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Solution:
Dockwater RD = 1.005,
Stbd WNA mark = 30mm below
Port WNA mark = 60mm below
Mean draft = (30 + 60) / 2
= 45mm
=4.5cm,below the water line.

Distance from WNA to water(W) = 50mm
= 5cm
So, sinkage available = (50 – 45)
= 5mm
= 0.5cm
We know that :
Water (W) is 1/48 of summer draft
= (1/48 x 8.4)
= 0.175m
= 17.5cm
Sinkage = (17.5 + 0.5 )
= 18cm
Total sinkage = (18 + 12.8)
= 30.8cm
As we know:
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.025 – 1.005 )x 16 / 0.025
= 12.8cm
Again, TPC = (A/100) x density
30 = (A/100) x 1.025
A = 30/ 1.025
A = 29.268 m2
Now , TPC for density of 1.005
TPC = (A/100) x 1.005
= (29.268 x 1.005)
= 29.41 t/cm.
Hence ,cargo can be loaded = (30.8 x 29.41)
= 905.82 t