sqa chief mates (unlimited) solved stability numerical (2005 - 2013)

SQA Stability – July 2005

1. A vessel is to transit a canal with a minimum clearance of 0.30m under a bridge, the underside of which is 20.24m above the waterline.

Present draughts in FW (RD 1.000): Forward 5.22m Aft 6.38m

The foremast is 112m FOAP and extends 25.92m above the keel. The aft mast is 37m FOAP and extends 26.94m above the keel. (Assume masts are perpendicular to the waterline throughout)

Using the Stability Data Booklet, calculate EACH of the following:

(a) The final draughts forward and aft in order to pass under the bridge with minimum clearance;

(b) The minimum weight of ballast to load in order to pass under the bridge with minimum clearance.

a) Air draught required = 20.24 – 0.30 = 19.94m

Foremast Aft Mast Height above the keel = 25.920m 26.940m Air draught required = 19.940m 19.940m ∴ Draught at Masts = 5.980m 7.000m

∴ Trim at Masts = 7.000 – 5.980 = 1.020m by stern

Correction for DraughtF

Correction for DraughtF = LBP – Dist. of Foremast from AP Trim at Masts Dist. of Foremast from AP – Dist. of Aft mast from AP

∴ Correction for DraughtF = (137.5 – 112.0) x 1.020 = 0.347m (112.0 – 37.0)

Correction for DraughtA

Correction for DraughtA = Dist. of Aft mast from AP Trim at Masts Dist. of Foremast from AP – Dist. of Aft mast from AP

∴ Correction for DraughtA = 37.0 x 1.020 = 0.503m (112.0 – 37.0)

Calculate required draft at perpendiculars

Foremast Aft Mast Required draught = 5.980m 7.000m Required correction = (-) 0.347m (+) 0.503m ∴ Draft at perpendiculars = 5.633m 7.503m

b) Calculate Initial Displacement

Initial trim = 6.380 – 5.220 = 1.160m by stern Initial AMD = (6.380 + 5.220) = 5.800m 2 With AMD, from hydrostatic tables, LCF = 68.57m TMD = 6.380 – (1.160 x 68.57) = 5.802m 137.5

∴ DisplacementTMDi = 11559 + (219 x 0.002) = 11563.38 t ≈ 11563 t 0.10

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Calculate Final Displacement

Final trim = 7.503 – 5.633 = 1.870m by stern Final AMD = (7.503 + 5.633) = 6.568m 2

With AMD, from hydrostatic tables, LCF = 67.90 – (0.11 x 0.068) = 67.83m 0.10 TMD = 7.503 – (1.870 x 67.83) = 6.581m 137.5

∴ DisplacementTMDf = 13102 + (222 x 0.081) = 13281.82 t ≈ 13282 t 0.10

Calculate minimum weight of ballast to load in order to achieve required trim

∴ Ballast to load = DisplacementTMDf – DisplacementTMDi

= 13282 – 11563 = 1719 t

2. A vessel is planning to enter dry-dock in dock water of relative density 1.015. Present draughts: Forward 3.00m Aft 3.80m KG 10.50m

a) Using the Stability Data Booklet, calculate the maximum trim at which the vessel can enter dry-dock so as to maintain a GM of at least 0.10m at the critical instant. Assume KM remains constant.

b) Explain why it is beneficial to have a small stern trim when entering dry-dock.

a) Calculate Initial TMD

Initial Trim = 3.800 – 3.000 = 0.800m Initial AMD = (3.800 + 3.000) = 3.400m 2 With AMD, from hydrostatic tables,

LCF = 70.20m TMD = 3.800 – (0.800 x 70.20) = 3.392m 137.5

Calculate Initial Δ, MCTC, LCF & GM

With TMD, from hydrostatic tables,

ΔFW = 6261 + (205 x 0.092) = 6449.6 t 0.10 ∴ ΔDW = 6449.6 x 1.015 = 6546.3 t MCTC = 139.4 + (1 x 0.092) = 140.3tm 0.10 ∴ MCTCDW = 140.3 x 1.015 = 142.4tm

LCF = 70.27 – (0.07 x 0.092) = 70.21m 0.10

KMT = 11.18 – (0.23 x 0.092) = 10.97m 0.10 KG = 10.50m ∴ GM = 0.47m

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Calculate required ‘Loss of GM’

Initial GM = 0.47m Required GM = 0.10m ∴ Reqd. Loss of GM = 0.37m

Calculate the required ‘P’ Force with the required ‘Loss of GM’

Loss of GM = P x KMT Δ ∴ P = 0.37 x 6546.3 = 220.8 t 10.97

Calculate the required trim with the ‘P’ Force

P = trim x MCTC LCF ∴ Reqd. Trim = 220.8 x 70.21 = 108.86cms = 1.089 m by stern 142.4

b) It is beneficial to have a small stern trim when entering dry-dock because this allows:

1) A more gradual transfer of weight and thus a more gradual reduction in stability. 2) The vessel’s stern to be used as a pivot to align the keel along the blocks. 3) Some of the shores to be positioned early, working from aft to forward. 4) The stern frame is specially strengthened to accept the forces exerted on it during dry-

docking but there will be a maximum limit that must not be exceeded. If the ‘P’ force becomes too great, structural damage will occur.

5) Greater the trim, greater the ‘P’ force and greater loss of GM. Therefore a slight stern trim is beneficial.

3. A vessel’s loaded particulars in salt water are as follows:

Displacement 16000 t Fluid KG 8.20m

Using the Stability Data Booklet, compare the vessel’s stability with ALL the minimum stability criteria required by the current Load Line Regulations, commenting on the result.

Δ = 16000 t KGF = 8.20m

Determine Angle of Flooding (θF) for Δ = 16000 t

θF = 44.5°

Derive the KN values from the Hydrostatic Particulars

Δ Angle of Heel (°)

(Tonnes) 12° 20° 30° 40° 50° 60° 75°

16000 1.73 2.98 4.40 5.60 6.35 6.83 7.00

Calculate the GZ values

Heel (°) KN (m) KG Sin θ (m) GZ (m) 12 1.73 1.70 0.03 20 2.98 2.80 0.18 30 4.40 4.10 0.30 40 5.60 5.27 0.33

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50 6.35 6.28 0.07 60 6.83 7.10 - 0.27 75 7.00 7.92 - 0.92

Calculate the vessel’s GMF From hydrostatic particulars, using Δ = 16000 t in SW

KMT = 8.33 + (0.01 x 24) = 8.33m 236 ∴ GM = 8.33 – 8.20 = 0.13m

Construct the GZ Curve

Calculate the area under the GZ Curve from 0° → 30°

Angle (°) GZ (m) SM Product 0 0 1 0

10 0.02 3 0.06 20 0.18 3 0.54 30 0.30 1 0.30 ∑ 0.90

Area 0° → 30° = 3 x 10 x 0.90 = 0.059 m-rad 8 57.3 Calculate the area under the GZ Curve from 0° → 40°


10 0.02 4 0.08 20 0.18 2 0.36 30 0.30 4 1.20 40 0.33 1 0.33 ∑ 1.97

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 10 20 30 40 50 60 70 80

GZ Curve

© ARNAB SARKAR



∴ Area 30° → 40° = 0.115 – 0.059 = 0.056 m-rad

Summary

Actual L.L. Regulation Complies or Not

Area 0° → 30° 0.059 m-r Not less than 0.055 m-r Yes



Max. GZ 0.33m At least 0.20m Yes

Max. GZ Angle 40° Equal to or greater than 30° Yes

GMF 0.13m Not less than 0.15m No

© ARNAB SARKAR

SQA Stability – November 2005

1. A vessel’s present particulars are as follows: Forward draught: 8.00m Aft draught: 9.00m in salt water

A total of 320 t of cargo, LCG 70.00m FOAP, is to be discharged immediately and then the vessel is to proceed to an upriver berth where the relative density of the dock water is 1.005.

During the passage the following items of deadweight are consumed:

90 t of Heavy Fuel Oil from No.3 DB Centre tank 18 t of Diesel Oil from No.4 DB Port tank 18 t of Diesel Oil from No.4 DB Starboard tank 12 t of Fresh Water from After Fresh Water tank

Using the Stability Booklet Data, calculate the draughts fore and aft, on arrival at the upriver berth.

1. Forward Draught SW = 8.000m Aft Draught SW = 9.000m

Calculate trim & AMD

Trim = 9.000 – 8.000 = 1.000m by stern AMD = (9.000 + 8.000) = 8.500m 2

Calculate LCF & TMD

LCF = 65.95m FOAP TMD = 9.000 – (65.95 x 1.000) = 8.520m 137.5

From hydrostatic tables, using TMD = 8.520m, calculate Initial Δ, MCTCSW & LCB

Δ = 18119 + (240 x 0.02) = 18167 t 0.10

MCTCSW = 206.4 + (1.3 x 0.02) = 206.7 tm 0.10

LCB = 69.36 – (0.05 x 0.02) = 69.35m FOAP 0.10

Calculate Initial LCG

1.000 x 100 = 18167 x (69.35 – LCG) 206.7 ∴ LCG = 68.21m FOAP

Calculate Longitudinal Moments about the LCG

Condition Weight (t) LCG (m) Longitudinal Moments (tm) Initial Δ 18167 68.21 1239171.07 Cargo discharged (-) 320 70.00 (-) 22400.00 HFO 3DB (C) (-) 90 57.02 (-) 5131.80 DO 4DB (P) (-) 18 35.66 (-) 641.88 DO 4DB (S) (-) 18 35.50 (-) 639.00 FW AFWT (-) 12 28.67 (-) 344.04 Final Δ 17709 68.33 1210014.35

∴ Final LCG = 68.33m FOAP

© ARNAB SARKAR


Final ΔSW = 17709 t ∴ Volume in DW = 17709 = 17620.9 m3 ≈ 17621 m3 1.005 ∴ ΔFW = 17621 t

From hydrostatic tables, using ΔFW = 17621 t, calculate TMD, MCTCDW, LCB & LCF

TMD = 8.40 + (179 x 0.10) = 8.476m 235

MCTCFW = 200.1 + (179 x 1.3) = 201.1 tm 235 ∴ MCTCDW = 201.1 x 1.005 = 202.1 tm

LCB = 69.40 – (179 x 0.04) = 69.37m FOAP 235

LCF = 66.02 – (179 x 0.07) = 65.97m FOAP 235

Calculate change of trim

trim = 17709 x (69.37 – 68.33) = 91.130 cms = 0.911 m by stern 202.1

COTA = 0.911 x 65.97 = 0.437 m 137.5

∴ COTF = 0.911 – 0.437 = 0.474 m

Calculate Final Draughts

Draught Fwd = 8.476 – 0.474 = 8.002 m Draught Aft = 8.476 + 0.437 = 8.913 m

2. A vessel is to load a cargo of grain (Stowage Factor 1.57m3/t). Initial displacement 6400 t. Initial KG 6.48m.

The tween decks are to be loaded as follows:

No.1 TD Full No.2 TD Part Full – Ullage 1.25m No.3 TD Part Full – Ullage 3.00m No.4 TD Full

The Stability Data Booklet provides the necessary cargo compartment data for the vessel.

a) With the aid of Maximum Permissible Grain Heeling Moment Table included in the Stability Data Booklet, determine whether the vessel complies with the minimum criteria specified in the International Grain Code (IMO).

b) Calculate the ship’s approximate angle of list in the event of the grain shifting as assumed by the International Grain Code (IMO).

2. a) Initial Δ = 6400 t

Initial KG = 6.48m SF = 1.57 m3/t

Calculate the Final Δ, KG & VHM

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Comp. Ullage (m)

Volume (m3)

SF (m3/t) Tonnes KG

(m) Vert. Mom.

(tm) VHM (tm)

Initial Δ 6400 6.48 41472

No.1 TD Full 1695 1.57 1080 11.26 12161 352

No.2 TD 1.25 1560 1.57 994 10.65 10586 1642

No.3 TD 3.00 758 1.57 483 10.13 4893 3339

No.4 TD Full 1674 1.57 1066 10.57 11268 604

No.1 Hold Full 2215 1.57 1411 5.09 7182 410

No.2 Hold Full 4672 1.57 2976 4.95 14731 1285

No.3 Hold Full 1742 1.57 1110 4.94 5483 475

No.4 Hold Full 3474 1.57 2213 4.95 10954 910

No.5 Hold Full 2605 1.57 1659 8.76 14533 455

Final Δ 19392 6.87 133263 9472

Calculate AHM

AHM = 9472 = 6033 tm 1.57

Calculate Maximum Permissible Heeling Moment

Δ = 19392 t KGF = 6.87m

Interpolation

Δ KGF = 6.80 KGF = 6.87 KGF = 6.90

19000 7577 7283.0 7157

19392 7428.3

19500 7770 7468.3 7339

Holding Δ = 19000 t constant, using variable KGF = 6.87m,

7577 – (420 x 0.07) = 7283.0 tm 0.10


7770 – (431 x 0.07) = 7468.3 tm 0.10

Holding KGF = 6.87m constant, using variable Δ = 19392 t,

7283 + (185.3 x 392) = 7428.3 tm 500

∴ PHM for Δ = 19392 t & KGF = 6.87m = 7428.3 tm

As AHM < PHM, vessel complies with the minimum criteria.

b) Calculate approximate angle of list in event of grain shift

Angle of Heel = 6033 x 12° = 9.7° 7428.3

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3. a) A vessel, initially upright, is to carry out an inclining test.

Present displacement 5700 t, KM 10.83m.

Total weights on board during the experiment:

Ballast 370 t, KG 3.47m, tank full. Bunkers 165 t, KG 3.98m, free surface moment 956 tm. Water 95 t, KG 4.44m, slack tank. Free surface moment 910 tm. Boiler water 19 t, KG 4.18m, free surface moment 102 tm. Inclining weights 56 t, KG 8.44m

A deck crane weights 19 t and still ashore will be fitted on the vessel at a KG of 9.74m at a later date.

The plumblines have an effective vertical length of 7.85m. The inclining weights are shifted transversely 7.0m on each occasion and the mean horizontal deflection of the plumb line is 0.65m.

Calculate the vessel’s lightship KG.

b) Explain why a vessel’s lightship KG may change over a period of time.

3. a) Δ = 5700 t KM = 10.83m Plumb length = 7.85m Deflection = 0.65m Trans shift = 7.00m

Calculate GMF

GMF = w x d x plum length = 56 x 7 x 7.85 = 0.83m W x deflection 5700 x 0.65

Calculate KGF

KM = 10.83m GMF = 0.83 m ∴ KGF = 10.00m

Calculating moments about the CL,

Condition Weight (t) KG (m) Vertical Moments (tm) Initial Δ 5700 10.00 57000

(-) Ballast (-) 370 3.47 (-) 1283.9 (-) Bunkers (-) 165 3.98 (-) 656.7 (-) FSM Bunkers (-) 956.0 (-) Water (-) 95 4.44 (-) 421.8 (-) FSM Water (-) 910.0 (-) Inclining Weight (-) 56 8.44 (-) 472.64 (+) Deck Crane 19 9.74 (+) 185.06 5033 52484.02

∴ Lightship KG = 52484.02 = 10.43m 5033

b) A vessel’s lightship KG may change over a period of time due to the following reasons:

Constant of the vessel keep changing due to accumulated sludge in fuel tanks, mud and rust in ballast tanks (un-pumpables)

Various stores remaining unconsumed might add to the constant. Any structural changes will affect the light ship KG and light ship displacement

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Lightship KG for a passenger vessel will change considerably over a period of time mainly because of the left over baggage etc. will accumulate over a period of time and add to the constant considerably.

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SQA Stability – March 2006

1. A box shaped vessel floating on even keel in salt water has the following particulars: Length: 170.0m Breadth: 22.0m Draught: 9.2m KG: 6.5m There is an empty forward end compartment, length 21.0m extending the full width of the vessel.

Calculate the draughts forward and aft, if this compartment is bilged. Volume of the vessel before bilging = 170 x 22 x 9.2 = 34408 m3 ∴ Displacement of the vessel = 34408 x 1.025 = 35268.2 t

Calculate Bilged TMD

Volume of bilged compartment = 21 x 22 x 9.2 = 4250.4 m3 Intact WPA = (170 – 21) x 22 = 3278 m2

∴ Sinkage = 4250.4 = 1.297m 3278

∴ Bilged TMD = 9.200 + 1.297 = 10.497m

Calculate BBH

BBH = 21 = 10.5m aft 2

∴ Trimming Moment = 10.5 x 35268.2 = 370316.1 tm

Calculate GML

KB = 10.497 = 5.25m 2

BML = (170 – 21)3 x 22 = 176.25m 12 x 34408

∴ KML = 176.25 + 5.25 = 181.50m ∴ GML = 181.50 – 6.50 = 175.00m

Calculate MCTC

MCTC = 35268.2 x 175.00 = 363.1 tm 100 x 170

Calculate COTF & COTA

∴ COT = 370316.1 = 1019.9cms or 10.199m by forward 363.1

∴ COTA = 10.199 x 74.5 = 4.470m 170

∴ COTF = 10.199 – 4.470 = 5.729m

Calculate Final Draughts

Forward Aft Bilged TMD 10.497m 10.497m COTF/A (+) 5.729m (-) 4.470m ∴ Final Draughts 16.226m 6.027m

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2. A vessel initially upright and on an even keel, has the following particulars:

Draught (in salt water): 6.80m Breadth: 20.42m KG: 7.88m Further particulars of the vessel can be found in the Stability Data Booklet.

The vessel’s heavy lift derrick is to be used to discharge a 60 tonne tank from a centreline position, KG 5.23m. The derrick head is 29.28m above the keel and 15.80m out of the centreline when plumbing over side.

a) Calculate the maximum list angle. b) Calculate the increase in draught when the vessel is at maximum list angle as

calculated in Q 2(a), assuming rectangular cross section midships. c) Calculate the maximum allowable KG prior to discharging the tank in order to

limit the list angle to 5°.

a) The maximum angle of list occurs when the tank will be just above quay hanging on the derrick. The KG of the tank will act on the derrick head.

Δ 6.80m in SW = 14115 t & KM 6.80m in SW = 8.36m

Calculate GGH

GGH = 60 x 15.80 = 0.067m 14115

Calculate moments about the keel

Weight Distance from keel Moments Initial Δ 14115 7.88 111226.2 Tank (-) 60 5.23 (-) 313.8 Tank (+) 60 29.28 (+) 1756.8 Final Δ 14115 112669.2

∴ Final KG = 112669.2 = 7.982m 14115

∴ GM = 8.36 – 7.982 = 0.378m

Tan θ List = 0.067 0.378 ∴ θ List = 10.1°

a) Draught when heeled = (6.800 x Cos 10.1°) + (20.42 x Sin 10.1°) 2

= 8.485m ∴ Initial Draught = 6.800m ∴ Increase in draught = 8.485 – 6.800 = 1.685m

b) Limitation of List = 5°

∴ Tan 5° = 0.067 GM ∴ Required GM = 0.766m ∴ Required KG = 8.36 – 0.766 = 7.594m

Assuming maximum allowable KG to be ‘x’ metres,

© ARNAB SARKAR


Calculate moments about the keel

Weight Distance from keel Moments Initial Δ 14115 x 14115x Tank (-) 60 5.23 (-) 313.8 Tank (+) 60 29.28 (+) 1756.8 Final Δ 14115 (14115x + 1443)

∴ Required KG prior discharging = (14115x + 1443) = 7.594 14115 Hence, required KG = 7.492m

3. A vessel is floating in dock water of RD 1.012. Present draughts are as follows:

Forward (mean) 6.500m; Midship (port) 7.160m; Midship (stbd) 7.140m Aft (mean) 7.700m;

The draught marks are displaced as follows: Forward : 2.40m aft of the FP Aft : 3.60m forward of AP Midship : 1.96m aft of amidships line

Using the Stability Data Booklet and Worksheet Q3, determine the vessel’s displacement.

Draught forward = 6.500m Trim = 7.700 – 6.500 = 1.200m by stern FP Correction = 2.40 x 1.200 = 0.021m 137.5 Draught at FP = 6.500 – 0.021 = 6.479m

Draught aft = 7.700m AP Correction = 3.60 x 1.200 = 0.031m 137.5

Draught at AP = 7.700 + 0.031 = 7.731m ∴ True Trim = 7.731 – 6.479 = 1.252m by stern

Draught (M) Port = 7.160m Draught (M) Stbd = 7.140m

Draught (M) Mean = 7.150m

Amidships Line Corr. = 1.96 x 1.252 = 0.018m 137.5 Draught at Midships = 7.150 – 0.018 = 7.132m ∴ Corr. (M) draught = 6.479 + (6 x 7.132) + 7.731 = 7.125m 8

From hydrostatic particulars using draught 7.125m

TPC = 23.19 + (0.07 x 0.025) = 23.21tm 0.10 LCF = 67.24 – (0.11 x 0.025) = 67.21m foap 0.10 Displacement = 14808 + (232 x 0.025) = 14866 t 0.10

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Distance of LCF from Midships = 68.75 – 67.21 = 1.54m 1st Trim Correction = 1.54 x 125.2 x 23.21 = 32.546 t 137.5

MCTC 6.632m = 178.3 + (1.6 x 0.032) = 178.8 tm 0.10 MCTC 7.632m = 193.9 + (1.5 x 0.032) = 194.4 tm 0.10 2nd Trim Correction = 50 x (1.252)2 x 15.6 = 8.892 t 137.5

∴ Corrected Δ = 14866 + 32.546 + 8.892 = 14907.438 t ∴ Dock Water Δ = 14907.438 x 1.012 = 14718.368 t 1.025

© ARNAB SARKAR


1. A box shaped vessel floating upright on an even keel in salt water has the following particulars:

LengthBP: 150.00m Breadth: 28.00m Even keel draught: 8.60m KG: 9.20m

The vessel has two longitudinal bulkheads each 9.00m from the side of the vessel. Calculate the angle of heel if an amidship side compartment 24.00m is bilged.

1. Calculate Sinkage and Bilged Draught

Volume of Buoyancy lost = Volume of Buoyancy gained or, (24 x 9 x 8.600) = {(150 x 28) – (24 x 9)} x Sinkage or, Sinkage = 1857.6 = 0.466m

3984 ∴ Bilged draught = 8.600 + 0.466 = 9.066m

Calculate new location of LCF from the side ‘XX’

Calculate moments of area about the axis ‘XX’

Area (m2) Distance from axis ‘XX’ (m) Moments (m4) Total Area 150 x 28 = 4200 14.0 58800 Bilged Area 24 x 9 = 216 4.5 (–) 972

3984 57828

∴ New location of LCF = 14.515m ∴ Distance BBH = 14.515 – 14.0 = 0.515m

Calculate ‘Moment of Inertia’ (I) about the new LCF

I LL = I XX – Ad2 or, I LL = [(150 x 283) – (24 x 93)] – [{(150 x 28) – (24 x 9)} x 14.5152] 3 3 or, I LL = (1097600 – 5832) – 839369.936 or, I LL = 252398.064 m4

Calculate bilged BM, KB, KM and GM

BM = 252398.064 = 6.988m (150 x 28 x 8.600) KB = 9.066 = 4.533m 2 ∴ KM = 4.533 + 0.593 = 11.521m ∴ GM = 11.521 – 9.200 = 2.321m

Calculate List

Tan θ = 0.515 2.321 ∴ List = 12.5°

2. Worksheet Q2 – Trim and Stability provides data relevant to a particular condition of

loading in a vessel in salt water. The Stability Data Booklet provides the necessary hydrostatic data for the vessel. By completion of the Worksheet Q2 and showing any additional calculations in the answer book, calculate each of the following: (a) The effective metacentric height; (b) The draughts forward and aft.

© ARNAB SARKAR


CONDITION: BALLAST

Comp. Grain Capacity (m3)

Stowage Factor (m3/t)

Weight (t)

KG (m)

Vertical Moment

(tm)

Free Surface Moment

(tm)

LCG foap (m)

Long. Moment (tm)

All holds 15875 1.94 8183 4.86 39769 68.1 557262

1 TD 1265 2.37 534 10.98 5863 114.0 60876

2 TD 1332 2.53 526 10.42 5481 95.6 50286

3 TD 1406 2.49 565 10.33 5836 74.0 41810

4 TD 1230 2.66 462 10.18 4703 55.0 25410

Oil Fuel 856 1786 868 31020

FW 84 618 75 4550

FSM

Lightship 3831 8.21 31453 61.66 236219

Δ 15041 6.41 96452 66.98 1007433

HYDROSTATICS True Mean Draught = 7.20 + (0.10 x 1) = 7.200m 234

LCB foap = 69.94m

LCF foap = 67.13m

LBP = 137.5m MCTC = 187.8 tm KMT = 8.33m GM = 1.92m

LCB from LCG = 69.94 – 66.98 = 2.96m (As LCB > LCG, therefore vessel is trimmed by the stern)

Trim between Perpendiculars: 15041 x 2.96 = 237.1 cm or 2.371m 187.8 COTA = 2.371 x 67.13 = 1.158m 137.5 COTF = 2.371 – 1.158 = 1.213m Draughts: Forward: 7.20 – 1.213 = 5.987m Aft = 7.20 + 1.158 = 8.358m

3. A vessel has the following particulars:

Displacement 17000 t, Even keel draught 8.03m, Maximum breadth 21.00m KG 8.00m KM 8.35m KB 4.17m (a) Explain why this vessel will heel on turning. (b) Calculate the angle and direction of heel when turning to starboard in a circle of

diameter 500m at a speed of 17.5 knots. Note: assume 1 nautical mile = 1852m, and g = 9.81 m/sec2

(c) Calculate the new maximum draught during the turn in Q3 (b), assuming the amidships cross-section can be considered rectangular.

(a) To turn a ship the rudder is turned and this causes a force to act on the ship towards the

centre of the turn. This is the centripetal force and it acts at the ship’s centre of lateral resistance which is assumed to be at the centroid of the underwater volume (B). The inertia of the vessel resists this change of direction with an equal and opposite force which acts at the ship’s centre of gravity (G) termed as the centrifugal force.

© ARNAB SARKAR


The vessel will be in equilibrium at its angle of heel when: The righting moment = The heeling moment

(b) BG = KG – KB GM = KM – KG = 8.00 – 4.17 = 3.83m = 8.35 – 8.00 = 0.35m

Speed during turning (v) = 17.5 Knots = 17.5 x 1852 = 9.00 m/sec 3600 Radius of turn = 500 = 250m 2

∴ Angle of heel during turning = tan θ = (9.00)2 x 3.83 9.81 x 250 x 0.35 or, θ = 19.9° to port

(c) Draught when heeled = (8.03 x Cos 19.9°) + (0.5 x 21.00 x Sin 19.9°) = 11.124m

© ARNAB SARKAR


1. A vessel is floating in salt water at draughts 7.80m forward and 8.24m aft. The vessel is to load cargo so as to finish on an even keel at a load draught of 8.68m. Two spaces available: No. 1 hold, LCG 118m foap; No. 5 hold, LCG 15m foap.

Using the Stability Data Booklet, calculate EACH of the following: (a) The total weight of cargo to load; (b) The weight of cargo to load in each compartment.

(a) Calculate AMD and Trim

AMD = 7.800 + 8.240 = 8.020m & Trim = 8.240 – 7.800 = 0.440m 2

Calculate LCF (from hydrostatic particulars) and TMD

LCF = 66.33 – (0.08 x 0.02) = 66.31m 0.10 ∴ TMD = 8.240 – (0.440 x 66.31) = 8.028m 137.5

Calculate initial Δ, MCTC and LCB

Initial Δ = 16922 + (239 x 0.028) = 16989 t 0.10 MCTC = 199.6 + (1.4 x 0.028) = 200.0 tm 0.10 LCB = 69.58 – (0.04 x 0.028) = 69.57m foap 0.10

Calculate initial LCG

0.440 x 100 = 16989 x (69.57 – LCG) 200.0 ∴ LCG = 69.05m foap

Calculate final Δ, MCTC, & LCB using Even Keel draught = 8.680m

Final Δ = 18359 + (242 x 0.08) = 18553 t 0.10 MCTC = 207.7 + (0.3 x 0.08) = 207.9 tm 0.10 LCB = 69.31 – (0.04 x 0.08) = 69.28m foap 0.10

∴ Total cargo loaded = 18553 – 16989 = 1564 t

Assume that ‘w’ tonnes of cargo is loaded in No.1 Hold, ∴ Cargo loaded in No.5 Hold = (1564 – w) tonnes.

Calculating longitudinal moments about the AP

Weight (t) LCG (m) Longitudinal Moments (tm) 16989 69.05 1173090.45 w 118.0 118w (1564 – w) 15.0 (23460 – 15w) 18553 (1196550.45 – 103w)

As the vessel is finally in even keel therefore, LCB = LCG = 69.28m foap

© ARNAB SARKAR


∴ 69.28 = (1196550.45– 103w) 18553 ∴ w = 862 t

∴ Cargo loaded in No.1 Hold = 862 t ∴ Cargo loaded in No.5 Hold = 1564 – 862 = 702 t

2. A vessel is floating at an even keel draught of 7.25m in salt water. KG 7.20m. An

amidship rectangular deck 28.00m long and extending the full breadth of the vessel is flooded with salt water to a depth of 0.50m.

Height of deck above keel 9.60m

Using the Stability Data Booklet, calculate EACH of the following: (a) The fluid GM (b) The angle of loll

(a) Calculate initial Δ (from hydrostatic particulars) using the Even Keel draught of 7.25m

Initial Δ = 15040 + (234 x 0.05) = 15157 t 0.10

Calculate weight of the salt water in the flooded compartment

Volume = 28.0 x 20.42 x 0.50 = 285.88 m3 ∴ Weight = 285.88 x 1.025 = 293 t

FSM caused due to water ingress = 28.0 x (20.42)3 x 1.025 x 15157 = 20364.2 tm 12 x 15157

Calculate the final KG after flooding by taking moments about the Keel

Weights (t) KG (m) Transverse Moments (tm) 15157 7.20 109130.4 293 9.85 2886.1 FSM 20364.2 15450 132380.7

∴ Final KGfluid after getting flooded = 8.568m

Calculate final KB & KM using final Δ = 15450 t

Final KB = 3.80 + (0.05 x 176) = 3.84m 233 Final KM = 8.33m

∴ Fluid GM = 8.33 – 8.568 = (-) 0.238m

(b) Calculate BMT

BMT = 8.33 – 3.84 = 4.49m

Calculate Angle of Loll after flooding

Tan θ = √ 2 x 0.238 4.49 ∴ Angle of loll = 18.0°

© ARNAB SARKAR


3. A vessel is floating in dock water of R.D. 1.016. Present draughts: Forward 7.50m; Midship (P) 8.44m; Midship (S) 8.38m; Aft 8.90m. The draught marks are displaced as follows: Forward: 1.86m aft of the FP. Aft: 2.50m aft of the AP. (The amidships draught marks are not displaced)

The Stability Data Booklet provides the necessary hydrostatic data for the vessel. By completion of the Worksheet Q3 and showing any additional calculations workbook, determine EACH of the following:

(a) The vessel’s displacement; (b) The maximum cargo to load for a Summer Load Line Zone. Note: assume no hog or sag in the Summer Load condition.

(a) Draught forward = 7.500m

Trim = 8.900 – 7.500 = 1.400m by stern FP Correction = 1.86 x 1.400 = 0.019m 137.5 Draught at FP = 7.500 – 0.019 = 7.481m


Draught at AP = 8.900 – 0.025 = 8.875m ∴ True Trim = 8.875 – 7.481 = 1.394m by stern



Amidships Line Corr. = Nil Draught at Midships = 8.410m ∴ Corr. (M) draught = 7.481 + (6 x 8.410) + 8.875 = 8.352m 8


TPC = 23.96 + (0.06 x 0.052) = 23.99tm 0.10 LCF = 66.10 – (0.08 x 0.052) = 66.06m foap 0.10 Displacement = 17639 + (239 x 0.052) = 17763.28 t 0.10 Distance of LCF from Midships = 68.75 – 66.06 = 2.69m

1st Trim Correction = 2.69 x 140.0 x 23.99 = 65.706 t 137.5


© ARNAB SARKAR


∴ Corrected Δ = 17763.28 + 65.706 + 9.384 = 17838.37 t ∴ Dock Water Δ = 17838.37 x 1.016 = 17681.7 t 1.025

(b) Summer load displacement = 19006 t ∴ Maximum cargo to load = 19006 – 17681.7 = 1324.3 t

© ARNAB SARKAR


1. A vessel is floating upright and is to load TWO weights using the ship’s own derrick. The maximum allowable list is 4 degrees. Initial draughts: 7.30m, forward and aft, in fresh water. Derrick head 26.5m above the keel. Two weights, each 42 tonne, are on the quay 17.5m from the vessel’s centreline. Stowage position on deck, KG 12.0m, 7.2m either side of the vessel’s centreline. The inboard weight is to be loaded first.

Using the Stability Data Booklet, calculate the minimum initial GM. Determine initial Δ and KM using initial draught of 7.30m in FW

ΔFW = 14901 t KM = 8.33m

Determine GM (worst condition)

Calculate Moments about the Centreline

Weights (t) Distance from CL (m) Moments about the CL (tm) 14901 0.00 0.00 42 7.2 302.4 42 17.5 735.0 14985 1037.4

∴ GGH = 1037.4 = 0.069m 14985 ∴ Tan 4° = 0.069 GM or, GM = 0.987m

Determine KG on completion of loading operations

KM = 8.33m (from hydrostatic particulars using Δ = 14985 t) GM = 0.987m ∴ KG = 7.343m

Determine KG prior commencing loading operations

Calculate Moments about the Keel

Weights (t) KG (m) Moments about the Keel (tm) 14985 7.343 110034.855 (-) 42 12.0 (-) 504.00 (-) 42 26.5 (-) 1113.00 14901 108417.855

∴ KG = 108417.855 = 7.276m 14901

Determine initial GM

Initial KM = 8.33m KG = 7.276m Initial GM = 1.054m

2. A vessel is planning to enter dry-dock in salt water.

Present draughts: Forward 4.00m Aft 5.60m KG 8.74m


© ARNAB SARKAR


(a) The maximum trim at which the vessel can enter dry-dock so as to maintain a GM of at least 0.15m at the critical instant. Note: Assume KM remains constant

(b) The weight of ballast to transfer from the aft peak to the fore peak in order to reduce the trim to the maximum allowable.

(a) Calculate TMD

Forward draught = 4.000m Aft draught = 5.600m

∴ AMD = 4.800m & trim = 1.600m by stern

Using AMD = 4.800m, from the hydrostatic particulars,

LCF = 69.29m foap ∴ TMD = 5.600 – (1.600 x 69.29) = 4.794m 137.5

Calculate Δ, KM, MCTC & LCF from hydrostatic particulars using TMD = 4.794m

Δ = 9420 + (218 x 0.094) = 9624.92 t 0.10 KM = 9.22 – (0.09 x 0.094) = 9.135m 0.10 MCTC = 156.0 + (0.9 x 0.094) = 156.846 tm 0.10 LCF = 69.35 – (0.06 x 0.094) = 69.294m foap 0.10

Calculate the ‘P’ force

Initial GM = 9.135 – 8.74 = 0.395m Reqd. GM = 0.15m ∴ Loss of GM = 0.395 – 0.015 = 0.245m

∴ 0.245 = P x 9.135 9624.92 or, P = 258.14 t

Calculate required trim during dry-docking

258.14 = trim x 156.846 69.294 ∴ Reqd. trim = 114.0cms or, 1.140m by stern

(b) Calculate the initial LCG of the vessel prior transferring ballast

Using TMD = 4.794m, from hydrostatic particulars,

Initial LCB = 70.90 – (0.04 x 0.094) = 70.862m foap 0.10 ∴ 1.6 x 100 = 9624.92 x (70.862 – LCG) 156.846 or, LCG Initial = 68.255m foap

Calculate the final LCG of the vessel after transfer of ballast to attain required trim

1.140 x 100 = 9624.92 x (70.862 – LCG) 156.846

© ARNAB SARKAR


or, LCG Final = 69.004m foap

Assume that ‘w’ tonne of ballast is transferred from aft peak to fore peak tank in order to attain the required trim.

Calculating longitudinal moments about the AP

Weight (t) LCG (m) Longitudinal Moments (tm) 9624.92 68.255 656948.91 (–) w 3.07 3.07 w (+) w 130.56 130.56 w 9624.92 (656948.91 – 127.49 w)

∴ 69.004 = (656948.91 – 127.49 w) 9624.92 or, w = 56.5 t ∴ Quantity of ballast to be transferred in order to attain the required trim = 56.5 t


Displacement 18000 tonne Fluid KG 8.10m

Using the Stability Data Booklet, sketch the vessel’s statical stability curve and compare the vessel’s stability with ALL the minimum stability criteria required by the current Load Line Regulations. Comment on the result.

3. Δ = 18000 t KGF = 8.10m


θF = 40.6°



(Tonnes) 12° 20° 30° 40° 50° 60° 75°

18000 1.75 2.93 4.27 5.36 6.12 6.67 6.92


Heel (°) KN (m) KG Sin θ (m) GZ (m) 12 1.75 1.68 0.07 20 2.93 2.77 0.16 30 4.27 4.05 0.22 40 5.36 5.21 0.15 50 6.12 6.20 - 0.08 60 6.67 7.01 - 0.34 75 6.92 7.82 - 0.90

Calculate the vessel’s GMF

From hydrostatic particulars, using Δ = 18000 t in SW

KMT = 8.39 + (0.02 x 122) = 8.40m 241

© ARNAB SARKAR


∴ GM = 8.40 – 8.10 = 0.30m




10 0.06 3 0.18 20 0.16 3 0.48 30 0.22 1 0.22 ∑ 0.88



10 0.06 4 0.24 20 0.16 2 0.32 30 0.22 4 0.88 40 0.15 1 0.15 ∑ 1.59


∴ Area 30° → 40° = 0.092 – 0.058 = 0.040 m-rad

Summary

Actual L.L. Regulation Complies or Not Area 0° → 30° 0.058 m-r Not less than 0.055 m-r Yes Area 0° → 40° 0.092 m-r Not less than 0.09 m-r Yes Area 30° → 40° 0.040 m-r Not less than 0.03 m-r Yes Max. GZ 0.22m At least 0.20m Yes

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0 20 40 60 80

© ARNAB SARKAR


Max. GZ Angle 30° Equal to or greater than 30° Yes GMF 0.30m Not less than 0.15m Yes

© ARNAB SARKAR



Present draughts in fresh water: Forward 5.42m Aft 6.56m The aft mast is 39m foap and extends 27.10m above the keel. The fore mast is 108m foap and extends 26.00m above the keel. (Assume masts are perpendicular to the waterline throughout)

Using the Stability Data Booklet, calculate EACH of the following: (a) The final draughts forward and aft in order to pass under the bridge with

minimum clearance; (b) The minimum weight of ballast to load in order to pass under the bridge with

minimum clearance.



∴ Trim at Masts = 7.120 – 6.020 = 1.100m by stern










Initial trim = 6.560 – 5.420 = 1.140m by stern Initial AMD = (6.560 + 5.420) = 5.990m 2 With AMD, from hydrostatic tables, LCF = 68.43m – (0.04 x 0.09) = 68.39m 0.10 TMD = 6.560 – (1.140 x 68.39) = 5.993m 137.5 ∴ DisplacementTMDi = 11778 + (219 x 0.093) = 11981.67 t ≈ 11982 t 0.10

© ARNAB SARKAR





∴ DisplacementTMDf = 13324 + (224 x 0.062) = 13462.88 t ≈ 13463 t 0.10

Calculate minimum weight of ballast to load in order to achieve required trim

∴ Ballast to load = DisplacementTMDf – DisplacementTMDi

= 13463 – 11982 = 1481 t

2. A vessel is floating upright in fresh water and is about to enter dry-dock. The vessel particulars are: KG 8.37m Forward draught 4.89m Aft draught 6.71m

(a) Using the Stability Data Booklet, calculate the vessel’s effective GM at the critical instant.

(b) Describe the methods of improving the initial stability is the GM at the critical instant is found to be inadequate.

(a) Calculate the TMD

AMD = 5.800m & Trim = 1.820m by stern

With AMD = 5.800m, from hydrostatic particulars,

LCF = 68.57m foap TMD = 6.710 – (1.820 x 68.57) = 5.802m 137.5

Determine initial Δ, MCTC & LCF using TMD = 5.802m

Initial Δ = 11559 + (219 x 0.002) = 11563.38 t 0.10 MCTC = 163.2 + (1.2 x 0.002) = 163.22 tm 0.10 LCF = 68.57 – (0.14 x 0.002) = 68.57m foap 0.10


P = 1.820 x 100 x 163.22 = 433.22 t 68.57 ∴ Displacement at critical instant = 11563.38 – 433.22 = 11130.16 t

Calculate KM using displacement at critical instant

KM = 8.67 – (0.04 x 6.16) = 8.67m 218

© ARNAB SARKAR


Calculate the loss of GM

Loss of GM = 433.22 x 8.67 = 0.325m 11563.38

Calculate GM at critical instant

Initial GM = 8.67 – 8.37 = 0.30m Loss of GM = 0.325m ∴ GM at C.I. = 0.30 – 0.325 = (–) 0.025m

(b) The trim can be reduced by transferring water, fuel or ballast forward. The initial GM can also be increased by reducing FSM, lowering weights, or by ballasting low or empty tanks.

3. A box shaped vessel floating on an even keel in salt water has the following particulars:

Length 130.00m Breadth 24.00m Draught 4.98m

There is a midship watertight compartment of length 20.00m, height 7.20m that extends the full width of the vessel and is filled with cargo of relative density 0.80 stowing at 1.70 m3/t.

If this compartment is bilged, calculate EACH of the following: (a) The final draught;

(b) The change in GM.

(a) Calculate the bilged draught

Volume of the box shaped vessel before bilging = 130 x 24 x 4.98 = 15537.6 m3 Volume of the bilged compartment = 20 x 24 x 4.98 = 2390.4 m3

Solid factor = 1 = 1.25 0.80 Permeability = 1.70 – 1.25 = 0.265 1.70 ∴ Sinkage = 2390.4 x 0.265 = 0.212m (130 x 24) – (20 x 24 x 0.265) Initial draught = 4.98m ∴ Final draught = 4.98 + 0.212 = 5.192m

(b) Before bilging

KB = 4.98 = 2.49m BM = 130 x (24)3 = 9.639m 2 12 x 15537.6 ∴ KM = 9.639 + 2.49 = 12.129m

After bilging

KB = 5.192 = 2.596m BM = (130 – 20 x 0.265) x (24)3 = 9.246m 2 12 x 15537.6 ∴ KM = 9.246 + 2.596 = 11.842m

∴ Change in KM = 12.129 – 11.842 = 0.287m decrease ∴ Change in GM = 0.287m decrease

© ARNAB SARKAR


4. A vessel plans to depart from port at:

Displacement 19000 t KG 7.80m

During the ensuing voyage the vessel will consume 420 t of fuel oil (KG 0.60m) and 35 t of fresh water (KG 7.35m) from full tanks causing a free surface moment of 1721 tm.

Using the Maximum KG Table in the Stability Data Booklet, determine the stability condition of the ship BOTH on departure and on arrival.

On departure

As per the Maximum KG Table in the Stability Data Booklet,

Displacement (t) KG (m)

19000 7.93

On departure, the vessel’s KG is 7.80m, therefore, it complies with the intact stability criteria as specified in the current Load Line Rules.

On Arrival

Calculate moments about the Keel to determine the final KG

Weights (t) KG (m) Vertical Moments (tm) 19000 7.80 148200 (–) 420 0.60 (–) 252 (–) 35 7.35 (–) 257.25 FSM (+) 1721 18545 149411.75

∴ Final KG = 8.057m

As per the Maximum KG Table in the Stability Data Booklet,

Displacement (t) KG (m)

19000 7.93

18545 ?

18500 8.02

∴ Maximum KG for Δ 18545 t = 8.02 – (0.09 x 45) = 8.01m 500 On arrival, the vessel’s KG is 8.057m, which is more than specified in the Maximum KG Table, therefore it does not comply with the intact stability criteria as specified in the current Load Line Rules.

© ARNAB SARKAR


1. A vessel is to load a cargo of grain (stowage factor 1.62 m3/t). Initial displacement 6300 tonne. Initial KG 6.50m. All five holds are to be loaded full of grain. The tween decks are to be loaded as follows:

No. 1 TD Part full – ullage 1.50m No. 2 TD Full No. 3 TD Full No. 4 TD Part full – ullage 2.75m


(a) Using the Maximum Permissible Grain Heeling Moment Table included in the Stability Data Booklet, determine whether the vessel complies with the minimum criteria specified in the International Grain Code (IMO).

(b) Calculate the ship’s approximate angle of heel in the event of the grain shifting as assumed by the International Grain Code (IMO).

(a) Initial Δ = 6300 t



Comp. Ullage (m)

Volume (m3)

SF (m3/t) Tonnes KG

(m) Vert. Mom.

(tm) VHM (tm)

Initial Δ 6300 6.50 40950

No.1 TD 1.50 1510 1.62 932 10.94 10196 1176

No.2 TD Full 1676 1.62 1035 10.78 11157 539

No.3 TD Full 1626 1.62 1004 10.59 10632 578

No.4 TD 2.75 900 1.62 556 10.20 5671 3278

No.1 Hold Full 2215 1.62 1367 5.09 6958 410

No.2 Hold Full 4672 1.62 2884 4.95 14276 1285

No.3 Hold Full 1742 1.62 1075 4.94 5311 475

No.4 Hold Full 3474 1.62 2144 4.95 10613 910

No.5 Hold Full 2605 1.62 1608 8.76 14086 455

Final Δ 18905 6.87 129850 9106

Calculate AHM

AHM = 9106 = 5621 tm 1.62


Δ = 18905 t KGF = 6.87m

© ARNAB SARKAR

SQA Stability – November 2007 Interpolation

Δ KGF = 6.80 KGF = 6.87 KGF = 6.90

19000 7577 7283.0 7157

18905 7233.2

18500 7307 7020.7 6898


7577 – (420 x 0.07) = 7283.0 tm 0.10


7307 – (409 x 0.07) = 7020.7 tm 0.10


7020.7 + (262.3 x 405) = 7233.2 tm 500

∴ PHM for Δ = 18905 t & KGF = 6.87m = 7233.2 tm


(b) Calculate approximate angle of list in event of grain shift

Angle of Heel = 5621 x 12° = 9.3° 7233.2

2. A box shaped vessel, length 90.00m, breadth 10.00m, depth 9.00m, is floating at a draught of 4.10m in salt water. Initial KG 3.70m.

(a) Calculate the angle of loll if 580 tonne of cargo, KG 7.80m is loaded on deck. (b) Calculate the GM at the angle of loll in Q2 (a).

(a) Calculate the displacement of the box shaped vessel

Initial Δ = 90.00 x 10.00 x 4.10 x 1.025 = 3782.25 t

Calculate moments about the Keel to determine the Final KG

Weights (t) KG (m) Vertical Moments (tm) 3782.25 3.70 13994.325 580 7.80 4524 4362.25 18518.325


Calculate the final draught of the box shaped vessel after loading the cargo

Final Δ = 4362.25 = 90.00 x 10.00 x d x 1.025 ∴ Final draught (d) = 4.729m

Calculate the final KM

KB = 4.729 = 2.365m 2 BM = 90.00 x (10.00)3 = 1.762m 12 x 90.00 x 10.00 x 4.729

© ARNAB SARKAR


∴ KM = 2.365 + 1.762 = 4.127m Calculate the Final GM after loading the cargo

Final GM = 4.127 – 4.245 = (–) 0.118m

Calculate Angle of Loll

Tan (Angle of Loll) = √ (2 x 0.118) 1.762 ∴ Angle of Loll = 20.1°

(b) GM at angle of Loll = 2 x 0.118 = 0.251m Cos 20.1°

3. A box shaped vessel floating on even keel in salt water has the following particulars:

Length 160.00m Breadth 22.00m Draught 9.10m KG 6.60m

There is an empty forward end compartment, length 20.00m extending the full width of the vessel.

Calculate the draughts forward and aft, if this compartment is bilged. Calculate the Bilged TMD of the box shaped vessel

Volume of the vessel = 160.00 x 22.00 x 9.10 = 32032 m3 ∴ Displacement of the vessel = 32032 x 1.025 = 32832.8 t

Volume of the bilged compartment = 20.00 x 22.00 x 9.10 = 4004 m3

Intact WPA = (160.00 – 20.00) x 22.00 = 3080 m2 ∴ Sinkage = 4004 = 1.300m 3080 Initial draught of the vessel = 9.100m ∴ Bilged TMD = 9.100 + 1.300 = 10.400m

Calculate the trimming moment

Horizontal shift of ‘B’ = 20 = 10.00m aft 2 ∴ Trimming Moment = 10.00 x 32832.8 = 328328 tm

Calculate Bilged GML

Bilged KB = 10.400 = 5.200m 2 Bilged BML = (140)3 x 22 = 157.05m 12 x 32032 ∴ Bilged KML = 157.05 + 5.2 = 162.25m ∴ Bilged GML = 162.25 – 6.6 = 155.65m

Calculate COT

MCTC = 32832.8 x 155.65 = 319.40 tm 100 x 160.0 ∴ COT = 328328 = 1027.9cms or 10.279m 319.40 LCF = (160 – 20) = 70.0m 2

© ARNAB SARKAR

SQA Stability – November 2007 ∴ COTA = 10.279 x 70.0 = 4.497m 160 ∴ COTF = 10.279 – 4.497 = 5.782m

Calculate the Final Draughts

Forward Aft Bilged TMD 10.400 10.400 COTF/A (+) 5.782 (–) 4.497 New Draughts 16.182m 5.903m

© ARNAB SARKAR


1. A vessel of length 152.00m, KG 8.28m, is floating upright in salt water at a True Mean Draught of 5.50m.

The vessel has a rectangular double bottom of length 22.00m, breadth 16.00m, depth 2.00m, which is subdivided by a longitudinal centreline division into port and starboard tanks of equal dimensions.

Using the Stability Data Booklet, calculate the angle of heel after partially filling the port side of this tank with 240 tonne of fuel oil, relative density 0.874.

Remarks: The length of the vessel though irrelevant is given wrong. It should be 137.5m instead of 152.0m

1. Using TMD = 5.500m, from the hydrostatic particulars,

Initial Δ = 11180 t

Determine the depth of the liquid in the tank after partially filling it up with 240 t of Fuel Oil

Breadth of the entire DB tank = 16.00m ∴ Breadth of each Port/Stbd tank = 8.00m

240 = 22 x 8 x d x 0.874 ∴ d = 1.560m ∴ KG F.O. = 0.780m

Determine FSM caused due to the Fuel Oil

FSM = 22 x (8)3 x 0.874 = 820.39 tm 12

Calculate Moments about the Keel to determine the Final KG

Weights (t) KG (m) Vertical Moments (tm) 11180 8.28 92570.4 240 0.78 187.2 FSM 820.39 11420 93577.99

∴ Final KG Fluid = 8.194m

Determine the Final KM using the Final Δ = 11420, from the hydrostatic particulars

Final KM = 8.67 – (0.04 x 18) = 8.67m 223

Determine GM Fluid

GM Fluid = 8.67 – 8.194 = 0.476m

Determine GGH

GGH = 240 x 4 = 0.084m 11420

Determine the Angle of Heel

Tan θ = 0.084 0.476 ∴ θ = 10.0° to port

© ARNAB SARKAR


2. A vessel’s present particulars are:

Forward draught 7.60m, Aft draught 8.80m in salt water.

A total of 300 tonne of cargo, LCG 74.00m foap, is to be discharged immediately and then the vessel is to proceed to an upriver berth where the relative density of the dock water is 1.007.

During the passage the following items of deadweight are consumed:

80 t of Heavy Fuel Oil from No.3 DB Centre tank 22 t of Diesel Oil from No. 4 DB Port tank 22 t of Diesel Oil from No.4 DB Starboard tank 15 t of Fresh Water from After Fresh Water tank

Using the Stability Data Booklet, calculate the draughts fore and aft, on arrival at the upriver berth.

Determine TMD of the vessel

AMD = 8.200m Trim = 1.200m by stern

From hydrostatic particulars, using AMD = 8.200m,

LCF = 66.17m foap ∴ TMD = 8.800 – (1.200 x 66.17) = 8.223m 137.5

Determine initial LCG of the vessel by using TMD = 8.223m from the hydrostatic particulars,

Δ = 17399 + (240 x 0.023) = 17454.2 t 0.10 MCTC = 202.4 + (1.4 x 0.023) = 202.7 tm 0.10 LCB = 69.49 – (0.04 x 0.023) = 69.48m foap 0.10

As vessel is initially trimmed by the stern, therefore LCB > LCG.

∴ 1.200 x 100 = 17454.2 x (69.48 – LCG) 202.7 or, LCG = 68.09m foap

Determine Final LCG on completion of upriver passage

Calculate Longitudinal Moments about the AP

Weights (t) LCG (m) Longitudinal Moments (tm) 17454.2 68.09 1188456.5 (–) 300 74.00 (–) 22200 (–) 80 57.02 (–) 4561.60 (–) 22 35.66 (–) 784.52 (–) 22 35.50 (–) 781.00 (–) 15 28.67 (–) 430.05 17015.2 1159699.33

∴ Final LCG = 68.16m foap

© ARNAB SARKAR


Calculate volume in FW using Final Δ = 17015.2 t

Volume in FW = 17015.2 = 16896.9 m3 1.007

Using Volume FW = 16896.9 m3, from hydrostatic particulars,

TMD = 8.10 + (0.10 x 154.9) = 8.167m 233

MCTC FW = 196.1 + (1.4 x 154.9) = 197.0 tm 233 ∴ MCTC 1.007 = 197.0 x 1.007 = 198.4 tm

LCB = 69.54 – (0.05 x 154.9) = 69.51 m foap 233 LCF = 66.25 – (0.08 x 154.9) = 66.20 m foap 233

Determine COT

COT = 17015.2 x (69.51 – 68.16) = 115.8cms or 1.158m by stern 198.4

Determine Final Draughts

COTA = 1.158 x 66.20 = 0.558m 137.5

∴ COTF = 1.158 – 0.558 = 0.600m

Forward Aft TMD 8.167m 8.167m COT F/A 0.600m 0.558m Final Draughts 7.567m 8.725m



Using the Stability Data Booklet, compare the vessel’s stability with all the minimum stability criteria required by the current Load Line Regulations.

3. Δ = 15000 t KGF = 8.10m


θF = 46.5°


Δ Angle of Heel (°) (Tonnes) 12° 20° 30° 40° 50° 60° 75°

15000 1.73 2.98 4.48 5.72 6.48 6.91 7.04


Heel (°) KN (m) KG Sin θ (m) GZ (m) 12 1.73 1.68 0.05 20 2.98 2.77 0.21 30 4.48 4.05 0.43

© ARNAB SARKAR


40 5.72 5.21 0.51 50 6.48 6.20 0.28 60 6.91 7.01 - 0.10 75 7.04 7.82 - 0.78



KMT = 8.34 - (0.01 x 192) = 8.33m 232 ∴ GMF = 8.33 – 8.10 = 0.23m




10 0.04 3 0.12 20 0.21 3 0.63 30 0.43 1 0.43 ∑ 1.18



10 0.04 4 0.16 20 0.21 2 0.42 30 0.43 4 1.72 40 0.51 1 0.51 ∑ 2.81

Area 0° → 40° = 1 x 10 x 2.81 = 0.163 m-rad 3 57.3

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 20 40 60 80

© ARNAB SARKAR

SQA Stability – March 2008 Calculate the area under the GZ Curve from 30° → 40°

∴ Area 30° → 40° = 0.163 – 0.077 = 0.086 m-rad

Summary

Actual L.L. Regulation Complies or Not Area 0° → 30° 0.077 m-r Not less than 0.055 m-r Yes Area 0° → 40° 0.163 m-r Not less than 0.09 m-r Yes Area 30° → 40° 0.086 m-r Not less than 0.03 m-r Yes Max. GZ 0.51m At least 0.20m Yes Max. GZ Angle 40° Equal to or greater than 30° Yes GMF 0.23m Not less than 0.15m Yes

© ARNAB SARKAR

SQA Stability – July 2008 1. A vessel’s present particulars are as follows:

A vessel is floating in salt water at a Forward draught of 8.300m and Aft draught of 8.500m. KG 8.00m. Vessel upright.

The vessel is to load bunkers of heavy fuel oil into No.3 DB port and starboard tanks and sail upright at a maximum draught of 8.500m.


(a) The maximum weight of bunkers to load. (b) The weight of ballast to transfer between the Aft Peak and the Fore Peak so that

the vessel sails on an even keel.

(a) Calculate TMD of the vessel

AMD = 8.400m Trim = 0.200m by the stern

From hydrostatic particulars, using AMD = 8.400m,

LCF = 66.02m foap

∴ TMD = 8.500 – (0.200 x 66.02) = 8.404m 137.5

Calculate Initial Δ, MCTC and LCB from hydrostatic particulars using TMD = 8.404m

Initial Δ = 17878 + (241 x 0.004) = 17887.64 t 0.10 MCTC = 205.1 + (1.3 x 0.004) = 205.2 tm 0.10 LCB = 69.40 – (0.04 x 0.004) = 69.40m foap 0.10

Calculate the Initial LCG

0.200 x 100 = 17887.64 x (69.40 – LCG) 205.2 ∴ LCG = 69.17m foap

From hydrostatic particulars determine Final Δ, MCTC and LCB using sailing draught of 8.500m E.K.

Final Δ = 18119 t MCTC = 206.4 tm LCB = 69.36m foap

Determine maximum weight of bunkers to load

∴ Maximum weight of bunkers to load = 18119 – 17887.64 = 231.36 t

(b) Assume that ‘w’ tonne of ballast is required to be transferred from aft peak to fore peak tank.

Calculating Longitudinal Moments about the AP

Weight (t) LCG (m) Longitudinal Moments (tm) 17887.64 69.17 1237288.06 231.36 57.87 13388.8 (–) w 3.07 (–) 3.07w (+) w 130.56 (+) 130.56 18119 1250676.86 + 127.49 w

© ARNAB SARKAR

SQA Stability – July 2008 Determine LCB from hydrostatic particulars using TMD = 8.500m

LCB = 69.36m foap

As vessel will sail out on an even keel draught of 8.500m, therefore LCB = LCG = 69.36m foap

∴ 69.36 = 1250676.86 + 127.49 w 18119 or, w = 47.5 t of ballast

2. Worksheet Q2 – Trim and Stability provides relevant to a particular condition of

loading in a vessel in salt water.

All holds and tween decks are to be filled with grain (S.F. 1.62 m3/t).

The Stability Data Booklet provides the necessary data for the vessel.

By completion of the Worksheet Q2 and showing any additional calculations in the answer book, calculate EACH of the following:

(a) The effective metacentric height; (b) The draughts forward and aft.

CONDITION: LOADED – GRAIN S.F. 1.62 m3/t



Weight (t)

KG (m)

Vertical Moment

(tm)

Free Surface Moment

(tm)

LCG foap (m) Long. Moment (tm)

All holds 14708 1.62 9079 5.64 51206 69.1 627359

1 TD 1695 1.62 1046 11.26 11778 115.5 120813

2 TD 1676 1.62 1035 10.78 11157 95.6 98946

3 TD 1626 1.62 1004 10.59 10632 74.1 74396

4 TD 1674 1.62 1033 10.57 10919 51.7 53406

Oil Fuel 913 1917 2184 31042

FW 86 634 75

Lightship 3831 8.21 31453 61.7 236373

Δ 18027 7.32 131955 69.06 1244885

HYDROSTATICS

True Mean Draught = 8.40 + (0.10 x 149) 241 = 8.462m

LCB foap = 69.40 – (0.04 x 149) 241 = 69.38m

LCF foap = 66.02 – (0.07 x 149) 241 = 65.98m

LBP = 137.5m

MCTC = 205.1 + (1.3 x 149) 241 = 205.9 tm

KMT = 8.39 + (0.02 x 149) 241 = 8.40m

GMF = 8.40 – 7.32 = 1.08m

© ARNAB SARKAR

SQA Stability – July 2008 Trim between Perpendiculars: 18027 x (69.38 – 69.06) = 28.02 cm or 0.280m 205.9 COTA = 0.280 x 65.98 = 0.134m 137.5 COTF = 0.280 – 0.134 = 0.146m Draughts: Forward: 8.462 – 0.146 = 8.316m Aft = 8.462 + 1.158 = 8.596m

3. A vessel is floating at an even keel draught of 8.600m in salt water. KG 7.40m.

A midship rectangular deck 29.00m long and extending the full breadth of the vessel is flooded with salt water to a depth of 0.40m.

Height of deck above keel 11.75m.


(a) The fluid GM; (b) The angle of loll. (a) Determine the Initial Δ of the vessel from hydrostatic particulars using TMD = 8.600m

Initial ΔSW = 18359 t

Determine the weight of salt water in the flooded deck

Weight of salt water = 29.00 x 20.42 x 0.40 x 1.025 = 242.8 t

Determine the FSM caused due to salt water ingress

FSM = 29.00 x (20.42)3 x 1.025 = 21092 tm 12 Calculate moments about the keel to determine the Final KG

Weights (t) KG (m) Vertical Moments (tm) 18359 7.40 135856.6 242.8 (11.75 + 0.20) 2901.46 FSM 21092 18601.8 159850.06


Determine the Final KM from the Final Δ

Final KM = 8.43 + (0.02 x 0.8) = 8.43m 242

Determine the Final GM Fluid

GM Fluid = 8.43 – 8.59 = (–) 0.16

(b) Determine the Final KB from the Final Δ

KB = 4.56 – (0.05 x 0.8) = 4.56m 242

BMT = 8.43 – 4.56 = 3.87m

Tan θ = √ (2 x 0.16) 3.87 ∴ Angle of Loll = 16.0°

© ARNAB SARKAR

SQA Stability – November 2008 1.(a) A vessel, initially upright, is to carry out an inclining test.

Present displacement 5530 t. KM 10.75m.


Ballast 350 t KG 3.52m Tank Full Bunkers 172 t KG 4.01m Free Surface Moment 897 tm Waters 93 t KG 4.46m Free Surface Moment 916 tm Boiler Water 16 t KG 4.19m Free Surface Moment 115 tm Inclining Weight 50 t KG 8.60m

A deck crane, weight 21 t and still ashore, will be fitted on the vessel at a KG of 9.90m at a later date.

The plumb lines have an effective vertical length 8.24m. The inclining weights are shifted transversely 7.3m on each occasion and the mean horizontal deflection of the plumb line is 0.68m

Calculate the vessel’s Lightship KG.

(b) Explain why a vessel’s Lightship KG may change over a period of time.

(a) Displacement = 5530 t KM = 10.75m Plumb length = 8.24m Deflection = 0.68m Trans shift = 7.3m

GMF = w x d x plum length = 50 x 7.3 x 8.24 W x deflection 5530 x 0.68

= 0.80m

KM = 10.75m GMF = 0.80m

Therefore, KGF = 9.95m

Calculating moments about the Keel,

Weights (t) KG (m) Vertical Moments (tm)

Δ Inclined 5530 9.95 55023.5 (-) Ballast (-) 350 3.52 1232.0 (-) Bunkers (-) 172 4.01 689.72 (-) FSM Bunkers 897.0 (-) Water (-) 93 4.46 414.78 (-) FSM Water 916 (-) Inclining Weight (-) 50 8.60 430.0 4865 t 50444 tm

∴ Lightship KG = 50444 = 10.37 m 4865

b) A vessel’s lightship KG changes over a period of time due to the following reasons:

Constant of the vessel keep changing due to accumulated sludge in fuel tanks, mud and rust in ballast tanks (un-pumpables)

Various stores remaining unconsumed might add to the constant. Any structural changes will affect the light ship KG and light ship displacement

© ARNAB SARKAR


Lightship KG for a passenger vessel will change considerably over a period of time mainly because of the left over baggage etc. will accumulate over a period of time and add to the constant considerably.

2. A vessel is to load a cargo of grain (Stowage Factor 1.65 m3/t). Initial displacement 6200 t Initial KG 8.50m All five holds are to be loaded full of grain.


No.1 TD Full No.2 TD Empty No.3 TD Part Full – Ullage 1.50m No.4 TD Full


Using the KN tables to construct a GZ curve, determine whether the vessel complies with the minimum criteria specified in the International Grain Code (IMO).

Initial Δ = 6200 t Initial KG = 8.50m SF = 1.65 m3/t


Comp. Ullage (m)

Volume (m3)

SF (m3/t) Tonnes KG

(m) Vert. Mom.

(tm) VHM (tm)

Initial Δ 6200 8.50 52700

No.1 TD Full 1695 1.65 1027 11.26 11564 352

No.2 TD Empty

No.3 TD 1.50 1399 1.65 848 10.45 8862 1900

No.4 TD Full 1674 1.65 1015 10.57 10729 604

No.1 Hold Full 2215 1.65 1342 5.09 6831 410

No.2 Hold Full 4672 1.65 2832 4.95 14018 1285

No.3 Hold Full 1742 1.65 1056 4.94 5217 475

No.4 Hold Full 3474 1.65 2105 4.95 10420 910

No.5 Hold Full 2605 1.65 1579 8.76 13832 455

Final Δ 18004 7.45 134173 6391

Using Final Δ = 18004 t, determine KM from the hydrostatic tables

KM = 8.39 – (0.02 x 126) = 8.38m 241

∴ GMF = 8.38 – 7.45 = 0.93m

Determine the Angle of Flooding

Angle of Flooding = 40.6°

© ARNAB SARKAR

SQA Stability – November 2008 Determine the GZ values from the KN table

Heel (°) KN (m) KG Sin θ (m) GZ (m)

0 0 0 0

12 1.75 1.55 0.20

20 2.93 2.55 0.38

30 4.27 3.73 0.54

40 5.36 4.79 0.57

50 6.12 5.71 0.41

60 6.67 6.45 0.22

75 6.92 7.20 - 0.28

Determine λ0 and λ40

λ0 = 6391 = 0.215m 1.65 x 18004

λ40 = 0.215 x 0.8 = 0.172m

Draw the GZ Curve

From the graph,

Angle of Heel after assumed grain shift = 12° Initial GM = 0.93m

Calculate the area (Residual Dynamical Stability) from 12° → 37°

Angle (°) GZ (m) SM Product 12.0 0.00 1 0.00 20.3 0.18 3 0.54 28.7 0.34 3 1.02 37.0 0.40 1 0.40

∑ 1.96

0.95

-0.10

0.10

0.30

0.50

0.70

0.90

0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 Angle of heel

GZ

(m)

GM

θ F 40.6 ° Max Ords between curves

12° 37 °

© ARNAB SARKAR


Area 0° → 30° = 3 x 10 x 1.96 = 0.128 m-rad 8 57.3

Check if vessel complies with the International Grain Code

The vessel complies with International Grain Code Regulations provided she is upright on sailing as Angle of Heel after assumed grain shift = 12° (complies) Residual Dynamical Stability = 0.128 m-rad (complies not less than

0.075 m-rad) Initial GM = 0.93m (complies not less than 0.3m)


Displacement 15000 t Even Keel draught 8.170m Max. Breadth 20.80m KG 7.88m KM 8.33m KB 4.15m

(a) Calculate the angle and direction of heel when turning to port in a circle of diameter 490m at a speed of 17.2 knots. Note: Assume 1 nautical mile = 1852m, and g = 9.81m/sec2

(b) Calculate the new maximum draught during the turn in Q 3(a), assuming the midship cross-section can be considered rectangular.

(a) BG = KG – KB GM = KM – KG

= 7.88 – 4.15 = 3.73m = 8.33 – 7.88 = 0.45m

Speed during turning (v) = 17.2 Knots = 17.2 x 1852 = 8.85 m/sec 3600 Radius of turn = 490 = 245m 2

∴ Angle of heel during turning = tan θ = (8.85)2 x 3.73 9.81 x 245 x 0.45 or, θ = 15.1° to stbd.

(b) Draught when heeled = (8.170 x Cos 15.1°) + (0.5 x 20.80 x Sin 15.1°) = 10.60m

© ARNAB SARKAR

SQA Stability – April 2009


Forward draught 8.180m Aft draught 9.420m at an upriver berth in fresh water.

The vessel is to proceed downriver to cross a sand bar at the river entrance where the relative density of the water is 1.020.

During the river passage the following items of deadweight will be consumed:

60 t of Heavy Fuel Oil from No. 3 DB Centre tank 21 t of Diesel Oil from No. 4 DB Port tank 21 t of Diesel Oil from No. 4 DB Starboard tank 18 t of Fresh Water from After Fresh Water tank

(a) Using the Stability Data Booklet, calculate the clearance of the vessel over the bar if the depth of water over the sand bar is 9.500m.

(b) State the maximum clearance over the sand bar if the vessel is brought to an even keel condition by internal transfer of ballast.

(a) Calculate TMD

AMD = 8.800m Trim = 1.24m by the stern

From hydrostatic particulars, using AMD = 8.800m

LCF = 65.74m foap ∴ TMD = 9.420 – (1.24 x 65.74) = 8.827m 137.5 Determine Initial Δ, MCTC and LCB using TMD = 8.827m

Initial ΔFW = 18383 + (237 x 0.027) = 18447 t 0.10 MCTCFW = 205.1 + (1.2 x 0.027) = 205.4 tm 0.10 LCB = 69.22 – (0.04 x 0.027) = 69.21m foap 0.10

Determine Initial LCG

1.24 x 100 = 18447 x (69.21 – LCG) 205.4 ∴ LCG = 67.83m foap

Calculate moments about the AP to determine the Final Δ and Final LCG

Weights (t) LCG (m) Longitudinal Moments (tm) 18447 67.83 1251260.01 (–) 60 57.02 3421.2 (–) 21 35.66 748.86 (–) 21 35.50 745.5 (–) 18 28.67 516.06 18327 1245828.39

Final Δ FW = 18327 t Final LCG = 67.98m foap

Determine volume in dock water density (1.020)

Volume = 18327 = 17967.65 m3 1.020

© ARNAB SARKAR


Determine the TMD, MCTC, LCB and LCF using ΔFW as 17967.65 t

TMD = 8.60 + (0.10 x 56.65) = 8.624m 236

MCTC = 202.6 + (0.3 x 56.65) = 202.7 tm 236 ∴ MCTC 1.020 = 202.7 x 1.020 = 206.8 tm

LCB = 69.31 – (0.04 x 56.65) = 69.30m foap 236

LCF = 65.87 – (0.06 x 56.65) = 65.86m foap 236

Determine COT

COT = 18327 x (69.30 – 67.98) = 117.0cms or 1.170m by stern (LCB > LCG) 206.8 ∴ COTA = 1.170 x 65.86 = 0.560m 137.5 ∴ COTF = 1.170 – 0.560 = 0.610m


Forward Aft TMD 8.624m 8.624m COT F/A 0.610m 0.560m Final Draughts 8.014m 9.184m

∴ Clearance of the vessel over the bar = 9.500 – 9.184 =0.316m

(b) Max. clearance over the sand bar if the vessel is in EK draught = 9.500 – 8.624 = 0.876m

2. A vessel is floating upright in salt water and is about to dry-dock.

The vessel’s particulars are:

Length 137.5m KG 8.34m

Present draughts: Forward 5.420m Aft 6.560m

(a) Using the Hydrostatic Particulars included in the Stability Data Booklet, calculate the vessel’s effective GM at the critical instant.

(b) Describe the methods of improving the initial stability if the GM at the critical instant is found to be inadequate.

(a) Calculate Initial TMD

Initial Trim = 5.420 – 6.560 = 1.140m Initial AMD = (5.420 + 6.560) = 5.990m 2 With AMD, from hydrostatic tables,

LCF = 68.43 – (0.04 x 0.09) = 68.39m foap 0.10 TMD = 6.560 – (1.140 x 68.39) = 5.993m 137.5

© ARNAB SARKAR


Calculate Initial Δ, MCTC, LCF

With TMD, from hydrostatic tables,

Δ = 12073 + (224 x 0.093) = 12281.32 t 0.10 MCTC = 168.5 + (1.3 x 0.093) = 169.7tm 0.10 LCF = 68.43 – (0.04 x 0.093) = 68.39m 0.10

Calculate the ‘P’ Force

‘P’ Force = 1.140 x 100 x 169.7 = 282.87 t 68.39

Calculate Virtual Δ

Virtual Δ = 12281.32 – 282.87 = 11998.45 t

Calculate the KMT using the virtual displacement

KMT = 8.59 – (0.04 x 150.45) = 8.56m 225

Calculate GM at C.I.

Initial GM = 8.56 – 8.34 = 0.22m Loss of GM = 282.87 x 8.56 = 0.197m 12281.32 ∴ GM at C.I. = 0.22 – 0.197 = 0.023m

(b) The trim can be reduced by transferring water, fuel or ballast forward. The initial GM can also be increased by reducing FSM, lowering weights, or by ballasting low or empty tanks.

3. A vessel is loading in dock water of R.D. 1.013.

Present draughts: Forward 7.940m Midship (Port) 8.610m Midship (Starboard) 8.550m Aft 9.120m

The draught marks are displaced as follows:

Forward: 1.66m aft of FP Aft: 2.14m forward of the AP The midship marks are not displaced.

The Stability Data Booklet provides the necessary hydrostatic data for the vessel.

By completion of the Worksheet Q3 and showing any additional calculations in the answer book, determine the vessel’s displacement. Draught forward = 7.940m


© ARNAB SARKAR






Amidships Line Corr. = Nil Draught at Midships = 8.580m ∴ Corr. (M) draught = 7.926 + (6 x 8.580) + 9.138 = 8.568m 8


TPC = 24.08 + (0.05 x 0.068) = 24.11tm 0.10 LCF = 65.95 – (0.08 x 0.068) = 65.90m foap 0.10 Displacement = 18119 + (240 x 0.068) = 18282.2 t 0.10 Distance of LCF from Midships = 68.75 – 65.90 = 2.85m 1st Trim Correction = 2.85 x 121.2 x 24.11 = 60.568 t 137.5


∴ Corrected Δ = 18282.2 + 60.568 + 6.570 = 18349.338 t ∴ Dock Water Δ = 18349.338 x 1.013 = 18134.52 t 1.025

© ARNAB SARKAR




The vessel has a centreline watertight bulkhead with an empty amidship side compartment of length 22.00m on each side of the vessel.

Calculate the angle of heel if ONE of these side compartments is bilged.


Volume of Buoyancy lost = Volume of Buoyancy gained or, (22 x 15 x 6.500) = {(140 x 30) – (22 x 15)} x Sinkage or, Sinkage = 2145 = 0.554m

3870 ∴ Bilged draught = 6.500 + 0.554 = 7.054m




3870 60525





BM = 288610.848 = 10.572m (140 x 30 x 6.500) KB = 7.054 = 3.527m 2 ∴ KM = 10.572 + 3.527 = 14.099m ∴ GM = 14.099 – 9.800 = 4.299m

Calculate List

Tan θ = 0.640 4.299 ∴ List = 8.5°

2. A vessel in salt water is trimmed 0.32m by the stern and heeled 8° to starboard and has the following particulars:

Displacement 10600 tonne KG 8.52m

Tunnel side tanks, port and starboard, are full of fuel oil and have centroids 4.30m each side of the centreline. No.3 DB tanks, port and starboard are empty and have centroids 4.80m each side of the centreline.

© ARNAB SARKAR



(a) The total weight of fuel oil to transfer from the Tunnel side tanks to No.3 DB tanks in order to bring the vessel to even keel.

(b) The final weight of fuel oil transferred into EACH of No.3 DB tanks in order to bring the vessel upright, assuming equal quantities are taken from the Tunnel side port and starboard tanks.

(a) Calculate initial MCTC of the vessel

MCTC = 160.8 + (1.0 x 84) = 161.2 tm 221 Calculate the initial trimming moment of the vessel

∴ Initial trimming moment = 0.32 x 100 x 161.2 = 5158.4 tm

Assume that ‘w’ tonne of fuel oil is needed to be transferred from the Tunnel side tanks to No.3 DB tanks in order to bring the vessel upright.

∴ 5158.4 = w (57.87 – 21.08) or, w = 140.2 t

∴ Total amount of fuel oil transferred to bring the vessel upright = 140.2 t ∴ Amount to be transferred from each Tunnel side tank = 70.1 t

Calculate initial KM of the vessel

KM = 8.86 – (0.05 x 84) = 8.84m 221

∴ GM = 8.84 – 8.52 = 0.32m

Calculate the initial listing moment of the vessel

Initial listing moment = 10600 x 0.32 x Tan 8° = 476.7 tm to starboard

Assume that ‘w’ tonne of fuel oil is needed to be transferred from No.3 DB tank (Stbd) to No.3 DB tank (Port) in order to bring the vessel upright.

∴ 476.7 = w x (4.8 + 4.8) or, w = 49.7 t

∴ Fuel oil transferred from No.3 DB (Stbd) to No.3 DB (Port) = 49.7 t ∴ Final quantity of Fuel oil transferred in No.3 DB (Stbd) = 70.1 – 49.7 = 20.4 t ∴ Final quantity of Fuel oil transferred in No.3 DB (Port) = 70.1 + 49.7 = 119.8 t

3. A vessel is to enter dry-dock in dock water of relative density 1.013.

Drafts: Forward 3.100m Aft 4.700m KG 9.55m

(a) Using the Stability Data Booklet, calculate the maximum trim at which the vessel can enter dry-dock so as to maintain a GM of at least 0.15m at the critical instant. Assume KM remains constant.

(b) Describe the measures to be taken to ensure that the stability of the vessel is adequate when the dock is flooded prior to the ship leaving the dock.

(a) Determine the TMD of the vessel

AMD = 3.900m & Trim = 1.600m by the stern

© ARNAB SARKAR


From hydrostatic particulars, using AMD = 3.900m

LCF = 69.88m foap ∴ TMD = 4.700 – (69.88 x 1.600) = 3.887m 137.5

From hydrostatic particulars, using TMD = 3.887m, determine Initial Δ, MCTC, LCF & KM

Initial Δ = {7296 + (209 x 0.087)} x 1.013 = 7575.04 t 0.10 MCTC = {144.2 + (0.9 x 0.087)} x 1.013 = 146.9 tm 0.10 LCF = 69.94 – (0.06 x 0.087) = 69.89m foap 0.10 KM = 10.25 – (0.14 x 0.087) = 10.13m 0.10

Determine Loss of GM

Initial GM = 10.13 – 9.55 = 0.58m Required GM = 0.15m ∴ Loss of GM = 0.58 – 0.15 = 0.43m

Determine the ‘P’ force

0.43 = P x 10.13 7575.04 ∴ P = 321.55 t

Determine required trim to enter the dry-dock

321.55 = trim x 146.9 69.89 ∴ Trim = 152.9cm or 1.529m

© ARNAB SARKAR


1. A box-shaped vessel floating on an even keel in salt water has the following particulars:

Length 120.00m Breadth 22.00m Draught 5.00m

There is a midship compartment of length 16.00m, extending the full breadth of the vessel and from the watertight tank top to the freeboard deck. Permeability of compartment 0.65. Height of watertight tank top 1.80m.

If this compartment is bilged, calculate EACH of the following:

a) The final draft b) The change in GM

a) Calculate Sinkage & Final draft of the box shaped vessel

Volume of the box-shaped vessel before bilging = 120 x 22 x 5.000 = 13200 m3 Volume of the bilged compartment = 16 x 22 x (5.000 – 1.800) = 1126.4 m3

Permeability of the bilged compartment = 0.65

Intact Water-Plane Area = (120 x 22) – (16 x 22 x 0.65) = 2411.2 m2

∴ Sinkage caused due to bilging = 1126.4 x 0.65 = 0.304m 2411.2 ∴ Final draft of the vessel after bilging = 5.000 + 0.304 = 5.304m

b) Before Bilging

KB = 2.50m

BM = 120 x (22)3 = 8.067m 12 x 13200

∴ KM = 2.50 + 8.067 =10.567m

After Bilging

Calculate KB by taking moments of volume about the Keel

Volume (m3) KB (m) Moments (m4) 120 x 22 x 5.304 2.652 37134.79 (–) 16 x 22 x (5.304 – 1.800) x 0.65 3.552 (–) 2847.7 13200.84 34287.09

∴ KB = 2.597m

BM = {120 – (16 x 0.65)} x 223 = 7.368m 12 x 13200 ∴ KM = 2.597 + 7.368 = 9.965m

∴ Change in KM = 10.567 – 9.965 = 0.602m decrease ∴ Change in GM = 0.602m decrease

2. A box shaped vessel, length 96.00m, breadth 11.00m, is floating at a draft of 4.20m in

salt water. Initial KG 4.18m.

a) Calculate the angle of loll if 620 t of cargo, KG 7.78m, is loaded on deck. b) Calculate the GM at the angle of loll in Q 2(a).

© ARNAB SARKAR


a) Calculate Initial Δ of the box shaped vessel

Initial Δ = 96.00 x 11.00 x 4.20 x 1.025 = 4546.08 t

Determine final KG of the box shaped vessel after loading cargo

Weights (t) KG (m) Moments (tm) 4546.08 4.18 19002.61 620 7.78 4823.6 5166.08 23826.21

∴ Final KG = 23826.21 = 4.612m 5166.08

Calculate the final draught after loading cargo

5166.08 = 96.00 x 11.00 x d x 1.025 ∴ d = 4.773m

Calculate final KM of the box shaped vessel

KB = 2.387m BM = 96 x (11)3 = 2.113m

12 x 96 x 11 x 4.773 ∴ KM = 2.387 + 2.113 = 4.500m

Calculate final GM of the box shaped vessel

Final GM = 4.500 – 4.612 = (–) 0.112m

Calculate Angle of Loll (θ)

Tan θ = √ 2 x 0.112 2.113 ∴ θ = 18.0°

b) GM at angle of loll = 2 x 0.112 = 0.236m Cos 18.0°

3. Worksheet Q3 – ‘Trim and Stability’ provides data relevant to a particular condition of loading in a vessel in salt water.

The Hydrostatic Particulars included in the Stability Data Booklet provides the necessary hydrostatic data for the vessel.


a) The effective metacentric height b) The draughts forward and aft

© ARNAB SARKAR


CONDITION: LOADED – GENERAL CARGO Length of vessel: 137.50m



Weight (t)

KG (m)

Vertical Moment

(tm)

Free Surface Moment

(tm)


All holds 14606 1.88 7769 5.86 45526 68.6 532953

1 TD 1365 2.22 615 10.98 6753 114.0 70110

2 TD 1332 2.36 564 10.50 5922 95.6 53918

3 TD 1435 2.14 671 10.37 6958 74.0 49654

4 TD 1230 2.28 540 10.28 5551 53.0 28620

Oil Fuel 856 1786 30074

FW 83 610 4590

Lightship 3831 8.21 31453 69.68 266944

Δ 14929 7.00 104559 69.45 1036863

HYDROSTATICS


LCB foap = 69.99 – (0.05 x 121) 232 = 69.96m

LCF foap = 67.24 – (0.11 x 121) 232 = 67.18m

LBP = 137.5m

MCTC = 186.2 + (1.6 x 121) 232 = 187.0 tm

KMT = 8.34 – (0.01 x 121) 232 = 8.33m

GMF = 8.33 – 7.00 = 1.33m

Trim between Perpendiculars: 14929 x (69.96 – 69.45) = 40.72 cm or 0.407m by stern 187.0 COTA = 0.407 x 67.18 = 0.199m 137.5 COTF = 0.407 – 0.199 = 0.208m Draughts: Forward: 7.152 – 0.208 = 6.944m Aft = 7.152 + 0.199 = 7.351m

© ARNAB SARKAR


1. A vessel, initially upright and on even keel, has the following particulars:

Draught (in salt water) 7.000m Breadth 20.42m KG 7.82m

Further particulars of the vessel can be found in the Stability Data Booklet.

The vessel’s heavy lift derrick is to be used to discharge a 58 tonne boiler from a centreline position, KG 5.30m. The derrick head is 30.00m above the keel and 16.00m from the ship’s centreline when plumbing overside.

a) Calculate the maximum list angle when the boiler is suspended by the derrick at its maximum outreach during discharge.

b) Calculate the increase in draught when the vessel is at the maximum list angle calculated in Q 1(a), assuming a rectangular cross section midships.

c) Calculate the maximum allowable KG prior to discharging the boiler in order to limit the list angle to 5°.

a) Determine the Initial Δ of the vessel from hydrostatic particulars


Calculate moments about the keel and the centreline to determine the Final KG & GGH

Weights (t) KG (m) Moments (tm) Dist. From CL Moments 14576 7.82 113984.32 - -

(–) 58 5.30 (–) 307.4 - - (+) 58 30.0 (+) 1740.0 16.0 928 14576 7.918 115416.92 0.064 928

∴ Final KG = 7.918m ∴ GGH = 0.064m

Initial KM = 8.340m (from hydrostatic particulars) Initial GM = 8.340 – 7.918 = 0.422m

∴ Maximum list angle = Tan θ = 0.064 0.422 or, θ = 8.6°

b) Draught when heeled = (7.000 x Cos 8.6°) + (20.42 x Sin 8.6°) 2

= 8.448m ∴ Initial Draught = 7.000m ∴ Increase in draught = 8.448 – 7.000 = 1.448m

c) Determine the GM when listed to 5°

Tan 5° = 0.064 ∴ GM = 0.732m GM

Determine the KG when listed to 5°

KM = 8.340m ∴ KG = 8.340 – 0.732 = 7.608m

Determine KG prior to commence operations

Weights (t) KG (m) Moments (tm) 14576 7.608 110894.208 (+) 58 5.30 (+) 307.4 (–) 58 30.0 (–) 1740.0 14576 7.510 109461.608

© ARNAB SARKAR


∴ Required KG prior to commence operations = 7.510m

2. a) A vessel, initially upright, is to carry out an inclining test. Present displacement 5300 t KM 10.96m

Total weights on board during the experiment: Ballast 390 t KG 3.45m Tank full Bunkers 175 t KG 4.01m Free Surface Moment 996 tm Water 102 t KG 4.45m Free Surface Moment 890 tm Boiler Water 20 t KG 4.19m Free Surface Moment 101 tm Inclining weights 48 t KG 8.42m

A weather deck hatch cover, weight 20 t, ashore for repair will be fitted on the vessel at a KG of 9.46m prior to sailing.

The plumblines have an effective vertical length of 8.00m. The inclining weights are shifted transversely 7.60m on each occasion and the mean horizontal deflection of the plumbline is 0.68m.

Calculate the vessel’s lightship KG.

b) List FIVE possible causes for a change to the vessel’s lightship KG over a period of time.

(a) Displacement = 5300 t KM = 10.96m

Plumb length = 8.00m Deflection = 0.68m Trans shift = 7.6m


= 0.81m

KM = 10.96m GMF = 0.81m




Δ Inclined 5300 10.15 53795 (–) Ballast (–) 390 3.45 1345.5 (–) Bunkers (–) 175 4.01 701.75 (–) FSM Bunkers 996 (–) Water (–) 102 4.45 453.9 (–) FSM Water 890 (–) Inclining Weight (–) 48 8.60 404.16 (+) Hatch Cover (+) 20 9.46 189.2 4605 t 49192.89 tm

∴ Lightship KG = 49192.89 = 10.682 m 4605


Constant of the vessel keep changing due to accumulated sludge in fuel oil tanks. Constant of the vessel keep changing due to accumulated mud and rust in ballast

water tanks (un-pumpables).

© ARNAB SARKAR


Various stores remaining unconsumed might add to the constant. Any structural changes will affect the light ship KG and light ship displacement Lightship KG for a passenger vessel will change considerably over a period of time

mainly because of the left over baggage etc. will accumulate over a period of time and add to the constant considerably.

3. A box shaped vessel floating on even keel in dock water of RD 1.015 has the following

particulars:

Length 130.00m Breadth 21.00m Draught 8.000m MCTC (salt water) 300

There is an empty watertight forward end compartment, length 10.00m, height 6.70m, extending the full width of the vessel.

Calculate the draughts forward and aft, if this compartment is bilged. Calculate the Bilged TMD

Volume of the vessel before bilging = 130 x 21 x 8 = 21840 m3 ∴ Displacement of the vessel before bilging = 21840 x 1.015 = 22167.6 t


Volume of the forward end compartment = 10 x 21 x 6.70 = 1407 m3 Intact water plane area = 130 x 21 = 2730 m2

∴ Sinkage caused to bilging = 1407 = 0.515m 2730 ∴ Bilged TMD = 8.000 + 0.515 = 8.515m

Calculate the Trimming Moment

Calculating moments of volume about the AP after bilging,

Volume (m3) Distance from AP (m) Moments (m4) 130 x 21 x 8.515 65.0 1510986.75 (–) 1407 125.0 (–) 175875 21838.95 1335111.75

∴ BH = 61.134m ∴ BBH = 65 – 61.134 = 3.866m ∴ Trimming Moment = 22167.6 x 3.866 = 85699.94 tm by forward

Calculate COT after bilging

COT = 85699.94 x 1.025 = 288.5cm or 2.885m 300 x 1.015 ∴ COTA = 2.885 x 65 = 1.443m 130 ∴ COTF = 2.885 – 1.443 = 1.442m

Calculate the Final Draughts of the vessel

Forward Draught = 8.515 + 1.442 = 9.957m Aft Draught = 8.515 – 1.443 = 7.072m

© ARNAB SARKAR


1. A vessel is upright, starboard side alongside, at a draught of 6.00m in salt water. KG 8.30m.

There is a 27 t boiler on deck, KG 7.78m, which is to be discharged using vessel’s crane the head of which is 26.0m above the keel. The boiler is to be lifted from a position on the vessel’s centreline and landed on a railway truck ashore. The distance of railway truck from vessel’s centreline is 23.30m.

Using the Hydrostatic Particulars included in the Stability Data Booklet, calculate EACH of the following:

a) The maximum angle of heel during discharge; b) The maximum angle of heel during discharge if the vessel was first listed 5° to port

prior to the discharge of the boiler.

a) Determine the Initial Δ and KM of the vessel from hydrostatic particulars,

Initial Δ = 12297 t KM = 8.52m

Calculate moments about the Keel and the CL of the vessel,

Weights (t) KG (m) Moments (tm) Distance from CL Moments (tm) (P) (S) (P) (S) 12297 8.30 102065.1 - - - - (–) 27 7.78 210.06 - - - - (+) 27 26.0 702.0 - 23.3 - 629.1 12297 102557.04 - 629.1

∴ Final KG = 8.340m ∴ GGH = 0.051m to starboard ∴ Final GM = 8.52 – 8.34 = 0.18m

Determine the maximum angle of heel during discharge

∴ Tan θ (angle of heel) = 0.051 0.18 ∴ Angle of Heel = 15.8° to starboard

b) Determine the GGH when initially listed to port 5°,

GGH = (8.52 – 8.30) x Tan 5° = 0.019m to port

Calculate moments about the Keel and the CL of the vessel,

Weights (t) KG (m) Moments (tm) Distance from CL Moments (tm) (P) (S) (P) (S) 12297 8.30 102065.1 0.019 - 233.64 - (–) 27 7.78 210.06 - - - - (+) 27 26.0 702.0 - 23.3 - 629.1 12297 102557.04 - 395.46

∴ Final KG = 8.340m ∴ GGH = 0.032m to starboard ∴ Final GM = 8.52 – 8.34 = 0.18m

Determine the maximum angle of heel during discharge

∴ Tan θ (angle of heel) = 0.032 0.18 ∴ Angle of Heel = 10.1° to starboard

© ARNAB SARKAR


2. A vessel, initially upright, is to carry out an inclining test.

Present displacement 4800 t, KM 10.58m.


Ballast 400 t KG 3.52m Tank full Bunkers 175 t KG 3.86m Free Surface Moments 997 tm Fresh Water 85 t KG 4.46m Free Surface Moments 810 tm Inclining weights 52 t KG 8.42m

At the time of the experiment the boilers were empty. They would usually contain a total of 24 t of water, KG 4.18m, with a free surface moment of 124 tm.

A deck crane, weight 18 t and still ashore will be fitted on the vessel at a KG of 9.85m before the next cargo operation.

The plumbline has an effective vertical length of 7.88m. The inclining weights are shifted transversely 7.20m on each occasion and the mean horizontal deflection of the plumbline is 0.66m.

a) Calculate the vessel’s lightship KG. b) Identify FIVE possible causes of a change in the vessel’s lightship KG over a period of

time. (a) Displacement = 4800 t KM = 10.58m

Plumb length = 7.88m Deflection = 0.66m Trans shift = 7.20m


= 0.93m

KM = 10.58m GMF = 0.93m




Δ Inclined 4800 9.65 46320 (–) Ballast (–) 400 3.52 1408 (–) Bunkers (–) 175 3.86 675.5 (–) FSM Bunkers 997 (–) Water (–) 85 4.46 379.1 (–) FSM Water 810 (–) Inclining Weight (–) 52 8.42 437.84 (+) Boiler Water (+) 24 4.18 100.32 (+) FSM Boiler Water 124 (+) Hatch Cover (+) 18 9.85 177.3 4130 t 42014.18 tm

∴ Lightship KG = 42014.18 = 10.173 m 4130


Constant of the vessel keep changing due to accumulated sludge in fuel oil tanks.

© ARNAB SARKAR


Constant of the vessel keep changing due to accumulated mud and rust in ballast water tanks (un-pumpables).

Various stores remaining unconsumed might add to the constant. Any structural changes will affect the light ship KG and light ship displacement Lightship KG for a passenger vessel will change considerably over a period of time

mainly because of the left over baggage etc. will accumulate over a period of time and add to the constant considerably.

3. A vessel is to load a cargo of grain (stowage factor 1.62 m3/t).

Initial displacement 5900 t, Initial KG 6.65m

All five holds are to be loaded full of grain. The tween decks are to be loaded as follows:

No. 1 TD Part Full – Ullage 1.75m No. 2 TD Full No. 3 TD Part Full – Ullage 2.50m No. 4 TD Full


a) Using the Maximum Permissible Grain Heeling Moment Table Included in the Stability Data Booklet, determine whether the vessel complies with the minimum criteria specified in the International Grain Code (IMO).

b) Calculate the ship’s approximate angle of heel in the event of the grain shifting as assumed by the International Grain Code (IMO).

a) Initial Δ = 5900 t



Comp. Ullage (m)

Volume (m3)

SF (m3/t) Tonnes KG

(m) Vert. Mom.

(tm) VHM (tm)

Initial Δ 5900 6.65 39235

No.1 TD 1.75 1416 1.62 874 10.88 9509 1372

No.2 TD Full 1676 1.62 1035 10.78 11157 539

No.3 TD 2.50 979 1.62 604 10.25 6191 3142

No.4 TD Full 1674 1.62 1033 10.57 10919 604

No.1 Hold Full 2215 1.62 1367 5.09 6958 410

No.2 Hold Full 4672 1.62 2884 4.95 14276 1285

No.3 Hold Full 1742 1.62 1075 4.94 5311 475

No.4 Hold Full 3474 1.62 2144 4.95 10613 910

No.5 Hold Full 2605 1.62 1608 8.76 14086 455

Final Δ 18524 6.92 128255 9192

© ARNAB SARKAR


Calculate AHM

AHM = 9192 = 5674 tm 1.62


Δ = 18524 t KGF = 6.92m

Interpolation

Δ KGF = 6.90 KGF = 6.92 KGF = 7.00

19000 7157 7070.0 6737

18524 6828.4

18500 6898 6816.2 6489


7157 – (420 x 0.02) = 7283.0 tm 0.10


6898 – (409 x 0.02) = 7020.7 tm 0.10


6816.2 + (253.8 x 24) = 6828.4 tm 500

∴ PHM for Δ = 18524 t & KGF = 6.92m = 6828.4 tm



Angle of Heel = 5674 x 12° = 10.0° 6828.4

© ARNAB SARKAR

SQA Stability – December 2010

1. A vessel is floating in salt water at draughts 7.640m forward, 8.180m aft.

The vessel is to load a cargo as to finish on an even keel at a draught of 8.500m. Two spaces available: No.1 hold and No.5 hold.


a) The total weight of cargo to load; b) The weight of cargo to load in each compartment.

1. Calculate AMD and Trim


Calculate LCF and TMD

LCF = 66.42 – (0.09 x 0.01) = 66.41m 0.10 ∴ TMD = 8.180 – (0.540 x 66.41) = 7.920m 137.5

Using Hydrostatic Particulars, calculate Initial Δ, LCB, & MCTC

Initial Δ = 16685 + (237 x 0.02) = 16732.4 t 0.10 LCB = 69.63 – (0.05 x 0.02) = 69.62m foap 0.10 MCTC = 198.2 – (1.40 x 0.02) = 197.9 tm 0.10


0.540 x 100 = 16732.4 x (69.62 – LCG) 197.9 or, LCG = 68.98m foap

Calculate the Final Δ, from hydrostatic particulars using draught 8.500m

Final Δ = 18119 t

a) Calculate total amount of cargo to load

Total amount of cargo to load = 18119 – 16732.4 = 1386.6 t ≈ 1387 t

Assume ‘w’ tons of cargo is loaded in No.1 Hold ∴ Weight of cargo loaded in No.5 Hold = (1387 – w) t

Calculate Longitudinal Moments about the AP

Weights (t) LCG (m) Longitudinal Moments (tm) Initial Δ 16732.4 68.98 1154200.952 No.1 Hold (+) w 114.48 (+) 114.48 w No.5 Hold (+) (1387 – w) 8.76 (+) 12150.12 – 8.76w Final Δ 18119 (1154200.952 + 105.72w)

∴ Final LCG = (1154200.952 + 105.72w) m foap 18119 Determine LCB from hydrostatic particulars, using draught 8.500m

LCB = 69.36m foap

As vessel will be in even keel on completion of loading, hence LCB = LCG = 69.36m foap

© ARNAB SARKAR


∴ (1154200.952 + 105.72w) = 69.36 18119 or, w = 969.9 t ≈ 970 t

b) ∴ Quantity of cargo to be loaded in No.1 Hold = 970 t ∴ Quantity of cargo to be loaded in No.5 Hold = 417 t



Using the Stability Data Booklet, compare the vessel’s stability with ALL the minimum stability criteria required by the current Load Line Regulations, commenting on the result.

2. Δ = 18000 t KGF = 8.20m


θF = 40.6°



(Tonnes) 12° 20° 30° 40° 50° 60° 75°

18000 1.75 2.93 4.27 5.36 6.12 6.67 6.92


Heel (°) KN (m) KG Sin θ (m) GZ (m) 12 1.75 1.70 0.05 20 2.93 2.80 0.13 30 4.27 4.10 0.17 40 5.36 5.27 0.09 50 6.12 6.28 - 0.16 60 6.67 7.10 - 0.43 75 6.92 7.92 - 1.00



KMT = 8.39 + (0.02 x 122) = 8.40m 241 ∴ GM = 8.40 – 8.20 = 0.20m

© ARNAB SARKAR





10 0.04 3 0.12 20 0.13 3 0.39 30 0.17 1 0.17 ∑ 0.68



10 0.04 4 0.16 20 0.13 2 0.26 30 0.17 4 0.68 40 0.09 1 0.09 ∑ 1.19


∴ Area 30° → 40° = 0.069 – 0.045 = 0.024 m-rad

Summary

Actual L.L. Regulation Complies or Not Area 0° → 30° 0.045 m-r Not less than 0.055 m-r No Area 0° → 40° 0.069 m-r Not less than 0.09 m-r No Area 30° → 40° 0.024 m-r Not less than 0.03 m-r No Max. GZ 0.17m At least 0.20m No Max. GZ Angle 30° Equal to or greater than 30° Yes GMF 0.20m Not less than 0.15m Yes

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0 20 40 60 80

© ARNAB SARKAR


∴ Vessel does not comply with the minimum stability criteria as required by the current Load Line Regulations.

3. A vessel is floating at an even keel draught of 7.80m in salt water. KG 7.14m.

A midship rectangular deck 29.00m long and extending the full breadth of the vessel is flooded with saltwater to a depth of 0.60m.

Height of deck above keel 9.50m.


a) The fluid GM; b) The angle of loll.

3. Calculate the vessel’s Initial Δ from hydrostatic particulars using draught of 7.800m


Calculate the amount of water ingress in the midship rectangular deck

Quantity of water ingress = 29.00 x 20.42 x 0.60 x 1.025 = 364.2 t

Calculate the Final Δ after the flooding

Final Δ = 16448 + 364.2 = 16812.2 t

Determine the Final KM & KB from hydrostatic particulars, using Final Δ 16812.2 t

Final KM = 8.35m

Final KB = 4.12 + (0.05 x 127.2) = 4.15m 237

Determine the FSM caused due to water ingress

FSM caused = 29 x (20.42)3 x 1.025 = 21091.518 tm 12

Determine the Final KG by taking moments about the keel

Weights (t) KG (m) Vertical Moments (tm) 16448 7.14 117438.72

(+) 364.2 9.50 3459.9 (+) FSM 21091.518 16812.2 141990.138

∴ Final KG = 141990.138 = 8.45m 16812.2

a) Determine Final GMF

Final GMF = 8.35 – 8.45 = – 0.10m

b) Determine Angle of Loll

BMT = 8.35 – 4.15 = 4.20m

Angle of Loll = √(2 x 0.10) 4.20 ∴ θ Loll = 12.3°

© ARNAB SARKAR



Forward draught 7.982m, aft draught 9.218m at an upriver berth in fresh water.

The vessel is to proceed downriver to cross a sand bar at the river entrance where the relative density of the water is 1.018.


56 t of Heavy Fuel Oil from No.3 DB Centre tank 24 t of Diesel Oil from No.4 DB Port tank 24 t of Diesel Oil from No.4 DB Starboard tank 15 t of Fresh Water from Forward Fresh Water tank

a) Using the Stability Data Booklet, calculate the clearance of the vessel if the depth of water over the sand bar is 9.440m.

b) State the maximum clearance over the sand bar if the vessel is brought to an even keel condition by internal transfer of ballast.

a) Calculate AMD and Trim



LCF = 65.87m TMD = 9.218 – (1.236 x 65.87) = 8.626m 137.5

Using Hydrostatic Particulars, calculate Initial Δ, LCB, & MCTC

Initial Δ = 17911 + (236 x 0.026) = 17972.36 t 0.10 LCB = 69.31 – (0.04 x 0.026) = 69.30m foap 0.10 MCTC = 202.6 – (0.03 x 0.026) = 202.6 tm 0.10


1.236 x 100 = 17972.36 x (69.30 – LCG) 202.6 or, LCG = 67.91m foap

Calculate Longitudinal Moments about the AP to determine the Final LCG

Weights (t) LCG (m) Longitudinal Moments (tm) Initial Δ 17972.36 67.91 1220502.968 HFO (–) 56 57.02 (–) 3193.12 FSM HFO (+) 1162.556 DO (–) 24 57.87 (–) 1388.88 FSM DO (+) 279.95 DO (–) 24 57.87 (–) 1388.88 FSM DO (+) 279.95 FW (–) 15 32.47 (–) 487.05 FSM FW (+) 29.522 Final Δ 17853.36 1215797.016

© ARNAB SARKAR


∴ Final LCG = 1215797.016 = 68.10m foap 17853.36

Determine volume using Final Δ

Volume 1.018 = 17853.36 = 17537.68 1.018

From hydrostatic particulars, determine TMD, MCTC, LCB & LCF using Final Δ

TMD = 8.40 + (0.10 x 95.68) = 8.441m 235 MCTCFW = 200.1 + (1.3 x 95.68) = 200.6 tm 235 MCTC1.018 = 200.6 x 1.018 = 204.2 tm

LCB = 69.40 – (0.04 x 95.68) = 69.38m foap 235 LCF = 66.02 – (0.07 x 95.68) = 65.99m foap 235

Determine CoT CoT = 17853.36 x (69.38 – 68.10) = 111.9cm or 1.120m by stern 204.2 ∴ CoTA = 1.120 x 65.99 = 0.538m 137.5 ∴ CoTF = 1.120 – 0.538 = 0.582m


Draught Forward = 8.441 – 0.582 = 7.859m Draught Aft = 8.441 + 0.538 = 8.979m

Determine the maximum clearance of the vessel over the sand bar

Clearance = 9.440 – 8.979 = 0.461m

b) Calculate the new AMD and Trim with present draughts



LCF = 66.02 – (0.07 x 0.019) = 66.01m foap 0.10 TMD = 8.979 – (1.120 x 66.01) = 8.441m 137.5

In even keel draught, vessel’s draught will be equal to TMD = 8.441m

Determine the maximum clearance of the vessel over the sand bar in Even Keel

Clearance = 9.440 – 8.441 = 0.999m


Length 100.00m Breadth 15.00m Depth 9.00m Draught 5.000m KG 5.40m

© ARNAB SARKAR


A midship watertight compartment 20.00m long and extending the full breadth and depth of the vessel is bilged. Permeability of the compartment is 0.85.

Calculate EACH of the following:

a) The new draught; b) The change in GM; c) The righting moment in the flooded condition for an angle of heel of 20°.

a) Volume of the box shaped vessel before bilging = 100 x 15 x 5 = 7500 m3

Volume of the bilged compartment = 20 x 15 x 5 = 1500 m3


Calculate Sinkage and bilged draught

Sinkage = 1500 x 0.85 = 1.024m (100 x 15) – (20 x 15 x 0.85)

∴ Bilged draught = 5.000 + 1.024 = 6.024m

b) Determine the KM prior bilging

KB = 5.000 = 2.500m 2 BM = 100 x 153 = 3.750m 12 x 7500 ∴ KM = 2.500 + 3.750 = 6.250m

Determine the KM after bilging KB = 6.024 = 3.012m 2 BM = {100 – (20 x 0.85) x 153 = 3.113m 12 x 7500 ∴ KM = 3.012 + 3.113 = 6.125m

Determine change in GM ∴ Change in KM = 6.250 – 6.125 = 0.125m decrease ∴ Change in GM = 0.125m decrease

c) Calculate GZ

GM = 6.125 – 5.400 = 0.725m Δ = 7500 x 1.025 = 7687.5 t

Using wall sided formula to determine GZ as the angle of heel is 20°

GZ = [0.725 + 3.113 x Tan2 20°] Sin 20° = 0.318m 2

∴ Righting Moment in flooded condition = 7687.5 x 0.318 = 2444.6 tm ≈ 2445 tm

3. A vessel completes under-deck loading in salt water with the following particulars: Displacement 15040 tonne KG 8.00m


© ARNAB SARKAR


During the passage, 200 tonne of Heavy Fuel Oil (RD 0.80) is consumed from No.3 DB tank centre which was full on departure (use VCG of tank for consumption purposes).

Calculate the maximum weight of timber to load on deck KG 12.60m assuming 15% water absorption during the passage in order to arrive at the destination with the minimum GM (0.05m) allowed under the Load Line Regulations.

Note: Assume KM constant

3. Assume that ‘w’ t of timber deck cargo is loaded on deck. Absorption of water during the passage by the timber deck cargo = 0.15 w t

Calculate the Initial KM of the vessel from hydrostatic particulars using Δ = 15040 t

Initial KM = 8.33m

Calculate the FSM for the HFO consumption

FSM HFO = 1142 x 0.80 = 913.6 tm

Calculate the final KG of the vessel by calculating moments about the Keel

Weights (t) KG (m) Vertical Moments (tm) 15040 8.00 120320

(+) 1.15w 12.6 14.49w (–) 200 0.60 (–) 120 (+) FSM (+) 913.6

(14840 + 1.15w) (121113.6 + 14.49w)

∴ Final KG = (121113.6 + 14.49w) m (14840 + 1.15w)

∴ Final GM = 8.33 – (121113.6 + 14.49w) = 0.05 (14840 + 1.15w) or, (121113.6 + 14.49w) = 8.28 (14840 + 1.15w) or, 121113.6 + 14.49w = 122875.2 + 9.52w or, 4.97w = 1761.6 or, w = 354.4 t

∴ Total weight of timber deck cargo loaded = 354.4 t

© ARNAB SARKAR



Present draughts in fresh water (RD 1.000): Forward 5.380m Aft 6.560m

The fore mast is 110m foap and extends 26.00m above the keel. The aft mast is 36m foap and extends 27.20m above the keel.


a) The final draughts forward and aft in order to pass under the bridge with minimum clearance;

b) The maximum weight of cargo that can be discharged in order to pass under the bridge with minimum clearance.

Note: Assume masts are perpendicular to the waterline throughout



∴ Trim at Masts = 6.340 – 5.140 = 1. 200m by stern










Initial trim = 6.560 – 5.380 = 1.180m by stern Initial AMD = (6.560 + 5.380) = 5.970m 2 With AMD, from hydrostatic tables, LCF = 68.43 – (0.04 x 0.07) = 68.40m foap 0.10 TMD = 6.560 – (1.180 x 68.40) = 5.973m 137.5

© ARNAB SARKAR

SQA Stability – July 2011 ∴ Displacement FW = 11778 + (219 x 0.073) = 11937.87 t ≈ 11938 t 0.10




∴ Displacement FW = 11559 + (219 x 0.012) = 11585.28 t ≈ 11585 t 0.10

Calculate maximum weight of cargo needed to be discharged in order to pass under bridge

∴ Weight of Cargo = 11938 – 11585 = 353 t

2. A vessel entered drydock for emergency repairs while carrying cargo.

Particulars on entry (salt water): Displacement 15716 t Trim 0.64m by stern KG 8.10m

While in drydock, 360 tonne of cargo, KG 3.80m, LCG 82.40m was discharged and is no longer on board. The vessel is now planning to leave the drydock.

a) Using the Stability Data Booklet, calculate the vessel’s effective GM at the critical instant on departure.

b) State the stability measures that should be considered prior to flooding the dock in order to ensure the safe undocking of this vessel.

a) Calculate the Final KG after discharging the cargo

Weights (t) KG (m) Vertical Moments (tm) 15716 8.10 127299.6

(–) 360 3.80 1368 15356 125931.6

∴ Final KG = 125931.6 = 8.201m 15356

Determine vessel’s initial LCB & MCTC from hydrostatic particulars, using Δ = 15716 t

LCB = 69.85 – (0.04 x 209) = 69.81m foap 235 MCTC = 190.9 + (1.5 x 209) = 192.2 tm 235

Calculate vessel’s initial LCG

0.64 x 100 = 15716 x (69.81 – LCG) 192.2 or, LCG = 69.03m foap

© ARNAB SARKAR


Calculate Final LCG of the vessel after discharging the cargo

Weights (t) LCG (m) Longitudinal Moments (tm) 15716 69.03 1084875.48

(–) 360 82.40 29664.0 15356 1055211.48

∴ Final LCG = 1055211.48 = 68.72m foap 15356

Determine vessel’s final LCB, LCF, MCTC & KM from hydrostatic particulars, using Δ = 15356 t

LCB = 69.90 – (0.05 x 82) = 69.88m foap 233 LCF = 67.03 + (0.11 x 82) = 67.07m foap 233 MCTC = 189.4 + (1.50 x 82) = 189.93 tm 233 KM = 8.33m

Calculate vessel’s final trim on departure

Trim on departure = 15356 x (69.88 – 68.72) = 93.8cm or 0.938m 189.93


P = 93.8 x 189.93 = 265.624 t 67.07

Calculate Loss of GM

Loss of GM = 265.624 x 8.33 = 0.144m 15356 Vessel’s virtual GM on departure = 8.33 – 8.201 = 0.129m ∴ Vessel’s effective GM at C.I. = 0.129 – 0.144 = – 0.015m

b) The drydock cannot be flooded with the vessel in this condition as her GM is negative. It is

necessary to take adequate measures to ensure that the vessel has sufficient positive stability at the critical instant during un-docking.

There are two options available (or a combination of both) to attain sufficient positive stability prior to un-docking:

Reduce the vessel’s KG by taking in ballast or loading deadweight items below VCG or remove deadweight items above VCG.

Reduce the vessel’s trim by removing deadweight items aft of the vessel’s LCG or by adding deadweight items forward of the vessel’s LCG.


Even keel draught 8.720m Maximum breadth 22.60m KG 7.96m KM 8.38m KB 4.18m

(i) Calculate the angle and direction of heel when turning to port in a circle of diameter 500m at a speed of 18.2 Knots.

(ii) Calculate the new maximum draught during the turn in Q 3(i), assuming the midships cross-section can be considered rectangular.

© ARNAB SARKAR


Note: Assume 1 Nautical Mile = 1852m and g = 9.81m/sec2

(i) Calculate BG and GM of the vessel

BG = 7.96 – 4.18 = 3.78m GM = 8.38 – 7.96 = 0.42m

Calculate the radius and velocity of turn

R = 500 = 250m 2 v = 18.2 x 1852 = 9.363 m/sec2 3600

Calculate the angle and direction of heel when turning

Tan θ = (9.363)2 x 3.78 9.81 x 250 x 0.42 ∴ θ = 17.8° to stbd.

(i) Calculate draught of the vessel when heeled

Draught when heeled = (8.720 x Cos 17.8°) + (0.5 x 22.6 x Sin 17.8°) = 11.762m

© ARNAB SARKAR


1. As a result of a collision a vessel is listed 8° to port in salt water. A deep tank is partially full of an oily water mixture, RD 0.95, which is to be fully discharged into a salvage barge alongside. The deep tank is rectangular and is 14.0m long and 18.0m wide.

Present displacement 16000 t KG 8.20m Weight of oily water mixture 846 t KG 3.20m

Using Hydrostatic Particulars contained in the Stability Data Booklet, calculate EACH of the following:

a) The final list of the vessel; b) The weight of ballast to transfer from a port tank to a starboard tank in order to bring

the vessel upright. The centroid of EACH tank is 5.00m from the centreline. a) Calculate the FSM

FSM = 14.0 x (18.0)3 x 0.95 = 6464 tm 12 x 16000

Determine the Final KG by taking moments about the Keel,

Weights (t) KG (m) Vertical Moments (tm) 16000 8.20 131200

(–) 846 3.20 2707.20 (+) FSM 6464 15154 122028.8

∴ Final KG = 122028.8 = 8.05m 15154

Calculate the Final GM after discharging the oily water mixture into the salvage barge

Initial KM = 8.33 + (0.01 x 24) = 8.33m (from hydrostatic particulars) 236 ∴ Final GM = 8.33 – 8.05 = 0.28m

Calculate Initial Listing Moment of the vessel

Initial GM = 8.33 – 8.20 = 0.13m ∴ Listing Moment = 16000 x 0.13 x Tan 8° = 292.3 tm to Port

Calculate the Final List Tan θ = GGH = Δ x GGH GM Δ x GM ∴ Tan θ = 292.3 15154 x 0.28 ∴ θ = 3.9° to Port

b) To bring the vessel to upright condition,

Port Moment = Stbd. Moment or, 292.3 = w x (5.00 + 5.00) or, w = 29.2 t

∴ Weight of ballast to transfer to bring the vessel upright = 29.2 t

© ARNAB SARKAR




The vessel has a centreline watertight bulkhead with an empty amidships side compartment of length 20.00m on each side of the vessel.

a) Calculate the angle of heel if ONE of these side compartments is bilged as a result of a collision.

b) Calculate the clearance over a sand bar in the bilged condition at the entrance of a port of refuge where the depth of water is 11.00m.

a) Calculate Sinkage & Bilged draught of the box shaped vessel

Volume of buoyancy lost = Volume of buoyancy gained or, 20.00 x 15.00 x 6.600 = {(120.00 x 30.00) – (20.00 x 15.00)} x Sinkage or, Sinkage = 1980 3300 ∴ Sinkage = 0.600m ∴ Bilged draught = 6.600 + 0.600 = 7.200m

Calculate the new location of LCF with respect to axis ‘XX’

Calculating Moments of area about the axis ‘XX’

Area (m2) Distance from axis ‘XX’ (m) Moments (tm) 120 x 30 = 3600 15.0 54000 (–) 20 x 15 = 300 7.50 2250 3300 51750

∴ New Location of the LCF = 51750 = 15.682m 3300 ∴ Distance BBH = 15.682 – 15.000 = 0.682m

Calculate Moment of Inertia about the axis ‘XX’

I LL = I XX – Ad2 = {LB3 – lb3} – {(L x B) – (l x b)} x d2 3 3 = {(120 x 303) – (20 x 153)} – {(120 x 30) – (20 x 15)} x 15.6822 3 3 = 1057500 – 811552.909 = 245947.091 m4

Determine bilged KB, BM & KM

KB = 7.200 = 3.600m 2 BM = 245947.091 = 10.351m (120 x 30 x 6.60) ∴ KM = 10.351 + 3.63 = 13.981m

Determine angle of heel after bilging

GM = 13.981 – 11.500 = 2.481m ∴ Tan θ = 0.682 2.481 or, θ = 15.4°

© ARNAB SARKAR


b) Draught when heeled = (7.200 x Cos 15.4°) + (0.5 x 30.0 x Sin 15.4°) = 10.925m

∴ Clearance over sand bar = 11.000 – 10.925 = 0.075m

3. A vessel is floating upright in dock water of RD 1.012 and is about to dry dock.

The vessel particulars are: Present draughts: Forward 5.340m Aft 6.660m KG 8.380m

Using the Stability Data Booklet, calculate the vessel’s effective GM at the critical instant.


AMD = 6.000m Trim = 1.320m


LCF = 68.39m foap TMD = 6.660 – (1.320 x 68.39) = 6.003m 137.5

Calculate Displacement, MCTC and LCF

Volume FW = 11997 + (220 x 0.003) = 12004 m3 0.10 ∴ ΔDW = 12004 x 1.012 = 12148 t

MCTC FW = 165.7 + (1.3 x 0.003) = 165.7 tm 0.10 ∴ MCTC DW = 165.7 x 1.012 = 167.7 tm

LCF = 68.39 – (0.09 x 0.003) = 68.39m foap 0.10

Calculate ‘P’ Force and Virtual Δ

P Force = 1.32 x 100 x 167.7 = 324 t 68.39 ∴ Virtual Δ = 12148 – 324 = 11824 t

Calculate KM at C.I.

Virtual Δ = 11824 t ∴ Volume DW = 11824 = 11684 m3 1.012 ∴ Δ = 11684 t

Determine KM from hydrostatic particulars, using Δ = 11684 t

KM = 8.59 – (0.04 x 125) = 8.57m 219

Calculate GM at C.I.

Loss of GM = 324 x 8.57 = 0.23m 12148 GM = 8.57 – 8.30 = 0.27m ∴ GM at C.I. = 0.27 – 0.24 = 0.04m

© ARNAB SARKAR


1. Worksheet Q1 – Trim and Stability provides data relevant to a particular condition of loading in a vessel in salt water.

All holds and tween decks are to be filled with grain (SF 1.65m3/t).



a) The effective metacentric height; b) The drafts forward and aft.

CONDITION: LOADED – GRAIN Length of vessel: 137.50m



Weight (t)

KG (m)

Vertical Moment

(tm)

Free Surface Moment

(tm)


All holds 14708 1.65 8914 5.64 50274.96 69.10 615957.40

1 TD 1695 1.65 1027 11.26 11564.02 115.50 118618.50

2 TD 1676 1.65 1016 10.78 10952.48 95.60 97129.60

3 TD 1626 1.65 986 10.59 10441.74 74.10 73062.60

4 TD 1674 1.65 1015 10.57 10728.55 51.7 52475.50

Oil Fuel 934 1956 2218 31106

FW 83 629 73 2538

Lightship 3831 8.21 31452.51 61.7 236372.70

Δ 17806 7.317 130290.26 68.92 1227260.30

HYDROSTATICS


LCB foap = 69.45 – (0.05 x 167) 239 = 69.42m

LCF foap = 66.10 – (0.08 x 167) 239 = 66.04m

LBP = 137.5m

MCTC = 203.8 + (1.3 x 167) 239 = 204.7 tm

KMT = 8.38 + (0.01 x 167) 239 = 8.39m

GMF = 8.39 – 7.317 = 1.073m

Trim between Perpendiculars: 17806 x (69.42 – 68.92) = 43.49 cm or 0.435m by stern 204.7 COTA = 0.435 x 66.04 = 0.209m 137.5 COTF = 0.435 – 0.209 = 0.226m Draughts: Forward: 8.370 – 0.226 = 8.144m Aft = 8.370 + 0.209 = 8.579m

© ARNAB SARKAR


2. A vessel is floating at draughts 7.800m forward, 8.560m aft in dock water of relative density 1.015.

The vessel is to load cargo so as to finish on an even keel at a load draught of 8.560m in dock water.

Two spaces are available: No.1 hold and No.5 hold.


a) The total weight of cargo to load; b) The weight of cargo to load in each hold.




LCF = 66.25 – (0.08 x 0.08) = 66.186m foap 0.10 ∴ TMD = 8.560 – (0.760 x 66.186) = 8.194m 137.5

With TMD = 8.194, enter Hydrostatic Tables,

Volume FW = 16742 + (233 x 0.094) = 16961 m3 0.10 ∴ Δ DW = 16961 x 1.015 = 17215 t MCTC FW = 196.1 + (1.4 x 0.094) = 197.4 tm 0.10 ∴ MCTC DW = 197.4 x 1.015 = 200.4 tm

LCB = 69.54 – (0.05 x 0.094) = 69.49m foap 0.10 LCF = 66.25 – (0.08 x 0.094) = 66.17m foap 0.10

Calculate LCG

0.760 x 100 = 17215 x (69.49 – LCG) 200.4 ∴ LCG = 68.61m foap

Calculate New Displacement with New TMD

TMD = 8.560m (as vessel will complete loading on an even keel draught) Volume FW = 17677 + (234 x 0.06) = 17817 m3 0.10 ∴ New Δ DW = 17817 x 1.015 = 18084 t

Calculate total weight of cargo to load

∴ Cargo to load = 18084 – 17215 = 869 t

b) Assume ‘w’ tons of cargo is loaded in No.1 Hold ∴ Amount of cargo loaded in No.5 Hold = (869 – w) tons

© ARNAB SARKAR


With TMD = 8.560m, calculate the final LCB

LCB = 69.36 – (0.05 x 0.06) = 69.33m 0.10 As vessel is in even keel, LCB = LCG = 69.33m

Calculate Moments about the LCG

Weights (t) LCG (m) Long. Moments (tm) Initial Δ 17215 68.61 1181121.15 No.1 Hold w 114.48 114.18w No.5 Hold (869 – w) 17.26 (14998.94 – 17.26w) Final Δ 18084 (1196120.09 + 97.22w)

∴ (1196120.09 + 97.22w) = 69.33 18084 or, w = 593 t

∴ Cargo to load in No.1 Hold → 593 t & Cargo to load in No.5 Hold → 276 t

3. A box shaped vessel of length 110.00m, breadth 16.40m, depth 9.10m is floating in salt water to an even keel draught of 5.90m.

a) Calculate the righting moment when the vessel is heeled to the angle of deck edge immersion if the KG is 6.60m.

b) Calculate the angle of loll if the KG is 6.80m.

a) Determine displacement of box shaped vessel

Displacement = 110 x 16.4 x 5.90 x 1.025 = 10910 t

Calculate Angle of Deck Edge Immersion

Freeboard of the box shaped vessel = 9.10 – 5.90 = 3.20m ½ Beam = 8.20m

∴ Tan θ DEI = 3.20 8.20 or, θ DEI = 21.3°

Calculate the KM for the box shaped vessel

KB = ½ Draught = 2.95m

BM = 110 x 16.43 = 3.80m 12 x 110 x 16.4 x 5.90 ∴ KM = KB + BM = 2.95 + 3.80 = 6.75m

Calculate GZ using the wall sided formula

GZ = {(6.75 – 6.60) + 0.5 x Tan2 21.3°} x Sin 21.3° ∴ GZ = 0.16m

Calculate the Righting Moment

Righting Moment = 0.16 x 10910 = 1745.6 tm

b) KG = 6.80m ∴ GM = 6.75 – 6.80 = – 0.05m

Tan θ Loll = √ (2 x 0.05) or, θ Loll = 9.2° 3.80

© ARNAB SARKAR


1. A vessel carrying timber on deck departs from port with a GM of 0.05m. The stability of the vessel deteriorates on passage and as a result the vessel settles to an angle of loll 12° to port. Even keel draught 5.700m in salt water.

Investigate the effect of ballasting No.2 DB tanks, filling the tanks one at a time, in the following order: (1) centre, (2) port, (3) starboard.

Using the Stability Data Booklet calculate EACH of the following:

a) The initial negative GM prior to ballasting; b) The angle of loll on commencing to ballast the centre tank (assume weight negligible); c) The GM when the centre tank is full; d) The angle of heel when the port tank, TCG 5.00m, is full.

a) Using E.K. of 5.700m, from hydrostatic particulars,

Displacement = 11625 t KM = 8.63m KB = 2.95m

∴ BM = 8.63 – 2.95 = 5.68m

Determine KG and GM

Tan 12° = √ (2 x GM) 5.68 ∴ GM = – 0.128m ∴ KG = 8.63 – (– 0.128) = 8.758m

b) Calculate Moments about the Keel to determine the KG prior commencing to ballast

Weights (t) KG (m) Vertical Moments (tm) 11625 8.758 101811.75 + FSMSW (1021 x 1.025) = 1046.525 11625 102858.275

∴ KG = 102858.275 = 8.848m 11625 KM = 8.630m ∴ GM = 8.630 – 8.848 = – 0.218m

∴ Tan θ Loll = √ (2 x 0.218) 5.68 or, θ Loll = 15.5°

c) Calculate Moments about the Keel to determine the KG when the centre tank is full

Weights (t) KG (m) Vertical Moments (tm) 11625 8.758 101811.75 + 277.8 0.59 163.902 11902.8 101975.652

∴ KG = 101975.652 = 8.567m 11902.8

Determine KM for Δ = 11902.8 t

KM = 8.59 – (0.04 x 55) = 8.580m 225 ∴ GM = 8.580 – 8.567 = 0.013m

© ARNAB SARKAR


d) Determine the Angle of Heel when the Port tank is full

Calculate Moments about the Keel to determine the KG

Weights (t) KG (m) Vertical Moments (tm) 11903 8.567 101973.001 + 228.6 0.60 137.16 12131.6 102110.161

∴ KG = 102110.161 = 8.417m 12131.6

Determine KM for Δ = 12131.6 t

KM = 8.55 – (0.03 x 58.6) = 8.542m 224 ∴ GM = 8.542 – 8.417 = 0.125m

Calculate Moments about the CL to determine the KG

Weights (t) Dist. from CL (m) Vertical Moments (tm) 11903 - -

+ 228.6 5.00 1143 12131.6 1143

∴ GGH = 1143 = 0.094m 12131.6

∴ Tan θ List = 0.094 0.125 or, θ List = 36.9°

2. A vessel’s present particulars are as follows:

Forward draught 8.240m, aft draught 9.240m at an upriver berth in fresh water.

The vessel is to proceed down river to cross a sand bar at the river entrance where the relative density of the water is 1.018.


58 t of Heavy Fuel Oil from No.3 DB Centre tank 19 t of Diesel Oil from No.4 DB Port tank 19 t of Diesel Oil from No.4 DB Stbd. tank 17 t of Fresh Water from After Fresh Water tank

Using the Stability Data Booklet, calculate the clearance of the vessel over the bar if the depth of water over the sand bar is 9.520m.




LCF = 65.81 – (0.07 x 0.04) = 65.78m foap 0.10 TMD = 9.240 – (1.000 x 65.78) = 8.762m 137.5

© ARNAB SARKAR


Calculate Initial Δ, MCTC, LCB and LCF, using TMD = 8.762m

Initial ΔFW = 18147 + (236 x 0.062) = 18293 t 0.10 MCTCFW = 202.9 + (2.20 x 0.062) = 204.3 tm 0.10 LCB = 69.27 – (0.05 x 0.062) = 69.24m foap 0.10 LCF = 65.81 – (0.07 x 0.062) = 65.77m foap 0.10


1.000 x 100 = 18293 x (69.24 – LCG) 204.3 ∴ LCG = 68.12m foap


Weights (t) LCG (m) Long. Moments (tm) Initial Δ 18293 68.12 1246119.16 (–) HFO 58 57.02 3307.16 (–) DO 19 35.66 677.54 (–) DO 19 35.50 674.50 (–) FW 17 28.67 487.39 Final Δ 18180 1240972.57

Determine Δ1.018

Final Δ = 18180 t ∴ Δ1.018 = 18180 = 17859 t 1.018

Calculate TMD, LCB, LCF and MCTC from hydrostatic particulars, using Δ = 17859 t

TMD = 8.50 + (0.10 x 182) = 8.578m 234 LCB = 69.36 – (0.05 x 182) = 69.32m foap 234 LCF = 65.95 – (0.08 x 182) = 65.89m foap 234 MCTC = 201.4 + (1.20 x 182) = 202.3 tm 234

Determine the Final Trim

Trim = 17859 x (69.32 – 68.26) = 93.58cms or 0.936m by stern 202.3 as (LCB> LCG)

∴ COTA = 0.936 x 65.89 = 0.449m 137.5 ∴ COTF = 0.936 – 0.449 = 0.487m


Forward Aft TMD 8.578m 8.578m COTF/A (–) 0.487m (+) 0.449m 8.091m 9.027m

© ARNAB SARKAR

SQA Stability – July 2012 Determine clearance above the sand bar

Clearance = 9.520 – 9.027 = 0.493m

3. A vessel is floating in dock water of R.D. 1.017.

Present draughts:

Forward 8.060m, Midship (Port) 8.620m, Midship (Starboard) 8.580m, Aft 9.260m

The draught marks are displaced as follows:

Forward : 1.32m forward of the FP Aft : 2.08m forward of the AP Midship : 0.66m of the amidship line

The Stability Data Booklet provides the necessary hydrostatic data for the vessel.

By completion of the Worksheet Q3 and showing any additional calculations in the answer book, determine the vessel’s displacement.

3. Draught forward = 8.060m






Amidships Line Corr. = 0.66 x 1.230 = 0.006m 137.5

Draught at Midships = 8.600 – 0.006 = 8.594m ∴ Corr. (M) draught = 8.048 + (6 x 8.594) + 9.278 = 8.611m 8


TPCSW = 24.13 + (0.05 x 0.011) = 24.14tm 0.10 LCF = 65.87 – (0.06 x 0.011) = 65.86m foap 0.10 Displacement SW = 18359 + (242 x 0.011) = 18386 t 0.10 Distance of LCF from Midships = 68.75 – 65.86 = 2.89m

∴ 1st Trim Correction = 2.89 x 123.0 x 24.14 = 62.41 t 137.5

MCTC 9.111m = 213.0 + (2.2 x 0.011) = 213.2 tm 0.10

© ARNAB SARKAR


MCTC 8.110m = 201.0 + (1.4 x 0.011) = 201.2 tm 0.10 ∴ 2nd Trim Correction = 50 x (1.230)2 x 12.0 = 6.600 t 137.5

∴ Corrected Δ = 18386 + 62.41 + 6.60 = 18455.01 t ∴ Dock Water Δ = 18455.01 x 1.017 = 18311 t 1.025

© ARNAB SARKAR


1. A vessel, length 137.5m arrives off an upriver berth in fresh water with the following particulars:

Forward draught 3.826m Aft draught 4.248m GM 0.15m

During berthing operations, a starboard side tank is damaged resulting in a loss of 120 t of water ballast, KG 9.12m, LCG 57.87m, TCG 5.10m and the creation of a (corrected) FSM of 298 m4.

Using the Hydrostatic Particulars contained in the Stability Data Booklet, calculate EACH of the following:

a) The resultant draughts, forward and aft; b) The resultant angle and direction of heel.




LCF = 69.81 – (0.07 x 0.037) = 69.78m foap 0.10 TMD = 4.248 – (0.422 x 69.78) = 4.034m 137.5

Calculate Initial Δ, MCTC, LCB and KM, using TMD = 4.034m

Initial Δ = 7713 + (210 x 0.034) = 7784.4 t 0.10 MCTC = 146.0 + (0.90 x 0.034) = 146.3 tm 0.10 LCB = 71.15 – (0.03 x 0.034) = 71.14m foap 0.10 KM = 9.96 – (0.13 x 0.034) = 9.92m 0.10 ∴ KG = 9.92 – 0.15 = 9.77m


0.422 x 100 = 7784 x (71.14 – LCG) 146.3 ∴ LCG = 70.35m foap


Weights (t) LCG (m) Long. Moments (tm) Initial Δ 7784 70.35 547604.4 (–) Ballast 120 57.87 6944.4 Final Δ 7664 540660

∴ LCG = 540660 = 70.55m foap 7664

Calculate TMD, MCTC, LCB, LCF and KM from hydrostatic particulars, using Δ = 7664 t

TMD = 3.90 + (0.10 x 159) = 3.976m 208 MCTC = 145.1 + (0.90 x 159) = 145.8 tm 208

© ARNAB SARKAR


LCB = 71.18 – (0.03 x 159) = 71.16m foap 208 LCF = 69.88 – (0.07 x 159) = 69.83m foap 208 KM = 10.11 – (0.15 x 159) = 10.00m 208

Calculate Trim

Trim = 7664 x (71.16 – 70.55) = 32.06cms or 0.321m by stern 145.8 as (LCB> LCG)

∴ COTA = 0.321 x 69.83 = 0.163m 137.5 ∴ COTF = 0.321 – 0.163 = 0.158m


Forward Aft TMD 3.976m 3.976m COTF/A (–) 0.158m (+) 0.163m 3.818m 4.139m

b) Calculate Moments about the Keel and CL

Weights (t) KG (m) Vertical Mom. (tm) Dist. from CL Trans. Mom.(tm) P S P S

7784 9.77 76049.68 - - - - (–) 120 9.12 (–) 1094.4 - 5.10 - 612 (+) FSM (+) 298.0 7664 75253.28 612

∴ Final KG = 75253.28 = 9.82m 7664 ∴ GGH = 612 = 0.080m 7664

Calculate Final GM

GM = 10.00 – 9.82 = 0.18m

Calculate the resultant heel

Tan θ Heel = 0.080 0.18 ∴ θ Heel = 24° to Port

2. A vessel is to enter drydock in dock water of relative density 1.007.

Drafts: Forward 3.220m Aft 4.920m KG 9.52m

a) Using the Stability Data Booklet, calculate the maximum trim at which the vessel can enter drydock so as to maintain a GM of at least 0.10m at the critical instant.

b) Calculate the weight of ballast to transfer from the After Peak to the Fore Peak in order to bring the vessel to the maximum trim calculated in Q2(a).



© ARNAB SARKAR



LCF = 69.81 – (0.07 x 0.07) = 69.76m foap 0.10 TMD = 4.92 – (1.700 x 69.76) = 4.058m 137.5

Calculate Initial Δ, MCTC, LCB, LCF and KM, from hydrostatic particulars,

Initial Δ FW = 7713 + (210 x 0.058) = 7835 t 0.10 MCTC FW = 146.0 + (0.90 x 0.058) = 146.5 tm 0.10 LCF = 69.81 – (0.07 x 0.058) = 69.77m foap 0.10 LCB = 71.15 – (0.03 x 0.058) = 71.13m foap 0.10 KM = 9.96 – (0.03 x 0.058) = 9.88m 0.10

Calculate the Loss of GM

KM = 9.88m KGF = 9.52m GMF = 0.36m

Required GM = 0.10m ∴ Loss of GM = 0.26m

Calculate the ‘P’ Force

0.26 = P x 9.88 7835 x 1.007 ∴ ‘P’ = 207.6 t

Calculate the maximum trim at which the vessel can enter the drydock

207.6 = Trim x 146.5 x 1.007 69.77 ∴ Max. Trim = 98.2cms

b) Assume that ‘w’ tons of water ballast needs to be transferred from the AP tank to the FP tank.

Determine the Initial LCG

1.700 x 100 = 7835 x 1.007 (71.13 – LCG) 146.5 x 1.007 or, LCG = 67.95m foap

Calculate Moments about the LCG

Weights (t) LCG (m) Long. Moments (tm) Initial Δ (7835 x 1.007) 67.95 536125.5 (+) Ballast + w 130.56 130.56 w (–) Ballast – w 3.07 – 3.07 w Final Δ 7890 (536125.5 + 127.49 w)

∴ LCG = (536125.5 + 127.49 w) 7890

© ARNAB SARKAR

SQA Stability – December 2012 ∴ (71.13 – 536125.5 +127.49 w)

98.2 = 7890 x 7890 146.5 x 1.007

or, w = 83.2 t

∴ Amount of water ballast to be transferred from AP to FP tank = 83.2 t

3. A box shaped vessel floating upright on an even keel in salt water has the following particulars:


The vessel has two longitudinal bulkheads each 6.00m from the side of the vessel.

Calculate the angle of heel if a midship side compartment 27.00m is bilged.


Volume of Buoyancy lost = Volume of Buoyancy gained or, (27 x 6 x 8.800) = {(162 x 30) – (27 x 6)} x Sinkage or, Sinkage = 1425.6 = 0.303m

4698 ∴ Bilged draught = 8.800 + 0.303 = 9.103m




4698 72414





BM = 339851.622 = 7.946m (162 x 30 x 8.800) KB = 9.103 = 4.552m 2 ∴ KM = 7.946 + 4.552 = 12.498m ∴ GM = 12.498 – 10.850 = 1.648m

Calculate List

Tan θ = 0.414 1.648 ∴ List = 14.1°

© ARNAB SARKAR


1. A box shaped vessel floating on even keel in dock water of R.D. 1.011 has the following particulars:

Length 120.00m Breadth 22.00m Draught 7.800m MCTC (salt water) 290

There is an empty watertight forward end compartment, length 12.00m, height 6.50m, extending the full width of the vessel.

Calculate the draughts forward and aft, if this compartment is bilged.

1. Calculate the Bilged TMD

Volume of the vessel before bilging = 120 x 22 x 7.800 = 20592 m3 ∴ Displacement of the vessel before bilging = 20592 x 1.011 = 20818.5 t


Volume of the forward end compartment = 12 x 22 x 6.500 = 1716 m3 Intact water plane area = 120 x 22 = 2640 m2

∴ Sinkage caused to bilging = 1716 = 0.650m 2640 ∴ Bilged TMD = 7.800 + 0.650 = 8.450m

Calculate the Trimming Moment

Calculating moments of volume about the AP after bilging,

Volume (m3) Distance from AP (m) Moments (m4) 120 x 22 x 8.450 60.0 1338480.0 (–) 1716 114.0 (–) 195624 20592 1142856.0

∴ BH = 55.500m ∴ BBH = 60 – 55.500 = 4.500m ∴ Trimming Moment = 20818.5 x 4.500 = 93683.25 tm by forward

Calculate COT after bilging

COT = 93683.25 x 1.025 = 327.5cm or 3.275m 290 x 1.011 ∴ COTA = 3.275 x 60 = 1.638m 120 ∴ COTF = 3.275 – 1.638 = 1.637m

Calculate the Final Draughts of the vessel

Forward Draught = 8.450 + 1.637 = 10.087m Aft Draught = 8.450 – 1.637 = 6.813m

2. A vessel completes under deck loading in salt water with the following particulars:

Displacement 14115 tonne KG 8.00m


During the passage 190 tonne of Heavy Fuel Oil (R.D. 0.81) is consumed from No.3 D.B. tank centre which was full on departure (use KG of tank for consumption purposes).

© ARNAB SARKAR


Calculate the maximum weight of timber to load on deck KG 13.20m assuming 15% water absorption during the passage in order to arrive at the destination with the minimum GM (0.05m) allowed under the Load Line Regulations.

Note: Assume KM constant.

2. Determine the KM of the vessel from hydrostatic particulars, using Δ = 14115 t

KM = 8.36m

Assume that ‘w’ tonnes of timber can be loaded to meet the required condition. Considering water absorption by the timber deck cargo, extra weight added = 0.15 w tons

Calculate Moments about the Keel to determine the Final KG

Weights (t) KG (m) Vertical Moments (tm) Initial Δ 14115 8.00 112920 Cargo w 13.2 13.2 w Water Absorption 0.15 w 13.2 1.98 w (–) HFO (–) 190 0.60 (–) 114 FSM HFO 1142 x 0.81 (+) 925.02 Final Δ (13925 +1.15 w) (113731.02 + 15.18 w)

∴ Final KG = (113731.02 + 15.18 w) m (13925 +1.15 w)

∴ 8.36 – 0.05 = (113731.02 + 15.18 w) (13925 +1.15 w)

or, 115716.75 +9.56w = 113731.02 + 15.18w

or, w = 353.3 t

∴ Weight of timber to load = 353.3 t

3. A vessel is to load a cargo of grain (stowage factor 1.65 m3/t). Initial displacement 6020 tonne Initial KG 6.68m

All five holds are to be loaded full of grain.


No.1 TD Full No.2 TD Part Full – Ullage 2.50m No.3 TD Part Full – Ullage 1.25m No.4 TD Full


a) Using the Maximum Permissible Grain Heeling Moment Table included in the Stability Data Booklet, determine whether the vessel complies with the minimum criteria specified in the International Grain Code (IMO).

b) Calculate the ship’s approximate angle of heel in the event of the grain shifting as assumed by the International Grain Code (IMO).

a) Initial Δ = 6020 t Initial KG = 6.68m

SF = 1.65 m3/t

© ARNAB SARKAR



Comp. Ullage (m)

Volume (m3)

SF (m3/t) Tonnes KG

(m) Vert. Mom.

(tm) VHM (tm)

Initial Δ 6020 6.68 40214

No.1 TD Full 1695 1.65 1027 11.26 11564 352

No.2 TD 2.50 1062 1.65 644 10.44 6723 3160

No.3 TD 1.25 1476 1.65 895 10.47 9371 1547

No.4 TD Full 1674 1.65 1015 10.57 10729 604

No.1 Hold Full 2215 1.65 1342 5.09 6831 410

No.2 Hold Full 4672 1.65 2832 4.95 14018 1285

No.3 Hold Full 1742 1.65 1056 4.94 5217 475

No.4 Hold Full 3474 1.65 2106 4.95 10425 910

No.5 Hold Full 2605 1.65 1579 8.76 13832 455

Final Δ 18516 6.96 128924 9198

Calculate AHM

AHM = 9198 = 5575tm 1.65


Δ = 18516 t KGF = 6.96m

Interpolation

Δ KGF = 6.90 KGF = 6.96 KGF = 7.00

19000 7157 6905.0 6737

18516 6660.7

18500 6898 6652.6 6489


7157 – (420 x 0.06) = 6905.0 tm 0.10


6898 – (409 x 0.06) = 6652.6 tm 0.10


6652.6 + (252.4 x 16) = 6660.7 tm 500 ∴ PHM for Δ = 18516 t & KGF = 6.96m = 6660.7 tm Vessel complies with the minimum criteria if she is upright during sailing.


Angle of Heel = 5575 x 12° = 10.0° 6660.7

© ARNAB SARKAR

sqa chief mates (unlimited) solved stability numerical (2005 - 2013)

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