Stability – I : Chapter 11

  1. On a ship of W 5000t , GM 0.3m, 20t was shifted transversely by 5m. Find the list.
Solution:

W = 5000t, GM = 0.3m,
Weight shifted (w) = 20t & d= 5m

Listing moment = (20 x 5 )
= 100tm

We know that :
Tanθ  = ( Listing Moment) /(W x GM)
= 100/(5000 x 0.3)
= 3.8 degree.

  1. On a ship of W 8000t, GM 2.0m, If the following transverse shifting were done , find the list:
    • 200cargo shifted 4m to stbd
    • 100t cargo shifted 2 m to port
    • 100t cargo shifted 4m to port
    • 50t stores shifted 20m to stbd

Solution:-

We can calculate Listing moment due to shifting of cargo
LM(1)  = ( weight x distance )
=   (200 x 4)
= 800 tm (S)

Again,   LM (2) = (Weight x distance )
= (100 x 2)
= 200 tm (P)

LM(3)  = ( weight x distance )
=   (100 x 4)
= 400 tm (P)

LM(4)  = ( weight x distance )
= (50 x 20)
= 1000 tm (S)

Final listing moment (LM) = LM(1) + LM(2) +LM(3)+ LM(4)
= 800(S) + 200(P) + 400(P) + 1000(S)
=1800(S) + 600(P)

So, final listing moments = (1800 – 600)
= 1200 tm (S)

Now, Tanθ  = (Final LM )/(W x GM)
= (1200/(8000 x 2 )
= 4.29 degree (S)

  1. If 200t cargo was shifted downwards by 10m and to starboard by 5m on a ship of W 10000t , KG 7.0m, KM 7.4m, find the list.
Solution :-

Given :
Cargo shifted (w)   = 200t,
Distance  = 10m,
d = 5m from CL to stbd
W = 10000 t ,
KG = 7.0m &
KM = 7.4m

Ship’s wtKGVM
10000 t7.0m70000 tm
200t (shift)10m(-) 2000 tm

     Final W =10000t                                             Final VM= 68000 tm

We know that:
Final KG = (final VM)/ (Final W)
Final KG = (68000/10000)
= 6.8m

Final GM= (KM – KG)
= (7.4 – 6.8)
= 0.6m

Listing moment (LM) = (weight x distance )
= (200 x 5)
= 1000tm

tanθ =  LM/(W x GM)
= 1000 /(10000 x 0.6)
= 9.46 degree

  1. A quantity of grain estimated to be 100t shifts transeversly by 12m and upwards by 1.5m, on a ship of W 12000t , GM 1.2m. Find the list caused.
Solution:

Grain shifted (w) = 100t,
Distance = 12m & transversely, d = 1.5( î)
W = 12000 t, GM = 1.2m

LM caused = (weight x distance)
= (100 x 12)
= 1200 tm

GG1 (î) = (distance x weight) /W
= (1.5 x 100) / 12000
= 0.0125 m

Final GM =  (GM – GG1)
= (1.2 – 0.0125)m
= 1.1875m

Tanθ = LM /(W x GM )
= 1200/(12000 x 1.1875)
= 4 degree 8’.

  1. A ship displaces 4950t and has KG 4.85m, KM 5.79m. cargo weighing 50t is loaded 1.25m above the keel and 4m port of the centre line . Find the list
Solution:

W = 4950 t ,
KG = 4.85m 
KM = 5.79m
Weight loaded (w) = 50 t
KG = 1.25m 
d = 4 m to CL to port

LM caused = (weight x distance )
= (50 x 4)
= 200 tm to port .

Ship’s wt
KG
VM
4950 t4.85m24007.5 tm
(+)50t1.25m(+ )62.5 tm

Final W = 5000 t                                                    Final VM = 24070 tm

We know that:
Final KG = (Final VM)/( Final W)
Final KG = (24070/5000)
= 4.814m

Final GM = (KM –KG)
= (5.79 – 4.814) m
= 0.976m.

Tanθ = LM/(Weight x GM)
= 200/ (5000 x 0.976)
θ     =    2.34 degree

  1. A weight of 100t is discharge from a position 2.45m above the Keel and 6m to port of the centre line of a ship of w 10000t, KM 8.25m, KG 7.45m. Find the list
Solution:

Weight discharged (w) = 100t,  & KG = 2.45m
Transverse distance = 6m to part from CL
W = 10000 t,
KM = 8.25 m,
KG = 7.45m

We know that:
LM caused = (weight  x distance)
= (100 x 6) tm
= 600 tm (P)

Ship’s wt
KG
VM
10000 t7.45m74500 tm
(-)100 t2.45m(-) 245 tm

Final W = 9900 t                                                     Final VM = 74255 tm

We know that:
Final KG = (Final VM)/ (Final W)
Final KG = (74255 / 9900)
= 7.5m

Final GM = (KM –KG)
= (8.25 – 7.50) m
= 0.75m

Tanθ =  LM /( W x GM)
= 600/(9900 x 0.75)
= 4.62 degree.

  1. A ship of 10000 t displacement, KG 8.3m carries out the following cargo operation:
Qty. (t)
LOaded or discharge
KG (m)
Distance off centre line
200D10.05m port
800D2.34m stbd
500D5.23m port
250L8.0Nil
250L12.0Nil
If the final KM is 9.6m, find the list.
Solution:
Ship’s wt
KG
VM
CL
LM
10000 t8.383000  
(-) 200(Dish)10.0(-) 20005P1000 P
(- ) 800(Dish)2.3(-)  18404S3200 S
(-) 500(Dish)5.2(-) 26003P1500 P
(+) 250(load)8.0(+) 2000NIL 
(+) 250(load)12.0(+) 3000NIL 

Final W = 9000 t                                Final VM =81560 tm                  Final LM = 700(S)

Final KG = (Final VM)/ (Final W)
Final KG = (81560 / 9000)
= 9.062 m

We know that:
GM = (KM – KG)
= (9.6 – 9.062)
= 0.538m

Tanθ   =   (Final LM)/(final W x GM)
= (700)/(9000 x 0.538)
=  8.2 degree

  1. A ship of W 9000t, KG 8.3m loads 600t of cargo (KG 4.0m, 3m to port of the centre line ) and discharge 400t of cargo ( KG 9.0m, , from 5m to port of the centre line ). 200t of cargo is then shifted upwards by 5m and to starboard by 8m. 300t of cargo is then is then shifted 1m downwards and 4m port .Find the list if the final KM is 8.95m.
Solution :
Ship’s wt
KG
VM
d
LM
 9000t8.374700
(+) 600t4.0(+) 24003(P)1800
(-) 400t9.0(-) 36005(P)2000
200 t5(+) 10008 (S)1600
300 t1(-) 3004(P)1200

Final W = 9200 t                                 Final VM = 74200 tm                                           FLM = 600(S)

Final   KG = (Final VM)/(Final W)
Final KG = (74200)/(9200)
= 8.065m

Final GM = (KM – KG)
= (8.95 – 8.065)
= 0.885m

tanθ = (600 )/(9200 x 0.885)
= 4 degree ( S)

  1. A ship of W 18000 t, KG 7.75m discharge 1500t (6.0m above the keel and 3m port of the centre line ) and loads 500t (10m above the keel and 4m port of the centre line ).
  • Cargo was then shifted as follows ;
  • 500t upwards 2m and to starboard 4m
  • 800 t downwards 2m and to port 3m.

If the final KM is 8.935 m, find the list .

Solution:

Ship’s wt
KG
VM
D
LM
18000t7.73m139500  
(-) 1500t6.0m(-) 90003P4500(S)
(+) 500t10m(+) 50004P2000(P)
 500t2m(+) 10004S2000(S)
800t2m(-) 16003P2400(P)

Final W = 17000                      Final VM = 134900 tm                       FLM =2100(S)

We know that:
Final KG = (Final VM) / (Final W)
Final KG = (134900/17000)
= 7.935m

Final GM = (KM –KG)
= (8.935 – 7.935)
= 1.0 m

We know that :
Tan θ  = Final LM/(Final W x GM)
=     2100 /(17000 x 1)
= 7 degree 2.6min

  1. A ship listed 8degree to port , displaces 12000t and has KM 7.54m and KG 6.8m . Find how many tonnes of SW ballast must be transfered from NO 2 port DB tank to NO2 stbd DB tank , to up right and vessel , if the tank- centre are 10m apart .
Solution :

List = 8 degree (P) ,
Displacement (W)  = 12000t,
KM = 7.54m &
KG = 6.8m

We know that:
GM = (KM – KG)
= (7.54 – 6.80)
= 0.74m

Again, listing moment (LM)= (W x GM x tanθ)
= (12000 x 0.74 x tan80)
= 1248 tm

We know that ,
Listing moment (LM) = (W x D)
1248 tm    = (W x 10m)
So,      W   = 124.8 t

Here to upright the vessel the lighting moment caused on particle must be same at stbd side. So we have to transfer 124.8t of sea water to  the stbd side of DB  tank to upright the vessel.

  1. A ship displacing 4000t has GM 1.0m (KM 10.0 m and KG 9.0 m) and is listed 10 degree to port .If 16000t of cargo is now loaded on the centre line and the final GM is 1.0m KM 9.0m and KG 8.0m state weather the list would change .if yes , find the new list .
Solution :

Displacement (W) = 4000t,
GM = 1.0m (initial) ,
list  = 10 degree (P)
W = 16000t, G
M = 1.0m (final)

We know that:
Listing moment (LM) = (W x GM tanθ)
= (4000 x 1 x tan 100)
= 705.3 tm

According to question ,  16000 t  of cargo loaded
So Final W = (16000 + 4000)
= 20,000t

But, Initial  GM = Final GM
Initial LM = Final LM

Again we know  that :
Tanθ = (FLM/(W x GM )
= 705.3/(20000 x 1)
= 2 degree 1 minute ( P).

Hence we can say that,(Yes) list would change due to the loading of cargo.

  1. A ship of W 10000t. GM 1.5m, is listed 5 degree to stbd . If cargo is shifted vertically until her final GM is 0.5m, state whether the list would change . if yes find the new list .
Solution :

Displacement (W) = 10000t,
GM = 1.5m,
list = 5 degree

Final GM = 0.5m

We know that :
Listing moment (LM)= (W x GM tanθ)
= (10000 x 1.5 x tan50 )
= 1312.33 tm (S)

According to question,  cargo shift vertically and now  GM becomes  0.5 m
Hence, Due to change of GM , list must be change .

We know that:
Tanθ  =  LM / (W x GM)
= 1312.33/(10000 x 0.5)
=    14.7dgree stbd

  1. A ship of W 8500t, KM 9.0m, KG 8.3 , is listed 8 to stbd . The following cargo operation were carried out :
  • 200 t discharge KG 4m from 5m stbd of CL .
  • 300t discharged KG 5m from 2m port of CL .
  • 100t loaded KG 2m, 4m to stbd of CL .
  • 200t shifted up by 2m and port by 3m.

If the final KM is 9.3m , find the final list .

Solution:

Ship’s wt
KG
VM
d
LM
8500t8.3m70550tm389.23tm 
(-) 200t4m(-) 800tm5S1000 P
(-)300t5m( -)1500tm2P600 S
(+)100t2m(+) 200tm4S400 S
200t( Shift)2m(+) 400tm3P600P

Final W = 8100t                                  Final VM = 68850tm                               FLM=289.23tm

We know that:  
Final GM = (Final VM )/(Final W)
Final KG = (68850 /8100)
= 8.5m

Final GM = (KM – KG)
= (9.3 – 8.5)
=      0.8m

We know that:
tanθ = Final LM/(W x GM)
=   289.23/(8100 x 0.8)
= 2degree 33.3minute

  1. A ship of 15000t displacement, KG 8.7m KM 9.5m is listed 10deg to port. The following cargo work was carried out:
  • 500t loaded KG 8.0m, 5m stbd of CL .
  • 300t discharged, KG 4.0m, 4 port of CL.

Find the quantity of SW ballast that must be transferred transversely to bring the vessel up right , the tank centres the being  12m apart.

NOTE : Since vessel is required upright , it is not necessary to calculate the final KG or GM unless specifically asked).

Solution:

Displacement (W) = 15000t,
KG = 8.7m, KM = 9.5m & list = 10 degree (P)
Final W = (15000 + 500) – 300  = 15200 t
Initial LM = 15000 x 0.8 x tan 10 = 2115.9 tm (P)

We know that:
Listing moment caused
LM(1) = (500 x 5)
= 2500 tm (S)

LM (2) = (300 x 4)
= 1200 tm (S)

Total listing moment = (2500 + 1200) tm
= 3700 tm (S)

According to question and calculation done above ,
Initial listing moment = 1584.1 tm (S)

So, maintain vessel upright same LM must be created on (P) side.
Again, we know that  LM = w x d
1584.1 tm   = (w x 12)
= (1584.1/12)
= 132 t

Hence , 132t SW should transferred to (P) side to keel the vessel upright.

  1. A bulk is carrier presently of 12250t, KM 9.8m , KG 9.0m has a list of 6dee to starboard .she then load 1250 t of ore (KG 8m , 2m to stbd of centre line ) and discharges 250t of ore (KG 2m 5m , from star board of centre line). 160t of SW ballast is then transferred from the stbd shoulder tank to the port DB tank vertically downwards by 9m and transversely  by 10m ) Find the final list assuming that they are no slack tanks given the final KM is 9.6m .
Solution:
Ship’s wt
KG
VM
D
LM
12250t9.0m110250 1116.57(S)
(+) 1250t8m(+) 100002S2500(S)
(-) 250t2m(-) 5005S1250(P)
160t (Shift)9m(-) 144010P1600(P)

Final W = 13250t                                     Final VM = 118310 tm                 FLM =766.57(S)

We know that :

Final KG = (Final VM) /(Final W)
= (118310/13250)
= 8.929 m

Again,

Initial LM = ( W x GM tanθ)
= (12250 x 0.8 x tan 6.50)
= 1116.57 (S)

GM = ( KM – KG)
= (9.6 – 8.929)
= 0.671m

tanθ = ( Final LM)/( W x GM)
= (766.57)/( 13250 x 0.671)
= 4.93 degree(S)

  1. From a ship of W 8000t KM 8.6m KG 8.0m some deck cargo was washed overboard KG 10m, 8m from the centre line .if the resultant list is 3 degree, find the quantity of cargo lost.
Solution:
Ship’s wtKGVMLM
8000 t864000 
X t10(-) 10X 

Final W = (8000 – X) t                                Final VM =(64000 – 10X)tm

We know that:
Final KG = (Final VM)/(Final W)
= (64000 – 10X)/(8000 –X)

Again, Final GM = (KM – Final KG)
= 8.6 – (64000 – 10X )/(8000 – X)
= 8.6 (8000 -X ) – (64000 – 10X) /(8000 –X)
= (68800 – 8.6X – 64000 + 10X)/(8000 – X)
= (4800 + 1.4X)/(8000 – X)

We know that:
Tan θ = (Final LM)/ (Final W x Final GM)

= 8X / ((8000 – X ) x (4800 + 1.4X)/(8000 –X))
= 8X /(4800 + 1.4X)
8X = tan 30 x (4800 + 1.4X)
= (251.56 + 0.073X)
(8X – 0 .073X) = 251.56

Hence, X = (251.56/7.927)
= 31.73t

  1. A ship of W 16000t, KM 7.5m, KG 6.0m , TPC 25, is listed 3 degree to port .Her present mean draft is 8.6m and she is to finish loading at 8.8m mean draft . Space is available 5m 0f the centre line, on either side. State how much cargo must be stowed on either side to finish upright
Solution :

W = 16000t,
KM, 7.5m KG = 6.0m,
Initial  GM = 1.5 m
TPC = 25,
List  = 3 degree P
Present draft = 8.6m
Final draft = 8.8

Sinkage available = (8.8 – 8.6)m
= 0.2m
= 20cm

Now , Cargo to load = (Sinkage x TPC)
= (20 x 25)
=  500 t

We know that :
LM  =(W x GM tanθ)

Again , Initial LM = (16000 X 1.5 X Tan 30)
= 1257.78 tm

Let  ‘X’  t 0f cargo will load on port side
Cargo to be load on stbd side = (500 – X) t
So, final LM caused due to loading of cargo would be

LM(1)   =  (Weight x distance)
= (X x 5)
=   5X (P)

LM(2)   =  (Weight x distance)
=   (500 –X) x  5
=   (2500 – 5X ) (S)

As per above calculation ,
Initial LM =    1257.78 ( P)

Hence  to keep vessel upright ,
LM(S)   =    LM(P)
(2500 – 5X)  = (1257.7 + 5X)
2500 – 5X – 5X   = 1257.7
(- 10X)  = (1257.7 – 2500)
X  = (1242.3/10)
= 124 .3 t

Hence cargo  loaded is  124.23 t on  port side
So, cargo loaded in stbd is (500 – 124.23)
= 375.7 t

  1. A ship displacing 12000t has KM 9.0m KG 7.25m. A 200t heavy lift is to be loaded by ship’s jumbo whose head is 24m above the Keel . Find
  • The list as soon as the derrick picks up the weight from the wharf on the stbd side with an outrech of 15m.
  • The list when the weight is placed on the upper deck KG 10m , 7m stbd of the centre line.

Solution:

Ship’s wt
KG
VM
LM
120007.25870000
(+)200 t244800(15 x200)s

Final W = 12200                                 Final VM = 91800 tm      FLM = 3000

We know that :
Final KG = (Final VM/Final W)
Final KG = (91800/12200)
= 7.524m

Again , we can calculate
GM = (KM – KG)
= (9.0 – 7.524)
= 1.476m

We know that :
tanθ     =  Final LM / (W x GM)
= 3000/(12200 x 1.4476)
=  9 degree 45min

Ship’s wt
FKG
VM
LM
120007.25870000
2001020001400 S

Final W= 12200 t                                  Final VM = 89200                 FLM 1400 S

We know that:

Final KG = (Final VM/Final W)
Final KG = (89200/12200)
= 7.311m

Again, Final GM = (KM –KG)
= (9.0 – 7.311)
= 1.689m

tanθ   =  1400/(12200 x 1.689)
= 3 degree 53.2min

  1. A ship of W 10000t, KM 7.3m , KG 6.8m is listed 5degre to port . A heavy lift weighing 100t, lying 6m to port 0f the centre line and KG 10.0 m, is to be shifted to the lower hold KG 2.0m on the centre line of the ship , by the ship’s jumbo derrick whose head is 28m above the keel. Find
  • The list as soon as the derrick takes the load .
  • The list when the derrick swings to the load to the centre line.
  • The list after the shifting is over.

Solution:

Displacement (W) = 10000t,
KM = 7.3m &
KG = 6.8m ,

List = 5 degree(P)
w = 100t, d = 6m to port of the CL KG = 10m & KG = 2.0m
Derrick head  = 28m .

We can calculate :
Initial LM = (W x GM x tanθ)
= (10000 x 0.5 x tan50)
= 437.44 P

Ship’s wt
KG
VM
LM
100006.86800437.44P
100t18m(+) 18000

Final W =10000t                                Final VM = 69800tm         FLM=437.44 P

We know that:
Final KG = (Final VM/Final W)
= (74800/10000)
= 6.98m

Again ,  Final GM  = ( KM – KG)
= (7.3 – 6.98)
= 0.32m

tanθ =LM/(W x GM)
= 437.44/(10000 x 0.32)
= 7.78 degree P

Case – 2

We know that :
List = ( 100 x 6)
= 600 tm (S)

Tan θ    = (LM/ (W x GM)
= 600/(10000 x 0.32)
= 10.6 degree

Initial LM as we calculated   = 437.44 tm(P)
LM caused = 600 tm (S)

Now resultant  listing moment = (600 – 437.44)
= 162.56 tm S

Again,Tanθ  = 162.56 /(10,000 x 0.32)
= 2.9 degree S

Case – 3

Ship’s wt
KG
VM
LM
100006.868000437.44 P
1008– 800600.00 S

Final W= 10000t                                     Final VM= 67200tm            FLM= 162.56 S

We know that:
Final KG = (Final VM/Final W)
= (67200/10000)
= 6.72m

Now,    Final GM = (KM – KG)
=(7.3 – 6.72)m
= 0.58m

Again, tanθ  = Final LM/ (W  x GM)
= 162.56 /(10000 x 0.58)
= 1.6 degree (S)

  1. A ship of W 13000t, KM 8.75m, KG 8.0m, has the list of 6degree to starboard . A heavy-lift weighing 150t lying on the upper deck 9m above the keel and 5m stbd of the centre line , is to be discharge using the ship’s jumbo derrick whose head is 22m above the keel . Calculate
  • The list as soon as the load is taken by the derrick.
  • When the load the hanging over the port side of the ship with an outreach of 10m from the centre line .
  • After discharging the heavy-lift.

Solution:

Given :

Displacement (W) = 13000t,
KM = 8.75m &
KG = 8.0m
Initial  GM =0.75m
List = 6 degree stbd ,
w  = 150t,
KG = 9m lying  = 5m to stbd ofC
derrick = 22m

Case – 1

Ship’s wt
KG
VM
LM
130008.0104,0001024.76
15013.0+ 19500

Final W= 13000t                                       Final VM= 105950tm       FLM =1024.76

We know that:
Final KG = (Final VM/Final W)
= (105950/13,000)
Final KG = 8.15m

Again we can calculate
Final GM  = (KM- KG)
= (8.75 – 8.15)
= 0.6m

Now,   Tan  =FLM /(W x  GM)
= 1024.76/(13000 x 0.6 )
= 7degree 48minute (S)

Case – 2

When  the load is hanging over the portside of the ship with on outreach of 10m from CL.
In this case the GM will be same but listing moment will change.

Listing moment caused =( W x d)
= (150 x 15)
= 2250 tm (P)

Hence, Final LM = (2250 – 1024.76)
= 1225.24 P

Tanθ = 1225.24/(13000 x 0.6)
= 8.9 degree P

Case – 3

After discharging the heavy lift.

Ship’s wt
KG
VM
LM
130008.01040001024.76(S)
(-) 150t9.0(-) 1350750 (P)

Final W= 12850                             Final VM = 102650 tm               FLM = 274.76(S)

We know that:
Final KG = (Final VM/ final W)
Final KG = (102650/12850)
Final KG = 7.988m

Now,Final GM =( KM  – KG)
=(8.75 – 7.988)
= 0.762m

We know that :
Tanθ = FLM/( W x GM)
= 274.76/(12850 x 0.762)
= 1.6 degree (S).

  1. A ship of 10000t displacement is floating in SW and has KM of 10.8m and KG of 9.0m .she is listing 10degre to stbd. she has two rectangular deep tanks, one on either side , each 12m long ,12m wide and 9m deep . The stbd tank is full of FW while the port one is empty . If FW is to be transferred from the stbd tank to the port one , find
  • The quantity of FW to transfer to bring the ship upright.
  • The list if one third of the original FW in the stbd tank is transferred to the port tank.

NOTE : Fluid GM should be used here.

Solution :

Displacement(W)  = 10000t,
KM = 10.8m, KG = 9.0M &  GM = 1.8M
List  = 10 degree  stbd

Volume of tank   = (L x B x H)
= (12 x 12 x 9 )m3

Initial LM = (W x GM x tan 100 )
= (10000 x 1.8 x tan 100)
= 3173.88tm

Case -1

To bring the ship upright the stbd LM must be equal to port side LM.

LM = (Weight  x Distance)
3173.88 tm  =  ( W x 12)

Hence W = (3173.88/ 12)
= 264.49 t

Case -2

Volume of the tank = (L x B x D)
= (12 x 12 x 9)
= 1296m3

Now, Weight of FW on stbd tank = (Volume x Density)
= (1296 x 1)
= 1296 t.

Now, calculating 1/3 of mass of FW
= (1296 x 1/3)
= 432t.

Ship’s wtKGVMLM
100009.0900003173.88 S
4326t(-) 25925184.00 P

Final W = 10000                       Final VM = 87408 tm           2010.12 P

Final KG = (Final VM/Final W)
Final KG =   (87408/10000)
= 8.741m

Solid GM = (KM – KG)
= (10.8 – 8.741)
= 2.059m

We know that :

FSC = (LB3 x di) / (12 x W)
= (12 x 12x 1) /(12 x 10000)
= 0.3456m

Fluid GM = (2.059 – 0.3456)

tanθ  = (2010.12/(10,000 x 1.7134)
=6.69 degree P

  1. On a ship 8000t displacement , 50t is shifted transversely by 4m. Find the list if the total FSM is 1216tm, KM 7.0m , KG 6.4m.
Solution :

W = 8,000t ,
w = 50t ,
transverse list =4m
FSM = 1216t ,
KM = 7.0m ,
KG = 6.4m

GM  = 7 – 6.4
= 0.6m

LM   =  W x D
= 50 x 4
= 200t

FSC = FSM / W
= 1216 / 8000
= 0.152m

Fluid GM = 0.6 – 0.152
= 0.448m

tanθ  = 200 / 8000 x 0.448
= 3.2deg11.6minut

  1. A ship has W 10000 t, KM 7.8m , KG 7.075m, and is upright . NO3 port and stbd DB tanks are full of HFO RD 0.95. Each tank is rectangular , 15m long ,12 wide and 2m deep . calculate the list when HFO is consumed from NO3 stbd until the sounding is 1.2m.
Solution :

W = 10000t,
KM = 7.8 ,
KG = 7.075,
GM = 7.8 – 7.075 = 0.725m ,
LM = 0
RD 0f HFO = 0.95

We know that, L x B x D = 15 x 12 x 2
Volume of stbd tank = 15 x 12 x 2
= 360m3

Wt of consumed HFO = 15 x 12 x 0.8 x 0.95
= 136.8t

LM = W x D = ? doubt why ‘d’ taken as (6m)

FSC = LB cub x di / 12W
= 15 x 12 x 0.95 / 12 x 9863.2
=  0.208 m

Fluid GM = 0.649 – 0 .208
= 0.441m

 

  1. A vessel displacing 9000t has KM 8.02m, KG 7.5m , and is up right . she loads 250t KG 12m, 3m to stbd of the centre line ; loads 1000t KG 3m, 1m to port of the centre line; discharges 250t KG 8m, 2m to stbd of the centreline.100t of cargo is then shifted transversly 3m to stbd . If the total FSM is 1200tm , Calculate the final list.
Solution :
Ship’s wt
KG
VM
D
LM
9000t7.5m67500 0
+ 250t12m+ 3,0003s750s
+ 1000t3m+ 3,0001p1000p
-250t8m– 2,0002 (s)500p
100       –3s300s

FW    = 10,000                 FKG = FVM / FW              71.500                          450p

FKG   = 71500 / 10,000
FKG = 7.15m

Final solid GM  = 8.02 – 7.15
= 0.87m

FSC  = FSM / FW
= 1200 / 10000
= 0.12m

Fluid GM   =  0.87 – 0.12
= 0.75m

Tanthi  =   LM / W x GM
= 450 / 10000 x 0.7
= 3.4p
= 3d26.

  1. A ship of 14000t displacement , KM 9.0m , KG 7.8m has a total FSM of 2100tm and is listed 8degre to port . How many tonnes must be shifted transversely by 10m to upright the ship?
Solution :

W = 14000t ,
KM = 9.0m KG = 7.8m ,
Initial GM = 9 – 7.8 = 1.2
FSM = 2100 tm thit = 8degp,
d = 10m

FSC = 2100 / 1400
=0.15

Fluid GM = 1.2 – 0.15
= 1.05m

LM = W x GM tanthi
= 1400 x 1.05 x tan thit
= 2065.1tm

To upright the vessel we have to shift ballast from to which is to the same LM of on stbd

We know that

LM = W x D
2065.1  = W x 10
W = 2065.1 / 10 = 206.51t to ss

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