# Stability – I : Chapter 4

###### Solution :

Given:-
RD of DW = 1.010,
Present Draft =     8.2m
Cargo can be load up to draft = 8.4m,
So, Sinkage = 0.2m = 20cm

TPC in SW = (A / 100) x (1.025)
40 = (40/100) x (1.025)
A = (40 x100)/ (1.025)
= 3902.44m2

Now, TPC in DW = (A/100) x (1.010)
= (3902.44 / 100) x(1.010)
=39.41 t/cm.

Cargo can be load = (39.41 x 20)
= 788.2 t

#### 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.

• sumit says:

many questions have wrong answers in various exercises,,at least book ke end me answers ko chk kar lete…anyways thanks for this website

• sumit says:

many questions have wrong answers in various exercises,,at least book ke end me answers ko chk kar lete…

• zafer says:

what is the draft of a box shape steel (L=4 meters , B= 4 meters , Depth = 0.60 meters)
floating in a salt water density 1.028 and the total weight of the steel box with steel on top of it is 1500 kg

• JJ says:

question 5 is wrong. you forget to take the u/w vol and multiply by the sw density.

• dibyendu bagchi says:

sir exercise no 4 problem no 7 draft you had taken is wrong it is not 4.1 it is 6.1 mtr

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