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

A PRACTICAL GUIDE
TO DRAFT SURVEYS
TRIMMING CALCULATIONS
ETC.
GLENN J SALDANHA
()

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 I - THE 6 - SIDED DRAFT ................ 3
CHAPTER II - THE DRAFT SURVEY ......... 11
CHAPTER III - TRIMMING ............................ 22

CHAPTER IV - HOG / SAG............................. 51

CHAPTER V - CONTROLLING DRAFTS ........ 59

CHAPTER VI - MAXIMUM DRAFTS ........... 63

2


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 A 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.

5


No matter what is taught in the various colleges of the world with respect to hydrostatic
draft, bulk carrier draft calculations the world over is based on the six-sided formula.
Drafts are read on all six sides Fp - Ford port
Fs - Ford stbd
Mp - Midship port
Ms - Midship stbd
Ap - Aft port
As - Aft stbd
After these drafts are read, the trim correction is applied to the ford & aft drafts & to the
midship drafts when relevant (on some vessels the midship draft marks are not exactly
at midships) - (same explained later) to convert these read drafts to drafts at the
perpendiculars We are now left with drafts – (where “c” stands for “corrected”)
Fpc, Fsc, Mpc, Msc, Apc, Asc - which are the same as above except that they are
corrected to the perpendiculars or midships as the case may be.
Using the mean of each set we get the Ford, Aft & Midship Drafts (Fpc+ Fsc) = F (Ford Draft)
(Mpc + Msc) = M (Midship Draft)
2

2
(Apc + Asc) = A (Aft Draft)
2

(F + A) = Mean (Mean Draft)
2

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
pre-calculation examples are shown for when the vessel is completing loading,
when in case of bulk carriers we almost always finish even keel, without trim, or
very close to even keel, which makes any inaccuracy caused by making the
assumption that the vessel is trimming about midships, so very small as to be
discarded. In any event pre-calculations are only that and the actual picture only
obtained, during & after loading.
Before any examples are given, the formula must be understood properly The Mean Quarter Mean Draft which is used for obtaining displacement concept must
be understood properly.
MQM = 6 M + F + A
or
MQM = 6 M + 2 Mean
8
8
And Hog / Sag = Difference between Midship Draft and Mean Draft
When Midship Draft is greater than Mean Draft, Vessel is Sagging.
When Midship Draft is less than Mean Draft, Vessel is Hogging.

Now it must be understood that if M= 17.0 meters and the vessel is sagging by 10 cms,
then the MQM is NOT 16.90 meters - to illustrate
Example 1:
Given that Midship Draft (M) = 17.0 meters, and vessel is sagging 10 cms
Vessel is sagging 10 cms, therefore the Midship Draft is 10 cms more than the Mean
Draft.
Therefore Mean Draft = M - 0.10 = 17.0 - 0.10 = 16.90 meters
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
Unpumpable 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
Displacement 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, (Midship Draft – Sag) = Mean Draft = 17 - 0.12 = 16.88
Now to calculate the fore & aft draft we assume that the v/l trims about the midships
and since v/l is to be trimmed 20 cms & is trimming equally ford & aft
& therefore
ford draft = 16.88 - (0.20 / 2) = 16.78
& aft draft = 16.88 + (0.20 / 2) = 16.98
MQM = 6 x 17.0 + 16.78 + 16.98 = 16.97 meters
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 = MQM = 6X + 2 (X - 0.08)
8
8
8(MQM) = 6X + 2X - 0.16
(8 x 15.85) = 8X - 0.16
(126.8) + 0.16 = 8X

Therefore X = Midship Drafts = 15.87 meters
V/l sagged 8 cms, Mean Draft = 15.87 - 0.08 = 15.79 = F = A (v/l finishing even keel)
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 Example 4 except vessel to finish hogged by 10 cms
Let Midship draft = X meters
As v/l hogged by 10 cms, Mean Draft = (X+0.10) meters
MQM = 6X + 2 (X+0.10) => MQM = 6X + 2X + 0.20
8
8
8 (MQM) = 8X + 0.20
8 (15.85) - 0.20 = 8X
X = 15.825 meters = 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 => MQM = 6X + 2(X-0.08)
8
8
MQM = 6X + 2X - 0.16 => 8 MQM +0.16 = 8X => 8(15.85) + 0.16 = 8X
8
Therefore X = 15.87 = Midship draft Sag = 8cms
Therefore Mean draft = 15.87 - 0.08 = 15.79 mts
Assuming vessel trims abt midships & trim = 14 cms F = 15.79 - 0.07 = 15.72
& 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 (from SILC)
Assuming that the sag remains same, & very likely it will,
MQM = 6X + 2(X - 0.08)
where X = Midship draft
8
=> 8(15.968) = 6X + 2X - 0.16 Therefore X = 15.988 = Midship draft
& Mean Draft = 15.988 - 0.08 = 15.908
Again assuming v/l trims around the midship we will have F=15.838 & A=15.978
However there will be a change of trim due change of density & this will have to be

applied to get the actual Ford & Aft Drafts in the DW of density 1.010.

If the vessel were to hog instead of sag, say hog by 10 cms then as in Example 5,
Assume X = Midship draft and Mean Draft = (X + 0.10) - all other calculations same.

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 (crest and trough reading each time - they

then discard the 2 extreme readings on either side of the mean & then take the average
of the remaining 16 readings - their argument is that if you take your time over the
reading, mentally & physically you lock onto a particular reading & stick with it If the level is oscillating between 17.0 & 17.8 meters & you feel it is 17.46 meters, you
mentally & physically lock onto 17.46 meters & discard anything else - I have found
this to be true - & the rapid 20 reading & noting method much better.
Remember also that the initial draft survey at the load port is basically to find out what
the constant is - more often than not in the case of a Panamax bulker will show a
negative constant in ballast with trim - no surveyor likes to record a negative constant &
some adjustments are made with the ballast to record a positive constant. This actually
results in the cargo figure after loading to show a little less than is actually on board.
This is not really bad, because then you do not finish the discharging and find a shortfall
of cargo - you normally land up with slightly more cargo, but remember this - when you
complete the discharging & you land up with a cargo figure greater than you have at the
loaded end, the surveyor is quite happy to make adjustments so that the figures tally.
However should you land up with slightly less cargo, at the discharging end, the
surveyor, who normally represents the receiver, who wants to pay less, is not likely to
make adjustments in order that the figures tally. So, do not be gung-ho at the loading
end to try & show more cargo than there actually is. The Surveyor might want to do
this, as he represents the shipper, who wants to sell more, but nobody gives you a medal
for having more cargo, and should you land up with less cargo at the discharge port, all
kinds of hell breaks loose.
Where you should try to show the maximum cargo, even insisting on recording a
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” and in the attached file.

CORRECTION TO THE PERPENDICULARS
For correction to the perpendiculars of the fore and aft drafts see Page 4.
If the midship draft is obtained by measuring the freeboard to the deckline then no
correction is necessary as the loadline disc, IN MOST CASES, can be considered to be
at the midlength of the vessel –

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, from amidships, as the deepest draft it is
added, and when the COF is on the other side of amidships as the deepest draft it is
subtracted.
This correction does not allow for the fact that, when a ship trims, the COF moves from
its tabulated position. Some ships have corrections for this, but when this is not
provided the following correction called the 2nd trim correction must be applied 2ND TRIM CORRECTION = (TRIM IN METERS) x (TRIM IN METERS) x 50 (dM/dZ)
(IN TONNES)
LBP (in meters)
Where dM/dZ is the difference between the MCT for a draft of 50 cm greater than the
corrected mean draft and 50 cm less than the corrected mean draft
I.E. if the corrected mean draft is 12.0 meters then dM/dZ would be the difference
between the MTC (Moment to change trim) at 12.50 meters and 11.50 meters.
This correction is ALWAYS ADDED to the displacement.
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 = TPC for the deepest draft amidships
TP2 = TPC for the shallower draft amidships
D1 = Deepest draft amidships (in METRES)
D2 = Shallower draft amidships (in METRES)

This correction is ALWAYS ADDED to the displacement because the effect of heel is
to increase the water plane area & so lift the ship out of the water.
CORRECTION FOR DENSITY
Almost all ships have their displacement tables tabulated for a Relative Density of 1.025
However what we are interested in finding is the actual displacement,
I.E. is the displacement in the dock water that the vessel is lying in.
TRUE DISPLACEMENT = R.D. OF DOCK WATER x SCALE DISPLACEMENT
R.D. USED FOR DISPLACEMENT SCALE

eg: On SILC if your draft is 15.00 meters & v/l is lying in DW of 1.017 then true
displacement would be calculated as follows:
Scale Displacement for 15.00 meters = 143291.2 tonnes
True Displacement =

SCALE DISPLACEMENT x R.D. OF DOCK WATER
R.D. FOR DISPLMNT SCALE

= 143291.2 x 1.017
1.025
= 14217.3 tonnes

The displacement now obtained is the true displacement, within the limits of accuracy
of the drafts and the ship’s stability data. While the actual calculation on all of our ships
is just a matter of feeding in the read drafts and the deductibles it is wise to know what
is actually being done - also not all surveyors are above board & some will try & pull
wool over your eyes in order to show more cargo at the loading port and less cargo at
the discharge port.

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 the
Deadweight.
2. As soon as possible after reading the drafts & density, sound all the fuel, ballast and
Fresh water tanks. Correct the soundings for list and trim and using the Calibration
Tables calculate the fuel, diesel, fresh water and water ballast from the deadweight.
The remainder represents the constant.
The value of the constant in the ship’s data represents the constant for a new ship.
However, as the ship ages its weight increases (primarily due to the reluctance of ship’s
officers to throw anything away.) In many cases the constant is too low. The tendency is
to try and show as small a constant as possible at the load port in order to show more
cargo on board - avoid this tendency as the surveyor at the disport is not going to be
charitable and will be trying to show as little cargo as possible & will be trying to
compute as large a constant as he or she can. The time to try and get as small a constant,
without going to ridiculously low levels, is when you are on a charter when the Owner
is getting paid on basis of freight.
(b) After Loading
Read the drafts and calculate the loaded displacement. Using the constant found
previously (prior loading), sound fuel, diesel, fresh water and ballast tanks, correct for
trim & list, use the calibration tables to find the quantity of fuel, diesel, fresh water and
ballast, subtract the lightship from the loaded displacement together with the deductibles
and you will have the cargo loaded.
AT THE DISCHARGING PORT

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

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 = B
Fuel Oil on completion of discharge
= FO2
Diesel Oil on completion of discharge
= DO2
Fresh Water on completion of discharge
= FW2
Ballast Water on completion of Discharge
= BW2
Constant
= K
Light Ship
= LS
Deductibles on completion of discharge

= b = FO2 + DO2 + FW2 + BW2

Displacement on completion of discharging = B = b + K + LS
Please note that K & LS are the same prior commencement & at completion of
discharge.

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 accepts a
full sounding but applies a trim correction to the sounding, (for the vessel’s current trim)
which results in the tank showing not completely full. This means it shows you have
less ballast on board & consequently less cargo - even though you know the tank is
completely full he does not accept this - to prevent this unless you can ensure that the

same surveyor is doing the initial & final draft survey at the disport & he will accept
that your tanks are completely full , do not fill the tanks completely full - keep them
slightly slack, actually sound the tanks, so that when he applies the trim correction this
will reflect the actual ballast in the tank & you will not have less cargo.
In many ways the method described here is used at the load port except for an
important difference - at the load port you compute your constant at the time of
initial draft survey. At that time to arrive at a constant satisfactory to you and the
surveyor some adjustments are made to the ballast quantity / density / drafts to
arrive at this figure - this is in order mainly not to get a negative constant or a
constant very different from the constant that is normal for your vessel.
If you used the same system (without making adjustments to arrive at a contrived
constant) you would get the actual cargo figure but a constant far removed from the
normal or a negative constant. Either of these is a red flag to anyone checking the
figures & therefore we adjust & arrive at a contrived constant. I would advise that you
do use this method when the cargo is being loaded on a charter where the Owners are
being paid for the freight.
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 and ballast, then
CARGO ON BOARD = D – C =
D - FO3 - DO3 - FW3 - BW3 - K - LS
Where D = displacement after loading.
Using this you should in theory be able to compute the cargo on board, but this assumes

that the value of the lightship is accurate and your computed constant at the load port is
accurate. This however is not always true so the cargo is actually calculated as follows Let Displacement after completion of loading = D
Fuel Oil on completion of loading
= FO4
Diesel Oil on completion of loading
= DO4
Fresh Water on completion of loading
= FW4
Ballast Water on completion of loading
= BW4
Constant
= K
Light Ship
= LS
Deductibles on completion of loading

= d = FO4 + DO4 + FW4 + BW4

Displacement on completion of loading = D = d + K + LS + CARGO
CARGO ON BOARD = D – d – K - LS

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

21


CHAPTER III - TRIMMING

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TRIMMING
This is probably the subject about which very little or almost nothing is taught in any
nautical school - have learnt the following method from the foremen in Brazil who are
loading at up to 16,000MT per hour and require to be spot on.
Am going to show various possible situations on the Cape-size vessel SILC at the stop
for draft check prior trimming and then trimming with different sets of holds.

CASE 1.
Assume the read drafts at the draft check to be
Fp = 16.70
Fs = 16.70
Mp = 16.82
Ms = 16.84
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

Mc = 16.83 Ac = 16.94

Corrected trim = 16.94 - 16.70 = 0.24m

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Vessel is in water of R.D. = 1.025 and is required to load such that its midship draft is
17.25 meters and finish with a stern trim of 42 cms. Max allowed draft = 17.50 meters.
Assume you are to trim with holds 4 & 8.
(This was an actual situation where we were loading in Richards Bay where max draft is
17.50 meters. Winter zone draft when passing Cape Finistere en route to Le Havre is
17.16 meters. The difference between 17.25 meters & 17.16 meters is the consumption

en route from Richards Bay to Cape Finistere and the 42 cms trim would after
consumption result in v/l arriving Le Havre even keel.)
Corrected drafts are Fc = 16.70
Mc = 16.83 Ac = 16.94
Midship draft is 16.83 meters and maximum midship draft is 17.25 meters.
Sinkage available = (17.25 - 16.83) = 0.42m = 42 cms
Just so that you do not overload assume that the vessel will sag a further 2 cms after
trimming.
Therefore sinkage available is 42 - 2 = 40 cms.
As TPC = 106.30 tonnes, therefore cargo to load = 106.30 x 40 = 4252 MT.
Final midship draft will be Midship draft + addl sag + sinkage = 16.83 + 0.02 +0.40 =
17.25m
From trimming tables at draft of 17.05 meters (mean between 16.83 and 17.25)
For every 100 MT
No. 4 Hold

sinkage ford

sinkage aft

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 = 4.57
(Disregard for the moment that #4 will trim the v/l by head & #8 trim by stern)
****
(In the case of No.4 hold as loading in #4 will cause both the fore and aft drafts to
increase and therefore the values in column 2 & 3 are positive, and the last column is
the DIFFERENCE between the two, the values 1.81 and 0.09 are subtracted to give
1.72 . In the case of No.8 hold, since the ford draft reduces and the aft draft increases,
the signs are different when you are looking for the DIFFERENCE, you add the
values 0.50 & 2.25 to get 2.85. Do go through all 9 cases so you fully understand this.)

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 & 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
Change of trim caused by #4 & #8

= 3620 x 1.72 = 1362.4 = 1362
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.

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