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
 GG_{1} = (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
GG_{1} = (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
Fwd  Aft  
Draft  5.23m  5.74m 
Rise  0.200m  .200m 
0.730m  +0.796m  
4.300m  6.336m 
Q3. A vessel of displacement 7200 tonnes. Length 120m, MCTC 110tm, KG 6.0m, TPC 16 tonnes, center of floatation 2 mts forward of midship ground on a rock 10mts. Abaft her forward perpendicular. Given initial draft F : 4.2m, A : 5.1m.

Find the fall in tide in cm, that will make the vessel unstable, given KM at that moment equal to 6.4 mts.

What will be the drafts forward and aft at that moment.
Solution –
 For vessel to unstable GM should be zero.
GG_{1} = 0.4m
So,
0.4 = P ⨯ 6.4/ 7200
P = 450 tonnes
We can calculate fall in tide –
Fall in tide = P/ TPC + ( P x AL / MCTC) + AL/ LBP
= 450/16 + 450 ⨯ 48/110 ⨯ 48/120
= 1.067m
 Trim caused = 450 ⨯ 48/110 ⨯ 100 = 1.964m
Ta = 1.014,
Tf = ( Tc – Ta)
= 0.95m
Rise = 450/1600
= 0.281m
Fwd  Aft  
Draft  4.200m  5.100m 
0.281m  0.281m  
3.919m  4.819m  
0.950m  +1.014m  
2.969m  5.833m 
Q4. A vessel of length 120m, Displacement 6400 tonnes, TPC 15 tonnes, MCTC 110tm, Km 6.4m, KG 5.1m, CF 2m for’ d of the midlength grounds on a rock 45m for’d of midlength the tide then by 80 cm. Calculate the residual GM then.
Solution –
We can calculate fall in tide –
Fall in tide = P/ TPC + ( P x AL / MCTC) + AL/ LBP
80 = P/15 + P ⨯ 43/110 ⨯ 43/120
So, P = 386.95 cm.
GG1 = 386.95 ⨯ 6.4/ 6400
= 0.3869m
Residual GM = 0.91m
Q5. A vessel of L = 140m, CF = 2m FWd of midship, KM = 6.1m, KG = 5.5m, MCTC = 130 mt, w = 5700t at a draft of F = 4.15m, grounds on a rpck 10 mts for’d of her stern. The tide then falls 80cm. Calculate her virtual GM then and her draft f & A, TPC = 16.
Solution –
We can calculate fall in tide –
Fall in tide = P/ TPC + ( P x AL / MCTC) + AL/ LBP
80 =P/15 + P ⨯ 58/130 ⨯ 58/140
P = 323.44 tonnes
GG_{1} = 0.346m
GM = 0.254m
Rise = P/ TPC ⨯ 100
= 0.202m
Trim caused = 323.44 ⨯ 58/130 ⨯ 100
= 144m
Ta = 0.701m,
Tf = ( Tc – Ta)
= 0.739m
Fwd  Aft  
Draft  4.15m  4.851m 
Rise  0.202m  0.202m 
+0.739m  0.701m  
4.687m  3.948m 
Q6. Vessel of length 120m, LCF = 2m, Aft of midship, Draft = 4.70m. For’d = 5.20 m, Aft TPC = 15 t, MCTC = 110 tm, KM = 5.9, KG = 5.1 m. Grounds on a rock 10 meters abaft her stern. The tide then falls 80cm. Calculate the drafts F & A then and state whether she would remain upright.
Solution –
Not possible to calculate displacement not given.
Q7. A vessel of displacement 12200 t, MCTC = 200 tm, COF = 3 m, for’d of midship, length 152 m, KG = 7.2 m, KM = 8.1 m has draft for’d 5.1m, A 5.2 m. This vessel runs aground on an isolated rock 11.5 m aft of for’d perpendicular. Find the virtual Gm and Draft for’d and Aft. Given TPC = 25t, fall in tide 1 meter.
Solution –
W = 12200, distance from COF of grounding location = 61.5m
We can calculate fall in tide –
Fall in tide = P/ TPC + ( P x AL / MCTC) + AL/ LBP
100 = P/25 + P ⨯ 58/200 ⨯ 58/152
P = 608.21 tonnes.
GG_{1} = (P ⨯ KM)/W
GG1 = 0.404
Virtual GM = 0.495m.
Rise = P/ TPC ⨯ 100
Rise = 0.243m
T_{C }= (P x D)/ MCTC
= (608.21x 58) /200
Tc = 1.870m,
We know that –
Ta = 0.972m,
Tf = ( Tc – Ta)
= 0.898m
Fwd  Aft  
Draft  5.1m  5.2m 
0.243m  0.243m  
0.898m  +0.972m  
3.959m  5.929m 
Q8. A vessel of displacement 15185t, MCTC = 378 tm, TPC = 36.9t, Draft for’d 4.4m, AFT = 4.7 m, Length 180m, runs lightly aground on an isolated wok 7.5m, abaft for’d end. Find the virtual GM and draft for’d and aft COF 4.06m for’d of midship. Given KG = 8.0m, KM = 8.5m, fall in tide 80cm.
Solution –
Distance from COF = 78.44m
We can calculate fall in tide –
Fall in tide = P/ TPC + ( P x AL / MCTC) + AL/ LBP
80 = P/36.9 + P ⨯ 78.44/378 ⨯ 78.44/180
P = 680.67 tonnes
GG_{1} = (P ⨯ KM)/W
= 0.381m
Virtual GM = 0.119m
Rise = P/ TPC ⨯ 100
Rise = 0.184m
T_{C }= (P x D)/ MCTC
= (680.67 x 78.44) / 378
Tc = 1.412m,
We know that –
Ta = ( Tc x AL ) / LBP
= (1.412 x
Ta = 0.674m,
Tf = ( Tc – Ta)
= 0.737m
Fwd  Aft  
Draft  4.4m  4.7m 
0.184m  0.184m  
0.674m  +0.737m  
3.542m  5.253m 
Q9. A vessel of Length L = 150 mtrs, CF = 2.5m forward of midship, KM = 6.0M, KG = 5.4m, MCTC = 120, W = 6000t, TPC = 15, ata draft of F : 4.25m, A : 5.00m ground on a rock at midship. The Tide then falls and draft at midship is observed to be 4.46m. Calculate her virtual GM then and her drafts forward and aft.
Solution –
Correction to aft draft = 0.75 ⨯ 75/150
= 0.375
Midship draft = 4.625
Fall in water at midship = (4.625 – 4.46)
= 0.165
16.5 = P/15 + P⨯ 75/120 ⨯75/150
P = 43.51 tonnes