trim & stability-basic
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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.
* Excerpt from H. Table based on 81,470 MT Δ:
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:
t = Δ x (LCG - LCB)
MTC x 100
= 81,470 x (-3.82 - -4.22)
1,007.98 x 100
= 32,588.00
100,798
t = 0.323 m. (By Stern) + = By Stern
- = By Head
2. Find the Draft Forward:
DFP = DCF- t
2
= 13.60 - 0.323
2
= 13.60 - 0.162
DFP = 13.438 m.
2. Find the Draft Aft:
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
KGo = 12.100 m. (actual KG) KM G GGo
KM = 13.380 m. - KGo
GMo = 1.2800 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.
+
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
t = 0.761 m. (By Stern) + = By Stern
- = By Head
2. Find the Draft Aft:
DAP = DCF+ t (reverse, instead of subtraction)
2
= 11.80 + 0.761
2
= 11.80 + 0.381
DAP = 12.181 m.
3. Find the Draft Forward:
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 2 = 9,836 MT (or 9,800 MT) . . . . .Vol. = Wt x SF (9,800 x 1.226 = 12,015 m³ - Volume )
• 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 4 = 9,301 MT (or 9,300 MT) . . . . .Vol. = Wt x SF (9,300 x 1.226 = 11,402 m³ - Volume )
. . . . .• 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
CONST.2 DW CONSTANT 2 100
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
TRIM AND STABILITY CALCULATION SHEET
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
WBT2 NO.2 WBT (P&S) 0
WBT3 NO.3 WBT (P&S) 0
WBT4 NO.4 WBT (P&S) 0
WBT5 NO.5 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
HOLD2 NO.2 C. HOLD 100
HOLD3 NO.3 C. HOLD 100
HOLD4 NO.4 C. HOLD 100
HOLD5 NO.5 C. HOLD 100
HOLD6 NO.6 C. HOLD 100
HOLD7 NO.7 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.
Total FSM = 16,480 MT-m.
* Excerpt from H. Table based on 78,724 MT Δ:
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:
t = Δ x (LCG - LCB)
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
t = 0.119 m. (By Stern) + = By Stern
- = By Head
2. Find the Draft Forward:
DFP = DCF- t
2
= 13.18 - 0.119
2
= 13.18 - 0.060
DFP = 13.120 m.
2. Find the Draft Aft:
DFP = 13.120 m.
t = +0.119 m.
DAP = 13.239 m.
3. Solve GGo or FSC:
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
KGo = 10.370 m. (actual KG) KM G GGo
KM = 13.340 m. - KGo
GMo = 2.9700 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.
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
7. Find the Draft Forward:
DFP = DCF- t
2
= 13.47 - 0.016
2
= 13.47 - 0.008
DFP = 13.462 m.
8. Find the Draft Aft:
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 1 = 7,556 MT (or 7,500 MT) . . . . .Vol. = Wt x SF (7,500 x 1.226 = 9,195 m³ - Volume )
• Wt. For Hold 2 = 12,817.1 x 58,600
Wt. For Hold 2 = 8,604 MT (or 8,600 MT) . . . . .Vol. = Wt x SF (8,600 x 1.226 = 10,544 m³ - Volume )
• Wt. For Hold 3 = 13,270.2 x 58,600
Wt. For Hold 3 = 8,908 MT (or 8,900 MT) . . . . .Vol. = Wt x SF (8,900 x 1.226 = 10,911 m³ - Volume )
• Wt. For Hold 4 = 12,118.9 x 58,600
Wt. For Hold 4 = 8,135 MT (or 8,100 MT) . . . . .Vol. = Wt x SF (8,100 x 1.226 = 9,931 m³ - Volume )
• Wt. For Hold 5 = 13,272.2 x 58,600
Wt. For Hold 5 = 8,909 MT (or 8,900 MT) . . . . .Vol. = Wt x SF (8,900 x 1.226 = 10,911 m³ - Volume )
• Wt. For Hold 6 = 12,799.5 x 58,600
Wt. For Hold 6 = 8,592 MT (or 8,500 MT) . . . . .Vol. = Wt x SF (8,500 x 1.226 = 10,421 m³ - Volume )
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;
%
CONST.1 DW CONSTANT 1 100
CONST.2 DW CONSTANT 2 100
PROV PROVISION 100
TOTAL CONST.=
(Capacity of Tk. x SF)
FOT1 NO.1 FOT
FOT2 NO.2 FOT
FOT3 NO.3 FOT
D. FOT DEEP FOT
FO SETT H.F.O. SETT/SERV.
TOTAL F.O. =
D. DOT DEEP DOT 96
K.G. V.MOMENT FSM
(MT) (m.) (MT-m.)
TRIM AND STABILITY CALCULATION SHEET
WEIGHT L.C.G. L.MOMENT
(MT-m.) (MT-m.)
Wt. x KGfr. H. Table
ITEM
570
9,992 13.63
Wt. x LCG fr. H. Table
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
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
WBT2 NO.2 WBT (P&S) 0
WBT3 NO.3 WBT (P&S) 0
WBT4 NO.4 WBT (P&S) 0
WBT5 NO.5 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
HOLD2 NO.2 C. HOLD 100
HOLD3 NO.3 C. HOLD 100
HOLD4 NO.4 C. HOLD 100
HOLD5 NO.5 C. HOLD 100
HOLD6 NO.6 C. HOLD 100
HOLD7 NO.7 C. HOLD 100
TOTAL CARGO =
LIGHTWEIGHT
(from ship's particulars)
DISPLACEMENT
From Preplan:
Δ = 70,274 MT
LCG = -3.82 m.
KG = 9.35 m.
Total FSM = 16,476 MT-m.
* Excerpt from H. Table based on 70,274 MT Δ:
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:
t = Δ x (LCG - LCB)
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
Table - based on Vol. Fig.)
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
DEADWEIGHT(total weight) (LMom./T.Wt.) (T. LMoment) (T.VMom/T.Wt) (T. VMoment) 16,476
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)
Δ LCG L. Moment KG V. Moment
70,274 -3.82 -268,690 9.35 656,938
(DWT+LShip) (LMom./T.Wt.) (T. LMoment) (T. VMom./ Δ) (T. VMoment)
95,622
t = 0.875 m. (By Stern) + = By Stern
- = By Head
2. Find the Draft Forward:
DFP = DCF- t
2
= 11.874 - 0.875
2
= 11.874 - 0.437
DFP = 11.437 m.
2. Find the Draft Aft:
DFP = 11.437 m.
t = +0.875 m.
DAP = 12.312 m.
3. Solve GGo or FSC:
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
KGo = 9.584 m. (actual KG) KM G GGo
KM = 13.31 m. - KGo
GMo = 3.726 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.
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
7. Find the Draft Forward:
DFP = DCF- t
2
= 12.13 - 0.798 Wt. To Shift =
2 DBH (Dist. bet. Holds)
= 12.13 - 0.399
DFP = 11.731 m.
8. Find the Draft Aft:
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
1. Find the mean draft:
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.
5. Find distance from LCF to LCG:
D = LCG - LCF (when LCG is measured from centerline, but if LCG is measured from AP…)
= 25.98 - -2.60
D = 28.58 m.
* If 'd' is negative (-), LCG of Total Weight is forward from LCF.
6. Find the change in Trim:
Tc =
= 2,680 x 28.58
883.45 x 100
Tc = 0.867 m.
7. Find trim forward and aft:
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
1. Find the mean draft:
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.
3. Find distance from LCF to LCG:
D = LCG - LCF (when LCG is measured from centerline, but if LCG is measured from AP…)
= 67.84 - 1.25
D = 66.59 m. LCG of Hold 7
* If 'd' is negative (-), LCG of Total Weight is forward from LCF.
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.
6. Find the change in Trim:
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
1. Find the mean draft:
mD = dF + dA
2
= 13.000 + 13.170
2
mD = 13.085 m.
2. Excerpt from H. Table based on 13.085 m. Mean Draft:
LCF = 1.08 m.
MTC = 993.53 MT-m.
3. Find distance from LCF to LCG:
D = LCG - LCF (when LCG is measured from centerline, but if LCG is measured from AP…)
= 67.84 - 1.08
D = 66.76 m. LCG of Hold 7
* If 'd' is negative (-), LCG of Total Weight is forward from LCF.
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.
6. Find the change in Trim:
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.
1. Find the mean draft:
mD = dF + dA
2
= 12.650 + 12.710
2
mD = 12.680 m.
2. Excerpt from H. Table based on 12.680 m. Mean Draft:
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.
7. Find trim forward and aft:
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.
1. Find the mean draft:
mD = dF + dA
2
= 13.260 + 13.340
2
mD = 13.300 m.
2. Excerpt from H. Table based on 13.300 m. Mean Draft:
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.
7. Find trim forward and aft:
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
%
CONST.1 DW CONSTANT 1 100
CONST.2 DW CONSTANT 2 0
PROV PROVISION 0
TOTAL CONST.=
(Capacity of Tk. x SF)
FOT1 NO.1 FOT (P/S)
FOT2 NO.2 FOT (P/S)
FOT3 NO.3 FOT (P/S)
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
WBT2 NO.2 WBT (P&S) 0
WBT3 NO.3 WBT (P&S) 0
WBT4 NO.4 WBT (P&S) 0
WBT5 NO.5 WBT (P&S) 0
TTL. BALLAST =
HOLD1 NO.1 C. HOLD 100
HOLD2 NO.2 C. HOLD 0
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
TRIM AND STABILITY CALCULATION SHEET
WEIGHT L.C.G.
fr. H. Table
FSM ( I )
ITEM (MT) (m.) (MT-m.)
0
24,840 16.30
Wt. x LCG fr. H. Table
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.
Table - based on Vol. Fig.)
0
33,000 -114.44 -3,776,520 13.98
0 0.00 0 0.00
461,373
0
HOLD3 NO.3 C. HOLD 100
HOLD4 NO.4 C. HOLD 0
HOLD5 NO.5 C. HOLD 100
HOLD6 NO.6 C. HOLD 0
HOLD7 NO.7 C. HOLD 100
HOLD8 NO.8 C. HOLD 0
HOLD9 NO.9 C. HOLD 100
TOTAL CARGO =
LIGHTWEIGHT
(from ship's particulars)
DISPLACEMENT
From Preplan:
Δ = 204,825 MT
LCG = -8.22 m.
KG = 13.95 m.
Total FSM = 6,542 MT-m.
* Excerpt from H. Table based on 204,825 MT Δ:
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:
t = Δ x (LCG - LCB)
MTC x 100
= 204,825 x (-8.221 - -9.309)
2,528.934 x 100
= 222,849.6
252,893.4
t = 0.881 m. (By Stern) + = By Stern
- = By Head
2. Find the Draft Forward:
DFP = DCF- t
2
= 18.322 - 0.881
2
= 18.322 – 0.441
DFP = 17.882 m.
3. Find the Draft Aft:
DFP = 17.882 m.
t = +0.881 m.
DAP = 18.763 m.
4. Solve GGo or FSC:
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
DEADWEIGHT(total weight) (LMom./T.Wt.) (T. LMoment) (T.VMom/T.Wt) (T. VMoment) 6,542
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)
Δ LCG L. Moment KG V. Moment
204,825 -8.22 -1,683,966 13.95 2,857,269
(DWT+LShip) (LMom./T.Wt.) (T. LMoment) (T. VMom./ Δ) (T. VMoment)
Δ
=
FSC = 0.032 m. (or GGo)
5. Solve for GM: M
KG = 13.950 m.
GGo = 0.0320 m. Go GM
KGo = 13.982 m. (actual KG) KM G GGo
KM = 19.112 m. KGo
GMo = 5.130 m. (actual GM)
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.=
(Capacity of Tk. x SF)
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
WBT2 NO.2 WBT (P&S) 9
WBT3 NO.3 WBT (P&S) 24
WBT4 NO.4 WBT (P&S) 21
TTL. BALLAST =
HOLD1 NO.1 C. HOLD 68
800
Total 14,300
TRIM AND STABILITY CALCULATION SHEET
(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
Wt. x LCG fr. H. Table
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.
Table - based on Vol. Fig.)
978
15,080 02,688 -45.23
HOLD2 NO.2 C. HOLD 79
HOLD3 NO.3 C. HOLD 79
HOLD4 NO.4 C. HOLD 81
TOTAL CARGO =
LIGHTWEIGHT
(from ship's particulars)
DISPLACEMENT
From Preplan:
Δ = 21,294 MT
LCG = -0.41 m.
KG = 5.96 m.
Total FSM = 8,448 MT-m.
* Excerpt from H. Table based on 21,294 MT Δ:
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:
t = Δ x (LCG - LCB)
MTC x 100
= 21,294 x (-0.41 - -1.11)
2,47.67 x 100
=14,905.6
24,767.0
t = 0.602 m. (By Stern) + = By Stern
- = By Head
2. Find the Draft Forward:
DFP = DCF- t
2
= 8.67 - 0.602
2
= 8.67 – 0.301
DFP = 8.369 m.
3. Find the Draft Aft:
DFP = 8.369 m.
t = +0.602 m.
DAP = 8.971 m.
4. Solve GGo or FSC:
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
DEADWEIGHT(total weight) (LMom./T.Wt.) (T. LMoment) (T.VMom/T.Wt) (T. VMoment) 8,448
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)
Δ LCG L. Moment KG V. Moment
21,294 -0.41 -8,651 5.96 126,984
(DWT+LShip) (LMom./T.Wt.) (T. LMoment) (T. VMom./ Δ) (T. VMoment)
FSC = 0.397 m. (or GGo)
5. Solve for GM: M
KG = 5.960 m. (KG SOLID )
GGo = 0.397 m. Go GM
KGo = 6.357 m. (actual KG) KM G GGo
TKM = 9.510 m. KGo
GMo = 3.153 m. (actual GM)
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)
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.
* 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.
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