Miscelleneous

C L DUBEY – EXERCISE – 09 (GROUNDING)

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 :
  1. The upthrust provided by the rock.
  2. The position with respect to AP where the grounding occurred.
  3. Her virtual GM (FI.) then
  4. 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

 FwdAft
Draft 5.23m5.74m
Rise-0.200m-.200m
 -0.730m+0.796m
 4.300m6.336m
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.
  1. Find the fall in tide in cm, that will make the vessel unstable, given KM at that moment equal to 6.4 mts.
  2. What will be the drafts forward and aft at that moment.
Solution –
  • For vessel to unstable GM should be zero.

GG1 = 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

 FwdAft
Draft 4.200m5.100m
-0.281m-0.281m
3.919m4.819m
-0.950m+1.014m
2.969m5.833m
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
GG1 = 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

 FwdAft
Draft 4.15m4.851m
Rise-0.202m-0.202m
 +0.739m-0.701m
 4.687m3.948m
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.
GG1 = (P ⨯ KM)/W
GG1 = 0.404
Virtual GM = 0.495m.

Rise = P/ TPC ⨯ 100
Rise = 0.243m

TC = (P x D)/ MCTC
= (608.21x 58) /200
Tc = 1.870m,

We know that –
Ta = 0.972m,
Tf =   ( Tc – Ta)
= 0.898m

 FwdAft
Draft 5.1m5.2m
 -0.243m-0.243m
 -0.898m+0.972m
 3.959m5.929m
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
GG1 = (P ⨯ KM)/W
= 0.381m
Virtual GM = 0.119m

Rise = P/ TPC ⨯ 100
Rise = 0.184m

TC = (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

 FwdAft
Draft4.4m4.7m
 -0.184m-0.184m
 -0.674m+0.737m
 3.542m5.253m
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

manish-mayank

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Manish Mayank

Graduated from M.E.R.I. (Mumbai). A cool, calm, composed and the brain behind the development of the database. The strong will to contribute in maritime education and to present it in completely different and innovative way is his source of inspiration.

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