# Stability – I : Chapter 5

A ship of 16000t displacement and TPC is 20 floating in SW at a draft of 8.0m . Find her draft in FW .

###### Solution:

Displacement (W) = 16000 t
TPC = 20 & SW draft = 8.0m

We know that:

Displacement when in SW = ( L X B x draft) x  1.025
Displacement when in FW  =( L X B X D) x 1

It is understood that displacement of ship will remain constant , as displacement is independent of change in density ,(is referred as MASS).

So, (L x B x 8) x 1.025 =  (L x B x draft) x 1
Hence draft  = ( 8 x 1.025)
= 8.2 m

###### 2nd  Method :

As we know that

FWA = (W/40TPC)
= 16000/(40 x 20)
= 20cm

We can calculated:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1) x 20 / 0.025
= 20c
=0.2m

So new draft of ship in FW = (8.0 +0.2)
= 8.2 m

###### Solution :

FWA = 180mm = 18cm

We can calculate :

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.008 – 1.025 ) x 18 / 0.025
=(0.017 x 18)/0.025
= 12.24c
= 0.12m

Here , Change in draft is  0.122m and it will be rise.

###### Solution :

FWA = 160mm = 16cm

We can calculate:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.010 – 1)x 16 /0.025
= 0.01 x 16 /0.025
= 6.4 cm
= 0.064 m

Here change in draft would lead to sinkage.

#### Amit Sharma

Graduated from M.E.R.I. Mumbai (Mumbai University), After a brief sailing founded this website with the idea to bring the maritime education online which must be free and available for all at all times and to find basic solutions that are of extreme importance to a seafarer by our innovative ideas.

• Joe Bee says:

A ship loads in fresh water to her salt water marks and proceeds along a
river to a second port consuming 20 tonnes of bunkers. At the second port,
where the density is 1016 kg per cu. m, after 120 tonnes of cargo have been