Trim & Stability-basic

Displacement

Principle of Flotation:

"When a body is floating in a liquid, the weight of liquid displaced equals to the weight of the body."

ARCHIMEDES' Principle states that when a body is wholly or partially immersed in a fluid, it suffers an apparent

loss of weight which is equal to the weight of fluid displaced. And that merchant ships are expected to float.

Weight becomes less

( i ) The volume of water displaced is the underwater volume of the ship.

( ii ) Buoyancy or displacement is the upward thrust experienced by the ship. When the ship is floating freely, its

displ. (or buoyancy) equals to its weight. The weight of the ship is therefore refered to as Displacement

( Δ ).

Important Terms

1. Displacement - the mass of the ship in tonnes. Technically, the mass of water displaced by a ship.

Δ = (L x B x draft) x Density of water displaced.

2. Lightship - the mass of the empty ship, without any cargo, fuel, lubricating oil, ballast water, freshwater in tanks,

consummable stores, passengers, crew and their effects.

3. Load Displacement (or Summer Displ.) - the mass of the ship when she is floating in saltwater (1.025) with her

summer loadline at the water surface.

4. Present Displacement - the mass of the ship at present. It is the sum of the light displacement of the ship and

everything on board at present.

5. Deadweight (DWT) - the total mass of cargo, fuel, freshwater, etc. that a ship can carry, when a ship is floating

in saltwater with her summer loadline at the water surface.

DWT of Ship = load displacement - lightship

6. Deadweight Aboard - the total mass of cargo, fuel, ballast, freswater, etc. on board at present.

DWT Aboard = present displacement - lightship

7. Deadweight Available - the total mass of cargo, fuel, ballast, freshwater, etc. that can be put on board at

present to bring her summer loadline to the water surface in saltwater.

DWT Available = load displacement - present diplacement

8. Water Plane Coefficient (Cw) - the ratio of the area of the water plane to the area of a rectangular having the

same length and maximum breadth.

Cw = Area of waterplane

L x B

9. Block Coefficient (Cb) - or coefficient of fineness of displacement, is the ratio of the underwater volume of the

TRIM & STABILITY

Water

Displaced

Overflow

ship at that draft to a rectangular box having the same extreme dimensions.

Cb = >> (can be used later for Squat computation);

L x B x draft * "Confined Water" Squat (m.) = 2 x Cb x Spd.²

100

* "Open Water" Squat (m.) = Cb x Spd.²

10. Tonnes per Centimeter (TPC) - the number of tonnes required to cause a ship's mean draft to sink or rise by

1 cm.

TPCFW = TPCSW x 1 TPCDW = TPCSW x d

and

11. Freeboard - the distance measured vertically downwards from the upper edge of the deckline to the upper

edge of the related loadline.

RB = total volume - underwater volume

12. Loadlines - lines marked on a vessel's side at mid-length to define the maximum drafts to which a vessel may

load in all sea areas, in rivers and harbors where the density of the water in which the vessel floats

is not equal to the density of the saltwater. Standards for loadlines are set by the International

Convention of Loadlines, 1966.

230 mm

540 mm

TF

F 230 mm

T

DWA

FWA

S

300 mm W

450 mm WNA

* To follow, simply do not immerse the said loadline marks (upper line edge) in a specific loadline zone.

Basic Formula and Conversions:

Δ

100

1.025 1.025

Mass

where:

Density = Mass and Volume = Mass

Freshwater Density = 1.0 kg. / L Saltwater Density = 1.025 kg. / L

= 1.0 gm. / cc. = 1.025 gm. / cc.

= 1.0 MT / m³ = 1.025 MT / m³

where:

1 cc. = 1 ml.

1000 cc. = 1 Ltr.

1 MT = 1000 kg. TON means Long Ton

1 MT = 2204.62 lbs. TONNE means Metric Ton1 LT = 2240 lbs.

1 LT = 1.016 MT

Hydrostatics Table Data

1. Draft 9. KB - Ht. Of Center of Buoyancy from Keel

2. Total Displacement 10. LKM - Longitudinal Ht. of Metacenter

3. Moulded Displacement 11. Cb - Block Coefficient

4. TPC - Weight to Change Trim by 1 cm. 12. CP - Prismatic Coefficient

5. LCB - Longitudinal Center of Buoyancy 13. CMID - Midship Coefficient

6. LCF - Longitudinal Center of Flotation 14. Cw - Water Plane Coefficient

7. MTC - Moment to Change Trim by 1 cm. 15. WPA - Water plane Area

8. TKM - Transverse Height of Metacenter 16. WSarea - Wetted Surface Area

Thickness of Shell

Shell

W L

Extreme Depth

Moulded Depth

Finding Deadweights

Ex. Load Displacement and Lightship from Ship's particulars are 83,246 MT and 9,502 MT respectively.

83,246 MT

9,502 MT

73,744 MT

Ex. At 1.025 Density, ship's draft is 13.50 m. Find DWT Aboard, DWT Avail. And TPC.

= Volume x Density

Volume Density

Extreme Draft

Breadth Moulded

Breadth Extreme

Load Δ =( - )

Lightship =

DWT =

80,820 MT from H. Table @ 13.50 m. draft

9,502 MT

71,318 MT

* DWT Avail. = DWT - DWT Aboard

= 83,246 MT - 80,820 MT

DWT Avail. = 2,426 MT (or remaining Cargo to FULL LOAD )

* TPC = 1.025 x AWP

100

= 1.025 x 6358 m²

100

TPC = 65.17 MT

Loadline

Ex. Nov. 15, 2007 - Houston, TX. to Hamburg, Germany

6 days (from Houston to Summer/Winter boundary - referrring to Loadline diagram)

Winter draft = 13.582 m. from

= 81,355 MT Ship's particulars

Burn-off = 360 MT (for 6 days)

= 81,715 MT

* Draft = 13.64 m. (from H. Table @ 81,715 MT)

FWA - change in draft when the ship goes from SW to FW and vice versa.

FWA

=

=

FWA = 0.313 m.

Ex. Present midship draft is 12.00 m. P/S in SW. Find new draft in FW.

from H. Table @

TPC = 64.15 MT 12.0 m. draft

FWA * FWA = 0.277 m. add bec. From

= 12.000 m. SW to FW

NEW Draft = 12.277 m.

=

=

FWA = 0.277 m.

DWA - change in draft when the ship goes from SW to DW and vice versa.

dw = 1.001 _ 1.0249

= 1.0251 _ 1.040

Present Δ =( - )

Lightship =

DWT Abrd=

Winter Δ

Dep. Δ

= Δ

40 x TPC x 100

81,715 MT

40 x 65.26 MT x 100

81,715 MT

261,040

Δ = 71,115 MT

= Δ+

40 x TPC x 100 Old middraft

71,115 MT

40 x 64.15 MT x 100

71,115 MT

256,600

= 0.99 _ 0.999

except 1.000 & 1.025

TF

F d = 1.000

T 0.025

d = 1.025-d

S d = 1.025

W

Therefore: dwa

FWA

= FWA x (1.025 - d)

dwa

Ex. Present midship draft in SW is 12.25 m. P/S. Find the new mid draft if DW Density is 1.004 MT/m³.

from H. Table @

TPC = 64.34 MT 12.25 m. draft

FWA

=

=

FWA = 0.283 m.

dwa =

= 0.283 X (1.025 - 1.004)

= 0.283 (0.021)

dwa = 0.232 m.

* dwa = 0.232 m. add bec. from SW

Old mid draft = 12.250 m. to Lower density

NEW Mid Draft = 12.482 m. 12.482 m.

Ex. Present midship draft in SW is 13.50 m. P/S. Find the new mid draft if she shifts to DW of 1.009 MT/m³.

from H. Table @

TPC = 65.17 MT 13.50 m. draft

FWA

dwa

=1.025 - d

0.025

dwa x 0.025

= FWA x (1.025 - d)

0.025

Δ = 72,718 MT

= Δ

40 x TPC x 100

72,718 MT

40 x 64.34 MT x 100

72,718 MT

257,360

FWA x (1.025 - d)

0.025

0.025

0.025

+

Δ = 80,820 MT

FWA

=

=

FWA = 0.310 m.

dwa =

= 0.310 X (1.025 - 1.009)

= 0.310 (0.016)

dwa = 0.198 m.

* dwa = 0.198 m. add bec. from SW

Old mid draft = 13.500 m. to Lower density

NEW Mid Draft = 13.698 m. 13.698 m.

Requiring WEIGHT of CARGO To Be Loaded

Ex.1. Nov. 15, 2007 - Houston, TX. To Hamburg, Germany

Cargo: COAL Stowage Factor (SF): 44 ft³ / LT

DW Density (Houston): 1.001 MT / m³

ROB Dep.: FO = 1,130 MT Constant: 152 MT

DO = 165 MT

FW = 275 MT

Consumption per Day: FO = 35 Tpd

DO = 5 Tpd to be known first

FW = 6 Tpd

Steaming time to Winter Zone: 7 days

- Please advise the weight of cargo.

Q1. What is the maximum permissible midship draft such that you are not overloaded when you go to sea?

Hamburg

Houston

Winter Draft as per ship's particulars = 13.582 m.

Winter Zone

7 days

1. Identify the Limiting Zone: WINTER (the zone with the least draft in the voyage ).

2. Identify Draft at Limiting Zone: Draft = 13.582 m. as per ship's

Δ = 81,355 MT particulars

TPC = 65.22 MT (from H. Table @ 13.58 m draft )

3. Find the Sinkage due to consumption:

a. Consumption = 7 days x (35 + 5 + 6) = 322 MT

b. Sinkage =

=

65.22 MT x 100 m.

= Δ

40 x TPC x 100

80,820 MT

40 x 65.17 MT x 100

80,820

260,680

FWA x (1.025 - d)

0.025

0.025

0.025

+

Florida

Consumption

TPC x 100

322 MT

=

Sinkage = 0.049 m.

4. Departure Mid Draft SW = Draft at Limiting Zone + Sinkage

= 13.582 m. + 0.049 m.

Dep. Mid Draft SW = 13.631 m.

where, Δ on Dep. = Δ at Limiting Zone + Consumption

= 81,355 MT + 322 MT

Δ on Dep. == 81,677 MT

TPC = 65.26 MT (from H. Table @ 13.361 m. )

5. Find Dock Water Allowance:

dwa =

= 81,677 x (1.025 - 1.001)

= 81,677 (0.024)

dwa = 0.300 m.

6. Find Departure Mid Draft in DW:

Dep. Mid Draft in DW = Dep. Mid Draft SW + dwa

= 13.631 m. + 0.300 m.

Dep. Mid Draft in DW = 13.931 m. (max. mid draft on dep. )

7. Get the Weight of Cargo to be Loaded:

Wt. Of Cargo = Δdep. - Lship - ROBDep - Constant - Safety Margin (SM)

= 81,677 - 9,502 - 1,570 - 152 - SM 2 to 3 cm. x (TPC)

= 70,453 MT

SM = 153 MT

Wt. Of Cargo = 70,300 MT (to be loaded)

* Upon having the knowledge of weight to be loaded, distributing of cargo and loading (calculation, loading

sequence and draft / trim check) will follow.

* Before target draft, finalization at ''30 cm. to 40 cm. x TPC'' before target draft.

Q2. On draft check, mid (P) = 13.610 m. and mid (S) = 13.590 m. Find the Remaining Cargo where mean

mid draft = 13.600 m.

8. Find Sinkage:

Sinkage = Dep. Mid Draft DW - Present Mid Draft

= 13.931 m. - 13.6000 m.

Sinkage = 0.331 m.

9. Finalize for Remaining Cargo:

Rem. Cargo = Sinkage x TPC x 100 x d

= 0.331 x 65.26 x 100 x 1.001

= 2,162

1.025

Rem. Cargo = 2,110 MT

NOTE: When voyage is from Winter Zone to Tropical or Summer Zone (i.e. Hamburg to Santos), limiting zone

will be WINTER and consumption will be Zero ( 0 ) since the least draft and consumption started imme-

diately from the Berth.

Ex.2. Jan. 12, 2008 - Santos, Brazil to Rotterdam, NL.

322

6,522 m.

Δ x (1.025 - d)

TPC x d x 100

65.26 x 1.001 x 100

6,532.6

-

1.025

1.025

Cargo: PELLETS Stowage Factor (SF): 48 ft³ / LT

DW Density (Santos): 1.020 MT / m³

ROB Dep.: FO = 1,430 MT Constant: 152 MT

DO = 165 MT

FW = 275 MT

Consumption per Day: FO = 35 Tpd

DO = 5 Tpd to be known first

FW = 6 Tpd

Steaming time to Winter Zone: 15 days

- Please advise the weight of cargo.

Q1. What is the maximum midship draft on departure?

Rotterdam

Winter Zone

Winter Draft as per ship's particulars = 13.582 m.

15 days

Santos

1. Identify the Limiting Zone: WINTER (the zone with the least draft in the voyage ).

2. Identify Draft at Limiting Zone: Draft = 13.582 m. as per ship's

Δ = 81,355 MT particulars

TPC = 65.22 MT (from H. Table @ 13.58 m. draft )

3. Find the Sinkage due to consumption:

a. Consumption = 15 days x (35 + 5 + 6) = 690 MT

b. Sinkage =

=

=

Sinkage = 0.106 m.

4. Departure Mid Draft SW = Draft at Limiting Zone + Sinkage

= 13.582 m. + 0.106 m.

Dep. Mid Draft SW = 13.688 m.

where, Δ on Dep. = Δ at Limiting Zone + Consumption

= 81,355 MT + 690 MT

Δ on Dep. == 82,045 MT

TPC = 65.29 MT (from H. Table @ 13.688 m. )

5. Find Dock Water Allowance:

dwa =

= 82,045 x (1.025 - 1.020)

=

Consumption

TPC x 100

690 MT

65.22 MT x 100 m.

690

6,522 m.

Δ x (1.025 - d)

TPC x d x 100

65.29 x 1.020 x 100

410.225

6,659.580

dwa = 0.062 m.

6. Find Departure Mid Draft in DW:

Dep. Mid Draft in DW = Dep. Mid Draft SW + dwa

= 13.688 m. + 0.062 m.

Dep. Mid Draft in DW = 13.750 m. (max. mid draft on dep. )

7. Get the Weight of Cargo to be Loaded:

Wt. Of Cargo = Δdep. - Lship - ROBDep - Constant - Safety Margin (SM)

= 82,045 - 9,502 - 1,870 - 152 - SM 2 to 3 cm. x (TPC)

= 70,521 MT

SM = 121 MT

Wt. Of Cargo = 70,400 MT (to be loaded)

* Upon having the knowledge of weight to be loaded, distributing of cargo and loading (calculation, loading

sequence and draft / trim check) will follow.

* Before target draft, finalization at ''30 cm. to 40 cm. x TPC'' before target draft.

Q2. On draft check, reading is 13.450 m. P/S. Find the Remaining Cargo

8. Find Sinkage:

Sinkage = Dep. Mid Draft DW - Present Mid Draft

= 13.750 m. - 13.450 m.

Sinkage = 0.300 m.

9. Finalize for Remaining Cargo:

Rem. Cargo = Sinkage x TPC x 100 x d

= 0.300 x 65.29 x 100 x 1.020

= 1,998

1.025

Rem. Cargo = 1,949 MT

NOTE: When voyage is from Winter Zone to Tropical or Summer Zone (i.e. Hamburg to Santos), limiting zone

will be WINTER and consumption will be Zero ( 0 ) since the least draft and consumption started imme-

diately from the Berth.

Center of Gravity (COG or G)

- the point through which the force of gravity may be considered to act vertically downwards, with a force

equal to the weight of the ship.

- LCG = Longitudinal Center of Gravity

- VCG = Vertical Center of Gravity ( or KG )

- TCG = Transverse Center of Gravity

The position of the COG of a ship depends on the distribution of weights or effects of adding a weight, removing

a weight or shifting a weight:

• when a weight is added (loaded), the COG of a ship moves directly towards the COG of the added weight.

• when a weight is removed (discherged), the COG of a ship moves away from the COG of the removed weight.

• when a weight already onboard is shifted, the COG of a ship moves in a direction parallel to that moved by the

weight.

The distance through which the COG would move has a formula of:

GG1 = W x d where:

∆ GG1 =the shift of COG in meters.

W = the weight loaded/discharged/shifted, in tonnes.

d = the distance between the COG of the ship and the COG of

the weight when loading/discharging, in meters.

-

1.025

1.025

∆ = the final displacement of the ship in tonnes,after the weight

has been loaded/discharged/shifted.

LBP = 216 m.

108 m. 108 m.

g g g g g g g

AP FP

KG = 10.60 m. + DBT Ht.

KG of partially loaded

LCG (H7) = +67.84 m. LCG (H1) = -84.12 m.

Japan, USA

China, Phil.

Europe,

Singapore, LCG (H1) = 192.12 m. (108 + 84.12)

China

Korea,

China LCG (H7) = -67.84 m. LCG (H1) = +84.12 m.

KG of Partially Loaded Hold (Leveled Cargo)

Ex.1. Load 8,000 MT of Cargo with a SF of 43 ft³ / LT to Hold 3. Find the KG then.

1. Find SF in m³ / MT:

= 43 ft³ / LT

35.8813 ft³ / LT

SF = 1.198 m³ / MT

2. Solve for the Volume:

Vol = Wt. x SF

= 8,000 MT x 1.198 m³ / MT

Vol = 9,584 m³

3. Use Hydrostatics Table - Volumetric Heeling Moment, Volume and KG 'Diagram or Table' based on VOLUME

figure and Cargo hold No. (9,584 m³ / No. 3 Cargo Hold)

KG = 8.200 m.

Note: when interpolation is needed, use the higher KG for allowance.

Ex.2. Load 9,000 MT of Cargo with a SF of 47 ft³ / LT to Hold 3. Find the KG and Ullage then.

1. Find SF in m³ / MT:

= 47 ft³ / LT

35.8813 ft³ / LT

SF = 1.310 m³ / MT

2. Solve for the Volume:

7 6 5 4 3 2 1

Country

Maker

Conversion

1 m³ / MT = 35.8813 ft³ / LT

1 ft³ / LT = 0.027868 m³ / MT

Conversion

1 m³ / MT = 35.8813 ft³ / LT

1 ft³ / LT = 0.027868 m³ / MT

Vol = Wt. x SF

= 9,000 MT x 1.310 m³ / MT

Vol = 11,788.87 m³

3. Use Hydrostatics Table - Volumetric Heeling Moment, Volume and KG 'Diagram or Table' based on VOLUME

figure and Cargo hold No. (11,789 m³ / No. 3 Cargo Hold)

KG = 9.600 m.

Ullage = 3.500 m.

Note: when interpolation is needed, use the higher KG for allowance.

KG of Partially Loaded Hold (Cargo of Different Heights)

Ex. Steel coils of 3,000 MT, CKD (600 MT) and steel pipes (1,200 MT) will be loaded in Hold No.1. Find the

LCG and KG as well.

H1 (LCG = -84.12 m.)

30 m.

KG = KG of Cargo +

DBT Ht.

12 m. Ht. (6 m. KG)

10 m. Ht (5 m. KG)

DBT Ht. = 2 m. 1 m.

10 m. 9 m.

V Moment

-74.12 m. (84.12 - 10 )

-85.12 m. (84.12 + 1 )

-93.12 m. (84.12 + 9 )

1. LCG = T. Lmoment

T. Weight

= 385,176

LCG = -80.245 m.

2. KG = V Moment

= 34,200

KG = 7.125 m.

Free Surface Moment

Ex. Loading 600 MT to FPT

1. Wt = Vol. x Density

Vol. =

=

Vol. = 585 m³

Cargo Weight LCG Diff. LCG L Moment KG

Type (in MT) (in meter) (Wt. x LCG)

6 3,600

st. coils 3,000 MT (-10 m.) -222,360

(KG + DBT Ht) (Wt. x KG)

7 21,000

8 9600

-385,176 7.125 m.

CKD 600 MT

st. pipes 1,200 MT (+9 m.) -111,744

(+1 m.) -51,072

34,200

4,800

T. Weight

4,800

Wt.

Total 4,800 MT -80.245 m.

Density

600 MT

1.025 MT/m³

15 m. 15 m.

8 m. Ht. (4 m. KG)

2. From H. Table - Tank Table (FPT), in volume 585 m³:

LCG = -102.26 m.

KG = 3.51 m.

Inertia (I-TR) = 1,323 m4

3. Free Surface Moment (FSM) = Inertia x Density

= 1,323 m4 x 1.025 MT / m³

FSM = 1,356 MT-m

NOTE:

1. To be safe, take the maximum Inertia of the tank (i.e. max. inertia of FPT = 3,030 m4), in order to

acquire the maximum FSM.

* Max. FSM = max. inertia x Density

= 3,030 m4 x 1.025 MT / m³

FSM = 3,106 MT-m (FPT)

2. By Graphical Method, the same procedure in finding KG of Cargo where Volume figure will be applied

in the graph.

Solving Trim, Dep. Drafts and GM from Preplan

(normally when an amount of cargo is booked for loading)

Ex. Preplan 1:

%

TRIM AND STABILITY CALCULATION SHEET

WEIGHT L.C.G. L.MOMENT K.G. V.MOMENT FSM

ITEM (MT) (m.) (MT-m.) (MT-m.) (MT-m.)fr. H. Table Wt. x LCG Wt. x KG

CONST.1 DW CONSTANT 1 100CONST.2 DW CONSTANT 2 100PROV PROVISION 100

TOTAL CONST.=

FOT1 NO.1 FOT 96

FOT2 NO.2 FOT 96FOT3 NO.3 FOT 96D. FOT DEEP FOT 96FO SETT H.F.O. SETT/SERV. 96

TOTAL F.O. =

D. DOT DEEP DOT 96DO SERV. DO SERV T. 96

TOTAL D.O. =

FWT1 NO.1 FWT 100FWT2 NO.2 FWT 100

TOTAL F.W. =

FPT FORE PEAK TK. 0WBT1 NO.1 WBT (P&S) 0WBT2 NO.2 WBT (P&S) 0WBT3 NO.3 WBT (P&S) 0WBT4 NO.4 WBT (P&S) 0WBT5 NO.5 WBT (P&S) 0TST5 NO.5 TST (P&S) 0APT AFT PEAK TANK 3CH 4 NO.4 FLOOD. TK. 0

TTL. BALLAST =

HOLD1 NO.1 C. HOLD 100HOLD2 NO.2 C. HOLD 100HOLD3 NO.3 C. HOLD 100HOLD4 NO.4 C. HOLD 100HOLD5 NO.5 C. HOLD 100HOLD6 NO.6 C. HOLD 100HOLD7 NO.7 C. HOLD 100

TOTAL CARGO =

LIGHTWEIGHT(from ship's particulars)

From Preplan 1:

Δ = 72,531 MT

LCG = -3.55 m.

KG = 10.55 m.

Total FSM = 20,182 MT-m.

* Excerpt from H. Table based on 72,531 MT Δ:

DCF (Draft at Center of Flotation) = 12.22 m.

LCB = -4.85 m.

LCF = 0.42 m.

MTC = 968.01 MT-m.

TKM = 13.30 m.

Illustration:

112 89.21 9,992 13.63

20530 -103.00 -3,090 19.00

1,527570

10 90.50152

905 20.50

894 9,841

517 -8.30 -4,291 0.85 439 4,835

251 1,4191,052 30.17

295 66.15 19,514 0.8531,739 0.85

295 89.58 26,426 15.9231 97.49 3,022 16.93

4,696 123525 4

2,190

164 92.74 15,209 16.4216 94.18 1,507 17.05

2,693 61273 1

180

118 103.02 12,156 17.41178 104.52 18,605 17.65

2,054 523,142 144

296

0 -102.21 0 8.97

0 19,8870 -84.00 0 9.10

0 3,1060 9,176

0 31,4120 -59.760 -22.78 0 8.60

0 8.07

0 -29.94 0 9.62

0 2,6980 68.00 0 2.71

0 12,5010 2,683

249 3,7020 68.59

20 103.94 2,079 12.470 17.73

0 -8.30 0 10.5120

0 67,162

88,1657,761 -84.12 -652,855 11.368,837 -59.41 -525,006 10.62

87,6549,150 -33.64 -307,806 10.56

93,84996,624

96,6358,356 -8.309,151 17.05 156,025 10.56

-69,355 10.49

8,825 42.81 377,798 10.608,111 67.84 550,250 11.14

93,54590,357

60,191

DEADWEIGHT(total weight) (LMom./T.Wt.) (T. LMoment) (T.VMom/T.Wt) (T. VMoment) 20,18263,029.00 -5.35 -337,176 10.54 664,347 (total moments

in comprnts

9,502 8.36 79,437 10.61 100,816with load)

DISPLACEMENT72,531 -3.55 -257,739 10.55 765,163

(DWT+LShip) (L. Mom./ Δ) (T. LMoment) (T. VMom./ Δ) (T. VMoment)

W LCG = -3.55 m. L

LCF = 0.42 m.

LCB = -4.85 m.

AP FP

where:

LCF = Longitudinal Center of Flotation. This is the center of waterplane area and the point where the

vessel trims.

LCG = Longitudinal Center of Gravity

LCB = Longitudinal Center of Buoyancy

MTC = Moment to Change Trim by 1 cm. Unit is MT-m.

* Note: If the LCB is located forward of LCG, the vessel is trimmed "By Stern".. if not, "By Head".

1. Solve for Trim:

t = Δ x (LCG - LCB)

MTC x 100

= 72,531 x (-3.55 - -4.85)

968.01 x 100

= 94,290.30

96,801

t = 0.974 m. (By Stern) + = By Stern

- = By Head

2. Find the Draft Forward:

DFP = DCF- t

2

= 12.22 - 0.974

2

= 12.22 - 0.487

DFP = 11.733 m.

3. Find the Draft Aft:

DFP = 11.733 m.

t = + 0.974 m.

DAP = 12.707 m.

4. Solve GGo or FSC:

FSC = Total FSM

Δ

= 20,182

72,531

FSC = 0.278 m. (or GGo)

LCF = 0.42 m.

5. Solve for GM: M

KG = 10.550 m.

GGo = 0.2780 m. Go GM

KGo = 10.828 m. (actual KG) KM G GGo

KM = 13.300 m. - KGo

GMo = 2.4720 m. (actual GM)

K

* Best "GM" depends on the type of vessel and ship's construction and as long as the vsl will be on the Safe

side.

Ex. 2. Practice

Δ = 81,470 MT

LCG = -3.82 m.

KG = 11.85 m.

FSC = 0.25 m.

Find Drafts and GM.


DCF = 13.60 m.

LCB = -4.22 m.

LCF = 1.39 m.

MTC = 1007.18 MT-m.

TKM = 13.38 m.

1. Solve for Trim:


MTC x 100

= 81,470 x (-3.82 - -4.22)

1,007.98 x 100

= 32,588.00

100,798


- = By Head


DFP = DCF- t

2

= 13.60 - 0.323

2

= 13.60 - 0.162

DFP = 13.438 m.


DFP = 13.438 m.

t = +0.323 m.

DAP = 13.761 m.

3. Solve for GM: M

KG = 11.850 m.

GGo = 0.2500 m. Go GM


KM = 13.380 m. - KGo


K


side.

+

KG

+

KG

Solving Trim and Dep. Drafts

(in this case, when LCG is measured from After Perpendicular)

Ex. 1. Δ = 69,830 MT

LCG = 112.00 m.

Find Drafts.

* Excerpt from H. Table based on 69,830 MT Δ (using LCG measured from AP):

DCF = 11.80 m.

LCB = 113.04 m.

LCF = 108.02 m.

MTC = 953.76 MT-m.

1. Solve for Trim:

t = Δ x (LCB - LCG) (reverse, instead of subtracting LCB from LCG)

MTC x 100

= 69,830 x (113.04 - 112.00)

953.76 x 100

= 72,623.20

95,376


- = By Head


DAP = DCF+ t (reverse, instead of subtraction)

2

= 11.80 + 0.761

2

= 11.80 + 0.381

DAP = 12.181 m.


DAP = 12.181 m.

t = 0.7610 m.

DFP = 11.420 m.

Distribution of Cargo and Dep. Drafts

Ex.1. Dec. 02, 2007 - Norfolk to Japan

Cargo: COAL Stowage Factor (SF): 44 ft³ / LT =

DW Density (Norfolk): 1.002 MT / m³

ROB Dep.: FO = 1,630 MT

DO = 165 MT

FW = 275 MT

Total = 2,070 MT

Constant: 152 MT

- Please advise Dep. Drafts.

* Before having the solution proper, acquire first the following:

a) Loadline check;

Lightship = (from ship's particulars)

Cargo =

ROB =

Constant =

Displ. (Δ) = (OK for winter Δ of 81,355 MT from ship's part.)

b) Capacity check;

Check hold capacity (total) against the cargo to be loaded…

where,

Total hold capacity = 87,298 MT while,

- (subtract, knowing that the trim is "By Stern")

44 ft³ / LT

35.8813

= 1.226 m³ / MT

9,502 MT

67,000 MT

2,070 MT

152 MT

78,724 MT

Cargo to be loaded = 67,000 MT… therefore, OK to load

table - hold info.

c) Cargo hold capacity check;

Wt. For Hold = Capacity of hold x Total Wt.

Total Capacity table - hold info.

i.e. • Wt. For Hold 1 = 11,256.3 x 67,000

Wt. For Hold 1 = 8,639 MT (or 8,600 MT) . . . . .Vol. = Wt x SF (8,600 x 1.226 = 10,544 m³ - Volume )

• Wt. For Hold 2 = 12,817.1 x 67,000


• Wt. For Hold 3 = 13,270.2 x 67,000

Wt. For Hold 3 = 10,185 MT( or 10,200 MT ) . . . . .Vol. = Wt x SF (10,200 x 1.226 = 12,505 m³ - Volume )

• Wt. For Hold 4 = 12,118.9 x 67,000


. . . . .• Wt. For Hold 5 = 10,186 MT ( or 10,200 MT ) . . . . .Vol. = Wt x SF (10,200 x 1.226 = 12,505 m³ - Volume )

• Wt. For Hold 6 = 9,823 MT (or 9,800 MT) . . . . .Vol. = Wt x SF (9,800 x 1.226 = 12,015 m³ - Volume )

Wt. For Hold 1 to Hold 6 = 57,900 MT

therefore,

Remaining Wt. for Hold 7 = Total Wt. - Wt. for Hold 1-6

= 67,000 MT - 57,900 MT

• Wt. For Hold 7 = 9,100 MT . . . . . . . . . Vol. = Wt x SF (9,100 x 1.226 = 11,157 m³ - Volume )

d) Prepare and fill-up the necessary figures in the Calculation Sheet;

%

CONST.1 DW CONSTANT 1 100


PROV PROVISION 100

TOTAL CONST.=

(Capacity of Tk. x SF)

FOT1 NO.1 FOT

FOT2 NO.2 FOT

FOT3 NO.3 FOT

WEIGHT L.C.G. L.MOMENT K.G.

87,298

87,298

(MT-m.) (MT-m.) (MT-m.)

87,298

87,298


fr. H. Table

V.MOMENT FSM

ITEM (MT) (m.)

570

9,992 13.63

Wt. x LCG fr. H. Table

1,527

Wt. x KG

112 89.21

10 90.50 905 20.50

30 -103.00 -3,090 19.00

152

205

340 4,835400 -8.30 -3,320 0.85

900 30.17 27,153 0.85

200 66.15 13,230 0.85

765 9,841

170 1,419

D. FOT DEEP FOT

FO SETT H.F.O. SETT/SERV.

TOTAL F.O. =

D. DOT DEEP DOT 96

DO SERV. DO SERV T. 96

TOTAL D.O. =

FWT1 NO.1 FWT 100

FWT2 NO.2 FWT 100

TOTAL F.W. =

FPT FORE PEAK TK. 0

WBT1 NO.1 WBT (P&S) 0





TST5 NO.5 TST (P&S) 0

APT AFT PEAK TANK 0

CH 4 NO.4 FLOOD. TK. 0

TTL. BALLAST =

HOLD1 NO.1 C. HOLD 100







TOTAL CARGO =

LIGHTWEIGHT

(from ship's particulars)

DISPLACEMENT

From Preplan:

Δ = 78,724 MT

LCG = -4.26 m.

KG = 10.16 m.



DCF = 13.18 m.

LCB = -4.41 m.

LCF = 1.14 m.

MTC = 995.96 MT-m.

TKM = 13.34 m.

TPC = 64.98 m.

1. Solve for Trim:


MTC x 100

= 78,724 x (-4.26 - -4.41)

995.96 x 100

1,592 123

508 4

100 89.58

30 97.49 2,925 16.93

8,958 15.92

1,630

2,463 61

256 1

150 92.74

15 94.18 1,413 17.05

13,911 16.42

165

2,002 52

2,824 144

115 103.02

160 104.52 16,723 17.65

11,847 17.41

275

0 0

0 0

0 -102.21

0 -84.00 0 9.10

0 8.97

0 -59.76 0 8.07

0 0

0 -22.78 0 8.60

0 0

0 0

0 0

0 -29.94

0 68.00 0 2.71

0 9.62

0 68.59 0 17.73

0 103.94 0 12.47

0 10.51

0 0

0 0

0 0

0 (Cargo Hold Heeling Mom.

0 -8.30

Table - based on Vol. Fig.)

8,600 -84.12 -723,432 10.78 92,708

9,800 -59.41 -582,218 10.04

92,721

10,200 -33.64 -343,128 10.08

98,398

102,816

102,816

9,300 -8.30

10,200 17.05 173,910 10.08

-77,190 9.97

9,800 42.81 419,538 10.10

9,100 67.84 617,344 10.70

98,980

97,370

67,000

DEADWEIGHT(total weight) (LMom./T.Wt.) (T. LMoment) (T.VMom/T.Wt) (T. VMoment) 16,480

69,222 -5.99 -414,530 10.10 699,030 (total moments

in comprnts

9,502 8.36 79,437 10.61 100,816with load)

Δ LCG L. Moment KG V. Moment

78,724 -4.26 -335,093 10.16 799,846

(DWT+LShip) (LMom./T.Wt.) (T. LMoment) (T. VMom./ Δ) (T. VMoment)

= 11,808.60

99,596


- = By Head


DFP = DCF- t

2

= 13.18 - 0.119

2

= 13.18 - 0.060

DFP = 13.120 m.


DFP = 13.120 m.

t = +0.119 m.

DAP = 13.239 m.


FSC = Total FSM

Δ

= 16,480

78,724

FSC = 0.210 m. (or GGo)

4. Solve for GM: M

KG = 10.160 m.

GGo = 0.2100 m. Go GM


KM = 13.340 m. - KGo


K


side.

5. Solve for draft on DW (Norfolk = 1.001 MT/m³) for Dock Water Allowance (dwa):

where:

t = 0.119 m.

DFP = 13.120 m.

DAP = 13.239 m.

Δ = 78,724 MT …. (produce Δ1 from 78,724 MT)

Δ1 = Δ x 1.025

d

=

=

Δ1 = 80,611 MT

* Excerpt from H. Table based on 80,611 MT Δ1:

DCF1 = 13.47 m.

LCB1 = -4.28 m.

LCF1 = 1.32 m.

MTC1 = 1,003.74 MT-m.

old LCG

6. Solve for Trim:

t = Δ1 x (LCG - LCB1)

MTC1 x 100

= 80,611 x (-4.26 - -4.28)

1003.74 x 100

= 1,612.22

100,374

+

KG

78,724 x 1.025

1.001

80,692

1.001

t = 0.016 m. (By Stern), Req'd. Trim + = By Stern

- = By Head


DFP = DCF- t

2

= 13.47 - 0.016

2

= 13.47 - 0.008

DFP = 13.462 m.


DFP = 13.462 m.

t = +0.016 m. Drafts @ 1.001 density

DAP = 13.478 m.

9. Solve for Dock Water Allowance:

dwa = Δ x (1.025 - d)

TPC x d x 100

= 78,724 x (1.025 - 1.001)

orig. TPC to check;

= DCF1 = 13.470 m. (new DCF)

DCF = 13.180 m. (previous DCF)

dwa = 0.290 m. dwa = 0.290 m.

* Effects of Change in Density

- Parallel sinkage (dwa)

- Change in Trim (reduced, bec. of LCB change due to density)

Ex.2. Oct. 04, 2007 - Norfolk to Panama Canal

Cargo: COAL Stowage Factor (SF): 44 ft³ / LT =

DW Density (Norfolk): 1.001 MT / m³

DW Density (Panama Canal): 0.9954 MT / m³

Maximum Draft @ Panama: 12.20 m. (draft limit @ destination)

ROB Dep.: FO = 1,630 MT Consumption: FO = 35 MT / day (35 x 8days = 280 MT cons.)

DO = 165 MT DO = 4 MT / day (4 x 8days = 32 MT cons.)

FW = 225 MT FW = 0 MT / day

Total = 2,020 MT

Constant: 152 MT

Steaming time to Panama Canal: 8 Days

- Please advise Wt. of Cargo and Dep. Drafts.

* Before having the solution proper, acquire first the following:

a) At 12.20 m. draft limit at destination, Displacement is 72,397 MT and LCB of -4.86 m. (from H. Table)

Production FOT1 MT-m

Deep DOT MT-m

MT-m

* wherein . . . from H. Table, Total Δ of 70,618 MT has a TPC of 64.08 MT

64.98 x 1.001 x 100

1,889.376

6,504.500

44 ft³ / LT

35.8813

= 1.226 m³ / MT

Weight LCG L. Moment

(in MT) (m.) (Wt. x LCG)

MT-m.(Δ x d / 1.025)

=72,397 x 0.9954 / 1.025 (70,306 MT)

S.W. Displacement @ Arrival

70,306 -4.86 -341,689

Tank Consumption and (Tank LCG)

280 -8.30 -2,324

(as per cons. for 8 days) (Tank LCG)

32 92.74 2,968

S.W. Displacement @ Dep. 70,618 -4.83 -341,045

(Total Δ) (L Mom/Wt.) (Total L. Moment)

a) Cargo to Load (assuming a Sag of 10 cm.)

where, Sag Allow. =

4 in case of Hogging:

= Hog Allow. = Deflection x 3 x TPC

4

Sag Allow. = 160.20 MT

Cargo to Load = Δdep. - Lship - ROBDep - Constant - Hog/Sag Allow. - Safety Margin (SM)

= 70,618 - 9,502 - 2,020 - 152 - 160 - SM

= 58,784 MT

SM = 184 MT

Wt. Of Cargo = 58,600 MT ( CARGO to be Loaded)

b) Loadline check;

Lightship = (from ship's particulars)

Cargo =

ROB =

Constant =

Displ. (Δ) = (OK for Tropical Δ of 85,136 MT from ship's part.)

c) Capacity check;

Check hold capacity (total) against the cargo to be loaded…

where,

Total hold capacity = 87,298 MT while,

Cargo to be loaded = 58,600 MT… therefore, OK to load

table - hold info.

d) Cargo hold capacity check;

Wt. For Hold = Capacity of hold x Total Wt.

Total Capacity table - hold info.

i.e. • Wt. For Hold 1 = 11,256.3 x 58,600


• Wt. For Hold 2 = 12,817.1 x 58,600


• Wt. For Hold 3 = 13,270.2 x 58,600


• Wt. For Hold 4 = 12,118.9 x 58,600


• Wt. For Hold 5 = 13,272.2 x 58,600


• Wt. For Hold 6 = 12,799.5 x 58,600


Wt. For Hold 1 to Hold 6 = 50,500 MT

therefore,

Remaining Wt. for Hold 7 = Total Wt. - Wt. for Hold 1-6

= 58,600 MT - 50,500 MT

Deflection x TPC

10 x 64.08

4

-

9,502 MT

58,600 MT

2,020 MT

152 MT

70,274 MT

87,298

87,298

87,298

87,298

87,298

87,298

• Wt. For Hold 7 = 8,100 MT . . . . . . . . . Vol. = Wt x SF (8,100 x 1.226 = 9,931 m³ - Volume )

e) Prepare and fill-up the necessary figures in the Calculation Sheet;

%



PROV PROVISION 100

TOTAL CONST.=


FOT1 NO.1 FOT

FOT2 NO.2 FOT

FOT3 NO.3 FOT

D. FOT DEEP FOT


TOTAL F.O. =

D. DOT DEEP DOT 96

K.G. V.MOMENT FSM

(MT) (m.) (MT-m.)


WEIGHT L.C.G. L.MOMENT

(MT-m.) (MT-m.)

Wt. x KGfr. H. Table

ITEM

570

9,992 13.63


1,527112 89.21

10 90.50 905 20.50

30 -103.00 -3,090 19.00

152

205

340 4,835400 -8.30 -3,320 0.85

900 30.17 27,153 0.85

2,070 123

200 66.15 13,230 0.85

765 9,841

170 1,419

0 4

130 89.58

0 97.49 0 16.93

11,645 15.92

1,630

2,463 61150 92.74 13,911 16.42


TOTAL D.O. =

FWT1 NO.1 FWT 100

FWT2 NO.2 FWT 100

TOTAL F.W. =

FPT FORE PEAK TK. 0






TST5 NO.5 TST (P&S) 0

APT AFT PEAK TANK 0

CH 4 NO.4 FLOOD. TK. 0

TTL. BALLAST =








TOTAL CARGO =

LIGHTWEIGHT


DISPLACEMENT

From Preplan:

Δ = 70,274 MT

LCG = -3.82 m.

KG = 9.35 m.



DCF = 11.87 m.

LCB = -5.01 m.

LCF = 0.06 m.

MTC = 956.22 MT-m.

TKM = 13.31 m.

TPC = 64.04 m.

1. Solve for Trim:


MTC x 100

= 70,274 x (-3.82 - -5.01)

956.22 x 100

= 83,626.06

256 115 94.18 1,413 17.05

165

1,915 52

2,030 144

110 103.02

115 104.52 12,020 17.65

11,332 17.41

225

0 0

0 0

0 -102.21

0 -84.00 0 9.10

0 8.97

0 -59.76 0 8.07

0 0

0 -22.78 0 8.60

0 0

0 0

0 0

0 -29.94

0 68.00 0 2.71

0 9.62

0 68.59 0 17.73

0 103.94 0 12.47

0 10.51

0 0

0 0

0 0

0 (Cargo Hold Heeling Mom.

0 -8.30


7,500 -84.12 -630,900 9.83 73,725

8,600 -59.41 -510,926 9.12

72,657

8,900 -33.64 -299,396 9.08

78,432

80,812

80,812

8,100 -8.30

8,900 17.05 151,745 9.08

-67,230 8.97

8,500 42.81 363,885 9.09

8,100 67.84 549,504 9.89

77,265

80,109

58,600


60,772 -5.73 -348,127 9.15 556,122 (total moments

in comprnts

9,502 8.36 79,437 10.61 100,816with load)


70,274 -3.82 -268,690 9.35 656,938


95,622


- = By Head


DFP = DCF- t

2

= 11.874 - 0.875

2

= 11.874 - 0.437

DFP = 11.437 m.


DFP = 11.437 m.

t = +0.875 m.

DAP = 12.312 m.


FSC = Total FSM

Δ

= 16,476

70,274

FSC = 0.234 m. (or GGo)

4. Solve for GM: M

KG = 9.350 m.

GGo = 0.234 m. Go GM


KM = 13.31 m. - KGo


K


side.

5. Solve for draft on DW (Norfolk = 1.001 MT/m³) for Dock Water Allowance (dwa):

where:

t = 0.875 m.

DFP = 11.437 m.

DAP = 12.312 m.

Δ = 70,274 MT … (produce Δ1 from 70,274 MT)

Δ1 = Δ x 1.025

d

=

=

Δ1 = 71,959 MT

* Excerpt from H. Table based on 71,959 MT Δ1:

DCF1 = 12.13 m.

LCB1 = -4.89 m.

LCF1 = 0.33 m.

MTC1 = 965.07 MT-m.

old LCG

+

KG

70,274 x 1.025

1.001

72,031

1.001

6. Solve for Trim:

t = Δ1 x (LCG - LCB1)

MTC1 x 100

= 71,959 x (-3.82 - -4.89)

965.07 x 100

= 76,996.13

96,507

t = 0.798 m. (By Stern), Req'd. Trim + = By Stern

- = By Head


DFP = DCF- t

2

= 12.13 - 0.798 Wt. To Shift =

2 DBH (Dist. bet. Holds)

= 12.13 - 0.399

DFP = 11.731 m.


DFP = 11.731 m.

t = +0.798 m. Drafts @ 1.001 density

DAP = 12.529 m.

9. Solve for Dock Water Allowance:

dwa = Δ x (1.025 - d)

TPC x d x 100

= 70,274 x (1.025 - 1.001)

orig. TPC to check;

= DCF1 = 12.130 m. (new DCF)

DCF = 11.870 m. (previous DCF)

dwa = 0.260 m. dwa = 0.260 m.

Trimming

. when the change in mean draft is only within 50 cms.

Sample Cases:

I. Loading 4,000 MT, where TPC is 64.98 MT.

II. Loading 3,000 MT, where TPC is 64.98 MT.

III. Loading 8,000 MT and discharging 5,000 MT.

=

=

Ex. 1. Departure Drafts : dF = 13.25 m. and dA = 13.65 m. Find Arrival Draft with the following consumption and

production:

Cons.: FOT2 = 400 MT

FOT3 = 200 MT

D. DOT = 100 MT

Formula when a WEIGHT needs to be shifted:

Tc x MTC x 100

where: Tc = Req'd. Trim - Preplan Trim

64.04 x 1.001 x 100

1,686.576

6,410.405

4,000 MT= 64.98 cm. (NOT TRIMMING)

64.98 MT

3,000 MT= 46.00 cm. (TRIMMING)

64.98 MT

8,000 MT-

5,000 MT

3,000 MT

3,000 MT= 46.00 cm. (TRIMMING)

64.98 MT

Prod.:FWT1 = 40 MT

FWT2 = 40 MT

1. Find the mean draft:

mD = dF + dA

2

= 13.250 + 13.650

2

mD = 13.45 m.

2. Excerpt from H. Table based on 13.45 m. draft:

TPC = 65.14 m.

LCF = 1.31 m.

MTC = 1003.2 MT-m.

3. Find total weights and moments:

4. Find Mean Rise or Sinkage (but in this case, this is "RISE" because Discharge is greater than (>) Loaded):

Rise =

=

Rise = 0.095 m.

5. Find distance from LCF to LCG:

D = LCG - LCF (when LCG is measured from centerline, but if LCG is measured from AP…)

= 42.37 - 1.31

D = 41.06 m.

* If 'd' is negative (-), LCG of Total Weight is forward from LCF.

6. Find the change in Trim:

Tc =

= 620 x 41.06

1003.2 x 100

= 25,457.2

100,320

Tc = 0.250 m.

7. Find trim forward and aft:

Tc

2

= 0.25

2

0.127 m.

8. Find the Final Drafts:

neg. (-) in order to Rise Rise =

Weight (MT) LCG Moment

Compt. (-) consumed (m.) (MT-m.)

(+) produced Tank LCG Wt. X LCG

FOT2 -400 30.17 -12,068

FOT3 -200 66.15 -13,230

D. DOT -100 92.74 -9,274

FWT1 +40 103.02 4,121

FWT2 +40 104.52 4,181

TOTAL -620 42.37 -26,270

(-)=Disch'd T. Mom./Wt. T. Moment

Wt.

TPC x 100

620

65.14 x 100

D = Af - Ag

Wt. x Dist.

MTC x 100

Tf / Ta =

Tf / Ta =

Forward. Aft

Dep. dF = 13.250 m. Dep. dA = 13.650 m.

Rise = -0.095 m. -0.095 m.

Tf / Ta = +0.127 m. Tf / Ta = -0.127 m.

* (Because disch. From Aft, Tf / Ta Fwd to be added and Tf / Ta Aft to be subtracted)

- Another method in getting new drafts is by using "Trimming Table" - change of draft due to add 100 MT weight;

Apply Old Draft… dF = m. Apply Old Draft… dA = m.

NEW DRAFT FWD = m. NEW DRAFT AFT = m.

* cdF100 = change of draft Forward by 100 MT.

* cdA100 = change of draft Aft by 100 MT.

Ex. 2. Present Drafts : dF = 9.15 m. and dA = 10.65 m. Find Arrival Draft with the following conditions:

Ballast in: WBT3 = 900 MT

WBT5 = 1,000 MT

Fill-in:FOT2 = 800 MT

FWT2 = 100 MT

Pump-out: APT = 120 MT


mD = dF + dA

2

= 9.150 + 10.650

2

mD = 9.900 m.

2. Excerpt from H. Table based on 9.900 m. draft:

TPC = 62.35 m.

LCF = -2.60 m.

MTC = 883.45 MT-m.

3. Find total weights and moments:

4. Find Mean Rise or Sinkage (but in this case, this is "SINKAGE" because Discharge is less than (<) Loaded):

Sinkage =

=

Fwd - cdF100 (cm.) Wt. x cdF100

NEW DRAFT FWD. = 13.282 m.

Aft - cdA100 (cm.) Wt. x cdA100Weight (MT)

NEW DRAFT AFT = 13.428 m.

Compt. (-) disch'd value from H. Table value from H. Table

4.730 -0.095

(+) loaded based on 13.45m. Mdraft Wt. x cdF100

2.960 -0.1180.075 -0.003

based on 13.45m. Mdraft Wt. x cdF100

(nearest or round-off) 100 MT x 100cm /MT(nearest or round-off) 100 MT x 100cm /MT

-3.080 0.031+

6.030

FOT2 -400

FOT3 -200 -1.740 0.035

-0.060+

6.540 0.026FWT1 +40 -3.600 -0.014

D. DOT -100

FWT2 +40 -3.670 -0.015 6.610 0.026

13.250 13.650

13.284 13.429

Weight (MT) LCG Moment

Compt. (-) consumed (m.) (MT-m.)

(+) produced Tank LCG Wt. X LCG

WBT3 +900 -22.78 -20,502

WBT5 +1,000 68.00 68,000

FOT2 +800 30.17 24,136

FWT2 +100 104.52 10,452

APT -120 103.94 -12,473

TOTAL 2,680 25.98 69,613

(+)=Loaded T. Mom./Wt. T. Moment

Wt.

TPC x 100

2,680

Sinkage = 0.430 m.



= 25.98 - -2.60

D = 28.58 m.



Tc =

= 2,680 x 28.58

883.45 x 100

Tc = 0.867 m.


Tc

2

= 0.87

2

0.433 m.

8. Sum of Drafts:

a) Rise = ( - ) , Sinkage = ( + )

b) If Trim by Stern, Tf/Ta Fwd = ( - ) , Tf/Ta Aft = ( + )

c) If Trim by Head, Tf/Ta Fwd = ( + ) , Tf/Ta Aft = ( - )

Sinkage Sinkage =

Trim by Stern

- Another method in getting new drafts is by using "Trimming Table" - change of draft due to add 100 MT weight;

Apply Old Draft… dF = m. Apply Old Draft… dA = m.

NEW DRAFT FWD = m. NEW DRAFT AFT = m.

* cdF100 = change of draft Forward by 100 MT.

* cdA100 = change of draft Aft by 100 MT.

Trimming with Two Compartments/Holds

Ex. 1. Loading operation almost completed with the remaining cargo of 2,000 MT. On draft check, dF = 13.15 m. and

dA = 13.55 m. Required trim at 0.28 m. Distribute the remaining cargo between Hold 1 and 7 to obtain the req'd.

trim.

* Note (Remaining cargo is based on the following):

62.35 x 100

D = Af - Ag

Wt. x Dist.

MTC x 100

Tf / Ta =

Tf / Ta =

Forward. Aft

Dep. dF = 9.150 m. Dep. dA = 10.650 m.

Sinkage = +0.430 m. +0.430 m.

NEW DRAFT AFT = 11.513 m.

Tf / Ta = -0.433 m. Tf / Ta = +0.433 m.

Fwd - cdF100 (cm.) Wt. x cdF100

NEW DRAFT FWD. = 9.147 m.

Aft - cdA100 (cm.) Wt. x cdA100Weight (MT)

Compt. (-) disch'd value from H. Table value from H. Table

5.650 0.565

(+) loaded based on 13.45m. Mdraft Wt. x cdF100

0.420 0.0382.720 0.245

based on 13.45m. Mdraft Wt. x cdF100

(nearest or round-off) 100 MT x 100cm /MT(nearest or round-off) 100 MT x 100cm /MT

-0.190 -0.015+

3.470

WBT3 +900

WBT5 +1,000 -2.270 -0.227

0.278+

7.750 0.078FWT2 +100 -4.280 -0.043

FOT2 +800

APT -120 -4.250 0.051 7.720 -0.093

9.150 10.650

9.161 11.515

I. Draft Survey

II. Shore Figure

III. Loadline (midship) Time for draft check: 30 cm. to 40 cm. to Target Draft

IV. Draft Limits


mD = dF + dA

2

= 13.150 + 13.550

2

mD = 13.350 m.

2. Excerpt from H. Table based on 13.350 m. Mean Draft:

LCF = -1.25 m.

MTC = 1,000.53 MT-m.



= 67.84 - 1.25

D = 66.59 m. LCG of Hold 7


Illustration:

W LCF = 1.25 m. L

D = 66.59m.

AP LCG = 67.84m. LCG = -84.12m. FP

DBH = 151.96 m.

4. Find the distance between holds:

DBH = LCG of Aft Hold - LCG of Fwd. Hold (if LCG is measured from midship)

= 67.84 - -84.12

DBH = 151.96 m. If LCG is measured from AP:

DBH = LCG of Fwd. Hold - LCG of Aft Hold

5. Find Present Trim:

Pt = dA - dF

= 13.55 - 13.15

Pt = 0.400 m.


Tc = Req'd Trim - Pt

= 0.280 - 0.400

Tc = -0.120 m.

7. Solve for Weight forward:

Wt. Fwd. = (D x Remaining Cargo) - (Tc x MTC x 100)

DBH

= (66.59 x 2,000) - (-0.120 x 1000.53 x 100)

151.96

D = Af - Ag

H7 H1

= 133,180 - -12,006.4

151.96

=

Wt. Fwd. = 955.400 MT (Weight for Hold 1)

8. Solve for Weight aft:

Wt. Aft = Remaining Cargo - Wt. Fwd.

= 2,000 - 955

Wt. Aft = 1,045 MT. (Weight for Hold 7)

Ex. 2. As per shore figure, the remaining cargo is 2,000 MT. On draft check, dF = 13.00m. And dA = 13.17 m.

Required trim at 0.30 m. Distribute the remaining cargo between Hold 2 and 7 to obtain the required trim.

* Note (Remaining cargo is based on the following):

I. Draft Survey

II. Shore Figure

III. Loadline (midship) Time for draft check: 30 cm. to 40 cm. to Target Draft

IV. Draft Limits


mD = dF + dA

2

= 13.000 + 13.170

2

mD = 13.085 m.


LCF = 1.08 m.

MTC = 993.53 MT-m.



= 67.84 - 1.08

D = 66.76 m. LCG of Hold 7


Illustration:

H2

W LCF = 1.08 m. L

D = 66.76m.

AP LCG = 67.84m. LCG = -59.41m. FP

DBH = 127.25 m.

4. Find the distance between holds:

DBH = LCG of Aft Hold - LCG of Fwd. Hold (if LCG is measured from midship)

= 67.84 - -59.41

DBH = 127.25 m. If LCG is measured from AP:

145,186.4

151.96

D = Af - Ag

H7

DBH = LCG of Fwd. Hold - LCG of Aft Hold

5. Find Present Trim:

Pt = dA - dF

= 13.17 - 13.00

Pt = 0.170 m.


Tc = Req'd Trim - Pt

= 0.300 - 0.170

Tc = 0.130 m.

7. Solve for Weight forward:

Wt. Fwd. = (D x Remaining Cargo) - (Tc x MTC x 100)

DBH

= (66.76 x 1,800) - (0.130 x 993.53 x 100)

127.25

= 120,168 - 12,915.89

127.25

=

Wt. Fwd. = 843.00 MT (Weight for Hold 2)

8. Solve for Weight aft:

Wt. Aft = Remaining Cargo - Wt. Fwd.

= 1,800 - 843

Wt. Aft = 957.00 MT. (Weight for Hold 7)

Change in Draft due to Change in Density

Ex. 1. Upon arrival Anchorage, drafts are dF = 12.65 m. and dA = 12.71 m. at density 1.022 MT / m³. Find the

new drafts if she shifts upriver to a density of 1.002 MT / m³.

* The tendency of shifting from higher to lower density is that draft forward will increase and draft aft decreases.


mD = dF + dA

2

= 12.650 + 12.710

2

mD = 12.680 m.


Δ = 75,492 MT

LCB = 4.64 m.

TPC = 64.65 m.

3. Find Sinkage:

Δ x (old 'd' - new 'd')

TPC x new 'd' x 100

= 75,492 x (1.022 - 1.002)

= 1,509.84

6,477.93

0.233 m.

107,252.1

127.25

Sinkage =

64.65 x 1.002 x 100

Sinkage =

4. Find new mean draft:

New mD = old mD + Sinkage

= 12.680 + 0.233

New mD = 12.913 m.

5. Excerpt from H. Table based on 12.913 m. new Mean Draft:

Δ1 = 77,001 MT simple interpolation: 12.913 m. to be interpolated, where..

LCB1 = -4.53 m. 12.91 m. from H. Table (3 to be converted into decimal = 0.3)

MTC1 = 988.62 MT-m. Δ1 = 76,982 + 0.3 x TPC

Δ1 = 76,982 + 0.3 x 64.65

Δ1 = 77,001 MT

6. Solve for corrected Trim:

Tc = Δ1 x (LCBo - LCB1)

MTC1 x 100

=

=

Tc = -0.086 m.


Tc

2

=

2

-0.043 m.

8. Sum of Drafts:

a) Rise = ( - ) , Sinkage = ( + )

b) Higher to lower density effect (Fwd draft increases (+) and Aft draft decreases (-))

Sinkage Sinkage =

Density effect

Ex. 2. Present drafts in DW density 1.018 MT / m³ are, dF = 13.26 m. and dA = 13.34 m. Find the new drafts in

density of 1.000 MT / m³.

* The tendency of shifting from higher to lower density is that draft forward will increase and draft aft decreases.


mD = dF + dA

2

= 13.260 + 13.340

2

mD = 13.300 m.


Δ = 79,519 MT

LCB = -4.35 m.

TPC = 65.05 m.

77,001 x (-4.64 - -4.53)

988.62 x 100

-8,470.110

98,862

Tf / Ta =

-0.086

Tf / Ta =

Forward Aft

Dep. dF = 12.650 m. 12.710 m.

Sinkage = +0.233 m. +0.233 m.

Tf / Ta = +0.043 m. Tf / Ta = -0.433 m.

NEW DRAFT FWD. = 12.926 m. NEW DRAFT AFT = 12.900 m.

NEW TRIM = 0.026 m.

3. Find Sinkage:

Δ x (old 'd' - new 'd')

TPC x new 'd' x 100

=

= 1,431.34

6,505.00

0.220 m.

4. Find new mean draft:

New mD = old mD + Sinkage

= 13.300 + 0.220

New mD = 13.520 m.

5. Excerpt from H. Table based on 13.520 m. new Mean Draft:

Δ1 = 80,950 MT

LCB1 = -4.25 m.

MTC1 = 1,005.07 MT-m.

6. Solve for corrected Trim:

Tc = Δ1 x (LCBo - LCB1)

MTC1 x 100

=

=

Tc = -0.080 m.


Tc

2

=

2

-0.040 m.

8. Sum of Drafts:

a) Rise = ( - ) , Sinkage = ( + )

b) Higher to lower density effect (Fwd draft increases (+) and Aft draft decreases (-))

Sinkage Sinkage =

Density effect

Interpolation (Sounding)

Ex. Sounding of DBT # 5 is 1.78 m. with a density of 1.025 MT / m³. Find the weight of the said ballast.

1. Convert the sounding from meter to cm. (table sounding is by cm.):

1.78 m. x 100 cm.

1 m.

2. From sounding table, apply the figures lower and higher from the value to be converted:

8.7

3

5 14.5

Sinkage =

79,519 x (1.018 - 1.000)

65.05 x 1.000 x 100

Sinkage =

80,950 x (-4.35 - -4.25)

1,005.07 x 100

-8,095.000

100,507

Tf / Ta =

-0.080

Tf / Ta =

Forward Aft

Dep. dF = 13.260 m. 13.340 m.

Sinkage = +0.220 m. +0.220 m.

Tf / Ta = +0.040 m. Tf / Ta = -0.040 m.

NEW DRAFT FWD. = 13.520 m. NEW DRAFT AFT = 13.520 m.

NEW TRIM = 0.000 m.

= 178 cm.

Sounding Volume

(in cm.) (in m³)

175 395.90

178 X

180 410.40

3 X

5 14.5

5X = 43.5

X = 43.5 / 5

X = volume is increasing, therefore add 8.7 to 395.90 m³ = 404.60 m³

3. Solve for Weight:

Wt. = Vol. x Density

= 404.6 x 1.025

Wt. = 414.715 MT

Draft Survey (Initial)

* Before anything else, order a responsible crew to take soundings of all tanks, freshwater, ballast, bilges and

others. Fuel oil tank soundings are generally made by the Engine Dept. Also obtain a hydrometer reading(S.G.)

the ship is floating.

Ex. 5.32 m. 5.32 m.

Deductibles:

Ballast = 20,900 MT (total wt. after sounding)

5.87 m. 5.86 m. F.W. = 200 MT

F.O. = 1,600 MT

D.O. = 150 MT

TOTAL = 22,850 MT

6.44 m. 6.44 m. Density = 1.015 MT / m³

* Solving the midship draft when using a sounding tape;

I. Summer Draft + Summer Freeboard = Ht. from Keel to Deckline

II. Ht. from Keel to Deckline - Dist. from Water from Deckline (by sounding tape) = MIDSHIP DRAFT

1. Draft at marks and mean draft:

F M A

Port --------- 5.32 m. 5.87 m. 6.44 m.

Stbd -------- 5.32 m. 5.86 m. 6.44 m.

Mean Draft = 5.320 m. 5.865 m. 6.440 m.

* Actual drafts are drafts on the Fwd / Aft Perpendicular (not the drafts on the draft marks).

2. Find apparent trim:

aT = dA - dF

= 6.440 - 5.320

=

8.7 m³ . . . .

aT = 1.120 m. (+, trimmed by the Stern)

Illustration:

mAP = 12.40 m. mFP = 11.25 m.

W L

DBM (Dist. bet. Marks) = 192.350 m.

AP FP

3. Find stem and stern correction:

a) Stem Corr'n = b) Stern Corr'n =

= 1.120 x 11.25 = 1.120 x 12.40

= =

Stem Corr'n = 0.066 m. Stern Corr'n = 0.072 m.

4. Find the actual drafts (drafts on the fwd/aft perpendicular):

a) dFP = dF - Stem Corr'n. (subtract bec. FP is higher than draft marks fwd esp. when trim is 'by stern')

= 5.320 - 0.066

dFP = 5.254 m.

b) dAP = dA + Stern Corr'n. (add bec. AP is lower than draft marks aft esp. when trim is 'by stern')

= 6.440 - 0.072

dAP = 6.512 m.

5. Find corrected/true trim:

cT = dAP - dFP

= 6.512 - 5.254

cT = 1.258 m. (true trim - in which correction already applied)

6. Find true mean draft:

True mD =

= 5.254 + 6.512

2

True mD = 5.883 m.

7. Find quarter mean draft:

Qm = (3 x dM) + True mD

4

= (3 x 5.865) + 5.883

4

= 23.478

4

Qm = 5.870 m.

LBP = 216.00 m.

aT x mFP aT x mAP

DBM DBM

192.35 192.35

12.600 13.888

192.35 192.35

dFP + dAP

2

8. As per Qm, look for the Displacement (Δ) in the H. Table:

In the table, draft available is only 5.86 m., therefore make the remaining digits into decimal ….

Qm = 5.86.95 TPC

Δ = 33,176 + (0.95 x 59.68)

Δ = 33,233 MT

TPC = 59.68 m.

LCF = -6.14 m.

9. Find the MTC Difference:

a) MTC at Qm + 0.50 = 5.87 + 0.50

MTC at Qm + 0.50 = 6.37 m. (proceed to H. Table)

6.37m. MTC = 790.98 MT-m.

b) MTC at Qm - 0.50 = 5.87 - 0.50

MTC at Qm - 0.50 = 5.37 m. (proceed to H. Table)

5.37m. MTC = 773.59 MT-m.

c) MTC Diff. = 790.98 - 773.59

MTC Diff. = 17.39 MT-m. (or dM/dZ = diff. in MTC per 1 m. draft)

10. Obtain trim corrections for Displacement:

a) 1st Corr'n = b) 2nd Corr'n = 50 x MTC Diff. x cT²

LBP

= 1.258 x 59.68 x -6.14 x 100 = 50 x 17.39 x 1.258²

= =

1st Corr'n = -213.414 MT 2nd Corr'n = 6.371 MT

Note: The 1st Corr'n can either be a plus (+) or minus (-) depending upon the location of the center of flotation

(LCF) and trim condition;

I. Vessel trimmed "By the Stern"

- LCF is fwd . . . . . Subtract

- LCF is aft . . . . . . Add

II. Vessel trimmed "By the Head"

- LCF is fwd . . . . . Add

- LCF is aft . . . . . . Subtract

11. Find Displacement at SW density:

Δ1.025 = Δ + 1st Corr'n + 2nd Corr'n

= 33,233 + -213.414 + 6.371

Δ1.025 = 33,026.00 MT

12. Find the new displacement:

New Δ = Δ1.025 x new 'd'

=

=

New Δ = 32,703.800 MT

13. Solve for Cargo / Constant (stores, crew, other small tanks, etc.)

Cargo / Constant = Δ - Lship - Deductibles

= 32,703.800 - 9,502.000 - 22,850.000

Cargo / Constant = 351.800 MT or 352.000 MT

cT x TPC x LCF x 100

LBP

216.00 216.00

-46,097.55 1,376.039

216.00 216.00

1.025

33,026.00 x 1.015

1.025

33,521.390

1.025

Note: If Cargo = 0.00 MT (no cargo yet), Constant = 352.00 MT

If Constant is known already, subtract the value of constant from the above formula for the cargo loaded.

Draft Survey (Final)

* Before anything else, order a responsible crew to take soundings of all tanks, freshwater, ballast, bilges and

others. Fuel oil tank soundings are generally made by the Engine Dept. Also obtain a hydrometer reading(S.G.)

where the ship is floating.

Ex. 12.48 m. 12.48 m.

Deductibles:

Ballast = 80 MT (total wt. after deballasting)

13.50 m. 13.50 m. F.W. = 270 MT

F.O. = 1,600 MT

D.O. = 100 MT

TOTAL = 2,050 MT

13.96 m. 13.96 m. Density = 1.016 MT / m³

* Solving the midship draft when using a sounding tape;

I. Summer Draft + Summer Freeboard = Ht. from Keel to Deckline

II. Ht. from Keel to Deckline - Dist. from Water from Deckline (by sounding tape) = MIDSHIP DRAFT

- Find the cargo loaded using the Constant 352.00 MT at initial survey.

1. Draft at marks and mean draft:

F M A

Port --------- 12.98 m. 13.50 m. 13.96 m.

Stbd -------- 12.98 m. 13.50 m. 13.96 m.

Mean Draft = 12.98 m. 13.50 m. 13.96 m.

* Actual drafts are drafts on the Fwd / Aft Perpendicular (not the drafts on the draft marks).

2. Find apparent trim:

aT = dA - dF

= 13.960 - 12.980

aT = 0.980 m. (+, trimmed by the Stern)

Illustration:

mAP = 12.40 m. mFP = 11.25 m.

W L

DBM (Dist. bet. Marks) = 192.350 m.

AP FP

3. Find stem and stern correction:

a) Stem Corr'n = b) Stern Corr'n =

= 0.980 x 11.25 = 0.980 x 12.40

= =

Stem Corr'n = 0.057 m. Stern Corr'n = 0.063 m.

4. Find the actual drafts (drafts on the fwd/aft perpendicular):

a) dFP = dF - Stem Corr'n. (subtract bec. FP is higher than draft marks fwd esp. when trim is 'by stern')

= 12.980 - 0.057

dFP = 12.923 m.

b) dAP = dA + Stern Corr'n. (add bec. AP is lower than draft marks aft esp. when trim is 'by stern')

= 13.96 - 0.063

dAP = 14.023 m.

5. Find corrected/true trim:

cT = dAP - dFP

= 14.023 - 12.923

cT = 1.100 m. (true trim - in which correction already applied)

6. Find true mean draft:

True mD =

= 12.923 + 14.023

2

True mD = 13.473 m.

7. Find quarter mean draft:

Qm = (3 x dM) + True mD

4

= (3 x 13.50) + 13.473

4

= 53.973

4

Qm = 13.493 m.

8. As per Qm, look for the Displacement (Δ) in the H. Table:

In the table, draft available is only 13.49 m., therefore make the remaining digit into decimal ….

LBP = 216.00 m.

aT x mFP aT x mAP

DBM DBM

192.35 192.35

11.025 12.152

192.35 192.35

dFP + dAP

2

Qm = 13.49.3 TPC

Δ = 80,755 + (0.3 x 65.17)

Δ = 80,775 MT

TPC = 65.17 m.

LCF = 1.33 m.

9. Find the MTC Difference:

a) MTC at Qm + 0.50 = 13.49 + 0.50

MTC at Qm + 0.50 = 13.99 m. (proceed to H. Table)

13.99m. MTC = 1,017.38 MT-m.

b) MTC at Qm - 0.50 = 13.49 - 0.50

MTC at Qm - 0.50 = 12.99 m. (proceed to H. Table)

12.99m. MTC = 990.81 MT-m.

c) MTC Diff. = 1,017.38 - 990.81

MTC Diff. = 26.57 MT-m. (or dM/dZ = diff. in MTC per 1 m. draft)

10. Obtain trim corrections for Displacement:

a) 1st Corr'n = b) 2nd Corr'n = 50 x MTC Diff. x cT²

LBP

= 1.100 x 65.17 x 1.33 x 100 = 50 x 26.57 x 1.100²

= =

1st Corr'n = 44.141 MT 2nd Corr'n = 7.442 MT

Note: The 1st Corr'n can either be a plus (+) or minus (-) depending upon the location of the center of flotation

(LCF) and trim condition;

I. Vessel trimmed "By the Stern"

- LCF is fwd . . . . . Subtract

- LCF is aft . . . . . . Add

II. Vessel trimmed "By the Head"

- LCF is fwd . . . . . Add

- LCF is aft . . . . . . Subtract

11. Find Displacement at SW density:

Δ1.025 = Δ + 1st Corr'n + 2nd Corr'n

= 80,775 + 44.141 + 7.442

Δ1.025 = 80,826.583 MT

12. Find the new displacement:

New Δ = Δ1.025 x new 'd'

=

=

New Δ = 80,116.886 MT

13. Solve for weight of Cargo Loaded:

Cargo Loaded = Δ - Lship - Deductibles - Constant

= 80,116.886 - 9,502.000 - 2,050.000 - 352.00

Cargo Loaded = 68,212.886 MT or 68,213.000 MT

Note: If Constant is known already, subtract the value of constant from the above formula for the cargo loaded.

cT x TPC x LCF x 100

LBP

216.00 216.00

9,534.37 1,607.485

216.00 216.00

1.025

80,826.58 x 1.016

1.025

82,119.808

1.025

Loading Sequence

* Normally, when a vessel is empty, the condition is “Hogging” because FPT and APT may be full. Therefore,

cargo holds within the midship section is technically appropriate to be loaded first.

Ex.1. From Haypoint to Kawasaki, cargo to be loaded is Iron Ore with an SF of 22 ft³ / LT, total weight is

179,000 MT. Trimming weight = 4,000 MT and trimming holds are Hold # 1 and Hold # 9.

Ship's Info.:

Ballast Pump rate = 2 x 2,500 = 5,000 MT / Hr. Note: when sounding is 0.35 m.,deballast-

Ballast stripping / Eductor = 160 MT / Hr. ing must be done by eductor.

Port Info.:

Loader Rate = 1 x 7,000 = 7,000 MT / Hr.

* 2 passes per hold except trimming

Limits:

SF = 100 % and BM = 100 %

Trim by Stern = 7.0 m.

Trim by Head = 2.5 m.

* Execute then the following distribution of weights in Loading Computer and take note of drafts/trim,shear forces

and bending moments as well as ballast operations ...

Ship: Load/Disch. Port: Max. Draft Avail (HW): Max. Air Draft in Berth:

Haypoint

Date: Max Sailing Draft: Min. Draft Avail (HW): Dockwater Density:

18.76 m.

Ass. SF of Cargo: No. of Loaders/Dischargers: Ballast Pumping Rate: Load/Disch Rate:

22 ft³ / LT 1 5,000 tph 7,000 tph

cargo loaded/load rate

Time

1.025

9 8 7 6 5 4 3 2 1

0.0 MT 0.0 MT 0.0 MT

33,000.1 MT 37,200.0 MT 33,000 MT38,600 MT

0.0 MT

37,200 MT

Calculated Values

Pour Cargo Req'd Air Draft

No. Hold (Hrs) Fwd Aft BM % SF% Draft Mid Trim

1 5 2.86 6.62 10.72 P:28.4 P:39.3 8.51

2 1 2.57 9.85 8.86 P:49.9 P:60.9 9.39

3 7 2.86 9.31 10.81 P:55.2 P:65.1 10.01

4 3 3.57 11.53 10.53 P:27.5 P:37.4 11.06

5 9 3.43 9.37 15.49 P:77.5 P:67.8 12.26

6 5 2.66 11.35 16.59 P:41.5 P:81.8 13.87

7 3 1.74 14.01 15.86 P:42.6 P:91.8 14.89

8 7 2.46 14.05 18.72 P:45.2 P:83.1 16.35

9 1 1.86 18.11 16.72 P:25.1 P:50.5 17.42

10 9 1.00 17.45 18.54 P:32.3 P:54.3 17.99

11 1 0.29 18.07 18.25 P:40.1 P:61.2 18.16

12 9 0.29 17.88 18.76 P:42.6 P:62.2 18.32

%



PROV PROVISION 0

TOTAL CONST.=


FOT1 NO.1 FOT (P/S)

FOT2 NO.2 FOT (P/S)

FOT3 NO.3 FOT (P/S)

D. FOT DEEP FOT


TOTAL F.O. =

D. DOT DEEP DOT 96


TOTAL D.O. =

FWT1 NO.1 FWT 100

FWT2 NO.2 FWT 100

TOTAL F.W. =

FPT FORE PEAK TK. 0






TTL. BALLAST =



Ballast Operations Draft Maximum

(in MT) Comments20,000 [PO] WBT 3 (P/S)

Weight

18,000 [PO] FPT, WBT 1 (P/S)

20,000 [PO] WBT 4 (P/S)

25,000 [PO] WBT 2 (P/S)

24,000 [PO] WBT 5 (P/S)

18,600

12,200

17,200

13,000

7,000

2,000 Fwd Trimming

2,000 Final Trimming

L.MOMENT K.G. V.MOMENT

(MT-m.) (MT-m.)

Total 179,000


WEIGHT L.C.G.

fr. H. Table

FSM ( I )

ITEM (MT) (m.) (MT-m.)

0

24,840 16.30


6,520

Wt. x KG

400 62.10

0 0.00 0 0.00

0 0.00 0 0.00

400

0

22,935 3,2601,000 95.87 95,870 22.94

1,600 104.47 167,144 18.62

947 11

380 115.30 43,812 15.45

29,795 1,308

5,869 1,859

947 11

43 110.10

43 106.50 4,580 22.03

4,734 22.03

3,066

2,554 91

212 1

117 124.37

10 124.95 1,250 21.22

14,551 21.83

127

7,588 0

7,588 0

335 131.53

335 131.53 44,023 22.67

44,023 22.67

669

0 0

0 0

0 -132.81

0 -113.43 0 10.49

0 6.78

0 -76.37 0 10.88

0 0

0 -25.76 0 10.85

0 0

0 0

0 0

0 24.95

0 72.40 0 11.36

0 10.96

(Cargo Hold Heeling Mom.


0

33,000 -114.44 -3,776,520 13.98

0 0.00 0 0.00

461,373

0








TOTAL CARGO =

LIGHTWEIGHT


DISPLACEMENT

From Preplan:

Δ = 204,825 MT

LCG = -8.22 m.

KG = 13.95 m.



DCF = 18.322 m.

LCB = -9.309 m.

LCF = -0.333 m.

MTC = 2,528.934 MT-m.

TKM = 19.112 m.

TPC = 122.622 m.

1. Solve for Trim:


MTC x 100

= 204,825 x (-8.221 - -9.309)

2,528.934 x 100

= 222,849.6

252,893.4


- = By Head


DFP = DCF- t

2

= 18.322 - 0.881

2

= 18.322 – 0.441

DFP = 17.882 m.


DFP = 17.882 m.

t = +0.881 m.

DAP = 18.763 m.


FSC = Total FSM

510,161

0

37,200 -64.37

0 0.00 0 0.00

-2,394,564 13.71

38,600 -13.13 -506,818 13.79

510,533

0 0.00 0 0.00

0.00

1,382,352 13.72

532,101

0

0

33,000 87.69 2,893,770

37,200 37.16

0 0.00 0

179,000

477,18014.46


183,262 -1,956,953 14.06 2,576,303 (total moments-10.68

in comprnts

21,563 12.66 272,988 13.03 280,966with load)


204,825 -8.22 -1,683,966 13.95 2,857,269


Δ

=

FSC = 0.032 m. (or GGo)

5. Solve for GM: M

KG = 13.950 m.

GGo = 0.0320 m. Go GM


KM = 19.112 m. KGo


K

* Best "GM" depends on the type of vessel and ship's construction & as long as the vsl will be on the Safe side.

Ex.2. In MV Olga from Santos, Brazil to Bizerte, Tunisia, cargo to be loaded is Raw Sugar (SF of 43 ft³ / LT).

Total weight = 14,700 MT, trimming weight = 400 MT and trimming holds are Hold # 1 and Hold # 4.

Ship's Info.:

Ballast Pump rate = 380 MT / Hr. Note: when sounding is 0.80 m.,deballast-

Port Info.: ing must be done by eductor.

Loader Rate = 1 x 2,000 = 2,000 MT / Hr.

Limits:

SF = 100 % and BM = 100 %

* Execute then the following distribution of weights in Loading Computer and take note of drafts/trim, shear forces

and bending moments as well as ballast operations ...

Ship: Load/Disch. Port: Max. Draft Avail (HW): Max. Air Draft in Berth:

Santos

Date: Max Sailing Draft: Min. Draft Avail (HW): Dockwater Density:

9.00 m.

Ass. SF of Cargo: No. of Loaders/Dischargers: Ballast Pumping Rate: Load/Disch Rate:

1.198 m³ / MT 1 380 tph 2,000 tph

cargo loaded/load rate

Time

Pour Cargo Req'd Air Draft

No. Hold (Hrs) Fwd Aft BM % SF% Draft Mid Trim

1 3 1.00 4.24 5.74 36 % 36 % 8.38 4.99 1.50

2 1 1.50 4.93 5.48 37 % 37 % 8.16 5.21 0.55

3 4 1.00 4.59 6.53 39 % 38 % 7.81 5.56 1.94

4 2 1.00 5.53 6.36 36 % 30 % 7.42 5.95 0.83

5 3 1.00 6.06 7.35 22 % 24 % 6.66 6.71 1.29

6 2 1.00 7.04 7.57 20 % 31 % 6.06 7.31 0.53

7 4 0.75 6.58 9.11 14 % 17 % 5.52 7.85 2.53

6,542

204,825

KG

MV Olga

1.017

9 8 7 6 5 4 3 2 1

3900 MT

2800 MT

Ballast Operations Draft Maximum

4000 MT

4000 MT

(in MT) Comments2,000 [PO] WBT 3 (P/S)

Calculated Values

Weight

2,000 [PO] FPT 50%, WBT 1 (P/S)

2,000 [PO] WBT 4 (P/S)

2,000 [PO] WBT 2 (P/S)

2,000

2,000 [PO] FPT

1,500

8 1 0.40 7.72 8.59 15 % 19 % 5.21 8.16 0.87

Note: 400 MT Balance for trimming

7.65

%

CONST.1 MEN & EFFECTS 0

CONST.2 STORES 0

CONST.3 OIL & W.IN E/R 0

PROV PROVISION 0

TOTAL CONST.=


FOT1 NO.2 DBFOT (P/S) 95

FOT2 NO.3 DBFOT (P/S) 91

FOT3 NO.4 DBFOT © 44

FO OT FO O'FLOW T (S) 1

TOTAL F.O. =

LO 1 LO ½ STOR. T. 30

LO 2 CYL..O. STOR. T. 79

LO 3 G/E LO STOR. T. 19

TOTAL L.O. =

D. DOT NO.5 DOT (P) 15

D.O.T. DOT (P/S) 12

TOTAL D.O. =

FWT1 FWT (P/S) 44

FWT2 APT © 71

TOTAL F.W. =

FPT FORE PEAK TK. 39

WBT1 NO.1 WBT (P) 3

WBT1 NO.1 WBT (S) 5




TTL. BALLAST =


800

Total 14,300


(MT-m.) (MT-m.) (MT-m.)

WEIGHT MID. G / LCG L.MOMENT K.G.

fr. H. Table

V.MOMENT FSM ( I )

ITEM (MT) (m.)

172

152 17.30


52

Wt. x KG

3.0 50.61

93.8 53.51 5,019 7.07

13.0 7.68 100 13.24

6.0 68.31 410 13.53

115.8

663 26

81

5,681

388 -20.75 -8,053 0.64

30 1,125

372 5.50 2,046 0.61

248 927

227 927

0 9

102 28.98

0 48.00 14 0.02

2,956 0.29

862 -3,037

43 4

219 12

4 56.15

22 51.95 1,122 10.13

247 9.78

3 0

26 1,386

0 54.48 16 9.37

4 48.03 187 0.17

43 59.05 2,563 8.34

1 19

362 50

47 2,750

43 63.39 2,738 9.65

217 68.50 14,858 9.67

417 46

2,097 1,655

260 17,596

360 -62.38 -22,432 3.31

2 310

14 -44.50 -601 0.05

1,190 156

1 269

19 443

25 -44.41

92 -20.55 -1,884 0.21

-1,124 0.09

246 5.57 1,372 0.50

242 31.83 7,693 0.42

-16,975

123 658

102 1,812

-121,578 5.61

(Cargo Hold Heeling Mom.


978

15,080 02,688 -45.23




TOTAL CARGO =

LIGHTWEIGHT


DISPLACEMENT

From Preplan:

Δ = 21,294 MT

LCG = -0.41 m.

KG = 5.96 m.



DCF = 8.68 m.

LCB = -1.11 m.

LCF = 2.22 m.

MTC = 247.67 MT-m.

TKM = 9.51 m.

TPC = 27.46 m.

1. Solve for Trim:


MTC x 100

= 21,294 x (-0.41 - -1.11)

2,47.67 x 100

=14,905.6

24,767.0


- = By Head


DFP = DCF- t

2

= 8.67 - 0.602

2

= 8.67 – 0.301

DFP = 8.369 m.


DFP = 8.369 m.

t = +0.602 m.

DAP = 8.971 m.


FSC = Total FSM

Δ

= 8,448

21,294

4,016 -21.11 -84,778 5.57 22,369 0

3,996 5.12 20,460 5.53

4,000 31.26 125,040 5.77

22,098 0

23,080 0

14,700 -60,856


16,990 -3.15 -53,456 5.22 88,679 (total moments

in comprnts

4,304 10.41 44,805 8.90 38,306with load)


21,294 -0.41 -8,651 5.96 126,984


FSC = 0.397 m. (or GGo)

5. Solve for GM: M

KG = 5.960 m. (KG SOLID )

GGo = 0.397 m. Go GM


TKM = 9.510 m. KGo


K

* Best "GM" depends on the type of vessel and ship's construction & as long as the vsl will be on the Safe side.

Stability

. The state of resisting change or of tending to return to original conditions after being disturbed.

M

Freeboard = Reserved Buoyancy

G

B Draft = Intact Buoyancy

K

Definition of Terms:

1. Heel - inclination caused by external forces.

2. List - inclination caused by internal forces.

3. Center of Gravity (G) - center of all download forces.

> it moves toward the weight loaded.

> it moves away from the weight discharged.

> it moves parallel to the direction of weight being shifted.

4. Center of Buoyancy (B) - center of all upward forces, before heeling (B1 is the COB after heeling).

> it moves toward the low side of an inclined vessel.

5. Metacenter (M) - limit where 'G' permits to have positive stability.

> position varies with the vessel's draft. ('meta' means change, so it’s a moving center)

6. Keel (K) - the baseline.

7. KB - Height of center of buoyancy above the keel.

8. KG - Height of center of gravity above the keel.

9. KGv - Virtual height of vessel's center of gravity above the keel.

> obtained by adding the correction for liquid free surfaces, and if applicable, the correction for vertical

shifting moments to the KG.

10. KM - Height of metacenter above the keel.

11. GM - metacentric height. Allowable GM = 30 cm. or 0.30 m.

12. BM - metacentric radius.

13. GZ - Righting Arm (lever). The horizontal distance between the force of buoyancy acting upwards through B

and the force of gravity acting downwards through G.

14. Righting Moment (RM) - the product of righting arm (GZ) multiplied by the Displacement (weight) of the vessel.

15. Moment - the product of a weight multiplied by a distance.

16. Heeling Moment - the moment resulting from a traverse shift of weight through a given distance which tends to

heel a vessel. Expressed as TON-METERS or FT-LONG TONS.

17. Volumetric Heeling Moment - product of a volume multiplied by a transverse distance. Expressed as M4 or FT

4.

Converted to Heeling Moment by dividing by the stowage factor (or multiplying by the density) of cargo.

18. Vertical Center of Gravity (VCG)- height of the center of gravity of a cargo compartment or tank above the keel.

19. Longitudinal Center of Buoyancy (LCB) - Longitudinal distance of center of buoyancy from midships. (some-

KG

WL

times measured from after perpendicular)

20. Longitudinal Center of Gravity (LCG) - Longitudinal distance of center of gravity from midships. (sometimes

measured from after perpendicular)

Equilibrium of Ships

1) Stable Equilibrium (positive GM) - 'G' is below 'M' / KG < KM. And Righting Lever (GZ) is on the lower part.

M

M

G G Z

B B B1

K

K

2) Neutral Equilibrium (GM = 0) - 'G' coincides with 'M' / KG = KM.

M G M G

B B B1

K

K

3) Unstable Equilibrium (negative GM) - 'G' is above 'M' / KG > KM. And Righting Lever is on the upper part.

G

Z G

M M

B B B1

K

K

List Caused by Negative (-) GM:

1. Removal of low weights

2. Addition of high weights

3. Shifting of weights upward

4. Due to free surface

Negative (-) GM can be recognized when,

1. Vessel will not remain upright and has a list either port or starboard.

2. Vessel flops to port or starboard.

3. Vessel posses a very long and slow rolling period.

CORRECTIVE ACTIONS

* It is SAFE to assume that the List is due to "negative GM" rather than due to an uneven weight distribution.

1. Press up all slack tanks located directly below the ship's Center of Gravity (COG).

2. And if this tank mentioned in No.1 is not on the centerline of the ship, then fill up the lower side tank first and

when Full, its counter part tank on the higher side. This is to avoid great effect of Free Surface.

3. Repeat action No.2 with another tank and so on until the ship becomes Stable.

4. If discharging or jettisoning of cargo is required, so then start from the higher side to lower side.

Stiff and Tender Vessel

1. vessel with an unduly large GM for her type, size 1. vessel with a small GM for her type, size and nature.

and nature.

2. angle and period of roll is small. 2. angle and period of roll is large.

3. rolling is violent and irregular. 3. rolling is smooth and regular.

4. has jerky movement, uncomfortable for people on 4. less uncomfortable for people onboard and if gen. cargo

board. Gen. cargo likely to break loose. is secured properly is less likely to break loose.

5. severe racking stresses on the hull. 5. less severe racking stresses on the hull.

6. bulk cargo less likely to shift as angle of roll is 6. bulk cargo more likely to shift as angle of roll is large.

small.

7. less possibility that ship becomes unstable due to 7. more possibility that ship becomes unstable due to

consumption of fuel or freshwater from DB Tanks consumption of fuel or freshwater from DB Tanks and

and also due to free surface effect of tanks in use. also due to free surface effect of tanks in use.

8. greater ability to withstand loss of GM, if any, 8. less ability to withstand such loss of GM, if any, caused

caused by bilging. by bilging.

9. greater ability to withstand transverse shift of 9. less ability to withstand transverse shift of cargo - list

cargo - list caused by such shift is small. caused by such shift is large.

List / Trim Correction

* LIST is the transverse inclination caused when the Center of Gravity (COG) of the ship is off the centerline.

(i) For a ship in static equilibrium, the COB and COG must be in a vertical line;

G

B

K

(ii) And when COG is moved out from the centerline of the ship, due to asymmetrical loading / discharging or

transverse shift of weights;

Wt.

G G1

B

K

(iii) Then the forces of buoyancy and gravity will form a couple which will cause the ship to incline until COB

comes vertically below the COG;

Stiff Vessel Tender Vessel

M Wt.

List (θ)

G G1

B B1

K

>> where in the Right Triangle, MGG1 or Angle of List (θ) …

Tan θ (List) = GG1 where: GG1 is the transverse shift of COG and,

GM GM is the final GM before listing.

and since GG1 = dw / Δ, the formula becomes

Tan θ (List) = d x w where: dw is the final listing moment

Δ x GM Δ is the final displacement and,

GM is the final GM

Ex. 1. On a ship of Δ 8,000 MT, KG 7.0 m., KM 7.5 m., 100 MT of cargo is loaded on the upper deck (KG 9.2 m.)

2.0 m. Port from the centerline. Find the list.

1. Find the final listing moment:

f.l.m. = dist. x wt.

= 2 x 100 (to port)

f.l.m. = 200 MT-m.

2. Find the final displacement (Δ):

Final Δ = 8,000 + 100

Final Δ = 8,100 MT

3. Find the final GM:

a) GG1 ↑ = d x w (listing moment)

= 2 x 100

GG1 ↑ = 0.025 m.

b) KG = 7.000 m.

GG1 ↑ = 0.025 m.

final KG = 7.025 m.

KM = 7.500 m.

GM = 0.475 m.

4. Find the list:

Tan θ = d x w

Δ x GM

= 200

=

Tan θ = (inv. tan)

List = 2.98°

Ex. 2. A ship of Δ 10,000 MT, KM 9.20 m., KG 8.30 m. and MTC of 200 MT-m., is listed 4˚ to port and trimmed

0.10 m. by the head. It is desired to bring the ship upright and trim 0.80 m. by the stern by shifting ballast

between No.1 DBT P/S (400 MT each) and No.5 DBT P/S (empty). The COG of each tank is 8.0 m. off the

Δ

8,100

8,100 x 0.475

200

3,487.50

0.052

centerline of the ship. The distance between the centers of No.1 and No.5 tanks is 95.0 m. Find how much

ballast should take place between the tanks and the final distribution, neglecting free surface correction.

- To bring the ship to the required trim,

1. Find the change of trim (cot):

Present Trim = 0.10 m. by the head

Required Trim = 0.80 m. by the stern

cot = 0.90 m. by the stern (or 90 cm.)

2. Find the trimming moment:

Trimming Moment = cot x MTC

= 90 x 200

Trimming Moment = 18,000 MT-m. by stern

3. Find the weight to shift:

Trimming Moment = w x d

therefore, w = trimming moment

d

w = 189.5 MT

- To bring the ship upright,

1. Find the transverse shift of COG:

Tan θ = GG1

GM

therefore, GG1 = Tan θ x GM

= Tan 4° x 0.90

GG1 = 0.0623 m.

2. Find the listing moment:

GG1 ↑ = d x w (listing moment)

therefore,

Listing Moment = Δ x GG1

= 10,000 x 0.0623

Listing Moment = 623. 0 MT-m.

Try using W = GM . D / Displ. X tan List

Free Surface

. The space where the liquid freely moves.

Free Surface Effect

If a tank is full of liquid, there is no movement of the liquid, where it can be treated exactly the same way as

any other weight on board concentrated at its center of gravity.

And when a vessel with a slack (partly full) tank rolls at sea, the surface of the liquid possesses Inertia, which

would move towards the lower side during each roll, thereby causing the angle / period of roll to increase. Then

because the vessel behaves as if her GM has been reduced, it is considered that a slack tank causes a virtual

(imaginary) loss of GM. This is then the Free Surface Effect.

Formulas for Free Surface Correction (FSC) / Constant (FSK) is used for computing the said loss of GM.

FSC = where:

r =

= 18,000

95.0

Δ

r L B³

420 x ∆ S.G. of Liquid Cargo in the Tank

S.G. of Water where Vsl Floats

FSK = L B³ L = length of compartment

420 B = breadth of compartment

FSC = FSK

∆

and, where:

Virtual Loss of GM V = ship's volume of displacement

d1 = S.G. of Liquid Cargo in the Tank

d2 = S.G. of Water where Vsl Floats

Ex.1. The liquid mud tanks on your vessel measures 11.00 m. length, 8.00 m. breadth by 2.83 m. depth. The

vessel's displacement is 3,044 MT and the S.G. of the mud is 1.90. What is the reduction in GM due to two

of these tanks being slack?

r = =

r =

FSC =

=

420 x 944

=

FSC = 0.026 m. (x 2 Tanks)

FSC = 0.053 m.

Ex.2. A ship of 10,000 MT ∆ is floating in dockwater od density 1.024 kg / m³ and is carrying an oil of density

0.84 in a double bottom tank. The tank is 25 m. long, 15 m. wide and is divided at the centerline. Find the

loss of GM due to this tank being slack?

Virtual Loss of GM =

=

12 x (10,000 / 1.024) x 1.024

=

(Volume = ∆ / Density)

= 0.591 m. (x 0.25 m. centerline)

Virtual Loss of GM = 0.148 m.

Intact Stability

Illustration:

M

θ where:

assumed KG = 0G Z KX = KG x sin θ

θ XN = KN - KX

since, GZ = XN …K x N

= L B³ x d1

12 x V x d2

S.G. of Liquid Cargo in the Tank 1.90

S.G. of Water where Vsl Floats 1.025

1.854

r L B³

420 x ∆

1.854 x 11.00 x 8.00³

10,441.73

396,480

L B³ x d1

12 x V x d2

25 x 15³ x 0.84 kg/m³

70,875

120,000

Ex.1. From Preplan 1: * KN figure to be taken from "Stability

Δ = 72,531 MT Cross Curves Table" as per Displace-

LCG = -3.55 m. ment against Heel Angles.

KG = 10.55 m. (KGSOLID)



DCF (Draft at Center of Flotation) = 12.22 m.

LCB = -4.85 m.

LCF = 0.42 m.

MTC = 968.01 MT-m.

TKM = 13.30 m.

* KGSOLID = 10.550 m.

FSC/GGo= 0.278 m.

KG = 10.828 m.

TKM = 13.300 m.

GM = 2.472 m.

(i) GZ = KN - (KG x sin θ) (v) GZ = KN - (KG x sin θ)

= 2.328 - (10.828 x sin 10°) = 10.162 - (10.828 x sin 50°)

GZ1= 0.447 m. (@ 10° Heel) GZ5= 1.867 m. (@ 50° Heel)

(ii) GZ = KN - (KG x sin θ) (vi) GZ = KN - (KG x sin θ)

= 4.704 - (10.828 x sin 20°) = 10.830 - (10.828 x sin 60°)

GZ2= 1.000 m. (@ 20° Heel) GZ6= 1.453 m. (@ 60° Heel)

(ii) GZ = KN - (KG x sin θ) (vii) GZ = KN - (KG x sin θ)

= 7.044 - (10.828 x sin 30°) = 10.976 - (10.828 x sin 75°)

GZ3= 1.630 m. (@ 30° Heel) GZ7= 0.517 m. (@ 75° Heel)

(iv) GZ = KN - (KG x sin θ) (viii) GZ = KN - (KG x sin θ)

= 8.912 - (10.828 x sin 40°) = 10.268 - (10.828 x sin 90°)

GZ4= 1.952 m. (@ 40° Heel) GZ8= -0.560 m. (@ 90° Heel)

* Heel in degrees (1 Radian) = 57.3° (constant) = 180 / PI (3.1416)

* Plot the said figures to acquire Curve of Intact Stability (GZ against every heel angle).

Curve of Intact Stability

GZ (m)

2.50 GM against constant 57.3°2.47 GM

Max. GZ = 42°

2.00

1.95 m.

1.87 m.

1.63 m.

1.50

1.45 m.

Angle for Max. GZ

1.00 1.00 m. is the guide for

GZ = KN - (KG x sin θ)

making a decision

to alter / weather

course.

0.50 0.52 m.

0.45 m.

1st separation (7°) is the

Initial Stability

0.00

10° 30° 50° 75°

Heel Angle (Heel in degrees)

* Establish then the Righting Levers (GZ) for every station or common interval (e.g. 0.3 m. at 7.5°, 0.73 m. at 15°

heel up to 30° heel) inorder to acquire Dynamical Stability at any angle of heel by SIMPSON'S RULE.

Ex. 2. Construct the GZ Curve of a ship displacing 72,531 MT, virtual KG 10.777 m. Calculate Dynamical Stability

at 30° and 40° heel. From KN curves, KN values for 73,000 ∆ (rounded-off) are obtained;

θ KN - (KG x sin θ) = GZ

10° 2.328 m. 10.777 x sin 10° 0.456 m.

20° 4.704 m. 10.777 x sin 20° 1.018 m.

30° 7.044 m. 10.777 x sin 30° 1.656 m.

40° 8.912 m. 10.777 x sin 40° 1.985 m.

50° 10.162 m. 10.777 x sin 50° 1.906 m.

60° 10.830 m. 10.777 x sin 60° 1.497 m.

75° 10.976 m. 10.777 x sin 75° 0.849 m.

90° 10.268 m. 10.777 x sin 90° 0.509 m.

* Heel in degrees (1 Radian) = 57.3° (constant) = 180 / PI (3.1416)

* Plot the said figures to acquire Curve of Intact Stability (GZ against every heel angle).

Curve of Intact Stability

GZ (m)

GM against constant 57.3°

2.50 2.523 GM

Max. GZ = 43°

2.00 1.99 m.

1.91 m.

1.66 m.

1.50 1.49 m.

Angle for Max. GZ

1.00 1.02 m. is the guide for

making a decision

to alter / weather 0.85 m.

course.

0.50

20° 40° 57.3° 60°

0.46 m.

1st separation (9°) is the

Initial Stability

0.00

10° 30° 50° 75°

Heel Angle (Heel in degrees)

* Establish then the Righting Levers (GZ) for every station or common interval (e.g. 0.3 m. at 7.5°, 0.73 m. at 15°

heel up to 30° heel) inorder to acquire Dynamical Stability at any angle of heel by SIMPSON'S RULE.

Grain Stability

Grain - the term covers wheat, maize (corn), oats rye, barley, rice, pulses (edible seeds such as peas, beans or

lentils), seeds and processed forms whose behavior is similar to that grain in natural state., the

Stability Requirements

1. the angle of heel due to the shift of grain shall NOT BE GREATER THAN 12˚ or in the case of ships

constructed on or after 1 Jan. 2004, the angle at which the deck edge is immersed, whichever is the lesser.

2. in the statical diagram, the net or residual area between the heeling arm curve and the righting arm curve up

to the angle of heel shall in all conditions of loading be NOT LESS THAN 0.075 m-radians.

3. the initial metacentric height after correction for the free surface effects of liquids in tanks, shall NOT BE

LESS THAN 0.30 m.

Since the weather, sea and even operating condition of the ship cannot be anticipated during the voyage, in spite

of the precautions being taken, the grain will shift wherein the grain center of gravity will move off the centerline of

the ship. The distance it moves multiplied by the weight of the grain constitutes to a force known as GRAIN

HEELING MOMENT.

To anticipate the said condition, IMO Grain Rules require that the grain be trimmed after it has been loaded. But

there is still a large of open space above the grain surface when the compartment is partly filled even the grain is

trimmed. Calculations using values in the''Filled, Untrimmed Compartment'' are wiser to use in initial calculations so

any differences will be on the safe side and remediable.

Filled Compartment with Untrimmed Ends:

Filled Compartment with Trimmed Ends:

20° 40° 57.3° 60°

On most Bulk Carriers, the grain is restrained against shift by graintight structure with slopes of 30˚ angle or

more where trimming is not really required unless the bulk grain is not filled up to the maximum extent of the hatch

opening.

Specially Suitable Compartment:

Upper Wing

Tanks

Acceptance of Vessels to Load Bulk Grain

1. Certificate of Readiness - a document issued by the National Cargo Bureau before a ship can load grain.

2. Document of Authorization - issued by the Administration of the country of registry. This shall accompany to

approved Grain Loading booklet to meet the requirements.

3. Change of Registry - when a vessel changes registry, Grain Loading booklet must be approved by the new

Administration as well as the new Document of Authorization issued.

4. Seaworthiness - vessel should have valid Cargo Ship Safety Construction Certificate.

5. Stability Calculations - calculations demonstrating that the vessel will comply with the stability requirements of

the Grain Loading booklet and must be presented to the attending Surveyor before a Certificate of

Readiness will be issued.

6. Fittings - all grain divisions for the particular stowage shall be grain tight, in sound condition and constructed in

accordance with the regulations.

7. Structural Defects - the boundaries of cargo compartment for loading shall be structurally intact and leak-free.

Preparing a Ship for Loading Grain

- Holds Cleanliness

. All spaces intended for grain be thoroughly clean, odor-free, free of loose rust and paint scale. Holds must

be swept, washed (if necessary) and dried.

- Bilges

. Bilges and/or drain wells must be cleaned and then sealed with burlap which is graintight but not watertight.

Bilge suctions and sounding pipes must be clear.

- Infestation

. Thorough inspections for any signs of insect or rodent infestation.

- Structural Integrity

. Cargo compartments shall be in sound and watertight. If necessary hose testing will do.

- Sheathing of Hot Bulkheads

. Required whenever bulk grain of any type is stowed adjacent to the bulkhead of a tank in which heated

liquid is carried.

- Tanks

. Operations leading to the said tanks (deep or wing tanks) shall be suspended. Heating of coil lines or

ballasting to be avoided.

- Electrical Wiring

. All electrical circuits in grain compartments shall be disconnected or defused.

Preparation of the Stability Calculation

1. The Quantity and Type of Grain to be loaded.

2. An accurate estimate of the Stowage Factor.

3. The Quantities of Fuel and Water on departure, daily consumption and the amount to be taken at bunkering port

during the voyage.

4. The Seasonal Zones to be traversed during the voyage.

5. The Quantities and Stowage of Other Cargo carried.

6. The Distance and Steaming Time to the port/s of discharge.

7. The Draft Restrictions which may be encountered during the voyage.

ARCHIMEDES' Principle states that when a body is wholly or partially immersed in a fluid, it suffers an apparent

dA = 13.55 m. Required trim at 0.28 m. Distribute the remaining cargo between Hold 1 and 7 to obtain the req'd.

Trim & Stability-basic

Documents

Transcript of Trim & Stability-basic