obs.2011.eocrevision 1

Revision Solution No. 1 (Heel due to Turning) A ship with a transverse metacentric height of 0.60m has a speed of 18 kts. Ship’s KG is 5.8m whilst the centre of lateral resistance (KB) is 3.7m. The rudder is put hard over to starboard and the vessel turns in a circle of radius 520m. Considering only the centrifugal forces involved, calculate the angle of heel due to turning. Take g as 9.81 m/s²

(Ans: 3.37° to port) .

Solution No.1

Ship speed in m/s = 18 x 1852/3600 = 9.26 m/s Tan θ = v2 x BG grGM = 9.262 x 2.1 9.81 x 520 x 0.6 = 0.0588 Heel = 3.37°

Ord SM A.func Ord M.func Ord I.func

0.00 1.00 0.00 0.00 0.00 0.00 0.009.70 4.00 38.80 9.70 376.36 9.70 3650.69

13.00 2.00 26.00 13.00 338.00 13.00 4394.0015.00 4.00 60.00 15.00 900.00 15.00 13500.0013.00 2.00 26.00 13.00 338.00 13.00 4394.009.70 4.00 38.80 9.70 376.36 9.70 3650.690.00 1.00 0.00 0.00 0.00 0.00 0.00

Total A.func 189.60 Total M.func 2328.72 Total I.func 29589.38

h = 3.00

Area = (h / 3) x Total A.func = (3.00 / 3) x 189.60= 189.6 m2

1st Moment about Tanktop (OZ) & Waterline: = h / 3 (Total M. Func / 2) = (3.00 / 3) x (2328.72 / 2) = 1164.36 m3

I about Tanktop (OZ) of Immersed Area below Tanktop (OZ) & Waterline: = h / 3 (Total I. Func / 3) = (3.00 / 3) x (29589.38 / 3) = 9863.13 m4

Revision Solution No.2 (Pressure & Thrust)

A bulkhead of a tank has depth of 15 metres. The vertical ordinates at equidistant intervals of 3.0 metres from port to starboard are as follows.

0 9.7 13.0 15.0 13.0 9.7 0

The tank is filled with salt water to a head of 2.2m above the top of the bulkhead. Calculate: 1. The position of the centre of area 2. The load on the bulkhead 3. The position of centre of pressure

= 9863.13 m

Depth of CG below Tanktop (OZ) & Waterline: = 1st moment about tanktop / Area (Total M func / Total A func) x ½= 1164.36 / 189.60 (2328.72 / 189.6) x ½

Ans 2.1 = 6.14 m 6.14

Depth of Centre of Pressure below Tanktop (OZ) & Waterline: (Total I func / Total M func) x ⅔(29589.38 / 2328.72) x ⅔

= 9863.13 / 1164.36 8.47= 8.47 m

= 8.34 m= 189.6 m2

= 1.025 t / m2

= Depth x Density x Area = 8.34 x 1.025 x 189.60

Ans 2.2 = 1620.80 tonnes

Theorem of Parallel Axis: ICG = IOZ - Area x (Depth of CG below OZ)² = 9863.13 - (189.60 x 6.14²)

2712.63

IWL = ICG + Area x (Depth of CG below WL)² = 2712.63 + (189.60 x 8.34²)

15900.37

Cente of Pressure Y = IWL / (Area x Depth of CG below WL) = 15900.37 / (189.60 x 8.34)

10.06 m below waterline2.20 m head of water

Ans 2.3 7.86 m below top of tank

Area Density

Load on Bulkhead

y =I about Tanktop below Tanktop & Waterline

1st Moment About Tanktop & Waterline

Depth of Centroid (CG) below Waterline

A bulkhead of a tank has depth of 15 metres. The vertical ordinates at equidistant intervals of 3.0 metres from port to starboard are as follows.

0 9.7 13.0 15.0 13.0 9.7 0

The tank is filled with salt water to a head of 2.2m above the top of the bulkhead. Calculate: 1. The position of the centre of area 2. The load on the bulkhead 3. The position of centre of pressure

Ord SM A.func Lever M.func Lever I.func

18.00 1.00 18.00 0.00 0.00 0.00 0.0017.60 4.00 70.40 1.00 70.40 1.00 70.4016.50 2.00 33.00 2.00 66.00 2.00 132.0015.00 4.00 60.00 3.00 180.00 3.00 540.0011.40 2.00 22.80 4.00 91.20 4.00 364.807.10 4.00 28.40 5.00 142.00 5.00 710.000.00 1.00 0.00 6.00 0.00 6.00 0.00

Total A.func 232.60 Total M.func 549.60 Total I.func 1817.20

h = 15.00 / 6 = 2.50

Area = (h / 3) x Total A.func = (2.50 / 3) x 232.60= 193.83 m2

1st Moment about Tanktop (OZ) & Waterline: = h2 / 3 (Total M. Func) = (2.502 / 3) x 549.60 = 1145.00 m3

I about Tanktop (OZ) of Immersed Area below Tanktop (OZ) & Waterline: = h3 / 3 (Total I. Func) = (2.503 / 3) x 1817.20 = 9464.58 m4

Revision Solution No.3 (Pressure & Thrust)

A bulkhead of a tank has depth of 15 metres. The tranverse width of the bulkhead, commencing at the upper edge are as follows. 18.0 17.6 16.5 15.0 11.4 7.1 0

The tank is filled with salt water to a head of 2.2m above the top of the bulkhead. Calculate: 3.1 The position of the centre of area below top of tank 3.2 The load on the bulkhead 3.3 The position of centre of pressure below top of tank

= 9464.58 mAlternative method

Depth of CG below Tanktop (OZ) & Waterline: = (1st moment about tanktop / Area) (Total M func / Total A func) x h= 1145.00 / 193.83 (549.60/232.60) x 2.5

Ans 3.1 = 5.91 m 5.91

Depth of Centre of Pressure below Tanktop (OZ) & Waterline: (Total I func / Total M func) x h(1817.20/549.60) x h

= 9464.58 / 1145.00 8.27= 8.27 m

= 8.11 m= 193.83 m2

= 1.025 t / m2

= Depth x Density x Area = 8.11 x 1.025 x 189.60

Ans 3.2 = 1610.72 tonnes

Theorem of Parallel Axis: ICG = IOZ - Area x (Depth of CG below OZ)² = 9464.58 - (193.83 x 5.91²)

2700.91

IWL = ICG + Area x (Depth of CG below WL)² 15443.17533 = 2700.91 + (193.83 x 8.11²)

15449.73

Cente of Pressure Y = IWL / (Area x Depth of CG below WL) = 15449.73 / (193.83 x 8.11)

9.83 m below waterline2.20 m head of water

Ans 3.3 7.63 m below top of tank

Area Density

Load on Bulkhead

=I about Tanktop below Tanktop & Waterline

1st Moment About Tanktop & Waterline

Depth of Centroid (CG) below Waterline

A bulkhead of a tank has depth of 15 metres. The tranverse width of the bulkhead, commencing at the upper edge are as follows. 18.0 17.6 16.5 15.0 11.4 7.1 0

The tank is filled with salt water to a head of 2.2m above the top of the bulkhead. Calculate: 3.1 The position of the centre of area below top of tank 3.2 The load on the bulkhead 3.3 The position of centre of pressure below top of tank

Solutions to Revision Question No.4 (Grain Loading)

A vessel has loaded grain, SF 1.55m³/t, to a displacement of 13,500 tonne. In loaded condition the effective KG is 7.12 m. All grain spaces are full, except No.3 tween deck, which is partially full. The tabulated transverse volumetric heeling moment are as follows. Cargo Spaces VHM (m4) No.1 hold 810 No.2 hold 1042 No.3 hold 1075 No.4 hold 1185 No.1 TD 723 No.2 TD 675 No.3 TD 403 The value of KG used in the calculation of the vessel’s effective KG were as follows. For lower holds, the centroid of the space. For tween decks, the actual Kg of the cargo 1.1 Using the Datasheet provided, determine the vessel ability to comply with the

statutory grain regulations. 1.2 Calculate the vessel’s approximate angle of heel in the event of a shift of grain

assumed in the grain regulations.

A vessel has loaded grain, SF 1.55m³/t, to a displacement of 13,500 tonne. In loaded condition the effective KG is 7.12 m. All grain spaces are full, except No.3 tween deck, which is partially full. The tabulated transverse volumetric heeling moment are as follows. Cargo Spaces VHM (m4) No.1 hold 810 No.2 hold 1042 No.3 hold 1075 No.4 hold 1185 No.1 TD 723 No.2 TD 675 No.3 TD 403 The value of KG used in the calculation of the vessel’s effective KG were as follows. For lower holds, the centroid of the space. For tween decks, the actual Kg of the cargo 1.1 Using the Datasheet provided, determine the vessel ability to comply with the

statutory grain regulations. 1.2 Calculate the vessel’s approximate angle of heel in the event of a shift of grain

assumed in the grain regulations.

A vessel has loaded grain, SF 1.55m³/t, to a displacement of 13,500 tonne. In loaded condition the effective KG is 7.12 m. All grain spaces are full, except No.3 tween deck, which is partially full. The tabulated transverse volumetric heeling moment are as follows. Cargo Spaces VHM (m4) No.1 hold 810 No.2 hold 1042 No.3 hold 1075 No.4 hold 1185 No.1 TD 723 No.2 TD 675 No.3 TD 403 The value of KG used in the calculation of the vessel’s effective KG were as follows. For lower holds, the centroid of the space. For tween decks, the actual Kg of the cargo 1.1 Using the Datasheet provided, determine the vessel ability to comply with the

statutory grain regulations. 1.2 Calculate the vessel’s approximate angle of heel in the event of a shift of grain

assumed in the grain regulations.

Spaces Wt S SF Volume VHM Factor Corr VHM GHMNo.1 Hold F 1.55 810.00 1.00 810.00 522.58No.2 Hold F 1.55 1042.00 1.00 1042.00 672.26No.3 Hold F 1.55 1075.00 1.00 1075.00 693.55No 4 Hold F 1 55 1185 00 1 00 1185 00 764 52

Cargo

A vessel has loaded grain, SF 1.55m³/t, to a displacement of 13,500 tonne. In loaded condition the effective KG is 7.12 m. All grain spaces are full, except No.3 tween deck, which is partially full. The tabulated transverse volumetric heeling moment are as follows. Cargo Spaces VHM (m4) No.1 hold 810 No.2 hold 1042 No.3 hold 1075 No.4 hold 1185 No.1 TD 723 No.2 TD 675 No.3 TD 403 The value of KG used in the calculation of the vessel’s effective KG were as follows. For lower holds, the centroid of the space. For tween decks, the actual Kg of the cargo 1.1 Using the Datasheet provided, determine the vessel ability to comply with the

statutory grain regulations. 1.2 Calculate the vessel’s approximate angle of heel in the event of a shift of grain

assumed in the grain regulations.

No.4 Hold F 1.55 1185.00 1.00 1185.00 764.52No.1 TD F 1.55 723.00 1.06 766.38 494.44No.2 TD F 1.55 675.00 1.06 715.50 461.61No.3 TD P 1.55 403.00 1.12 451.36 291.20Displ 13500 6045.24 3900.15

From Datasheet: Using Δ = 13500 t and KG 7.12 mMaximum permissable grain heeling moment = 4072.40 t-mActual grain heeling moment = 3900.15

A vessel has loaded grain, SF 1.55m³/t, to a displacement of 13,500 tonne. In loaded condition the effective KG is 7.12 m. All grain spaces are full, except No.3 tween deck, which is partially full. The tabulated transverse volumetric heeling moment are as follows. Cargo Spaces VHM (m4) No.1 hold 810 No.2 hold 1042 No.3 hold 1075 No.4 hold 1185 No.1 TD 723 No.2 TD 675 No.3 TD 403 The value of KG used in the calculation of the vessel’s effective KG were as follows. For lower holds, the centroid of the space. For tween decks, the actual Kg of the cargo 1.1 Using the Datasheet provided, determine the vessel ability to comply with the

statutory grain regulations. 1.2 Calculate the vessel’s approximate angle of heel in the event of a shift of grain

assumed in the grain regulations.

Actual grain heeling moment 3900.154.1 Since the actual grain heeling moment is less than the maximum permissable grain heeling

moment, vessel comply with the statutory grain regulation.4.2 The approximate angle of heel = GHM x 12° = 3900.15 x 12° = 11.49°

Max Permissable 4072.40

A vessel has loaded grain, SF 1.55m³/t, to a displacement of 13,500 tonne. In loaded condition the effective KG is 7.12 m. All grain spaces are full, except No.3 tween deck, which is partially full. The tabulated transverse volumetric heeling moment are as follows. Cargo Spaces VHM (m4) No.1 hold 810 No.2 hold 1042 No.3 hold 1075 No.4 hold 1185 No.1 TD 723 No.2 TD 675 No.3 TD 403 The value of KG used in the calculation of the vessel’s effective KG were as follows. For lower holds, the centroid of the space. For tween decks, the actual Kg of the cargo 1.1 Using the Datasheet provided, determine the vessel ability to comply with the

statutory grain regulations. 1.2 Calculate the vessel’s approximate angle of heel in the event of a shift of grain

assumed in the grain regulations.

Solutions to Revision Question No.5 (Grain Loading)

MV Del Monte, a 7-holds bulk carrier, is scheduled to load a consignment of wheat in bulk at Alumar, Brazil.

Cargo information and loading distribution are as follows.

Hold Weight (tonnes) SF Stowage 1 5450 1.67 Filled 2 5930 1.67 Filled 3 6300 1.67 Filled 4 6600 1.67 Filled 5 5240 1.67 Partly Filled 6 6600 1.67 Filled 7 6300 1.67 Filled

Upon completion of loading, the following ship’s figures are obtained.

Other weights/deductibles onboard = 4800 tonnes FSM = 900 t-m

Lightship = 10533 tonnes Solid KG = 9.50 m

Note: For vertical shift of G due to grain shift, use a factor of 1.06 for filled

compartments and 1.12 for partly-filled compartments.

Required:

a) Using the KN and HM curves provided, draw the curve of statical stability for the departure condition. The curve should include the heeling moment curve for grain shift.

b) Determine the angle of heel due to grain shift using the following method. i. By estimation, using the curve of statical stability. ii. By calculation, given the maximum allowable heeling moment at

departure condition is 39800 t-m. c) Determine if vessel comply with the statutory grain regulations

MV Del Monte, a 7-holds bulk carrier, is scheduled to load a consignment of wheat in bulk at Alumar, Brazil.

Cargo information and loading distribution are as follows.

Hold Weight (tonnes) SF Stowage 1 5450 1.67 Filled 2 5930 1.67 Filled 3 6300 1.67 Filled 4 6600 1.67 Filled 5 5240 1.67 Partly Filled 6 6600 1.67 Filled 7 6300 1.67 Filled

Upon completion of loading, the following ship’s figures are obtained.

Other weights/deductibles onboard = 4800 tonnes FSM = 900 t-m

Lightship = 10533 tonnes Solid KG = 9.50 m

Note: For vertical shift of G due to grain shift, use a factor of 1.06 for filled

compartments and 1.12 for partly-filled compartments.

Required:

a) Using the KN and HM curves provided, draw the curve of statical stability for the departure condition. The curve should include the heeling moment curve for grain shift.

b) Determine the angle of heel due to grain shift using the following method. i. By estimation, using the curve of statical stability. ii. By calculation, given the maximum allowable heeling moment at

departure condition is 39800 t-m. c) Determine if vessel comply with the statutory grain regulations

MV Del Monte, a 7-holds bulk carrier, is scheduled to load a consignment of wheat in bulk at Alumar, Brazil.

Cargo information and loading distribution are as follows.

Hold Weight (tonnes) SF Stowage 1 5450 1.67 Filled 2 5930 1.67 Filled 3 6300 1.67 Filled 4 6600 1.67 Filled 5 5240 1.67 Partly Filled 6 6600 1.67 Filled 7 6300 1.67 Filled

Upon completion of loading, the following ship’s figures are obtained.

Other weights/deductibles onboard = 4800 tonnes FSM = 900 t-m

Lightship = 10533 tonnes Solid KG = 9.50 m

Note: For vertical shift of G due to grain shift, use a factor of 1.06 for filled

compartments and 1.12 for partly-filled compartments.

Required:

a) Using the KN and HM curves provided, draw the curve of statical stability for the departure condition. The curve should include the heeling moment curve for grain shift.

b) Determine the angle of heel due to grain shift using the following method. i. By estimation, using the curve of statical stability. ii. By calculation, given the maximum allowable heeling moment at

departure condition is 39800 t-m. c) Determine if vessel comply with the statutory grain regulations

MV Del Monte, a 7-holds bulk carrier, is scheduled to load a consignment of wheat in bulk at Alumar, Brazil.

Cargo information and loading distribution are as follows.

Hold Weight (tonnes) SF Stowage 1 5450 1.67 Filled 2 5930 1.67 Filled 3 6300 1.67 Filled 4 6600 1.67 Filled 5 5240 1.67 Partly Filled 6 6600 1.67 Filled 7 6300 1.67 Filled

Upon completion of loading, the following ship’s figures are obtained.

Other weights/deductibles onboard = 4800 tonnes FSM = 900 t-m

Lightship = 10533 tonnes Solid KG = 9.50 m

Note: For vertical shift of G due to grain shift, use a factor of 1.06 for filled

compartments and 1.12 for partly-filled compartments.

Required:

a) Using the KN and HM curves provided, draw the curve of statical stability for the departure condition. The curve should include the heeling moment curve for grain shift.

b) Determine the angle of heel due to grain shift using the following method. i. By estimation, using the curve of statical stability. ii. By calculation, given the maximum allowable heeling moment at

departure condition is 39800 t-m. c) Determine if vessel comply with the statutory grain regulations

Cargo

MV Del Monte, a 7-holds bulk carrier, is scheduled to load a consignment of wheat in bulk at Alumar, Brazil.

Cargo information and loading distribution are as follows.

Hold Weight (tonnes) SF Stowage 1 5450 1.67 Filled 2 5930 1.67 Filled 3 6300 1.67 Filled 4 6600 1.67 Filled 5 5240 1.67 Partly Filled 6 6600 1.67 Filled 7 6300 1.67 Filled

Upon completion of loading, the following ship’s figures are obtained.

Other weights/deductibles onboard = 4800 tonnes FSM = 900 t-m

Lightship = 10533 tonnes Solid KG = 9.50 m

Note: For vertical shift of G due to grain shift, use a factor of 1.06 for filled

compartments and 1.12 for partly-filled compartments.

Required:

a) Using the KN and HM curves provided, draw the curve of statical stability for the departure condition. The curve should include the heeling moment curve for grain shift.

b) Determine the angle of heel due to grain shift using the following method. i. By estimation, using the curve of statical stability. ii. By calculation, given the maximum allowable heeling moment at

departure condition is 39800 t-m. c) Determine if vessel comply with the statutory grain regulations

Hold Wt S SF Volume VHM Factor Corr VHM GHM1 5450.00 F 1.67 9101.50 1800.00 1.06 1908.00 1142.512 5930.00 F 1.67 9903.10 3300.00 1.06 3498.00 2094.613 6300.00 F 1.67 10521.00 4000.00 1.06 4240.00 2538.924 6600.00 F 1.67 11022.00 3000.00 1.06 3180.00 1904.195 5240.00 P 1.67 8750.80 40330.00 1.12 45169.60 27047.666 6600.00 F 1.67 11022.00 3000.00 1.06 3180.00 1904.197 6300.00 F 1.67 10521.00 4000.00 1.06 4240.00 2538.92

Cargo

MV Del Monte, a 7-holds bulk carrier, is scheduled to load a consignment of wheat in bulk at Alumar, Brazil.

Cargo information and loading distribution are as follows.

Hold Weight (tonnes) SF Stowage 1 5450 1.67 Filled 2 5930 1.67 Filled 3 6300 1.67 Filled 4 6600 1.67 Filled 5 5240 1.67 Partly Filled 6 6600 1.67 Filled 7 6300 1.67 Filled

Upon completion of loading, the following ship’s figures are obtained.

Other weights/deductibles onboard = 4800 tonnes FSM = 900 t-m

Lightship = 10533 tonnes Solid KG = 9.50 m

Note: For vertical shift of G due to grain shift, use a factor of 1.06 for filled

compartments and 1.12 for partly-filled compartments.

Required:

a) Using the KN and HM curves provided, draw the curve of statical stability for the departure condition. The curve should include the heeling moment curve for grain shift.

b) Determine the angle of heel due to grain shift using the following method. i. By estimation, using the curve of statical stability. ii. By calculation, given the maximum allowable heeling moment at

departure condition is 39800 t-m. c) Determine if vessel comply with the statutory grain regulations

Total 42420.00 65415.60 39171.02Lightship 10533.00 Solid KG = 9.50Deductibles 4800.00 FSC = 0.02Displacement 57753.00 Fluid KG = 9.52

MV Del Monte, a 7-holds bulk carrier, is scheduled to load a consignment of wheat in bulk at Alumar, Brazil.

Cargo information and loading distribution are as follows.

Hold Weight (tonnes) SF Stowage 1 5450 1.67 Filled 2 5930 1.67 Filled 3 6300 1.67 Filled 4 6600 1.67 Filled 5 5240 1.67 Partly Filled 6 6600 1.67 Filled 7 6300 1.67 Filled

Upon completion of loading, the following ship’s figures are obtained.

Other weights/deductibles onboard = 4800 tonnes FSM = 900 t-m

Lightship = 10533 tonnes Solid KG = 9.50 m

Note: For vertical shift of G due to grain shift, use a factor of 1.06 for filled

compartments and 1.12 for partly-filled compartments.

Required:

a) Using the KN and HM curves provided, draw the curve of statical stability for the departure condition. The curve should include the heeling moment curve for grain shift.

b) Determine the angle of heel due to grain shift using the following method. i. By estimation, using the curve of statical stability. ii. By calculation, given the maximum allowable heeling moment at

departure condition is 39800 t-m. c) Determine if vessel comply with the statutory grain regulations

λ0 = VHM = 65415.60 = 0.678251SF x W 1.67 x 57753.00

λ40 = 0.8 x λ0 = 0.542601

Heel (θ) KN KG.Sinθ GZ Ans: 5.20 0 0 0 From curve, angle of heel = 11.9°5 1.17 0.90 0.27 The approximate angle of heel = GHM x 12°

10 2.30 1.79 0.51 Max Permissable12 2.76 2.14 0.62 = 39171.02 x 12°15 3.53 2.66 0.87 39800.0020 4.78 3.51 1.27 = 11.81°25 6.03 4.33 1.7030 7.33 5.11 2.22 Ans: 5.335 8.45 5.84 2.61 The angle of heel is less than 12°.40 9.28 6.52 2.76 ∴ Vessel comply with the grain regulation.50 10.38 7.71 2.6760 10.92 8.62 2.3075 10.75 9.40 1.3590 9.78 9.44 0.34

Ans: 5.1

11.9°11.9°

Revision Solution No.6 (Unresisted Rolling)

A vessel has the following particulars. Displacement = 9000 tonne Natural rolling period = 15 sec GM = 1.20 m The following cargo operation has taken place. 2000 tonne loaded at 4.0m above ship’s KG 500 tonne discharged at 3.0m below ship’s KG Assume that KM remains constant, determine the new natural rolling period and discuss if the final condition results in a “stiff ship” or “tender ship”

A vessel has the following particulars. Displacement = 9000 tonne Natural rolling period = 15 sec GM = 1.20 m The following cargo operation has taken place. 2000 tonne loaded at 4.0m above ship’s KG 500 tonne discharged at 3.0m below ship’s KG Assume that KM remains constant, determine the new natural rolling period and discuss if the final condition results in a “stiff ship” or “tender ship”

Displacement = 9000 tGM = 1.20 m

GG1 = (w x d) / W G1G2 = (w x d) / W= (2000 x 4) / 11000 = (500 x 3.7273) / 10500= 0 7273 m = 0 1775 m

A vessel has the following particulars. Displacement = 9000 tonne Natural rolling period = 15 sec GM = 1.20 m The following cargo operation has taken place. 2000 tonne loaded at 4.0m above ship’s KG 500 tonne discharged at 3.0m below ship’s KG Assume that KM remains constant, determine the new natural rolling period and discuss if the final condition results in a “stiff ship” or “tender ship”

= 0.7273 m = 0.1775 m

GG2 = 0.7273 + 0.17750.9048 m

GM = 1.2000 mG2M = 0.2952 m

A vessel has the following particulars. Displacement = 9000 tonne Natural rolling period = 15 sec GM = 1.20 m The following cargo operation has taken place. 2000 tonne loaded at 4.0m above ship’s KG 500 tonne discharged at 3.0m below ship’s KG Assume that KM remains constant, determine the new natural rolling period and discuss if the final condition results in a “stiff ship” or “tender ship”

T = 2πK / √gGM= 2 x 3.1429 x K / (√ 9.81 x 1.2)= 6.2858K / 3.431

K = (3.431 / 6.2858) x 15= 8.1875 m

Mass moment of inertia (I) = M K²

A vessel has the following particulars. Displacement = 9000 tonne Natural rolling period = 15 sec GM = 1.20 m The following cargo operation has taken place. 2000 tonne loaded at 4.0m above ship’s KG 500 tonne discharged at 3.0m below ship’s KG Assume that KM remains constant, determine the new natural rolling period and discuss if the final condition results in a “stiff ship” or “tender ship”

Mass moment of inertia (I) = M.K²Original I = 9000 x 8.1875²

= 603316.4063

Loaded Weight (I) = 2000 x 4² = 32000 t-m²

Di h (I) 500 3 7273²

A vessel has the following particulars. Displacement = 9000 tonne Natural rolling period = 15 sec GM = 1.20 m The following cargo operation has taken place. 2000 tonne loaded at 4.0m above ship’s KG 500 tonne discharged at 3.0m below ship’s KG Assume that KM remains constant, determine the new natural rolling period and discuss if the final condition results in a “stiff ship” or “tender ship”

Discharge mass (I) = 500 x 3.7273²= 6946.3826 t-m²

Net (I) = 25053.6174 t-m²

New (I) of ship = 603316.4063 + 25053.6174= 628370.0237 t-m²

A vessel has the following particulars. Displacement = 9000 tonne Natural rolling period = 15 sec GM = 1.20 m The following cargo operation has taken place. 2000 tonne loaded at 4.0m above ship’s KG 500 tonne discharged at 3.0m below ship’s KG Assume that KM remains constant, determine the new natural rolling period and discuss if the final condition results in a “stiff ship” or “tender ship”

A vessel has the following particulars. Displacement = 9000 tonne Natural rolling period = 15 sec GM = 1.20 m The following cargo operation has taken place. 2000 tonne loaded at 4.0m above ship’s KG 500 tonne discharged at 3.0m below ship’s KG Assume that KM remains constant, determine the new natural rolling period and discuss if the final condition results in a “stiff ship” or “tender ship”

New I2 about new CG = New I about original CG - (W x CG²)= 628370.0237 - (10500 x 0.9048²) = 619774.0617 t-m²

I2 = M.K²K² = 619774.0617 / 10500

= 59.02610K = 7.6828 m

New T = 2πK / √gGM= 2 x 3.1429 x 7.6828 / (√ 9.81 x 0.2952)= 28.38 secs

Notes: Stiff ship - roll period - as low as 8 secTender ship - roll period - 25 to 35 secComfortable roll period - between 15 and 25 sec averaging out to 20 sec

Revision Solution No.7 (Inclining Experiment)

A ship of 18500 tonne displacement is inclined by moving a 10 tonne weight through a distance of 25 m transversely. The deflections of two pendulums were as follows.

Forward pendulum Length 18 m Deflection 16.0 cm Aft pendulum Length 18 m Deflection 15.8 cm

To bring the ship to light displacement, the following weights are moved. Weights on: Four sets of cargo winches each 8 tonne Kg 20.3m Weights off: Inclining test weight 10 tonne Kg 18.8m Six sets of shore generator each 4 tonne Kg 19.0m Required Calculate the GM and angle of heel obtained from the inclining test. Calculate the ship’s final light displacement and VCG, given her KM was 9.2m

A ship of 18500 tonne displacement is inclined by moving a 10 tonne weight through a distance of 25 m transversely. The deflections of two pendulums were as follows.

Forward pendulum Length 18 m Deflection 16.0 cm Aft pendulum Length 18 m Deflection 15.8 cm

To bring the ship to light displacement, the following weights are moved. Weights on: Four sets of cargo winches each 8 tonne Kg 20.3m Weights off: Inclining test weight 10 tonne Kg 18.8m Six sets of shore generator each 4 tonne Kg 19.0m Required Calculate the GM and angle of heel obtained from the inclining test. Calculate the ship’s final light displacement and VCG, given her KM was 9.2m

GM = GG1 x AB BC

= w x d x ABW BC

A ship of 18500 tonne displacement is inclined by moving a 10 tonne weight through a distance of 25 m transversely. The deflections of two pendulums were as follows.

Forward pendulum Length 18 m Deflection 16.0 cm Aft pendulum Length 18 m Deflection 15.8 cm

To bring the ship to light displacement, the following weights are moved. Weights on: Four sets of cargo winches each 8 tonne Kg 20.3m Weights off: Inclining test weight 10 tonne Kg 18.8m Six sets of shore generator each 4 tonne Kg 19.0m Required Calculate the GM and angle of heel obtained from the inclining test. Calculate the ship’s final light displacement and VCG, given her KM was 9.2m

W BC= 10 x 25 x 18

18500 x 0.159= 1.5298 m

KM = 9.2000 mKG = 7.6702 m

Angle of heel (θ): Cot θ = AB/BC

A ship of 18500 tonne displacement is inclined by moving a 10 tonne weight through a distance of 25 m transversely. The deflections of two pendulums were as follows.

Forward pendulum Length 18 m Deflection 16.0 cm Aft pendulum Length 18 m Deflection 15.8 cm

To bring the ship to light displacement, the following weights are moved. Weights on: Four sets of cargo winches each 8 tonne Kg 20.3m Weights off: Inclining test weight 10 tonne Kg 18.8m Six sets of shore generator each 4 tonne Kg 19.0m Required Calculate the GM and angle of heel obtained from the inclining test. Calculate the ship’s final light displacement and VCG, given her KM was 9.2m

Angle of heel (θ): Cot θ = AB/BC= 18/0.159= 113.2075

θ = 0.5061°

Wt KG Moments18500 7.6702 141898.1132

32 20 3000 649 6000

A ship of 18500 tonne displacement is inclined by moving a 10 tonne weight through a distance of 25 m transversely. The deflections of two pendulums were as follows.

Forward pendulum Length 18 m Deflection 16.0 cm Aft pendulum Length 18 m Deflection 15.8 cm

To bring the ship to light displacement, the following weights are moved. Weights on: Four sets of cargo winches each 8 tonne Kg 20.3m Weights off: Inclining test weight 10 tonne Kg 18.8m Six sets of shore generator each 4 tonne Kg 19.0m Required Calculate the GM and angle of heel obtained from the inclining test. Calculate the ship’s final light displacement and VCG, given her KM was 9.2m

32 20.3000 649.600010 18.8000 188.000024 19.0000 456.0000

18498 7.6713 141903.7132

Ans: During inclining experiment, GM = 1.53 m, angle of heel (θ) = 0.51°S CG

A ship of 18500 tonne displacement is inclined by moving a 10 tonne weight through a distance of 25 m transversely. The deflections of two pendulums were as follows.

Forward pendulum Length 18 m Deflection 16.0 cm Aft pendulum Length 18 m Deflection 15.8 cm

To bring the ship to light displacement, the following weights are moved. Weights on: Four sets of cargo winches each 8 tonne Kg 20.3m Weights off: Inclining test weight 10 tonne Kg 18.8m Six sets of shore generator each 4 tonne Kg 19.0m Required Calculate the GM and angle of heel obtained from the inclining test. Calculate the ship’s final light displacement and VCG, given her KM was 9.2m

Ship's light displacement = 18498 tonne, VCG 7.67 m

A ship of 18500 tonne displacement is inclined by moving a 10 tonne weight through a distance of 25 m transversely. The deflections of two pendulums were as follows.

Forward pendulum Length 18 m Deflection 16.0 cm Aft pendulum Length 18 m Deflection 15.8 cm

To bring the ship to light displacement, the following weights are moved. Weights on: Four sets of cargo winches each 8 tonne Kg 20.3m Weights off: Inclining test weight 10 tonne Kg 18.8m Six sets of shore generator each 4 tonne Kg 19.0m Required Calculate the GM and angle of heel obtained from the inclining test. Calculate the ship’s final light displacement and VCG, given her KM was 9.2m

obs/2011

Revision Solution No. 8 (Angle of Loll) A box-shaped vessel 50m x 9m x 6m floats at an even keel draft of 2.5m in salt water. KG = 4.2m. Calculate the angle of loll. BM = B2/12d = 9 x 9 12 x 2.5 = 2.7000 m KB = 1.2500 m KM = 3.9500 m KG = 4.2000 m GM = - 0.2500 m Tan θ = √(2GM/BM) = √(2 x 0.2500 / 2.7000) = 0.4303

θ = 23.28°

9

L x B x di = (L x B x db) - (l x b x h)ρ120 x 20 x 6 = (120 x 20 db) - (15 x 20 x 4)0.4

14400 = 2400db - 4802400db = 14880

db = 14880/2400= 6.2000 m

LCB’ = Total Long Moment @ AP / Total Volume= 810000/13920= 58.1897 m

BB’ = (L/2) - LCB'= (120/2) - 58.1897= 1.8103 m

BML = BL³ OR L² / 12d12V (80 x 80) / (12 x 4)

= 20 x 120 x 120 x 120 133.333 m12 x 120 x 20 x 6

= 34560000172800

= 200 m

MCTC = W x BML

100L= (L x B x Draft x Density) x BML

100L= (120 x 20 x 6 x 1.025) x 200

100 x 120= 2952000

12000= 246 t-m

54000112.548013920 810000

A box-shaped vessel 120m long and 20m beam floats at an even keel draft of 6m and has a compartment 15m long at the extreme fore-end. The bilged compartment has a watertight flat 4m above the keel and permeability of 40%. Calculate the final drafts if this compartment is bilged.

Volume LCBs Long Moment @ AP14400 60 864000

Change of Trim = W x BB'MCTC

= ( L x B x di x density ) x BB'MCTC

= (120 x 20 x 6 x 1.025) x 1.8103246

= 26720.69246

= 108.62069 cm= 1.0862069 m by the stern

COD F & A = COT / 2= 108.6207/2= 54.3103 cm= 0.5431 m

Forward AftBilged Draft 6.2000 m 6.2000 mChange of Draft due to Trim 0.5431 + 0.5431 -Final Drafts after Bilging 6.7431 m 5.6569 m

S Solution No.10

11

Initial GM = 0.50 mVirtual GM= 0.30 mVirtual Los= 0.20 mVirtual Loss (MM1) = P x KM / W

P = 0.2 x 3000 / 6.0 = 100 tonnesP = MCTC x t/lt = 100 x 48 / 40 = 120 cm

Virtual Los = 0.50 – 0.30 = 0.20 mVirtual Loss (GG1) = P x KG / W - P

W-P = P x 5.5 / 0.2 0.2(3000 – P) = P x 5.5 600 – 0.2P = 5.5P

5.7P = 600 P = 105.26 tonnes P = MCTC x t/l t = 105.26 x 48 / 40

= 126.31 cm

A ship 3000 tonnes displacement is 100m long, KM = 6m, KG = 5.5m, CF = 2m aft of amidships and MCTC = 40 t-m. Calculate the maximum trim for the ship to enter drydock of GM at the critical instant before the ship takes the blocks overall is to be not less than 0.3m