GIÁM ĐỊNH MỚN NƯỚC VÀ TÍNH TOÁN HÀNG HÓA (A PRACTICAL GUIDE TO DRAFT SURVEYS TRIMMING CALCULATIONS)

On a Cape-size vessel A = 1.00 metres and LBM = 248.50 meters Since Correction to be applied to the ford read draft marks = a = A x Ta LBM Therefore a = 1 x Ta 248.5 And a = 0.004

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INDEX

CHAPTER I - THE 6 - SIDED DRAFT 3

CHAPTER II - THE DRAFT SURVEY 11

CHAPTER III - TRIMMING 22

CHAPTER IV - HOG / SAG 51

CHAPTER V - CONTROLLING DRAFTS 59

CHAPTER VI - MAXIMUM DRAFTS 63

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CHAPTER I - THE 6 - SIDED DRAFT

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In the above diagram

L.B.P = Length between Perpendiculars

L.B.M = Length between Marks

C = LCF = Longitudinal Center of Floatation - distance fm midships

c = difference between draft at LCF & draft at midships

dF = Ford Draft mark

dA = Aft Draft Mark

FP = Forward Perpendicular

AP = Aft Perpendicular

WL = water line when even keel

W1L1 = water line when trimmed (in this case by stern)

T = TRIM (at the perpendiculars)

Ta = APPARENT TRIM (at the marks)

A = Distance between the ford draft mark & ford perpendicular

a = Difference between draft at ford mark & draft at ford perpendicular

B = Distance between the aft draft mark & aft perpendicular

b = Difference between draft at aft mark & draft at aft perpendicular

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On a Cape-size vessel

A = 1.00 metres and LBM = 248.50 meters

Since Correction to be applied to the ford read draft (marks) = a = A x Ta

LBM

Therefore a = 1 x Ta

248.5

And a = 0.0040241 x Apparent trim

And since B = 10.5 meters

Therefore b = 10.5 x Ta

248.5

And b = 0.0422535 x Apparent Trim

When the Draft Marks are on the Perpendiculars then no correction is to be applied

If the ford draft mark is aft of the ford perpendicular and the aft draft mark is ford of the aft perpendicular as in the diagram and on most but not all vessels then,

“a” will be subtracted from the read ford draft to get the draft ford and

“b” will be added to the read aft draft to get the aft draft

WHEN THE VESSEL IS TRIMMED BY THE STERN

WHEN A VESSEL IS TRIMMED BY THE HEAD, the correction sign will be

- because the loadline mark is exactly at midships and the midship draft marks are actually displaced either just forward or just aft of midships

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No matter what is taught in the various colleges of the world with respect to hydrostatic draft, bulk carrier draft calculations the world over is based on the six-sided formula Drafts are read on all six sides -

Fp - Ford port Fs - Ford stbd Mp - Midship port Ms - Midship stbd

Ap - Aft port As - Aft stbd

After these drafts are read, the trim correction is applied to the ford & aft drafts & to the midship drafts when relevant (on some vessels the midship draft marks are not exactly

at midships) - (same explained later) to convert these read drafts to drafts at the perpendiculars -

We are now left with drafts – (where “c” stands for “corrected”)

Fpc, Fsc, Mpc, Msc, Apc, Asc - which are the same as above except that they are

corrected to the perpendiculars or midships as the case may be

Using the mean of each set we get the Ford, Aft & Midship Drafts -

(Fpc+ Fsc) = F (Ford Draft) (Mpc + Msc) = M (Midship Draft)

2 2

(Apc + Asc) = A (Aft Draft) (F + A) = Mean (Mean Draft)

2 2

Now to get the draft, reqd for calculating displacement, known as Mean Quarter Mean

(MQM) draft we use the six - sided formula -

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The 6-sided formula is central to all calculations on bulk carriers & must be dinned into your head and understood completely

In the first place it is used in all actual all draft calculations, but can also be used in calculation of cargo figures as the following examples will show -

pre-It is to be remembered that using the 6 sided formula for actual draft calculation is completely accurate, the following examples, which are useful in pre calculation, makes one assumption which is theoretically incorrect, and that is, that we are assuming the vessel is trimming about midships - in actual fact the vessel trims about the Centre of Flotation which is not necessarily midships, but the

pre-calculation examples are shown for when the vessel is completing loading, when in case of bulk carriers we almost always finish even keel, without trim, or very close to even keel, which makes any inaccuracy caused by making the assumption that the vessel is trimming about midships, so very small as to be discarded In any event pre-calculations are only that and the actual picture only obtained, during & after loading

Before any examples are given, the formula must be understood properly -

The Mean Quarter Mean Draft which is used for obtaining displacement concept must

be understood properly

MQM = 6 M + F + A or MQM = 6 M + 2 Mean

8 8

And Hog / Sag = Difference between Midship Draft and Mean Draft

When Midship Draft is greater than Mean Draft, Vessel is Sagging

When Midship Draft is less than Mean Draft, Vessel is Hogging

Now it must be understood that if M= 17.0 meters and the vessel is sagging by 10 cms,

then the MQM is NOT 16.90 meters - to illustrate

Example 1:

Given that Midship Draft (M) = 17.0 meters, and vessel is sagging 10 cms

Vessel is sagging 10 cms, therefore the Midship Draft is 10 cms more than the Mean Draft

Therefore Mean Draft = M - 0.10 = 17.0 - 0.10 = 16.90 meters

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Example 2:

Consider a Cape-size bulk carrier, SILC, of about 151,000 dwt

Given that max draft is 17.5 meters, by experience you expect that the vessel with that cargo, loaded in all 9 holds will sag 10 cms, and you wish to sail out even keel - you will by the formula get Midship draft = 17.50 meters (v/l sagged & even keel), Mean draft = 17.40 meters and as v/l to finish even keel, F = 17.40m A = 17.40m and therefore,

To expand: the density of Dockwater = 1.023 lightship = 18643 K= 400

(FO: 2000 DO: 100 FW: 200 Unpumpable Ballast: 100)

Max draft = 17.00 meters V/l to finish with 20 cms trim by stern 12 cms of sag

To calculate Ford & Aft drafts and MQM to obtain displacement & therefore cargo

To get max cargo have the Midship draft equal to the max draft - remember that in the formula the midship draft is multiplied 6 times, the fore & aft draft only once so it is much better to have the midship draft maximum

As v/l is sagging 12 cms, (Midship Draft – Sag) = Mean Draft = 17 - 0.12 = 16.88

Now to calculate the fore & aft draft we assume that the v/l trims about the midships and since v/l is to be trimmed 20 cms & is trimming equally ford & aft

& therefore ford draft = 16.88 - (0.20 / 2) = 16.78

& aft draft = 16.88 + (0.20 / 2) = 16.98

MQM = 6 x 17.0 + 16.78 + 16.98 = 16.97 meters

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Example 4:

Cargo to load = 130,000 MT FO: 1800 MT DO: 100 MT FW: 250 MT

U/P ballast: 100 MT LTSHIP: 18643 K: 400 Density: 1.025

V/l to sail out even keel, sagged by 8 cms, calculate F, A, M

Displacement = 130000 + 1800 + 100 + 250 + 100 + 18643 + 400

= 151293

For SILC, displacement of 151293 in 1.025 corresponds to draft of 15.85 meters Therefore MQM is 15.85 meters

Let Midship Draft = X meters

Since v/l sagged 8 cms, Mean Draft = (X - 0.08) meters

From Formula MQM = 6M + 2Mean = MQM = 6X + 2 (X - 0.08)

8 8

8(MQM) = 6X + 2X - 0.16

(8 x 15.85) = 8X - 0.16

(126.8) + 0.16 = 8X

Therefore X = Midship Drafts = 15.87 meters

V/l sagged 8 cms, Mean Draft = 15.87 - 0.08 = 15.79 = F = A (v/l finishing even keel)

Same conditions as in Example 4 except vessel to finish hogged by 10 cms

Let Midship draft = X meters

As v/l hogged by 10 cms, Mean Draft = (X+0.10) meters

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Example 6:

MQM = 15.85m Sag: 8 cms Trim: 14 cms by stern RD: 1.025

Midship draft = X mts Mean draft = (X - 0.08) mts (sag of 8 cms)

MQM = 6M + 2Mean => MQM = 6X + 2(X-0.08)

8 8

MQM = 6X + 2X - 0.16 => 8 MQM +0.16 = 8X => 8(15.85) + 0.16 = 8X

8

Therefore X = 15.87 = Midship draft Sag = 8cms

Therefore Mean draft = 15.87 - 0.08 = 15.79 mts

Assuming vessel trims abt midships & trim = 14 cms F = 15.79 - 0.07 = 15.72

Therefore MQM in 1.010 = 15.968 mts (from SILC)

Assuming that the sag remains same, & very likely it will,

MQM = 6X + 2(X - 0.08) where X = Midship draft

8

=> 8(15.968) = 6X + 2X - 0.16 Therefore X = 15.988 = Midship draft

& Mean Draft = 15.988 - 0.08 = 15.908

Again assuming v/l trims around the midship we will have F=15.838 & A=15.978

However there will be a change of trim due change of density & this will have to be applied to get the actual Ford & Aft Drafts in the DW of density 1.010

If the vessel were to hog instead of sag, say hog by 10 cms then as in Example 5,

Assume X = Midship draft and Mean Draft = (X + 0.10) - all other calculations same

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CHAPTER II - THE DRAFT SURVEY

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THE DRAFT SURVEY

Always keep the 6-sided formula in mind when the draft survey is being conducted Remember not to create friction between yourself & the draft surveyor, but at the same time remain firm In some ports such as Hay Point, the draft surveyor’s read draft is taken as the official draft by the port authorities, & in Hay Point if you are overloaded

by even 1 cm they may refuse to sail the vessel, & with the swell you encounter it can

be a notoriously difficult place to read the draft

Ideally both of you should be able to read the draft exactly the same, but this is not always the case If there is a difference, be prepared to compromise a little on the fore & aft drafts, but not on the midship drafts - remember the 6 sided formula - the midship draft is multiplied 6 times, the fore & aft only once each Remember also, that very often the surveyor at the load port is often representing the shipper & would like to show more cargo whereas the surveyor at the discharge port very often represents the receiver & might like to show less cargo, so keep this in mind & guard against it

In places where there is a swell have asked the surveyors the best way to get an accurate draft - the best way I have found was taught me by surveyors in Australia - they very rapidly read the draft 20 times, noting the (crest and trough reading each time - they then discard the 2 extreme readings on either side of the mean & then take the average

of the remaining 16 readings - their argument is that if you take your time over the reading, mentally & physically you lock onto a particular reading & stick with it -

If the level is oscillating between 17.0 & 17.8 meters & you feel it is 17.46 meters, you mentally & physically lock onto 17.46 meters & discard anything else - I have found this to be true - & the rapid 20 reading & noting method much better

Remember also that the initial draft survey at the load port is basically to find out what the constant is - more often than not in the case of a Panamax bulker will show a negative constant in ballast with trim - no surveyor likes to record a negative constant & some adjustments are made with the ballast to record a positive constant This actually results in the cargo figure after loading to show a little less than is actually on board This is not really bad, because then you do not finish the discharging and find a shortfall

of cargo - you normally land up with slightly more cargo, but remember this - when you complete the discharging & you land up with a cargo figure greater than you have at the loaded end, the surveyor is quite happy to make adjustments so that the figures tally However should you land up with slightly less cargo, at the discharging end, the surveyor, who normally represents the receiver, who wants to pay less, is not likely to make adjustments in order that the figures tally So, do not be gung-ho at the loading end to try & show more cargo than there actually is The Surveyor might want to do this, as he represents the shipper, who wants to sell more, but nobody gives you a medal for having more cargo, and should you land up with less cargo at the discharge port, all kinds of hell breaks loose

Where you should try to show the maximum cargo, even insisting on recording a

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CALCULATION

PROCEDURE:

DRAFTS - Drafts are to be read on all 6 sides, if possible - they are then corrected to

the perpendiculars and to midships (if required) - see Page 4

DENSITY - As soon as possible after reading the drafts obtain the density of the dock

water This should be done without any delay as the density at many ports varies with the tide Preferably take 3 samples, always on the off-shore side of the vessel Ideally, you should lower the container (preferably with a perforated lid) to the maximum draft

& pick it up at a constant speed

Most instruments are calibrated for water in vacuum & so 0.0011 and 0.002 should be subtracted for glass and brass instruments respectively to allow for the different buoyancy of water in air

Glass instruments are more accurate than brass ones Also, as the instruments are not being used at their calibration temperatures, further corrections supplied with the instruments must be used

Most important - remember that there is a difference between a draft hydrometer and a loadline hydrometer - surveyors in Australia and South Africa use the loadline hydrometer which is theoretically the correct one - using this you get Seawater density

of about 1.023 & not 1.025 as you would get on the draft hydrometers normally carried

- as we normally load in Australia & South Africa we should not have a problem because when we reach the discharge port & the surveyor there uses the reading that

we get we get on our hydrometer we land up with more cargo - when they use the reading that they get in Australia & South Africa do not argue with them because they are right - I do not know where else they do this but as those 2 are export countries you should not have a problem - where you would have to be careful is if your discharging

in one of these countries - you would land up with less cargo when you arrive if you have not made the correction in the load port Remember though that when you load right up to the loadline assuming the density to be 1.025 as read from the draft hydrometer, you are actually overloaded by a factor of density of 0.002 when using a glass draft hydrometer

PLEASE SEE THE EXPLANATION OF THIS TOPIC IN THE PUBLICATION, “BULK CARRIER PRACTICE” and in the attached file

CORRECTION TO THE PERPENDICULARS

For correction to the perpendiculars of the fore and aft drafts see Page 4

If the midship draft is obtained by measuring the freeboard to the deckline then no correction is necessary as the loadline disc, IN MOST CASES, can be considered to be

at the midlength of the vessel –

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CORRECTION FOR HULL DEFORMATION

If the vessel is neither hogged or sagged (at amidships) then the midship drafts will be the mean of the fore & aft drafts but this is very seldom the case - the best formula used for calculating the draft making allowance for the various hull deformations for a bulk carrier or tanker has been found to be the 6 sided formula where

MQM = FORD DRAFT + AFT DRAFT + (6 x AMIDSHIPS DRAFT)

FIRST TRIM CORRECTION = TRIM (in cms) x LCF(in meters) x TPC

(IN TONNES) LBP (in meters)

In the above formula LCF is the distance of the Centre of Flotation (COF) from

amidships

In some ships the LCF is given in the hydrostatic tables from the aft perpendicular

Remember in this formula, the LCF is the distance of the COF from amidships

Also on some ships the sign (-) indicates the LCF is aft of midships, on others the sign (-) indicates the LCF is ford of midships - please make sure you know exactly what the sign convention in your ship means

This correction is also known as the ‘layer correction’ and is applied as follows - when the COF is in the same direction, from amidships, as the deepest draft it is added, and when the COF is on the other side of amidships as the deepest draft it is subtracted

This correction does not allow for the fact that, when a ship trims, the COF moves from its tabulated position Some ships have corrections for this, but when this is not provided the following correction called the 2nd trim correction must be applied -

2ND TRIM CORRECTION = (TRIM IN METERS) x (TRIM IN METERS) x 50 (dM/dZ)

(IN TONNES) LBP (in meters)

Where dM/dZ is the difference between the MCT for a draft of 50 cm greater than the corrected mean draft and 50 cm less than the corrected mean draft

I.E if the corrected mean draft is 12.0 meters then dM/dZ would be the difference

between the MTC (Moment to change trim) at 12.50 meters and 11.50 meters

This correction is ALWAYS ADDED to the displacement

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CORRECTION FOR HEEL

Vessel should be upright for draft survey but when it is not the following correction

must be applied

CORRECTION FOR HEEL IN TONNES = 6 x (TP1 - TP2) x (D1 - D2)

Where TP1 = TPC for the deepest draft amidships

TP2 = TPC for the shallower draft amidships

D1 = Deepest draft amidships (in METRES)

D2 = Shallower draft amidships (in METRES)

This correction is ALWAYS ADDED to the displacement because the effect of heel is

to increase the water plane area & so lift the ship out of the water

CORRECTION FOR DENSITY

Almost all ships have their displacement tables tabulated for a Relative Density of 1.025

However what we are interested in finding is the actual displacement,

I.E is the displacement in the dock water that the vessel is lying in

TRUE DISPLACEMENT = R.D OF DOCK WATER x SCALE DISPLACEMENT

R.D USED FOR DISPLACEMENT SCALE

eg: On SILC if your draft is 15.00 meters & v/l is lying in DW of 1.017 then true

displacement would be calculated as follows:

Scale Displacement for 15.00 meters = 143291.2 tonnes

True Displacement = SCALE DISPLACEMENT x R.D OF DOCK WATER

R.D FOR DISPLMNT SCALE

= 143291.2 x 1.017

1.025

= 14217.3 tonnes

The displacement now obtained is the true displacement, within the limits of accuracy

of the drafts and the ship’s stability data While the actual calculation on all of our ships

is just a matter of feeding in the read drafts and the deductibles it is wise to know what

is actually being done - also not all surveyors are above board & some will try & pull wool over your eyes in order to show more cargo at the loading port and less cargo at the discharge port

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The weight of the cargo on board is found from the displacement as follows : -

1 AT THE LOADING PORT

(a) Before Loading Cargo

The initial draft survey taken before loading cargo is, as said earlier, to determine the vessel’s constant In truth, the value of the vessel’s constant should not matter, as it does not actually come into play, when calculating the cargo, (as will be shown later) but every draft survey will be used to determine the constant - this is because no surveyor likes to have a negative constant

1 From the calculated displacement subtract the light displacement to obtain the

Deadweight

2 As soon as possible after reading the drafts & density, sound all the fuel, ballast and Fresh water tanks Correct the soundings for list and trim and using the Calibration Tables calculate the fuel, diesel, fresh water and water ballast from the deadweight The remainder represents the constant

The value of the constant in the ship’s data represents the constant for a new ship

However, as the ship ages its weight increases (primarily due to the reluctance of ship’s officers to throw anything away.) In many cases the constant is too low The tendency is

to try and show as small a constant as possible at the load port in order to show more cargo on board - avoid this tendency as the surveyor at the disport is not going to be charitable and will be trying to show as little cargo as possible & will be trying to compute as large a constant as he or she can The time to try and get as small a constant, without going to ridiculously low levels, is when you are on a charter when the Owner

is getting paid on basis of freight

(b) After Loading

Read the drafts and calculate the loaded displacement Using the constant found previously (prior loading), sound fuel, diesel, fresh water and ballast tanks, correct for trim & list, use the calibration tables to find the quantity of fuel, diesel, fresh water and ballast, subtract the lightship from the loaded displacement together with the deductibles and you will have the cargo loaded

AT THE DISCHARGING PORT

Repeat the draft survey prior commencing discharging and after completion of discharge and you will be able to get the cargo discharged Remember you will get your constant on completion of discharge - at the initial draft survey at the disport, using the Bill of Lading figures, you will only get an idea of what your constant is

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At the discharge port cargo is calculated as follows-

Let Displacement prior commencement of discharge = A

Fuel Oil prior commencement of discharge = FO1

Diesel Oil prior commencement of discharge = DO1

Fresh Water prior commencement of discharge = FW1

Ballast Water prior commencement of discharge = BW1

Constant = K

Light ship = LS

Deductibles on arrival = FO1 + DO1 + FW1 + BW1 = a

Displacement prior commencement = A = a + K + LS + Cargo on Board

CARGO ON BOARD = A - (a + K + LS)

Opening the bracket

CARGO ON BOARD = A - a - K - LS

As a = FO1 + DO1 + FW1 + BW1

CARGO ON BOARD = A - FO1 - DO1 - FW1 - BW1 - K - LS

Using this you should in theory be able to compute the cargo on board, but this assumes that the value of the lightship is accurate and your computed constant at the load port is accurate This however is not always true so the cargo is actually calculated as follows -

Let Displacement after completion of discharge = B

Fuel Oil on completion of discharge = FO2

Diesel Oil on completion of discharge = DO2

Fresh Water on completion of discharge = FW2

Ballast Water on completion of Discharge = BW2

Constant = K

Light Ship = LS

Deductibles on completion of discharge = b = FO2 + DO2 + FW2 + BW2

Displacement on completion of discharging = B = b + K + LS

Please note that K & LS are the same prior commencement & at completion of

discharge

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Ideally, if there were no changes to fuel, diesel, fresh water and ballast between

commencement and completion of discharge, cargo on board would be a simple matter

of A- B But as all these values change,

CARGO ON BOARD = ((A - (a + K + LS))) - ((B - (b + K + LS)))

Opening the single brackets

CARGO ON BOARD = ((A - a - K - LS)) - ((B - b - K - LS))

Opening the double brackets

CARGO ON BOARD = A - FO1 - DO1 - FW1 - BW1 - B +FO2+DO2+FW2 + BW2

Please carefully see the signs on all the values in the above equation - the tendency

is to try and show less ballast on completion of discharge - please remember that less ballast at the completion of discharge will result in LESS CARGO

A problem sometimes occurs on completion of discharge, when the ballast tanks are absolutely full and the vessel is trimmed by the stern - a shrewd surveyor then accepts a full sounding but applies a trim correction to the sounding, (for the vessel’s current trim) which results in the tank showing not completely full This means it shows you have less ballast on board & consequently less cargo - even though you know the tank is completely full he does not accept this - to prevent this unless you can ensure that the same surveyor is doing the initial & final draft survey at the disport & he will accept that your tanks are completely full , do not fill the tanks completely full - keep them slightly slack, actually sound the tanks, so that when he applies the trim correction this

will reflect the actual ballast in the tank & you will not have less cargo

In many ways the method described here is used at the load port except for an important difference - at the load port you compute your constant at the time of initial draft survey At that time to arrive at a constant satisfactory to you and the surveyor some adjustments are made to the ballast quantity / density / drafts to arrive at this figure - this is in order mainly not to get a negative constant or a constant very different from the constant that is normal for your vessel

If you used the same system (without making adjustments to arrive at a contrived constant) you would get the actual cargo figure but a constant far removed from the normal or a negative constant Either of these is a red flag to anyone checking the figures & therefore we adjust & arrive at a contrived constant I would advise that you

do use this method when the cargo is being loaded on a charter where the Owners are being paid for the freight

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At the load port to get an accurate amount of cargo you could use the same method, to

calculate the cargo loaded, eliminating the lightship and constant, but in the case of the load port at the time of initial survey, adjustments are made to either the drafts, ballast

or soundings to compute the constant Remember, that once this is done then the method

of calculating the cargo on board is the same as that at the discharging port

Let Displacement prior commencement of loading = C

Fuel Oil prior commencement of loading = FO3

Diesel Oil prior commencement of loading = DO3

Fresh Water prior commencement of loading = FW3

Ballast Water prior commencement of loading = BW3

Constant = K

Light ship = LS

Deductibles on arrival = FO3 + DO3 + FW3 + BW3 = c

Displacement prior commencement of loading = C = c + K + LS

As c = FO3 + DO3 + FW3 + BW3

In theory, if there are no changes to fuel, fresh water and ballast, then

CARGO ON BOARD = D – C = D - FO3 - DO3 - FW3 - BW3 - K - LS

Where D = displacement after loading

Using this you should in theory be able to compute the cargo on board, but this assumes that the value of the lightship is accurate and your computed constant at the load port is accurate This however is not always true so the cargo is actually calculated as follows -

Let Displacement after completion of loading = D

Fuel Oil on completion of loading = FO4

Diesel Oil on completion of loading = DO4

Fresh Water on completion of loading = FW4

Ballast Water on completion of loading = BW4

Constant = K

Light Ship = LS

Deductibles on completion of loading = d = FO4 + DO4 + FW4 + BW4

Displacement on completion of loading = D = d + K + LS + CARGO

CARGO ON BOARD = D – d – K - LS

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Ideally, if there were no changes to fuel, diesel, fresh water and ballast between

commencement and completion of discharge, cargo on board would be a simple matter

of D - C But as all these values change,

CARGO ON BOARD = D - FO4 - DO4 - FW4 - BW4 - C + FO3+DO3+FW3+ BW3

Once again it appears that the constant and lightship are eliminated, but

remember you have adjusted the draft, the ballast soundings or the density which has changed the initial displacement C or the initial ballast BW3 to arrive at a contrived constant

The constant is calculated at the initial draft survey - less ballast shown at the

initial draft survey will show a larger constant and more ballast shown at the

initial draft survey will result in a lesser constant

As I have repeated over and over again do not reduce the constant, even though less constant means more cargo - should you then get the real constant at the

disport ( which you do by using this same method without making any adjustments)

you will end up with less cargo and real problems

IF ONE UNDERSTANDS THE DRAFT SURVEY AND HOW TO GO ABOUT IT

& IF YOUR DRAFT READINGS AND SOUNDINGS ARE ACCURATE YOU SHOULD NEVER HAVE PROBLEMS WITH THE CARGO TO LOAD, CARGO LOADED OR CARGO DISCHARGED AND NO DRAFT SURVEYOR WILL BE ABLE TO PULL WOOL OVER YOUR EYES

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CHAPTER III - TRIMMING

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TRIMMING

This is probably the subject about which very little or almost nothing is taught in any nautical school - have learnt the following method from the foremen in Brazil who are loading at up to 16,000MT per hour and require to be spot on

Am going to show various possible situations on the Cape-size vessel SILC at the stop for draft check prior trimming and then trimming with different sets of holds

CASE 1

Assume the read drafts at the draft check to be

Fp = 16.70 Fs = 16.70 Mp = 16.82 Ms = 16.84 Ap = 16.93 As = 16.93 Therefore,

Aft trim correction = Apparent trim x Dist fm aft mark to aft perpendicular

Length between marks

= Apparent trim x 10.50 = Apparent trim x 0.042

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Vessel is in water of R.D = 1.025 and is required to load such that its midship draft is 17.25 meters and finish with a stern trim of 42 cms Max allowed draft = 17.50 meters

Assume you are to trim with holds 4 & 8

(This was an actual situation where we were loading in Richards Bay where max draft is 17.50 meters Winter zone draft when passing Cape Finistere en route to Le Havre is 17.16 meters The difference between 17.25 meters & 17.16 meters is the consumption

en route from Richards Bay to Cape Finistere and the 42 cms trim would after

consumption result in v/l arriving Le Havre even keel.)

Corrected drafts are Fc = 16.70 Mc = 16.83 Ac = 16.94

Midship draft is 16.83 meters and maximum midship draft is 17.25 meters

Sinkage available = (17.25 - 16.83) = 0.42m = 42 cms

Just so that you do not overload assume that the vessel will sag a further 2 cms after trimming

Therefore sinkage available is 42 - 2 = 40 cms

As TPC = 106.30 tonnes, therefore cargo to load = 106.30 x 40 = 4252 MT

Final midship draft will be Midship draft + addl sag + sinkage = 16.83 + 0.02 +0.40 = 17.25m

From trimming tables at draft of 17.05 meters (mean between 16.83 and 17.25)

For every 100 MT sinkage ford sinkage aft difference = CHANGE OF TRIM

No 4 Hold 1.81 0.09 1.81 - (0.09) = 1.72

No 8 Hold - 0.50 2.35 2.35 - (-0.50) = 2.85 ADDING THE VALUES FOR CHANGE OF TRIM FOR 100MT IN EACH = 4.57

(Disregard for the moment that #4 will trim the v/l by head & #8 trim by stern) ****

(In the case of No.4 hold as loading in #4 will cause both the fore and aft drafts to increase and therefore the values in column 2 & 3 are positive, and the last column is the DIFFERENCE between the two, the values 1.81 and 0.09 are subtracted to give 1.72 In the case of No.8 hold, since the ford draft reduces and the aft draft increases, the signs are different when you are looking for the DIFFERENCE, you add the values 0.50 & 2.25 to get 2.85 Do go through all 9 cases so you fully understand this.)

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We have a 24 cm trim by stern presently and require to finish with a 42 cm trim by stern

- therefore you have to trim the vessel by a further 42 - 24 cms = 18 cms by stern

First tackle this -

To trim the vessel by stern & you are trimming with 4 & 8 you would have to load in #8

Use the formula - Trim required in cms x 100

Trim caused by 100 MT

(From the values lifted from the trimming tables listed above)

Trim required in cms x 100 = 18 x 100 = 631.58 MT = 632 MT

Trim caused by 100 MT 2.85

Therefore 632 MT in #8 will be required to finish with the required trim

Therefore we have to load 4252 - 632 = 3620 MT distributed in holds #4 & #8.without causing any further change in trim

To load in #4 without any change in trim we load

Weight to load x Change of trim caused by #8 = 3620 x 2.85 = 2257.5 = 2258

Change of trim caused by #4 & #8 4.57

To load in #8 without change of trim we load

Therefore you will have to load 2258 MT in #4 and (632 + 1362) = 1994 MT in #8 to finish with the required drafts - you can check this - see below

Which is a 42 cms trim by stern

Midship draft from above = 17.25M

PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT

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CASE 2

Assume the read drafts to be

Fp = 17.00 Fs = 17.00 Mp = 17.21 Ms = 17.21 Ap = 17.29 As = 17.29 R.D = 1.025 V/l to finish even keel Max.draft: Summer = 17.525m

Trimming with #3 and #7

Apparent trim is 17.29 - 17.00 = 0.29 cms by stern

Ford trim corrn = Apparent trim x 0.004 = 0.29 x 0.004 = 0.001 which is negligible Aft trim corrn = Apparent trim x 0.042 = 0.29 x 0.042 = 0.01

Corrd drafts are Fc = 17.00 Mc = 17.21 Ac = (17.29 + 0.01) = 17.30

CORRECTED TRIM = 17.3 - 17.0 = 0.30m BY STERN

V/l to load to max draft of 17.525 meters - let us assume that there will be 1 cm further sag

Therefore sinkage available is 17.525 - (17.21 + 0.01) = 0.305 = 30.5 cms

Since TPC = 106.35 cargo to load = 30.5 x 106.35 = 3244 MT

Final midship draft will be Midship draft + sag caused + sinkage

= 17.21 + 0.01 + 0.305 = 17.525m

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From Trim tables for draft of 17.37 (mean between 17.21 & 17.525)

No 3 Hold 2.35 - 0.48 2.35 - (-0.48) = 2.83

No 7 Hold 0.07 1.79 1.79 - (-0.07) = 1.72 ADDING THE VALUES FOR CHANGE OF TRIM FOR 100MT IN EACH = 4 55

(Disregard for the moment only that #3 will trim the v/l by head & #7 trim by the stern)

(PLEASE SEE NOTE MARKED **** IN CASE 1)

Present trim is 30 cms by stern & we require to finish loading, even keel -

Therefore change in trim should be 30 cms by head

First tackle this - loading in #3

To trim by head by 30 cms = Trim required in cms x 100 = 30 x 100 = 1060

Trim caused by 100 MT 2.83

Therefore (3244 - 1060) = 2184 must be distributed in #3 & #7 without any further change in trim

Therefore you will have to load (1060 + 826) = 1886 MT in #3 and 1358 MT in #7 to finish with the required drafts - you can check this

This is EVEN KEEL

Midship draft from above = 17.525m

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CASE 3

Fp = 17.25 Fs = 17.25 Mp = 17.19 Ms = 17.19 Ap = 17.00 As = 17.00 R.D = 1.023 V/l to finish even keel Max.draft: - 17.50m

Apparent trim is 17.25 - 17.00 = 0.25 cms by head

Corrd drafts are Fc = 17.25 Mc = 17.19 Ac = (17.00 - 0.01) = 16.99

CORRECTED TRIM = 17.25 - 16.99 = 0.26m BY HEAD

V/l to load to max draft of 17.50metres - let us assume that there will be a 1 cm hog caused by the quantity of cargo loaded in #3 & #8 during trimming

Therefore sinkage available is 17.5 - (17.19 - 0.01) = 0.32 = 32 cms

Since TPC = 106.35 x 1.023 / 1.025 = 106.14=>

Cargo to load = 32 x 106.14 = 3396MT

Final midship draft will be Midship draft - hog caused + sinkage

= 17.19 - 0.01 + 0.32 = 17.50m

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From Trim tables for draft of 17.35 ( mean between 17.19 & 17.50)

No 3 Hold 2.385 - 0.475 2.385 - (-0.475) = 2.86

No 8 Hold -0.495 2.34 2.34 - (-0.495) = 2.835 ADDING THE VALUES FOR CHANGE OF TRIM FOR 100MT IN EACH = 5.695

(Disregard for the moment only that #3 will trim the v/l by head & #8 trim by

stern)

(PLEASE SEE NOTE MARKED **** IN CASE 1) In this case as the signs are

opposite for both holds the values will be added to get the Difference

Present trim is 26 cms by head - we require to finish even keel -

Therefore change in trim should be 26 cms by stern

To trim by stern by 26 cms = Trim required in cms x 100 = 26 x 100 = 917.1 = 917

Therefore (3396 - 917) = 2479 must be distributed in #3 & #8 without any further

change in trim

Weight to load x Change of trim caused by #8 = 2479 x 2.835 = 1234 MT

Therefore you will have to load 1234 MT in #3 and (917 + 1245) = 2162 MT in #8 to finish with the required drafts - you can check this

This is EVEN KEEL

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CASE 4

Fp = 16.81 Fs = 16.81 Mp = 17.07 Ms = 17.07 Ap = 17.23 As = 17.23 R.D = 1.0215 V/l to finish 70 cms by stern Reqd midship draft by calculation =17.45 Depth available at this port 22.0 meters

V/l to load to max midship draft of 17.45meters - let us assume that there will be a 3 cm sag caused by the quantity of cargo loaded in #4 & #7 during trimming

Therefore sinkage available is 17.45 - (17.07 + 0.03) = 0.35 = 35 cms

Since TPC =106.35 x 1.0215 / 1.025 = 105.99

 cargo to load= 35 x 105.99 = 3710MT

Final midship draft will be = Midship draft + sag caused + sinkage

= 17.07+ 0.03+ 0.35 = 17.45m

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From Trim tables for draft of 17.26 (mean between 17.07 & 17.45)

No 4 Hold 1.81 0.09 1.81 - (0.09) = 1.72

No 7 Hold 0.07 1.79 1.79 - (0.07) = 1.72 ADDING THE VALUES FOR CHANGE OF TRIM FOR 100MT IN EACH = 3.44

(Disregard for the moment only that #4 will trim the v/l by head & #7 trim by

stern)

(PLEASE SEE NOTE MARKED **** IN CASE 1) In this case as the signs are same

for both holds the values will be subtracted for both to get the Difference

Present trim is 43.7 cms by stern - we require to 70 cms by stern -

Therefore change in trim should be 26.3 cms by stern

To trim by stern by 26 3cms = Trim required in cms x 100 = 26.3 x 100 = 1529

Therefore (3710 - 1529) = 2181 must be distributed in #4 & #7 without any further change in trim

Therefore you will have to load 1091 MT in #4 and (1529 + 1090) = 2619 MT in #7 to finish with the required drafts - you can check this

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CASE 5

Fp = 16.75 Fs = 16.75 Mp = 17.03 Ms = 17.03 Ap = 17.20 As = 17.20 R.D = 1.022 V/l to finish 60 cms by stern Maximum draft at load port 17.60m V/l required to sail out with 60 cms stern trim to arrive disport even keel

Drafts by calculation prior to loading F: 17.00 M: 17.30 A: 17.60 in R.D 1.022 with no hog or sag Trimming with #4 and #7

By pre-calculation v/l should finish with drafts of F: 17.00 M: 17.30 A: 17.60

Using the 6-sided formula the MQM = F + A + 6M = (17.0 + 17.6) + 6(17.3) = 17.30

8 8

However present drafts at the stop for trimming are F: 16.75 M: 17.03 A: 17.219 I.E we have a sag of 4.55 cms - assume that the cargo loaded for trimming will cause a further sag of 1.45 cms Also the consumption of fuel from the load port to disport by your estimation will result in a further sag of 1 cm - remember that you should have incorporated this further sag of 1 cm in you pre-calculation, so we will assume that you have already taken this into account when in your pre-calculation you have pre-calculated a departure load port midship draft of 17.30 meters

The max.draft for the load port has been declared to be 17.60 meters By your pre-cal you have found that to arrive at the disport at the required draft, your midship draft at the load port should be 17.30m & you should be trimmed 60 cms by stern to arrive even keel

To calculate the final drafts use the 6-sided formula -

You already know that the midship draft is to be 17.30 meters

Now you going to finish loading with a sag of 4.55(present) + 1.45(caused by the

trimming cargo) = 6 cms

Therefore you should finish with drafts of

(M - 0.06) - (0.6/2) = (17.30 - 0.06) - 0.3 = 17.24 - 0.30 = 16.94 mts = Ford draft (M - 0.06) + (0.6/2) = (17.30 - 0.06) + 0.3 = 17.24 + 0.30 = 17.54 mts = Aft draft

This will mean that your MQM = 16.94 + 17.54 + 6(17.30) = 17.285

8

This means that you will have loaded (1.5 x TPC) MT less than you have asked for, so it would be wise to always estimate some sag in your pre-calculation

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