Miscelleneous

C L DUBEY – EXERCISE – 08 (DRY DOCKING)

  1.    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 –

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 BB1 =
(change in v/w vol.) ⨯ (Distance between LCB + LCF) /Final volume

We can calculate, v/w vol. in 1.005
= 19200/1.005
= 19104.477m2

v/w vol. in 1.020
= 19200/1.020
= 18823.529m3

BB1 = 280.95 ⨯ 8/18823.529
= 0.119m

Since COF is abaft midship vessel will him by stern

Change in trim = (W⨯BB1/ MCTC) 
= 19200 ⨯ 0.119/252.76 ⨯ 100

MCTC in 1.025 = 254
MCTC in 1.020 = 252.76
Trim caused ( TC) = 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
Drafts9.42m     10.88m
DWA-0.09m    -0.09m
 9.33m   10.79m
   -0.047m+0.043m
 9.283m10.833

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
  1. 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 –

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

GG1 = P ⨯ KM/W
= 245.03 ⨯ 6.9/6220
= 0.272m

Virtual  GM = (Old GM – GG1)
                = (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
  1. 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 –
  • Trim = 3.60m, d = 74m

P = (trim ⨯ MCTC)/d
= (360 ⨯ 290)/74
= 1410.8 tonnes

GG1 = (P ⨯ KM)/W
= (1410.8 ⨯ 9.4)/19500
= 0.680m

Old GM = 0.5m,
GG1 = 0.680m

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
  1. 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 –

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, P1 = 376.64 tonnes

GG1 = 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
  1. 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 –

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
GG1 = 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, Tf = 0.578m

 FwdAft
Draft  3.9m5.2m
 -0.154m-0.154m
 +0.578m-0.546
 4.324m4.50m
  1. 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 –

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

P = 280 ⨯ 161/72.1
= 625.24 tonnes

GG1 = 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 10

= 24.14 tonnes m.
  1. 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 –

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

GG1 = 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, Tf = 0.783

 FwdAft
Draft

 

3.400m5.800m
-0.166m-0.166m
+0.783m-0.834m
4.017m4.800m
Therefore, Draft = 4.017m fwd
                                   = 4.800m aft
  1. 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 –

P = 60 ⨯ 155/80
= 116.25 tonnes
Old GM = 0.4m

GG1 = P ⨯ KM/W
= 116.25 ⨯ 7.2/11000
= 0.076m

New GM = (old GM – GG1)
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
  1. 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 –

GM should be 0.3m, then GG1 = 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 GG1 = 0.50m,
P = 305 tonnes
Trim = 3.00m
Ta + Tf = 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
  1. 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.
Solution –

Question same as Q. No. – 8

  1. 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 –

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 –
GG1 = 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

Ta = 0.477m, Tf = 0.509m

       Fwd   Aft
Draft 8.00m9.38m
-0.153m-0.153m
+0.509m-0.477
8.356m8.750m
  1. 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 –
  • 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

GG1 = 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

GG1 = P ⨯ KM/W
= 0.870m

Old GM = 0.6m
Therefore, New GM = (old GM – GG1 )
                                         = -0.270m
  1. 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 –

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
    Fwd    Aft
Draft

 

3.72m5.70m
-0.122m-0.122m
+ 0.532-0.577m
4.130m5.000m
  1. 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 –

In this question KMT,MCTC & TPC is not given
Since this is a box shaped vessel both can be found.
MCTC = W ⨯ GML/100 ⨯ LBP
TPC = A/100 ⨯ density of water displaced

KMT = KB + BM
BMT = b2/12D
= 16 ⨯ 16/12 ⨯ 3.5
= 6.095
KM = (6.095 + 1.75)m
= 7.845m

For TPC
Area of water plane = (80 ⨯ 16)m2
                                           = 1280m2
TPC = 1280/100 ⨯ 1.025
        = 13.12 tonnes / cm

For MCTC
GML = (KML – KG)
& KML = KB + BML
BML=   I/V
I = lb3/12 for long   bl3/12

V = volume = l ⨯ b ⨯ d
= b ⨯ l3/12 ⨯ (l ⨯ b ⨯ d)
= l2/12d
= 80 ⨯ 80/12 ⨯ 3.5
= 152.38m

KB = depth/2
= 3.5/2
= 1.75m

KML = (1.75 + 152.38)m
= 154.13m

Therefore, GML = 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

GG1 = P ⨯ KM/W
= 0.367m

GM old = (KM – KG)
= (7.845 – 4.2)m
= 3.645m

GG1 = 0.367m
Virtual  GM = (3.645 – 0.367)m
= 3.278m
  1. 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 –

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
  1. 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 –
  • 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
  • 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 
  • When unstable GG1 = 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
  1. 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 –

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

GG1 = P ⨯ KM/W
= P ⨯ KM/(W – P)
= 1.301m

  1. 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.
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
  • 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
  • When draft is 7.8m

Reduction in hydrostatic draft = rise = (8.205 – 7.8)m
= 0.405m
0.405 = P/4800
P = 1944 tonnes

GG1 = P ⨯ KM/W
= 0.417m

 Therefore, New GM = 0.423m
  • When she become unstable GG1 = old GM

= (8.2 – 7.46)/Old GM
= P ⨯ 8.2/38700
P = 3492.43 tonnes

Rise = P/TPC ⨯ 100
= 0.727m

  1. 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 –

Old GM = KM – KG
= (7.8 – 7.4)m
= 0.4m

New GM = 0.1m ( As per question )
We know that –
GG1 = P ⨯ KM/W
0.3 = P ⨯ 7.8/26240
P = 1009.23

Trim = 1009.23 ⨯ 88/300 ⨯ 100
= 2.960m by stern
  1. 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 –

Old GM  =  (7.12 – 6.96)m
                 = 0.16m

GG1 permissible = 0.16m
0.16 = P ⨯ 7.12/21480
P = 482.696 t

Now,  trim = P ⨯ d/ MCTC ⨯ 100
= 1.54m by stern
  1. 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 –

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

GG1 = P ⨯ KM/W
= 0.199m
  1. 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 –

Change in trim = 3.3m
P = 3.3 ⨯ 350 ⨯ 100/110
= 1050 tonnes

GG1 = P ⨯ KM/W
= 1050 ⨯ 9/43200
= 0.218m

New virtual GM = (0.40 – 0.48)m
                            = 0.182m
  1. 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 –

Present GM = 0.36
Required = 0.15m

GG1 = 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

manish-mayank

About the author

Manish Mayank

Graduated from M.E.R.I. (Mumbai). A cool, calm, composed and the brain behind the development of the database. The strong will to contribute in maritime education and to present it in completely different and innovative way is his source of inspiration.

Leave a Comment

From Marine Enclyopedia to MARINE GYAAN ACADEMY in Lucknow - Preparing YOUTH to become SEAFARERS ( for IMU-CET and various courses) From Marine Enclyopedia to MARINE GYAAN ACADEMY in Lucknow - Preparing YOUTH to become SEAFARERS ( for IMU-CET and various courses) From Marine Enclyopedia to MARINE GYAAN ACADEMY in Lucknow - Preparing YOUTH to become SEAFARERS ( for IMU-CET and various courses) From Marine Enclyopedia to MARINE GYAAN ACADEMY in Lucknow - Preparing YOUTH to become SEAFARERS ( for IMU-CET and various courses)