at a draft of forward 3.9m and Aft 5.2m. the water level in the dock is 6m above the top of the blocks. Calculate the virtual GM and drafts F & A, when the water level is lowered by 1.5m.**

**Solution –**

**Solution –**

**Draft Fwd = 3.9 m,**

** Draft aft = 5.2m**

** Corn to aft draft = 0.631m**

** Hydrostatic draft = 4.568m**

** Now, Water level = 4.5m.**

**Since water level is almost same to hydrostatic draft we will consider that ship has taken block aft but not overall.**

**Reduction in aft draft = 70cm**

** Or, 70 = P/14 + P ⨯ 68/130 ⨯ 68/140**

**Or, P = 215.06 tonnes**

** GG _{1} = P ⨯ KM/W**

**= 215.06 ⨯6.1/5210**

**= 0.252m****GM = (0.3 – 0.252)m**

** = 0.048m**

**Now Trim caused = 215.06 ⨯ 68/130 ⨯ 100**

** = 1.125m,**

** Ta = 1.125 ⨯ 68/140**

** = 0.546m, T _{f }= 0.578m**

| Fwd | Aft |

Draft | 3.9m | 5.2m |

| -0.154m | -0.154m |

| +0.578m | -0.546 |

| 4.324m | 4.50m |

**A ship of displacement 8487t, LBP 140m, draft For’d 3.0m and Aft 5.8m, KG 8.6m, FSM 800 tm, LCF 72.1m from ap, MCTC 161 tm enters a salt water drydock. Calculate the GM (fluid) and the moment of statical stability at 1° heel at the critical instant of taking the blocks all over, given the KM at this instant is 9.57m.**

**Solution –**

**Solution –**

**Once vessel has taken block allover change in trim = 2.8 m**

**P = 280 ⨯ 161/72.1**

** = 625.24 tonnes**

**GG _{1} = P ⨯ KM/W**

**= 625.24 ⨯ 9.57/8487**

**= 0.705m**

**GM = (0.868 – 0.705)**

** = 0.163m**

**Moment of statical stability = (W ⨯ GZ)**

** = W ⨯ GM SinӨ**

** = 8487 ⨯ 0.163 ⨯ Sin 1 ^{0}**

**= 24.14 tonnes m.**

**A vessel, with following particulars, enters a seawater drydock: length B.P = 140m, draft Fwd : 3.4m, Aft : 5.8 m, KG = 8.0m, KM = 9.02m, displacement = 8930 t, TPC = 21.9, LCF = 2m from A.P, MCTC = 162.5 tm. Find the virtual GM and the drafts F & A, when the level of water has fallen one metre after the stern has taken to the blocks, given the KM at this displacement = 9.18m.**

**Solution –**

**Solution –**

**Clearly stern will take water level equal aft draft i.e 5.8m**

**Present level = 4.8m**

** Corn to aft draft = 1.238m, Hd( Hydrostatic draft ) = 4.562m**

** Since level ˃ Hd means stern has taken to block while forward end is shall afloat**

** Reduction in aft draft = 1.0m = 100cm**

** or, 100 = P/21.9 + P ⨯ 72.2/162.5 ⨯ 72.2/140**

** Or, P = 363.90 t**

**GG _{1} = P ⨯ KM/W**

**= 0.374m****Virtual GM = ( old GM – Virtual loss of GM )**

** = (old GM – 0.374)**

** = (1.18 – 0.374)**

** = 0.806m**

**Trim caused = 1.617m, **

** Rise = 0.116m,**

** Ta = 0.834, T _{f }= 0.783**

| Fwd | Aft |

Draft
| 3.400m | 5.800m |

-0.166m | -0.166m | |

+0.783m | -0.834m | |

4.017m | 4.800m |

**Therefore, Draft = 4.017m fwd**

** = 4.800m aft**

**Therefore, Draft = 4.017m fwd**

**= 4.800m aft**

**A vessel of length 180m, displacement : 11000t, LCF : 80m, TPC : 22t, MCTC : 155tm, KM : 7.2m, KG : 6.8m, at a draft of F : 6.10m, A : 6.70m, is dry docked. Calculate her residual GM and the drafts F & A at the critical instant.**

**Solution –**

**Solution –**

**P = 60 ⨯ 155/80**

** = 116.25 tonnes**

** Old GM = 0.4m**

**GG _{1} = P ⨯ KM/W**

**= 116.25 ⨯ 7.2/11000**

**= 0.076m****New GM = (old GM – GG _{1})**

**New GM = 0.324m****Corn to aft draft = 0.6 ⨯ 80/180**

** = 0.267m**

**Hd (Hydrostatic draft )= 6.433m**

** Rise = 116.25/2200**

** = 0.053m**

**Therefore , Draft F & A = 6.380m**

**Therefore , Draft F & A = 6.380m**

**A vessel of displacement 4740t, mean draft 4.22m, length 120m, KM 6.0m, KG 5.5m, CF amidships, TPC 10t, MCTC 79tm, is to be dry docked. If the GM on taking the blocks F & A is to be not less than 0.3m, what is the maximum permissible trim by the stern? At what drafts F and A would she become unstable?**

**Solution –**

**Solution –**

**GM should be 0.3m, then GG _{1} = 0.20m**

**We know that –**

**0.20m = P ⨯ 6.0/4740**

**P = 158 tonnes**

**We have P ⨯ d = (trim ⨯ MCTC)**

**= 158 ⨯ 60/79****Trim = 120cm**

** = 1.2m**

** Therefore, maximum permissible trim = 1.2m by stern**

**For unstable GG _{1} = 0.50m,**

**P = 305 tonnes**

**Trim = 3.00m**

**Ta + T**_{f}= 1.5**Rise = P/ TPC ⨯ 100**

** = 305/1000**

** = 0.305m**

**Draft should be fwd = (4.22 – 1.5)m**

** = (2.72 – 0.305)m**

** = 2.415m**

**Aft = (4.22 + 1.5)m**

** = (5.72 – 0.305)m**

**= 5.415m**

**= 5.415m**

**A vessel about to dry dockn is in the following condition :**

**draft : For’d 6.10m, Aft 6.70m**

** KMₒ = 7.20m, KGₒ = 6.8m**

** MCTC = 155 tm, TPC = 22t, LCF = 80m for’d of AP**

** Length = 180m, Disp = 11000 tonnes**

** Find : (a) GM of the vessel at critical instant**

** (B)The righting moment of 1° heel**

** (c)The draft For’d and at the critical instant.**

**draft : For’d 6.10m, Aft 6.70m**

**KMₒ = 7.20m, KGₒ = 6.8m**

**MCTC = 155 tm, TPC = 22t, LCF = 80m for’d of AP**

**Length = 180m, Disp = 11000 tonnes**

**Find : (a) GM of the vessel at critical instant**

**(B)The righting moment of 1° heel**

**(c)The draft For’d and at the critical instant.**

**Solution –**

**Solution –**

**Question same as Q. No. – 8**

**A vessel with length of 185m, displacement of 29000 is floting in a graving dock with a depth of 10m at drafts F 8.00 and A 9.38m. KM = 11.50m, KG = 10.95m, MCTC = 410 t/m, TPC = 29.6, LCF = 89.5m. find the effective GM and draft fore and aft of the vessel after the water level has dropped by 1.25m.**

**Solution –**

**Solution –**

**After level dropped by 1.25m, **

** New water Level = (10 – 1.25) m**

** = 8.75m**

**Correction to aft draft = (1.38 ⨯ 89.5)/ 185**

** = 0.668m**

**(hydrostatic draft ) Hd = 8.712m**

** Clearly since level of water 8.75m ˃ Hd 8.712m**

**It means stern has taken to the block and fwd end is still afloat.**

** Reduction in aft draft = 0.63m**

** Or, 63 = P/29.6 + P ⨯ 89.5/4100 ⨯ 89.5/185**

** Or, P = 451.96 tonnes**

**We know that –**

** GG _{1} = P ⨯ KM/ W**

**= 451.96 ⨯ 11.50/29000**

**= 0.179m****Therefore, GM = 0.371m**

** Trim caused = Tc = 451.96 ⨯ 89.5/410 ⨯ 100**

** = 0.986m**

**Rise = 451.96/29.6 ⨯ 100**

**= 0.153m**

**Rise = 451.96/29.6 ⨯ 100**

**= 0.153m**

**Ta = 0.477m, Tf = 0.509m**

| Fwd | Aft |

Draft | 8.00m | 9.38m |

-0.153m | -0.153m | |

+0.509m | -0.477 | |

8.356m | 8.750m |

**A vessel of w = 6500T, l = 120m, CF = 2mfwd of midship, MCTC = 110mt, TPC =- 15T, KM = 6.7m, KG = 6.1m at a draft of F = 4.00m A = 6.2m is floating in a drydock where the W/L is 6.5 mtrs above the top of the blocks. Calculate the residual GM of the vessel when (a) W/T is lowered by 1m. (b) it is lowered a further 1m.**

**Solution –**

**Solution –**

**When level lowered by 1m**

**Present level = 5.5m**

** Correction to aft draft = 2.2 ⨯ 62/120**

** = 1.137m**

** Hd ( hydrostatic draft ) = 5.063m**

** Since present level is 5.5m it means stern has taken to block and forward end is still afloat.**

** Reduction in aft draft = 0.7m**

** Or, 70 = P/15 + P ⨯ 62/110 ⨯ 62/12**

** Or, P = 195.6 tonnes**

**GG _{1} = P ⨯ KM/W**

**= 0.202m****GM = 0.398m**

**When water level is further lowered by 1.0m**

**Level = 4.5m**

** Since water level < hydrostatic draft ship has taken over block over all.**

** Reduction in Hydrostatic draft = (5.063 – 4.5)m**

** = 0.563m**

**Or, 56.3 = P/15**

** Or, P = 844.5 tonnes**

**GG _{1 }= P ⨯ KM/W**

**= 0.870m****Old GM = 0.6m**

** Therefore, New GM = (old GM – GG**_{1} )

** = -0.270m**

**Old GM = 0.6m**

**Therefore, New GM = (old GM – GG**

_{1})**= -0.270m**

**A vessel of L = 145m, CF = 3 mts forder of *, MCTC = 150 mt TPC = 18 t, KM = 7.5m, KG = 6.9, w = 11200 at a draft of F = 3.72m, A = 5.70m is floating in a drydock where depth of water is 6.5m above the top of the blocks. Calculate the vessel’s residual GM and her draft’s F & A, when the W/T is lowered by 1.5m.**

**Solution –**

**Solution –**

**When level is lowered by 1.5m then water level = 5.0m**

** Correction to aft draft =( trim ⨯ LCF)/ LBP**

** = 1.031m**

** Hydrostatic draft = 4.669m**

**Since water level is greater than 4.669m it means stern has taken to block while forward end is still afloat.**

**Reduction in aft draft = 70cm**

** Or, 70 = P/18 + P ⨯ 75.5/150 ⨯ 75.5/145**

** Or, P = 220.4 tonnes**

**GG1 = P ⨯ KM/W**

** = 0.147m**

**Virtual GM = (0.6 – 0.147)m**

** = 0.147m**

**Trim caused = Tc = 220.4 ⨯ 75.5/150**

** = 110.93**

** Or = 1.109m**

**Rise = 220.4/18 ⨯ 100**

** = 0.122m**

**Ta = 1.109 ⨯ 75.5/145**

** = 0.577m**

**Tf = 0.532m**

**Tf = 0.532m**

| Fwd | Aft |

Draft
| 3.72m | 5.70m |

-0.122m | -0.122m | |

+ 0.532 | -0.577m | |

4.130m | 5.000m |

**A box shaped vessel 80 × 16m at draft of 3m (F), 4m (A) KG = 4.2m is drydocked. Calculate the residual GM on taking blocks F & Aft. KG = 4.2m.**

**Solution –**

**Solution –**

**In this question KM _{T},MCTC & TPC is not given**

**Since this is a box shaped vessel both can be found.**

**MCTC = W ⨯ GM**_{L}/100 ⨯ LBP

**TPC = A/100 ⨯ density of water displaced****KM _{T }= KB + BM**

**BM**_{T}= b^{2}/12D

**= 16 ⨯ 16/12 ⨯ 3.5**

**= 6.095**

**KM = (6.095 + 1.75)m**

**= 7.845m****For TPC**

** Area of water plane = (80 ⨯ 16)m ^{2}**

^{ }= 1280m^{2}

**TPC = 1280/100 ⨯ 1.025**

**= 13.12 tonnes / cm****For MCTC**

** GM _{L }= (KM_{L} – KG)**

**& KM**_{L}= KB + BM_{L}

**BM**_{L}= I/V

**I = lb**^{3}/12 for long bl^{3}/12**V = volume = l ⨯ b ⨯ d**

** = b ⨯ l ^{3}/12 ⨯ (l ⨯ b ⨯ d)**

**= l**^{2}/12d

**= 80 ⨯ 80/12 ⨯ 3.5**

**= 152.38m****KB = depth/2**

** = 3.5/2**

** = 1.75m**

**KM _{L }= (1.75 + 152.38)m**

**= 154.13m****Therefore, GM**_{L} = 149.93 = 150

**Therefore, GM**

_{L}= 149.93 = 150**W = (80 ⨯ 16 ⨯ 3.5 ⨯ 1.025)**

** = 4592 tonnes**

**MCTC = 4592 ⨯ 150/80 ⨯ 100**

** = 86.1 tm**

**P = 100 ⨯ 86.1/40**

** = 215.25 tonnes**

**GG _{1} = P ⨯ KM/W**

**= 0.367m****GM old = (KM – KG)**

** = (7.845 – 4.2)m**

** = 3.645m**

**GG**_{1} = 0.367m

** Virtual GM = (3.645 – 0.367)m**

** = 3.278m**

**GG**

_{1}= 0.367m**Virtual GM = (3.645 – 0.367)m**

**= 3.278m**

**A vessel of length 140m, W = 19200t, KM = 6.7m, KG = 6.2m, CF = 4m fwd of midship, MCTC = 145 Tm is to be drydocked. Calculate the max. permissible tm. By the stern to ensure that she would have a positive GM of 30 cm on taking blocks F & A.**

**Solution –**

**Solution –**

**GM now = 0.5m**

** GM required = 0.3m**

** GG1 = 0.2m**

**We know that –**

** GG1 = P ⨯ KM/W**

** 0.2 = P ⨯ 6.7/19200**

** P = 573.13 tonnes**

**Therefore, permissible P = 573.13 tonnes**

** We know P ⨯ d = trim ⨯ MCTC**

** = 573.13 ⨯ 74/145 ⨯ 100 = trim in metres**

**= 2.925m by stern**

**= 2.925m by stern**

**A vassel of length 84m, displacement 3000t, TPC = 12t, KM = 4.5m, KG = 4.05m, MCTC = 80 TM, CF = 2m abaft midship at a draft of 4m (F) 4.5m (A) is drydocked. Calculate (1) the residual GM in taking blocks all over. (2) her drafts F & A then (3) her draft F&A when she becomes untable.**

**Solution –**

**Solution –**

**After taking block allover change in trim = 50cm**

**P = 50 ⨯ 80/40**

** = 100 tonnes**

** GG1 = P ⨯ KM/W**

** = 100 ⨯ 4.5/3000**

** = 0.15m**

** Residual GM = 0.3m**

**Residual GM = 0.3m**

**Original hd (Hydrostatic draft) = Aft draft minus correction.**

**Correction to aft draft = trim ⨯ LCF/LBP**

** = 0.5 ⨯ 40/84**

** = 0.238m**

**Hd = (4.5 – 0.238)m**

** = 4.262m**

** Rise = P/ TPC ⨯ 100**

** = 100/1200**

** = 0.083m**

**Therefore, Fwd & Aft = 4.178 **

**Therefore, Fwd & Aft = 4.178**

**When unstable GG**_{1}= 0.45

**0.45 = P ⨯ 4.5/3000**

** P = 300 tonnes**

** Rise = P/ TPC ⨯ 100**

** = 0.25**

**Draft Fwd & Aft when unstable = (4.262 – 0.25)m**

** = 4.012m **

**Draft Fwd & Aft when unstable = (4.262 – 0.25)m**

**= 4.012m**

**A ship LBP 240m, W 20000 tonnes, KG 9.0m, TPC 30, LCF 5m abaft of amidships is being dry-docked drawing 6m forward 8m aft. When just landing flat on the blocks the draught is 6.2m. calculate the loss of GM at this time.**

**Solution –**

**Solution –**

**Correction to aft draft = ( 2 ⨯ 115)/240**

** = 0.958m**

**Hd (Hydrostatic draft ) = 7.042m**

** After landing flat = draft is 6.2m**

** Change in Hydrostatic draft = (7.042 – 6.2)m**

** = 0.842m**

**Rise = P/ TPC ⨯ 100**

** = 0.842m**

**P = 2526 tonnes **

**P = 2526 tonnes**

**GG1 = P ⨯ KM/W**

** = P ⨯ KM/(W – P)**

** = 1.301m**

**A ship length 200m LCF 5m forward of amidships is being docked at draughts F 7.35m A 9.15m. W 38700 tonnes, TPC 48, MCTC 360, KG 7.46m.**

**Find : (i) The GM when flat on the blocks if KM then is 8.20m.**

** (ii) The draught when flat on the blocks.**

** (iii) The GM when the draught is 7.8m if KM is then 8.30m.**

** (iv) The draught at which the ship becomes unstable.**

**Find : (i) The GM when flat on the blocks if KM then is 8.20m.**

**(ii) The draught when flat on the blocks.**

**(iii) The GM when the draught is 7.8m if KM is then 8.30m.**

**(iv) The draught at which the ship becomes unstable.**

**Solution –**

**Solution –**

**When flat on blocks change in trim = 1.800m**

**P = 1.8 ⨯ 360/105**

** = 6.171m ⨯ 100**

** = 617.1 tonnes**

** GG1 = P ⨯ KM/W**

** = 617.1 ⨯ 8.20/38700**

** = 0.131m**

** New GM = 0.609m**

**New GM = 0.609m**

**Correction to aft draft = 1.800 ⨯ 105/200**

**= 0.945m**

** Hd = 8.205m**

** Rise = P/TPC ⨯ 100**

** = 0.128m**

**Therefore, New GM = (8.205 – 0.128)m**

** = 8.076m**

**Therefore, New GM = (8.205 – 0.128)m**

**= 8.076m**

**When draft is 7.8m**

**Reduction in hydrostatic draft = rise = (8.205 – 7.8)m**

** = 0.405m**

** 0.405 = P/4800**

** P = 1944 tonnes**

**GG _{1} = P ⨯ KM/W**

**= 0.417m**** Therefore, New GM = 0.423m**

**Therefore, New GM = 0.423m**

**When she become unstable GG**_{1 }= old GM

**= (8.2 – 7.46)/Old GM**

** = P ⨯ 8.2/38700**

** P = 3492.43 tonnes**

**Rise = P/TPC ⨯ 100**

** = 0.727m**

**A ship is being docked at a displacement 26240 tonnes. LBP 184m, LCF 4m abaft of amidships. KG 7.4 m, KM 7.8m, TPC 38, MCTC 300. Calculate the maximum allowable stern trim to ensure GM is at least 0.1m when landing overall.**

**Solution –**

**Solution –**

**Old GM = KM – KG**

** = (7.8 – 7.4)m**

** = 0.4m**

**New GM = 0.1m ( As per question )**

** We know that –**

** GG _{1} = P ⨯ KM/W**

**0.3 = P ⨯ 7.8/26240**

**P = 1009.23****Trim = 1009.23 ⨯ 88/300 ⨯ 100**

** = 2.960m by stern**

**Trim = 1009.23 ⨯ 88/300 ⨯ 100**

**= 2.960m by stern**

**A ship LBP 170m, displacement 21480 tonnes, KG 6.96m, KM 7.12m is being dry-docked. Calculate the maximum allowable stern trim to ensure stability is not negative when landing flat overall. TPC 30, MCTC 260, LCF 2m abaft of amidships.**

**Solution –**

**Solution –**

**Old GM = (7.12 – 6.96)m **

** = 0.16m**

**GG _{1 }permissible = 0.16m**

**0.16 = P ⨯ 7.12/21480**

**P = 482.696 t****Now, trim = P ⨯ d/ MCTC ⨯ 100**

** = 1.54m by stern**

**Now, trim = P ⨯ d/ MCTC ⨯ 100**

**= 1.54m by stern**

**A ship length 200m, W 1400 tonnes is being docked at draughts 6.30m forward 7.60m aft LCF is 4m abaft of amidships. TPC 18, KM 8.80m. When just landing flat on the blocks the draught is 6.8m. Calculate the loss of GM.**

**Solution –**

**Solution –**

**Correction to aft draft = 1.3 ⨯ 96/200**

** = 0.624m**

**Hd( hydrostatic draft)= 6.976m **

** Change in draft = rise = 0.176m**

** 0.176 = P/18 ⨯ 100**

** P = 316.8 tonnes**

**GG**_{1} = P ⨯ KM/W

** = 0.199m**

**GG**

_{1}= P ⨯ KM/W**= 0.199m**

**Calculate the GM taking the blocks overall when drydocking a (damaged) vessel, given W 43200 tonnes, LBP 210m, LCF 5m abaft of amidships and draughts F 13.14m A 9.84m. KG 8.60m, km9.00m, TPC 36, MCTC 350 t-m.**

**Solution –**

**Solution –**

**Change in trim = 3.3m**

** P = 3.3 ⨯ 350 ⨯ 100/110**

** = 1050 tonnes**

**GG _{1} = P ⨯ KM/W**

**= 1050 ⨯ 9/43200**

**= 0.218m****New virtual GM = (0.40 – 0.48)m**

** = 0.182m**

**New virtual GM = (0.40 – 0.48)m**

**= 0.182m**

**A ship LBP 180m, W 18000 tonnes is damaged forward and trimmed by the head. Calculate the maximum allowable trim by thr head to ensure that there is a positive GM of 0.15m when taking the blocks overall given that the blocks have a run down (towards the gate) of 10cm per 100m and that the ship is being docked bow in. KG 7.20, KM 7.56m, MCTC 210 t-m and LCF 8m abaft of amidaships.**

**Solution –**

**Solution –**

**Present GM = 0.36**

** Required = 0.15m**

**GG _{1} = 0.21m**

**0.21 = P ⨯ KM/W**

**P = 500 tonnes****Now, P ⨯ d = trim ⨯ MCTC**

** Trim = 500 ⨯ 98/210 ⨯ 100**

** Run down = 0.1m per 100m will provide 0.18m trip**

**Therefore, trim = (2.33 – 0.18)m**

** = 2.15m by head**

**Therefore, trim = (2.33 – 0.18)m**

**= 2.15m by head**