C L DUBEY – EXERCISE – 08 (DRY DOCKING)

Therefore, New GM = (8.205 – 0.128)m
= 8.076m

• When draft is 7.8m

Reduction in hydrostatic draft = rise = (8.205 – 7.8)m
= 0.405m
0.405 = P/4800
P = 1944 tonnes

GG₁ = P ⨯ KM/W
= 0.417m

 Therefore, New GM = 0.423m

• When she become unstable GG₁ = old GM

= (8.2 – 7.46)/Old GM
= P ⨯ 8.2/38700
P = 3492.43 tonnes

Rise = P/TPC ⨯ 100
= 0.727m

19. A ship is being docked at a displacement 26240 tonnes. LBP 184m, LCF 4m abaft of amidships. KG 7.4 m, KM 7.8m, TPC 38, MCTC 300. Calculate the maximum allowable stern trim to ensure GM is at least 0.1m when landing overall.

Solution –

Old GM = KM – KG
= (7.8 – 7.4)m
= 0.4m

New GM = 0.1m ( As per question )
We know that –
GG₁ = P ⨯ KM/W
0.3 = P ⨯ 7.8/26240
P = 1009.23