Stability – I : Chapter 11

Case – 3

After discharging the heavy lift.

Ship’s wt
KG
VM
LM
13000 8.0 104000 1024.76(S)
(-) 150t 9.0 (-) 1350 750 (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

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