Practical Guide to Draft Surveys

A PRACTICAL GUIDE TO DRAFT SURVEYS TRIMMING CALCULATIONS ETC. GLENN J SALDANHA ([email protected])

PLEASE PRINT THIS OUT USING BOTH SIDES OF THE PAPER - HAVE ADJUSTED THE FILE TO ALLOW EASIER READING WHEN PRINTED & FILED THIS WAY

INDEX CHAPTER

TOPIC

PAGES

I

THE 6-SIDED DRAFT

II

THE DRAFT SURVEY

11 - 21

III

TRIMMING

22 - 51

IV

HOG / SAG

52 - 56

V

CONTROLLING DRAFTS

57 - 59

VI

MAXIMUM DRAFTS

61 - 63

2

3 - 10

CHAPTER I - THE 6 - SIDED DRAFT

3

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 ^M = Angle of Trim As all are similar triangles with the same ^M : T = Ta = c = a = b = LBP LBM LCF A B Since Ta = a then a = A x Ta LBM A LBM &

Tan ^M

Ta = b then b = B x Ta LBM B LBM In the formula for correction to be applied to the ford draft A is a constant value on EACH SHIP as is LBM as is B

4

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 VESSEL IS TRIMMED BY THE HEAD the correction sign will be opposite.

After reading the drafts and applying the draft correction then one will proceed to calculate the Mean Quarter Mean Draft using the 6 sided formula Many surveyors do not worry about the correction to the read drafts when the trim is very small. Actually, in theory, there is a correction to be applied to the read midship drafts as well - unless the midship draft is computed by measuring the freeboard from the deck line or computing the draft by measuring the distance from the Loadline Mark to the water line - because the loadline mark is exactly at midships and the midship draft marks are actually displaced either just forward or just aft of midships but as this distance is so small & the correction so small I have never known anyone to apply a correction to the read midship draft mark.

5

No matter what is taught in the various colleges of the world wrt 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 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) 2

(F + A) = 2

Mean (Mean Draft)

Now to get the draft, reqd for calculating displacement, known as Mean Quarter Mean (MQM) draft we use the six - sided formula (6M + F + A) 8

= MQM

OR 6 x Midship draft + Ford Draft + Aft Draft 8

= Mean Quarter Mean Draft

Now as (Ford Draft + Aft Draft) = Mean Draft OR ( Ford Draft + Aft Draft) = 2 x Mean Draft 2 Therefore the 6 Sided formula is also :

(6M + 2Mean) = MQM 8 OR

6 x Midship Draft + 2 x Mean Draft 8

=

6

Mean Quarter Mean Draft

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 precalculation of cargo figures as the following examples will show 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 recalculation 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 Therefore as MQM = 6M + 2Mean 8 MQM = 6M + 2(M - 0.10) = 6(17.0) + 2(17.0 - 0.10) 8 8 = 102 + 2 (16.9) = 102 + 33.8 8 8 Therefore MQM = 16.975 meters AND THIS MUST BE UNDERSTOOD COMPLETELY BEFORE PROCEEDING FURTHER 7

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, MQM = (6 x 17.50) + 17.40 + 17.40 = 17.475 meters 8 Therefore you will get the displacement for 17.475 meters, apply the Dock water correction, reduce the lightship, constant and deductibles and you will get the cargo to be loaded. To expand : the density of Dockwater = 1.023 lightship = 18643 K= 400 (FO: 2000 DO: 100 FW: 200 U/Pumpable Ballast : 100) Deductibles = 2400 Displacement in SW at 17.475 Dock water 1.023 , displ in DW

= 169560.15 = 169560.15 x 1.023 1.025 Displ in DW = 169229.3 Lightship = - 18643 Deadweight = 150586.3 K + deductibles = - 2800 CARGO TO LOAD = 147786.3

Example 3: 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, 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 8

8

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 = 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) To confirm in reverse MQM = 6M + 2Mean = 6(15.87) + 2(15.79) 8 8 MQM = 15.85 meters Example 5: Same conditions as in Eg 4 except vessel to finish hogged by 10 cms Let Midship draft = X metres As v/l hogged by 10 cms, Mean Draft = (X+0.10) metres MQM = 6X + 2 (X+0.10) = 6X + 2X + 0.20 8 8 8 ( MQM) = 8X + 0.20 8 (15.85) - 0.20 = 8X X = 15.825 metres = M Since v/l hogged by 10 cms & even keel Mean Draft = F = A = 15.825 + 0.10 = 15.925 You can confirm same by formula MQM = 6M + F + A = 6 x 15.825 + 15.925 + 15.925 8 8

9

= 15.85

Example 6: MQM = 15.85m

Sag: 8 cms Trim: 14 cms by stern

Midship draft = X mts

RD: 1.025

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

MQM = 6M + 2Mean = 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 & A = 15.79 + 0.07 = 15.86 Example 7: If the v/l, in Eg.6 was then to go into RD less than 1.025, say 1.010, you would have to first convert the MQM of 15.85 into the MQM of that density At MQM 15.85 mts displacement in SW = 151293 Equivalent displacement in 1.010 = 151293 x 1.025 = 153539.9 1.010 Therefore MQM in 1.010 = 15.968 mts 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 and 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.

10

CHAPTER II - THE DRAFT SURVEY

11

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 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 i.e. 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, as nobody gives you a medal for having more cargo, but should you land up with less cargo all kinds of hell breaks loose. Where you should try to show the maximum cargo, even insisting on recording a negative constant, is when you are loading when the vessel is on a direct charter with Owners - i.e. Owners are getting paid on basis of freight - then every tonne on board is more money in the bank for the Owners. 12

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. CORRECTION TO THE PERPENDICULARS For correction to the perpendiculars of the ford 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 - please note that even though the midship draft marks are slightly displaced from the center line - most surveyors disregard this.

13

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) 8 TRIM CORRECTION When a ship is trimmed the calculated mean draft is not the same as the true mean draft measured at the LCF. To correct the displacement to that corresponding to the true mean draft the following correction(s) are applied: FIRST TRIM CORRECTION (IN TONNES)

= TRIM (in cms) x LCF(in meters) x TPC 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 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 it s 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. When there is very little trim many surveyors ignore the 2nd Trim Correction.

14

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 TP2 D1 D2

= = = =

TPC for the deepest draft amidships TPC for the shallower draft amidships Deepest draft amidships ( IN METRES) Shallower draft amidships ( IN METRES)

This correction is ALWAYS ADDED to the displacement because the effect of heel is to increase the waterplane 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 metres & v/l is lying in DW of 1.017 then true displacement would be calculated as follows: Scale Displacement for 15.00 metres = 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.

15

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 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 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 figure, you will only get an idea of what your constant is.

16

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17

At the discharge port cargo is calculated as followsLet Displacement prior commencement of discharge Fuel Oil prior commencement of discharge Diesel Oil prior commencement of discharge Fresh Water prior commencement of discharge Ballast Water prior commencement of discharge Constant Light ship

= = = = = = =

A FO1 DO1 FW1 BW1 K 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 Fuel Oil on completion of discharge Diesel Oil on completion of discharge Fresh Water on completion of discharge Ballast Water on completion of Discharge Constant Light Ship Deductibles on completion of discharge

= = = = = = =

B FO2 DO2 FW2 BW2 K LS

= 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.

18

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 - a - K - LS - B + b + K + LS CARGO ON BOARD = A - a - B + b THUS THE CONSTANT AND LIGHTSHIP DO NOT COME INTO THE PICTURE AT ALL. CARGO ON BOARD = A - a - B + b 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 sounds the tanks and applies a trim correction to the sounding 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 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.. 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.

19

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 Fuel Oil prior commencement of loading Diesel Oil prior commencement of loading Fresh Water prior commencement of loading Ballast Water prior commencement of loading Constant Light ship

= = = = = = =

C FO3 DO3 FW3 BW3 K 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, 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

20

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 - (d + K + LS) ))

-

(( C - (c + K + LS) ))

opening the single brackets CARGO ON BOARD = (( D - d - K - LS )) - (( C - c - K - LS )) opening the double brackets CARGO ON BOARD = D - d - K - LS - C + c + K + LS CARGO ON BOARD = D - d - C + c 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 DRFAT 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.

21

CHAPTER III - TRIMMING

22

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 Therefore, F = 16.70 M = 16.83 A = 16.93

Ap = 16.93

As = 16.93

Ford trim correction = Apparent trim x Dist fm ford mark to ford perpendicular (see page 4) Length between marks = Apparent trim x

Aft trim correction

1 248.5

= Apparent trim x 0.004

= Apparent trim x Dist fm aft mark to aft perpendicular Length between marks = Apparent trim x 10.50 248.5

= Apparent trim x 0.042

Apparent trim = 0.23m Apparent trim x 0.004 = 0.23 x 0.004 = 0.0009 = corrn to ford draft which is negligible Apparent trim x 0.042 = 0.23 x 0.042 = 0.01 = corrn to aft draft(to be added) Therefore corrected drafts are Fc = 16.70 Corrected trim = 16.94 - 16.70 = 0.24m

23

Mc = 16.83

Ac = 16.94

Vessel is in water of R.D. = 1.025 and is required to load such that it s midship draft is 17.25 metres 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 metres, Winter zone draft when passing Finistere en route to Le Havre is 17.16 metres. The difference between 17.25 metres & 17.16 metres is the consumption en route from Richards Bay to 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 metres ( 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 ford 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.)

24

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 and 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 Trim caused by 100 MT

= 18 x 100 2.85

= 631.58 MT = 632 MT

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 Weight to load x Change of trim caused by #4 = 3620 x 1.72 = 1362.4 = 1362 Change of trim caused by #4 & #8 4.57

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 Ford Aft 16.70 16.94 Loading 2258 MT in #4 0.409 0.02 After loading in #4 17.109 16.96 Loading 1994 MT in #8 - 0.10 0.469 On completion of trimming 17.009 17.428 which is a 42 cms trim by stern Midship draft from above = 17.25M PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT.

25

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

26

from Trim tables for draft of 17.37 ( mean between 17.21 & 17.525) for every 100 MT

sinkage ford

sinkage aft

difference = CHANGE OF TRIM

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 stern) (PLEASE SEE NOTE MARKED **** IN CASE 1) Present trim is 30 cms by stern & we require to finish 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 To load in #3 without any change in trim we load Weight to load x Change of trim caused by #7 = 2184 x 1.72 = 825.6 = 826 Change of trim caused by #3 & #7 4.55 To load in #7 without change of trim we load Weight to load x Change of trim caused by #3 = 2184 x 2.83 = 1358.4 = 1358 Change of trim caused by #3 & #7 4.55 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 Ford Aft 17.00 17.30 Loading 1886 MT in #3 0.443 - 0.09 After loading in #3 17.443 17.21 Loading 1358 MT in #7 0.01 0.243 On completion of trimming 17.453 17.453 which is EVEN KEEL Midship draft from above = 17.525m PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT. 27

CASE 3. Assume the read drafts to be 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 Trimming with #3 and #8. Apparent trim is 17.25 - 17.00 = 0.25 cms by head Ford trim corrn = Apparent trim x 0.004 = 0.25 x 0.004 = 0.001 which is negligible Aft trim corrn = Apparent trim x 0.042 = 0.25 x 0.042 = 0.01 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

28

from Trim tables for draft of 17.35 ( mean between 17.19 & 17.50) for every 100 MT

sinkage ford

sinkage aft

difference = CHANGE OF TRIM

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. First tackle this - loading in #8. To trim by stern by 26 cms = Trim required in cms x 100 = 26 x 100 = 917.1 = 917 Trim caused by 100 MT 2.835 Therefore (3396 - 917) = 2479 must be distributed in #3 & #8 without any further change in trim To load in #3 without any change in trim we load Weight to load x Change of trim caused by #8 = 2479 x 2.835 = 1234 MT Change of trim caused by #3 & #8 5.695 To load in #8 without change of trim we load Weight to load x Change of trim caused by #3 = 2479 x 2.86 = 1245 MT Change of trim caused by #3 & #8 5.695 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 Ford Aft 17.25 16.99 Loading 1234 MT in #3 0.294 - 0.059 After loading in #3 17.544 16.931 Loading 2162 MT in #8 - 0.107 0.506 On completion of trimming 17.437 17.437 which is EVEN KEEL Midship draft from above = 17.50m PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT. 29

CASE 4. Assume the read drafts to be 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 Trimming with #4 and #7. Apparent trim is 17.23 - 16.81 = 0.42 cms by stern Ford trim corrn = Apparent trim x 0.004 = 0.42 x 0.004 = 0.002 which is negligible Aft trim corrn = Apparent trim x 0.042 = 0.42 x 0.042 = 0.017 Corrd drafts are Fc = 16.81 Mc = 17.07 Ac = (17.23 + 0.018) = 17.247 CORRECTED TRIM = 17.247 - 16.81 = 0.437m BY STERN 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

30

from Trim tables for draft of 17.26 ( mean between 17.07 & 17.45) for every 100 MT

sinkage ford

sinkage aft

No. 4 Hold

1.81

0.09

difference = CHANGE OF TRIM 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. First tackle this - loading in #7. To trim by stern by 26 .3cms = Trim required in cms x 100 = 26.3 x 100 = 1529 Trim caused by 100 MT 1.72 Therefore (3710 - 1529) = 2181 must be distributed in #4 & #7 without 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 #7 = 2181 x 1.72 = 1091 MT Change of trim caused by #4 & #7 3.44 To load in #7 without change of trim we load Weight to load x Change of trim caused by #4 Change of trim caused by #4 & #7

= 2181 x 1.72 = 1090 MT 3.44

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 Ford Aft 16.81 17.248 Loading 1091 MT in #4 0.197 0.01 After loading in #4 17.007 17.258 Loading 2619 MT in #7 0.018 0.469 On completion of trimming 17.025 17.727 which is 70 cms by stern Midship draft from above = 17.45m PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT. 31

CASE 5. Assume the read drafts to be 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 pre-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. Apparent trim is 17.20 - 16.75 = 0.45 cms by stern Ford trim corrn = Apparent trim x 0.004 = 0.45 x 0.004 = 0.002 which is negligible Aft trim corrn = Apparent trim x 0.042 = 0.45 x 0.042 = 0.019 Corrd drafts are Fc = 16.75 Mc = 17.03 Ac = (17.20 + 0.019) = 17.219 CORRECTED TRIM = 17.219 - 16.75 = 0.469m BY STERN 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 precalculated a departure load port midship draft of 17.30 metres. The max.draft for the load port has been declared to be 17.60 metres. 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 metres. 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 This calculation of MQM in this case has no meaning with regards to the trimming but have just shown it to tell you to always take into account hog or sag in your pre-calulation so as not to load less than the cargo you asked for. 32

Sinkage available = 17.30 - (17.03 + .0155) = 0.2545 = 25.45 cms Since TPC= 106.35 x 1.022 / 1.025= 106.04 cargo to load= 25.45 x 106.04= 2699MT Final midship draft will be Midship draft + sag caused + sinkage = 17.03+0.0155+ 0.2545 = 17.3m from Trim tables for draft of 17.165 ( mean between 17.03 & 17.30) 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. 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 46.9 cms by stern- we require to be 60 cms by stern Therefore change in trim should be 13.1 cms by stern. First tackle this - loading in #7. To trim by stern by 13.1cms = Trim required in cms x 100 = 13.1 x 100 = 762 Trim caused by 100 MT 1.72 Therefore (2699 - 762) = 1937 must be distributed in #4 & #7 without 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 #7 = 1937 x 1.72 = 969 MT Change of trim caused by #4 & #7 3.44 To load in #7 without change of trim we load Weight to load x Change of trim caused by #4 = 1937 x 1.72 = 968 MT Change of trim caused by #4 & #7 3.44 Therefore you will have to load 969 MT in #4 and (762 + 968) = 1730 MT in #7 to finish with the required drafts - you can check this Ford Aft 16.75 17.219 Loading 969 MT in #4 0.174 0.009 After loading in #4 16.924 17.228 Loading 1730 MT in #7 0.012 0.310 On completion of trimming 16.936 17.54 which is 60 cms by stern Midship draft from above = 17.30m 33

CASE 6. Assume the read drafts to be Fp = 17.03 Fs = 17.03 Mp = 17.20 Ms = 17.20 Ap = 17.29 As = 17.29 R.D. = 1.025 V/l to finish with max. trim by stern. Maximum draft at load port 17.75 m V/l required to sail out at summer mark of 17.525 m. Trimming with #3 and #8. Apparent trim is 17.29 - 17.03 = 0.26 cms by stern Ford trim corrn = Apparent trim x 0.004 = 0.26 x 0.004 = 0.001 which is negligible Aft trim corrn = Apparent trim x 0.042 = 0.26 x 0.042 = 0.010 Corrd drafts are Fc = 17.03 Mc = 17.20 Ac = (17.29 + 0.010) = 17.30 CORRECTED TRIM = 17.30 - 17.03 = 0.27m BY STERN We have a sag of 3.5 cms - assume that the cargo loaded for trimming will have a hog effect of 1.5 cms. Therefore you will complete loading with a sag of (3.5 - 1.5)= 2.0 cms As you will be loading to summer mark of 17.525 metres THIS will be the midship draft. I MAKE THE ASSUMPTION HERE THAT AS LONG AS YOU DO NOT SUBMERGE YOUR LOADLINE MARK YOU CAN HAVE GREATER DRAFTS AT THE ENDS - THIS IS ALMOST ALWAYS ACCEPTED BUT THERE ARE PORTS WHERE THEY STIPULATE THAT THE DRAFT AT NO POINT SHOULD EXCEED THE SUMMER MARK WHEN IN THE SUMMER ZONE. As your midship draft will be 17.525 metres and you will finish with a sag of 2.0cms, your mean draft will be (17.525 - 0.020) = 17.505 metres. As the maximum draft will be 17.75 metres and you require maximum stern trim your aft draft will be 17.75 metres. Therefore your trim will be (17.75 - 17.505) x 2 = 0.49 metres. Assuming that you are trimming about midships you should end up with a ford draft of 17.75 - 0.49 = 17.26 metres. Present midship draft = 17.20 metres Hog caused by loading trimming cargo = 0.015m V/l to load to summer mark of 17.525 metres Therefore sinkage available = 17.525 - (17.20 - 0.015) = 0.34 metres = 34cms As TPC=106.34 and sinkage = 34 cms cargo available for trimming= 106.34 x 34= 3616MT Final midship draft will be Midship draft - hog caused + sinkage = 17.20 - 0.015+ 0.34 = 17.525m

34

from Trim tables for draft of 17.3625 ( mean between 17.20 & 17.525) for every 100 MT No. 3 Hold

sinkage ford 2.385

sinkage aft - 0.475

difference = CHANGE OF TRIM 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 27 cms by stern - we will finish 49cms by stern Therefore change in trim should be 22 cms by stern. First tackle this - loading in #8. To trim by stern by 23cms = Trim required in cms x 100 = 22 x 100 = 776 MT Trim caused by 100 MT 2 .835 Therefore (3616 - 776) = 2840 must be distributed in #3 & #8 without any further change in trim To load in #3 without any change in trim we load Weight to load x Change of trim caused by #8 = 2840 x 2.835 = 1414 MT Change of trim caused by #3 & #8 5.695 To load in #8 without change of trim we load Weight to load x Change of trim caused by #3 = 2840 x 2.86 = 1426 MT Change of trim caused by #3 & #8 5.695 Therefore you will have to load 1414 MT in #3 and (776 + 1426) = 2202 MT in #8 to finish with the required drafts - you can check this Ford Aft 17.03 17.30 Loading 1414 MT in #3 0.337 - 0.067 After loading in #3 17.367 17.233 Loading 2202 MT in #8 - 0.109 0.515 On completion of trimming 17.258 17.748 which is as required - TRIM = 0.49m Midship draft from above = 17.525m

35

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36

CASE 7. Assume the read drafts to be Fp = 16.89 Fs = 16.89 Mp = 17.00 Ms = 17.00 Ap = 17.04 As = 17.04 R.D. = 1.023 Maximum draft at load port 17.50 mts. V/l required to arrive at disport with max draft of 17.20 m. Charter party stipulates maximum cargo = 145000 MT Total change of trim on voyage from load port to disport including change due to consumption of fuel & fresh water & change of trim due to change of density = 52 cms Sinkage due to change of density at disport = 2 cms Sag due to consumption of fuel and fresh water on voyage = 1 cm Rise due to fuel and fresh water consumption on voyage = 9 cms. Trimming with #4 and #8. At load port : FO: 2179.0 DO: 87.0 FW: 250 U/pump ballast: 93.0 ltship: 18643 Deductibles = (FO + DO + FW + BALLAST) = 2609

K:393

In almost all cases trimming is governed by drafts not quantity of cargo to be loaded where the draft is a limiting factor - max.draft at the load port or max draft at the disport or the draft when v/l crosses into Winter zone from Summer Zone or when it crosses from Tropical Zone into Summer zone etc. However you can also have a maximum quantity of cargo to load - very often the C/P states something like 135,000 +/-5% MOLOO which means you can load anywhere between128250 & 141750 but not less than 128250 & not more than 141750 - in the Case under study I have given a situation where you are not permitted more than 145000. Also just for practice & understanding have stated a max draft at the load port & the discharge port. Normally you would be told the density of the disport & then you would have to calculate the rise or sinkage between the load port & disport due to the change of density and also the change of trim due to the change of density - you would have to apply the fuel consumption to compute your change of trim on the voyage etc. - but as the object of this exercise is TRIMMING I have given an example where all the required changes have been pre-calculated.

37

From the information given we have Max cargo to load = 145000 MT Max draft at any point at the load port = 17.50m Max draft at disport = 17.20m Trim change between load port & disport = 52 cms - as we are aiming for the max draft at the disport we will require to arrive there even keel , therefore trim on departure load port should be 52 cms by stern. Change in midship draft on voyage from load port to disport = Rise due to consumption - sag caused by consumption - sinkage caused by change of density at disport = 9cms - 1 cm - 2 cms = 6 cms. Therefore as we will be sailing out with some sag & there will be an increase in sag on voyage & we will arrive at the disport even keel, the max draft at the disport will be amidships and therefore the max POSSIBLE midship draft at the load port = Max.draft at the disport + Rise on voyage = 17.20 + 0.06 = 17.26m Read drafts to be Fp = 16.89 Fs = 16.89 Mp = 17.00 Ms = 17.00 Ap = 17.04 As = 17.04 Apparent trim is 17.04 - 16.89 = 0.15 cms by stern Ford trim corrn = Apparent trim x 0.004 = 0.15 x 0.004 = 0.0006 which is negligible Aft trim corrn = Apparent trim x 0.042 = 0.15 x 0.042 = 0.006 Corrd drafts are Fc = 16.89 Mc = 17.00 Ac = (17.04 + 0.006) = 17.046 CORRECTED TRIM = 17.046 - 16.89 = 0.156m BY STERN We have a sag of 3.2 cms - assume that the cargo loaded for trimming will have a sag effect of 0.8 cms. Therefore you will complete loading with a sag of (3.2 + 0.8) = 4.0 cms. Midship draft max. will be 17.26 mts - if reqd to load to this draft then mean draft will be 17.26 - sag = 17.26 -0.04 = 17.22 m If we assume that she trims about the midship & we require to be trimmed 52 cms by stern our departure drafts would then be F: 16.96 M: 17.26 A: 17.48 However it is not necessary we will reach these drafts as this time we are also limited by a maximum quantity of cargo we can load.

38

At the stop for trimming the corrected drafts are F: 16.89 M: 17.00 A:17.046 By the 6-sided formula - the MQM = 16.89 + 17.046 + 6 ( 17.00) = 16.992 8 Displacement in SW for 16.992 m draft = 164419.41 Trim correction = + 10.25 Displacement corrected for trim = 164429.66 Correction for Density (1.023) = - 320.85 True Displacement in DW = 164108.81 Lightship = - 18643.00 Deadweight = 145465.81 Constant = 393.00 145072.81 Deductibles = - 2609.00 Cargo on Board at trimming stop = 142463.81 Required to load = 145000.00 Cargo available for trimming = 2536.19 First we will have to check whether after loading this 2536.19 MT what the sinkage will be - add the estimated sag caused by same (of 0.8 cm) and check whether the midship draft will be within the allowed calculated maximum of 17.26 mts. Sinkage = W / TPC (TPC = 106.35 x 1.023 / 1.025) = 106.14 Sinkage = 2536.19 / 106.14 = 23.895 cms Sag caused by trimming = 0.8 cms Total sinkage = 23.895 + 0.8 = 24.695 cms = 0.247 m Therefore final midship draft = 17.00 + 0.247 = 17.247m As max permissible midship draft is 17.26m, this is fine. Therefore we can load the 2536.19MT and therefore lift the max. of 145000. We are attempting to sail out with a trim of 52 cms by stern, within the limitation of a max.draft of 17.50 mts. As we will be sagged 4 cms and the midship draft will be 17.247 mts our mean draft will be 17.247 - 0.04 = 17.207 & assuming we trim about midships we will finish with Ford draft of 16.947 and Aft draft of 17.467 meters which is permitted.

39

from Trim tables for draft of 17.1235( mean between 17.00 & 17.247) for every 100 MT

sinkage ford

sinkage aft

No. 4 Hold

1.81

0.09

difference = CHANGE OF TRIM 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 = 4.55 (disregard for the moment only that #4 will trim the v/l by head & #8 trim by stern) (PLEASE SEE NOTE MARKED **** IN CASE 1) Present trim is 15.6 cms by stern - we require to finish 52cms by stern Therefore change in trim should be 36.4 cms by stern. First tackle this - loading in #8. To trim by stern by 36.4cms =Trim required in cms x 100 = 36.4 x 100=1277.19MT Trim caused by 100 MT 2.85 Therefore (2536.19 - 1277.19) = 1259 must be distributed in #3 & #7 without 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 = 1259 x 2.85 = 785.1 = 785 Change of trim caused by #4 & #8 4.57 To load in #8 without change of trim we load Weight to load x Change of trim caused by #4 = 1259 x 1.72 = 473.8 = 474 Change of trim caused by #4 & #8 4.57 Therefore you will have to load 785 MT in #3 and (1277.19 + 474) = 1751.19MT in #8 to finish with the required drafts - you can check this Ford Aft 16.89 17.046 Loading 785 MT in #4 0.142 0.007 After loading in #4 17.032 17.053 Loading 1751.19 MT in #8 - 0.088 0.412 On completion of trimming 16.944 17.465 which is 52 cms by stern Midship draft from above = 17.247m PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT. 40

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41

CASE 7A. All conditions same as in Case 7 except that the limiting draft at the disport is 17.15m Due to the above the max. midship draft at the load port = Max. midship draft at disport + Rise on Voyage - Sag caused by consumption Sinkage due to change in density Max. midship draft at Load port = 17.15 + 0.06 = 17.21m From Case 7 we have found that we are required to load 2536.19 MT at the trimming stage to lift the max. cargo - however if we do that we find our midship draft to be 17.247m which is greater than the max. permissible midship draft at the load port. Since max. midship draft is to be 17.21m, present midship draft is 17.00m and estimated sag caused by trimming to be 0.008m therefore the sinkage available is 17.21 - (17.00 + 0.008) = 0.202m TPC = 106.35 x 1.023 / 1.025 = 106.14 Therefore Cargo available for trimming = 20.2 x 106.14 = 2144 MT. We will go about the trimming to complete with 52 cms trim by stern as before. In the above you have reached the trimming stage and found you can load less than the maximum - sometimes you reach the trimming stage & find you can load more than you had calculated prior arrival load port. (please note this is different from & not more than maximum) To illustrate You are loading on the Cape-size at Richards Bay where the max draft is 17.50 meters. We were given a stowage factor of the coal such that by our calculation we were going to have all the holds volumetrically full and would finish well below the draft of 17.50 meters. By careful monitoring of the loading and several on-the-run draft checks we found that the cargo was stowing much better than the S/F declared by the shipper and we were actually able to lift as much as 2498 MT of cargo more than we had asked for , on the basis of the estimated S/F given us by the shipper and when we reached the trimming stage we made our calculations based on finishing with midship drafts of 17.50metres, even keel with some sag. In this case we were loading for Rotterdam, which has no draft restriction for a v/l this size. So, sometimes at the trimming stage you are able to calculate how much more you can load rather than how much less.

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from Trim tables for draft of 17.105( mean between 17.00 & 17.21) 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 = 4.55 (disregard for the moment only that #4 will trim the v/l by head & #8 trim by stern) (PLEASE SEE NOTE MARKED **** IN CASE 1) Present trim is 15.6 cms by stern - we require to finish 52cms by stern Therefore change in trim should be 36.4 cms by stern. First tackle this - loading in #8. To trim by stern by 36.4cms = Trim required in cms x 100 = 36.4 x 100 =1277 Trim caused by 100 MT 2.85 Therefore (2144 - 1277) = 867 must be distributed in #4 & #8 without 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 = 867 x 2.85 = 540.7 = 541 MT Change of trim caused by #4 & #8 4.57 To load in #8 without change of trim we load Weight to load x Change of trim caused by #4 = 867 x 1.72 = 326.3 = 326 MT Change of trim caused by #4 & #8 4.57 Therefore you will have to load 541 MT in #4 and (1277 + 326) = 1603MT in #8 to finish with the required drafts - you can check this Ford Aft 16.89 17.046 Loading 541 MT in #4 0.098 0.005 After loading in #4 16.988 17.051 Loading 1603 MT in #8 - 0.080 0.377 On completion of trimming 16.908 17.428 which is 52 cms by stern Midship draft from above = 17.210m PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT. 43

CASE 7B. (When following this case please disregard Case 7A.) All conditions the same as in Case 7 except that the limiting draft at the loadport is 17.40m The limiting draft at the disport remains 17.20m. We will be trying to load the maximum cargo and also as the limiting draft at the disport is still 17.20m we will be still required to arrive at the disport even keel. Since the change of trim on the voyage remains 52 cms we will still have to have an effective 52cms trim by stern after loading. If we load to the max. of 145,000 MT our midship draft we have found to be 17.247m. As we will be sagged by 4 cms our Mean draft would then be 17.247 - 0.04 = 17..207m & assuming we trim about midships and that our max. draft can be 17.40 mts our trim will be (17.40 - 17.207) x 2 = 0.386m & the ford draft would 17.014m. However we are required to have a trim of 52 cms in order that we arrive at the disport even keel. In order to arrive at the disport even keel we would have to fill some water in the afterpeak such that it will result in (52 - 38.6) = 13.4 cms of trim. From trimming tables for this draft - 100MT in the afterpeak will cause a change of trim of 6.11cms - Therefore to cause a trim of 13.4 cms we will have to fill (13.4 x 100)/8 = 219 MT in the afterpeak. However when we fill this water in the afterpeak this would result in a sinkage of W/TPC = 219 / 106.14 = 2.06 cms = 0.021m. As the midship draft without filling the afterpeak was 17.247m the effective midship draft would be17.247 + 0.021 = 17.268m. However we have found out that the max. midship draft we could have was 17.26m. Therefore we would be 17.268 - 17.26 = 0.008m = 0.8cms overloaded. We would therefore have to load 0.8 x 106.14 = 84.91 MT less during trimming. From Case 7 we have 2536.19 MT for trimming - we would therefore trim with (2536.19 - 84.91) = 2451 MT. Also note that we would fill the 219 MT required in the afterpeak only after sailing & would thus calculate our drafts without considering this quantity and working to finish with a trim of 38.6cms. Midship draft at trim stop = 17.00 mts Sag caused by trimming cargo = 0.008m Sinkage caused by trimming cargo = 2451.28 / 106.14 = 0.231m Midship draft on sailing = 17.239 m (if we had filled the 219MT in the afterpeak prior sailing it would have been 17.26m but we cannot prior sailing as we are limited by a max.draft of 17.40m & so will fill the 219MT in the afterpeak after sailing.) As we would be sagged by 0.04m our Mean draft would be 17.239 - 0.04 = 17.199m we are looking to trim vessel by 0.386m which means our aft draft would be 17.199 + (0.386/2) = 17.392m which is permissible.

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from Trim tables for draft of 17.1195( mean between 17.00 & 17.239) for every 100 MT sinkage ford No. 4 Hold

1.81

sinkage aft 0.09

difference = CHANGE OF TRIM 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 = 4.55 (disregard for the moment only that #4 will trim the v/l by head & #8 trim by stern) (PLEASE SEE NOTE MARKED **** IN CASE 1) Present trim is 15.6 cms by stern - we require to finish 38.6cms by stern Therefore change in trim should be 23 cms by stern. First tackle this - loading in #8. To trim by stern by 23cms = Trim required in cms x 100 = 23 x 100 = 807 MT Trim caused by 100 MT 2.85 Therefore (2451 - 807) = 1644 must be distributed in #4 & #8 without 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 = 1644 x 2.85 = 1025.3 = 1025 Change of trim caused by #4 & #8 4.57 To load in #8 without change of trim we load Weight to load x Change of trim caused by #4 = 1644 x 1.72 = 618.7 = 619 Change of trim caused by #4 & #8 4.57 Therefore you will have to load 1025 MT in #4 and (807 + 619) = 1426MT in #8 to finish with the required drafts - you can check this Ford Aft 16.89 17.046 Loading 1025 MT in #4 0.186 0.009 After loading in #4 17.076 17.055 Loading 1426 MT in #8 - 0.071 0.335 On completion of trimming 17.005 17.39 which is 38.5 cms by stern as required. Midship draft from above = 17.239m PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT.

From the above 9 cases you will assume that trimming is really quite easy and it 45

really is, if you are careful and know what you are doing - remember that the change of trim values per 100MT in the trimming tables assumes that the weight is loaded or discharged from the middle of the hold and you know that it would be impossible to load all the cargo for trimming in the centre of the hold - what you should ensure is that the cargo for trimming is distributed evenly across the hold - in the athwartship and fore & aft directions. You should then be able to get the trimming as accurately as in the formula. Remember that the vessel is almost always hogged or sagged and hogs or sags under the effect of cargo loaded or discharged in different holds - always play safe - if your maximum draft is to be at midships assume she will sag a bit due to the trimming cargo, i.e. if the sag before trimming is about 3 cms, assume that after trimming it will sag further, if the maximum draft is aft assume she will hog a bit after trimming, i.e. if prior trimming you are sagging 3 cms, after trimming, she will sag 1 cm. Of course once you are comfortable with the way your v/l behaves, you can estimate whether she will hog or sag depending which holds you are using. In Case 3, even tho max draft is to be at midships, I am confident that on the Cape-size, trimming with 3 & 8 will produce a hog effect - but please do remember that the ship is not a violin and the hog & sag effect is not always instantaneous - if you load at #1,2 or 9 on the Cape-size you will probably see the hog effect come into play almost instantaneously. However, should you load in 3 or 8 it will be much more gradual similarly the sag effect will be much slower when loading in 4 or 7 and much faster when loading in 5 or 6. Remember it is not always instantaneous - an example to illustrate Voyage 8 on SILC - loading coal at R/Bay - about 3500 MT for trimming at the stop for trimming - trimming with #4 & #8. Voyage 9 similar situation. On Voyage 8 the loading sequence was such that the run before the trim stop called for loading 5450 MT in #6 hold - #6 is just abaft midships - we were sagged at this point after trimming at #4 & #8 we ended up with a sag greater than at the trim stop. On Voyage 9 the loading sequence was such that the run before the trim stop called for loading 6600 MT in #2 hold - at the trim stop we were sagged - after trimming at #4 & #8 with almost the same quantities for 4 & 8, as in Voyage 8, we found that the sag had decreased. This seems to support the idea that the v/l hogs & sags over a period of time - what my conclusion is, is that in Voyage 8, the run in #6, was causing the v/l to sag & this sagging does not stop immediately you stop loading at #6 - the v/l takes some time to complete this sag effect. Conversely the v/l was undergoing a hog effect when loading at #2, in Voyage 9, and this does not stop immediately on stopping loading at #2.

Also, this problem can be experienced especially when loading in alternate holds invariably when loading alternate holds you end up with a hog for the vessel - to 46

compound that you probably will be trimming with #3 & #7 which means that at the trim stop, with loading complete at the end holds - 1 & 9 - you will be even more hogged at the trim stop than at completion - after loading the trimming cargo at #3 & #7 you should be able to reduce this hog - but the difficulty you encounter is that it is extremely difficult to predict what amount this hog effect or any sag effect actually is however if you trim as suggested below you should be able to eliminate any estimation errors. This is the 2 drop method for trimming to which I will refer to a little later. In all the cases I have explained - I have shown you calculations right down to the absolute maximum - i.e. if the max.draft was 17.50 metres - the calculations were right down to 17.50 metres - there is absolutely nothing wrong with this method & you should be able to get right down to the absolute maximum but prudence would suggest that you always have a little in hand - let s say about 1 cm at least on the Cape - size which translates to abt 100 tonnes & about 1.5 cms on a Panamax which also translates into 100 tonnes. In my early days as a Chief Officer, after I was comfortable with this trimming calculation method, I would reach the trimming stage, keep the required safety margin in hand, and do the trimming in 1 drop, i.e. if I was trimming with #3 & #7 - I would calculate the quantities to load in #3 & #7 & load them - was later advised & realised that it is much better to do the trimming in 2 drops i.e. stop for trimming - make your calculations for #3 & #7, then remove a total of about 700 tonnes from the figures - do ONE drop in #3 & #7, do another draft check, then do a final drop in #3 & #7. This method is infinitely better - this small amount of cargo in the 2nd drop eliminates any error caused by the error in the loader - when the shore side is required to load a certain quantity, very unlikely you will get that exact figure. Secondly, your estimation of the sag or hog caused cannot be far wrong with the final drop of abt 700 tonnes - also if you have underestimated the effect of hog or sag at the first trimming drop you can still make a correction - besides that if you have erred the other way with your hog/sag estimation - i.e. on the safe side - you will be able to load a little more - for eg if you had estimated that the v/l would sag 3 cms more during the first drop & reduced your quantity in line with that, & you actually sagged only 1 cm more, you could then load a little more cargo However, in ports where they will absolutely not allow you to sail even with 1 cm excess, such as Hay Point or Richards Bay I would strongly advise that you keep something in hand even at the 2nd drop.. There are still many possibilities that will cause you to be overdraft - chief among them is that even though you do your very best to ensure there is absolutely no list you are 1 cm in excess on 1 side & have no sinkage available to correct this - always play safe - REMEMBER nobody gives you a medal for loading extra cargo - but you will face all sorts of trouble if you are over draft. Choosing which holds to use for trimming is a matter of personal preference but I will list some of the factors I would take into consideration Firstly, in any type of loading I would avoid trimming with the end holds - 1,2 & 47

9 on the Cape-size and 1,2 & 7 on the Panamax - loading in these holds definitely cause the v/l to hog - the problem is that it is impossible to predict how she will or how much she will hog - i.e. if you are loading at #1 this will cause the ford draft to increase more than the trimming tables suggest, but the decrease in the aft draft due to the hog effect will be much less & it is impossible to predict how much at each end. When sagging the effect is at one point - midships. Also because loading in #5 & #6 on the Cape-size or #4 & #5 on the Panamax would cause the v/l to sag I would not trim with these holds if I could help it. On the Cape-size would prefer to trim with any combination of #3 & 4 ford & #7 & 8 aft. Ideally if loading at alternate holds would trim with #3 & #7. On the Panamax would prefer to trim with #3 & #6, at a push would trim using #2 & when loading alternate holds would probably trim with #3 & #7 but in no event would I trim with #1. Examination of the trimming tables would show that almost all vessels have ONE hold ford of midship where loading would cause a very negligible change in the aft draft and also ONE hold aft of midships where loading would cause a very negligible increase in the ford draft - on Panamax size (7 holds) the likely holds are #3 & #6 - on the Cape size (9 holds abt150,000 dwt) it is likely #4 & #7 & hence I would prefer to do my trimming with these sets of holds, if circumstances permit. Considering the Cape-size Trimming with #4 & #7. You have reached the draft check stop prior trimming - you read drafts and make you calculations to load quantities in #4 & #7. Let us say that these figures are 1500 MT in #4 & 2000 MT in #7. I would then load abt 1200 MT in #4, then 1600MT in #7 and keep the loader at #7. Then I would read the drafts and make calculations to load certain quantities in #7 & #4 to reach the final required drafts. Assume these quantities to be 440MT in #7 & 320 MT in #4. Now the trimming tables tell us that for every 100 MT loaded in a) #4 the ford draft increases by 1.81cms & aft draft increases by 0.09cms Therefore for the 320 MT to be loaded in #4 the increase in ford draft would be 5.8 cms and there would be no increase in the aft draft. b) #7 the ford draft increases by 0.07cms & aft draft increases by 1.79cms Therefore for the 440 MT to be loaded in #7 the increase in ford draft would be nil and the increase in the aft draft would be 7.9 cms.

Assume the vessel is finishing even keel with 17.00 metres & 5cms sag i.e.midship draft of 17.05metres which is the maximum possible.. 48

At this final drop I would have someone ensuring the v/l is to be upright, we would load first in #7, with the Mate watching the aft draft on the shore side. The Mate is kept informed all the time about the quantity remaining - when the loader states that there is 300 MT to go, the Mate reads the draft. Knowing that the remaining 300 MT is going to cause an increase in the aft draft of = 5.4 cms, he can then monitor if all is well - if the reqd aft draft is 17.00 mts and he reads the draft to be 16.946 with 300 MT to go, all is well - if however the draft is 16.966 then he knows that he is going to finish over draft so he can stop the loading and make adjustments. By trimming with #4 & #7 he knows loading in #4 will not change the aft draft so at this stage - when loading at #7 - he only has to bother about the aft draft. When the 440 MT is complete at #7 and you have reached the reqd draft of 17.00 metres aft - this now removes the problem of the aft draft. Immediately loading at #7 complete - the Mate would make his way quickly forward and read the ford draft - if all is going well the ford draft should be 17.00 - 0.054 = 16.942 metres - he would then make his way to the midship draft on the shore side, with one person on the offshore side on the ladder, checking the midship draft. Now all you are to worry about is the midship draft - you have people checking on either side with one person in contact with the loader to ensure you complete absolutely upright. Now the midship draft is 17.01metres - you are going to load 320 MT in #4 which will result in a sinkage of 3 cms & assuming this will cause a sag of 1 cm you will finish with 17.05 metres. In the example of the trimming operation above, I have again finishing to the absolute maximum - if you are loading at one of the extremely strict ports would as before suggest you keep something in hand. In the above example I have shown the situation when the vessel is finishing even keel, but you maybe finishing with a trim, say by stern, of 50cms. Now we know that on SILC Ford trim correction = Trim x 0.004 = 0.50 x 0.004 = 0.002 which is negligible Aft trim correction = 0.50 x 0.042 = 0.02m No problem at the ford draft but the draft at the aft perpendicular is 0.02m more than at the aft draft mark, which means that the Mate monitoring the aft draft should ensure that the read draft aft should not be more than 17.00 - 0.02 = 16.98 metres.

49

In the example on the previous page have shown a situation where you are trimming with 4 & 7 which do not change the draft at one end. However you may be trimming with #3 and #8 with say, 320MT in #3 & 440MT in #8 Trimming tables tell us that for every 100MT a) in #3 the ford draft increases by 2.35cms & the aft draft reduces by 0.48cms therefore loading the 320 MT in #3 would result in the ford draft increasing by 7.5cms and the aft draft would decrease by 1.5 cms. b) in #8 the ford draft decreases by 0.50cms & the aft draft increases by 2.34cms therefore loading the 440 MT in #8 would result in the ford draft decreasing by 2.2cms and the aft draft would increase by 10.3 cms. Now when you are doing the final drop the Mate at the aft draft mark would, if you were finishing even keel, allow the aft draft to reach 17.00 + 0.015( the decrease in aft draft caused by loading at #3) =17.015 on completion at #8.This would result in the aft draft being 17.00 m as required on completion of loading at #3. If the vessel was to finish with a 50 cms trim then he would ensure that the aft draft, would on completion of loading at #8, be 17.00 + 0.015 - 0.02(diff between read draft & draft at perpendicular) = 16.995 This would result in the aft draft (at the perpendicular) to be 17.00 meters on completion of loading at #3. Each of the combination of holds you would use for trimming would have it s pros & cons #4 & #7 by virtue of changing the draft only ford and only aft respectively make it easy for monitoring the draft during trimming. However both 4 & 7 would cause the vessel to sag so your sag caused by the trimming cargo would be quite a bit. Also, 4 & 7 change the trim only very slightly - about 1.8cms per 100MT - therefore you would have problems if you reach the trimming stage with a trim far different from what you require to finish with. For eg: in Case 7A when you had to change the trim by 36.4 cms at the trimming stage you had to load 1277MT in #8 to achieve that - were you trimming with #7 you would have to load 2034MT to achieve the same. #3 & #8 would both work to slightly hog the vessel & it would be easier to correct any trim using #3 & 8 rather than 4 & 7. When possible, i.e. when not loading alternate holds a combination - such as 4 & 8 sometimes works best - but the choice is yours. Also have found that about 2000MT, on the Capesize, is quite sufficient for trimming it is not too large, such that the hog or sag caused becomes a very big factor, also not so large as the shore loader error begins to make a big difference & also large enough to make any changes of trim you might require. On the Panamax I would be happy to trim with about 1400-1600 MT.

50

Also, I would advise that you make your cargo plan such that in the pre-calculation when you reach the trimming stage you have the same trim as when you complete loading - it is a pipe dream to think that when you actually make the stop for draft check & trimming that you will have the trim you desire - but this way you only have to make the correction in trim caused by the loader. If however you plan to finish with 50 cms by stern and your plan calls for you to reach the trimming stage with 25 cms of trim - then during trimming you would be not only trying to change the trim to the 50 cms but also making the correction to the trim caused by the loader which would be so much more difficult if, say, you reached the trimming stage with 5cms trim. Also, it is not necessary that should you ,say, make a plan to trim the v/l with #4 & #8 and plan for 2000MT for trimming, that in your plan necessarily you need to have 1000 in #4 & 1000 in #8. It s quite okay to have, say, 1200 in #4 & 800 in #8 even in your plan.

Probably the most important part of trimming is to make absolutely sure that each time you stop loading at a hold during trimming you are absolutely upright. Say, you are trimming with #4 & #8 using the 2 drop method. At the first drop you load at (a) #4 then (b) at #8. Then after a quick draft check and calculation in the 2nd drop you load at (c)#8, then at (d) #4. Be sure that at the end of each one i.e. at the end of (a), end of (b), (c) & (d) you are absolutely upright. A list of 0.25 degrees translates into almost 25 cms difference in the midship draft.

51

CHAPTER IV - HOG / SAG

52

HOG / SAG What your hog/sag will be is very difficult to predict - however having loaded these ships over the years have observed out the following On one Panamax I served on - Navios Bulker if you are using the old loadicator - a new loadicator has been fit in the 2nd half of 1998 - but the old one exists - if you can distribute your weights such that as per the OLD loadicator the Bending Moment at Midships in the Harbour Condition shows 8% sag you will end up with no hog or sag at midships. For every 2% the loadicator shows on either side of this 8 % you should get either 1 cm of hog or sag - i.e. if the loadicator shows 6% sag at midships you should end up with a 1 cm hog at midships. If it shows 12% sag at midships you will probably end up with a 2cm sag at midships. On another Panamax - Lucky Bulker I think you have to distribute the weights such that as per the loadicator the Bending Moment at Midships in the Harbour Condition is about 8-9 % sag to get no hog or sag at midships - I was last on the Lucky in 95and I am not sure if I am absolutely right but even if I am not I would not be far wrong. On the 2 sister Cape-size I sailed on you require to distribute the weights such that as per the loadicator the Bending Moment at Midships in the Harbour Condition is between 15-16% sag - to get no hog or sag at midships - then every about 1.6% on either side of the 15-16% translates into 1 cm of hog or sag at midships.. PLEASE REMEMBER THAT IN ALL CASES WE ARE TALKING ABOUT THE LOADICATOR INFORMATION OF THE BENDING MOMENT IN HARBOUR CONDITION & this is limited to individual vessels & loadicators. If you had noticed I have kept mentioning no hog or sag at MIDSHIPS - it is important that I am talking only about no hog or sag at midships and it is here only that it is important - the vessel s drafts can only be read at the ford and aft and at the midships on both sides - if you can distribute the weights such that after reading the drafts you have no hog or sag at the midships then for all means and purposes you have no hog or sag. Have loaded the Cape-size more than twice and finished with no hog or sag by calculation - I knew full well that the v/l was sagged over the length of the v/l, but with my cargo distribution I had no hog or sag at the midship draft marks - the strain gauge clearly showed the v/l to be sagged but the drafts did not & since the only way you can calculate hog or sag is by reading the drafts at the 6 draft marks to all means & purposes we had no hog or sag.

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You will not always be able to distribute the cargo such that you will end up with the figures mentioned above & achieve no hog or sag - at midships - when you are loading alternate holds you will almost definitely end up hogged - or when you are loading all holds and are volumetrically full or very close to you don t have much choice on how to distribute the cargo - but what you can do is that within the limitations of the particular circumstance you work, to get as close to the above figures as you can, or the figures that by experience you have determined for your vessel/loadicator.. We have found how to be able to make the distribution of cargo work for us in that there is no hog or sag at the draft marks & therefore we have loaded the max.cargo & also have actually a little more cargo on board than the Surveyor s figure shows so that we have no problem at the discharge port. However this same situation sometimes can work against you - when you are in the ballast condition on both Panamax & Cape-size you will be hogged over the length of the v/l - but if you are in the Heavy Ballast condition - i.e. with the hold in and also all the ballast tanks you will also in all probability be hogged over the length of the v/l, but because the ballast hold is just off the midships you will be sagged at the midships even though you may be hogged over the length of the v/l. - which means that the read midship draft is greater than it actually should be. As we use the 6-sided formula this results in the Initial Displacement being higher than it actually is, and there is no correction factor that can be applied here, and the v/l is said to be doubly deflected which results in the constant being very much higher by calculation than it actually is if you are in heavy ballast on arrival at the load port and you get a very high constant be sure to tell this to the surveyor and try & bring the constant down nearer normal levels. Also, temperature effects the hog or sag. If you read the drafts at night, and make no changes at all after that and read the drafts again at midday, it is quite likely that the drafts will not be exactly the same as at night, because the hog or sag has changed. This is best illustrated by vessels plying in the Great Lakes, where they are working at a max. draft and cannot allow this changed hog or sag to increase their draft & so have the deck water running at all times over the main deck in order that the entire shell plating is at the Lake water temperature. Where this may affect you is at ports such as Richards Bay, where this has actually happened to some vessels - the max. draft you are permitted to sail with is 17.50 meters, when your summer draft permits & you are allowed to have your midship draft to a max. of the Summer mark when that is below 17.50 metres. It can happen there that you finish loading at night, working to your max. drafts permitted & you & the draft surveyor agree on the drafts & these are all within permissible limitshowever you are unable to sail immediately for whatever reason -such as too much swell at the entrance & you only sail at midday - if somebody from the port authority then reads the draft prior sailing you might have exceeded the max draft - only when you explain the day/night effect to them will they permit you to sail - on the other hand if at the time of completion of loading your midship draft has exceeded your summer loadline they will not permit you to sail even though your MQM may be equal to or less than the summer mark - no allowance is made for hog or sag at the completion of loading by the port - they will only permit you the day/night effect allowance in case you cannot sail on completion. For this reason, keep something in hand when transiting the Panama Canal. A very important thing to remember is that the while the loadicator calculates & shows 54

the ford & aft draft, (from which you can calculate the mean draft) it cannot take into account hog or sag or anything else - it therefore shows mean, midship & MQM as the same. Eg. if the loadicator shows F: 16.0 meters A: 17.0 meters - it is showing all figures on basis of displacement at the mean draft of (16.0+17.0) / 2 = 16.50 metres. If you are hogged or sagged it cannot take that into account & therefore as far as the loadicator is concerned your mean draft is your midship draft & when that happens so is the MQM. Now if you assume that the vessel is going to sag say 10 cms you will have to adjust your figures to show the right displacement - remember displacement is got from MQM. Say your midship draft is to be 17.50 metres & you expect to sag 10 cms & are at even keel.. Your first instinct would then be to work your figures such that the loadicator shows ford draft to be 17.40 & aft draft to be 17.40 metres. THIS IS WHERE YOU ARE WRONG. If you used the 6 sided formula with drafts of F:17.4 M:17.5 A: 17.4 you would get an MQM of (( (17.4 x 2) + (6 x 17.5) )) / 8 = 17.47. What you would then seek to do is to adjust your weights to get F: 17.47 A: 17.47 on the loadicator. You would of course declare your drafts to be 17.4 / 17.5/ 17.4 If you expect to be trimmed 50 cms with a midship draft of 17.50 metres and 10cms sag, you would work your weights such that the LOADICATOR shows drafts of F: 17.47 - 0.25 = 17.22 & A: 17.47 + 0.25 = 17.72 Of course your declared drafts would be worked on the basis of Midship - sag = Mean & then trimmed about the Mean which would be 17.5 - 0.10 = 17.40 = Mean & 17.4 - 0.25 = 17.15 = ford draft & 17.65 = aft draft.

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CHAPTER V - CONTROLLING DRAFTS

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CONTROLLING DRAFTS When you start making your calculations for a particular voyage it is most important to know where to start i.e. where your controlling draft is. Sometimes this is quite easy. 1. Say, you are loading at Tubarao for Rotterdam when it is summer in the Northern Hemisphere. Draft available at Tubarao is 22m, at Rotterdam 23m, your summer draft is 17.52 meters - simple you can load in Tubarao up to the summer mark assuming Tubarao is in the Summer Zone - if it is in the Tropical Zone you will be able to load up to Summer Loadline plus the consumption in the Tropical zone till you reached the Summer Zone. 2. Say, you are in Tubarao, loading for Rotterdam, when it is winter in the Northern Hemisphere. - you can load such that when you enter the Winter Zone off Cabo Finistere you are at the Winter Mark of 17.16 metres. Your controlling draft would therefore be 17.16 metres when entering the Winter Zone & you would have to start your calculations from that point. It would then work out that your sailing draft from Tubarao would be Winter Draft + consumption from Tubarao to Finistere. When calculating I would start at Finistere and then work backwards to Tubarao 3. Say, you are loading in Tubarao, then discharging at a port where the allowed draft is 16.00 metres - simple - your controlling draft would be at the disport - you would base your cargo to arrive even keel at the disport with 16.00 metres draft after calculating what your fuel etc. ROB would be at the disport. 4. You are loading at Newcastle, Australia, where max. draft is 15.50m, your summer draft is 17.52 metres, for Rotterdam where max draft is 23.0m. Simple your controlling draft is Newcastle & you would start your calculations to sail out of Newcastle even keel, then work for arrival Rotterdam, taking ballast to avoid arriving down by the head.

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5. You are on a Panamax size where you are required to load as follows Disport 2 is Eregli, Turkey where max draft is 11.5 metres. Your summer draft is 13.25 metres. Loading at Saldanha Bay, South Africa. Disport 1 is Constanza, Romania. You have to load for Constanza all you can, that is possible, assuming you are loading maximum for Eregli. Max draft in Constanza 17m. All ports & the entire route in the summer zone. Max draft in Saldanha Bay 20m. In this case the initial controlling draft is Eregli. You would have to start your calculations based on first arriving at Eregli with an even keel draft of 11.50 metres after calculating what your fuel etc. ROB would be at Eregli.. Then work your fuel consumption backwards to sail out of Saldanha Bay, at the summer loadline - you would then work out the cargo you can load for Constanza, & calculate the trim you would have to sail out of Saldanha Bay with, such that after fuel consumption en route, & discharging in Constanza you would arrive at Eregli with the even keel draft of 11.50 metres - your arrival draft & trim at Constanza in this case is not really a problem.

6. You are loading on the Cape-size, summer draft = 17.52m, at Tubarao, where max.draft is 23m, for Pohang where max. draft is 17.40metres. Consumption of fuel en route causes a rise of 15cm. Both ports & route in the summer zone. Here even tho the draft available in Tubarao is much higher than in Pohang, the departure draft in Tubarao is the controlling draft. You are permitted to load up to your summer mark of 17.52 metres & no more - when you do that in Tubarao - your consumption en route to Pohang cause a rise of 15 cms - thus your draft arrival Pohang would be 17.52 - 0.15 = 17.37m. Of course you would have to adjust your trim departure Tubarao, to arrive, after the fuel consumption, even keel at Pohang.

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7. Cape-size situation - summer draft = 17.52 metres - all ports & route in summer zone. 1st loading port - Sept Isles, Canada - iron ore lumps. Max draft at Seven Islands less than summer draft. 2nd loading port Saldanha Bay, South Africa. - iron ore pellets. max draft 20m. 1st discharge port Kimitsu, Japan - discharge the Saldanha Bay cargo. Max draft 20m 2nd discharge port Nagoya, Japan - limiting draft, reqd to disch the Sept Isles cargo. Required to load max cargo for Nagoya - no bunkering en route. This was a situation on SILC (actually the cargo was iron ore concentrates - 2 types). Now, in this calculation, you would have to start with Nagoya, after calculating what your arrival Nagoya bunkers etc.would be then work the fuel etc. consumption backwards to S.Isles, via Kimitsu & Saldanha Bay, & then see whether with the max. Nagoya cargo on board, you were within the limiting draft in S.Isles -it was - but if it were not then the limiting draft would be at S.Isles - in this case it was at Nagoya - then after calculating your weights at S.Isles, work your fuel consumption to Saldanha Bay on basis of that work your figures to sail out of Saldanha Bay at your summer draft, then discharge the Saldanha Bay cargo at Kimitsu, then after consumption of fuel Kimitsu, to arrive at the max. draft at Nagoya.

Please remember there will be probably be density changes which you have to take into account. Also vessel must be within allowable stress limits at all times.

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CHAPTER VI - MAXIMUM DRAFTS

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WHAT ARE YOUR MAXIMUM DRAFTS? Consider this situation on the SILC C/P terms require you to load a cargo of iron ore from Tubarao, Brasil to Pohang, Korea. Max. draft in Tubarao - 23.0 meters, max.draft in Pohang - 17.40 meters. Consumption of fuel from Tubarao to Pohang = 1656 MT. Increase in FW = 100MT. Therefore, difference in displacement Tubarao to Pohang = 1656 - 100 = 1556 MT less. Notwithstanding the available depth of water at Tubarao, the v/l s summer draft is 17.52 meters. Max.draft at Pohang = 17.40 meters. Difference = 0.12 meters = 12cms TPC = 106.38 Therefore difference = 12 x 106.38 MT = 1276.56MT But difference in displacement will be 1556 MT. Thus, since the vessel cannot load more than the summer draft the controlling draft in this case will be departure Tubarao - when the vessel can load to the summer draft - if you can adjust your trim then you will arrive at Pohang, in this case, less than 17.40 meters. By calculation we have found that to arrive at Pohang even keel, after sailing out of Tubarao at summer draft, the drafts by the loadicator, departure Tubarao are as follows F: 17.12m M: 17.52m A: 17.92m With these drafts after our consumption of fuel & increase of fresh water, the arrival drafts at Pohang are computed to be F: 17.38 M: 17.38 A: 17.38. Relative Density at both ends taken to be 1.025. Thus while loading/trimming what would be the maximum drafts you could load to at ford, at midships and at aft? In this case the midship draft is quite simple - it is the summer loadline so come what may you can load upto a maximum of 17.52m draft at midships. Your computed drafts tell you that with a ford draft of 17.12 meters with your F.O/FW you will arrive in Pohang with a ford draft of 17.38. Similarly you would arrive with an aft draft of 17.38 after leaving Tubarao with an aft draft of 17.92. Max.draft at Pohang is 17.40 meters & therefore in theory you would be able to load to a max.draft of 17.14 meters ford & 17.94 meters aft - but I would highly recommend that the 2 cms you have in hand you keep in reserve & not load, in this case, to more than 17.12m ford & 17.92 m aft. This means that even if your ford draft is 17.00 you cannot load to an aft draft more than, in theory, 17.94, & in practice 17.92. Or for that matter, you cannot load more than to a ford draft of 17.14m, in theory, & 17.12 meters in practice, even if the aft draft is 17.80 meters. With your arrival Pohang midship draft, only 2 cms less than the maximum you do not have much sinkage, if any available, to fill ballast to change the trim.

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Also if you were hogged, you would not be able to fill more cargo, such that the ford & aft draft exceed 17.12 & 17.92 respectively, even if your midship draft does not reach 17.52m. Should you be hogged say 5 cms, your max. ford draft would still be 17.12 meters, your max aft draft would still be 17.92 metres. Therefore your max. mean draft would be (17.12 + 17.92) / 2 = 17.52m With a 5cm hog your midship draft would be 17.52 - 0.05 = 17.47m Using the 6-sided formula, your MQM would be 17.48m - this means less cargo, but there is nothing you can do about it as otherwise you will be over draft at the ford & aft ends arrival Pohang. Please remember that if the density at Tubarao is not 1.025 then you have to make allowance for that.

Now consider this situation - say you are on a vessel where the summer draft is 17.25 meters. You are loading at a port where max draft is 22m. The max draft at disport is 17.10 meters, By calculation you have found that to arrive at the disport at an even keel draft of 17.10m you have to sail out of the load port with drafts F: 17.00 A: 17.20 M: 17.40. In the previous example you had a situation where the max. midship draft was clear-cut - you could only load to your summer draft. In this case you find that you are going to finish with a sag - say, your drafts at completion are F: 16.95 M: 17.20 A: 17.35. You know that you have not reached the max. possible ford or aft drafts, but can you load anymore - loading would cause your midship draft also to increase, and you know that your present midship draft is 17.20 metres & the summer draft is 17.25 metres. The answer is NO - if you do so, while you will comply with your summer loadline, you will be over draft at the disport.

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