ship stability tutorials-mca oow unlimited written exam-nuri kayacan

Dec 2012

Answers Prepared by MCA OOW Nuri KAYACAN

NOTES ON

SHIP STABILITYFOR PHASE s/CLASS 4 POST/OOW

/ıo-ı

Gla S gg,H,,,Çtg}J,çge

Dec 2012


The crane driver now lowers the block so that itbecomes halfsubmerged in the dock water which has a density of 1.020 Vm3.

DOCKWATERDENSITY1.020 t/mr

SA'O ',, 1, ,,,"', ',,,, . : , ::: :: ::;:;::,i,, :,:, j : ,

Wrt'at,Ioad(mass) wilı the gauğe noi indicaİe? ,,'',',::' . '

Basic principles (MARFrev.72108102)

A'nswbrrnuın"k is now displacing a uolume of wialer wlıere:

Voilume of lıjdter clisplaced =_(İx: 2m x hm)

Mass of water displaced : Volume x Density of the dockwaleı1

, =4m3x 1.020t/nf: 4.08 t whiclı represents tlıe upthrust of 'lhe

buoyancyforce (Bfl created by the displacedwater.

Tlıerefore: Mass of block = 62.72 tUnlhrust due to Bf : 4.08 tGause reading : 58.64 ı

BUOYANCYFORCE ACTING AT CENTROIDoF UNDERWATER VOLUME ( 4.08 0

WEIGHT FORCE ACTING AT CENTRE OFGRAVITY OF THE BLOCK (62.72 t)

Basic principles (MARPrev.l2l08l02)

Dec 2012


The densily of any given substance is it's muss per unİt volume.

This can be expressed as DENSITY: MASSVOLUME

I9jlhip st1_bjlity purposes the.u-nits commonly used aıe:

Mass: tonnes (t)

!'oıu-"-. cubic metres (m3)

Density: tonnes per cubic metre (t/m3)

Rearranging the above formula gives:

VOLUME: MASSDENSITY

and MASS=VOLUMExDENSITY

BASIC PryryCryLJS

The densily of any given substance is it's muss per unİt volume.

This can beexpressed as: ffi ^I VoLUME ı N

-

I9jlhip st1_bjlity purposes the.u-nits commonly used aıe:

Mass: tonnes (t)

!'oıu-"-. cubic metres (m3)

Density: tonnes per cubic metre (t/m3)

Rearranging the above formula gives:

l VOLUME: MASS ı

l DENsITY ı

anci:

Density of water in which a ship tvpically floatsA ship is presumed to always float in water which lies in the

following density range:

FİESE-WATER'6ır7 - 1.000 t/m3 toS,eır WATER (Sıil1:Water that lies between these two extremes is termed DOCKtyATER (DW).

Basic principles (MAR IRev. l2l08/02)

BASIC PRINCIPLES The laws governinq flotationTwo laws need to be considered:

J.

,r

Archimedes' principle;The law offlotation.

l. Archimede's principlethat when a body is wholly or partially immersed in

a liquid, it experİences an upthrust (apparent loss of mass- lermed Buoyancy force (Bİ)) equal to tlıe muss of liquiddisplaced

Consider a block of steel measuring 2m x 2m x 2m that hasa density of 7 .84 tlm3.

ship's side in air?

Basic principles (MARPiev.12/08102)

Dec 2012


Law of flotationŞtates that every Jloating body displaces iı's ownof ıhe liqaid in which itfloals.

The displacement of a ship (or any floating object) is defined as

ıhi- number of ıonnes of water it displaces. It is usual toconsider a ship displacing salt water of density I.025 tlm3,however, fresh water values of displacement (1.000 t/m3) areoften quoted in ship's hydrostatic data.

The u,olume of displacement is the underwater volume of a ship

1float i.e. the volume below the waterline.

To calculate the displacement (W) of a ship the following needsto be known:

The volume of displacement (V)The density of the water in ı'}t lctı it floats (p)

Since: MASS: VOLUME X DENSITY

the mass, or displacement, of a ship is calculated by:

DISPLACEMENT = VOL. OF DISPL, X WATER DENSITY

t.e.

Basic principles (MARRev.72108102)

To calculate the displacement of a box-shaped vessel

Consider the vessel shown.

VOLUME OF DISPL.: LENGTH x BREADTH x DRAUGHT

Vıox:LxBxd

Therefore:

DISPLACEMENT : VOL. OF DISPL. x WATER DENSITY

Wro* =(LxBxd)x

SAOap e d u q ş ş e: I' ı h aı'.;h as',.. ats at a draught of 4i2

Basic principles (MARRev. l2l08/02)

Dec 2012


w : ' ;,''Whuı'Ioad (m'ass)'',will :driie,t

'iow.'Iowers'th'e'blo;bk s in'thı'dock,water?

DOCKWATERDENSITY7.020 tlm3

Basic principles (MARRev. 12108/02)

Th now ter , :: :,,, . ,

:,,:,,Vo ater mS

Anlswil',,:;t:,, :, ::: ,

:'8 m1

Therefoie:

BUOYANCY FORCE ACTING AT CENTROIDoF UNDERWATER VOLUME ( 8.16 t)

DOCKWATERDENSITY1.020 t/mr

WEIGHTFORCEACTING AT CENTRE OFGRAVTTY OF THE BLOCK (62.72 t)

Basic principles (MARRev. l2l08/02)

Dec 2012


Basic principles (MARPıev.12/08102)

lt

.6

Dec 2012


Answer, W"o*= (Lx Bx d) x p

.1 ,' Wnox =(80 x I6,x 4.2) x 1.025.,.,; , ,Wnox,= {l!o.oJ

r :, '

To calculate the displacement of a shipSince a ship is not box-shaped, a factor known as the blockcoeflicient (Cu) needs to be considered.

The block coefficient (C) of a ship is the ratio ofy4ğ-e-7waıq4 yolqmg to the volume of the circumscrİbing block.

Cr= vLxBxd

Therefore: Wrro =(LxBxdxCr)x

Basic principles (MARRev. 12108102)

An alternative solution might be as follows:

Answer, ], IuınexDeın;siıy.: ', ',.

Disp oİdisii. ) iotn, auorrıy'

.:. '11400:VoL of ctİspl. x 1.025 ' :

.'

1.025

?-.'LB:',.'.,:ll. .. ,..,:'. . .- . : ,.:.

A

t' 140x18x5;60

C-: ,0.788 ,

..:.....-...--::..REMEMBER THAT C|HAS NO.UNITS,IT IS A nerıoı

Basic principles (MARRev. 12108102)

10

Dec 2012


3. (a)

o)

(c)

(d)

Defıne üe term 'Block coefEıcient'

A ship has a length at the waterline of 56.2 m and breadth of 11.6 mand floats at a draught of 3.64 m in salt water. If the vessel'sdisplacement is 2096 tonnes calculate the block coeffıcient.

Calculate the TPCsw of this vessel given that Cw is 0.822.

How much more czırgo may be loaded so that the shipdraught of 3.96 m in salt water?

(17s.8 t)

[Jse the hvdrostatİc partİculars data sheet for the following examples4. A ship arrives in port and has a draught of 5.20 m in salt water. How much

caİgo must be discharged so that the ship may sail with a draught of 4.60 m.(1314 t)

5. A ship has an arrival draught of 4.90 m in dock water RD 1.006.(a) What is the aırival displacement?

(9674.3 t)

(b) 2174 t of cargo is loaded. What will be the final draught of this shipwhen it enters salt water RD 1.025.

(5.800 m)

6. A ship displaces 10,516 t in salt water. 960 tonnes of cargo is loaded.Calculate the final draught in salt water using:(a) the displacement and draught scales only;

(b) the appropriate TPC value.

7. A ship has a draught of 6.00 m in dock water RD 1.004. How much cargo maythe ship load to ensrıre that the morimum draught on sailing is 6.46 m in saltwater. (1292.7 t)

(0.862)

(s.4e3)sails with a

(s.63s m)

(s.633 m)

2Basic principles (MAR Rcv 22108103

Dec 2012


GLASGOW COLLEGE OT NAUTICAL STUDIES

BASIC PRIT*CIPLES

l' {9} Statt Arçhimfüıeı'fficiple"

tb} A b-Işck of şteçI tıaş dİrıçnşions qs şIıoıvtr üEd hgs a deıuiş of 7.96t/ııt]

Calculde the mışg oftfu bloc&-(+v.760 il

ThĞ blösh i* panİElly imınets+d tö r dğpü of 36 cnİ İn do+kugter RD l.00t. whaİ load iı tonnes uıill regıst€r gD thff Graffigrugr?

fdJ.JsJ fJ(iii) The crane driver now lowers the block so that it is fully

immersed, what load in tonnes will now register on the cranegauge?

(41.712 t)

A box-shaped vessel has the following particulars:length 86 m, breadth 18.2 m, depth 10 m and floats at a draught of 3.6 m insalt water (RD 1.025).(a) Calculate the displacement of the vessel.

(s77s.6 t)(b) What draught will the vessel float at if it is towed into dock water RD

r.0t2?(j.646 m)

(c) 300 tonnes of rock ballast is now loaded into the vessel whilst floatingin üe dock water. Calculate the new volume of displacement.

(6003.6 m3)

(c) How much more rock ballast must be loaded so that the vessel willfloat at it's moıimum permitted draught of 5.62 m in salt water (RDr.02s)?

(2940.7 t)

(d) How much rock ballast must be loaded to sink the vessel over a sandbank as part of a coastal protection scheme?

(i)

tİü

lm

Basic principles (MAR Rev 2?J08103

(7027 t)

Dec 2012


lİ €.

Thg hydpmeter is used todetermine the relative density offluids, including the dock water inwhıcn the vesseI is floating andliquid cargo densities. It maY benteGssiıy when in poıt to calculatethe dock water allowance by use ofthe formula :

DWA=FWA(1025-D)25 1

Where D =- dock water densiğ.This will then be used to ensure thatlhg ship js correctly loaded to hermarks.Ships iydrometers are-usually madeof polished steel or brass, thoughthey can also be made from glass. Abulb weighted with lead shot ormercury acts to keep the graduated stem upright. The operation of thehydrometer is based upon the Laws of Floatation, where the mass of thehydrometer is constant.Tb determine the density of dock water the following procedure should beqdopted:

Graduated scale(1000 -7}25tg/m3)

Float Chamber

Lead shot

,/ Ensure the hydrometer is free from any damage,/ li p13ctlcalüre severa!samples from various locations around the

vessel.,/ laught dock water samPle

) rece-ptacte, Jpp !!ıe miQ thatin the üater until it-isat th , and

inat tne samp]e point is c|ear fr-om any gverboa1d dischar-geş, Theremust be sutficienf depth of liquid to ensure the hydrometer floatswithout interference.

/ F_l__oa! the hydrometer in the water (once the sample is still). .Givethe instrument a slight twist to break surface tension and releaseany trapped air.

/ Take a ieading from the scale, at the water level, once theinstrument has settled, allowing for any meniscus.

./ Usıng the DWA formulabalculA{e the DWA.

Dec 2012


' HYDRoSTATıG PARTıGULARS

)RAUGH]

m

DISPL.

t

DISPL.

t

TPC

t

TPC

t

MCTC

t-m

MCTC

t-m

KMt

m

KB

m

LCB

foap

m

LCF

foap

mSW

RD 1.025

FW

RD 1.000

SW

RD 1.025

FW

RD r.000

swRD r.025

FW

RD 1.000

7.00 r4576 14220 23.t3 22.51 184.6 ı80.l 8.34 3.64 70.03 67.35

6.90 t4345 13996 23.06 22.50 t 83.0 ı78.5 8.35 3.58 70.08 67.46

6.80 l4l ı5 t3771 22.99 22.43 181.4 177.O 836 3,53 70.t2 67 -57

6.70 ı3886 I3548 22.92 22.36 t79.9 t 75.5 8.37 3.48 70.16 67.68

6.60 t3657 13324 22.85 22.29 178.3 174.O 8.38 3.43 70.20 67.79

6.50 13429 13102 22.78 22.23 r 76.8 172.5 8.39 3.38 70.24 67.90

6.40 13201 t2879 22.72 22.17 175.3 171.0 8.41 3.33 702E 58.00

6.30 t2975 r2558 22.66 22.t1 173.9 ı69.6 843 3.28 10.32 6E.10

6.20 12748 12437 22.60 22.05 172.5 r68.3 8.46 3.22 70.35 68.20

6.r0 t2523 12217 22.54 2t.99 t7l.I 167.0 8.49 3.17 70.38 68.30

6.00 12297 1t997 22.48 2t.93 ı69.8 t65.7 8.s2 3.tl 70.42 68.39

5.90 t2073 l177E 22.43 2t.87 r68.5 164.4 8.55 3.06 70.46 68.43

s.80 ı t848 il559 22.37 21.82 167.3 t63.2 8.59 3.01 70.50 68.57

5.70 1t625 tt342 22.32 21.71 ı66.ı t62.t 8.53 2.95 70.53 68.65

5.60 11402 t1t24 22.26 21J2 t65.0 ı6ı.0 8.67 290 10.51 68.?3

5.50 lıt80 ı0908 22.2t 21.66 163.9 ı60.0 8.7 r 2.85 70.60 68.80

5.40 r0958 t069ı 22.15 2t.61 162.9 t58.9 8.76 2.80 10.64 68.88

5.30 10737 10476 22.10 2t.56 ı6l.8 t57.9 8.8r I 2.74 I 70.68 68.95

5.20 t05t6 10260 22.05 21.51 ı60.8 156.9 8.86 2.69 70.72 69.02

5.10 10296 10045 22.00 21.46 t59.8 155.9 8.92 2.63 70.75 69.09

5.00 10076 9830 2t.95 21.41 158.8 ı 54.9 8.98 2.58 70.79 69.16

4.90 9857 9616 21.90 2t.36 157.9 ı54.0 9.06 2.53 70.82 69.23

4.80 9638 9403 21.85 2t.32 ı56.9 l53.ı l 9.t3 2.48 70.86 69.29

4.70 9420 l 9ı90 21.80 2t.27 156.0 I rs2.2 9.22 2.43 70.90 69.35

4.60 9202 8978 21.75 2t.22 155.1 | rsr.3 9.30 2.38 70.93 69.42

4.50 8985 8766 2t.70 21.17 1542 150.5 9.40 2.32 70.96 69.48

4.40 8768 8554 21.65 21.12 153.3 t49.6 9.49 2.27 71.00 69.55

4.30 8552 E3M 2t.60 21.07 152.4 r48.7 9.60 2.22 7t.04 69.62

4.20 8336 8 t33 2t.55 21.02 ı5t.5 147.8 9.71 2.t7 71.08 69.68

4.ı0 8l2l 7923 21.50 20.91 150.6 t46.9 9.83 2.t2 71.t2 69.74

4.00 7906 7713 21.45 20.93 149.7 ı46.0 9.96 2.07 71.15 69.81

3.90 7692 7505 2t.40 20.88 ı48.7 t45.r l0.ı l 2.01 7ı.ı8 69.88

3.80 7478 1296 2t.35 20.83 147.8 144.2 10.25 ı.96 7r.22 69.94

3.70 7265 7088 2 t.30 20.78 ı46.8 r43.3 ı0.4l l.9l 7t.25 70.00

360 7052 6E80 21.24 20.72 r45.9 142.3 r0.57 1.86 71.29 70.07

3.50 6840 6673 2t.19 20.67 144.9 t 41.3 ro.76 1.8 t 71.33 70.14

TI{ESE FTYDROSTATIC PARTICULARS HAVE BEEN DEVELOPED WITH TTIE VESSEL FLOATING ON EVEN

KEEL.

Dec 2012


Ex.Calculate the Dock Water Allowance (DwA) of a vesseı, FWA = 280mm,

floating in dock water with a density of 1.011.

,/ DWA = FWA(1025 - D)25 ___

80(1025 - 101 1)t25

owl = ıss.emm

A hydrometer has a bulb with a volume of 24cm3 and a uniform stem with a crossseciıonaı area of o.22cm2. When floating in SW (1.025i 4.6cm of the stem isimmersed.Calculate the length of stem that will be immersed when the hydrometer is

floating in FW' (7.442cm)

A hydrometer has a weighted bulb radius 2.5cm and a uniform stem with a crossseciıonal area of 0.36cm2. When floating in FW 7cm of the stem is immersed.calculate the length of stem immersed when it is floating in sw.

(5.056cm)

Dec 2012


Answer

The muss of the additional slice of displuced water is thesame us the added weight 'w'.

Since: Mass : Volume x Density

then: AQded dişnl1cement -:

Vol' of slice x density.

If the wPA is assumed to not significantly change between

Added displacement: (WPA(m2) x 1 cm) x density;

Added displacement: WPA(mr) " (m) r density;100

Thus, the formula for TPC is given by:=--

SAO

Calculute the TPC for ship wİth a wuterplane area of 1500m2 when it is flouting in:

(a) salt wuter; (b) fresh wuterl @) dock water of RD 1.005.

TPC:WPAxp100

2TPC (MAR Rev. 03/01/01)

Dec 2012


TONNES PER CENTIMETRE IMMERSION (TPC)

The TPC for any given draught is the weight which must beloaded or discharged to change the ship's mean draught by onecentimetre.

Consider thea waterplane

ship shownarea (WPA)

floating in salt water (RD 1.025) withat the waterline as shown.

A weight 'w'increases by 1

tonnes is loaded on deck so that the mean draughtcm.

+

SAO

What is the mass of the udditional 'slice' of displaced waterequul to?

TPC (MAR Rev. 03/01/01)

Dec 2012


DRAUGHT

m

DISPL.

t

DISPL.

t

MCTCt-m

MCTC

t-m

KMt

m

KB

m

LCB

foap

m

LCF

foup

mSW

RD 1.02s

FW

RD 1.000

SW

RD 1.025

SW

RD 1.025

FW

RD 1.000

700 r4576 14220 23.13 1J.57 184.6 t80 I 8.34 3.64 70.03 67.35690 14345 t3996 B.A6 r83 0 t78 5 8.35 358 70.08 67.46680 l4l l5 t3771 22.99 ?2,43 181.4 177.0 8.36 353 70.12 67.57670 t3886 r 3548 22.92 179.9 175.s 8.37 348 70.16 67.68660 t3657 13324 22.8s ln ao 178 3 174.0 838 3.43 70.20 67.796s0 13429 13102 22.78 176 8 t72.5 839 338 70.24 67.906.40 13201 12879 22.72 ll5 3 t7t 0 841 J.JJ 70 28 68 00630 t2975 t2658 22.66 173.9 169.6 843 328 70.32 68. l06.20 t2748 t2437 22.6A 22.45 172.5 168 3 8.46 322 70.35 68.20610 12s23 12211 7? \4 21.99 171 1 167 0 849 3.17 70.38 68.306.00 12297 11997 22.ü 2t.93 169.8 16s.7 8.52 311 70.42 68.395.90 12073 11778 168 5 164 4 85s 3.06 70.46 68.435.80 I 1848 I 1559 22.37 21..82 167.3 163.2 859 301 70.50 68.57s.70 t162s 11342 21.77 166 I 162 1 863 2.9s 70 53 68.65560 11402 11124 n,26 21.12 165 0 161 0 8.67 2.90 70.57 68.73550 r1180 I 0908 22.2t 21.66 163.9 160 0 871 2.85 70.60 68.80540 109s8 10691 22.15 2l'6İ 162.9 ls8 9 876 280 70.64 68.885.30 10737 10476 22.10 161 8 157 9 8 81 2.74 70.68 68.9s5.20 10516 10260 22.05 21.51 160 8 t56 9 8.86 269 70.72 69.02

5.10 10296 I 0045 22.00 21.16 159 8 155.9 8.92 2.63 70.75 69.09

500 10076 9830 21,95 ?1.41 158 8 1549 898 258 70.79 69.16

4.90 9857 9616 2t.90 t57.9 t54.0 906 2.53 70.82 69.23

480 9638 9403 2I.E5 1s6.9 153 1 913 2.48 70 86 69.29

470 9420 9190 21.80 r56 0 152.2 9.22 243 70.90 69.35

460 9202 8978 2İ'75 155 I t51.3 930 238 70.93 69.42

4.50 8985 8766 21.70 2t-17 154 2 t50 5 940 2.32 70.96 69.48

440 8768 8554 21.65 2L..12. 153.3 149.6 949 2.27 71.00 69.55

430 8552 83M 2tffi 152 4 t48.7 960 2.22 71.04 69.62

420 8336 8133 21.55 2t.02 r 51.5 t47.8 971 2.17 71.08 69.68

410 8ı2l 7923 2l:50 rs0 6 t46.9 983 2.t2 71.12 69.74

400 7906 7713 21.45 149.7 146.0 996 2.0'7 71.15 69.81

390 7692 750s 2t.40 ?ü.88 487 t45 1 10.1 I 2.01 71.18 69.88

380 7478 7296 2t.35 47.8 t44 2 t0.25 1.96 71.22 69.94

370 7265 7088 2İ.3o 20.78 46.8 143 3 l0 41 r91 71.25 70.00

3.60 7052 6880 21.24 459 1423 10 5',7 1.86 71.29 70 07

3.50 6840 6673 2t.le 24.67 144.9 141 3 10.76 l8t 7t.33 70 14

THESE FIYDROSTATIC PARTICIJLARş'P€VPFppıpR@I'pry9/UI7Uııyı VESSEL FLOATING oN EVEN 4KEEL.

Dec 2012


AnswerTPC-ru.x P

100

u) TPC - 7fl x 1.025 : 75.375100

b) TPC: !fl x 1.000 = 75.000100

c) TPC : 1500 x 1.005 : 75.075100

Consideration of the TPC formula indicates that:*- -iPc inıcreases with WPA und for a normul ship-shape

the WPA wiII increase witlı druught.,< TPC increases with density. Two vulues of TPC are often

quo(ed in ship's hydrostatic data, TPCswund TPCFW

TPC values for the draught range of a ship allow us to calculate

how much cargolballast etc. to load or discharge to achieve a

iequired draJght.

SinkagelRise (cms) :ğTPC

where 'w' represents the total weight that is loaded or

discharged to change the draught of the ship'JTPC (MAR Rev. 03/01/01)

Dec 2012


AnswerMethod IInitial draught 5.10 m Arr: 102g6 tRequired draught 6.40 m Arr= 13201 tCurso to load - 2905 t

Method 2Initial draught 5.70 m TPCsw: 22.00Required draught 6.40 m TPCsw:22.72

Meun TPCsw:22.00 * 22.72 : 22.362

Sinkage (cms) : 6.40 m - 5.10 m : 7,30 m = 130 cms

Sinkuge (cms) _ wTPCSW

Curgo to loud (w) : Sinkuge x Mean TPCsw

: 730 x 22.36 : 2906.80 t

The answers mayvalue of TPC it is

differ slightlyassumed that

because in using the meanthe TPC value will change

linearly between the range of draughts concerned.

SAOHad the ship been floating in fresh water (Frl/), would theamount of cürgo to load be the Süme to uchieve the requireddraught of 6.40 m in fresh wüteF.

6TPC (MAR Rev. 03/01/01)

Dec 2012


SAOA shİp has an initiul mean druughtand is required to complete loudingm. Using the hydrostutic particularscargo that must be loaded.

of 5.10 m in salt wuterwith a druught of 6.40culculute the amount of

The answer may be calculated in two ways:

Method 1

1. Read off the displacements (SW) for both the initial and

required final draughts.2. Subtract the smaller from the larger.

3. Result equals the amount to load.

Method 21. Read off the TPC ,, values for both the initial and required

final draughts.2. Calculate the mean TPCsw value.3. Calculate the required change in draught; in this case

sinkage.4. Use the formula:

Sinkage/Rise (cms) _ wTPC

to find '\M', the amount to load.

TPC (MAR Rev. 03/01/01)

Dec 2012


AnswerMethod 1Initial draught 5.10 mRequired draught 6.40 mCARGO TO LOAD

Method 2Initial druught 5.10 m TPCFW:27.46Required druught 6.40 m TPCFW:22.77

Mean TpC_,,,= 21.46 + 22.17 _ 2I.BIs,

Sinkage (cms) - 6.40 m - 5.70 m :1.30 m:130 cms

Sinkage (cms) -TPCFIT

Curgo to loud (w) : Sİnkage x Mean TPCFW

:130 x 21.815:2835.95 t

2k The displacement for the correct density must be used in allcalculations.

"k The TPC for the density in which the ship is loading inshould be used in calculations.

* It is usual to calculate the amount to load on the basis of the

required salt wuter draught.

TPC (MARRev.03/01/01)

Dec 2012


Coefficient of fineness of the water-plane area (C*)Is defined as the ratio of the ship's water-plane areato the areağ a rectangle having the Same length and breadth of the ship atthe waterline in question.

IWaterlinebreadth

I

Waterline length

Cw:WPALxB

Since the ship's WPA is less in area than the rectangle formedaround it , the vulue of Cıa must ulways be less than 7.00.

Basic principles (MARRev. 03/01/01)

2

Dec 2012


FORM COEFFICIENTS

Form coefficients are ratios which numerically compare the

llipts underwater form to that of a regular shape (such as arectangle or box-shape).

They ate primarily used aİ the design stage, prior toconstruction, to determine factors such as resistance to forwardmotion İhat the ship will experience during operation, this thenbeing used to determine the ship's power requirements/engine(s)size.

Design coefficients o{primary concern are:

- CoeİIİcient of Jineness of the wuter-plune areü (Cw)

Block coefficient (C)

Midship s co efJicient (C n)

- Longitudinal prismutic cofficient (Cp)

_1 Jl

Basic principles (MARRev.03/01/01)

Dec 2012


Midships coefficient (Cnn)

fhe midships ioefficient- (cr) of a ship at any draught is theratio of the underwater transverse area of the midships secti_on

tg the product of the breadth and draught (the surroundingrectangle.

raught

Bread

CM : Underwater transverse area of mid@ (A_)Breadth x Draught

Similarly, the vulue of Cıımust always be less than 7.00.


4

Dec 2012


Block coefficient (Cr)The block coeffıcient (Cr) of a ship is the ratio of theunderwater volume of a ship to the volume of thecircumscribing block.

\cs: Volume of displacement

LxBxd

Therefore: Displacement.r,, - (Lx B x d x Cr) x P

Since the ship's volume of displacement is less than the volumeof displacement of the surrounding block, the vulue of C o must

alwuys be less than 7.00.

Block coefficient is an important factor when the assigned

ne"UoarO of a ship is being calculated (see 'Calculation andAs si;ignment of Freeboard' notes.)


Dec 2012


Longitudinal prismatic coefficient (Cr)The longitudinal prismatic coefficient (Cr) of a ship at anydryught is the ratio of the underwater volume of the ship to the

volume of the prism formed by the product of the transverseare?of the midships s_ection and the waterline length.

C, : Volume of dispıacem-rVolume of prism

cp: Volume of displacement of shiWaterline length x Area of midship section (Am)

This coefficient gives an indicationchanges at the ends. SimilarlY, the

Iess thun 7.00.

of how much the ship's formvulue of Cp must ulwaYs be

Basic principles (MARRev.03/01/01)

5

Dec 2012


Dock water allowance (DWA)T-he bock Water Allowance (DWA) of a ship is the number ofmillimetres by which the mean draught changes when a shippasses from salt water to dock water, or vice-versa, lyhen the shipis loaded to the summer displacement.

The DWA is a fraction of the FWA and is found by the formula:

(.'\ DWA (mm): FWA x25

NoteThe densities are multiplied by 1000 to simplifiı the formula.

The same formula can be easily modified to calculate the changein draught if the ship passes from dock water of one density todock water of another.

DWA(mm):FWAtW25

Load Lines (MAR Rev.rzl08l02)

4

Dec 2012


Fresh water allowance (FWA)The Fresh Water Allowunce (FWA) of a ship is the number ofmillimetres by which the mean draught changes when a shippasses from salt water to fresh water, or vice-versa, when the shipis loaded to İhe summer displacement.

The FWA is found by the formula:

FWA (mm): A Summer4TPCsw

If the load line marks are considered, the top of the Summermark and the top of the Fresh mark act as the limits of a scale ofdensity that would appear on a hydrometer (an instrument forrneasuring liquid density).

The ship behaves exactly üS ü very large hydrometer!

ı.000 (Fw)

Tr r r -05

FWA - r r -10 C

- r r -15

rrr-20

r.ozs (sw)

Load Lines (MAR Rev.r2108102)

aJ

Dec 2012


LOAD LINE MARKS. STARBOARD SIDE(Dimensions in mm)

lll- 300 i

FWA(mm) :Asummer x= SUMMERDRAUGHT4TPCsw 4g

'Always look forward to summer! '

230 i-230-*lSTATUTORYFREEBOARI)

RD 1.000 (FW)

RD 1.02s (SW)

x

LI50I

WWNA*

* Ship. 100 m or less

Load Lines (MAR Rev. 03/01/01)

Dec 2012


The load lines for the starbourd side of a ship are shown.

NoteThe spacings between the load lines are measured from the top

ğse-ot one line to the top edge ofthe other!

The assigned (Summer) freeboard is measured from the top edgegf the plimsoll line (which corresponds to the top edge of the

Şgq-.r line) to the top edge of İhe deck line.

NOT TO SCALE

FWA (mm) = PISPL. summer4TPCsw

X = SUMMER DRAUGHT48

'Afways looJç forward fç sıımnıerİ'I

I

ASSIGNEDSUMMER

FREEBOARD

RD {.000 (FW}

, I RD 1.02s (sw)

i<-3oo---'+i

x+{50t

WWNA *

450--+l * shıps 100 m or less

Load Lines (MAR Rev.r2l08l02)

2

Dec 2012


LOAD LINESThe M. ^S. (Load Line) Regulations 1998 (Amended 2000)require that all UK registered ships be assigned a freeboard and acorresponding set of load lines to be marked pernanently on theship's side.

The assigning authority, usually a classification society such as

'Lloyds Register of Shipping' will issue a Load Line CertiJicute.

Important factors that are taken into account include:

* Ship stability and reserve buoyancy;* Structural strength;* Hatchways;* Machinery space ope ings;* All openings in the freeboard deck;*

lree_iqg norts;* Protection of crew;{< Ship type (A or B).

(and many others also.)

Two ship types are considered:

Tupe AIs a ship designed to cawy only liquid curgoes in bulk (tankers).

Tvpe BAny other type of ship (bulk curuier, contuiner ship, general

c_qlgq qtc.)

Load Lines (MAR Rev.r2l08l02)

Dec 2012


LOAD LINE CALCULATIONS

SAO

A ship flouts in SW at the summer displacement of 1680 tonnes.If the TPCSW js 5. 18, how much witl the druughı chunge by ifthe ship is towed ıo a berth where the density of the water is1.000 t/m3 ?

Answer

rn moving from SW to FW the ship will experience sinkage byan ümount equal ıo the FWA.

FWA (mm) = DISPL. Summer4TPCSW

FWA: 1680 : 87.7 mm4 x 5.18

The draught will increase by 81.1 mm!

Load Line calculations(MAR Rev. 13/11103)

Dec 2012


LOAD LINE CALCULATIONS

Introduction

Most ships will be assigned a

_cöriesponding set of load lines.marked on each side of the shipexempt from these requirements).

minimum freeboard and a

These will be permanently(Certain classes of ship are

Load lines assigned to a ship correspond to ocean areas or'zones'. Oceans around the world are divided into these zones interms of both geographical location and time of year (season). Byensuring that the appropriate seasonal load line mark is not

submerged at sea in salt water (RD 1.025) the ship will alwayshave the necessary reserve buoyancy to ensure seaworthiness.

To ensure that the appropriate load line is never submerged at sea,

it is essential that the learner has a thorough knowledge of the

load line markings, their spacing and dimensions. The ability to

perform calculations to determine the maximum amount to load is

also important, especially to the ship owner, as the absolute

maximum cargo in terms of weight should be carried whenever

possible. It is also essential that the ship is never'overloaded', os

contravention of the conditions of load line assignment will arise,

resulting in the ship being unseaworthy with respect to legislativerequirements.

Load Line çalculations(MAR Rev. l3llll03)

Dec 2012


AnswerDWA (mm) : FWA x (RDpwr- RDor)

Therefore: DWA (mm) = 260 x (1016 - 1004125

DWA = 724.8 mm

The draught will decreuse by 125 mm since the shİp İs movinginto more dense water!

Answers need only be to the nearest mm!

Load Line calculations(MAR Rev. 13/11/03)

4

Dec 2012


SAO

A ship hus a FWA of 200 mm. Calculute the change in druughtthat will occur if the ship proceeds from SW to a berth where theRD of the dock water is 1.018.

AnswerDWA (mm) : FWA x (1025 - RD dock water)

25

Therefore: DWA (mm) : 200 x (1025 - 10181

25

DWA: 56 mm

The draughı will increase by 56 mm!

Load Line calculations(MAR Rev. I3ltll03)

Dec 2012


Answer

Qn reuching seqwater of greater density, the ship will be light ofthe summer marks as shown below.

MORE CARGO COALD HAVE BEEN LOADED!

- -REQ'D (SW)

1.02s (sw)

To avoid this situation but to also ensure that too much cargo isnever loaded, the amount to safely load can be readily calculated.

The uim of the problem is to ensure that on proceeding to sea theship rises to the desired seasonal loud line murk. This is achievedby considering the Fresh Water Allowance or Dock WaterAllowance as appropriute in the culculation.


6

Dec 2012


Typical load line calculations

When loading a ship it is desirable to load as much cargo as

possible. If lhe ship is being loaded in water thaİ is less dense than

salt water, such as dock water, then allowance must be made forthe ship rising oıt of the water on reaching the Sea, salt water

density being I.025 tlm3 .

Consider the following situation:

A ship is louding in the Summer zone in dock water RD 1.012. Itcan legully load so that the salt wuter wuterline is level with the

lop edge of the Summer Load Line.

Consider the situution where the officer in churge loads cürgo

until the dock water wuterline is level with the Summer loudline!

r.0r2 (Dw)

will be the situation when the ship reaches the sea?

Load Line calculations(MAR Rev. 13lIIl03)

Dec 2012


3. culculute the maximum amount thut cun stilt beloaded in dock wuter, ignoring rny allowances for fuelor other items.

Permitted sinkage (cms) : wTPC

Therefore: w : Permiffed sinkage (cms) x TpC

lYote thut TPC must be coruected for the density of the dockwater!

4. Muke allowance now for items other than cargo thatmust be loaded.

Total that can be loaded 345.1 tonnesFuel 26.0 tonnesMaximum cargo to load 379.7 tonnes

NOTEHad the gİven TPC not been converted for the density of thedock water, the total thut could be loaded would have workedout us:

w - 16.1 x 27.82 : 351.3 tonnes;

resulting in the shİp beİng OVERL0ADED BY 6.2 T0IY]YES!

Load Line calculations(MAR Rev. 13/11103)

Dec 2012


Example IA ship hus a summer loud draught of 5.80 m, FWA 140 mm andTPC of 21.82. The ship is loading rt u berth in dock water RD1.007 und the present druught İs 5. 74 m. Calculate the maximumümount of cargo that can still be loaded for the ship to be ut theSummer loud line murk on reaching the sea ullowing for 26tonnes of fuel still to be louded prior to suiling.

The following procedure and layout should be followed exactly.

7. Calculute DWA (to the neurest mm).

DWA (mm) : 140 x (1025 - 1007) : 100.8 mm = 101 mm25

2. Culculute the 'permitted sinkuge' in dock wüleF.

Always start with the required load line draught undwork as follows:

Required Summer draught (I.025) 5.800 m

DWA +0.101 mRequired draught (1.007) 5.901 mInitial draught (1.007) 5.740 mPermitted sinkage (1.007) 0.161 m

Load Line çalculations(MAR Rev. 13ltll03)

7

Dec 2012


Answer1. Calculate DWA

DWA (mm) = 100 x (1025 - 1002t :92 mm25

2. calculute the 'permitted sinkage' in dock wuteF.

Required Winter draught (1.025) 4.320 mDWA 10.092 mRequired draughtIniıiul druught

(1.002) 4.412 m(1.002) 4.300 m

Permitted sinkage (1.002) 0.112 m

3. Calculate the maximum umount that can still beIoaded in dock wüteF.

Permitted sinkage (cms) - w

Therefore:

TPL

w : Permitted sinkage (cms) x TPC

Total thut cün be loaded 236.5 tonnes

Load Line calculations(MAR Rev. 13llrl03)

10

Dec 2012


SAOA ship is flouting in dock water RD 1.002 at a draught of 4.30 m.

How much more cargo must be louded to ensure that the shipwill be at the Winter load line murk gİven that the Wİnterdruught corresponding to the winter displucement is 4.32 m andthe TPC is 27.60 and the FWA is 100 mm.

Note that the TPC value given will ulways be the one thatcorresponds to salt wuter for the waterline which is beİng loadedto.

Load Line calculations(MAR Rev. l3llrl03)

9

Dec 2012


X: 0.143 m0.12 m

Port WL

(Not to scale)

3. Starting with a known druught (Summer) calculute thedraught on eüch side by applying the distunces in thesketch.

STBDSummer draughtLine thickness

Draught each side

6.860-0.143

+0.0406.7 57 m

4. Culculate initial mean draught.

Initial mean draught (RD 1.006) _ 6.715 + 6.757 - 6.736 m

Load Line calçulations(MAR Rev. 13/1 Il03)

t2

Dec 2012


Sometimes a question may be a little more difficult whereby aknowledge of the load line dimensions is essential. It is essentialthat a sketch he drawn to fully understund what is being asked!

Example 2A ship is floating in dock water RD 7.006. The wuterline to portis 12 cm below the lower edge of the '^S' mark and on thestarbourd side is 4 cm above the upper edge of the 'W' mark. Ifthe summer displacement is 21620 tonnes (correspondİng to adraught in salt water of 6.86 m, TPC 18.6), how much cargoremains to be loaded to ensure that the ship will be ut the Wintermark in salt wüteF.

7. Identify the load lines that üre mentioned in thequestion ('S' and 'W' in this case); sketch them andenter all known dimensions, calculatİng them aS

necessary.

Thickness of the lines: 25 mm (2.5 cms; 0.025 m)

Distance between Winter and Summer load lines (X):

X - Summer draught - 6.86 : 0.143 m48 48

2. Druw ü sketch (Port or Sturboard side: İt does notmatter!)

Load Line calculations(MAR Rev. 13/1 Il03)

11

Dec 2012


5. Culculate DWA (in this case FWA must Jirst becalculuted.

FWA (mm) - DISPL. Summer _ 6T620 _ 290.6 mm4TPCsw 4 x 18.6

DWA (mm):290.6 x (1025 - 1006):220.8 mm x221 mm25

6. Calculate the 'permitted sinkuge' in dock water

Required Winter draught (1 .025) 6.717 mDWA +0.221 mRequired draught (1.006) 6.938 mInitial draueht (1.006) 6.736 mPermiffed sinkage (1.006) 0.202 m

7. Culculute the maximum ümount that cun still beloaded in dock water.

Permiffed sinkage (cms) : wTPC

Therefore: w: Permiffed sinkage (cms) x TPC

NOTESteps 5-7 apply to all load line questions and should be strictlyfollowed!


13

Dec 2012


7. The hydrostatic particulars for a ship give the following data for the Summermark:

Displacement 35800 tDraught 12.0 mFreeboaıd 2400 mmTPC 42

The ship is loading up river in dock water RD 1.012 and at a stage in loadingthe freeboards to port and starboard are 2600 mm and 2580 mm respectively.Calculate how-much cargo to load so as to be at the tropical marks in saltwater if 150 tonnes of fresh water have to be taken on board before sailing and50 tonnes of fuel are to be consumed on passage down river to the sea.

(2184 t)

Sketch a hydrometer suitable for use on board a ship and describe how andunder what circumstances it should be used.

A ship is loading in a dock up-river in water RD 1.004 and is required toproceed to an explosives anchorage at the mouth of üe river in order tocomplete loading a further 450 t. The summer freeboard in SW is 4990 mmand the winter penalty is 190 mm. FWA is 180 mm and TPC for the loadedsunımer draught is 29. Calculate the freeboard which the ship must be at onleaving the dock berttı, if it is to be at the winter load line on reaching the opensea if it is anticipated that 60 tonnes of fuel will be consumed on passage fromthe berth to the sea'

er66 mm)

A small ship has a swnmer displacement of 6220 t, summer freeboard in SWof 1072 mm which corıesponds to a srrııımer draught of 5.808 m. Whenfloating at the sunımer load line the waterline length and breadth are 95.0 mand 13.0 m respectively, Cw being 0.750.

(a) If the ship has to load to the tropical maıks in salt water find thedraught to which it should be loaded in a port where the dock waterRD is 1.008.

(6.0a m)(b) Sketch the load line marks for this ship as they would appear on the

Port side indicating all variable and fixed dimensions showing allcalculations as appropriate.

8.

9.

10.

MAR.LOADLINE CALCULATIONS.tutorial

MAR Rev l2ll2l00 2

Dec 2012


SHıP

STABıLıTY I01

(

Dec 2012


where v is the volume of th9 transferred wedge of buoyancy and

V is the ship's volume of displacement.

Initial TransverseMetacentre (MAR Rev.

o7/0r/01)

If B is plotted for several small angles of heel it may be assumedthat it follows the arc of a circle centred at M.

BM is termed the metacentric radius anğ the height of the initialtransverse metacentre (KM) may be gılgglqled thuş;

KM=KB+BM


07lovor)

('

Dec 2012


Calculatins KM for box-shaped vessels

For a box-shaped vessel on an even keel:

and;

where I is the moment of inertia (second moment of area) of the

WPA and V is the volume of displacement of the box-shaped

vessel.

For a box-shaped vessel: I: !ğ312

where L and B are the length and breadth of the WPArespectively.

Therefore:

Thus: Wro*=draught * IJB32 lzLB,d

Note that this formula can obviously be simplified further!


0710t/01,

SAOProve that the KM of a box-shaped vessel changes wİthdraught as shown below for the range of draaghts 1.00 m to15.00 m given that: length: 100 m

breadth: 20 m.

zuase the formulae to jİnd values of KB and BM and lhen sumthese toJind the correspondİng KM values.


07/ou0t)

Dec 2012


AnswerThe values for KM are shown below having been calculatedasing: KM:KB +BM where:

KB: draught : and BM = B-

)raught (m KB (m) BM (m) KM (m)

1 0.5 33.33 33.83

KMREDUCING

I

I

2 1 {6.67 17.67

3 1.5 11.11 12.61

4 2 8.33 f 0.33

5 2.5 6.67 9.17

6 3 5.56 8_56

7 3.5 4.76 8.268 4 4.17 8.179 4.5 3.70 8.20

KMINCREASING

{0 5 3.33 8.3311 5.5 3.03 8.53

12 6 2.78 8.78

13 6.5 2.56 9.06

14 7 2.38 9.38

15 7.5 2.22 9.72

SAQPlot the valııcs of KM with lhe X'axİs labelled 'draught' andthe Y-axİs labelled 'KM' and 'KG'. PIot a value of KG = 9.00 m

on the graph.


o7l0l/ot)

The graph is plotted at the end of these notes.

SAOase the graph you have draıın to deİermİne thefollowing:(a) The minimuın value of KM and the draughı at whİch

it occurs;(b) The range of draughts at whİch the vessel wilI be

unstable;(c) The righting moment when the box is heeled to an

angle of 5" when the uprİght draughl İS 3.00 m in saltwater (RD 1.025).


07/0t/0t)

Dec 2012


Answer(a)

(b)

(c)

The minimum value of KM = 8.00 m and occars at

draught= 8.00 m.

The range of draughts at which the vessel will be

unstable İs between 5.20 m and 12.50 ınAt a draught of 3.00 m KM was calculated to be 12.61

m

KM 12.61mKG 9.00 mGM 3.61m

Displacement:L x B x d x densitY

Displacement: 100 x 20 x 3,00 x 1.025

Displacement = 6150 tonnes

GZ=GM x Sin0

GZ:3.6lx Sİn 5"

GZ:0.31463....

Righting momenı= GZ x DisplacemenlRighting Moment=0.31463.. x 6150

Righıing moment= 7M-Therefore:


07/0v01)

F'ACTORS AFX'ECTING KM

from the

f buoYancY is transferred

-9gPşıng B to move fuıther

BI


oT lov0r)

Dec 2012


Drausht (I)isnlacement)

M

B

2.

Consider the formula: BB.:vxbb.I

-t

V

a!.rhe--ıe4--qa1ıgl1t !he-y9_lg19 o{ !b" transferred wedge o{buoyancy (v) represents a smaller part of the total volume of

-disFlacern€nTcif 6ö ahip (v) than-at-the iight draught.


07/0t/01)

I I

Dec 2012


TRANSVERSE STATICAL STABILITY:--

momentarilY at res1l.

It is the relative positions of the centre of gravity (G) and the

centre of buoyancy (B) as the ship is heeled to a particular

angle that determines how stable a ship is.

Transverse statical stabilitY(MAR Rev. 03/03101)

SAO

WitI ıhis ship heelfurther or return to the upright if the

erternal force İs remov ed?

Transverse statical stability(MAR Rev. 03/03/01)

Dec 2012


Answer

The ship will return to the apright.

The horizontal separation of the lines of action of Wf and

g i is suöh that a r İ g hl İ ng !e-v 9 r- (G-1) q9w e1ists,


Righting lever (GZ) increases to some maximum value and

then decreises ui the ship pıogressivöIy heels further.


Dec 2012


Rishtins moment

The righıing momenl ğ gıv given a_ngle 9f heel is found----

RIGHTING MOMENT = GZ X DISPLACEMENT

The rightİng moment at any angle of heel represents the

ğ-1gglg9-'rpin 'still water' conditions.

Transverse statical stabilitY(MAR Rev. 03103101)

WIs defined as the ooint of intersection of the lines of action**-l-_--_*".q@-üqlı}9'-^ş!_,p-

-ış " iılte_iil9ıılIt

fo10' heel approximately).

It's height is quoted in relation to the keel - KM.


Dec 2012


ryIş lbç_ v_qütı-q3________________!-a1ş!1_1-99-b-e!'w99ı_L!-9-"hip's 9e1ty of gravity

(G) and the metacentre (Io. ':i 'l l / (' ) ıı '')

)(-,)

SAO

If the centre of gravily (G) of the ship shown wos

iıghu, would the shİp be more ol lesş stable?

Transverse statical stabiliğ(MAR Rev. 03103101)

Answer

The ship would be less stable. Consİder the same shipwİth G now at G,.

G lı-ı

I

Rİghıİng tever (GZ) has reduced to GrZ, as a result ofthe upward movement of the centre of gravity (G).

Therefore, available righıing moment wil.l also be

reduced.

Similarly, if G was lowered the righting lever (GZ) (and

available rightİng moment) would be increased.

Transverse statical stability(MAR Rev. 03103/01)

Dec 2012


In triangle GZM: Sin 0: OPP : GZHTP GM

Therefore:

Having found GZ:

GZ=GM x Sİn0 a

RM: GZ X DISPLACEMENT

Note (,r'" /g)

The formula for GZ can only be used atsmiıll angles of heel.

Transverse statical stabilitY(MAR Rev. 03/03/01)

ŞlaD&_çgiliurıum

A ship is iı stable equİlİbrİum if, when heeled by an externalforce to a small angle, it returns to the upright when the forceis removed.

In this condition the ship has apositİve iniliql GM.

Transverse statical stability(MAR Rev. 03103/01)

t0

Dec 2012


Transverse staticat stabilitY(MAR Rev. 03/03/01)

Answer

Possibly. As the shİp heels furİher over the centre ofbuoyancy (B) wiII move outward as lhe underwaler volumeclıanges shape. Provİded that tlıe centre of baoyancy canmove sulftciently outboard to attain a new posİtionvertically below G the capsizing lever will disoppear andthe shİp wİll come to rest at an angle of loII. If the centreof gravily wus very hİgh then the ship woald capsize.

SAO

[hen the vessel comes to reş! at on qnglg of loll will-İı be.

!!ı1e ot a small or a large angle?


l2

Dec 2012


Answer

I['|he me-tagentre (M) is considered, it is now no longer aİM. It has moued to a posİtİon on the same yerhİcal as G-aiİ

may be consİdered to be at the same posiıİon as G. II is

ryow lermed a 'pro-melacentre2 or a |ınovİng metacenılre'

@!. fhere re the ship must now be lying at a 'large'angk.

B

SAO

If ıhe ship İs heeled beyond the angle of loll what willhappen?


l3

Answer

The centre of buoyancy (B) will move outboard of thecentre of grauiıy (G) and a posİtive righting lever (GZ) willtake e ect to right the shİp back to the angle of loll.

Transverse statical stability(MAR Rev. 03/03/01,

Dec 2012


Transverse statical stabiliğ(MAR Rev. 03/03/01)

Answer

Eventually the centre of huoyancy (B) will move oatboud ofthe centre of gravity (G) and rİghtİng levers wİll becomepositİve to right the shİp back ıo some İndetermİnate smallangle.


t6

Dec 2012


THE GZ CURVE _

- AD is a line drawn as a tangent to the GZ cqve which emanates flom thg

brigin at A.

- AD cuts a [ine drawn yertigatly upwards from the

radian).

* BC is a line very close to the origin at angle 0'.

heel scale at 57.3' (one

E

(57.3.)

(ı ",.ı.n )

- Triangles ABC and,ADE.:j1,li.... DE:BC /

AE AC -'

GZ=GM x Sin 0

and Sin 0 is approximately equal to 0. (where 0. is measured

in radians and 57.3': I radian.)

Consider, for example SinZ" :0'0348994;

2o expressed in radians = 0.034904.

. DE = GMg^ .'. DE: GMlo"

Transverse statical stabilitY(MAR Rev. 03/03/Ot)

t7

Dec 2012


CT]RVES OF STATICAL STABILITY FOR VARYINGCONDITIONS OF STABILITY

A ship is in a stable condition of stability if, when heeled by an

ixteiniı 1orci İn Sıı#!u*t-9r to a small angle of-İnclination, iı

1. ,/

condition

returns taıhd when theforce is removed'

iorrıj". the ship shown being progressively inclined from the

upright.

Fieure I

KM - KG: GM; which isposiıive'

GZ is zero.

GZ curves for different

conditions of stabilitY(MARRev. t8/09/01)

Fisure 2

lhe s!ip- ı! now heeled by an external force to a small angle ofinclination.

GZ is positive, which will act to right the ship when the force is

removed.

A typical curve of statical stability for a stable ship will be as

shown. Figures I and 2 are related to the curvo as indicated'

GZ curves for differentconditions of stability(MAR Rev. 18/09/01)

Dec 2012


Cıırve of staticat sıabıııtv for a STABLE sHlP

o.7

0.6

0-5

o.4

0.3

GZo-2

(m)0.1

0

-0.'ı

-o.2

-0.3

GZ curves for differentconditions of stability(MARRev. t8/09/01)

2. ı' Curve of statical stability for a shin in a neutralcondition of stabilitv

A ship İs in a neutral condition of Stahilia ıf' when heeled hyan external force in still ıiater to a small angle of inclinatİon,it comes to rest aı Some indeterminate angle of heel witlıİn

ryg!!4! 8! € ! of i n c l İ n ati o n.

Consider the ship shown being progressively inclined from theupright.

Figure 1

[Y - 5c: o, GM:0GZ is zero.

wf

G M

B

K

GZ curves for differentconditions of stability(MARRev. l8/09/01)

Dec 2012


Fisure 2

Bf

wf

I G M

I

K


conditions of stabilitY(MARRev. r&l09l0l)

Figure 3

Jh9 ship_ls now heeled beyond small angles of heel.

GZ becomes positive and the curve now departs from the

iangential line drawn from the origin. The initial transverse

metacentre no longer applies to the ship since it is now heeled

to a large angle. --

wf

A ğpical curve of statical stability for a ship in a neutral

condition of stability will be as shown. Figures 1,2 and 3 are

related to the curve as indicated.

( ii

(

GZ curves for differentconditions of stability(MARRev. 18/09101)

Dec 2012


Curve of stAtleal stabilitv for a NEUTR

0.3

o.25

o.z

0.'t5

GZ o'1

(m) o.oı

0

-0.05

-o.l

-0.'15

Heel(deg.)

I

Fig. I

I l

i do I ,b \.br - -r- - L - J - - - \ r -r r r r \rrrrrYı _ _l- - ts - f - - _ - ı -trrtllıIll

J

4

Note

Since GM : 0, the x-axis of the graph is also the tangent along

which the GZ curve initially follows.


3. ' Curve of statical stability for a ship in an unstablecondition

A ship is İn an unstable condition if, when heeled by ane_rylernal force in still water to a small ansle, it continues to

lgeı!yrther when tlıe externalforce is removed.

Consider the ship shown being progressively inclined from theupright.

Fisure 1

KM - KG: GM; which is anegative value.

GZ is zero.

(>(

wf

G

M

B

K

GZ curyes for differentconditions of stability(MAR Rev. 18/09/01)

Dec 2012


Fisure 2

-GZ iş negative; itrepresents a capsizing |ever'

(If the external force is now removed the ship would continue

,9h?:lfuıther over!)

wf

I Z

M

G

I

I


conditions of stabilitY(MAR Rev. 18/09101)

Fisure 3

The ship continues to heel

vertically below G as shown.

restat anangleofloll.

over until BIt is now that

attains a positionthe ship comes to

When lying at an angle of loll:

GZ is zero.

The angle of loll is a large angle of heel since the line of action

of the buoyancy force (Bf) is.no longer passing through the

initial transverse metacentre (M).


l0

I

Dec 2012


Fieure 4 İ

ı( -'ı;l- ]r {'ir;l i :i :'ı l

If thg s!ip_ is heeled further by an external force B moves

-outbo4rd of G.

GZ is now positive which will act to right the ship back to the

angle of lolL

n,;

Gı'.

- a"l

Bf

Z

ll,K

wf

A typical curve of statical stability for a ship in an unstable

condition of stabiliğ will be as shown. Figures 1,2,3 and 4 are

related to the curve as indicated.


Curve of statical stabilitv for an UNSTABLE ship

0.25

0.15

0.1

GZ0-05

(m)

0

-0.05

-o.1. . ıi

-0.15

Heel (deg.)

Note

Since GM is negative, the tangent along which the GZ curveinitially follows runs below the base. As the vessel heels over tolarger angles of inclination the GZ curve departs from thetangent and where it crosses the base is the angle of loll(approximately 11o in this case). In theory this could be to theport or the starboard side since G is assumed to be on thecentre-line whereby port and starboard moments are equal.

l l9v ı \P t l i+ı),,iitial GM l- '

_ ı__r_lttti

ılı llllitt- l I__i ı-ltllttl;

. ' ,......' '' .' .. .'j

trt

I

-+--t--F-

Fig.3

(0,-ouı)

i-,ı ,

-ri

il.';l 1

\t

|ıi iN.!__r

Co:H- li'l+ ı{o UıGM

[ig.2

7I

--r-f-T-T--t----r'ttltt I


t2

Dec 2012


4. , Curve of statical sta

When a ship is listed th9 celtre of gravity of the ship is off the

centre-line to port or starboard by a distance we have termed

GGr'

Fiqure 1

Consider the ship shown that has G off the centreJine to

starboard that is initially in the upright condition-

GG" repres ents a capsizİng lever; a negatİve value of GZ-

GZ curves for differentconditions of stabilitY(MAR Rev. 18/09/01)

l3

Fisure 2

A' üç s!ııp liştş over the capsizing lever caused by G being offthe centre-line becomes less and less.

GrZ is negative.

GZ curves for differentconditions of stability(MAR Rev. 18109/01)

t4

Dec 2012


Figure 3

GZ is zero.

t

ll

\ruI

GZcurves for differentconditions of stabilitY(MAR Rev. 18/09/01)

l5

X'isure 4

A typical curve of statical stability for a listed ship will be as

shown. Figures l, 2, 3 and 4 are related to the curve as

indicated.

GZ curves for differentconditions of stability(MAR Rev. I8/09/01)

t6

Dec 2012


Curve of statical stabilitv for a LISTED ship

o.25

o.2

0.15

0.t

GZ0-05

(m)

0

-0.05

{.1

-0.15

Hee! (deg.)

Note

At 0o heel, the GZ value is negative by an amount equal to the

distance that G is off the centre-line (GGn)- This causes the

base of the graph to be dropped vertically to coincide with the

new origin - the green line being shown as the base of the

graph. The initial GM used to produce the tangent to the curve

at the origin is measured from the new base as shown'

The angle of list is identified as the point on the curve where it

crosses the original base ofthe curve as shown.


conditions of stabilitY(MAR Rev. l8l09l0l)

17 GZ curves for differentconditions of stability(MAR Rev. l8/09/01)

l8

7,.;-

Dec 2012


PRODUCING A CTIRVE OF'STATICAL STABILITYAsaihip-tr-eeliR-cön5tamlymövEs;-if 's-p-oIitiönis-depenctenton:* the displacement (draught) of the ship;* the angle ofheel at anY instant.

KN Curves (MAR Rev.03/0r/0t)

If the figure is considered:

Sine 0: OPP Sine 0: Correction to KNrryP

Therefore:

and:

KG

Correçtion to KN: KG x Sine 0

GZ = KN - (KG Sine 0) \Cross curves of stability (KN curves) are provided by the builderto allow GZ values to be determined for any value ofdisplacement and KG. Sometimes the values may be tabulated.

It is usual that KN values are given for angles of heel at 10" orl5o intervals.

An example of such values is given;

KN Curves (MAR Rev.03/01/0r)

Dec 2012


DISPLACEMENT (t)

15000 1.72 2.98 4.48 5.72 6.48 6.91 7.05

14500 1.73 2.98 4.51 5.79 6.58 6.9s 7.08

14000 r.74 2.98 4.55 5.85 5.68 7.00 7.10

ı3500 1.15 2.99 4.58 s.90 6.73 7.08 7.13

13000 L.77 3.00 4.62 s.93 6.78 7.14 7.16

12500 1.78 3.03 4.63 s.98 5.83 7.18 7.18

12000 1.78 3.05 4.65 6.O4 6.88 7.20 7.20

11500 1.80 3.12 4.70 5.10 6.93 7.25 7.22

11000 1.82 3.15 4.75 6.15 6.98 7.30 7-24

10500 1.83 3.19 4.79 6.18 7.02 7.35 7.27

10000 1.86 3.23 4.83 6.22 7.07 7.40 7.30

9500 1.93 3.28 4.91 6.25 7.tt 7.45 7.35

9000 2.00 3.36 s.00 6.28 7.t8 7.50 7.40

8500 2.O5 3.43 5.04 6.32 7.20 7.55 7.4t

8000 2.to 3.52 s.10 6.36 7.22 7.60 7.42

7500 2.17 3.62 5.18 6.38 7.24 7.65 7.46

7000 2.22 3.70 5.25 6.40 7.26 7.70 7.50

6500 2.32 3.85 5.J) 6-43 7.21 7.70 7.51

6000 2.42 4.00 5.45 6.48 7.28 7.70 7.52

s500 2.57 4.15 5.5s 6.53 7.29 7.68 7.51

5000 2.72 4.32 5.6s 6.58 7.30 7.66 7.50

tcıvALUEsARbmırtIND TDrcD TRIM

JLLAND FORECASii

IATED FORVESSI

ti'l

|TLE O

)LON,

NLY.

\N EV EN KEEL

Notewhen KN values are tabulated as shown interpolation for

displacements other than those stated may be done, but it should

be Lorne in mind that the rate of change of KN will not be linear.

KN Curves (MAR Rev.03/01/01)

Procedure for constructing a cuıve of statical stabilitv

1. Determine the ship's displacement and effective KG for thecondition being considered (effective KG being that takinginto account free surfaces in tanks).

2. From hydrostatic data find value of KM for ship'sdisplacement.

3. Find GMrr,o using: GM= KM-KG

4. Enter KN tables (or curves) and obtain KN value infor each angle of heel given.

metres

6.

7.

5. Using: determine GZvalues for angles of heel given.

Plot the GZ values.

Before joining all the points on the curve construct a verticalat 57.3" and from the base upwards mark off the value of the

effective GM (using the GZ scale). From this point draw a

straight line to the origin of the curve to be drawn. This willindicate the initial trend of the curve at small angles of heeland will assist in sketching the actual curve between the

origin and the first plotted GZ value.

(GZ and GM are closely related at small angles of heel)

KN Curves (MAR Rev.o3l01/01)

Dec 2012


|ntıct stabilitv reqıirements - M.S. (Load Lİne| Regulalİons1998

GZ(m)

* Area 0o to 30o to be not less than 0.055 m-r;

't Area 0o to Xo to be not less than 0.09 m-r;* Area 30o to Xo to be not less than 0.03 m-r;* Xo is equal to 40o or any lesser angle at which

progressive down-flooding would take place;rr Maximum GZto be not less than 0.20 m and to occur

at an arıg|e ofheel not less than 30o;* Initial GM to be not less than 0.15 m.

KN Curves (MAR Rev.03/0r/ot)

Procedure to verifv that a ship's loaded condition comnlieswith legislation req uirements

Consider the following example:

A ship has a displacement of 12000 t,KG 8.22 m and a KM of8.54 m. Using the KN values provided determine whether theship's loaded condition complies with the requirements of theM.S. (Load Line) Regulaıions 1998.

l. Determine the GZ values.

KG i 8.22 KM i8.s4 GM 0.32

HEEL 0 t2 20 30 40 50 60 75

KN (m)(KG Sin Heel

0.000.00

1.78

t.7r3.052.81

4.6s4.11

6.04

5.28

6.886.30

7.207.12

7.207.94

GZ(m\ 0.00 0.07 o.24 0.54 0.76 0.58 0.08 -0.74

2. Plot the GZ curve.

Obtain a value for 10o heel so thatGZ values are available at10" intervals up to 40o heel.

3. Using Simpson's rules calculate the areas under the curve(0o- 30o and 0"- 40o and 30'- 40')

KN Curves (MAR Rev.03/0t/01,

Dec 2012


Area 0"- 30"

Heel i GZhrdl i SM i Area Fn.

Area:3/8 x 10157 .3 x I.47 : 0.096 m-r (0.055)

Area 0"- 40"

Heel0

0.07

SMI4

Area Fn.

0

l00

i 0.28

20: 0.24 2 : 04ü30i40i

0.54 4 i 2.L6

0.76i I 0.76

Area: l/3 x 10/57.3 x 3.68 : 0.214 m-r (0.090)

Area 30"- 40"Area: 0.214 - 0.096 : 0.118 m'r (0.03)

Max GZ and angle at which it occurs

Satisfies the requirements.

Effective GMKM - KG: GM;8.54 - 8.22 : 0.32 m (0.1 5) SHIP COMPLIES

KN Curves (MAR Rev.03/01/01)

Intıctstahilitv reqlirements - Deoartment of the Envİronmentand Regions

0f

The ratio of the area under üe righting nm (GZ) curve to that under the windheeling arm curve to be nol less lhan 1.4.

These areas are to be measured from the upright position to an arıgle of heelnot exceeding either the angle ofprogressive down-flooding (0f) or the secondintercept ofthe righting arm and heeling ilm cırrves whİchever is lesser.

Minimum GM to be nol less than 0.30 m..

Minimum GZ for any angle of heel &,

up to the angle ofprogressive down-flooding 0Ç or;angle of maximum righting lever, or;l!",whichever İs leasl to be given by:

Mİnimum GZ=0.5 x 0.3 Sin 0.

Angle ofsteady wind heel not to exceed l5o.(Based on a wind velocity of 51.5 m/sec (100 Knots)

KN Curves (MAR Rev.03/0r/0r)

4

Dec 2012


ship in a stable condition heeled to a small angle

Z

B1

Dec 2012


SHıP

STABıLıTY I01(_

ı

Dec 2012


CENTRE OF GRAVITY (G)

we the vessel and

The position of G will not move as the ship heels (provided that*ggtl'-ggıgıgıl':4-te-e1"--ryiihln}-g-ş]ıiP j.ı{alwayİaöısverlicallv downwards!

wten weights are ş-4ı[!g ! :ı-L. gL ]9 ryLq d- olJiŞ ğ grye J!--G- w i\move. Whenever G is caused to move the 'shift of G' must be-calculated.

Centre of gravity/buoyancy(MAR Rev. 03l0ll0l)

Effect of shiftins a weisht already on board

parallel to and inthe sameof the

w

K

.:wxdI

-

w 0.

'w' - (A) iı-ıtç* şbıp]şdisplacement

The verlİcal posilİon oJ G İs expressed İn terms of metres abovethe keel (KG).

SAO

A ship displaces 5000 ı and has an iniliol RG of 4.5 m.Cqlcaıaıe |he Jİnal KG iİ a weİghl of 20' t,İİ moved verlİcallyu.pi,ııardı from the lowqr hold (Kg 2;0.m) ıo':ıhe rypper dech, ([g6.5 m)

'd' is the distance throughwhich the weight is s-tüftedl---

Centre of gravity/buoyancy(MAR Rev, 03/0t/01)

(_'

Dec 2012


Answer

'GGı--a-'I*! = 20 x (6'5 - ?'0),'= 0,018 m

w s000

Inİıİal KG :4,500 m

GG, = 0a!ğJLFINAı KG = 4,578 m

.

KG has increased which would make the ship less stshle'

Centre of gravitY/buoYancY(MAR Rev. 03/01/01)

Effect of loadinq a weisht!+psğ^lgyqrL@

lga494_U9ıght-@.

I

.=wxd|

-W+w

'w'is?ıs een Gofthe ship and g- of the loa^dgğ

.weight.'W' (A) is the ship's initialdisplacement.

SAO

A ship displaces 12500 t and has an initial KG of 6.5 m.

Calculaıe the !İnal KG ,f 1000 t of cargo is loaded into lheIower'hold at Kg 3.0 m.

Centre of gravity/buoyancy(MAR Rev. 03/0ı/01)

Dec 2012


Answer

GG,=wx d=!W!)=0.259mW+ w 12500+1000

InitİalKG :6.500 mGG, :0:259 mFINAL KG :6.241m

KG has'decreased which would make the shİp more stable.

Centre of gravityöuoyancy(MAR Rev. 03/01/01)

from the centre of sraviW of the di

GG.=wxdI

-

W-w

'w' is the weight discharged.id' il-the aıisfahTğ-b€ffi-een G-of -ffi- shib-ana s ;f.---i[entW' (A) is the ship's initialdisplacemerıt.

- _ '

SAO

A ship displaces 18000 t and has an iniıinl KG of 5.30 ınCalculate ıhe JİnaI KG ıf 10000 t of cargo is discharged fromthe lower hold (Kg 3;0 m).

Centre of gravity/buoyancy(MAR Rev. 03/01/01)

I

Dec 2012


Answer

GGı:Lğ_!: !W0)=2.875mW-w 18000-10000

Iniıial KG :5.300 mGG, =2.875 mFINAL KG = 8.175 m

KG has increased whİch makes the ship less şlable.

It would be very tedious to do a calculation for every singleweight that was either shifted, loaded or discharged from the ship.

Lır".tıqg-g_._qgı!s qboyJ [he kgel' ge 9şe!-!o d9j9rmine !tıe-final KG of the ship.

MOMENTS (t-m) = WEIGHT (t) x DISTANCE (m)


Moments about the keel

Example

A ship displaces 10000 t and has a KG of 4.5 m.

The following cargo is worked:

Load:

Discharge:

shift:

170 tat Kg 6.0 m;

730 tatKg3.2 m.

68 t from Kg 2.0 m;

100 t from Kg 6.2 m.

86 t from Kg2.2 m to Kg 6.0 m.

Calculate the final KG.

FINAL KG: SUM OF THE MOMENTS

FINAL DISPLACEMENTN

WEIGIII (t) KG (m) MOMENTS (cm)slip-ft)L9-1d (+-)

Load (+)

10000.00 4.s0 45000.00120.00

730.006.00 720.003.20 2336.00

Dİscharge QP"ış"çlıı"r"gç ()"

* D_|ş.gh3_g_e_ (;)*Load (+)

68.00-ı00.00

-86.00

2.006-,.?9

ü.ü

-ı36.00-620.00

-189.2086.00 6.00 5ı6'00

F'INAL ı0682.00 4.4s9 47626.80

Centre of gravitylbuoyancy(MAR Rev. 03/0ı/0l)

Dec 2012


SAO

A ship displaces 14200 t and has a KG of 6.22 m.

Thefollowİng cargo İs worked:

Load: 768 t at Kg 7.20 m;

Discharge: 348 tfrom Kg 4.62 m;

266 tfrom Kg 5.36 m

Shifı: 188 tfromKg 8:00 mto Kg 3.60 ın

Calc ulate, th e Jİn al KG,

Answer

WEIGIIT (t) KG (m) MOMENTS (t-m)

Ship (+) 14200.00 6.22 88324.00

Load (+) ı68.00 7.20 1209.60

_D_iş_s'l"r"rg-"_ (:)

Dhcharge (-):348:90_-256.00

4.62 -1607.76

5.36 -1425.76

-1504.00676.80

I P-işg!,ı__._c"- (:)*Load (+)

-188.00 8.00188.00 3.60

FINAL ı3754.00 6.229 85672.88


The position oJ B _has nothing to do with the disposition ofiveights within the ship.

The centre o=| eravjtJ (G) !s qsşg-mgd !9 19main in the -same

place(proviled weights do not shift within the ship) as the ship heelsbui ıhe cen7ie of buoyanğ ioistaiıtıy'moies ai ıııe shıp pitches,ioÜSvndhaanöS:****'

For a box-shaped vessel on even keel KB is halfthe draught.


l0

5

Dec 2012


bg-@

FR-EE SURFACE EFFECT

M

\zv

Iıi

ıiı|ıi

,LI

R--.g,

$ ship has a slack tankas shown i.e [g_tank it=_only part full.

External forces cause the ship to heel to some angle and liquidin the tank moves from-th+Lğh slde to the low side (gg').

This. causes G to move off-t[e-cglıge-line to G,-

As a result of the shift of liquid, righıing lever is reducedfrom

97-'o 9,"ıFree surface effect (MAR

Rev.03/01/01)

The righting lever GrZ, is the same as

the GZ that would have existed had Gbeen raised to Gu.

GGu represents the virtual rise of Gthat results from the free surface effectof the slack tank.

Thus:

GM isthe solid GM;

G!ıı

When calculating the GM of the ship it is important that theeffects of free surfaces are considered i.e.

IT TS ALWAYS THE FLUID GM THAT MUST BEbEfBnMıNED ro rı.rE ACCoIINT oF THE REDUCTIoNIN GZ VAL_UES THAT aırsps FROM LIQUIDMOVEMENT WITHIN THE SHIP AS IT IS HEELED

NoteWhen the ship is in the upright condition the centre of gravity(G) will move back to it's original position at G. Hence, it istermed a 'virtual rise' of G since it does not actually move up to

Gv.

Gv

ıı

G.)VI isthe effective orfluid GM. Z

zr

Free surface effect (MARRev-03/01/01)

Dec 2012


virtual rise of G

GG.,=lb3 x dt12Y ds

JI

where: 99, i'J!ç yirtua-trise of G in metres;

I is the tank length;tiJ ttıednk breadfh;

at ii tıı" density of the liquid in the tank;

ds is the density of the-water in wnichthe ship floats

(.n25 t/fi);-and,

Since: A:Vxds

it follows thaİ (r r ı: )

li( ül_iiil

('ii;ı'^ıj

GGu =ğ x dt12L

displacement is 8,80 m.

Free surface effect (MARRev-03/01/01)

Answer

1. Taking momenıS about tlıe keel, calculate the new solidKG.

WEIGHT(t) KG (m) MOMENTS (t-m)

shipft)Carso oil (+)

10s00.00 7.60 79800.00

9600.00 8.00 76800.00

FINAL 20100.00 7.791 ı56600.00

('t,'' "i ."-''ı'

Calculate the effect offree surfoce. '; L

GGv=-!b3- x dt.!i lz\

1 ._'.

= 30 x 203 x 0.86 = 0.856 m12 x20100

Calculate the solİd GM and then apply the free surfocecorrection to obtain theJluid GM.

KM 8.8007.791SOLID KG

SOLID GMFSE (GGv)

1.0090.856

T'LTJID GM 0.153 (Answer)

Free surface effect (MARRev.03/01/01)

Dec 2012


öalculctions _

The moment of inertia (ğ, often termed the second moment ofarei,-o{thehie liquid suiface must first be determined.

For a rectangular free liquid surface: I lb3 jn4)t2

If the value of I is multiplied by the liquid density then a value

of 'Free Surface Moments' (FSM's) (t-m) is obtained.

FSM's (t-m; = lb3ldt12

In the previous free surface effect formula:

GG,r: Jğ| x dt12L

GG.,: FSM'sDisplacement

Final KG = Sum of momentsDisPlacement

it is evident from the above that the greater the value ofthe free

surface moments, the greater the loss of GM (GGv) and the

greater the value of the effective KG.

In calculating tlıe effective GM it is us

for free surfaces by incorporating the

where they must alwaYs beADDED.

Therefore:

Since:

Free surface effect (MARRev.03i01/01)

Consider the previous example.

t. Calculate the FSM's using: FSM'ş (t-m) : Ib3 X dıI2

2. Taking moments about the keel, also adding the FSM's inthe 'momenls' colıımn', calculate ıhe/Iuid KG.

Apply thefluid KG value to thefinal KM to obtain thefi'nal

fluid GM.

(Answer)

NoteIn tank sounding/ullage tablesfree surface momenls (t-m) maybe tabulated for an assumed value of liquid density.If the actual

]qüa--{qıılty !i different -then the FSM valueş must be

corrected.

KMFLTJID KG

8.8008.647

FLUID GM 0.1s3


Dec 2012


Factors affecting free surface effectIf the free surface formula is ion-ği?i'ered it is obvious that the

breqdlh of the tank is. the most impgrtlnt factor.

If a tank is_subdivided_the loss of GM can be greatly reduced'

Consider the rectangular tank shown below.

1. With no subdivision '"

.

If the loss of GM due to the free surface of the tank is 0'64 m

and the solid GM was found to be 0.70 m, say, the effective GMwould be:

Solid GMFSE (m)

Effective GM 0.060 m


2. With a sinsle centreJine division r r

: E

-(.'.(

The volume of liquid that has moved is halved.The distance that the total volume of liquid has movediJ ai.o halved.

ry!ü ! single centre-Iine division the FSE is reduced to aquarter of th|e orİğİnal valub

0.64:0.16 mT Solid GMFSE (m)

0.700 m0.160 m


Free surface effect (MARRev.03i0ll01)

Dec 2012


3 s {-r==-

3. With two subdivisions , ; ,a)

\

The volume of liquid that has moved is now a third.

The distance that the to'idlvolume of liquid has moved

is also only a third.

With the tank equally subdivided with two longitudinal

bulkheads ıhe FSE İs retruced Io a niith of ıhe orİginal value'

0.64 :0.071 mT Solid GMFSE (m)

0.700 m0.071 m



With three suhdivisions i.e. dividing the tank into fourimpiiımeits, FSE would be reduced to one sixteentlı and so

9L

It is usual to subdivide tanks into three compartments at most as

the benefit of any further subdivision would improve effectiveGM by a less significant amount each time (with the exceptionof product carriers/chemical carriers etc. where a ship isdesigned to carry a wide range of cargoes at any one time).

Typical oil lanker lank arrangemenl


l0

Dec 2012


Points to note

- Fr* trrk to be considered subdivided it must be fıtted with

an 'oil-tight' or 'water-tight' bulkhead ensuring that there is

no possibiliğ of liquid transfer. This means that any valves

in between the-tq!ş_ry9tb-e capable of being fully shut'

A 'wash plate' is fitted to prevent damage-to internal tank

plating that may be caused by wave action withİn the tank'


ll

The two tanks shown are exactly the same in size and both

hold liquid of the same density but are filled to differentlevels.

Tlıefree surfuce momentsfor each are the Same.

>ı/''


12

6

Dec 2012


Consider a ship floating upright, G and B on the centreline with a

yr-eighf 'w' on one side-.

List(MAR)

The weight ''w' is shifted transverselycausinğ G to mo'e off the centre-line as

shown. The ship lists over, coming to rest

with the centre of buoyancy, B, verticallyba;w ahö .*ti" of gravity, ,o*€ffi €.,

d

;ru

List(MAR)

Dec 2012


The angle at the metacentre in the right angled triangle GG,M is

the list (termed 0).

Tan 0rr.r= OPP: GG,ADJ GM

Therefore:

Tan 0161: sırGM

,(

For the above formula to be true, the list must be restricted to a

small angle, i.e. the metacentre is assumed to be in a fixed

poiition *iahTn small angles of inclination..l

SAO .

A shİp inİtially uprİght dİsplaces 12000 t and has KG 6.7 m andKM 7.3.m- A weiğht oİ 60 t already on board is shİfted 14 m

horİzontally acİoss the deck- Calculate the resulting angle ofIİSt.

List(MAR)

Answer

KM 7.j m GGu:w x d GGn- ğ0x 14 :0.07 mKG 6-7 m w 12000GM 0.6 m

Tfln0,,o-= GG- = 0.07 = 0.11666 List:6.65"Lto I

-ttGM 0.6

If a weight is loaded or discharged then both the veıtical andhorizontal components of the shift of G must be considered andthö final GM must be used to calculate the final list.

, ,l

1. Calculate GGu, and hencethe final GM.

2. Calculate GGr.

3. Calculate the list.

ı',

SAOA shİp displaces 6400 t and haş KG 4.6 m and KM 6.5 m, Aweİght, of 80 t İs loaded dn deck at Kg 10.2' m, 6.2 m off thece ntre-Iİne to starbo ard. CaIc ulate the final,: Iist,

List (MAR)

Dec 2012


AnswerGGr=w x d

W*wGGv= 80x (10.2 - 4.0 = 0.069 m

6400 + 80

KM= 6.500 mInİıİal KG 4.600 m,

Inİtial GM I.900 mGG"(uil 0.069 mFinaı GM 1.83I ry

GGu= & GGr= 80'x t6.2 = 0.077 m

W*w 64A0,+80

Tanax1ş7: GGu :0M- = 0.04205 List= 2.4" StbdGM'^nı 1.83t

lt is easier to solve list problems using moments.

l. Take moments about the keel to determine the final KG:

Final KG = Sum of moments about keel (t-m)

Final DisPlacement (t)

2.

5.

Calculate the final GM: GM: KM. KG

Take moments about the centre-line to calculate the final

distance that G is off the centre-line, GGr:

GGu = Sum of moments about centre-line (t-m)

Final DisPlacement (t)

4. Calculate the list: Tan 0rrr.: GGni i-l

List (MAR)

SAa',; .' _ ı;- . ] '.l .* :

A ship:dİsplaces 8000 tonnes, RG 7.60 m and İs inİtİally aprİght.Thefollowing cargo is worked:

Load: ', 300 ! atKg 0,60 m,6.1 mto port oJCL; :

'i'; 250 t aı Kg 6.10 m' 7.6 m to sıbd of CL;Disch:: .,, 50 tfrom Kg 1.20 m, 4.6 m to pott of CL;

,,,, 500 tfrom Kg 12.60 m; 4:6 m to stbd of CL.

Calculate:thefinal angle of list on completİon of cargo.

Answer

)'| ( '-f t:, ,ç (

List(MAR)

Dec 2012


SUSPENDED WEIGIITS

when the weight is plumbed over the side a larger than normal list

will also occui and certain precautions will have to be taken to

ensure that the maximum list is restricted to an acceptable limit-

Suspended Weights (Rev.

03/09101)


03l09lor)

Dec 2012


the weight is lifted clear of the tank top the centre ofthe weight moves vertically up to it's point ofat- gr This results in a corresponding vertical

movement of G to Grn, causing an increase in KG.

7X

I

I

Suspended Weights (Rev.03109/0t)

GG,,=wxdY-

w

where 'd'-iş the ğistange through which the weight is effectively

.$ifted upwards from it's initial stowage position to the derrickhead.

ExampleA ship has a displacement of 10516 t, KG 8.20 m and KM 9.00ın A weight of 86 t İn the lower hold, Kg 3.40 m, is lifıed by theship's heauy lİft derrick, the head of which iS 22.00 m above the

keeI.

(a) Calculate the GM when the weight İs suspended.

(b) Calculate theJinal GM when İIıe weİght is reslowed inthe tween deck at Kg 8.50 m.

Suspended Weights (Rev.03109101)

4

Dec 2012


GG, :ü:W

86 x (22.00 - 3.40ı = 0.152 m10s16

KM 9.000 m

GM when weight susPendedlQlfu!fu

This is lhe minimum GM during tlıe liftİng operalion'

(b) To calculate the final GM when the weiglıt has been

shifted treal as a normal single weight problem -simply shifı ıhe weightfrom it's initigl stowage

position (Kg 3.a0 m) to il's final stowage position (Kg

8,50 m) i.e. ignore the derrick.

86 x (8.50 - 3.40 : 0.042 m10s16

Iniıial KGGG,KGwhen

Initial KGGG,FinalKGKMFINAL GM

8.200 m0.152 m

weight suspended 8.352 m

GG,,:wxd=W


03/09101)

The previous example was very simple as there was no horizontalshift of G involved as would be the case when loading ordischarging a weight.

Wlen lgading or discharging weights using ship's lifting gear thefollowing must be considered:

(t)

(2)

the increase in KG/decreqse in GM wlıen the weight issuspended at the derrick/crane head;

the maximum angle of list that wiII occur when thederrİck or crane is plumbed over the shİp's side.


03l09lot)

Dec 2012


Consider the movement of the ship's centre of gravity (G) when a

weight is lifted offthe quay on one side of the ship and stowed in

the lower hold on the centre-line. (For the purpose of this

explanation the ship is zol shown to list.)

1. Derrick plumbed over tlıe weight.

Suspended Weights (Rev.03109/01)

2. Derrick picks the weight up off the quay.G moves to G1, directly towards the centre of gravity ofthe loaded weiğht - the point of suspension (gı).The movement GG, has two components:GGr; whİch cüuses an increuse in KG/decrease İnGM.GGu: which causes the ship to list.

Suspended Weights (Rev.03t09t0t)

Dec 2012


o*n**rrrr*,towards/ı Eı

Tan 0r*.rsr = GG,n

GM,'*n'


03/09/01)

!t Derrick swings inboard to plumb stowage position on

the centre-line.G, moves to G, as the weight is swung inboard from g,

to gr. Ship becomes upright.

Suspended Weighs (Rev.

03l09lol)

Dec 2012


4. Weight landed in the lower hold.

The weight is refuovgd from the derrick head at g, as it

is landed in the lower hold and finally acts at Eı-GzmovGs to Gj-

Note that the weight is loaded below the initial centre ofgravity of the ship, -hence the final position of G, G,must be lower. KG is reduced overall.

ıEı

Gı

Suspended Weights (Rev.03/09/or)

The following example illustrates a typical list question involvinga heavy lift being loaded. The calculation may done by one of twomethods:

Formula method.Taking momenis-abouı tlıe keel and the centreline,

(1)(ı

Both methods are shown, however, the moments method is much

simpler and will be adopted for all other examples in this section.

Suspended Weights (Rev.03/09/ot)

12

Dec 2012


7.15 m and KM 7.98 ın A weight ofe quay which is 15.0 m to starboard

of ıhe ship's centreline. Iİ the head of the derrick is 27'0 m

ibor" lhe keel when topped to it's maximam extent for the

I fıİng o p eralİo n, calc uI ate :

(a) the GM when the weight is suspended;

(b) the maximum angle of list;

(c) the Jinal angle of list if ıhe weight is placed on deck at

Kg 10.40 m, 5.0 m off the centrelİne to starboard

(d) the weİght of ballast to transfer between two double

bofiom tanks, each having iİs centre of gravity 4'0 moffthe cenlrelİne, to bring the ship uprİght'

(Assume KM rernains constant)


03/09/ol)

Solutİon (Method 1|

(a) To calculate the GM when tlıe weighl is suspended

Load tlıe weİglıt at tlıe derrick lıead.

GG,,: ıyıJI GG"= 40 x (27.00 - 7.1fl = 0.089 mW*w

Iniıial KGGG,

7.150 m0.089 m

Tan 0**us7: GGnGM,,,,

8850 + 40

Tan 9rn*. Lrsr: L067 : 0.090420.741

Maximum list= 5.2" Stbd.

MaxİmumKG 7.239 mKM 7.980 mMINIMUM GM 0.741 m (GM when the weİght is suspended)

(b) Calculate the distance that G is off the centrelinewhen the weight is suspended over the quay (CCr)-

GGu:wxd GGn: 40x15 : 0.067mW*w 8850 + 40


03/o9iol)t4

Dec 2012


(c) Calculate the final angle of list when the weight isplaced on deck.(Simply load tlıe weight on deck, ignoring the derrickas in a normal single weight probleın)

GGv:40 x (10.40 - 7'15) : 0'015 m

8850 + 40

7.150 m0.015 m7.165 m7.980 m0.815 m

=wxd GG-=40x5 =0.022mW*w 8850 + 40

GG.,: w x d/-W+w

Initial KGGGnFinal KGKMFinal GM

GG,

Tan 011s7: GG.n Tan O"rrr: 0.022 : 0'02700GMor*nr. 0.815

FİnaI list: 1.5" Stbd.

Suspended Weights (Rev.03/09/ol)

l5

(d) Calculate weight of ballast to transfer from Stbd. DBtank to Port DB tank.

In tlıefinal condition G is off the centreline by 0.022 m (GG*)

GG..:wx d 0.022-- w x I w:0.022x 8890 : 24.5 tD

W 8890 I

Transfer 24.5 t

Suspended Weights (Rev.03/09iol)

I6

Dec 2012


Solution Method 2l

(a) To calculate the GM when tlıe weiglıt is suspended'

Load the weİglıt at tlıe derrick lıead. Take momenls

aboul the keel.

weisht (t) Ks (m) noments (t-m

8850-0 7.15 632',77.s

40.0 27.00 1080.0

8890;0-- +.239- - -643515

GM when weİght suspended 0.741 m (Mİnimum GM)

Take moments aboııt the centreline to determine GGff

Dist off Port stbd

weisht (t) CL(m) moments (t-m noments (t-m

8850.0 000 0.0 00

40.0 I5.00 600.0

-8890tr fJ67 _(€0J-

Tan 9,^*."tsr= 0,067. : 0.090420.741

KM 7.980

KG 7.239

GM 0.741


03/09/01)

t7

(c) Calculate the Jinal angle of list when the weighl is

weisht (t Ks (m) moments (t-m'

8850.0 7.15 63277.5

40.0 ı0.40 415.0,8890i 7.165 636935-

Take moments about the centreline to determine GG*

Dist off Port stbdweisht (t) CL(ml moments (t-m moments (t-ml

8850.0 0.00 0.0 0.0

40.0 500 200.0

8890.0 ' 0.02> 200;0'

= Kn Tan 0r,rr= 0.022 : 0.02700GMr,rnt 0.815

KM 7 980

KG 7 165

GM 0.8I5

Suspended Weights (Rev.03t09101)

t8

Dec 2012


(d) Calcalate weight of ballast to transfer from Stbd. DBtank to Port DB tank.

In theJİnal condition G is offıhe centreline by 0.022 m (GG*)

GG..:wxd 0.022: wx I w=0.022x 8890 : 24.5tD-

8890

Transfer 24.5 t

Suspended Weights (Rev.03/09/01)

t9

The key point here is to firstly identifu the situation during thelifting operation which will create the maximum list. Drawing adiagram will help.

Consider the following example.

ExsmpleA ship dİsplacing 16200 t is uprİght and lıas a 90 t weight ondeck at Kg 13.0 m, 6.00 m to port of the centre-line. This weightis lo be dİscharged into a lighter on lhe port side, 14.00 mfromthe centre-Iine usİng lhe ship's heavy lift derrick. If the angle oflist İs not ıo exceed 8o at any time during the operation,calculate the maximum allowable KG prior lo dİsclıarge giventharKMİs9.60m. (., ' :i l i l''ı'' ' : li] ..l] Il' i 'i 1'| ,;]

Suspended Weights (Rev.03/09/0t)

20

10

Dec 2012


Solution

Maximum list will occur when the weight is suspended,at the

derrick head and the derrick is plumbed over the port side. t , ,,)

ııIııI

GM will have it's minimum value.

GHG Tan Orn*.ror: Es

GM''n'ıtI

IııııI

ı


03/09/01)

Calculate the distance that G will be off ıhe centre-Iine (GGH)

when the shİp is at it's maximum angle of list.

GGn :wxd GGH: 90x(14-61 : 0.044m16200 | ,

Maximum allowed lİst is 8'. GGH iS 0.044 m. This allows the

GM to be calculated.

TanO**usr= GG.n

GMrr*

Tan 8":9WGM,,,

Therefore: GMr,u:0.044 = 0.313 m v

Tan 8o

A minimum GM of 0.313 m is requİredwhen the weight İs plumbed over the sideat Rg 27.0 m

KM 9.600 mMinİmumGM 0.313 mMaxİmumKG 9.287 m '


03/09/01)

22v

t1

Dec 2012


Havİng calculated the moximum allowed KG when the weight issuspended from the derrick head (Kg 27.0 m), shifı ıhe weight

back to it's original stowage position on deck (Kg 13.0) to

calculate the maximum initial KG requiredfor the operalİon.

Take moments about the keel.

i ': : r.. ı,li------\ :i i'(' t:;)-j ton>

weisht (t) Ke(m) moments (t-ml

t6200 9.287 150449.4

-90.0 27.00 -2430.0

900 t3 00 t 170.0

ffiE

TIıe İnitial KG must not be greater than 9.209 m in order to limitthe list to 8". - 5-r' l"utt'.'- ] r,ıı^" .': ------ . -'.- _, _],-;,),..

,' Z*J \'ı'!ıf

Suspended Weights (Rev.03/0910t)

23

t2

Dec 2012


During the course of a voyage the stability of the ship should beclosely monitored. It is recommended that a calculation of fluİdGM and a coıTesponding GZ curve be produced foı the woıstanticİpated condition. Calculations should be done for bothdeparture and anticipated arrival conditions, these beingadjusted.to suit any changes that may take place as the voyageprogtesses.

An angle;of loll situation may arise in ships carrying timberdeck cargoes. Deck cargoes will absorb moisture causing G torisd.'fuel and water will also be consumed from low down inthe ship raising G further due to the removal of weight from lowdown in the ship and the introduction of free surfaces in tanksthat may have been initially full. These effects should beaccounted for. Poor tank management may cause excessive freesurface moments, leading to a similaı loss of stability'

Unfortunately things do not always go to plan and mistakes maybe made in calculations. Any number of things might happensuch as a collision or a fire where water is introduced into acompartment for fire fighting purposes.

Whatever the cause, a situation might arise whereby you findthat the ship is lying at an unexpected angle of inclination.

It is not possible to ascertain easily whetlıer a ship is listed orlolled and since the remedial action for each case is verydifferent it is essential that the cause of the inclination be

carefully investi

Correcüng an angle ofloll(MAR Rev. 09110/04)

The following procedures should be carefully observed:

1. Alter course to pul the ship's heod into lhepredominanı waveş.

If the ship is in a lolled situation it is essential that the ship stayslolled to the same side. Wave action may cause the ship to rollthrough the vertical to loll on the other side. This is a dangeroussituation since the ship will heel from the vertical of it's ownaccord and the momentum it will have in lolling over to theother side may be suffıcient to capsize it. In any event, the shipwill initially heel beyond the angle of loll before settling at theangle of loll whereby cargo shift may result which will worsenthe situation further.

2. Check that port and starboard lisıİng momenıS alelhe same.

By verifying tank soundings and checking for cargo shift itshould be possible to account for any listing moments that maycause the ship to be in a listed situation. If it is calculated thatthere are no net listing moments then a case of instabiliğ maybe assumed and the ship will be lying at an angle of loll.

3. Checkfor slack ıanks

In carrying out (2) above it should also be evident if there areexcessive free surface moments causing a loss of GM suffıcientenough to make the ship unstable. In this case a loll situationmay be confirmed.

Correcüng an angle ofloll(MAR Rev. 09ll0l04,

Dec 2012


4. Take aclİon to loweı G (reduce KG)

It would be impractical to consider shifting weights on board

using ship,s lifting equipment at sea. If the ship has high ballast

tanı<i tııat are fuil then these may be emptied, dİschargİng

ballast from the high sİde tank frst - the greaİer verlİcal

dİstancZ between G of ıhe shİp and g of ıhe weİghı heİng

dİschargedwİll ensure thaı the greatesl lowerİng of G wİll take

place İn ıheftsı İnstance-

once the high side tank is empty the one on the low side may

then be emptied.

Correcting an angle ofloll(MAR Rev. 09110104)

5. '

Minİmise Free surfaces

Hauing sounded all the tanks any that are slack will be

identified. Minimise the loss of GM due to free surface effect

by topping up low down ballast tanks and transferring fuel as

necessary. This action alone may remedy the situation'

6, Bollast lanks low down İn ıhe shİp

select a set of suitably subdivided double bottom tanks to

ballast. Ideally start with tanks that have the smallest free

surface areas to minimise the effects of free surface whilst

filling. The order of filling is as follows and must be strictly

adhered to:

(a) Start by filling the centre tank (No' 1\ as

shown. Because of ıhe introduction of more freesurfoces whİlstJİtling the sİtuation will İnitİolly

Correcting aıı angle ofloll(MAR Rev. 09110104)

4

2

Dec 2012


(b) When the first tank is completely full,@.tank (No. 2t.

(c) When the low side tank is full, fill the final tank (No.3)

(d) If G is lowered sufficiently then the ship shouldcomplete in an uprİghı condİtİon (having initially

. .verified that the port and staüoard moments were üe'., saıne).

The movement of G after completely filling each of the tanks isas shgyn (ignoring the upward movement of G which arises asa result of the introduced free surface at intermediate stages offilling the tanks).

If the situation is not remedied then a second set of tanks mustbe chosen for fılling, the process is repeated.

tan

)G(

G.

\

c.: ,.G.fı

3 ı 2

Correcting an angle of loll(MAR Rev. 09110104)

Note

once u loll situalİon iİ conftrmed oqly,. ever JİIl one, tonk..at,-otİme.

I

Always JiIl !g",'i4q lonks Jİrst (or cen|p tank and then lowside). I

ftrstIf ıhqr..ç ıİ a!ıy d9ı1bı as to whe|her;the,shİB

iınqyş'*'ume İl İs a IoIIed iıunııi,n aiaaöıİon

low down in the ship) may be discha-rged,

To ticqt,q loll silugtion İn lht sameway woald,I1ave dİsoslro4ş-conseqaences for reasons alreağ Ç.fl'toiqed!

Correcting an angle ofloll(MAR Rev. 09110104)

Dec 2012


1.

2.

3.

i).'

TUTORIAL OUESTIONS

Calculate the effect on G of shifting 250 t of cİırgo doş,nward through 15 m from a posiüonon deck to the lower hold in a ship displacing 22000 t.

(0.182 m down)

A ship displaces 24300 t and has a KG of 5.60 m. Calculate üe fiıal KG if 180 t is loadedon deck at Kg 9.60 m.

(5,629 m)

A ship displaces 12260 t and has KG 6.68 m. A weight of 34 t is discharged from the lowerhold, Kg2.2 m. Calcüate the final KG.

(6.6e2 m)

A ship has an.iniüaı KG of 6.20 m. If the displacement is 23360 t, what.is the maırimı:maıııoıınt of cargo üat may be loaded on deck in a position Kg 10.40 m to ensrııe that the finalKG does not exceed 6.46 m?

, (1541.5 t)-

A ship displaces 12300 t and has KG 5.84 m. What is the maırimırm amorınt of cargo üatcan be discharged fronı the lower hold, Kg l.80 m, to ensııre ttıat ttıe finaı KG does notexceed-6.00 m?

(

5.

6.

, (468.6 t)

I

A ship displaces 2730 tonnes and has aKG of 6.00 m. The ship then loads the following:540 toruıes at 5.0 m above the keel;370 tonnes at 8.5 m above the keel;110 tonnes at 10.4 m above the keel;850 tonnes at 4.6m above the keel.

Calculate the'finalKG.(5.9j0 m)

7. A Ioadeütighter displaces 856 tonrıes and has a KG of l.5 m. Find üe new KG after thefollowing weigbts have been discharged:

160 tonnes from 2.5 m above the keel;40 tonnes from 3.7 m above üe kee[;

395 tonnes from 1.2 m above the keel.(1.00a m)

8. A ship leaves port with a üsplacement of 9060 tonnes and a KG of 5.2 m. During the

voyage üe following is consıımed:oilfuel: 200 torınes from 0.8 m above the keel;

320 tonnes from 0.7 m above the keel;Stores: 98 tonnes from 9.5 m above ttıe kee[;

FW: 87 tonnes from 10.0 m above the keel.What will be the KG on arrival at port of destination?

(5.377 m)

-ş$o

KG CALCULATI6NS (Formulae meüod and moments) - Tutorial Questions (Rev. 03/09/02)

Dec 2012


4285 tonnes, KG 6- O m. The following is loaded:ove the keel;ove the keel;ve üe keel.

ö"H: ffi fifr' of 6'5 m and a dısd.a9ement of 6020 tonnes. FindLoad: -- 50 ıg weights:

g5 bove the keel;

22 bove üe keel;Dischaıge: ;; bove the keeİ

700 toıuıes from 2.6fr;rş:fr:I:İ;

,r*.rrr"*:;;;:worked:Loadr

l62m;.56 m;

Discharge: 6 m;

Shift: 236 tonnes tom rğ 'o.,lİ""a position on deck Kg l2.2 m.

5.64 m. Cargo is worked as follows:

;

t

m..

.can-be loadecl on deck at Kg

|^ı

t0. Q.a59 n)

üe new KG after

(6.2s9 m)

ı2.00 m to ensrıre..

Q0r.8 t)

('ı

()

j cALcULATIoNs (Forııulae method and ınoments) _ Tutorial Questions (Rev. 03/09/02)

3ı

(, ,

Dec 2012


SOLUTIONS

1. GGv:wx d GGv:250x 16 GGv=0.182mdownward22000

2. GGv: w x d GGv: 180 x (9.60 - 5.60) GGv: 0'029 m up

W + ır 24300 + l80

Initiat KG 5.600CıGv (up) 0.029FINAL KG 5.629

3. GGv:wxcl GGv:34x(6.68'2'20\ GGv:0'012mupW-w 12260-34

/'- Lıitiaı KG 6.680I' GGv (uP) 0.012FINAL KG 6.592

4. Initial KG 6-200Fihal KG 6.460Moı. GGv (up) 0.260 .

GGv: w x d 0.260: w x (10.40 - 6..20) Solve the equation for wW+w 23360+w

0.260 x (23360* w) : w x (10.40 - 6.20)

6073.6 + 0.260w = 4.2w 6073.6= 4'2v'ı _ 0'26w

6073.6:3.94w w = 1541.5 tonnes

\ ). Inıtial KG- 5.840Final KG= 6.000Maır. GCıv (.up)

-== 0.160

GGv=wx d 0.160:wx (5.84 - 1.80) SolveüeequationforwW-w 12300-w

0.160 x (12300 - w) :w x (5.84 - 1'80)

1968 - 0.160w: 4.04w 1968 = 4'04w + 0'16w

1968:4.20w w=468.6tonnes

w

KG CALCULATI9NS (Formulae metlıod and moments) - Tutorial Questions (Rev. 03/09/02) 3L GF

Dec 2012


,-.i -"': l:f . ..,

: i*'' - - '

6. Take moınents abut üe keel.

7. Take moments about the keel.

Take moments'about the keel.

9. Take moments aboutthe keel.

lı

(

(

('

Itreight (t) Kg hilonenb (tm)

Itrİtisı Ditpl ?734 6.00 I63EOLOAI} 5{) 5.00 2700LOAI} 3TO E.50 3145LOAI} ıı0 10.40 t ı44LoAIı 850 4.60 39t0EıNAL 4fi{il} 5.99) 2.n79

w€ight (t) Kg l\ilomrnb (tm)

Inİtiül Dİsul 9060 5.20 471l2 \

DISCH -200 0.80 -160DISCH -320 0.70 -224DISCH -98 9.50 -931 i

DISCH -87 10.00. -870FINAL 8355 s.377 44927

Moments (tm)

KG CALCULATIONS (Formulae method aııd rnoments) _ Tutorial Questions (Rev. 03/09/02) 33*

Dec 2012


. u.r., .-.i r1;. :.. - , :

t.lrli,

Iu. Take popents aboutthe keel.I

(

(

'.-.Itu

ı 1. Take moments aboutttıe keel.

3+KG CALclJLATIoNs (Formulae ınethod and moments) _ Tutorial Questions (Rev.03/09/02) 6

Weight (t) Ks Moments (tm)

Initial Displ 6020 6.50 39130

LOAI) 500 2.50 1250

LOAI) 850 5.00 4250

LOAI) 220 8.40 r848

DISCH -330 5.50 -1815

DISCH -700 2.60 -1820

F'INAL- - 6s60_- 6.531 42843

Weight (t) Kg Moments (tm)

tnitial.Dİsnl 18940. 6.22-_ 117806.8:,

LOAI) 200 8.62 1724LOAI) 188 2.56 481.28

LOAI) 46 3.46 1s9.16

LOAI) 236 12.20 2879.2

DISCH -562 7.68 -4316.16

DISCH -236 4.20 -99t.2\

FINAL 18812 6.259 117743.08

Dec 2012


iİ

7-ıL2. Take moments about the keel, let x = cargo to load on deck at Kg t2.00 m. The fınal KG is \;

known which is 5.80 m.

Weİght (t) Ks Moments (tm)

Initial Displ r6420 5.64 92608.8

LOAI) 1500 6.50 9750LOAI) 1200 5.00 6000

LOAI) 900 4.20 3780LOAD 1000 8.20 8200DISCH -220 1.50 -330

Deck cargo x 12.00 lZxFINAL 20800 + x 5.800 120008.8 + 12x

Finat KG = Moments'Finaı Displacement

Therefore: 5.80: 120008.8 + l2x=20800 + x

5.80 x (20800 * x): 120008.8 +12x ı20640 + 5.8x: 120008.8 + l2x

120640 - 120008.8 : t2x- 5.8x 631.2:6.2x x = 101.8 t

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KG CALCULATIONS (Formulae method and rnoments) - Tutoriat Questions (Rev.03/09/02) 3s*,

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Dec 2012


n'?

--ASGOW COLLEGE OF NAUTICAL STTJDIESrl

I

1.

KG CALCT]LATIONS

A ship displaces 2730 awıes and has a KG of 6.00 m. She then loads the following:540 tonnes at 5.0 m above üe keel;370 toıınes at 8.5 m above the keel;110 tonnes at 10.4 m above the keel;850 tonııes at 4.6 m above the keel.

Calculate the fınal KG.(s.9i m)

A loaded lighter displaces 856 tonnes and has a KG of 1.5 m. Find üe new KG after thefollowing weights have been discharged:

160 tonnes from 2.5 m above üe keel;40 tonnes from 3.7 m above the keel;395 tonnes from- l.2mabove the keel.

2.

(-(1.00a n)

3. A ship leaves port with' a displacement of 9060 tonnes and a KG of 5.2 m. Dıuing.the,voyage the folloıuing is consıımed: I

Oil tuel:

Stores:FW:

200 tonnes from 0.8 m above the keel;320 tonnes from 0.7 m above the keel;98 tonnes from 9.5 m above the keel;..

t7 tonnes fiom l0.0 m above üe keel.What uıill be the KG on arrival at port of destination?

The original displacem.ent of a ship was 4285 tonnes, KGweights:

800 tonnes at 3.6m above the keel;440'tonnes'at'1.0 m above the keel;1 10 tonrıesat'S,8 m abovete keel;630 tonnesat.3.O mabove-the keel.

Find thenewKG.

A ship has o KG of 6.5 m and a displacement of 6020loading and dischargıng the following weights:Load: 500 tonnes at2.5 m above the keel;

850 toıınes at 5.0 m above the keel;220 torınes at 8.4 m above üe keel;

Discharge: 330 tonrıes from 5.5 m above the keel;700 tonnes from 2.6 m above üe keel;

6.0 m. She loads the following

(5.459 m)

Find the new KG after

(6.531m)

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4.

(

5.

MAR Rev l2ll2l00 3LE

Dec 2012


6. Find üe new KG of a lighter which has loaded and discharged üe following weights: (i

7.

Dischaıged:

Loaded:

IüAR/I(G cALcULATIoNs

tüARRev |2/I2l00

The original displacement and KG were 650 tonnes and 2.0 m respectively.(1.730 m)

The light displacement of a ship is 2875 tonnes. 390 tonnes is loaded 7.0 m above the keeland 7l0 torıııes at2.5 m above the keel. If the KG was then 5.2 m what was the light KG?

(s.622 m)

A ship displaces 18940 tonnes, KG 6.22 m. Calculate üe finat KG if the following cargo isworked:Load:

Discharge:shift:

140 tonnes from 2.5 m above the keel;270 tonnes from 1.4 m above the keel;215 tonnes at l.0 m above üe keel.

200 tonnes at Kg 8.62 m;188 tonnes at Kg 2.56 m;46 toruıes at Kg 3.46 m;562 toruıes from Kg 7.68 m; ( 'i

236 tonnes from Kg 4.20 m to a position on deck Kg 12.2 m.

(ı

9.

(6.259 m)

A ship displaces 16420 tonnes, KG 5.64 m. Caıgo is worked as follows:Load: 1500 t at Kg 6.50 m;

1200 t at Kg 5,00 m;900 at Kg 4.20 m;1000 t at Kg 8.20 m;.

Discharge: 220tfromKg 1.50 m.what is the maııimrım arıorınt of caıgo that can be loaded on deck at Kg 12.00 m to ensruethat the final KG does not exceed 5.80 m?

Q01.8 t)

A ship has an iniüal displacement of 14000 tonnes, KG 5.90 m. Cargo is then worked as

follows:Load: 220t at Kg 3.20 m;

86 t at Kg 8.00 m;-

Discharge: 500 t from Kg 6.20 m.-A weight of 146 tonnes is then shifted from the lower hold, Kg 2.20 m, to the upper deck,Kg 10.60 m.Calculate the maximıırı height at which a fınal weight of l00 toruıes may be loaded toensure that the fınaı KG does not exceed 5.96 m.

Q.61a m)

10.

3lş

l'ı

Dec 2012


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GLASGOW COLLEGE OF NAUTICAL STUDIES

KG. GM & SUSPENDED WEIGHT PROBLEMS INVOLVING WRTICAL COMPONENTS

1. A ship of 1200 tonnes displacemenq KG 3.1 m,loads cargo as follows:230tat aKg 5.3 m;420tat aKg 3.3 m;240tataKg3.l m.

The KM after loading is 3.7 m.(a) Calculate the KG of üe ship on completion of cargo.

(b) Calculate the final metacentric height.(3.382 n)

(0.318 m)

2. A vessel has a displacement of 32450 tonnes, KG 8.23 m, KM 8.97 m (constant) and is toload a quantity of deck cargo Kg 15.0 m. Calculate the quantity of deck cargo that can be'loaded so ttıat the GM sailing will be 0.4 m.

(17r5.9 t)

3. Tht folİowing information relates to a box-shaped.vessel fioating in dock rı,ater. RD-I.025i-.LBP 70 m,B 12 m, Depth 7 m,KG 4.2 m, BM 3.3 m, Draughts (even keel) 3.64 m.A ,weight of 76 tonnes, Kg T.4 m is to be lowered vertically to a posİtion in the lo'ıver holdwhere its Kg will be 2.5 m.Calculate üe GM of the vessel after the weighthas been moved.

Q.0an)

4. A vessel has a displacement of 14400 tonnes. KM 8.0 m; KG 7.28 m. A heavy lift of 85

tonnes, Kg 2.0 in is to be moved vertically upwards and re-stowed in the tween declq Kg 9.0

miThe vessel's own heavy lift derrick is to be used for the operaüon, wiü its head 20.0 m above

the keel.Calculate: (a) the minimum GM;

(0.614 m)

(b) the fınal GM.(0.679 m}

5. A vessel displacing 4500 tonnes has a constantKM 6.2 m and present KG 5.8 m. The vessel

loads 50 t of cargo on deck at Kg 8.5 m and then moves 100 t of cargo from the tween deck(Kg 5.5 m) to the lower hold (Kg 2.3 m), all weights being on the centre line.

Calculate ttıe final GM.(0.44 m)

6. A vessel is to use her oıvn derrick to move cargo already on boaıd,-the derrick head being 25

m above üe keel. 8 packing cases, each weighing 5 t are to be moved individually from aposition 3 m above the keel to a position 8 m above the keel.

The vessel's present displacement is 5000 tonnes, KM 7.0 m and KG 6.2 m.

Calculate: (a) the least GM

(b) üe fınal GMQ.7a3 n)

(0.76 m)

KC, GM & SUSPENDED WEIGHT PROBLEMS INVOLVINGz2toa0B)

VERTICAL COMPONENTS OF SHIFT OF G ONLY (Rcv. 1

3€

Dec 2012


fl

E.

Aıı upright vessel displaces 8,500 tonnes and has a KM of l0.2 m (assume coıstan$ -T9 1

KG oi g.s m. A 40 t weight is to be discharged fiom a position on the cente-line at a Kğ of

, 5.0 m using the vessel's own derrick. The derrick head is 25 m above the keel.

Calculate the GM when:(a) theweightis liftedjustclearofthetanktop.

(b) t}ıe weight is finally. discharged ashore.

A vessel displacing 9420 tonnes has a.KM of 9.0 m (assumed constant) and a KG of 8.5 m.

Weights are then loaded and discharged as follo*s:" --q 3000 tba!ıJ

_a's.i. dilcharged;

'

1400 t gargo ' Kg 4.5 m loaded;.. 2000 t cargo Kg I0.$ m loaded;

!000 t.fuel ' Kg 2.0 m loaded.

Calculate the fınal GM of thovessel (answer to three döcimal places).(0.s80 m)

A vessel is initially displacing 5000 t. KG 7.90 m; KM E.90 m (cons!ant). A 30 t weight is to

L" aı'"ır"'ged from a iosition on the cente_line, Kg 4.5 m, using the vessel's own deırick.

The denick head -is 30 m above the keel.

Calculate the GM wheıi:-(a)

ln" *rı*n is lifted just clear of it's initial stowage position;

(b) the weight is fınally discharged ashore.

(0.i06 n)

(0.377 m)

(0.847 m)

(0.9795 m)

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9

& SUSPENDED WEıGIIT PRoBLEMS INVoLVING

Zq t

KC, GM2A0U03)

VERTICAL COMPONENTS OF SHIFT OF G ONLY (RCV.

Dec 2012


iIGLASGOW COLLEGE OF NAUTICAL STUDIES

.lı.t

; SHIFT OF G/KG PROBLEMS

SECnoN a: SHIFT o{ G

(

1. Calculate the shift of G which results fiom loading 500 tonnes of cargo at a height of 3 m'above the c'entre bf graviğ of the ship, if the initial displacementtis 7000 tonnes. Is this a rise

a fall in KG? Is it an increase or a decrease in GM?i . (0.2 m)

Z. A ship displacing 1i000 tonnes has KG 7.15 m and KM 8.0 m before discharging 1000

. ' tonnes of cargo. from Kg 2.15 m. Calculate the new GM. Could the ship sail safely with üıis'.-lalu-e-bTcM?--- t l-

. e.35 n)

3. A ship displacing 5400 tonnes has a GM of 0.25 m before taking on 600 tonne! of bumkers at

Kg 0.5 m, when the GM rises to 0.85 m.

. What must have been the original KG?(6.s m)

4. A.ship has displac"-.nt 15550 tonnes and GM 0.65 m. 500 tonnes of cargo is transferred''' fromNo.. ZLP^to the foredeck (a vertical rise of 8.25 m): What is the new GM?(0.385 n)

. ,ErcSECTIONB:SIMPLEKG .. . . ,

A ship has displacement |24oo tonnes KG 7.24 m on aırival at a port where Fgo ı, worked

as foliows: \

Discharged 1000 tonnes fiom Kg 8.12 m;

Discharged 1500 tonnes frorn Kg 4.56 m;

T-aaded'2200 toruıbs at Kg 6.42 m.Calculatç üe KG on departııre.

(7.3sm)

W1iat,change.in KG,will occuria!ı a result.of üe following,work taking place aboard a ship

displacing 12000 tonnes andwithKG of 6.E7 m?=

Loaded 1525 tonnes at Kg 8.23 m;Loaded 876 tonnes at Kg 3.69 m?

f.. the following information calculate the KG on completion of loading:'

Length of vessel 80 m; Breadü 1l.7 m; Block coefficient 0.8.

Draught of vessgl: 4 m in water of RD 1.024.Initial GM 0:84 m. KM 5.8 m (assume constant).

Cargo to load 170 tonnes atKg 6.2 m.

(Scn drop)

(5.025 m)

From the following information c.alculate the quantity of deck cargo to load so that the vessel

will sailwith a.GM of 0.5 m;

Displacemönt priorto loading deck caıgo 22450 tonnes.

Initial KG 8.37 m.Kg of üeck cargo 15.0 m. KM 8.97 m constant.

(34s.8 0

5.

6

7.

8.

SHIFT OF G/KG PROBLEMS (MAR Rcv' 23.rcA0E) 4o.:l'B

Dec 2012


g. From the following inforrıation calculate üe quantity of deck caıgo to load so that üıe vessel 1^ ı

will sail wiüı a GM of 0.42 m:Displacement of vessel prior to loading: 4200 tonnes, KG 5.85 mCargo to load below deck 6200 tonnes Kg 5.95 m.Kg of deck cargo9.7 m. Assııme KM 6.5 m constant.

(489.s 0

10. From the following details calculate üe GM of üe vessel at the time of sailing:Displacement prior to commencing cargo 8000 tonnes; KG 7.0 m.Cargo to discharge: 1000 tonııes, Kg 5.5 m;Cargo to load: 900 tonnes, Kg 3.0 m;Cargo to load: 600 tonnes, Kg 5.0 m.Assume KM7.25 m constant.

(0.6a m)

I l. From the following details calculate the weight of cargo to load so that the vessel will sailwith a GM of 0.35 m:Displacementpriorto loading 1800 tonnes KG t.3 m. ( I

Cargo loaded: 800 tonnes at Kg 4.88 m; I

Space available forremaining cargo-Kg 6.8 m.Assume KM constanttbroughout at 5.0 m.

(207:4 t)

12. From the following information:(a) Calculate the GM on completion of loading iÇ prior to loading the displacement was

5400 tonnes KG 4.5 m, cargo-loaded was 550 tonnes at Kg 4.3 m and.-420 tonnes-at:K95.7 m, KM 4.85m and free surface effect is 0.15 m.

Q.Ia m)

(b) Is this a safe GM? Give reasons for your answer.

(,

SHIFT OF G/KG PROBLEMS (It ARRc,v.23l0t/03) *ı g

(ı

Dec 2012


(

2.

J.


SEIFT OF G. KG AND GM

'Single weİght'formulae to be ıısefor each questionAssume KIı,I to be constaııt in each question.

1. Calculate the effect of taıısferring250 tonnes of cargo from deck stowage to lower holdstowage (aveıtical shift of 16 m) in a ship displacing 20,000 tonnes.

(0.20 n doıı,n)

A l2O tonne heary lift resting initially at a Kg of 2.5 m is to be restowed on deck at Kg 12.5m. What rise of G will occur as a resulğ given üat the ship's displacement is 24000 tonnes.

(0.05 mııp)

If the lift in the previous question is to be restoıred using the ship's heavy denich üıe head ofwhich is 25 m above the keel, what will be the effect on the GM when the load is lifted justclear of the deck of the hold?

(0.1125 mdeuease)

What will be the effect orı G of traıısı'erriııg8O'torırıesof cargofrorıtNo:3'TD'(Kg l0 m;6 m-:to port of CL) to No.3 LH(Kg 2 m;.4 m to starboard of CL) in a ship of displacement 16000.tonnes?

(0.04 m down: 0.05 m to stbd)

A ship displacing 12100 tonnes has KG 7.0 m and KM 7.8 m. 150 tonnes of cargo is loadedatKgl2 m. Calculatethe new GM.

(0.73e n)

A ship has KG 7.12m and KM 7.79 m while displacing 8250 tonnes. What will bethemetacentic height after 200 tonnes of cargo is loaded at Kg 2.19 m?

(0.787 n)

Where should the cargo in question 6 have been loaded if the final GM was required to be 75cms?

(K9.3,740 m)=

In a vessel dispJacing 90O0-tonnes'with KG 5.75 m, KM 7.90 m, 500 tonnes of cargo isdischarged frorn Kg 2.5 m. What is the final GM?

(0.900 n)

A ship displaces 12200 tonnes with KG 7.22 m and KM 8.00 m. 700 tonnes aıe offJoadedfiom K94.5 m. Findthe GM.

(0.614 m)

A ship has displacement 9250 tonnes, KG 7.4 m, KM 8.0 m. From what height above thekeel should 500 tonnes be discharged in order to complete with a GM of 70cm?

(9.15 m)

4.

(

5.

6.

7.

8

9.

10.

Shift of G, KG and GM (MAR Rev. 2UOA03),

4>

Dec 2012


GLASGOW COLLEGE OF NAUTICAL STT.'DTJES

i)KG/GM PROBLEMS INCLUDING FREE SITRFACE EF 'ECT

I

t . Calculate the GM of a vessel on aırival at the discharging port from the foltowing details:Prior to loading displacement 6050 tonnes KG 5.65 m.

2

Bunkers used on passage 250 tonnes Kg 0.9 m.KM on arrival 7.00 m.

(0.96 m)

Calculate the amount of deck cargo to load in port 'A' so that the vessel will arrive in port 'B'with aGMof 0.5 m.At port'A', prior to loading deck cargo, displacement is 11800 t KG 7.90 m. Estimatedconsumption on passage:

Cargo to loadCargo to loadBunkers to loadStores to load

I 15 tonnes fuel155 tonnes fuel85 torınes F'W-25 tonnes stores

2606 tonnes Kg 4.03 m;I100 tonnes Kg 8.15 m;600 tonnes Kg 1.00 m;250 tonnes Kg 9.95 m.

Kg 1.10 m;Kg 6.60 m;Kg 7.88 m;Kg 10.0 m.

(l

4.

5.

Position of deck cargo: Kg 12.80 mAssume KM 8.6 m remains constant.

', (287.5 t)

Assuming üat a certain timber cargo will absorb l0% of its own'_weight in moistııre during avoyage calculate üe amount of timber to load on deck at Kg 8.58 m in port 'A' so üat üevessel will reach port'B' with an effective GM of 0.35m. i

I

Atport'A' priortotoading deck cargo, displacement is9275tKG 5.50 m. \

Estimated consumption on passage I 12 tonnes fuel and water from Kg 0.75 d, free surfaceeffectcaused is 0.114 m.ArrivalKM 6.ll m.

Qso t)

Calculate üe change,in transverse metacentric height of a box shaped vessel of 15 m breadthi,-and a salt water draught of 8 m on passing from wateç of relative densiğ l.025 to water ofrelative density 1.000.

(increase 0.43 m)

A vessel KG 4.0 m displacing 8250 tonnes has a righting lever of 0.2 m at 100 heel. Onpassage a rectangular DB tank containing 120 tonnes of fuel Kg 0.5 m is half consumedresulting in a fiee surface effect of 0.12 m.Calculate üe righting moment in the arrival condition for a heel of l0o.

(1433.6 t-m)

çKG/GM PROBLEMS INCLIJDING FREE suRrAcE ErrpcT (İı{AR Rcv. 23108/03) 4g

(i

Dec 2012


::'1-,i:ti:.'İ':' .' :''.

. :;..,,

6. (a) Show clearly that if a tank is subdivided by a longifudinal bulkhead into two eqııalparts the loss of metacentic height due to the free surface effect will be reduced to aquarter of that of the undivided tank.

(b) A vessel of 8000 tonnes displacement in salt water has a double bottom tank eqııallysubdivided into trree parts by tıı,o longifudinal bulkheads. The overall dimensions ofthe tank are l5.0 m long and 21 m in breadth. Find the virtııal loss of GM if tlıe tanksare half full of tuel oil ofRD 0.95.

(0.153 n)

7. (a) The effect on stability at sea of free surface in a slack tank is governed by:(a) the vessel's displacement

and (b) the relative densiŞ of the liquid in üe tank. Explain cleaıly why this is

((b) The free surface moment of a double bottom tank for liquids of relative density 1.0 is

350 tonnes-metres.Calculate the reduction:in the GM of a vessel whose displacement is 2552 tonneswhen the tank is partially filled with fuel oil of relative density 0.875.

(0.12 n)

t

?i:.,:

KG/GM PROBLEMS INCLUDING FREE SLJRFACE EFFECT (MAR Rev. 23108/03) 4+E

Dec 2012


(ıii

GTASGOW COLLEGE OF NAUTICAL STTJDIES

KG & GM: SUSPENDED WEIGIITS

l. A vessel has a displacement of 14400 tonnes. KM 8.0 m, KG 7.28 m. A heaıry liftof 85 tonnes, Kg 2.0m, is to be moved vertically upwards and restowed in thetween declç Kg 9.0 m. The vessel's own heavy lift derrick is used for üeoperation, with its head 20.0 m above the keel.Calculate:(a) the minimum GM;(b) the final GM.

(0.614 n; 0.679 m)

2. (a) List the precautions to be taken before discharging a heavy lift using thevessells own derrick.-.

O) An upright-vessel-displaces,.8500 tonnes and has a KM of 10.2 m (assumeconstant) and a KG of 9.8 m. A 40 tonnes weight is to be discharged from' aposition on the centreline.at a Kg of 5.0 m, using üe vessel's owır denick. Thederrick head is 25.0 m above the keel.Calculate-the GMwhen:

(ı) üe weight is lifted just clear of the tank top;(ii) the weight-is finally discharged ashore.

(0.306 m; 0.377 m)

3. A vessel is to use her own derrick to move cargo already on board, the derrickhead being 25 m above the keel. 8 packing cases, each weighing 5 tonnes, are tobe moved individually from a position 3 m above üe keel to a position 8 m abovethe keel.The vessells present displacement is 5000 tonnes.KM: 7.0 m KG: 6.2 mCalculate:,.(a) the least GM;(b) the final GM.

(0.743 m; 0.760 m)

4. A vessel is initially displacing 5,000 t. KG = 7.90 m; KM = 8:90 m (constant).A 30 t weight is to be discharged from a position on the centreline, Kg 4.50 m,

using the vessel's own derrick. The derrick head is 30 m above the keel.

Calculate the GM when:(a) the weight is lifted just clear of its initial stowage position;(b) the weight is fınally discharged ashore.

(0.847 m; 0.975 m)

()

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MARRev l2JlA00

1ç

(ı

Dec 2012


(

(

5 A ship displaces 9862 \KG 5.32 m and is to dischar ge a364t weight from thelower hold, Kg 2.|2 m, using the ship's heavf lift crarıe. Calculate the maııimumheight of the crane head permissible to ensure that the maııimum KG during theoperation does not exceed 6.00 m.

Q0.5a4 n)

A ship displaces 16220 t, KG 6.34 m and is to discharge a220 t weight from thelower hold, Kg 3.00 m, using the ship's heaw lift craııe. Calculate the moıimumheight of the crane head permissible to ensure that the maırimum KG during theoperation does not exceed 6.64 m.

(25.118 m)

A heary lift barge displaces 12000 t and has a KG of 6.42 m. A load of 388 t is tobe discharged from a position Kg 4.60 m. If the KM for the current condition is7.44 m, calculate the maximum pemıissible height of the barge's crane-head toensure üat the GM does not fall below 0.400 m during the lift.

(23.775 m)

TvIAR/KG & GM:SUSPENDED WEIGHTS

7.

MARRev l2ll?J00

+6

Dec 2012


GLASGOW COLLEGE OF NAUTICAL STTJDIES

CALCULATING GM & SUSPENDED WEIGHTS1. A ship displaces 12000 tonnes, KG 3.1 m, and loads as follows:

230tatKg5.3 m;420tatKg3.3 m;240taİKg3.l m.

Given that üe KM for final displacement is 3.70 m, calculate:

(a) the KG of the ship on compleüon of cargo;

(b) the final metacentric height.(0.593 m)

Z. A ship displaces 4500 tonnes, KG 5.8 m. 50 tonnes of cargo is loaded on deck at Kg8.5 rnand then 100 torırres is moved from the tween deck (Kg 5.5 m) to the lower hold(Kg2.3 m), all weights being on üe centre-line. Calcüate the final GM assuming that

the KIvI is constaııt at6.2 m.Q.aal m)

3. A ship displaces 9420 tonnes, KG 8.5 m. Weights are loaded and discharged as

follows:Dischaıge: 3000 t of ballast from Kg 5'3 m;

1400 t of caıgo at Kg 4.5 m;

2000 t of cargo at Kg 10.4 m;10001of fuel atKg 2.0 m.

Calculate the final GM if the KM is assumed to be:constant at 9.0 m'(0:j80 m)

4. A box_shaped vessel has length 70 m, breadü !2 m, depü 7 m, KG 4-2 m arıd BM3.3 m. It is floating at an even keel draught of 3.64 m in salt water. A weigbt of 76

tonnes, Kg7.4m is to be lowered to a position in the lower hold, Kg 2,5 m. Calculate

the GM after the weight has been moved.(1.03e m)

5. A ship displaces 32450.tonnesr=KG 8.23.m and.KM 8^97 m (constant) and is to-load=-

deck iargo at Kg 15.0 m. Calculate the amoıınt of cargo to load on deck to ensure that

the sailing GM is 0.40 m. (I7IS.' t)

6. A ship initiauy displaces 5000 t, KG 7.90 m, KM 8.90 m (assume constant). A 30

tonne weight is to üe discharged from a position on the centreJine, Kg 4.5 m, ısingthe ship's own deırick. If the head of ıe deırick is to be 30 m above üe keel dııring

the operation, calculate the GM when:

(a) the weight is üfted just cleaı of it's initid stowage position;

oı the weight is finally discharged ashore'(0.847m, 0.975 m)

t\

(1

4ı

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Dec 2012


. -jı'.-..a ]...'j-, . r.1'l 1l

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::fi;i

A ship dişlaces ı4400 tonnes, KG 7.28 m, KM 8.0 m. A heaıy lift of 85 tonnes, Kg2.0 m is to be restowed in the tween deck in a position Kg 9.0 m, The ship's ownheavy lift crane is to be used, it's head being 20 m above the keel. Calculate:(a) ftg minimum ğ\ıl;(b) the final GM.

(0.614n 0.679n)

A ship mııst ııse it's orı,n denick to move cargo alıeady on board, üe head of thederrick being 25 m above üe keel dııring üe operation 8 packing cases, eachweighing 5 tonııes are to be moved indiüduatly from a posiüon Kg 3.0 m to aposition Kg 8.0 m. If the ships cıırrent displacement is 5000 tonnes, KG 6.2 m andKM 7.0 m, calculate:(a) the least GM;(b) üe final GM.

(0.743 m,0.760 m)

An upright ship displaces-,8s00 tonnes:and has a K]vt of l0.2 m (assııme constant),,=

and a of KG 9.8 m. A 40 tonne weight is to be discharged from a position on thecente-line at a Kg of,5.0 m using the,.shiplıown*crane*-thç_head_of which iı25,m-*aboye üe keel. Calculate.the_GM -when:= -

(a), the weight is liftedjustpleaıof the-tanktop;(b) the weight is finally discharged ashore.

(0.306m,0,377m)

A heavy lift ship displaces.l6040 tonnes and-has KG'4.22 m and KIvI 5,14 m. A lift of640 tonnes is to be dischaıged using the ship's two heavy lift cranes from a positionon deck Kg 4.64 m. Calculate üe maıdmıım perrıissible height of the heads of bothcranes dı:ring üe lifting operation to ensı:re that the GM does not fall below 0.40 m.

'.:.:tgtk

7.

8.

9.

10.

(1 7.67 m)

MAR/GM & SUSPEI.{DED WEIGHTS.Tutorial

48

Dec 2012


(

GLASGOW COLLEGE OF NAUTICAI STIJDIES

LIST (OOW LEVEL)

List caused b}ı a shift ofweİght1. A ship has displacement 3500 t; KM 3.8 m and KG 3.15 m. Calculate the list caused

by shifting 15 tonnes athwartships through a distance of 14 m.(s.3")

2. A ship has displacement 6500 t and metacentric height 0.55 m. Find the list producedby shifting 30 t through 6 m transversely in the ship.

(2.88)

3. A ship of 1000 t displacement has a weight of 5 t moved 4.6 m across her hold. If sheis initially upright with a GM of 0.305 m, find the resultant list.

(4.31)

4. An upright vessel displacing 11850 t has KM 8.4 m and KG 7.3 m. A locomotiveweighing 150 t is loaded on deck at Kg 14.8 m, 6.5 m off the centreline. Calculate theangle of list.

(4.62)

5. Aı upright vessel has a displacement of 15500 q KG 8.1 m and GM 0.7 m. 500 t oftween deck cargo is discharged from Kg 10.4 m at a distance of 2.4 m from thecentreline. Calculate the list caused.

(s.88)

6. A ship of 4000 t displacement, KG 7.4 m, KM 7.51 m is upright. The followingchanges in loading then take place:

8030 t of cargo loaded at Kg 6.35 m on the CL;530 t of bunkers loaded at Kg 7.68 m on the CL;85 t machinery loaded at Kg 14.lm, 5.8 m to starboard of CL;80 t ballast loaded at Kg 1.0 m, 3.8 m to port of CL.

Calculate ttıe angle of list and direction-(1.1" Stbd)

7 . A vessel of 15500 t displacement, KG 6.0 m, is listed 3" 50' to port. Calculate thefınal angle of list if she loads üe following:

250 t Kg 6.0 m, 4.0 m to port of CL;400 t Kg 7 .0 m,5.0 m to stbd. of CL;600 t Kg 5.0 m, 3.0 m to port of CL;500 t Kg 9.0 m, 6.0 m to stbd. of CL.

Assume KM 7.50 m constant.(1.3'Stbd)

8- A vessel of 15000 t displacement loads:600 t at Kg 4.0 m, 11.0 m to port of CL;250 tatKg 6.0 m, 8.0 m to stbd. of CL;

and discharges:350 t at Kg 8.0 m, 9.0 m to port of CL;450 t at Kg 5.0 m, 4.0 m to stbd. of CL.

Calculate the angle of list if the vessel was initially upright with KG 5.0 m. FinalKM:6.423 m.

(8.06" Port)(

LIST (OOW LEVEL) (MAR Rev. 24/08103

Dec 2012


t

i 1.

12.

9. A ship of displacement 7500 t, KM 7.25 m, KG 6.5 m is upright before the followingchanges occur:

Cargo discharged: 600 ! Kg 4.4m,3.Zmto stbd. of CL;Cargo loaded: 100 t Kg 1.8 m, 5.0 m to port of CL;

940t, Kg 4-9 m,1.9 m to stbd. of CL.

Assume KM is constant. Calculate the angle of list when the ship completes cargo.(5.4. port)

10. A ship of 8000 t displacement is listed 6o to starboard. Calculate the weight of cargoto load in the port tween deck if Kg of the cargo is to be 8.0 m, the space availablels5 m to port of the CL, KM is 5.5 m and the KG is initially 4.5m.The ship is tocomplete loading upright.

(t68.I6 t)

A vessel of 4000 t displacement, KM 4.25 m, KG 3.5 m, is listed 3.5o to starboard. If600 t of cargo are still to be loaded and there is space in No.3 TD (Kg 6.5 m and 5.5m to port of cL) and in No.4 LH (Kg 3.5 m and 4.0 m to stbd. of cL); calculate theweight to stow in each space to ensure that the vessel completes loading upright.

(271.9 t Port; 328.1 t Stbd.)

A vessel of 500 t displacement, KG 3.5 m, KM 4.0 m, is listed 4" to port. If 20 t ofcargo is stillto be loaded atKg 5.0 m,4.0 m to port of the CL, calculate howtheremaining cargo of 50 t should be stowed to bring the vessel upright. The availablespaces are atKg 3.0 m, 3.0 m to port of CL and at Kg 5.5 m,4.5 m to stbd. of CL.

(17 t Port; 33 t Stbd.)

HEAW LIFT DERNCK1. A ship of 10000 t displacemen! KG 8.0 m, KM 9.0 m, is upright and berthed

starboard side to the quay. A load of 60 t stowed on deck at Kg 10.0 m and 6.5 m offthe CL to port is to be discharged by the ship's heavy lift derrick, the head of which is16 m above the keel. The load is to be landed on the quay 11 m out from the cL.Calculate the maximum angle of heel during the operation and the final angle of listafter discharge. Assume KM constant.

(6.22o Stbd; 2.21" Stbd.)

A ship of 10000 t displacement, KM 8.1 m (constant) and KG 6.8 m has to load 2 x50 t loads on deck, Kg 10.5 m, and 7.1 m off the cL to port and starboard. The headof the ship's derrick lifting the loads is 20 m above the keel and has an outreach of10.5 m. Calculate the maximum angle of list during the operation if the inshore loadis placed aboard fırst.

(4.1)

A vessel of 4000 t displacement heels 10o when a load of 30 t is lifted from a quay bya ship's derrick. If üe GM before lifting the load is 0.75 m, KM is constant at7.75 mand the derrick head is 20.5 m above the keel, calculate the outreach ofthe derrick

(1s.396 m)

A vessel of 2000 t displacemen! GM 0.6 m, lifts a 15 t piece of machinery with herheavy lift deırick from the lower hold (I(e 2.0 m) and lands it in a barge alongside.Kg of the derrick head is 14 m aıd the oukeach to the barge is 8.0 m. Find themaximum list.

(6.7)

(ı\2

3.

4.

LIST (ooW LEVEL) (MAR Rev.24/08/03

Dec 2012


NOTES ON

SHIP STABILITYFOR PHASE s/CLASS 4 POST/OOW

LO-4

ÜİJİ/'f fofuaıooqft4 PoİT ı'uJ -€QYİ

Gla S gg,H,',Çg,J,gge

Dec 2012


LONGITUDINAL STABILITY (1)

TrimI-s_the difference in cms between the forward and aft draughts,as measured at the forward and aft perpendicularsresoectivelv.

The ship above has draughts F 2.20 m A 2.68 m.The trim of the ship is: 2.68 -

2.200.48 m by the stern;

48 cms by the stern.

The ship above has draughts F 2.70 m A 2.32 m.

The trim of the ship is: 2.70 -

2.320.38 m by the head;

38 cms by the head.

Longitudinal Stability(MAR Rev 03/01/01)

Dec 2012


Loneitudinal centre of eravitv (LCG)['-ihç -posi1ion of the ship's centre of gravity relative to the

length of the Ship, Terqıgd Grin djagrpmş. ] i ' ' -5i--: l-ı ''i "/^:

!:ly----'-'-'| -'. ;{ 1 t i 1{-' ''ı';1r'' ii} '- j

Lonsitudinal centre of buo ancv (LCBIs the position of the ship'slength of the ship. Is termed

: :) ( .''r(( 'i I ';; l '

Longitudinal MetacentreIS the point of intersection of the lines of action of buoyancyforce acting through the LCB when the ship is in the initialeven keel condition and subsequently trimmed conditions. Istermed M, in diagrams.

Loneitudinal Metacentric heieht (GMr)Is the vertical difference between the centre of gravity and thelongitudinal metacentre. Termed GM, in diagrams.

Consider the ship shown.

centre of buoyancy relative to theB, in diagrams. '' ü

1'


2

Dec 2012


Consider a ship initially on even keel. A weight already on

board is moved aft through'd' metres. This causes Gı, to move

to Grr.G,G,,-wxd

w


Dec 2012


Rearranging this gives: GrG, x \tr/ - w x d

Trimming moment : G,Grı X W : w x d t-(

The ship trims until both LCB and LCG are in the same verticalagain.

Chanee of trim (COT)Is the difference in cms between the trim in the initial conditionand the trim in the final condition.

SAOA ship has the following initial druughts:

F 6.00 m A 5.86 mundJinul draughts: F 5.66 m A 6.20 mafter cürgo on bourd is shifted.Calculate the change of trim that has occıırred.


4

Dec 2012


Answer

Initiul druughts:F 6.00 m A 5.86 m: Trim - 0.14 m bY HEADFinal draughts:F 5.66 m A 6.20 m: Trim - 0.54 m bv STERNTherefore: Chanse of trim - 0.68 m by STER {

:68 cms by STERJY

Moment to chanse trim bv one centimetre (MCTCThis is the trimming moment (w x d) required to change the

ship;s trim by exactly 1 cm. It is tabulated in the ship'shydrost atic particulars and used to determine the change in trimwhen -cargo is either shiffid, loaded or discharged.

COT (cms) : fv t d : Trimming momentMCTC MCTC

SAO

A weight of 150 tonnes is moved uft

the MCTC for the current draught

Jinat trim of the ship if the initialsteFn.

by a distance of 20 m. Ifis 250 t-m determine thetrim was 0.20 m bY the


Dec 2012


Answer

COT(cm{:wxd :150x20 :12cmsMCTC

COT - 0.120 m

250

Initial trim: 0.200 m by STERIVCOT: 0.120 m-further bv the STERIYFINAL TRIM 0.320 m bv STERN

Longifudinal centre of flotation (LCF or F)Is the geometric centre of the ship's water-planeparticular draught and is the point about which thetrim.It'ş positiqn will change with d1aught.

area at aship will

Longitudinal Stability(MAR Rev 03i01/01)

6

Dec 2012


The position of the LCF determines how the change(COT) will be apportioned between the forward

draughts.ir l |; i\ 1 \,, ,ı.,.)'

Ship with LCF amidships

of trimand aft

If LCF amidships then: Ta : Tf: COT2

where: Ta : change of draught aft due to trim; andTf - change of draught forward due to trim.

SAOA ship flouts ut draughts F 6.50 m und A 6.80 m. Determinethe linal draughts if 25 tonnes is moved 45 m forward given

that MCTC is 172.5 t-m and the LCF is umidships.


7

Dec 2012


AnswerCoT= wxd : 25x45 =10cms -0.100m ( -:ı...-,

MCTC 112.5

Ta : Tf - 0.100 -- +/- 0.050 m2

Weight is moved forward so the ship will trim by the HEAD.

Initiul draughts F 6.500 A 6.900Tfhn- :, ".! ı'^;t .['i | ; J/F l + 0.050 - 0.050FINAL F6.550m A6.750m


Dec 2012


Ship with LCF not amidshipsIn-Ihi-S öase ihö chanğe of trlm (CoT) wi|l have to_ be

ifurnoiid to i6e foiward and aft draughts according to the

position of the LCF within the ship's length.

If the similar triangles are considered then:

a: f andTa*Tf:COT {:'

Ta Tf

Therefore: Ta: ,a x COTLBP

and Tf: f x COTLBP n..

'' '' ııı'{ '''1f

|-rıı:{ ,'"i,, a'

SAOA ship hus initial druughts F 10.25 m and A 10.15 m. Aweight of 95 tonnes is moved aft through a distance of 42 m.

Culculate the Jİnal druughts given that LBP is 100 m, LCF is

48 mfoap und MCTC is 285 t-m-

Longitudinal Stability 9

(MAR Rev 03/01/01) {

Dec 2012


AnswerCOT= wxd:95x42 :74cms

MCTC 285Tu: 48 x 74 : 6.7 cms - 0.067 m

100TT= 52 x 74 = 7.3 cms - 0.073 m

100Weight is moved aft so the ship wiII trim by the STERN.

Initial draughts F 10.250 A 10.150Trim - 0. 073 + 0.067FIIYAL F70.177m A10.217m

Longitudinal Stability(MAR Rev 03/01i01)

10

Dec 2012


The effect of loadine and dischareine weishts

T. effect of bodily İinkuge/rise must be taken into account:

Sinkage/Rise cms _ wTPC

The following procedure should be followed:1. Loadldischarge the weight from the LCF, calculating the

sinkage/rise using the TPC value given.2. Calculate the COT by moving the weight from the LCF

position to it's actual loaded/discharged position.3. Find TalTf by apportioning the COT according to the

position of the LCF.4. Apply both the sinkage/rise and TalTf to the initial

draughts to determine the final draughts.

SAO

A ship 100 m in length floats ut draughts F 7.00 m und A 6.80

m. Calculate the final druughts if 150 t is louded 20 m foupgiven that TPC is 15 und MCTC is 150 t-m and LCF is 45 m

foap.

TIP

Alwuys draw a sketch to help you picture what is happening!


11

Dec 2012


150 t

45m

ü

AnswerSİnkage: w : 150 : 10 cms - 0.100 m

TPC 15

COT: wxd _IS0x(45_201 :25cmsMCTC 150

Ta:45 x 25: 11.25 cms - 0.113 m100

Tf : 55 x 25 = 13.75 cms - 0.137 m100

Weight is louded aft of the LCF so the ship wiII trim by theSTER]Y.

Initial druughts F 7.000 A 6.800Sinkage

Trim

+ 0.100 + 0.1007.1 00 6.900

- 0.137 + 0.113FIIYAL F6.963m A7.013m


T2

Dec 2012


Loadine/dischareine multiple weishtsA tabular approach needs to be adopted where moments aretaken about the LCF.

Consider the following example:

A ship 120 m in length floats at druughts F 6.24 m and A 6.36m. LCF is 54 mfoap, TPC 14.2 and MCTC 116 t-m.

The following cürgo is worked:Load 120 t lcg 10.0 mfoup;Load 68 t lcg 86 m foap;Dischurge 36 t lcg 22 mfoup;Dischurge 48 t lcg 60 m foap.

Calculate the Jİnal draughts.


13

Dec 2012


TRıM - MULTıPLE

Enter data

LBP = 120 mLCF = 54 m foap

TPC 14.2MCTC 116 t-m

Weight (t) Dist from LCF

Sinkage = gTPC

Sinkage = 7.3 COT = Trimming momentMCTC= 0.073 m

COT = 19.3 cms

Ta = 8.7Tf = 10.6 cms = 0.106 m

6.313-0.106


t4

Dec 2012


Most trim problems are straight forward provided that youunderstand the information that is being given and can recognisethe formula to which it belongs.

Sinkage/Rise cms : wTPC

COTcms: wxd - TrimmingmomentMCTC MCTC

Apportion COT to forward and aft draughts using:

Ta: a x COT and Tf: f x COTLBP LBP

NOTEIn practice, the mean value of TPC must be used to determine

the sinkage/rise of the ship. Similarly, the meün values ofMCTC and LCF must be used when calculating the change oftrim. The change of trim is then apportioned to the final

waterline using the Jinul LCF. If a 'hydrostatic particulqrs'table is not given, then it has to be assumed that the values ofTPC, MCTC and LCF position do not significantly change i.e.

they remain constant for the range of draughts concerned.


15

Dec 2012


GLASGOW COLLEGE OF NAIJTICAL STI.JDIES

TEE TRIIE MEAN DRAUGET : DISPI,ACEMENT WEEN OUT OF TRIMA stıip's displacement scale is usually drawn up for even keel conditions. If the ship has a

rcentre of flotation otıidships, such a displacement scale will always give the correctdisplacernent when referred to with üe arıidshiPs mean draııght, regardless of the ship'strim. However, the centre of flotation is seldom arıidships. When this is the case, thedisplacement given by the scale for the (amidships) mean draııght will only be correct ifthe ship is on an even keel. If the ship is trimmed üe'amidshiPs meao draııght is not a truemeasure of displacement on the scde.

The sketch shows a ship on an even keel waterline WıLı for which the displacement is Wtonnes. F is abaft of amidships. If a weight is moved aft, without change of displacement,the ship rotates about F to some new waterline such as W2L2. The draught amidshipschanges (reduces in the case shown). If the displacement scale is now referred to with theneıı, atidships mean draught the value obtained wilı be the displacement up to the evenkeel waterline WıLı. This is less than the true displace,rıent CüD bv the weight of the layerbetween WıLı and WıLı. The draııght at F (if referred to the displacement scale) will ğvethe correct displacement ıınd for this reason the draııght at F is termed the Tıııe MeanDraaght @,D). ı-

l--e-Jw,

V,

}J3, 3

2

ı.L

L

The displacement scale should be referred to with the draught at F and this draught iseither'.

i) draught forward + trim between forward end and F, or;

ii) draught amidships + trim between amidsiıips anti F.

ie. True mean &oııght @,ID) : dıaughtfwd + (J_ x Trim) or;L

Tııe mean &aught @,ID) : fuaughı amidships + ( d x Trim)L

ı

o(ı

I

r t,I

tr rİ"r,yeı-Fıı,r Ev€ıı KEELIL

Tr^o lş* L

Dec 2012


Alternatively, the true displacement is the displacement given by the scalg if entered withthe amidships mean draııgfrt,plus the weight ofthe layer between WıLı and w3L3 (shownshaded). This is termed the layer conection The displacerıent of the layer correction isgiverı by:

w(tonnes) : x x TPC

x Tim) x TPC

The true mean draught (at F) is greater than the arithmetic mean draught (at arıidships) if(i) F is abaft of amidships and the ship is trimmed by the stern or (ü) F is forward ofamidships and the ship is trimmed by the head; and layer corrections are then additive(see sketches below).

These layer (shown shaded) ueadditive to the mean draught

The true mean draught (at F) is /ess than the arithmetic mean (at amidship$ if (i) F is abaftof amidships and the ship is trimmed by head, or (ü) F is forward of amidships and theship is trimmed by sterıı, and layer coıTedions are then sııbtractive (see sketches below).

: (-d-

L

ücorrections

üdisplacement.

! coıgcr €.(.

ıI

ıI

*

These layer corrections (shown shaded) are subtracfve from the mean draughtdisplacement.MAR\TMD.DOC

Dec 2012


(

a

'l


TRIJE MEA_ıI DRAUGHT AND DISPLACEMENTl. A ship length LBP l22 m has a displacement 10000 tonnes on an even keeı

draııght 7.62 nı TPc l9.7, LCF is 4.6 m abaft arıidships. Find the displacementat a draught 7.01 m Fwd 8.23 m Aft.

(Ans 10090.6 t)

2. A ship length l50 ıİ', TTc 20, LCF 5m üaft of amidships is at a draught 6.80 mforward and 8.00 m aft. Calcıılate the quantity to load to put the ship on an evenkeel draught of 7.70 m.

(fuis 520 tonnes)

. 3. Estirrıate the displacement and weight of cargo on board a ship LBP 150 rİı' LCF 2m abaft of amidships at dratıghts 6.50 m F and 9.80 m A when there are 300

. tonnes of stores, fuel etc. on board, given the following extracts fiom hydrostaücdata:

Draııght Displacement8.40 1ı0908.00 10570

(Light) z.so 3700(Ans Displ. 10885t; Cargo 6886t)

4. A ship has a loaded SW displacement 12000 t at a SW draught 8.54 m even keel.LBP l22 rrı, T?c 18.23, LCF l.5m abaft amidships. At present the ship is indock water density l0l5 kğm3 at draughts 7.27 mFwd and 9.30 m Aft. Find thequantity to load to put ship to the loaded even keel draught.

(Ans 536.4 tonnes)

5. A ship length 120 m designed to float at a trim 0.60m by the stern is floating atdraughts 4.2O mforward and 6.00 m aft. The displacement from the scale for amean draught 5.00 m is 5800 tonnes. TPC 12, LCF '4 m abaft amidships.Estimate the displacement.

(Aııs 5968 tonnes)

MAR\TMD.DOC

Dec 2012


GLASGOW COLLEGE OF NAUTICAL STTJDIES

LONGITIJDINAL STABILITY (1): TRIIVI TTIIORIAL SffiET I

1. Calculate üe change of trim which results when a weight of 50 tonnes is tarısferıed 20 mforward aboard a ship with MCTC 200 t-m.

(Scn by lnd)

2. A ship has a trim of l0 cm by the head before bansferring 200 tonnes of oil from No.2 DB toNo.4 DB (a distance of 52 m). Calculate the new trim, given MCTC lt9.l t-m.

(4Scmbystem)

3. A ship floats at draughts F 6.84 m and A 7.l4 m. Calculate the final trim and fınal draughsafter a weight of 42 tonnes is moved a distance of 60 m fonıaıd, given MCTC l05 t_m andLCF is amidships.

(6cm by stem; F 6.96 m, A 7.02 m)

4. A vessel has draughts F 6.90 m and A 7.00 m, MCTC l80 t-m. After transfening fiıel oil 60m further aft the draughts became F 6.76 m and A 7.14 m. Calculate the weight of oilfransfened.

(84 tonnes)

5. A ship has MCTC 250 t_m. What weight would have to be moved ürough a distance of 20 min order to change the trim by 30 cm?

(375 tonnes)

5. A ship is on even keel at a draught of 7.32 m. what distance must a weight of 120 tonnes bemoved if the ship is to be trimmed 15 cm by the stern? Given MCTC 240 t-m.

(30 m)

6. e'inıp floats at draughts F |2.24 m and A l2.l8 m. Calculate the weight of fuel to transferfrom No. I DB (lcg 182.5 m foap) to No. 7 DB (1c926.5 m foap) in order to achieve a finaltrim of 20 cm by the stem.Determine also the final draughts. Given LCF amidships and MCTC 200 t-m.

(33.3 ıonnes; F ]2.11 m' A I2.3I m)

8. Prior to sailing a ship has draughts F 9.92 m and A 10.87 m. Calculate the weight of fuel totransfer from No. 7 DB to the settling tanks (a distance of 52.4 m) to reduce the trim to 32 cmby the stern and determine the resulting draughts if the MCTC is 227.5 t-m and LCF isamidships.

(273.5 ıonneg; F 10.235 m, A 10.555 m)

g. A ship floats at draughts F 10.25 m and A ı0.l5 m. A weight of 95 tonnes is tansferred aft adistance of 42 m. Calculate the resulting draughts if LBP 100 m, LCF is 48 m foap and

MCTC is 285 t-m.(F 10.177 m, A 10.217 m)

10. A ship completes cargo with draughs F 12.24 m and A 13.24 m. To reduce the trim to 0.5 m

by the stern prior to departure a quantity of fuel is to be transferred forward a distaııce of E0

m. MCTC 400 t-m, LBP is 200 m and LCF is 95 m foap.(a) Calculate the quantiğ to transfer.(b) Determine the final sailing draughts.

(250 tonnes; F 12.50 m, A 13.00 m)

Dec 2012


11.

t2.

13.

t4.

15.

16.

17.

ı8.

A ship has draughts F 6.66 m and A7 .44 m. it is required to cross a shoal ıvhere the depth athigh water is 7.60 m with an underkeel clearance of 0.3 m. What quantiğ of fuel sııouıd betransfened through a distance of 32 m in order to achieve the required reduction in draughtaff?Find also the final forward draught given LBp E0 m, LCF 3E m foap and MCTC 96 t-m.

(88.4 tonnes; 6.815 m)

A vessel has draughts F 12.00 m and A 12.80 m. On passage 540 tonnes of fuel oil areconsumed from tanks 120 m abaft of amidships. Calculate the arrival draughts assuming LCFamidships, TPC 50 and MCTC 840 t-m.

(F 12.278 n A 12.306 n)

A ship floats at draughts F 4.30 m and A 4.80 m. The following cargo is loaded:55 tonnes 68 m foap;100 tonnes 38 m foap.

Calculate the draughts on completion of loading given LBP 108 m, TPC t6, MCTC I l8 t-mand LCF 52 m foap.

(F 4.374 m, A 4.918 m)

Prior to working cargo a vessel floats at draughts F 5.00 m and A 5.10 m. The followingcargo is worked:

Load 300 tonnes 40 m forward of amidships;Load 150 tonnes l0 m forward of amidships;Load 200 tonnes 50 m abaft of amidships;Discharge 400 tonnes 20 m forward of amidships.

calculate the new draughts given LCF amidships, TPC 25 and MCTC zzs t-m.(F 5.00 m, A 5.30 m)

From the following information calculate the draughs fore and aft when 50 tonnes of fuel istransferred aft through a distance of 67 m.Initial draughts F 3.01 m and A3.74 m. LBP I l0 m, LCF 53 m foap and MCTC 67 t-m.

çr 2.75 m, A 3.98 m)

A vessel has the following characteristics:Draughts F4.05 m and A 4.60 m.LBP 60 m. LCF 28 m foap. TPC 7. MCTC 2l t-m.

The fore peak tank is to be filled with 35 tonnes of water. Calculate the final draughts afterfilling the fore peak tank assuming that the tank's lcg is on the FP.

(F 4.38 m, A 4.40 m)

A vessel floating at draughts F 4.10 m and A 4.85 m has the following characteristics:LBP 100 m. LCF 45 m foap. TPC 12. MCTC 40 t-m.Cargo to load: 100 tonnes 75 m foap;

50 tonnes 20 m foap.Calculate the draughts forward and aft on completion of loading.

(F 4.47 m, A 4.78 m)

A vessel is floating at draughts F 2.89 m and A 4.36 m. LBP 56 m, LCF 26 m foap, TPC 6and MCTC 20 t-m.The following cargo is loaded: 308 tonnes 46 m foap;

200tonnes7mfoap.Calculate the final draughts on completion.

g 4.37 m, A 4.66 m)

Dec 2012


19. From the following information calculate the final draughts on completion of loading:Initial draughts F 6.20 m and A 6.00 m.Cargo to load: 100 tonnes 120 m foap;

I 50 tonnes 30 m foap.LBP 150 m. LCF 80 m foap. TPC 25. MCTC 120 t-m.Assume that TPC and MCTC remain constant over üe range of draughts involved.

F 6.16 m A 6.25 n)

From the following information calculate the final draughts on completion of discharge:Initial draughts F 5.35 m and A 6.24 m. LBP I l0 m.Caıgo to discharge: 450 tonnes 75 m foap;

5 l0 tonnes 40 m foap.LcF 60 m foap. TPc ı2 (average). MCTC 48 t-m (average).

F 4.Eg m, A 5.05 m)

A vessel floating at draughts F 7.25 m and A 8.45 m in salt water has an after Peak tank wiüıit's lcg on the AP. The lcg of the fore peak tank is 165 m from that of the after peak ank. Thevessel must cross a bar with a depth of 8.50 m while maintaining an underkeel clearaııce of0.5 m. The following hydrostatic data applies:

MCTC 200 t-m.LBP 175 m. LCF 85 m foap.Calculate:(a) The weight of water ballast to transfer from the after peak to üe fore peak to enable

the bar to be crossed with the required cleaıance.(b) The final draughts forward and aft'

(I12.3 tonnes; F 7.73 m, A g.00 m)

From the given information calculate:(a) the weight of water to transfer from the after peak to the fore peak tank to enable a

- vessel to cross a bar with the maximum under keel clearance. Bar depth 9.25 m.(b) the clearance overthe bar.jş Initial draughts F E.45 m and A 8.90 m. Lcg of fore Peak tank is l70 m forward of üe

AP. Lcg of after peak tank is 5.0 m aft of the AP. MCTC 210 t-m. Bar depth 9.25 m.LCF amidships.

(54.0 tonnes; 0.575 m)

From the data given calculate the quantiğ of cargo to discharge and üe draughts forward andaft for dry-docking.LBPI60 m.Initial drafu F 6.25 m and A 6.75 m.Space available for discharge: No. 2 hold, lcg 130 m foap.LCF amidships. TPC 24.44. MCTC I l0 t-m. Trim required for dry{ocking is 1.5 m by the

stem. Assume that TPC and MCTC remain constant over the range of draughts involved.(220 tonnes; F 5.66 m, A7.16 m)

A ship 140 m long arives offa port with draughts F 5.70 m and A 6.30 m. LCF is 67 m foap.TPC 30. MCTC 420 t-m. It is required to reduce the draft aft to 6.0 m by running water intothe fore peak tank (lcg 67 m forward of amidships). Find the minimum amount of water toload and also give the final draught forward.

(215.4 tonnes,5.96 m)

From the following information calculate the draughts fore and aft when the fore peak is filledwith 35 tonnes of water:

Initial draughts F 4.05 m and A 4.60 m.LBP60 m. Fore peak ank lcg 58 m foap.LCF 28 m foap. TPC 7.0. MCTC 2l çm.

F 4.37 m, A 4.42 m)

20.

21.

1',)

23.

24.

25.

Dec 2012


Glasgow College of Nautical StudiesSchool of Nautical Studies

sTABlLlw 1 TRıM - TUToRıAL SHEET - 2

1. From the İnformation tabulated calculate the draughts fore and aft when cargo hasbeen completed:

lnitial draughts: forward 2.89 m, aft 4.36 mLength BP: 56mCargo to load: 308 tonnes 46m forward of APCargo to load: 200 tonnes 7m fonıyard of APLCF: 26 m fonıvard of APTPC: 6MCTC: 20

2. From the information tabulated calculate the final draughts fore and aft afiercompletion of loading:

lnitial draughts: 4.10m for'd 4.85m aftCargo loaded: 100 tonnes 75m fo/d of APCargo loaded: 50 tonnes 20m for'd of APLength BP: 100 mTPC: 12MCTC: 40LCF: 45 m from AP

3. From the information tabulated calculate the draughts fore and aft when the forepeak is filled with 35 tonnes of water:

lnitial draughts: forward 4.05m, aft 4.60mLength BP: 60mFore peak tank: cg 58m forward of APLCF: 28m fonnıard of APTPC: 7MCTC: 21

4. From the following information calculate:

a. The amount of cargo to load to bring the ship to an even keel draughtb. The final draughts forward and aft

Length BP: 160mlnİtial draughts: Forward 6.25m Aft 7.05mSpace available for cargo: No. 2 hatch Cg 125m forward of APCF: amidshipsTPC: 24.44MCTC: 110

STAB t Tutorial/MR/AC 126 -09.02

Dec 2012


Gı_AsGoW CoLLEGE oF NAUTıCAL STUDIES

TRıM US|NG HYDRoSTATıC DATA ' TUToRıAL 3

Use the hydıostatic particulars data sheet to answerthese guesfıbns

1. A shİp with LBP 1M m arrives in port on an even keet draught of 5.30 m. Thefollowİng cargo is then worked:

Discharge 1560 t from lcg 81 m foap;Load 1700 t at lcg 112 m foap;Load 2100 t at lcg 41 m foap.

Calculate the final draughts forward and aft.(F 6.194 m, A 6.395 m)

2. A ship LBP 148 m has draughts F 3.92 m and A4.26 m. The following cargo ısworked:

Discharge 216 t from 1c422 m foap;Load 400 t at lcg 40 m foap;Load 600 t at lcg 52 m foap.

CalcuIate the final draughts fonııard and aft.(F 3.874, A 4.985 m)

3. A ship LBP 148 m has draughts F 5.22 m and A 6.00 m. The following cargo isworked:

Load 652 t at Kg 6.50 m, lcg 24 m foap;Discharge 194 t from Kg 6.4, Jcg 22 m foap;Discharge 362 t from Kg 2.8 m, lcg 130 m foap;.-n.İ Discharge 145 tfrom Kg 8.9 m' lcg 88 m foap.

".Calculate the final draughts fonıard and aft.(F 3.736 m, A 7.246 m)

4. A ship arrives in port on an even keel draught of 5.90 m in salt water. LBP is 150m. A weighl oİ 224 t is to be loaded in order that the ship sails with a trim of 0.50m by the stem in salt water.(a) Calculate the position foap to load the weight.(b) Determine the final draughts.

(30.38 m foap; F 5.728 m A 6.228 m)

5. At the start of a cargo, a ship is floating on an even keel draught of 6.20 m in saltwater LBP 140m:The following cargo operations are planned:

Load 1800 t at lcg 107.0 m foap;Load 1500 t at lca 42.0 m foap;Discharge 1640 t from72.O m foap.

Calculate the anticipated draughts in SW on completion of cargo operations.(F 7.627 m A 6.276 m)

TRIM USING HYDROSTATIC DATA (MAR Rev. '14111/02)

Dec 2012


6. A ship anives in port with draughts F 5.20 m A 5.80 m in SW. LBP 137.6 m.Cargo is worked as follows:

Discharge 1350 t from lcA 90.0 m foap;Discharge 800 t from lcg 70.0 m foap;Discharge 720 t from lcg 32.0 m foap;Load 1050 t at 90.0 m foap;Load 700 t at 70.5 m foap;Load 620 t at 28.5 m foap.

Calculate the final displacement and draughts on completion of cargo.(10680 t; F 4.826 m A 5.725 m)

7. A ship LBP 140 m has a summer load displacement oİ 14115 tonnes. ln itscurrent partly loaded condition the ship floats at draughts F 5.26 m A 5.48 m inSW. The ship is to complete loading at the summer displacement with a trim of0.50 m by the stern. The remaining cargo is to be loaded into two holds:

No. t hold lcg 116.0mfoap;No. 4 hold lcg 32.0m foap.

Calculate each of the following:(a) the quantity to load in each of the holds;(b) the final draughts in SW.

(1331.2 t in No. 1, 1887.7 t in No. 4, F 6.541 m A7.041 m)

8. Aship is floating at draughts F 4.600 m A5.460 m in SW. Atotal oİT72toİcargo is to be loaded in a position to keep the draught aft constant. LBP 146 m.Calculate each of the following:(a) the distance from AP to load the cargo;(b) the final draught forward.

(84.45 m foap; 5.340 m)

TR|M USıNG HYDRosTATlc DATA (MAR Rev. 14/11/02)

Dec 2012


GI.A*S@W COLLEGE OF NAUTICAL STTJDIES

TRIM MEASI.JRED BY TIIE DIFFERENCE IN POSITIONS OF IJCB AND LCGFor a ship to be in longinıdinal equilibriıım LCB and LCG mııst be in the same

longittıdinal posiüon. If there is a change in üe longitııdinal distribution of weight anew posiüon of LCG mııst arise; this position can be foıınd by taking momen8 ofweigbt (longifudinally) about aıy clıosen datıım _ possibly üe aft peqpendioılar. TheEimnıing moment şrhich aıises can be measıırEd in terms of üe separatioıı betwecnLCG and LCB.

The sketch shoş,s a ship, dişlacing W toıuıes, originaııy on an even keel (wL), withcentres of buoyaıcy and gravity at B and G respectively ie. in üe same vertical line.If a weight is now moved aft. G moYes aft to G1 as shcişın.

A trimrıiıg couple then exists, üe moment of ş,hich is (w X GGı) or (Şl x BGı)tonnes-metres. This causes üe ship to trim by üe stern, tipping aıoııııd F on toş,aterline WıLı (and B will then move aft to become vertically below G1 şıhen aıequilibriıırı state will again be achieved).

The change of trim from üe even keel condition carı be found as folloş,s;

Change of trim = moment = W x BGı , ofi w(LcB-LcG}MCTC MCTC MCTC

It is important to realise that we arc trying to find the change of trim üıat occurs fıomthe even teeJ condition.

BF

Ytıvl

L,

( ııı}ı'?oııaıJ B

oQı<ıBs,'-(.t

ııl AL(rıt

I F Q'ı- e1 /

)g -/

hlt

Dec 2012


ı

t,Examole IA ship dispıacing 10000 tonnes on an even keeı draught 7.60 metr,es has LcFamidships and LCB 2 m forward of amidships. Shifting a weight aft resutts in LCGbeing 1.5m forçrard of amidships. McTc = 250 tonnes-metıts. Calculate üe fınaldraughts.

Trimming moment = W(LCB'LCG)= 10000 x 0.5 = 5ü)0 toınes-metcs

Change of trim = moment = 50ü) = 20 cmMCTC ?50

büt LcF is amidships,Tr=T.=10cnFinal draught F = 7.50 m, A = 7.70 m.

Example 2A ship lengü t20 a, arrives in port at draııghb F 5.00 m A 5.20 m. The hydrostaticdaıa for adraııght 5.10 m (even keel) aıt:W = 7000 tonnes, LcF amidships, LCts 1.0 m forş,ard of amidships aııd McTc 1o0t-m.The ship noşı loa.ls 1000 toıınes, LcG 50 m forward of AP. The daa for adişlacement of 8000 tonnes aıı:Draqght 6.20 m, LcF amidships, LCB 0.5 m forward of aınidships, MCTC 110 t-m.Calculate üe final draughs.

Becaııse tlıe vessel İs İnİtİally 20 cm by tlıe stern tlıe LCG mııst İnİtblty be abd1t tlıeeven kcel posİtbn of LCB.

Change of trim = W(LCG-LCB) 20 = 7000(LCB-LCG)MCTC ı00

(LCB-LCG) = 0-857 m

LCG mııst originally h 0.7143 m forı'ard of amidships.Takiıg moments about aft perpendioılan

2000 =7000

wt70m1m08000

LCG = 474998 = 59.375 mfoap8000

2

Dec 2012


but for loaded shiP LCB = 60.5 m foap (assııning even keel)(LCB-LCG) = 1.15 n

Becaııse LCG İs abaft LCts trim is by üe st€E.

Chaıge of trim (from even keel) = wtlcEl'c(İ},MCTC

= EO@-EJJaı = 8l.82 cm by sem'. ıı0

Equivaleot even keel draııght is ğven as 6.20 m ond LcF İs amidqhiPs,F = 6.20-0.fi9 = 5.791m3A = 6.20 + 0.409 = 6.609 m.

Dec 2012


GT.ASGOW COLLEGE OF NAI..ITCAL STTJDIES

CALCULATION OF DRAUGHTS BY CONSIDERING LCB A}ID IfGı. The hydrostatic partioılars for a ship LBP ı35 m aİe:

LCB(m) LCF(m)Draueİıt(d Dişl(il fBC McTc from amidshiPs from amidshiPs

8.1 tilt4 22 180 0.58F 2.05A8.4 t6072 ?2.2 ı88 0.47F z.n^8.7 L6735 22.4 tn 0.36F 2.52A

Estimate üe dişlacement and position of LCG when draughts aıe7.95 forı'aıdand 9.05 aft.

(16336 tonne, 0.85 m aft of arİıidships)

2. A üip at dişlacemeııt|n$ t is on even keel.I,CB = 3.70 abaİt of amidships. LBP = 156 m.She loads 3300 t lcg 3 m forı,aıd of arıidships and 180 t|cg20 m abıftamidships. For dişlaceme^tl6220 t the equivalent even keel draught is 7.30 m,LCB = 2.25 m abaft amidships, LCF = 5.70 m abaft amidships and MCTC 250t-8.Calculate üe fiııal dnughts.

(F 7.207 m A 7.380 m)

3. A ship lengü.140 m has sw draughts 7.35 m forward and 7.65 m aft.Hydıostaüc data for a mean draught 7.50 m aıe:Displacement 13000 tonne,McTc 150 tonne-metrıe,LCB 71.38 m fonvard of AP,LCF 70.0 m forc/ard of AP.Calculate the draughts afrcr discharging 3500 toıuıes of cargo lcgII.2 m forırardof AP and 3000 toırne lcg 35 m forş,ard of AP.

Hydrostatic paaioılaıs for a displacement 6500 tonne are:Draught 3.7 m,LcB 72.20 m forqıard of AP,McTc = 110 toıuıe-metıp,LCF = 71m forward of AP.

(Note to soluüoıı: original separation BG = 0.346 moriginaı LCG = 7t.034 rn forward of APFinaı LCG = 65.607 m foruıaıd of APFinal separation BG = 5.593 mFinaı Trim 389.6 m by stern)

N{AR\srAB33.Doc (F 1.780 m A 5-676 m)

Dec 2012


2.

Gı-AsGow CoLLEGE oF ı{AUTıCAL sTUDıEs

1. A ship LBP 146 m anives in port with an even keel draught of 5.40 m. Cargois worked as follows:

Load 200 t at lcg 42 m foap;Discharge 50 t from lcg 53 m foap;Load 320 t at lcg 100 m foap;Load 197 t at lca 60 m.

Calculate the final draughts.(F 5.819 m, A 5.595 m)

A ship LBP 142 m has an even keel draught of 4.20 m. The following cargo isloaded:

262tat lcg 36 m foap;304 t at lq 122 m foap;24tal lcg 140 m foap.

Calculate:(a) the final draughts;(b) the ballast to transfer between the

tank through a distance of 134.6minimum.

(F 4.769 m, A 4-189 m; 66.4 t)

3. A ship LBP 152 m floats in SW at draughts F 4.70 m and A 5.30 m. Calculatethe final draughts if cargo is loaded as follows:

460 t at lcg 40 m foap;300 t at lcg 92 m foap;63 t at lcg 76 m foap.

(F 4.882 m, A 5.830 m)

4. A ship LBP 143 m has draughts in SW of F 5.28 m and A 4.42 m. Cargo isworked as follows:

Discharge 200 t from Icg 103.6 m foap;Load 484 t at lca 60.6 m foaP;Load 73 t at lcg 96.4 m.

(a) Calculate the draughts on completion;

iUi Calculate where to load an additional heavy lift oJ 220 t on deck in

order that the ship completes with a trim of 0.10 m by the stern.(F 5.144 m, A 4.865 m: 41.28 m foaP)

fore peak tank and the aft Peakm to reduce the draught to a

TR|M BY (LcB-LcG) METHoD UslNG HYDRosTATlc DATA (ıüAR Rev. 15/1ll02)

Dec 2012


HYD ROSTATIC P ARTICUTARS

DRAUGHTm

DISPL.

t

DISPL.

t

TPC

t

TPC

t

MCTCt-m

MCTCt-m

KMtm

KBm

LCBfoap

m

LCFfoap

mSW

RD 1.025

FWRD 1.000

SW

RD 1.025

FWRD 1.000

SW

RD 1.025

FW

RD 1.000

7.00 14576 t4220 23.13 22.57 84.6 80 I 8.34 3.64 70.03 67.3s6.90 14345 13996 23.06 22.50 83.0 785 8.35 3.s8 70.08 67.46

6.80 t4t l5 13't71 22.99 22.43 81.4 77.0 8.36 3.53 '70.12 67.57

6.70 l3 886 t3548 22.92 22.36 79.9 75.5 8.37 3.48 70.16 67.68

6.60 13657 13324 22.85 22.29 78.3 74.0 8.38 3.43 70.20 67.79

6.s0 13429 13102 22,78 22.23 768 72.5 8.39 3.3 8 70.24 67.90

6.40 13201 t2879 22.'72 22.t7 75.3 7t.0 8.41 J.JJ 70.28 68.00

6.30 2975 12658 22.66 22.t1 73.9 69.6 8.43 3.28 '70.32 68. l06.20 2748 12437 22.60 22.0s 172.5 68.3 846 3.22 70.35 68.20

5.10 2523 12217 22.54 2r.99 171.1 67.0 8.49 317 70.38 68.30

6.00 2297 11997 22.48 21.93 r 69.8 65.7 8.52 3.11 70.42 68.39

5.90 2073 11778 22.43 21.87 168.5 64.4 8.55 3.06 70.46 68.43

5.80 l 848 I 1559 22.37 21.82 167.3 63.2 8.59 3.01 70.50 68.57

s.70 t62s 11342 22.32 21.77 66r 62.1 8.63 2.95 '70.53 68.6s

5.60 1402 11124 22.26 21.72 65.0 61.0 8.67 2.90 70.57 68.73

5.50 I 180 10908 22.21 2t.66 63.9 60.0 8.71 2.85 70.60 68.80

5.40 0958 10691 22.15 2t.61 62.9 58.9 8.'76 2.80 '70.64 68.88

s.30 0737 10475 22.10 21.56 61.8 5',7.9 8.81 2.74 70.68 68.9s

5.20 05 l6 t0260 22.0s 21.51 60.8 56.9 8.86 2.69 70.72 69.02

5.10 t0296 10045 22.00 21.46 59.8 55.9 892 2.63 70.75 69.09

5.00 t0076 9830 21.9s 2t.4r 58.8 54.9 8.98 2.58 70.79 69.76

4.90 9857 9616 21.90 21.36 57.9 54.0 9.06 2.53 '70.82 69.23

4.80 9638 9403 21.85 21.32 s6.9 53. I 913 2.48 70.86 69.29

4.',70 9420 9190 21.80 21.27 s6.0 52.2 9.22 2.43 70.90 69.35

4.60 9202 8978 21.75 21.22 55. I 51.3 9.30 2.38 '70.93 69.42

4.50 898s 8766 21.70 21.t7 54.2 50.5 9.40 2.32 70.96 69.48

4.40 8768 8554 21.65 21.17 53.3 49.6 9.49 2.27 71.00 69.s5

4.30 8552 8344 21.60 21.07 52.4 48.7 9.60 2.22 71.04 69.62

4.20 8336 8133 21.55 21.02 51 .5 47.8 9.71 2.17 71.08 69.68

4.r0 8t2l 7923 21.50 20.97 50.6 46.9 9.83 2.12 71.t2 69.74

4.00 '7906 7713 21.45 20.93 r49.7 46.0 9.96 207 71.\s 69.81

3.90 7692 7505 21.40 20.88 r48.7 45.r r0.ll 2.01 71.18 69.88

3.80 74'78 7296 21.35 20.83 147.8 44.2 10.25 1.96 71.22 69.94

3.70 '7265 7088 21.30 20.78 146.8 43.3 10.41 1.91 '71.2s 70.00

3.60 '7052 6880 21.24 20.72 145.9 42.3 10.57 1.86 71.29 70.07

3.50 6840 6673 21.t9 20.67 144.9 413 r0.76 181 71.33 70.14

THESE HYDROSTATIC PARTICULARS FIAVE BEEN DEVELOPED WITH THE VESSEL FLOATING ON EVEN

KEEL.

Dec 2012


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