A ship of 7500 tonnes displacement has KM 8.6m, KG 7.8m and 20m beam. A quantity of deck cargo is lost from the starboard side (KG 12m, and centre of gravity 6m in from the rail). If the resulting list is 3 degrees 20 minutes to port, find how much deck cargo was lost.

A ship of 7500 tonnes displacement has KM 8.6m, KG 7.8m and 20m beam. A quantity of deck cargo is lost from the starboard side (KG 12m, and centre of gravity 6m in from the rail). If the resulting list is 3 degrees 20 minutes to port, find how much deck cargo was lost.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 14.7

 

Solution:

Initial GM = KM – KG = 8.6 – 7.8 = 0.8 m

Let, x tons of deck cargo were lost

d = 12-7.8 = 4.2 m

GG₂ = w*d/7500-w

=> GG₂ = x*4.2/7500-x

 

d = beam/2 -6 = 20/2 – 6 = 4 m

G₂G₁ = w*d/7500-w

=> G₂G₁ = x*4/7500-w

 

G₂M = GG₂ + GM

         = 4.2x/7500-x + 0.8

          = 6000+3.4x/7500-x

 

Tan (3.33⁰) = (4x/7500-x) / (6000+3.4x/7500-x)

=> x = 91.9 tons

OR

w	KG	TCG	Moment (keel)	Moment (LCG)
7500	7.8	0 (p)	58500	0
-w	12	-4 (s)	-12w	+4w

 

GG₁ = 4w/7500-w

 

KG` = 58500-12w / 7500-w

 

GM = KM – KG`

=> GM = 8.6 – (58500-12w/7500-w)

=> GM = 6000+3.4w/7500-w

 

Tan 3.33⁰ = GG₁/GM

=> 0.058 = 4w/6000+3.4w

=> 349.46 + 0.198w = 4w

=> w = 91.9 tons