seminar workshop on “trim and stability” for bulk...

Volume 1, Issue 1

Seminar Workshop on “Trim and Stability” for Bulk Carrier

May 10, 2006

P O L A R I S J U N I O R M A R I N E D E C K O F F I C E R F A M I L I A R I Z A T I O N C E N T R E

V o l u m e 1 , I s s u e 1 P a g e 2


STABILITY NOMENCLATURE

K = Keel

G = Center of gravity

B = Center of Buoyancy

M = Metacentre

ø = Angle of Heel

BM = Metacentric Radius

GM = Metacentric Height

GZ = Righting Lever measured from G

KB = Height of Center of Buoyancy from keel

KG = Height of Center of Gravity from keel

KM = Height of Metacenter from keel KM = KG + GM KM = KB + BM GZ = KN - KG x sin ø where KN can be found from KN curves Righting Moment = Δ x GZ where Δ = displacement

CALCULATION OF KG

KG = Total Vertical Moment of Weights about keel [metre.tonnes] Δ [tonnes] GG1 = Moment of Weight,W shifted over Distance, D [metre.tonnes] Δ [tonnes] : vertical shift of G

CALCULATION OF KM

KM = KB + BM KB = 0.53 x Draft [metre] BM = 2nd moment of waterplane area = I [metre] volume of displacement V where I = L x B3 [metre4] for a rectangular barge 12

CALCULATION OF GM

GM = KM - KG


VIRTUAL LOSS OF GM DUE TO FREE SURFACE GGv = s.g. of Liquid in the Tank x I x 1 s.g. of Water in which vessel floats V n2 where GGv = virtual rise in G or deduction in G I = 2nd moment of the free surface about the centre line = L x B3 [metre4] for a rectangular compartment 12 L = Length of the Tank [metre] B = Breadth of the Tank [metre] V = Volume of the Tank [metre3] n = number of longitudinal compartments into which the tank is subdivided LARGE ANGLE STABILITY - where the force of buoyancy can no longer be considered to act vertically upwards through the initial metacentre, M GZ = (GM + ½BM.tan2 ø) sin ø : for ship's side that are "WALL-SIDED" in the vicinity of the water line Note:- for small angle stability, the term ( ½BM.tan2 ø) becomes negligible � GZ = GM.sin ø

CHANGE OF TRIM - the difference between initial trim and final trim i.e. change of draught forward + change of draught aft Trimming Moment = Weight x Distance shifted = W x d [tonnes.metre] Change of Trim, t = Trimming Moment [metre] 100 x MCT.1cm MCT.1cm = Moment To Change Trim by 1 cm = Δ x GM L [tonnes.metre] 100 x L ≅ Δ x BM L [as GML is small when compared with BML] 100 x L where Δ = displacement [tonnes] GM L = Longitudinal Metacentric Height BM L = height of the longitudinal metacentre, ML above centre of buoyancy, B GML = KB + BML - KG [metre]


where BML = long. 2nd mmt of waterplane about centre of flotation, F volume of displacement = I L [metre] V Change of draught aft, ta = l a x change of trim [metre] L Change of draught forward, tf = l f x change of trim [metre] L Change in mean draught = Weight loaded or discharged [metre] TPC TPC = Tonnes per Centimetre Immersion = Aw x ρ 100 where Aw = area of waterplane [metre2] = L x B x Cw (waterplane area coefficient) ρ = density of sea water [tonnes per metre3]

LARGE CHANGE IN DISPLACEMENT Trimming Moment = Δ x (longitudinal separation LCG and LCB) where LCG = Longitudinal centre of gravity [metre] LCB = Longitudinal centre of buoyancy [metre]

CHANGE IN DENSITY Change in Mean draught due to change in density = Δ x (ρ1 - ρ2) Aw (ρ1.ρ2) Trimming Moment = Δ x (horizontal shift of LCB) or (mass of layer of water added or removed due to change in density) x (horizontal distance between initial LCB & final LCF of waterplane)


DRAFT SURVEY

SURVEY PROCEDURE

This Survey Procedure is International Standard for any type of ship. The ship is first surveyed light, to calculate the constant. It is then re-surveyed after loading to determine the weight of cargo.

APPARENT TRIM

The Forward (Fwd), Aft (Aft), and Midships (Mid) drafts are read at both Port (P) and Starboard (S) marks. The P and S readings are added, and the result divided by two.

The Aft draft is subtracted from the Fwd draft, and the result is Apparent Trim. If Trim is positive, the ship is trimmed By the Head; if Trim is negative, the ship is trimmed By the Stern.

Fwd Draft = Fwd(P) + Fwd(S) 2

Aft Draft = Aft(P) + Aft(s) 2

Mean Mid = Mid(P) + Mid(s) 2

Trim = Fwd – Aft

DRAFT CORRECTIONS TO THE PERPENDICULARS

The After Perpendicular is a right angle line to the keel passing through the rudderpost; it is also the first frame marked «0» on the vessels drawings.

The Forward Perpendicular is a right angle line to the keel cutting the vessel’s Summer waterline at the stern. The vessels stability information is calcu-lated on the drafts measured at the perpendiculars; as the draft marks rarely coincide with these lines, a draft as read must be corrected.

If the marks are not on the perpendiculars, the vessel usually has a tabulated plan in her hydrostatic books. However, some of the older vessels do not have their tabulation and it is therefore necessary to work out the correction to be applied by referring to the vessels capacity plan and measuring the horizontal distance between the draft marks and the perpendiculars of the waterline.

The correction is calculated as follows:

Aft Perpendicular Corr = 7.10 x 1.75(trim)= 0.0971 cm (+) 128.0

Fwd Perpendicular Corr = 1.21 x 1.75(trim)= 0.0165 cm( - )

The rule to apply the correction:

Trimmed by the STERN: Forward Correction = MINUS ( - ) After Correction = PLUS ( + ) Trimmed by the HEAD : Forward Correction = PLUS ( + ) After Correction = MINUS ( - )

The above corrections are applied to the forward and after drafts read.

Fwd 2.64 m Aft 4.30 m — Corr 0.0165 + Corr 0.0971 Corrected Draft 2.6235 Corrected Draft 4.3971

V o l u m e 1 , I s s u e 1 Corrected Trim: Aft P a g e 8

CORRECTED DRAFT

Aft 5.019 = Aft - Fwd = Corr. Trim Fwd 2.361 Corrected Trim 2.658

NOTE: This value is used in Trim Correction Formulas to adjust the displacement

MEAN DRAFT CORRECTION ( M / M / M )

The Quarter Mean ( QM ) or Mean Draft Corrected for Deformation must be solved next. Use the corrected draft values

First calculate the Fwd/Aft Mean Draft. Add Fwd to Aft, and divide the result by two:

Fwd /Aft Mean = Fwd + Aft 2 Next, calculate the Mean of Mean Add the Fwd/Aft Mean (calculated in 3.10.) to the Mid Mean, and divide the result by two:

Mean of Mean = Fwd/Aft Mean + Mid Mean 2

Now calculate the QM Add the Mean of Mean to the Mid Mean, and divide the result by two

NOTE: The Mid Mean is applied twice, first in calculating the Mean of Mean, and second in calculating the QM.

QM = Mean of Mean + Mid Mean 2

EXAMPLE :

FWD P 2.377 AFT P 5.017 MIDSHIP P 3.59 S 2.377 S 5.017 S 3.72 4.754 10.034 7.31

FWD= 4.754 / 2 = 2.377 AFT= 10.034 / 2 = 5.017 MEAN-MID = 7.31 / 2 = 3.655

TRIM = AFT - FWD (vice versa) = = 5.017 - 2.377 (vice versa) = - 2.64 Trim by the Stern ( Apparent )

DRAFT CORRECTION

Corrections for the Fwd and Aft Drafts (Fwd corr. and Aft corr.) and Corrected Trim must be calculated. The corrections values are different for each ship, and are found in the Stability Manuals. If required, they can be calculated from the formula given in page 9 figure.

EXAMPLE:

Fwd Correction Value =(distance from Fwd Draft to Fwd Perp) (distance between Fwd and Aft Drafts)

Fwd Correction = Fwd Correction Value x Trim ( Apparent )= 0.006037 x 2.64 = 0.016( - )

Aft Correction Value = (distance from Aft Draft to Aft Perp) (distance between Fwd and Aft Drafts)

Aft Correction = 0.034716 x 2.64 = 0.91 (+)

Corrected Draft:

Fwd = 2.377 Aft = 5.017 Corr = -0.016 Corr = + 0.091 2.361 5.108


Corrected Trim: Aft Draft - Fwd Draft = Corr Trim (CT)

Aft = 5.108 Fwd = - 2.361 2.747

Note: this value used in the Trim correction Formulas to adjust the displacement.

EXAMPLE:

Mid Mean = 3.655 M

Fwd + Aft = 2.361 M + 5.019 M = 7.38 M

Fwd and Aft Mean = 7.38 = 3.69 M 2

Fwd and Aft + Mid Mean = 3.69 +3.655 7.345

Mean Mean of Means = 7.345 = 3.672 M 2

Mean Mean of Means + Mid Mean = 3.672 + 3.655 = 7.327 M

Quarter Mean = 7.327 = 3.663 M 2

QM = 3.663 M

The value for QM is used throughout the remaining Draft Survey Calculations.

NOTE : QM is the same as M/M/M

M/M/M = ( Fwd + 6 x Mid + Aft ) / 8 = 3.663 M

(7.10) - represents the distance the draft marks are from the perpendiculars.

(128.0) - represents the length of the vessel between the draft marks

(trim) - the difference between the forward and after drafts.

Refer to the vessel's Stability & Hydrostatic Manuals and Tables for the following values:

TPC: tonnes per Centimetre Immersion MTC: Moment to change Trim One Centimetre LCB: Longitudinal Centre of Buoyancy LCF: Longitudinal centre of Flotation KB: Transverse centre of Buoyancy TKM: Transverce Metacentric Height

Interpolation

Calculate the Displacement Correction ( DISP. Corr.)

a) Subtract the nearest smaller Draft from the calculated QM.

V o l u m e 1 , I s s u e 1 P a g e 1 0

b)Multiply the result by 100 to convert Meters to Centimetres.

c) Multiply this by the TPC for the displacement.

d) This correction is added to the displacement given for the nearest smaller draft.

NOTE: Refer to Figures 13 and 14 for sample Hydrostatic Tables.

Draft Remainder (cm) = Draft remainder x 100

DISP. Corr. = TPC x Draft remainder (cm)

Displacement = DISP. + Disp. Corr. = Actual Displacement

EXAMPLE:

a) Draft remaining = 3.6635 - 3.66 = 0.0035 M b) Draft remaining = Draft Remainder(M) x 100 = 0.0035 M =0.35cm c) Displacement Correction = TPC x Remaining draft (cm)

= 17.66 x 0.35 cm = 6.181 MT

d) DISPL.CORRECTED = DISPL.(at SMALLER DRAFT ) + correction= 7587.00 + 6.181= 7593.181 MT (corrected)

TRIM CORRECTION

Trim Correction values for a given Displacement is tabulated in the Ship Stability Manual. Even if these are readily available, the following formulas should be studied in order that the principles governing a Draft Survey are fully understood.

Before calculating the First Trim Correction, Corrected Trim (CT) (Ref. 3.3) must be converted from meters to centimetres. Multiply CT (in) by 100 to get centi-metres.

To calculate the First Trim Correction, multiply TRIM by TPC, then multiply the product by the longitudinal Centre of Flotation (LCF) x 100. Then, divide the final product by the Length Between Perpendiculars (LBP).

CT = CT x 100

First Correction = TRIM x TPC x LCF x 100 LBP

Second Correction = TxT x +/-50 cm x MTC diff. LBP

The first correction can be either positive (add), or negative (subtract), depending on the location of the LCF and the trim condition. (It's mean sign of LCF and TRIM )

VESSEL TRIMMED BY THE STEM LCF is Forward (Fwd) (+) ADD Trim Correction LCF is Aft (Aft) (-) SUBTRACT Trim Correction

VESSEL TRIMMED BY THE STERN LCF is Fwd (—) SUBTRACT Trim Correction LCF is Aft (+) ADD Trim Correction

The second Trim Correction is required when the Trim is greater than the LBP divided by 100. It may be applied without adverse effect at smaller trims.

The second correction is always (+) (additive) regardless of the trim or other factors.


Before calculating the Second Trim Correction, MTC differ-ence, sometimes referred to as dM/dZ, must be found.

ADD 50 cm to the Quarter Mean Draft ( QM ) to find the corresponding MTC from the vessel’s Hydrostatic book.

SUBTRACT 50 cm from the Quarter Mean Draft (QM) to find the corresponding MTC from the Vessels Hydrostatic book.

The difference between 3.21.1 and 3.21.2 is the MTC difference, or dM/dZ.

EXAMPLE:

(1)First correction: Trim = 2.74 M ( By STERN "+" ) TRIM x LCF x TPC x 100 LBP

(+)2.74 x (—)4.53 x 17.65 x 100 = 159.909 (-) MT= 159.91(-) 137.00

2) Second Correction: TxT x +/-50 x MTC diff LBP

MTC diff.:

a) Q M + 50cm = MTC ( Found in Ship’s book)

b) Q M - 50cm = MTC ( Found in Ship’s book)

MTC diff = ( a) — ( b)

a) QM = 3.675 + 0.50 4.175 MTC = 169.4

b) QM = 3.675 - 0.50 3.175 MTC =160.7

(a) = 169.4 (b) = - 160.7 MTC diff = 8.7

7.5 x 50 x 8.7 = 23.81 + MT 137

(3) DISP Corrected for TRIM:

1st correction: a) 7587.00 b)- 159.91 =c) 7427.09 (-)

2nd Correction: c) 7427.09 d) + 23.81

SPECIFIC GRAVITY CORRECTION

A Specific Gravity (Sg) of 1.025 is generally assumed for SeaWater in calculating Displacement (DISP). Because the Sg. is almost never exactly 1.025, Sg. correction must be calculated.


Sg. is always minus if the measured Sg. is 1.025 or less.

Sg. is plus if the measured S9. is 1.026 or more.

Calculate the Sg. correction by subtracting the measured density from 1.025, divide this by 1.025 and then multi-ply that answer by the DISP.

Sg. corr. = 1.025 - Measured Density x DISP. 1.025

EXAMPLE:

Measured Density = l.020.4 1.025 — 1.0204 x 7450.9 = 32.71 1.025 DISP. corr. for Trim 7450.90 Density Corr. (Sg.) — 32.71 DISP. Corr. for Density 7418.19

VESSEL’S CONSTANT

Subtract-ing the Lightship, weights, ballast and consumables from the Displacement solves the Constant of an unladen vessel.

Tank tables or graphs should be available so the tank soundings can be converted from measure to volumes and corrected for trim.

Volume multiplied by Sg. equals weight. A Sg. of 1.000 is used for Fresh Water, and for Salt Water Ballast a Sg. of 1.025 is used. Therefore, one cubic meter of Fresh Water equals one Metric Tonne and one cubic Meter of SeaWater equals 1.025 Metric Tonnes. The Chief Engineer is obliged to supply the Sg. of the various fuel oils on board. It is good practice, if possible, to measure the Sg. at the same time the tanks is being sounded.

WEIGHTS

FUEL OIL 545.86 MT

DIESEL OIL 100.70 MT

LUBE OIL 21.00 MT

FRESH WATER 401.00 MT

DRINK WATER NIL

BOILER WATER NIL

BALLAST WATER 1870.84 MT

SLUDGE (BILGE) 5.50 MT

STORES, etc. NIL

CONSTANT 200.42 MT = TOTAL WEIGHT: 3145.32 MT


NETT DISP. 7418.19 MT

-TOTAL WEIGHT - 3145.32 MT

NETT DISP LIGHTSHIP 4272.87 MT

FINAL SURVEY

The Final Survey follows the same procedure as the Initial Survey. Total cargo equals DISP. minus Lightship Weight.


AT A DRAFT OF 3,25 M, AN ADDITIONAL 25 TONNES OF FRESH WATER IS TAKEN ABOARD. FIND NEW DRAFT.

ANSWER :

FIND APPROPRIATE VALUE TPC FOR DRAFT = 3,25M. IT'S 8,61 TONNES. SINKAGE EQUALS TONNAGE ADDED THEN DI-VIDED BY TPC. NEW DRAFT EQUALS OLD DRAFT PLUS SINKAGE.

SINKAGE = LOADED WEIGHT = 25,00 = 2,90 CM = 0,290 M T P C 8,61

NEW DRAFT = OLD DRAFT + SINKAGE = 3,25 + 0,29 = 3,54 M

A SHIP OF 1000 TONNES DISPLACEMENT IS FLOATING IN SEA WATER. WHAT WILL THE CHANGE OF DRAFT BE WHEN SHE SAILS INTO RIVER WATER ? T P C AT LOAD LINE IS 23,5 TONNES.

ANSWER :

1. APPARENT CHANGE IN TONNAGE = Sg FW = 1,00 = 0,98 OR 98 % Sg SW 1,03

2. DISPL. -- ( DISPL. x Sg FW ) = 1000 -- ( 1000 x 0,98 ) = 20 Sg SW

3. INCREASE IN DRAFT = 2. = 20 = 0,85 CM ( + ) T P C 23,5

A SHIP HAS A FRESH WATER ALLOWANCE OF 2 CM AND THE HARBOUR SPECIFIC GRAVITY IS 1,010. HOW MUCH CAN THE LOADLINE BE SUBMERGED BEFORE PROCEEDING TO SEA ?

ANSWER :

VALUE OF SUBMERGE = ( 1,025 -- MEASURED Sg ) * FWA = ( 1,025 -- Sg FRESH WATER )

= ( 1,025 -- 1,010 ) * 2 = 0,015 * 2 = 0,03 = 1,3 CM ( 1,025 -- 1,000 ) 0,025 0,025

A SHIP IS 65 M LONG; 10 M IN BREADTH AND HAS A DRAFT OF 5 M TO THE TOP OF THE WEATHER DECK MID-SHIP. THE FREEBOARD MIDSHIP IS 0,8 M AND THE COEFFICIENT OF DISPLACEMENT IS 0,65. WHAT IS HER DIS-PLACEMENT IN SALT WATER ?

ANSWER :

1. DRAFT -- FREEBOARD = " XXX " = 5 M -- 0,8 M = 4,2 M

2. DISPL. IN SALT WATER = " XXX " * L * B * COEFF. OF DISPL. = 4,2 * 65 * 10 * 0,65 1,025 1,025 = 1731,219 MT

DRAFT AND STABILITY PROBLEMS


A SHIP DISPLACED 6342 cu.m WHEN Sg IS MEASURED AT 1,018. DISPLACEMENT CHANGES TO 6358 cu.m . WHAT IS THE NEW Sg ?

ANSWER :

NEW Sg = OLD DISPL. * MEASURED Sg = 6342 * 1,018 = 1,015 NEW DISPL. 6358

A SHIP IS 65 M LONG; 10 M IN BREADTH AND IS FLOATING IN SALT WATER AT DRAFT OF 5,48 M. COEFFICIENT OF FINESSE ( SOMETIMES REFFERED TO AS A BUCK COEFFICIENT ) OF HER WATERPLANE AT THAT DRAFT IS 0,758. a ) FIND HER TPC b ) WHAT WILL THE DRAFT BE IN Sg. OF 1,018 ?

ANSWER :

a ) TPC = L * B * COEFF.OF FINESSE = AREA = 138 *17 *0,758 = 17,78 100 100 100 b ) NEW DRAFT = ( OLD DRAFT * OLD Sg ) = 5,48 * 1,025 = 5,52 M NEW Sg 1.018

A SHIP FLOATS AT A DRAFT OF 6,88 FWD AND 6,93 AFT. HER MTC IS 105 TM AND THE CENTRE OF FLOTATION IS ON THE LONDITUDINAL CENTRE LINE. FIND THE CHANGE IN TRIM AND THE NEW DRAFT IF A WEIGHT OF 40 TON-NES IS SHIFTED 60 M FORWARD.

ANSWER :

1. CHANGE OF TRIM = MOMENT TO CHANGE TRIM = 40 * 60 = 22,85 CM MTC 105,00

2. 22,85 = 11,425 CM = 0,114 M 2,00

3. FWD DRAFT = 6,88 + 0,114 = 6,994 M AFT DRAFT = 6,93 - 0,114 = 6,816 M

A SHIP HAS SUMMER DRAFT OF 8,15 M AND FRESH WATER ALLOWANCE OF 17 CM. TO WHAT DRAFT MAY SHE LOAD IN DOCK WATER OF A DENSITY OF 1,007 ?

ANSWER :

1. ( 1,025 -- Sg OF DOCK WATER ) * FWA = ( 1,025 -- 1,007 ) * 17 = 0,018 *17 = 12,24 CM ( 1,025 -- Sg FRESH WATER ) 0,025 = 0,1214 M

2. ALLOWABLE DRAFT AT DOCK WATER = SHIP'S DRAFT + 1. = 8,15 + 0,1224 = 8,2724 M

A SHIP HAS DISPLACEMENT OF 2000 TONNES. FIND THE CHANGE OF HER CENTRE OF GRAVITY IF A WEGHT OF 100 TONNES SHIFTED ACROSS HER HOLD 12 METRES ?

ANSWER :

1. WEIGHT * SHIFTING = 100 * 12 = 0,60 M DISPLACEMENT 2000


A WEIGHT OF 550 TONNES IS LOADED ON A SHIP. THE CG OF THE WEIGHT IS LOCATED 3 M FROM THAT OF THE SHIP. IF THE SHIP'S ORIGINAL DISPLACEMENT WAS 3250 TONNES, WHAT WILL BE THE CHANGE OF THE CG ?

ANSWER :

550 * 3 = 1650 = 0,434 M 3250 + 550 3800

A SHIP AND HER CARGO DISPLACE 8400 TONNES. IF A WEIGHT OF 120 TONNES IS REMOVED FROM A POINT 35 M FROM THE ORIGINAL CG, WHAT WILL BE THE CHANGE OF THE CG ?

ANSWER :

120 * 35 = 4200 = 0,51 M 8400 -- 120 8280

A WEIGHT OF 35 TONNES IS LOADED INTO A VESSEL. THE NEW DICPLACEMENT IS 2550 T. THE OLD KG WAS 4,52 M. WHAT IS THE NEW KG UNDER THE FOLLOWING CONDITIONS ? ( a ) LOADED 6,2 M ABOVE THE CG ? ( b ) LOADED 6,2 M BELOW THE CG ?

ANSWER :

KG = 35 * 6,2 = 0,085 M 2550

( a ) LOADED ABOVE CG =4,52 + 0,08 = 4,60 M ( NEW KG )

( b ) LOADED BELOW CG = 4,52 -- 0,08 = 4,44 M ( NEW KG )

A VESSEL DISPLACED 3250 TONNES, AND HER KG IS 6,8 M. 150 TONNES ARE LOADED WITH THE CG 2,4 M ABOVE THE KEEL. WHAT IS THE NEW KG ?

ANSWER :

1. (6,8 -- 2,4 ) * 1150 = 4,4 * 1150 = 1,15 M ( 3250 + 1150 ) 4400

2. NEW KG = 6,8 -- 1,15 = 5,65 M

A DERRICK IS USED TO LIFT A 12 TONNES WEIGHT FROM THE HOLD. THE HEADS OF THE DERRICK IS 15 M ABOVE THE ORIGINAL CG OF THE WEIGHT. DISPLACEMENT IS 3200 TONNES. WHAT IS THE SHIFT OF CG ?

ANSWER :

12 * 15 = 0,056 M UPWARDS 3200


A VESSEL DISPLACED 4600 TONNES, AND HER KG IS 4,6. A WEIGHT OF 25 TONNES IS LIFTED FROM THE LOWER HOLD AND STOWED ON THE DECK WITH THE VESSEL'S DERRICK. THE HEAD OF THE DERRICK IS 21,5 M ABOVE THE KEEL. THE CG OF THE WIGHT WAS 1,4 M ABOVE THE KEEL WHEN IT WAS IN THE HOLD, AND IS 10,4 M ABOVE THE KEEL WHEN ON DECK. FIND THE FOLLOWING :

( a ) WHAT IS THE NEW KG WHEN THE WEIGHT IS SUSPENDED FROM THE DERRICK ? ( b ) WHAT IS THE NEW KG WHEN THE WEIGHT IS LANDED ON THE DECK ?

ANSWER :

( a ) ( 21,5 -- 1,4 ) * 25 = 20,1 * 25 = 0,109 4600 4600

NEW KG = 4,6 + 0,11 = 4,71 M

( b ) ( 10,4 -- 1,4 ) * 25 = 9,0 * 25 = 0,049 4600 4600

NEW KG = 4,6 + 0,05 = 4,65 M

20 TONNES ARE LIFTED BY A DERRICK, THE DERRICK HEAD IS 15 M ABOVE THE ORIGINAL POSITION OF THE WEIGHT. VESSEL'S DISPLACEMENT IS 1750 TONNES. KG IS 3,2 M. FIND THE FOLLOWING :

( a ) WHAT IS THE NEW KG WHEN THE WEIGHRT IS 1,3 M ABOVE THE KEEL ? ( b ) 10,4 M ABOVE THE KEEL ?

ANSWER :

( a ) 20 * 15 = 0,171 ( b ) THE SAME. WEIGHT SUSPENDED FROM 1750 A DERRICK ACTS AT THE DERRICK HEAD.

3,2 + 0,17 = 3,37 KG

A VESSEL DISPLACED 2635 TONNES & A KG OF 4,28; LOADS A WEIGHT OF 35 TONNES; THE DERRICK WITH IT'S HEAD ABOVE THE KEEL IS USED. WHEN THE WEIGHT IS STOWED IN THE HOLD, THE FINAL POSITION OF IT'S CG IS 2,8 M ABOVE THE KEEL. FIND THE FOLLOWING :

( a ) WHAT IS THE KG WHEN THE WEIGHT IS SUSPENDED FROM THE DERRICK ? ( b ) WHAT IS THE KG WHEN THE WEIGHT IS LOADED IN THE HOLD ?

ANSWER :

( a ) NEW DISP. = OLD DISP. + WEIGHT = 2635 + 35 = 2670

SHIFT OF CG = WEIGHT * HEIGHT DERRICK ABOVE THE KEEL = 35 *17 = 0,22 DISPLACEMENT 2670

NEW KG = OLD KG + SHIFT OF CG = 4,28 + 0,22 = 4,50 M

( b ) NEW DISP. = OLD DISP. + WEIGHT = 2635 + 35 = 2670

? = 35 * 2,8 = 0,04 2670

? = 4,28 -- 0,04 = 4,24 M


DOUBLE BOTTOM TANK CAN HOLD 420 TONNES OF SEA-WATER. WHEN FULL IT'S CG IS 0,8 M ABOVE THE KEEL. WITH THE TANK EMPTY, DISPLACEMENT IS 3700 TONNES & KG IS 8,78 M. WHAT IS THE KG WHEN THE TANK IS FILLED ?

ANSWER :

d = OLD KG -- CG OF FULL TANK = 8,78 -- 0,8 = 7,98 NEW DISP. = OLD DISPL. + WEIGHT = 3700 + 420 = 4120

420 * 7,98 = 0,813 4120

NEW KG = 8,78 -- 0,813 = 7,97 M

VESSEL'S TANK HOLDS 192 TONNES OF WATER; IT'S CG IS 41,3 M FROM THAT OF THE SHIP. IF THE DISPLACE-MENT IS 2974 TONNES WHEN THE TANK IS FULL, WHAT IS THE SHIFT IN G WHEN IT IS PUMP OUT ?

ANSWER :

NEW DISPL. = OLD DISPL. -- WEIGHT = 2974 -- 192 = 2782 TONNES

SHIFT OF G = WEIGHT * CG OF TANK = 192 * 41,3 NEW DISPLACEMENT 2782

THE VESSEL TRANSFERRED 135 TONNES OF FUEL FROM THE FOREPEAK TO AN AFTER TANK. THE DISTANCE BETWWEN THE TANK'S CG IS 125 M; DISPLACEMENT IS 7620 TONNES. WHAT IS THE SHIFT IN G ?

ANSWER :

SHIFTING = WEIGHT * DISTANCE = 135 * 125 = 2,21 M AFT DISPLACEMENT 7620

THE SHIP'S LCF IS 4 M AFT OF MIDSHIPS. WEIGHTS ARE SHIFTED AFT TO CHANGE THE TRIM 0,4 M. LBP IS 130 M. FORWARD DRAFT WAS 5,97 M. AFT DRAFT WAS 6,13 M. FIND NEW DRAFT.

ANSWER :

FORWARD DRAFT = 5,97 -- 0,4 = 5,77 M 2

AFTER DRAFT = 6,13 + 0,4 = 6,33 M 2

MEAN DRAFT = 5,77 + 6,33 = 6,05 M 2

BY PROPORTIONAL METHODS

CHANGE FORWARD = 69 * 0,4 = 0,21 130

CHANGE AFT = 61 * 0,4 = 0,19 130

FORWARD DRAFT = 5,97 -- 0,21 = 5,76 M

AFTER DRAFT = 6,31 + 0,19 = 6,32 M

MEAN DRAFT = 5,76 + 6,32 = 6,04 M


A PLANK WEIGHT 25,09 KG. WHAT WILL BE THE SHIFT OF IT'S CENTRE OF GRAVITY IF A WEIGHT OF 17,24 KG IS PLACED A DISTANCE OF 3,66 M FROM IT'S ORIGINAL CG ?

ANSWER :

( a ) THE CG OF A BODY MOVES PARALLEL TO THE SHIFT OF THE CG OF ANY WEIGHT SHIFTED WITHIN IT.

( b ) THE DISTANCE IT MOVES IS EQUAL TO THE WEIGHT SHIFTED, MULTIPLIED BY THE SHIFT OF IT'S CG; ALL DIVIDED BY THE TOTAL WEIGHT OF THE BODY.

17,24 * 3,66 = 63,09 = 1,49 M 25,09 + 17,24 42,33

A BEAM CARRIES 2438 KG AT A DISTANCE OF 1,83 M FROM ONE END & THE CENTRE OF GRAVITY OF THE WHOLE MASS IS 6,09 M FROM THE END. IF THE WEIGHT OF THE BEAM ALONE IS 4267,47 KG, WHAT IS THE DIS-TANCE OF THE CG OF THE ORIGINAL BEAM FROM THE END ?

ANSWER :

GG = w * d = 2438,55 * 4,26 = 2,43 W 4267,47

2,434 + 6,09 = 8,524

FIND THE TONNES PER CENTIMETER IMMERSION OF A BOX SHAPED VESSEL 64 M LONG BY 10,668 M BREADTH.

ANSWER :

T P C = 1,025 * AREA OF THE WATERPLANE ( A ) = 1,025 * 10,668 * 64,0 = 6,99 100 100

VESSEL'S LENGHT IS 115,82 M, T P C IS 14,96 TONNES, M T C IS 330,7 TONNES. CENTRE OF FLOTATION ON THE CENTRE LINE.THERE IS 57,9 TONNES REMOVED 39,62 M AFT OF THE STERN. ORIGINAL DRAFTS ARE 5,23 M FWD & 5,33 M AFT. FIND NEW DRAFTS.

ANSWER :

LENGTH = 115,82 = 5791 -- 39,62 = 18,29 M 2

MOMENT TO CHANGE TRIM = W * D = 57,9 * 18,29 = 1058,99

CHANGE OF TRIM = 1059,99 = 3,20 CM 330,7

REMOVAL OF WEIGHT THE VESSEL WILL RISE = W = 57,9 = 3,6 CM T P C 14,96

ORIGINAL DRAFT RISE FWD = 0,23 M AFT = 5,33 M 5,30 5,298 TRIM -0.036 + 0,036 5.162 M 5,334 M


INTERPOLATION: GIVEN IS A DISPLACEMENT 25138.86

ANSWER :

Displ. Draft

24712.2000 5.8000

25138.8600 ????

25169.2100 5.9000

X = 426.66

0.1000 457.01

X = 426.66 x 0.1000 = 0.09335

457.01

5.8000 + 0.09335 = 5.89335 = 5.894

STABILITY CALCULATION FORMulas

Draft Displ TPC MCTC TKM LCB LCF

5.8000 24712.2000 45.7100 473.7600 15.9900 93.3200 92.3100

5.8934 25138.8600 45.6260 475.2911 15.8313 93.3013 92.2633

5.9000 25169.2100 45.6200 475.4000 15.8200 93.3000 92.2600

x 0.1000

426.66

457.01


FRESH WATER ALLOWANCE = (FWA) is the number of millimeters by which the mean draught changes when a ship passes from salt water to fresh water, or vice-versa, when the ship is loaded to the Summer displacement.

FWA (mm) = DISPL @ Summer

4xTPC @ Summer

EXAMPLE: A ship floats in SW at the Summer displacement of 1680 tones. If the TPCSW is 5.18, how much will the draught change by if the ship is towed to a berth where the density of the water is 1.000 t/m3 ?

ANSWER :

FWA (mm) = 1680 . = 81.1 mm

4 x 5.18

= The draught will increase by 81.1 mm

DOCK WATER ALLOWANCE = (DWA) of a ship is the number of millimeters by which the mean draught changes when a ship passes from salt water to dock water, or vice-versa, when the ship is loaded to the Summer displacement.

DWA (mm) = FWA x (RD dw1 - RD dw2 )

25

Note: RD = relative density

dw = dock water

EXAMPLE: A ship has a FWA of 200 mm. Calculate the change in draught that will occur if the ship proceeds from SW to a berth where the RD of the dock water is 1.018.?

ANSWER :

DWA (mm) = 200 x (1025 - 1018) = 56 mm = 5.6 cm

25

= The draught will increase by 5.6 cm


EXAMPLE: A ship is loaded to its summer displacement and is to proceed down river from a berth where the dock water RD is 1.004 to another berth where the dock water RD is 1.016. If the FWA is 270 mm, calculate the change in draught that will occur and state whether it is an increase or a de-crease.

ANSWER :

DWA (mm) = 270 x (1016 - 1004) = 129.6 mm = 12.96 cm

25

= The draught will decrease by 12.96 cm since the ship is moving into more

dense water

CHANGE OF TRIM DUE TO CHANGE OF DENSITY

FORMULA :

COT (cm) = DISPL x ( RD1 - RD2 ) x (LCF - LCB)

RD1 x MCTC x RD2

RD1

EXAMPLE: A ship is loaded to its summer displacement of 27803 and is to proceed on a river with a density of 1.006 from the outer anchorage which is at sea. Her LCF is 83.67 fm AP, LCB is 86.30 fm AP and MCTC of 377. Find the change of Trim.

ANSWER :

COT (cm) = 27803 x ( 1.025 - 1.006 ) x ( 83.67 - 86.30) = - 3.27 cm

1.025 x 377 x 1.006

1.025

Note: Because the answer is negative the change of trim will be 3.7 cms by the head.


CHANGE OF TRIM DUE TO CONSUMPTION FOR THE VOYAGE

FORMULA :

COT (cm) = CONS x (LCG foap - LCF foap)

MCTC

Note: CONS = Total weight of consumption for the whole voyage (FO+DO+FW)

LCGfoap = final LCG of the consumption measured from after perpendicular

LCFfoap = LCF (Hydrostatic table) from Departure and Arrival Displacement

MCTC = MTC (Hydrostatic table) from Departure and Arrival Displacement

EXAMPLE: A ship is loaded to its summer displacement of 54130 and her mean LCF is 85.615 fm AP, mean LCB is 86.30 fm AP, mean MCTC of 618.302 and with LBP of 175 mtr.

Consumption:

Wt LCG

FO 720 17.60

DO 72 68.25

FW 150 81.37

ANSWER :

Wt LCGfm mid Moments

FO 720 17.60 12672.00

DO 72 68.25 4913.79

FW 150 81.37 12220.50

TTL 942 31.64 29806.29

LCG foap = ( LBP ÷ 2 ) - LCGfm mid = ( 175 ÷ 2 ) - 31.64 = 55.86

COT (cm) = 942 x ( 55.86 – 85.615 ) = - 45.33 cm

618.302

Note: Because the answer is negative the change of trim will be 45.33 cms by the head.


WEIGHT TO SHIFT TO SAIL ON REQUIRED TRIM

FORMULA :

WEIGHTTO SHIFT = TRIMTO CHANGE x (MTC) x 100

DISTBETWEEN HOLDS

EXAMPLE: A ship is loaded to its summer displacement of 54130 with a present draft Fwd 12.198 and Aft 11.689. Her MTC 618.3019, COTCONSUMPTION 45.33 cm by head, COTDENSITY 11.55 cm by head, Dist fm Cargo Hold # 1 to Cargo Hold # 5 is 109.05 mtr. Find the weight to shift so as to depart compensating the trim caused by consumption and density difference and arrived even keel at next port.

ANSWER :

Present Trim = 11.689 - 12.198 = - 0.5090 mtr

COT due to consumption = (+) - 0.4533 mtr

COT due to change of density = (+) - 0.1155 mtr

COT total needed - 1.0778 mtr

WEIGHTTO SHIFT = 1.0778 x 618.3019 x 100 = 611.11 MT

109.05


CALCULATING FINAL DRAFT WHEN WEIGHT IS SHIFTED

FORMULA :

Tf = f x wtto shift x distfm hold - to hold

MCTC x LBP

Ta = a x wtto shift x distfm hold - to hold

MCTC x LBP

Where:

Tf = change in draught forward due to trim

Ta = change in draught aft due to trim

f = length of ship forward of the LCF

a = length of ship aft of the LCF

EXAMPLE: A ship floats at draughts F 12.198 m and A 11.689 m. Determine the final draughts if 611.11 tones is moved from Hold #1 to Hold # 5 with a distance of 109.05 m, given that MTC 618.3019 t-m, LBP 175 m and LCF 85.6147 m foap.

ANSWER :

a = 85.6147 m

f = LBP - LCF = 175 - 85.6147 = 89.3853

Tf = 89.3853 x 611.11 x 109.05 = 5956774.53 = 55.05 cms

618.3019 x 175 108202.83

Ta = 85.6147 x 611.11 x 109.05 = 5705495.93 = 52.73 cms

618.3019 x 175 108202.83

Calculating Final Draughts:

Initial draughts F 12.198 m A 11.689 m

Trim (-) 00.550 m (+) 00.527 m

Final draughts 11.648 m 12.216 m

Note: Weight is moved aft so the ship will trim by the STERN


TRIMMING

EXAMPLE: A ship floats at draughts F 11.41 m and A 11.86 m in salt water and is to complete load-ing to the summer displacement of 11.938 m. The remaining cargo is to be loaded in Hold #1 ( with LCGfoap 151.34 m ) and Hold # 5 ( with LCGfoap 42.29 m ) . How much cargo must be loaded into each hold for the ship to complete cargo with a trim of 0.612 m by stern?

ANSWER :

AMD (M) = FWD Draft + AFT Draft = 11.41 + 11.86 = 11.635 m

2 2

Hydro Data at AMD: Displ = 52615.90; MCTC = 613.1050; LCFfoap = 85.7590

TMD (M) = dA - (dA - dF ) x LCF foap = 11.86 - (11.86 - 11.41) x 85.759

LBP 175

= 11.6395 m

Hydro Data at TMD: Displ = 52638.35; MCTC = 613.1815; LCFfoap = 85.7563

Hydro Data at Limit Draft: Displ = 54129.83; MCTC = 618.2984; LCFfoap = 85.6148

MEAN Data: MCTC = 615.7398; LCFfoap = 85.6856

Cargo to Load = Displ @ Limit - Displ @ Present TMD

= 54129.83 - 52638.35 = 1491.38 mt

COT = Required Trim - Initial Trim = 0.612 - 0.45 = 0.1620 m by stern

Trimming Moment = COT in m x MCT mean = 615.7398 x 16.20

= 9974.98

Cargo for Hold 5 = (Cargo to Load x Hold 1 dist fm LCF mean) + Trimming Moment

Distance between Holds

= (1491.38 x 65.65) + 9974.98 = 989.308 mt

109.05

Cargo for Hold 1 = Cargo to Load - Cargo for Hold 5 = 502.07 mt


APPARENT MEAN DRAFT (AMD)

FORMULA :

AMD (M) = FWD Draft + AFT Draft

2

Note: dF = Draft Fwd; dA = Draft Aft

TRUE MEAN DRAFT (TMD)

FORMULA :

TMD (M) = dA - (dA - dF ) x LCF foap

LBP

Note: LCF foap = LCF of AMD from Aft perpendicular

TONNES PER CENTIMETER IMMERSION (TPC)

FORMULA :

TPC = WPA ( m2 ) x DENSITY ( t/m3 )

100

EXAMPLE: Calculate the TPC for a ship with a water-place area of 1500 m2 when it is floating in: (a) Fresh Water; (b) Dock Water of RD 1.005; (c ) Salt Water

ANSWER :

(a) TPC = 1500 x 1.000 = 15.000

100

(b) TPC = 1500 x 1.005 = 15.075

100

(c ) TPC = 1500 x 1.025 = 15.375

100


LOADING/DISCHARGING PROBLEM

EXAMPLE: A vessel has an initial mean draught of 4.40 m in salt water and is required to complete loading with a draught of 6.70 m. Calculate the amount of cargo that must be loaded.

Hydrostatic data: @ 4.40 m, Displ = 13200 t, TPC = 31.78 t/cm

@ 6.70 m, Displ = 20610 t, TPC = 32.66 t/cm

METHOD 1 (by Displacement) :

4.40 m = 13200 t

6.70 m = 20610 t

Cargo to load = 7410 t

METHOD 2 (by TPC) :

Sinkage/Rise = 6.70 m - 4.40 m = 2.30 m = 230 cm

Mean TPC = 31.78 + 32.66 = 32.22 t/cm

2

Cargo to load = Sinkage/Rise x mean TPC

= 230 cm x 32.22 t/cm

= 7410.6 t

EXAMPLE: A vessel arrives in port with a mean draught of 5.30 m in dock water RD 1.016. Hoe much cargo may be loaded till 5.70 m draught in the dock water?

Hydrostatic data: @ 5.30 m, Displ = 16080 t, TPC = 32.07 t/cm

@ 5.70 m, Displ = 17370 t, TPC = 32.22 t/cm


METHOD 1 (by Displacement) :

5.30 m = 16080 t x 1.016 = 15939 t

1.025

5.70 m = 17370 t x 1.016 = 17217 t

1.025

Cargo to load = 17217 t - 15939 t = 1278 t

METHOD 2 (by TPC) :

Sinkage/Rise = 5.70 m - 5.30 m = 0.40 m = 40 cm

TPC @ 5.30 = 32.07 x 1.016 = 31.79 t/cm

1.025

TPC @ 5.70 = 32.22 x 1.016 = 31.94 t/cm

1.025

Mean TPC = 31.79 + 31.94 = 31.865 t/cm

2

Cargo to load = Sinkage/Rise x mean TPC

= 40 cm x 31.865 t/cm

= 1274.6 t


The following points should be carefully noted:

The notations assigned to the load lines are as follows:

S : Summer

W : Winter

T : Tropical

WNA : Winter North Atlantic

F : Fresh

TF : Tropical Fresh

Each load line indicates the minimum freeboard that applies to the seasonal zone and/or area as stipulated in the International Convention on Load Lines, 1966.

The ship will be loaded to the appropriate load line when the waterline is level with the top edge of the mark con-cerned when floating in salt water (RD 1.025).

The spacing between the load lines are measured from the top edge on one line to the top edge of the other.

The assigned (Summer) freeboard is measured from the top edge of the Plimsoll line (which corresponds to the top edge of the Summer line) to the top edge of the deck line.

The “WNA” load line mark is only assigned to ships that are 100 meters or less in length. Ships over 100 m will load to the “W” mark as appropriate.

With the exception of “FWA” and “X”, all dimensions are the same for all ships, regardless of size of ship.

Load lines should be clearly and permanently marked on the ship’s side; dark on light background or vice-versa.


EXAMPLE: A vessel has a Summer load draught of 11.938 m FWA 270 mm and TPC of 50.012. The ship is loading at a berth in dock water RD 1.007 and the present draught is 11.80 m. Calculate the maximum amount of cargo that can still be loaded for the ship to be at Summer load line mark on reaching the sea allowing for 200 tones of diesel oil still to be loaded prior to sailing.

ANSWER :

1st. Calculate DWA (to the nearest mm)

DWA (mm) = 270 x ( 1025 - 1007 ) = 194.4 mm = 194 mm

25

2nd. Calculate the “permitted sinkage” in dock water. Always start with the required load line draught and work as follows:

Required Summer draught (1.025) 11.938 m

DWA (+) 0.194 m

Required draught (1.007) 12.132 m

Initial draught (1.007) 11.800 m

Permitted sinkage (1.007) 0.520 m

3rd. Calculate the maximum amount that can still be loaded in dock water, ignoring any allowances for fuel or other items.

Permitted sinkage (cms) = w ÷ TPC

Therefore: w = Permitted sinkage (cms) x TPC

= 52.0 cm x 50.012 x 1.007 = 2554.95 tonnes

1.025

Note: TPC must be corrected for the density of the dock water.

4th. Make allowance now for items other than cargo that must be loaded

Total that can be loaded * 2554.95 tonnes

Diesel still to load 200.00 tonnes

Maximum cargo to load 2354.95 tonnes

*Note: Had the given TPC not been converted for the density of the dock water, the total that could be loaded would have worked out as :w = 52 x 50.012 = 2600.62 tones; resulting in the ship being OVERLOADED BY 45.674 tones!


EXAMPLE: A vessel is floating in dock water RD 1.002 at draught of 11.50m. How much more cargo must be loaded to ensure that the ship will be at the Winter load line mark given that the Winter draught corresponding to the winter displacement is 11.69 m and the TPC is 49.88 and the FWA is 270 mm.

Note that the TPC value given will always be the one that corresponds to salt water for the waterline that is being loaded to.

ANSWER :

1st. Calculate DWA (to the nearest mm)

DWA (mm) = 270 x ( 1025 - 1002 ) = 248.4 mm = 248 mm

25

2nd. Calculate the “permitted sinkage” in dock water.

Required Winter draught (1.025) 11.690 m

DWA (+) 0.248 m

Required draught (1.002) 11.938 m

Initial draught (1.002) 11.500 m

Permitted sinkage (1.007) 0.438 m

3rd. Calculate the maximum amount that can still be loaded in dock water, ignoring any allowances for fuel or other items.

Permitted sinkage (cms) = w ÷ TPC

Therefore: w = Permitted sinkage (cms) x TPC

= 43.8 cm x 49.880 x 1.002 = 2135.72 tonnes

1.025

Total that can be loaded 2135.72 tonnes.


CURVES OF STATICAL STABILITY

(GZ CURVES)

The curve of statical stability, or GZ curve as it is most commonly referred to, is a graphical repre-sentation of the ship’s transverse statical stability.

Transverse statical stability is the term used to describe the ability of a ship to return to the upright, when it has been forcibly heeled by an external force and is momentarily at rest when floating in still water.

At any angle of heel, it is the horizontal disposition of G and B that determines the GZ value.

As a ship progressively heels over the righting lever, GZ, increases to some maximum value and then decreases until at some angle of heel it becomes negative i.e. it becomes a capsizing lever.

Calculating the value of GZ, at specified angles of heel for a ship’s particular condition of loading, will allow a curve of statical stability, or GZ curve, to be produced.

The greater the values of GZ, the greater will be the area under the curve. Minimum standards with respect to the area under the curve. Minimum standards with respect to the area under the curve (and other criteria) are specified in the “Code on intact stability (IMO)” and these are incorporated in the government legislation of most countries that adopt the IMO conventions.

KN values

As a ship heels the centre of buoyancy (B) constantly moves, it’s transverse position being dependent on:

the volume of displacement (and draught) of the ship;

The angle of heel at any instant.

θ

Correction to KN


The GZ value is predominantly dependent on the ship’s KG. Because of the many possible positions of G, it is convenient to consider the GZ that would exist if G were at the keel, termed KN, and to make a correction for the actual height of G above the keel.

If the figure is considered:

Correction to KN = KG x Sin θ

GZ = KN - ( KG x Sin θ )

Cross curves of stability (KN curves) are provided by the shipbuilder to allow GZ values to be deter-mined for any value of displacement and KG. Alteratively, KN values may be tabulated.

It is usual that KN values are given for angles of heel at 10° or 15° intervals.

CALCULATING THE MOMENT OF STATICAL STABILITY

AT SMALL ANGLES OF HEEL

A small angle of heel is often considered to be any inclination of the ship up to approximately 10°.

GZ = GM x Sin θ

Righting Moment = GZ x Displacement

(The above formula for GZ can only be used for small angles of heel.)


IMO STABILITY CRITERION

The area under the righting lever curve (GZ curve) should not be less than 0.055 metre-radian up to 30° angle of heel and not less than 0.09 metre-radian up to 40° or the angle of downflooding θf, if this angle is less than 40°. Additionally, the area under the righting lever curve (GZ curve) between the angles of heel of 30° and 40° or between 30° and θf, if this angle is less than 40°, should not be less than 0.03 metre-radian.

The maximum righting arm should occur at an angle of heel preferably exceeding 30°, but not less than 25°.

The initial metacentric height GM should not be less than 0.15 m.

(θf is an angle of heel at which openings in the hull, superstructures or deck-houses which cannot be closed weathertight immerse. In applying this criterion, small open-ings through which progressive flooding cannot take place need not be considered as open.)


COMPUTING ANGLE OF DECK EDGE IMMERSION

FORMULA :

Tan θ = (Depth - Draught)

½ Breadth

EXAMPLE: A vessel has a length of 182.79 m, breadth 31 m and depth of 16.76 m with a draught of 11.98 m. What is the angle of heel that will immerse the deck into the water?

ANSWER :

Tan θ = (16.760 - 11.938) = 17.28 deg

½ 31

COMPUTING MID DRAUGHT ON SEA SIDE DURING PORT STAY DAILY SURVEY

FORMULA :

correction = Tan List x ½ Breadth

Draughtsea side = Draughtwharf side ± (correction x 2)

Note: If vessel list to wharf side, correction is (-) minus.

If vessel list to sea side, correction is (+) plus.

EXAMPLE: A vessel has with breadth of 31 m, a wharf side draught of 8.42 m and vessel listing about 3 deg to sea side. What is the draught reading on the sea side?

ANSWER :

correction = Tan 3° x ½ 31 = 0.812 m

Draughtsea side = 8.42 m + ( 0.812 x 2) = 10.044 m


Material Approx. stowage factor (m3/mt)



ALFALA 1.39 TO 1.97 FERROCHROME, exothermic

0.18 TO 0.26 PHOSPHATE, uncal-cinedPIG IRON

0.700.30

ALUMINA 0.92 TO 1.28 FERROMANGANESE 0.18 TO 0.28 POTASH 0.77 TO 1.03

ALUMINA, calcined (CALCINED CLAY)

0.61 FERROMANGANESE, exothermic

0.18 TO 0.28 POTASSIUM SUL-PHATE

0.90

ALUMINA SILICA 0.70 FERRONICKEL 0.24 PUMICE 1.90 TO 3.25

ALUMINA SILICA, pellets

0.78 TO 0.84 FERTILIZER WITHOUT NITRATES, (Non-Haz)

0.90 TO 1.40 PYRITE (containing copper and iron)

0.33 TO 0.50

AMMONIUM NITRATE FERTILIZER (Non-Haz)

0.83 TO 1.00 FISH MEAL (anti-oxidant treated)

1.7 PYROPHYLLITE 0.50

AMMONIUM SUL-PHATE

0.95 TO 1.06 FLY ASH 1.26 QUARTZ 0.60

ANTIMONY ORE (STIBNITE) and Residue

0.34 TO 0.42 GRANULATED SLAG 0.90 QUARTZITE 0.64

BARYTES 0.34 GYPSUM 0.67 TO 0.78 RASORITE (ANHYDROUS)

0.67 TO 0.78

BAUXITE 0.72 TO 0.84 ILMENITE SAND 0.31 TO 0.42 RUTILE SAND 0.39

BORAX ANHYDROUS (Crude or Refined)

0.78 IRON ORE 0.29 TO 0.80 SALT 0.81 TO 1.12

BORAX (PENTAHYDRATE CRUDE, “Rasorite 46”)

0.92 IRON ORE PELLETS 0.24 TO 0.53 SALT CAKE 0.89 TO 0.95

CALCIUM NITRATE FERTILIZER

0.90 TO 0.95 IRON PYRITES 0.40 SALT ROCK 0.98 TO 1.06

CARBORUNDUM 0.56 IRONSTONE 0.39 SAND 0.50 TO 0.98

CEMENT 0.67 TO 1.00 LABRADORITE 0.60 SCRAP METAL varies

CEMENT CLINKERS 0.61 TO 0.84 LEAD ORE 0.24 TO 0.67 SEED CAKE 1.39 TO 2.09

CHAMOTTE 1.50 LIMESTONE 0.67 TO 0.84 SODA ASH(dense and light)

1.03 TO 1.67

CHROME ORE (CHROMIUM ORE)

0.33 TO 0.45 MAGNESIA 0.5 STAINLESS STEEL GRINDING DUST

0.42

CHROME PELLETS 0.60 MAGNESITE, natural 0.70 STONE CHIPPINGS 0.71

CLAY 0.66 TO 1.34 MANGANESE ORE 0.32 TO 0.70 SUGAR 1.00 TO 1.60

COKE (coal origin) 1.25 TO 2.93 MARBLE CHIPS 0.85 TO 1.00 SULPHATE OF POTASH AND MAGNESIUM

0.89 TO 1.00

COLEMANITE 0.61 MILORGANITE 1.53 SUPERPHOSPHATE 0.84 TO 1.00

COPPER GRANULES 0.22 TO 0.25 MONOAMMONIUM PHOSPHATE

1.21 SUPERPHOSPHATE, triple granular

1.17 TO 1.23

COPPER MATTE 0.25 TO 0.35 MURIATE OF POTASH 0.81 TO 1.12 TACONITE PELLETS 1.53 TO 1.67

CRYOLITE 0.70 PEANUTS (in shell) 3.29 TALC 0.64 TO 0.73

DIAMMONIUM PHOS-PHATE

1.20 PEBBLES (sea) 0.59 TAPIOCA 1.36

DOLOMITE 0.56 TO 0.65 PELLETS (concentrates)

0.47 UREA 1.17 TO 1.56

FELSPAR LUMP 0.60 PERLITE ROCK 0.98 TO 1.06 VERMICULITE 1.37

FERROCHROME 0.18 TO 0.26 PHOSPHATE, deflouri-nated

1.12 WHITE QUARTZ 0.61

PHOSPHATE ROCK, calcined

0.64 TO 1.26 ZIRCON SAND 0.36



Packaging Material Approx. stowage factor (m3/mt)

Packaging

ACETONE 1.4 DRUM AUTO PARTS 3.86 CASE

ACORN KERNEL MEAL

1.56 BAG BALL STONE 0.77 BULK

ACORN KERNELS 1.67 BAG BALL STONE 0.97 BAG

ACORN KERNELS 1.49 BULK BARITE 0.34 BULK

AIR RIFLES 3.1 CASE BARLEY 1.59 BULK

ALUMINA 0.61 BULK BASE OIL 1.46 DRUM

ALUMINA SILICA 0.7 BULK BASIC SLAG 0.87 BULK

ALUMINIUM DROSS 0.81 BULK BATH TOWELS 3.1 CASE

ALUMINIUM FERRO-SILICON POWDER

0.72 BULK BAUIXTE 1.1 BULK

ALUMINIUM INGOTS 0.88 BULK BEANS-HARICOT 1.45 BULK

ALUMINIUM NITRATE 0.6 BULK BEANS-HORSE 1.52 BULK

ALUMINIUM ORE 0.78 BULK BEARING 0.71 WODDEN BOX

ALUMINIUM SILICON POWDER

0.72 BULK BEE HONEY 1.38 DRUM

ALUMINIUM TEA POT 7.65 CASE BEER 1.32 CASE

ALUNITE 0.72 BULK BEET PULP 3.1 BAG

AMMONIUM BICAR-BONATE

1.4 BAG BEET PULP 1.6 BULK

AMMONIUM CHLO-RIDE

1.34 BAG BENTONITE 1.14 BULK

AMMONIUM NITRATE 0.92 BULK BENZYL CHLORIDE 1.4 DRUM

AMMONIUM NITRATE FERTILIZER (NON-

HAZARDOUS)

0.92 BULK BICYCLES 4.25 CASE

AMMONIUM NITRATE FERTILIZER(TYPE A)

1 BULK BITER APRICOT KER-NELS

2 BAG

AMMONIUM NITRATE FERTILIZER(TYPE B)

1 BULK BLACK LEAD PENCILS 2 CASE

AMMONIUM SUL-PHATE

1.01 BULK BLACK MELONSEEDS 2.32 BAG

ANEUROLEPIDIUM 7.5 BUNDLE BLENDE 0.55 BULK

ANTIMONY ORES 0.38 BULK BONES MEAL 2.27 BAG

ANTIMONY RESIDUE 0.38 BULK BOOTS 7.08 CASE

ANTI-MOSQUITO INCENSE

7.71 CASE BORAX 1.19 BAG

APPLE RINGS 2.97 CASE BORAX 0.78 BULK

ARGENITE 0.71 BULK BORAX ANHYPRUS 0.78 BULK

ASBESTOS 1.25 BAG BORIC ACID 1.8 BAG

ASBESTOS-CEMENT CORRUGATED

SHEETS

1.56 CASE BORNITE 0.45 BULK




Packaging

BOWN AND FEATH-ERS

4.53 CASE CEMENT CLINKER 0.73 BULK

BRISTLES 2.04 CASE CERARGYRITE 0.71 BULK

BROAN BEAN 1.5 BULK CERUSSITE 0.45 BULK

BROWN MANGANESE 0.51 BULK CHALCOCITE 0.45 BULK

BRUSHES 3.15 CASE CHALCOPYRITE 0.45 BULK

BUCKWHEAT 2 BAG CHALKS 2.3 CASE

BUCKWHEAT 1.72 BULK CHARCOAL 2.7 BULK

BUTTER 1.72 DRUM CHLORIDE OF LIME 1.4 DRUM

CALCINED PYRITE 0.45 BULK CHLORINATED PAR-AFFIN

1.6 PLASTIC DRUM

CALCIUM CARBIDE 1.34 DRUM CHLOROPICRIN 1.46 DRUM

CALCIUM CARBON-ATE PRECIPTATED

1.34 BAG CHOPSTICKS 2 CASE

CALCIUM NITRATE 0.9 BULK CHROME PELLETS 0.6 BULK

CALCIUM NITRATE FERTILIZER

0.93 BULK CHROMIC IRON ORE 0.39 BULK

CAMPHOR 2 CASE CHROMIUM ORE 0.39 BULK

CANDLE 1.6 BAG CIGARETTES 4.21 CASE

CANNED BEEF 1.67 CASE CLAY 0.82 BULK

CANNED CHICKEN 1.64 CASE CLAY 1.15 BAG

CANNED DUCK OR GOOSE

1.67 CASE CLOCKS 3.1 CASE

CANNED FISH 1.6 CASE CLOVES 3.1 CASE

CANNED FRUIT 1.6 CASE COAL 1.26 BULK

CANNED PORK 1.6 CASE COAL 1.2 BULK

CAP PAPER 2.4 TRAY COAL 1.09 BULK

CARBORUNDUM 0.56 BULK COAL 1.34 BULK

CARPET 2.12 BUNDLE COAL 1.26 BULK

CASTING IRON PIPES 1.4 BUNDLE COAL 1.2 BULK

CASTOR BEANS 2.2 BAG COAL 1.16 BULK

CASTOR OIL 1.67 DRUM COAL SLURRY 1.07 BULK

CAUSTIC SODA 1.26 DRUM COAL TAR 1.46 BULK

CEMENT 0.82 BULK COBALT GLANCE 0.51 BULK

CEMENT 0.85 BAG COBALT ORE 0.52 BULK




Packaging

COCK 43-10 1.06 WOODEN BOX DRIED CHILLIES 6.29 CASE

COKE 1.5 BULK DRIED POTATO SLICES

2 BAG

COKE 1.7 BULK DRIED SILK WERM CHRYSALIS MEAL

1.95 BAG

COKE 1.9 BULK DRIED TURNIP 2.7 BAG

COLEMANITE 0.61 BULK DRY BATTERIES 1.43 CASE

COPPER GRANULES 0.24 BULK DRY COPRA 2.02 BULK

COPPER MATTE 0.3 BULK EARTHY COBALT 0.6 BULK

COPPER NICKEL 0.45 BULK ELECTRIC FANS 2.4 CASE

COPPER ORE 0.45 BULK ENAMEL WARES 3.4 CASE

COPPER ORE CON-CENTRATE

0.45 BULK EQUAL ANGLES BARS 0.7 BUNDLE

COPPER PRECIPI-TATES

0.67 BULK ETHYL ALCOHOL 2 PLASTIC DRUM

COPPRA MEAL 1.8 BULK FEATHER ORNA-MENTS

8.5 CASE

COTTON BED SHEETS 3.1 BUNDLE FEATHERS MEAL 4.53 BAG

COTTON SEED MEAL 1.8 BULK FEEDING STUFFS 2.27 BAG

COTTON SEEDS OIL 1.81 DRUM FELDSPAR 0.6 BULK

COTTON SHEETING CLOTH

1.34 BUNDLE FERBERITE 0.74 BULK

COVELLITE 0.45 BULK FERROCHROME 0.22 BULK

CRINDING WHEEL 1.11 WODDEN BOX FERROMANGANESE 0.22 BULK

CRYOLITE 0.7 BULK FERROPHORPHORS 0.2 BULK

CUBE SUGAR 1.67 CASE FERROSILICON 0.6 BULK

DEAD MAGNESITE 0.82 BULK FERTILIZER WITHOUT NITRATE

1.15 BULK

DEAD MAGNESITE 1.33 BAG FIRE BRICKS 1.1 BUNDLE

DEFORMED BARS 0.65 BUNDLE FIRE CLAY 1.5 BULK

DIAMMONIUM PHOS-PHATE

1.2 BULK FISH MEAL 1.34 BULK

DIRECT REDUCED IRON

0.5 BULK FISH MEAL 2 BAG

DOLOMITE 0.61 BULK FLOURSCENT TUBE BALLAST

1.86 CASE

DOOR LOCKS 1.2 CASE FLUOSPAR 0.7 BGD

DOOR PULL 1.2 CASE FLY ASH 1.26 BULK

DOWN BIB COCKS 1.2 CASE FOLDING CHAIR 6.23 CASE

DRAWN NAIL WIRE 1.1 ROD FOOT BALL 9.91 CASE




Packaging

FOUNDRY SAND 0.75 BULK GRASS MEAL 2.7 BAG

FOUNTAIN PENS 2 CASE GRAY COBALT 0.52 BULK

FRESH APPLE 3.26 CASE GREEN BEEN 1.73 BAG

FRESH EGGS 2.55 CASE GROUND TILES 1.1 CASE

FROSEN BEEF BONE IN

2.6 CASE GRUDE MEDICINES 2.83 BAG

FROSEN BEEF BONE-LESS

2.12 CASE GYPSUM 0.76 BULK

FROSEN CHICKEN 2.4 CASE GYPSUM 1.22 BAG

FROSEN DONKEY 1.84 CASE HAND BAGS 2.3 CASE

FROSEN DUCK 2.55 CASE HARDBOARD 2 BUNDLE

FROSEN FISH 2.2 CASE HAZELNUT KERNELS 2.12 BAG

FROSEN HORSE 1.84 CASE HEMATITE 0.55 BULK

FROSEN MUTTON 2.83 CASE HEMP SEEDS 2.12 BAG

FROSEN PORK BONE-LESS

2.18 CASE HOCKSAW BLADE 0.85 CASE

FROSEN PRAWNS 2.4 CASE HORSE TAIL HAIRS 2.18 CASE

FROSEN RABBITS 2.2 CASE HUBNERITE 0.74 BULK

FURFURAL 1.51 DRUM HYROCHORIC ACID 1.8 PLASTIC DRUM

GALENA 0.45 BULK ILMENITE 0.36 BULK

GARLIO 2.5 CASE IRON ORE 0.55 BULK

GARMENTS 5.14 CASE IRON ORE CONCEN-TATE

0.55 BULK

GELATINUM 2 CASE IRON ORE PELLETS 0.39 BULK

GINSENG 3.1 CASE IRON OXIDES 0.45 BULK

GLASS MIRRORS 1.84 CASE IRON OXIDES 1.34 DRUM

GLASS SHEETS 1.4 CASE IRON SPONGS 0.45 BULK

GLASS WARES 2.55 CASE IRON SWARF 0.45 BULK

GLUTEN PELLETS 1.8 BULK IRONSTONE 0.55 BULK

GOAT WOOL 5.1 CASE JELLEY FISH 1.84 CASE

GRANITE 0.77 BULK JUTE BAGS 2.7 BUNDLE

GRANULATED SLAG 0.9 BULK JUTE CLOTH 2.27 BUNDLE

GRAPHITE IN FLAKES 2 BAG KAINITE 3.6 BULK

GRAPHITE IN POW-DER

1.13 BAG KRAFT PAPER 2.27 ROD




Packaging

LEAD CHROMATE ORE

0.39 BULK MALACHITE 0.4 BULK

LEAD NITRATE 0.8 BULK MANGANESE ORE 0.51 BULK

LEAD ORE 0.45 BULK MANGANIC CONCEN-TRATE

0.45 BULK

LEAD ORE CONCEN-TRATE

0.4 BULK MANGANITE 0.51 BULK

LEAD SULFATE ORE 0.45 BULK MANIOC 1.74 BULK

LIME 0.76 BULK MARBLE BLOCKS 0.77 BULK

LIME 1.1 BAG MATCH 3.96 CASE

LIMONITE 0.55 BULK MECHANICAL PRINT-ING PAPER

2.7 TRAY

LINSEED MEAL 1.8 BULK METHYL ALCOHOL 1.54 DRUM

LINSEED OIL 1.72 DRUM MILK POWDER 2.12 CASE

LINSEEDS 1.93 BAG MILORGANITE 1.53 BULK

LITHIUM 0.71 BULK MIXED ANIMAL HAIR 4.53 CASE

LITHOPONE 1.26 BAG MONOAMMONIUM PHOSPHATE

1.21 BULK

LOCKS 1.33 CASE MOONSTONE 0.74 BULK

LOGS 3.1 BULK MOTOR 1.24 WODDEN BOX

LUMP SUGAR 1.61 CASE MURIATE OF POTASH 0.87 BULK

MAGNESITE 1.5 BAG MUSIOAL INSTRU-MENT

2.8 CASE

MAGNESITE 0.61 BULK NAIL CLIPPERS 1.1 CASE

MAGNESIUM CAR-BONATR ALIGHT

1.26 BAG NAPHTHALENE BALLS 2.36 CASE

MAGNESIUM CHORIDE

1.61 BAG NEPHELINESYENITE 0.45 BULK

MAGNESIUM NITRATE 0.82 BULK NEWSPRINT PAPER 2.8 ROD

MAGNETITE 0.55 BULK NEWSPRINT PAPER 2.7 TRAY

MAGNISIUM SUL-PHATE

1.29 BAG NICKEL ORE CON-CENTRATE

0.45 BULK

MAIZE 1.6 BGD NIGER SEED MEAL 1.8 BULK

MAIZE 1.36 BULK NOTE BOOKS 2 CASE

MAIZE BROKEN 1.61 BAG NPK 1.5 BAG

MAIZE COB MEAL 3.54 BAG NPK FERTILIZER 0.99 BULK

MAIZE EMBRYO 1.78 BAG OATS 1.88 BULK

MAIZE EMBRYO CAKE 1.81 BAG OIL CAKE 1.8 BULK

MAIZE OIL 1.67 DRUM ONION BULBS 2.12 CASE




Packaging

OTHER KINDS OF WINE

1.36 CASE POSHAN BAUXITE 0.7 BULK

OXILIC ACID 1.54 BAG POTASH 0.9 BULK

PADDY 2 BAG POTASSIUM CHLO-RATE

1.47 DRUM

PADDY 1.5 BULK POTASSIUM FELSPAR SAND

0.74 BULK

PALM KERNEL MEAL 1.8 BULK POTASSIUM NITRATE 0.88 BULK

PARAFFIN WAX 1.53 BAG POTASSIUM SUL-PHATE FERTILIZER

0.9 BULK

PEACH KERNELS 2.4 BAG POTATO 1.95 BAG

PEANUT KERNELS 1.84 BAG POYVINYL CHORIDE 1.34 BAG

PEANUT MEAL 1.8 BULK PRINTED MATTERS 2.27 CASE

PEANUT OIL 1.59 DRUM PRINTED SHEETING CLOTH

1.7 BUNDLE

PEANUTS 3.29 BULK PRITING INK 1.31 DRUM

PEBBLES 0.59 BULK PSILOMELANE 0.51 BULK

PELLETS 0.47 BULK PUMICE 2.58 BULK

PENCIL PITCH 1.46 BULK PUMPKIN SEEDS 2.4 BAG

PERLITE ROCK 1.02 BULK PVC RESINS 1.46 BAG

PETROLEOM OIL 1.4 DRUM PYRIDNE 1.47 DRUM

PETROLEOM OIL 1.34 BULK PYRITE 0.45 BULK

PETROLEUM COKE 1.46 BULK PYRITE ASHES 0.45 BULK

PETROLEUM COKE 1.46 BULK PYRITE CINDERS 0.45 BULK

PHENOL 1.46 DRUM PYRITES 0.45 BULK

PHENOLIC MOULDING POWDER

1.8 DRUM PYROLUSITE 0.51 BULK

PHOSPHATE ROCK 0.95 BULK PYROPHYLITE 0.5 BULK

PHOSPHATE ROCK 0.7 BULK QUARTZ 0.62 BULK

PHOSPHORIC ACID 1.77 DRUM RABBIT HAIR 5.38 CASE

PIANOS 3.1 CASE RADIO 2.8 CASE

PIG IRON 0.4 BULK RANGA OF PAINTS 1.04 DRUM

PINENUT KERNELS 2.12 BAG RAYON GOODS 2.55 BUNDLE

PING PONG BALL 9 CASE RAYON SPUN YARN 2.55 BUNDLE

PITCH PRILL 1.46 BULK RED BRICKS 1.1 BUNDLE

PLYWOOD 2.55 BUNDLE RED GINSENG 2.97 CASE




Packaging

RED LEAD 1.06 BAG SLEEPERS 2 BULK

REFINED SALT 0.99 BAG SLIG 0.55 BULK

RESIN 2.5 bag SLIVER ORE 0.71 BULK

RICE 1.46 BAG SMALL RED BEANS 1.56 BULK

RICE BRAN MEAL 1.8 BULK SMALL RED BEANS 1.6 BAG

RICE STRAW 5 BUNDLE SMALL RED BEANS PASTE

1.89 CASE

ROSIN 1.89 DRUM SODA ASH 1.9 BAG

ROUND IRON WIRE NAIL

1.1 CASE SODA ASH 1.84 BAG

RUTILE SAND 0.39 BULK SODA FELSPAR SAND 0.74 BULK

RYE 1.3 BULK SODIUM ALGINATE 1.74 DRUM

SACCHARIN SODIUM 1.33 DRUM SODIUM ASH 1.35 BULK

SALT 1.2 BAG SODIUM BICARBON-ATE

1.34 BAG

SALT 0.96 BULK SODIUM CHLORATE 1.4 BAG

SALT CAKE 0.93 BULK SODIUM MDYBDATE 1.63 DRUM

SALT MUSHROOMS 1.72 CASE SODIUM NITRATE 0.88 BULK

SALTED CASINGS 1.93 DRUM SODIUM NITRATE AND POTASSIUM

NITRATE

0.88 BULK

SALTED VEGETABLE 1.6 DRUM SODIUM NITRITE 1.4 BAG

SAWDUST 2.9 BULK SODIUM SULTHATE 1 50KG´�

SCHEELITE 0.74 BULK SORGHUM 1.59 BAG

SCISSORS 1.19 CASE SORGHUM 1.3 BULK

SEED CAKE 1.8 BULK SOYABEAN 1.4 BULK

SELL OF BUCKWHEAT 7.15 BAG SOYABEAN 1.61 BAG

SHOE-TACKS 1.1 CASE SOYABEAN OIL 1.53 DRUM

SIDERITE 0.45 BULK SOYABEAN OIL 1.15 BULK

SILICA 0.8 bulk SOYBEAN MEAL 1.8 BULK

SILICA SAND 0.74 BULK STEEL CHANNELS 0.88 BUNDLE

SILICOMANGANESE 0.22 BULK STEEL FLANGE 0.85 CASE

SILK FLOWERS 8.5 CASE STEEL HEAD RIVATS 0.9 GUNNYSACK

SILK GOODS 3.4 BUNDLE STEEL HEXAGONAL NUTS

1.1 GUNNYSACK

SKIN PLATES 3.4 CASE STEEL I BEAMS 0.82 BUNDLE




Packaging

STEEL PLATES 0.48 BUNDLE TOILET PAPER 5.1 ROD

STEEL ROUND BARS 0.6 BUNDLE TOILET PAPER 9.9 CASE

STEEL SQUARE BARS 0.6 BUNDLE TRICHLORETHYIENE 1.57 DRUM

STEEL SWARF 0.45 BULK TUNGSTEN 0.74 BULK

STEEL TUBES 1.1 BUNDLE TURPENTINE 1.56 DRUM

STEEL WINDOW SEC-TIONS

0.7 BUNDLE TYRES FOR TRUCKS 5.71 BULK

STEEL WIRE ROPE 1.6 ROD UREA 1.37 BULK

STONE IN GRAINS 0.93 BAG VANADINITE 0.65 BULK

SUIT-CASES 6.23 CASE VANADIUM ORE 0.65 BULK

SULPHUR 1.13 BAG VERMICULITE 1.37 BULK

SULPHUR 0.74 BULK VE-SU 2.12 CASE

SUN-FLOWER EXPEL-LER

2 BAG VE-TAIN 2.18 CASE

SUNFLOWER SEED MEAL

1.8 BULK WAX CRAYONS 1.4 CASE

SUN-FLOWER SEEDS 2.72 BAG WHEAT 1.35 BULK

SUN-FLOWER SEEDS 3.4 CASE WHEAT 1.46 BAG

SUPERPHOSPHATE 0.92 BULK WHEAT BRON 3.1 BAG

SUPERPHOSPHATE TRIPL

1.2 BULK WHITE OIL 1.63 DRUM

SWEET POTATOES 1.74 BULK WHITE SUGAR 1.53 CASE

SWITCHES 1.69 CASE WILD SESAME SEEDS 1.95 BAG

TABLE VICE 1.17 WODDEN BOX WILD VEGETABLE 1.4 DRUM

TACONITE 0.45 BULK WILLOW ARTICLES 8.8 CASE

TALC 0.69 BULK WOLFRAMITE 0.74 BULK

TALC IN POWDER 1.07 BAG WOODCHIPS 2.8 BULK

TANKAGE 1.5 BULK WOODPULP PELLETS 3 BULK

TEA 3.43 CASE WRITING INK 1.52 CASE

TERRAZZO TILES 1.05 CASE WRITING PAPER 2.5 TRAY

THERMOS 5.1 CASE YELLOW PETROLEUM JELLY

1.31 DRUM

TIMBER 2.55 BUNDLE ZINC CHLORIDE 1.21 BAG

TNT 1.6 WODDEN BOX ZINC CRUDE 0.45 BULK

TOBACCO 4.1 BUNDLE ZINC ORE 0.45 BULK




Packaging

ZINC ORE CONCEN-TRATE

0.45 BULK ZINC SLUDGE 0.45 BULK

ZINC RESIDUES 0.45 BULK ZIRCON SAND 0.36 BULK

ZINC SINTER 0.45 BULK

V o l u m e 1 , I s s u e 1 N o t e s

seminar workshop on “trim and stability” for bulk...

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