Q1. A vessel of length 142m, displacement 16130t, KM 8.24m, KG 7.26m MCTC 192 mt, TPC 23.4t, LCF 71m, FSM 1372 mt, at a draft of F 6.70m, A 8.86m, ground on a rock. The draft then F 5.90m, A 9.30m.
Calculate :
-
The upthrust provided by the rock.
-
The position with respect to AP where the grounding occurred.
-
Her virtual GM (FI.) then
-
The rise of tide required for her to refloat.
Solution –
- Draft before Grounding = 6.70m, Aft = 8.86m
Trim = 2.16m
Correction to aft draft = 2.16 ⨯ 71/142
= 1.08m
∴ Hydrostatic draft = 7.78m
Draft after grounding = 5.90 m fwd
Trim = 3.4m
Correction to aft draft = 3.4 ⨯ 71/142
∴ Hydrostatic draft = 7.60m
Rise = P/ TPC ⨯ 100
P = (0.18 ⨯ 23.4 ⨯ 100)
= 421.2 tonnes
- Trim before grounding = 2.16m,
Trim after grounding = 3.40m
Change in trim = 1.24m
1.24 = 421.2 ⨯ d/192 ⨯ 100
d = 56.52 m from COF
From AP distance = 71.00 + 56.52 = 127.52m
- GG1 = (P ⨯ KM)/W
= 421.2 ⨯ 8.24/16130
= 0.215m
New GM = 0.765m
FSC = 1372/(16130 – 421.2)
= 0.0873m
GM (fluid) = 0.678m
- Rise of tide required = P/ TPC + P ⨯ d/ MCTC ⨯ d/ LBP
= 421.2/23.4 + 421.2 ⨯ 56.52 ⨯ 56.52/192 ⨯ 142
= 18 + 49.35
= 67.35cm
Q2. A vessel of length 140m is floating at draft forward 5.23m Aft 5.74m runs aground lightly on a rock 12m abaft the stern. The tide falls by 80cms. Given the following data estimate the virtual GM of the vessel and her draft forward and aft after the fall of tide. KG 5.80m, KM 6.40m, MCTC 130tm, TPC 18t, AF 73m, Displacement 18500t.
Solution –
Distance of grounding point from COF = 55m
We have P = trim ⨯ MCTC/d
Change in draft = P/TPC + P ⨯ d/MCTC ⨯ d/LBP
So, 80 = P/18 + P ⨯ 55/130 ⨯ 55/140
Or, P = 360.74
GG1 = (P ⨯ KM)/ W
= 0.125m
Virtual GM = GM – Virtual loss of GM( GG1)
= 0.475m
Also,
Trim caused = Tc = (P ⨯ d)/ MCTC ⨯ 100
= 1.526m
Ta = 0.796m,
Tf = 0.730m
Rise = P/ TPC ⨯ 100
= 0.200 tonnes
