Book-Merchant Ship Stability

MERCHANTSHIPSTABILITYSEVENTH EDITION (METRIC)

By H.J. PURSEY EXTRA MASTERFORMERLY LECTURER TO THE SCHOOL OF NAVIGATION,

UNIVERSITY OF SOUTHAMPTON

Revised by J. WARDLE, EXTRA MASTER

||l uring the past few years there have beenl/considerable changes in the approach to shipstability, so far as it affects the merchant seaman. Themost obvious of these is the introduction of metricunits, in addition, examination requirements havebeen increased and recommendations for a standardmethod of presenting and using stability informationhave been produced, which will undoubtedly bereflected in the various examinations,

fhis seventh edition has been designed to meet allI these requirements. Basic information contained

in early chapters has been retained for the benefitof anyone who is not familiar with the subiect. Theremainder of the text has been re-arranged andexpanded, as desirable, to lead into new materialwhich has been introduced, there is also a chapteron stability information to illustrate the standardmethod of presentation.

fhe theory of stability has been covered up to theI standard required for a Master's Certificate and

includes all that is needed by students for OrdinaryNational Diplomas and similar courses. This has beencarefully linked up with practice, since the connectionbetween the two is a common stumbling block.Particular attention has also been paid to matterswhich may be misunderstood, or not fully appreciatedby seamen.

4

rsBN 0-8517

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MERCHANT SHIP STABILITY

MERCHANT SHIPSTABILITY

(METRTC EDTTTON)

A Companion to " Merchant Ship Construction"

BY

H. J. PURSEY, Extra MasterFomrarly lzcturer to the School of Navigation

Unive rsity of S outhanpton

This Edition Revised by

Mr. James Wardle. Extra Master.

GLASGOW

BROVIN, SON & FERGUSON, UtD.,N,c,urIcAL PUBLISHERS

4-ro DlnNnv SrnEer

Copyright in all countries signatory to the Berne ConventionAll rights reserved

First EditionSixth EditionRevisedReprintedReprintedReprintedSeventh Edition

1945r97719831992199620012006

ISBN 0 85 I 74 728 0 (Seventh Edition)ISBN 0 85 I 74 442 7 (Revised Sixth Edition)

@2006-BROWN, SON & FERGUSON, LTD., GLASGOW G4l 2SDPrinted and Made in Great Britain

INTRODUCTIONDLIRING the past few years there have been considerable changes in the

4proach to ship stability, so far as it affects the merchant searnan. The mostobvious of these is the introduction of metic unie. In addition, the Departmentof Trade have already increased their examination requirements: they have alsoproduced recommendations for a standard method of presanting and using stabilityinformation, which will undoubtedly be reflected in the various examinations.

This revised edition has been designed to meet the above-mentioned requirements.Tbe basic information contained in the early chapters has been retained for thebenefit of those who are not familiar with such matters, The remainder of the texthas been re-arranged and expanded, as desirable, to lead into the new material whichbs been introduced; whilst a new chapter on stability information has been added toillustrate the Department of Trade recommendations.

The theory ofstability has been covered up to the standard required for a Master'sC€rtificate and includes all that is needed by students for Ordinary NationalDiplomas and similar courses. This has been carefully linked-up with practice,fince the connection between the two is a conrmon stumbling block. Particulardention has been paid to mafters which are commonly misunderstood, or not fullyapreciated by seamen.

H. J. P.Southampton, 1982.

CONTENTS

CHAPTER I-SOME GENERAL INFORMATION

Increase of pressure with depthEfect of water in sounding pipes

The Metric System

Ship dimensions

The Law ofArchimedes

PAGE

I22344

455666

889

l0l2l3t3l3l5l5l6l8

t920

2527282931

35

Floating bodies and the density ofwater

DecksShip tonnagesGrain and bale measurementDisplacement and deadweightDraft.FreeboardLoadlines

CHAPTER 2-AREAS AND VOLUMESAreas ofplane figuesSurface areas and volumesAreas ofwaterplanes and other ship sectionsSimpson's First RuleSimpson's Second Rule -The 'Five-Eight Rule'Sharp-ended waterplanesUnsuitable numbers of ordinatesVolumes ofship shapesHalf-intervalsCoefficients of6nenessWetted surface

CHAPTER 3-FORCES AND MOMENTSForceMomentCentre of gravityEtrect ofadded weights on centre ofgravityThe use of moments to find the centre of gravityof an areaTo find the centre ofgravity ofa waterylaneTo find the cenhe ofbuoyancy ofa ship shape -The use ofintermediate ordinatesAppendagesInertia and Moment ofinertia and radius ofgyrationEouilibrium

VI CONTENTS

CTIAPTER,I-DENSITY,DEADWEIGIITANDDRAFT PAGEEfi€ct ofdensity on draft - 37Tonnes p€r ceNrtimetre irnmersion - 39I,oading to a given loadline - - 40

CHAPTER s-{ENIRE OF GRAVTTY OF SHIPSCente ofcravity ofa ship-G -KGShiftofG -KG for any condition of loadingDeadweight momentReal and virtual cenbes of gravityEfect oftatrks on G

CIIAPTER HENTRES OF BUOYANCY AND FLOTATIONCentre ofbuoyancy-8 - 49Cenhe offlotation-F - 49ShiftofB - - 50

CHAPTER 7-THE RIGIITING LEVERAND METACENTREEqui l ibr iumofships----53The righting lever-{Z - 55Tbe metacentre-M - - - - 55Metacentric height-CM - - - 55Stable, uqstable and ueuhal equilibrium - - 55Inngitudinat netacentric height-GM, - - 56

CIIAPTER 8-TRANSVERSE STATICAL STABILITYMom€ot of statical stability - - 57Relation between GMand GZ - - 57Ini t ia lstabi l i tyandrangeofstabi l i ty---5?Calculation ofa ship's stability - 58Calculation ofBiv - - 58Th€ hcliniry Experiment - - - 60Staticsl stability at small angles ofheel - - 62GZ by the Wall-Sided Formula - - 62Lotl, or list - - 63HeelduetoGbeingoutof thecente- lne----63Lotl due to a negative GM - - O

CHAPTER g-FREE SUFJACE EFFECTThe effect offree surface ofliquids - 69Free surface effect when tanks are filled or emptied - - 71Free surface in divided tanks - - 72Free surface moments - - 74

42424244454647

CONTENTS

CHAPTER IO-TRANSVERSE STATICAL STABILITY IN PRACTICE PAGEFactor afecting statical stability - 75Placing ofweights - - '17

Stiffand tender ships - - - 77Unstable ships - 79Ships in ballast - - 80The effect ofwinging out weights - 8lDeck cargoes - 82Free liquid in tanks - - 83Free surface effect in oil tankers - 84

CHAPTER I I-DYNAMICAL STABILIry

Dynamical stability - - 85Dynamical stability from a curve of statical stability - 85Calculation ofdynamical stability - 87

CHAPTER I2-LONCITUDINAL STABILITY

Longitudinal metacentric height-GMCalculation ofSMr

Change ofdraft due lo change of trimDisplacement out ofdesigned trimMoment to change trim by one centimeteThe effect ofshifling a weightEfect ofadding weight at the centre offlotationModerate weights loaded offthe cenhe offlotationLarge weights loaded offthe centre offlotationTo obtain special trim or draftUse of moments about the after perpendicular

CHAPTER I3-STABILITY CI'RVES AND SCALESHydrostatic curvesThe deadweight scaleHy&ostatic particula$Curves of statical stabilityCross curvesEffect ofheight ofG - -r(N curvesThe Metacentric Diagram -

The efect ofbilging

CHAPTER I,'-BILGING OF COMPAMMENTSa compamnenr - -

PermeabilityBilging an empty compartment amidshipsBilging an amidships comparhnent, wilh cargoBilging an empty comparhncnt, not amidships

vll

89909l9295969899

l0l104106l l l

l l4l l5115116tt'7119t2lt2l

123123t24125126128Effect ofa watertieht flat

Viii CONTENTS

CHAPTER IHTABILITYAND THB IJOAD LINE RI,,LESSt$ility rcquircmcob

PACEl3 lt32ttzr37138

r42t45145t45147t47I'E1,18

150150151l5ll5 lr52t52t53t53

lnfcnation to bc n4plicd to sbipeTtc Sbbility Informdio Boo&laThc usc of maximun dradwcight mmcnbSinplif cd stability idomrtion

CI{APIER I6_MISCELLANEOUS MATTERSOrydoclcing 8nd gondingTbc cfioor ofdcosity m stlbilityThc cffcct ofdcnsity on dni of shipDoriyrtion of thc Acch-wrtcr sllowanc€Rcrcrvc buoyancyLogiodid NfhcadcBulkhcrd subdivisioo 0d sb€crPlaerc on bullficads

CHAPIER I7_ROLLINGThc fouation of wrvocTtc Trochoidd ltcoryThc pcriod ofwavccTbc pcriod ofa rhipSyac,bmirmU[rcaistcd mlingRcsistrlc€s to lollingThc cfrccts ofbilp koclrCues for hcavy mlling

CHAPTIR IHUMMARYAbbrwiatims t54

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FcnulacDcdnitionshoblcos

DEADWEIGIIT SCALE, T{YDROSTAilC PARTICI,'LARS ANDIIYDROSTANCCI,.'RVES - S€clion at cod ofbook


CHAPTER 1

SOME GENERAL INFORMATION

The Metric System

Length.-The basic unit of length is the metre. I mehe (m) = l0 decimetres(dm) : 100 centimetres (cm) = 1000 millimetres (mm).

Weight.--One metric ton, or tonne (t), is the weight of 1 cubic metre of fieshwater. I tonne: 1000 kilogrammes (ke) = 1,000,000 grans (g). One gram is theweight of I cubic centimetre of fresh water.

Volume.-Is measured in cubic metes (m3), or cubic centimetres (cc, or cnrl).Area.-Is measured in square metres (m2), or square centimetres (cm2).

Force.-Is usually measured, in stability, in tonnes or kilogrammes. To indicatethat it is a force or weight, as distinct fiom mass, an .f, may be added; e.g. .tonnes

/', or "kilogrammes l"Moment -Is measured in tonne-metres, (tonne-m).

Pressure.-Is force per unit area. May be given as tonnes per squarc metre(tonnes /m2), or as kilogrammes per square centimetre (kglcm2).

Density.-Is mass per unit volume. For our purpose it can be regarded as theweight ofone cubic metre or ofone cubic centimetre ofa substance. We may expressit as either:

Grams per cubic centimehe (g/cm3).

Kilograms per cubic metre ftg/m3).Tonnes per cubic metre (tonnes/rd).

Relative Density.-Was fonnerly called ,.specifc Cravity". It is the ratio

between the density of a substance and the density of fresh water.

Relative Dursitv = density gf-sulstanceoenslty ot lresn water

\- I

2 MERCHANT SHIP STABILITY

Water. The following values are used throughout this book:-

Fresh water Salt water

Density (g/cmr or tonnes/mr)

Density (kg/m3)

Relative density

Weight per cubic metre (tonnes) 1 000

Weight per cubic metre (kilograms) 1000

Volume per tonne (mr) 1.000

FIG, 1

1.000

r 000

1.000

1025

1025

1.025

1.025

1025I

0 9156 or -I UZ)

A

B

D

Weight in tonnes : Volume (mr) x Relative density.

Increase of Pressure with Depth. The pressure on a horizontal surfacewhich is under water is equal to the weight ofa column of water above it.

Consider Fig. I which represents a column of water having anarea of one square metre. Let A, B, C, D, E and F be points one metreapart vertically. The volume of water above B is one cubic metre;above C, two cubic metres; above D, three cubic metres; and so on.If d is the density of the water in tonnes per cubic metre, the weightabove B will be d tonnes; above C will be 2d tonnes; above D will be3d tonnes; and so on. We can see from this that ifpoint I is at the seasurface, then the pressure at a depth of, say, .4F metres, will be lF x dtonnes per square metre.

The pressure at any depth, in tonnes per square metre. is equal tod T/m3 times the depth (in metres.) Since water exerts pressure equallyin all directions, this pressure will be the same horizontally, vertically,

or obliquely. We can say, then, that if a horizontal surface ofarea A square metres isplaced at a depth ofD metres below the surface of liquid density d, then:-

Pressure : d x, tonnes per square metre

Total force (thrust) on the area:d x I xrtonnes

The Effect of Water in Sounding Pipes, etc. When water rises in soundingpipes or air pipes to a height above the top of a tank, pressure is set up on the tank-top. The actual weight of water in the pipe may be small, but its effect may beconsiderable. Water exerts pressure equally in all directions and so the pressure atthe bottom of the pipe is transmitted over the whole ofthe tank{op. This pressurewill not depend on the actual weight of water in the pipe, but on the height of waterand will be the same whatever the diameter ofthe pipe. For this reason, tanks should

-

I

hb

t

hEt'5

SOME GENERAL INFORMATION 3

not be left '?ressed up" for long periods, because this can exert considerable stresson the tanldop.

Example.-A rectangular double bottom tank is being filled with seawater. Ifthe water is allowed to rise in the sounding pipe to a height of 7 metres above thetank top, find the presswe on the tank top.

Pressure = dD: l'025 x 7 :7'175 t/m'z

The Law of Archimedes.-A body immersed in a liquid appears to suffera loss in weight equal to the weight of liquid which it displaces. A floating bodydisplaces its own weight of water.

A block of iron, one cubic metre in size and of density 8 tonnes/m3 weighs8 tonnes in air. Ifplaced in fresh water it would displace one cubic metre of water,which weighs one torme; so the weight ofthe block would thus appear to be 7 tonneswhen it was under water.

If we now take the block and make it into a hollow, sealed box, its weight inair will remain the same but its volume will increase. If placed in watet it woulddisplace more of the water and its apparent weight will decrease accordingly. Forinstance, ifthe box were 3 cubic metres in volume it would displace 3 cubic metres ofwater (or 3 tonnes), so that its apparent weight in fresh water would now be 5 tonnes.

If we increase the volume of the box stillfurther, it will displace still more water and itsapparent weight under water will decrease stillmore.

Eventually, when the volume of thebox became greater than 8 cubic metres, anequivalent volume of water would weigh morethan the box. So if the box were now placedunder water, it would be forced upwards, andpartially out of the water until the upwardforce exactly equalled the weight of the box.In other words, the box would rise until it floated at such a draft that it would displaceits own rveight of water.

Application to Ships. -A ship may be regarded as a closed iron box, so thattwo conclusions can be drawn from a study ofthe last section:-

(a) So long as the weight of the ship does not exceed the weight of its ownvolume of water, it will float.

(b) The draft at which it floats will be such that the weight of water displacedwill be equal to the weight of the ship.

b

lvv,is

rgt-belll

t€

E[td

I to{{Es

FIG.2

I totf.Es

: . : : : : l

: t .1/ / )t'!.,21vi

\-____


Floating Bodies and the Density of Water.-Weight is equal to volumemultiplied by density and therefore volume is equal to weight divided by density. Inthe case ofa floating body:-

volume of water di.otu..d = w"f8ht ?f

*ltlt ditPl"t"d' Densrty ot tne water

The volume of water displaced is equal to the underwater volume ofthe bodyand since a floating body displaces its own weight of water, weight ofthe body canbe substituted for weight of water displaced.

Underwater volume = Weight of bodyDensity of the water

If the weight ofthe body remains the same, it can be seen that the underwatervolume must vary inversely as the density ofthe water. In other words, if the densityofthe water changes:-

New volume disolaced Old Densitv-oia;ffiTsftA = N"-E..tty

Ship Dimensions.-The following are the principal dimensions used inmeasuring ships.

Lloyds' Length is the length of the ship, measured from the fore side of thestem to the after side ofthe stem post at the summer loadline or the centre line oftherudder axis ifno stern frame is fitted, or 96 per cent ofthe length overall whicheveris the greater.

Moulded Breadth is the greatest breadth ofthe ship, measured from side to sideoutside the frames, but inside the shell plating.

Moulded Depth is measured vertically at the middle lenglh ofthe ship, from thetop of the keel to the top ofthe beams at the side ofthe uppermost continuous deck.

The Framing Depth is measured vertically from the top ofthe double bottom tothe top ofthe beams at the side ofthe lowest deck.

Decks.-The Freeboard Deck is the uppermost complete deck, havingpermanent means of closing all openings in its weather portion.

Ship Tonnages.-These are a measure of space within the ship; one ton beingapproximately equal to 100 cubic feet or 2'83 cubic metres. Tonnages are not ameasure of weight ofa ship.

Gross Tonnage is a number representing the total volume of all enclosedspaces within the ship.

lL- I


Nett Tonnage is another nurnber that represents the eaming capacity of a shipaccording to a formula. Its value depends on

Gross tonnage

Volume of cargo spaces

Moulded depth and moulded draft

Number ofpassengers in cabins sleeping eight or less

Number of other passengers

Nett Tonnage must not be less than 30olo ofGross Tonnage.

Net tonnage is also referred to as Universal Measurenent System Net or UMS Net.Gross and Net tonnage are legal and commercial values, and as such have liftle to dowith the study of stability.Gross tonnage is used to determine what safety equipment is carried on a vessel.Net tonnage is used as the basis ofport dues and other shipping taxes.

Grain and Bale Measurement.-These terms are often found on the caoacitvplans of ships and refer to the volume ofthe cargo spaces.

Grain Measurement is the space in a compartment taken right out to the ship'sside and sometimes up between the beams. In other words, it is the amount ofspacewhich would be available for a bulk cargo such as grain.

Bale Mersurement is the space in a compartment measured to the inside of thespar ceiling, or, if this is not fitted, to the inside ofthe fiames and to the underside ofbeams. It is the space which would be available for bales and similar cargoes.

Displacement,-Is the actual weight of the ship and all aboard her at anyparticular time. Since a floating body displaces its own weight of water, this meansthat displacement is equal to the weight ofwater displaced by the ship.

Light Displacement is the weight ofthe ship when she is at her designed lightdraft. It consists of the weight of the hull, machinery spare parts and water in theboilers.

Loaded Displacement is the weight of a ship when she is floating at hersummer draft in salt water.

Deadweight.-This is the weight of cargo, stores, bunkers, etc., on board aship. In other words, it is the difference between the light displacement and thedisplacement at any particular draft. When we say that a ship is of so many tonnesdeadweight, we usually mean that the difference between her light and loadeddisolacements is so manv tonnes.

\


Draft,-This is the depth ofthe bottom ofthe ship's keel below the surface ofthe water. It is measured at the forward and after perpendiculars. When the draftsat each end are the same, the ship is said to be on an even keel. When they diffeqthe ship is said to be trimmed by the head, or by the stem, according to which is thegreater ofthe two drafts.

Mean Draft is the mean of the drafts forward and aft and measured atmidlength.

Freeboard,-Statutory Freeboard is the distance from the deck-line to thecentre of the Plimsoll mark. The term "Freeboard" is often taken to mean thedistance from the deckline to the water.

Loadlines.-The loadlines and deck line must be painted in white or yellow ona dark background, or in black on a light background.

The deck-line is placed amidships and is 300 millimetres long and 25millimetres wide. Its upper edge marks the level at which the top of the freeboarddeck, ifcontinued outward, would cut the outside ofthe shell plating.

A loadline disc, commonly called "the Plimsoll mark", is placed below thedeckline. The distance from the upper edge of the deckline to the cenre of the

-F

(!Ec|( L'd.) disc is the statutory summer freeboard. 540millimetres forward of the disc are placedthe loadlines, which mark the drafts to whichthe ship may be loaded when at sea and incertain zones. All lines are 25 millimetresthick and their upper edges mark the levelto which they ref'er. The following are themarks required for steam-ships:-

S the "summer loadline"-is levelwith the centre ofthe disc.

,/ the "winter loadline"-is placedbelow the summer loadline at a distance ofone forty-eighth of summer draR.

Z the "tropical loadline"-is placedabove the summer loadline at a distance ofone forty-eighth of summer draft

IItNA-:he "riiinter North Atlantic loadline"-is placed 50 millimetres belowthe winter loadline. It is only marked on ships which are 100 metres or less in length.

F The "fresh-water (surnmer) loadline"-indicates the draft to which theship can be loaded in fresh water, if she is to rise to her summer loadline at sea.

le---rso"" . - - -1

FIG.3l-2to^an


Its distance above the summer loadline (S) is called the "Fresh Water Allowance"A

and is found by the formula ji millimetres, where / equals the displacement at

summer draft and T is the "tonnes per centimetre immersion" at that draft.

TF---Tll,e "tropical fresh water loadline"-is the fresh-water line to which theship can be loaded, in order that she may come to her tropical mark when she reachesthe sea. It is found as for F, but is measured above the tropical loadline.

Timber Loadlines.-These are marked abaft the loadline disc. They show thedrafts to which a ship so marked may load when carrying a deck cargo of timber,provided that the cargo is stowed according to special nrles and to a certain minimumheight.

IS-The "summer timber loadline"-This is a little above the ordinary sumrnerline. The reduced freeboard is an allowance for the extra buoyancy provided by thetimber deck cargo. If a ship, loaded to timber loadlines, lost her deck cargo, shewould come back approximately to her ordinary summer loadline.

FIG.4

LT--:t"he "tropical timber loadline"-is placed above the summer timber loadline, at a distance ofone forty-eighth of timber summer draft.

LW- :t"he "winter timber loadline"-is placed below the summer timberloadline at a distance equal to one thirty-sixth of timber summer draft.

LWNA---The "winter North Atlantic timber loadline"-is placed level with theordinary WNA line in ships of 100 metres or less in length. Ifa vessel is longer than100 metres, then there is no LWNA.

ZF-The "fresh water timber loadline" found by allowing the timber freshwaterallowance above the LS line.

LTF-The "tropical fresh-water timber loadline" is found as above, butmeasured from the IT line.

CIIAPTER 2

AREASANDVOLUMESAreas of Plane Figures.-The areas of certain common plane figures are often

used in stability calculations.Square.-Where a is the length of each side:-

Area = a2

Rectangle.-Where a and 6 are the lengths of the sides:-Area = axb

Triangle.-Where a, 6 and c arethe lengths ofthe respective sides; lr the perpendicularheight; and o the angle between a and 6. -

**=t1UIea = abx;ias

FIG.5vThsrss=|(a+D+c)

_-1 Trapezium, or Trapezoid.-is a four-sided figure, having trvo of

.f,_ _l-_JU its sides parallel.

- r - I Where a and b are the lengths of the parallel sides, and l the

FlG. 6 perpendicular distaDce between them:-

xea=!@+b)

Circle.-Where r is the radius; and where z is equal to 3 . 1 4 I 6, or approximately fArea = xr2

Circumference = 2nr

Surface Areas end Volumes.-The following are often useful:-Cube.-Where a is the length of each edge:-

Arca = Js(s - axs -D)(s -c)

Surface area = 6a2 Volume = a3

AREASANDVOLUMES

Box shapes.-Where a, b and / are the lengths of theedges:

Surface area = 2(al + bl + ab)

Volume = abl

Wedges and prisms.-Where I is the area of either end and/ the length:-

Volume = l/

Sphere.-Where r is the radius:-

Stxface area = 4xr2

volu,ne = 4o=t'J

Hollow Sphere.-Where ,, is the intemal radius and R the extemal radius:-

t_nt ^_rJVolume of materi

!G'-")Cylinder.-Where r is the radius and / the length:-

Strfrce area = 2tt r(r + l)

Volume = zr2l

Hollow round section.-Where R is the extemal radius, r the intemal radius and /6e lenglh:-

Area of cross section = r R2 - t 12

= tt(R2 - 12)

Volume of materi al =rl(R2 -12\

Arers of Waterplanes and Other Ship-Sections.-These cannot usually befomd with any degree of accuracy by simple mensuration, but there are severalmethods which may be used to find them. We need only concem ourselves here with-Simpson's Rules" and the "Five-Eight Rule".

\

I

IO MERCHANT SHIP STABILITY

Simpson's Rules were designed for finding the area under two types ofcurve.The first and the Five Eight rule are used when the curve is a parabola. The secondrule is used when the curve is a cubic curve. These curves are similar to the shapes ofthe edges of water-planes and other ship-sections and we can use them to find areasand volumes of ship shapes with sufficient accuracy for practical purposes. All rulesare equally accurate but the first is usually used for preference.

The preliminary steps in calculating the area of a waterplane or section areas follows. A number of equidistant points are taken along the centre line andperpendiculars are dropped from these points to meet the curved sides. The lengthsof these perpendiculars are measured and also the distance between them. Theperpendiculars are called "Ordinates" and the distance between them, the "CommonInterval". The latter is usually denoted in formulae as "ft".

Figure 8 represents awaterplane. In this case. thecentre line (18) is divided into

I six equal parts, each havinga length of h (the commoninterval). The ordinates areHHr JJp KKp etc. I and B arealso ordinates, although in thiscase they have no length.

It will be noticed that halfofthe figure has been drawn in plain lines and half indotted lines. The perpendicular distances shown in theplain lines (Cll, DJ, ErK, etc.) forthe half-waterplane are usually called "HalfOrdinates", in order to distinguish them.

When a ship's plans are drawn, they usually show only the half-waterplane. It iseasier, in practice, to measure the half-ordinates from the plans, to put them throughthe Rules and then to double the half-area so found to give the whole area.

The halfordinates, put through the rules, give the area ofthe ha lf-waterp lane: theordinates will give the area ofthe whole waterplane,when put through the same rules.

Sirnpson's First Rule.-ln its simplest form, this rule states:- The areabetween any three consecutive ordinates is equal to the sum ofthe end ordinates, plusfour times the middle ordinate, all multiplied by one+hird of the common interval.

Consider in Fig. 9 thearea contained between thehalf-ordinates t and v. If thecommon interval is l, this areaeouals:- h' j ( r+4u+v)

.J

H,' i ..--;;' ' - , ! - - - - -LJr Kr

- -" ; '

FIG.8

h L h h

-"--rn L

E

FIG.9

e.dDflsls

AREASANDVOLTJMES 11

The total area of the half-waterplane can be obtained by finding, in the sameway, the areas between v and .r, and x and z, and taking the sum of the three.

hArea between t and v =::(, + 4ll + v)

:h

Area between v an6x=iQ+4w+ x)

LArea between x and z = i(x + 4y + zl

; . h. h.Total Area =

iQ + lu + v) +

i(v+4w+.x) +, (x + 4y + z)

n= iU + 4u + 2v + 4w + 2x + 4y + z)

The numbers by which the successive halfordinates, or ordinates, are multiplied(in this case l, 4,2,4,2,4, - - - - l) are called "Simpson's Multipliers".

From the above, we can see two things about this rule:-

(") It can be used when, and only when, an odd number of ordinates aretaken.

(b) The area is found by multiplying successive ordinates, including the ends,by the multipliers l, 4,2, 4,2, 4, - - - - I, adding the results togethe4 and thenmultiplying by one-third ofthe common interval.

Example.-Use Simpson's First Rule to find the area of a waterplane whichhas the following half-ordinates, spaced 12 metres apart:-l'7; 5'9:'7'0; 5'2; l3meues.

Since there are 5 ordinates we can use Simpson's First Rule

rEdlsE

ll

aEb

rsn1€

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inrtrls

F,ei.!EIEE

E

!

= \1t. t + 1txs.9) + (2x 7. 0) + (4xs'2) + l. 3)

11"11a16 =f x61.4

Arez=2x?x6l-4

= 491 . 2 square metres

Han arer = *{a

+ 4b+2c + 4d + e}

Ordinate Multiplier Productt .7 I I ' t

5.9 23.67.0 l4'05.2 20.81.3 I 1.3

61.4

12 MERCTIANT SHIP STABILITY

Slmlxon'r Second Rule.-The area between any four consecutive ordinates isequal to the sum of the end ordinates, plus three times each ofthe middle ordinates,all multiplied by three-eighths ofthe common interval.

Consider in Fig. 9, the area between the half-ordinates r and w. Ifthe commoninterval is i, this area is equal to:-

Area between , and * =1h( +3u+ 3v + u)d

The arca b€tween w and z can be found in the same way and added to the above,to give the total area ofthe half-waterplane.

ArcaMween t andw =lh? +3u+ 3v + w)U

Area between w and z =1hfu+3x+3v + z\U

Total Area = +r(t +3u +3v +2tt+3x+3y + z)U

The conclusions we can draw for this rule arc:-(a) The Rule can be used when, and only when, the number ofordinales isfourior four plus some multiple of three (e.9., 4, 7, 10, 13, 16, etc).

(b) The arca is found by multiplying successive otdinates, including the ends,by the multipliels I, 3, 3, 2, 3, 3, 2, - - - - l, adding the results together and thenmultiplying by three-eighths ofthe common interval.

Example.-Find, the area of a waterplane, using Simpson's Second Rule. Thecommon interval is 15 metres and the ordinates are:{;9'0; l3'3; l4'7; l2'8:,7'5;0'6 metres.

psa=]n6+Zb+3c+2d +3e+3f + g)6

Ordinate Multiplier Product0.0 I 0.09.0 27.013.3 J 39.914.7 2 29.412.8 38.47'5 3 22.50.6 I 0'6

157.8

6rcs =$xtsxl57.S

= 887 . 6 square metres

AREASANDVOLUMES 13

The *'Five-Eight Rule."-This may be used to find the area between twoconsecutive ordinates. We must know the length of one other equally spacedordinate. next to the area which we wish to measure.

The rule is that tfie area is equal tofve times one end ordinate, plus eight timesthe other minus the known external ordinate (in that order), all multiplied bv one-twelfih of the common interval.

Suppose that we wish to find the area between.r and y in Fig. 10. z is theextemal ordinate and I the common interval.

FtG. 10

fhen the rrlea = #(5x +8y - z)

Similarly, the area between y and z would be equal to:-

${sz +8t - x)

Example.:Tbree ordinates, a, b and c, are 12 metres apart and have lengths of29'2,33'5 and 37'6 metres, respectively. Find the area contained between a and D.

First nart area = ]x5x8l .7

= 136 2

Second part area =frx5x2l 9=9 I

Total area = 145 . 3

Sharp-ended Waterplanes.-In the above rules, the ends of the Waterplaneshave been considered as squared-off, but if they are pointed, the rules still apply.The end ordinates are then taken as 0, but are put through the multipliers in theordinary way.

Unsuitable Numbers of Ordinates.-It sometimes happens that a number ofordinates must be used which will not respond to any ofthe above rules. In this case,the area is found in two parts, which are later added together.

For example, if there were eighteen ordinates, neither of Simpson's Ruleswould give the area directly. We could in this case find the area within the firstnine ordinates by the First Rule, then add to it the area witlin the remaining ten(remember that the ninth ordinate would be taken twice) found by the Second Rule.Alternatively, we could find the area between the first seventeen ordinates by theFirst Rule and that between the remaining two ordinates by the Five-Eight Rule, lateradding them together.

t4 MERCHANT SHIP STABILITY

For example, if there were eightordinates, as shown in Fig. ll,neither the First nor the Second Rulewould give the area directly. In thiscaae, we could find, say, the areabetween a and g (seven ordinates) bythe First Rule; the remaining area,

benreen g and i, by the Five-Eight Rule. Altematively, we would use, say, the SecondRule to fnd the area between a and d (fow ordinates); then the First Rule to fnd thearea between d and g (five ordinates). The following example will illustrate this.

Example.-A waterplane has ordinates, spaced 5 metres apart, of lengths0; 3'l; 4.9; 6.3; 6.1; 4.8;2.9;0.7 metres. Find its area.

First method

Ordinate Multiplier Product0.0 0'03.1 4 12.44.9 2 9.86.3 4 25'26.1 2 12.24.8 4 19.22.9 I 2.9

81.7

Second method

Ordinate Multiplier Product6.3 I 6.36.1 4 24.44.8 2 962'9 4 I1.60.7 I o.7

52.6

Fint nart area = |

x5x8l.7 =136.2 m2

Second part area = $xSx2l'9=9.1m2Total area = 145.3 m2

Ordinate Multiplier Product0.7 ) 3.52.9 8 23.24.8 - l -4.E

21.9

Ordinate Multiplier Product0.0 I 0.03.1 5 9'34.9 J 14.76.3 I 6.3

30.3

First oart area = 1x5x30. 3 = 56. 8 m2.U

Second part area = ]x52.6=87.7 m2

Total area = 144. 5 m'

Note the small discrepancy in the answers above, which is due to shape of theobject being not a parabola or a cubic curve.

AREASANDVOLUMES 15

Volumes of Ship Shapes.-The ship is divided up into a number of equallyspaced sections, the area ofeach of which is found by Simpson's Rules. The volumeis found by putting these areas through the rules in the same way as ordinaryordinates. The sections may be either vertical or horizontal, as convenient. Whengrcat accuracy is required, the volume may be found by both methods, one beingused to check the other.

Fig. 12 a shows how the calculation may be done by using the vertical sectionsB, C, D, E and F. The area ofeach section is found by the Rules in the ordinary wayand the volume can then be calculated by Simpson's First Rule, thus:-

Volume = 3(l

+4 B + 2C + 4D +2E + 4F + G)

Fig. 12 b shows how the same volume could be found by using horizontalsections. Where A is the deck area, G the area at the keel, and B, C, D, E andF,theareas of intermediate sections:-

L

Volure = +(,{ + 4 B + 2C + 4D + 2E + 4F + G)J

Half-Intervals.-Near the ends of a ship, where the curvature is great, wesometimes bring in intermediate ordinates, halfivay between the main ordinates, inorder to obtain greater accuracy.

Consider Fig. 13,which shows a waterplanewith an intermediateordinate midway betweena and b. The intewals axand bx re each )h .

I t - r!c t I

Lr i

4.:::z. ' - . ' - . ' .11...".,,-;28

h

d

h h

!.

,f,

alaI TL IL

FtG. 13

16 MERCHANTSHIPSTABILMY

|.Area between a anat=-{@+ax+b)

=l(1"+z-lt)Area between 6 zrlaf =

l{b + lc +2d + 4e + f)

The areas could be calculated separately n"d the results added togetherafterwards. Usually, however, we combine the calculations by adding together theformula as shown below. There are advantages to be gained by using this combinedmethod of calculation when fnding centres of gravity of shapes.

r""r "*

= f (j" + zx +rlr + k + 2d + 4e + f)

If there are in0ermediate ordiaates at the other end of the waterplane, they areteated in the same way. For instance, if we had the ordinate y midway between/and g:-

rotar area = +(+a + 2x + t+b + 4c + za + + + rl I + zt + I sl

Ordinate Multiplier hoduct1.6 % 0.89.4 2 18.8

13.6 l% 20.426.5 4 106.025.1 2 s0.210.6 4 42.42.1 I 2.1

240.7

Example,-A waterplane has ordinates,20 metres apa6 of 1.6, 13.6,26.5,25.1, 10.6,and 2.1 metes. There is also an intermediateordinate, midway b€tween the flst two, of 9.4metes. Find the area-

L,rotdrca=l(240.7)

=lx2no.t1605 sq. mefes

Coelficient of Fineness of the MldshipsSection.-This is the ratio between the actualunderwater area of a midships section and thatof a rectangle of the same depth and width. InFig. 14 which represents such a section, thecoefficient would be:-

Shaded ArcaAreg of ABW FlG. 14

AREASANDVOLUMES t7

Coelncient of Fineness of rWaterplane.-This is the ratio between thearea ofthe waterplane and that ofa rectangle ofthe same length and breadth.

In Fig. 15 the coefrcient offineness ofthewaterplane shown is:-

FrG. 1s

Block Coelficient of Fineness of Displecement.-This is usually referred toby seamen as the "Coefficient of Fineness". It is the ratio between the underwatervolume of tlte ship and that of,{ box shape, having the same length, breadth and

FtG. 16mean draft. In Fig. 16 the shaded volume represents the underwater part ofl shipand the dotted lines, the box shape mentioned above. The coefficient is then:-

Prismatic Coelfcient of Fineness of Displecement -This is the ratio betweenthe underwater volume ofthe ship and that ofa prism having the same lenglh as theship and the same cross-section as her midships section.

- ' l r

FFtG. 17

In Fig. 17, the shaded area represents the underwater part of the midshipssection; the plain lines, the underwater part ofthe hull; and the dotted lines, the prismdescribed. The prismatic coefficient of fineness ofdisplacement is then:-

Shaded AreaYolume of ABCDEFG

Undelvater volume ofthe shipVolume of the prism ABCDEF

18 MERC}IANT SHIP STABILITY

Wetted Surface.-This is the surface area of the underwater part of theship's hull. It is of geat importance to naval architects, since it is one of the factorsdetermining the resistance to the movement of the ship through the water

It is difficult to calculate the area of wetted surface accurately although it canbe found very closely by taking the underwater girths ofthe ship at regular intervalsand then putting the "ordinates" so found through Simpson's Rules. Alternatively, itmay be found by approximate formulae, such as:-

Wetted surface = Z ll.1d +(CxB) |Wbere I = Length of the ship.

d = Mean draft.

B = Breadth ofthe ship.

C = Block coefficient of fineness of displacernent.

CHAPTER 3

FORCESAND MOMENTSThe stability of ships depends entirely on forces and moments, so that to

undentand stability properly, it is necessary to have a general knowledge ofmoments.

Force.-This is any push or pull exerted on a body. Whan a force is beingconsidered, thrce things must be taketr into account:-

l. The amount offorce applied.2. The point at which the force is considered to be applied.3. The direction in which the force acts.A force pushing on one side has the same effect as an equal force pulling on the

opposite side. A point always tries to move directly away from a force pushing at it,

E+i----?FlG. 18

or directly towards a force pulling on it. In Fig. I 8 the force .r,pushing on the point,{, has exactly the same effect as an equalforce pulling in the direction y. The point will try to move inthe direction ly.

Resultrnt Forces.-Any number of forces may act on one point and theircombined effect u.ill be the same as that of a single force acting in one padiculardirection. This imaginary force is called the "resultant force".

If two forces act in one shaight line and in the same direction on one point, the

^ -g--+ - resultant force is equal to the sum oftle forces and

tr- -- - -- --)1-- - 5->acs in the same direction. In Fig. 19 the forces.r-5,4 and/ produce the resultant force R, which is equal

FlG. 19 to (.r + /), which acts in the same direction andwhich tries to move the point in the direction z.

If two forces act in opposite directions on one point, the resultant force willbe equal to the difference of the forces and its direction will be the same as that ofthe greater of the two. In Fig. 20, r reprcsents the greater of two forces, .r and y,

acting in opposite directions on the point I . The<-7---'

?i a- resultant force will be (.r -y), and will act in the

FlG.20 il:T[!}#lt will trv to move the point in


act at an angle to each otheq the resultant will depend onthe amount and direction of the forces causing it. Theresultant is found by using the "Parallelogram ofForces",the principle of which can be seen from Fig. 21. Let r andy be two forces acting on a point A. Let the amounts ofthe forces be represented by the lenglhs of BA and CA,respectively. Draw CD parallel to AB and BD parallelto AC. T"he diagonal DA will represent the amount anddirection of the resultant R.

When two forces

CDFtG.21

We sometimes find one force pushing and the other pulling on a point at anglesto each other. In such a case, one of the forces must be treated as a push or a pull inthe opposite direction, so that both are regarded as pushing, or both pulling. This is

allowable, since, as we have seen, a push in one direction isequivalent to an equal pull on the opposite side.

Suppose force .r to be pulling and force y to be pushingon the point ,4, as in Figure 22. Force x can be transformedinto force z, exeding a push equal to x on the opposite side ofI . The parallelogram can then be completed, and the resultantforce, R, found in the same way as in the last example.

If scale drawings and measurements are not sufficiently accurate, the resultantcan be calculated from the parallelogram by trigonometry.

Moment.-When a body is free to tum about some point and a force is appliedto it at some other point, the body will try to revolve. A body turns because there is amoment applied to it. The value of the moment dep€nds on two things:

(a) The amount of theforce.

(b) The length of the lever on which the force acts: that is, the perpendiculardislance between the direction of the force and the point about which the bodytends to rotate.

Suppose that the body shown in Fig. 23 is free torotate about the point C, and that a force,;r, is applied at thepointl. Then the moment trying to tum the body about Cis the effect of the force ;r, acting on the lever 8C.

When a lever is used to lift a weight, the "lifting: power" depends on the length of the lever and on the

amount offorce applied to it. Moment is a form ofleverage,so we can meiuiure it in a similar way:-

Moment = Force x lensth of lever.

FORCESANDMOMENTS 21

Various quantities are used to measure moments. In ship-stability, where theforces are large, the force is usually measured in tonnes, the lever in metres and themoment in "Tonne-metres".

Properties of a Moment.-Moments can be taken about an imaginary turningpoint, or about one side ofan area, even though the body is not actually free to turnabout that point or side. In other words, we could say that if the body were freeto tum about a certain point or side, the moment would be so much. Consider a

L force ofx tonnes, applied at a point P, in a directionperpendicular to the surface of the area KZ,l0y'. If weassume that the area were free to tum about the pointC, the moment is x x PC tonne metres. If we assumethat the area can turn about the side Kiy', the momentabout the side is r x PB tonne-metres. where PB is theperpendicular distance fiom P to KN. Similarly, themoment about ,<Z would be .r x PI' tonne-metres.

This property ofmoments is very important in ship stability, as we shall see later.

Combinations of Moments.-When several moments are considered to act onone point or on one side ofan area, their effects may be combined and considered asthat of a single moment. This resultant moment must, of course, be regarded as if itwere produced by a different force and./or lever. For this purpose, we take the sumof moments which act in one direction, or the difference of moments which act inopposite directions. This is best seen from examples:-

AnlT-TTF FrG.2s

Consider two men working at a capstan. Letthem push with forces of x and y kilogrammesrespectively, on the capstan bars AC and BC, asin Fig. 25. Then the moment produced by thefirst man to tum the capstan about its centre,C is .r " lC kg-metres. That produced by thesecond man is y x 3g kg-metres. Since theyare both trying to tum the capstan in the samedirection. the total moment to tum it will be@ x Aq + 0/ x BC) kg-metres.

Now, suppose that the man at B were totum around and push with the same force, butin the opposite direction; that is, against the manat l. The total moment to tum the capstan hasnow become (x x Aq - 0 " BC) kg-metres.(Fig.26.)

c.-----

FtG.26


Suppose that there are four men working atthe same capstan, three pushing in one directionand one in the other, against them. (Fig. 27.)

The moments in one direction are (.r x lC1,(w x DC) rnd (z x EC) kg-metres. The momentin the other direction is (y x ,C) kg-metres. Thetotal moment to tum the capstan is:-

(x, AQ+(w" DC)+(z x EC)-(y " BC)kg-metres.

Finally, let us go one step firther andconsider the same capstan and men as inthe last example, but with a rope around thecapstan barrel, pulling against the men with aforce ofp kilograrrmes, as in Fig. 28.

The moments trying to turn the capstan inone direction are now (x x AC), (w x DC') and(z x EC) kg-metres. ln the other direction theyte O x Bq and (p x RC) kg-metres. So thetotal moment to tum the capstan is now:-

l(x x AC) + (wx DC) + (z " Eql-Ky x BC) +(P " RC)l kg-meres.

From the above we can see that the final moment about a point or side ofan areais the sum ofthe moments trying to produce rotation in one direction, minus the sumofthe moments trying to produce rotation in the opposite direction.

Example l.-A force of 60 kilogrammes is applied to one end of a lever, whichis 150 centimetres long. What is the moment about the other end?

Moment = force x distance = 60 x 150 = 9000 kg-centimetres.

Example 2.-Three men are working at a capstan and push on bars 3 metreslong with forces of 70, 95 and 86 kilogrammes respectively. Find the moment to tumthe capstan.

Moment produced by fint man = 70x3 = 210 kg-metenMoment produced by second man = 95 x 3 = 285 kg-meters

Moment produced by third man = 86 x 3 = 258 kg-metersTotal moment to tum the capstan = 753 kg-meters

FORCESANDMOMENTS 23

Example 3.-A see-saw is exactly balanced about its centre line. Weights of50 kilogrammes and 90 kilogrammes are placed on one side at distances of3 metresand 7 metres, respectively from the centreJine. What weight must be placed on theother side, at a distance of6 metres from the centre, to cause the see-saw to balanceonce more?

Moment of weights on one side about centre-line : (50 x 3) + (90 x 7)Ionne-meres.

Let -r be the weight on the otler side.Then, moment of weight on the otler side, about the cente-line = 6:. For the

see-saw to balance, these moments must be equal.So.

6r = (50x 3) + (90x 7)

__ (50x 3) + (90x 7)--6

= 130 kilograms

Couple.-Ttvo equal and opposite forces which act atdifferent points on a body are called a "couple". The momentproduced equals the product of one of the forces and theperpendicular distance between them.

Moment = Fd tonne-metr€s

Centre of Gravity.-This is the point about which a body or area wouldbalance. It may be regarded as the geometrical centre ofany area, or the centre ofallthe weight ofa body. In the case ofa body, the force of gravity is considered to actvertically downwards through it, with a force equal to the weight ofthe body.

Centre of Gruvity of en Aree.-The centres of gravity of certain areas areused in ship-stability, the principal ones being as follows:-

B Circle.-The cente of gravity of a circle is at its centre.Square or Rectangle.-The centre of gravity of either

of these is at the intersection of the diagonals. When we are= considering the stability of box-shaped ships, we always regard

the centre of gavity as being at the intersection of the centre-lines.

^ For example, the centre of gravity (G) of the rectangle

@FtG.29

D ,llCO, is at thi intersection of the longitudinal centreline wxand the transverse centre-line vz.

t

FtG.30

1^ MERCHANT SHIP S'IABILITY

Triangle.- Draw a median- that is, a line from onecomer to bisect the opposite side. The centre of gravity is onthis median, at a distance of one-third of its length from thebisected side.

In the triangle ABC (Fig. 31), let AD be the median,bisecting the side BC. Then the centre of gr4vity will be at G andGD will equal one-third of ,{D. (Or, AG =

f 1D.; Altematively,draw a second median, CE, and G will be the point at which lDand CE intersect.

BDC

FtG.31

Quadrilateral.-Draw the diagonals lC andBD. Let them intersect at .r. Measure off Ay ar:dDz, so that Ay equals Cx and Dz equals ,Br. Find thecentre of gavity of the triangle ryz and this will bethe centre of gravity of the quadrilateral.

Centre of Gravity of a Body.-It is often necessary to find the centre ofgravity of a homogeneous body-that is, a solid body composed of the samematerial throughout. The three types of body, for which this is usually required arebox shapes, prismatic wedges and ship shapes.

For convenience of calculation, we alwaysconsider the position of the centre of gravity ofsuch bodies in the transverse and longitudinaldirections separately: that is, as it would be seenfrom the end or from the side respectively. Fig.33 illustrates this. Let G be the centre of gravityof tlre box shape ABCDEFH shown in Fig. (a).If we were considering this, we should fust findG ransversely by looking directly at the end,ABCD, as nFig. @) and considering its positionin the plane ofthat area. We should then find thelongitudinal position of G, by looking directly atthe side CDE4 as illustrated in Fig. (c).

Box shapes.-We can see from thelast figure, that the centre of gravity of ahomogeneous box shape is at the centreline andat the half depth ofthe body, both longitudinallyand transversely.

(do

c

iG

icI

(cl

FORCES AND MOMENTS 25

"A:*,, FrG'34

Prismatic Wedges,-{onsider theprismatic wedge shown in Fig. 34 (a). ln atransverse direction, the position of the centreof gravity will be that of the triangle lBC.In other words, it will be on the median lDand at a distance of two-thirds of AD fromthe point l. Longitudinally, it will be at thehalf-length, and at the same height as in thetransverse direction.

Elfect of Added Weights on Centre of Gravity.-{onsider a seesaw,consisting of a plank balanced over a block of wood. When the block is directlyunder the centre of gravity ofthe plank (G), the see-saw will balance.

\,

Ifa weight, w, were now placed on the end l, that end would move downwards.We should probably say that this happens because the end I has become heavier thanthe other. Let us see what has really happened here. We regard the force of gravityas acting vertically downwards through the centre of gravity of a body. Beforethe weight was added, this force acted directly over the block CD, and the plankbalanced because there was no moment to tip it. The centre of gravity ofa body isthe centre ofall the weight in it and since the added weight becomes, in effect, a partofthe plank, G moves along to G,, toward the endA. The force of gravity now actsvertically downward through G,, producing a moment to cause the end A to movedownwards. This moment will be the product ofall the weight in the body and ofthelever GG.. so that if Zis the total weisht after w has been added:-

The moment = WxGG,.

26 MERCIIANTSHIPSTABILIry

Since the see-saw balanced before the weigbt was added, the above efect mustbe entirely due to the moment ofu about G. So we can also say that ifg is the centreof gravity of the weight, w:-

The momeot = wx Ggl

and thus ZxGG' =wxcg

From this, we can derive a formula for tle shift of the cenae of gra.vity ofa bodywhen a weight is added to it:-

TWo things are obvious from this:-(a) The centre of grnity of a body will move directly towards the eentre ofgrovity of any h,eight added to it.

@ The distance it will move will equal the moment of the added weight about thecmtre ofgrovity ofthe body, divided by the total weight afer the weight has been aMed.

The Effect of RemoviDg Welghts.-Suppose that in the last example, we hadsawn a piece off tbe end A, of the see-saw, instead of adding weight. The end Bwould then move downwards, because the cenfie ofgravity ofthe plank would movetowards it. If we use a similar process ofrearnning to that which we used in the lastsection, we shall see that the moment ofthe weight removed must, agqin, be equal tothe total moment caused by the shift ofthe centre of gravity ofthe plank. Thus, if 7is the toal weight after the piece has been sawn off:-

Two rhings are again obvious:-(a) The centle of grovity ofa body will move directly away from the centreof gravity of any weight removed from it.

@ The distance it will move will equal the monent of the weight rcmovedabout the centre ofgravity of the body, divided by the total weight rcmaining afterthe weight has been removed.

The, Elfect of Shiftitrg Weights.--Consider aflat plate ABCD,having a weight w placed on it. Letg be the cente of gravity of this weight, and G thecentre of gravity of plale and weight together ll't Wbe the total weight of the whole na,cs.

If we remove the weight from the plate, G vdllmove directly away to G,. If we then replac€ theweight so that its centre of gravity is at g,, the cente

cG,=|9fg

Gr{

"a\FlG. 36

FORCES AND MOMENTS

of gravity of the whole will now move from G, to G, in the direction G,g,. We canshow, by working out the moments, that:-

(a) ggr is parallel to Gq

@ lV x GG2: w\ggl

We can say from the above that:-(t) The centre of gravity of a body moves parallel to the shift of the centreof gravity of any weight shifted v,ithin it.

ft) The distance it moves is equal to the weight shifted, multiplied by the shiftof its cente of gravity, all divided by the total weight of the body.

The Use of Moments to find the Centre of Gravity of a Body.-It has alreadybeen shown that for a plank of weight 7 kilos and a line l8:-

Moment about AB : Wx kg-metres.Conversely, if we know, or can find theweight and moment of the plank about lB,we can find r, since

- _ Moment about ,4.B

The Use of Moments to find the Centre of Gravity of an Area.-Areas areconsidered to have moments, but this is more difficult to visualise because an areahas no actual weight. However, if we think of an area as a very thin "plank", ofinfinitesimal weight, we could then imagine it as having a very small moment andthe above formula would apply to it. In this case, there will be unit weight per unitarea: that is, each square metre of the area will weigh, say, p grams.

Total weight of area = p x area in square metres.

Now, if the plank in the last section were so thin as to become an area only theposition of its centre of gravity, relative to l,B could be found, as follows:-

27

Weight of plank (lll = Area of plank x p

Moment of plank about AB = Moment of arca about ABx p

,, Moment of planl about ,48Now' 'r =

Z

^ Moment of aret rbont ABx p

so' t = Arca of plank x p

^ 'oment of area about ,4.Bp cancels out leaving,r=oo=ffiffiffi.g

The moment of an area is tlte area times a distance


To Find the Centre of Grrvity of r Waterplane.-The position of the centreof gravity of a waterplane can be found from the above formula if its area and itsmoment about one end or one ordinate are calculated.

The area can be found by means of Simpson's Rules, as shown in Chapter 2.The moment could be found by dividing the waterplane into a series of nanow

transverse strips. The area of each srip multiplied by its distance from the chosenordinate will give its moment: and the moment of the whole waterplane will be thealgebraic sum ofthe moments ofall the strips.

For example, in Fig. 38, the moment of the strip shown about,4B would be:-Area of strip x mean distance from lB

Ifthe whole waterplane were divided into a series of such strips:-Moment of waterplane = Algebraic sum of moments of all strips, i.e. Sum of all

moments to the right oflE - sum of all moments to the left oflB. Distance of centreof gravity of waterplane from lB: =

H#il#The centre of gravity will lie on that side of,4B which has the greater moment.In practice, the above would be a laborious procedure if it were carried out in

full, but fortunately Simpson's Rules provide us with a convenient short method.When these Rules are used to find an area, they divide the waterplane up, in effectinto a number of narrow strips and then give the sum ofthe areas ofall the strips. If,when using the Rules, we multiply the product for each ordinate by its distance fromthe chosen ordinate, it will give us a product for the moment. The algebraic sum ofthe products of moments for all the ordinates multiplied by one-third of the commoninterval will equal the sum ofthe moments ofall the strips: that is, it will give us themoment ofthe area about the chosen ordinate.

This can best be seen from an example. Let us find the position ofthe centre ofgravity ofa waterplane,250 metres long, which has the following ordinates, spaced25 metres apart, from forward to aft: 0'4:7'8: l7'2:21'1:27'5:30'0:29'3:28'2:22'5:l5'6: l'0 metres. Put these ordinates through Sirnpson's Rules, as follows, to find theposition of the centre of gravity relative to the mid-ordinate.

FORCESAND MOMENTS 29

No. ofOrd.

Length Mult'r Productfor Area

Interval fromMid-Ord.

Product forMoment

I2J

45

0.47.8

17.221.127.5

14242

0.45 l 'z

34.484.455.0

J

2I

2.0124.8103.2168.8)) .u 453'8 for'd

o 30.0 120.0 0 0.0789l011

29.328.222.515.61.0

242

I

58.6112.8

45062.4

1.0

I2J

45

)E.b

225.6135.0249.6

5'0 673'8 aft605.2 220'0 aft

(Strictly speaking, this should be the distance from the mid-ordinate but itmakes for more simple arithmetic if we take the intervals from the mid-ordinate andthen multiply afterwards by the common interval, as shown.)

6rss = {x Produc, 1o, *"u = {x 6 05-2 = 5043.3 m23J

Moment about mid-ordinate = Product for moment x 4 x ftJ

=220x?x2s=4ss33

Distance of centre of eravitv from miJ-ordinate =

V-o-.n, - o'fffArea

=t1ffi=e u9 metres

The greater moment is abaft the mid-ordinate, so the centre of gravity will be9'09 metres abaft that ordinate.

To Find the Centre of Buoyancy of a Ship Shape.-In Chapter 2 it has been shown how we can obtain the volume ofa ship shape,

by putting cross-sectional areas through Simpson's Rules as ifthey were ordinates.Similarly, if we put cross-sectional areas through the process described in the lastsection, we can obtain the position ofthe centre of gravity ofa homogeneous shipshape. The centre ofgravity ofa ship's underwater volume is the centre ofbuoyancy.So if we take a series of equally-spaced sections for the ship's underwater volumeand put them through the Rules, we shall obtain the fore and aft position ofthe centreofbuoyancy. Similarly, a series ofequally-spaced waterplanes, put through the Ruleswill give the vertical position ofthe centre ofbuoyancy.


Example.-A ship's underwater volume is divided into the following verticalcross-sections, from forward to aft, spaced 20 metres apart: l0; 91; 164;228; 265;292;273;240; 185; I I l; 67 square metres. Ifthe same underwater volume is dividedinto waterplanes, 2 metres apart, their areas, from the keel upwards are: 3001'27M;3ll0; 3388; 3597;3759;3872 square rnehes. Find the position of the centre ofbuoyancy (a.) fore and aft, relative to the mid-ordinate, (b) vertically, above the keel.

No. ofOrd. Area Mult'r hoduct for

Volume Interval hoduct forMoment

t0 I l0 ) 502 9l 4 3& 4 1456J l& 2 328 J 984

J)e 4 912 2 18245 265 2 530 I 530 4844 for'd6 292 4 I168 0 07 546 I 5468 240 4 960 19209 185 2 370 -' 1l l0l0 l l l ' t44 4 1776l l 67 I 67 ) 335 5687 aft

5699 843 aft

Volume = 4 x product for volume = 2t 56990 = 37993 rn:

14orn"n = prodult for moment x{ xh= 8$x-4x20=11240033

Position of centre of buoyancy = m

= W

=''ll$rth:t

No. ofOrd.

Area Mult'rProduct for

VolumeInterval

Product forMoment

I 300 I 300 0 02 2704 108r6 1 10816J 3l l0 2 6220 2 124404 3388 4 13552 J 40656) 3597 2 7194 4 287766 3759 4 15036 5 751E07 3872 I 3872 o 23232

56990 l9l100

FORCESAND MOMENTS

Volume = t x product for volume = 2156290 =37gg3 ^z

Moment = product for moment x {

x n = |.)ttOix3xz = 254800

Position ofcentre ofbuoyancv = Yolnent = 2!!=89 = a.ll m above keelVolume 37993

The Use of Intermediate Ordinrtes.-If we are given intermediate ordinateswhen finding the centre of gravity of an area or volume, we use the same basicmethod as above. In this case, however, the multipliers are modified as shownin Chapter 2; whilst the intervals for the moments are measured in intervals andhalf intervals. This is best seen from examples.

Example I .-A waterplane is 60 metres long and has ordinates, from forward toaft of l'0, 6'9, ll'1, ll'2, 8'9 and 5'0 metres. There is also an intermediate ordinateof3'8 metres, midway between the first two forward ones. Find the area and positionof the centre ofgravity of this waterplane.

Let us take moments about the 1 1.2 metres ordinate.

No. ofOrd.

Length Mult'r Product forArea Interval Product for Moment

I 1.0 Y2 0.5 J 1.5

l% 3.8 2 / .o 2% 1902 6.9 t% l0'4 2 20.8J 111 4 44.4 I 44'4 85'7 for'd4 |.2 2 0 05 8.9 4 Jf .o I 35.66 ) 'u 1 5.0 , t0.0 45'6 aft

r25.9 40'l for'd

Common intewal th1 = @ =l 2 metres

Nea = 4 x producr tor area =Px125.9 = 503.6 m2JJ

Moment : Product for mome ntxlx h = 40.lxQxn = i-9i2a.SJJ

position of centre of sraviw = Mgment = 1??! '9Area 503.6

= 3 .8 metres (forward of I 1 . 2 rn ordinate)

JT

the moments about the keel.No. oford. Area Mult'r Product for Volume Interval Product for Moment

I a^ Vz t2 0 0t% 153 2 306 % ls32 ,+00 tk 600 I 600J 807 4 3228 2 u564 to32 I 1032 J 3096

5178 10305

32 MERCHAI.TT SHIP STABILITY

Example 2.-Avessel has waterplanes, 2 metres apart from the keel upward,ot 24,400,807 and 1032 square metres. An intermediate waterplane I metre abovethe keel has an area of 153 square metres. Find the displacement and i(8, when thevessel is floating at the uppermost waterplane.

yoluns = { x product for volume = 1x5fi8 =3452 m3

Displacement = iolumex I . 025 = 3425 x I . 025 = 3539 tonnes

Moment = product for Moment x { x h =l03}5x?x2=i.i7a0

*=m=ffilQ=3.eemetresAppendages.-An appendage is a small area (or volume) adjacent to the main

area (or volurne), which has for some reason not been measured into the main one.Its effect on the centre of gravity can be found by applying its area (or volume) andmoment about the choseo ordinate. to those for the main area.

For example, suppose that the waterplane given in the example on waterplanes,had an appendage of arca 245 square metres, with its centre of gavity 135 metesabaft the mid-ordinate:-

Main area wasAppendageTotal ares

5043 m2245 m2

5288 m2

:245 x 135Total Moment

Moment of main area was 45833 aftMoment of appendage = area x distance mid-ordinate

3307578908

aftaft

Position of centre of gravity of area and appendage, abaft the mid-ordinate =

W=#=l4.emetres.The effect of an appendage on the centre of gravity of a homogeneous ship

shape can be calculated in the same way.

FORCESANDMOMENTS 33

Inertia.-A stationary body resists any attempt to move it and a moving bodyany attempt to change its speed or direction. This property is called "inertia" and acertain amount of force must be exerted to overcome it. If we consider what wouldhappen if we tried to play football with a carnon ball, it should be obvious thatthe greater the weight of the body, the greater will be its inertia. Thus, the weightof a body gives a measure of its inertia so far as ordinary non-rotational motion isconcemed.

Moment of Inertia and Radius of Gyration.-It has been shown earlier in thischapter that in ordinary motion, the behaviour of a body depends on the amounts oftheforces applied to it: but that where a tuming or rotational movement is attempted, thebehaviour ofthe body depends on the moments ofthe forces applied. In a somewhatsimilar way, although the inertia of ordinary motion is govemed by weight, theinertia of rotational motion is govemed by a quantity called its "moment of inertia",or "second moment of area". There is this difference, however, that both inertia andmoment of inertia are independent ofthe forces applied to the body. Roughly speakingwe may say that in the case of ordinary motion, the greater the weight, or inertia, thegreater the resistance ofthe body to being moved; in the case ofrotational motion, thegreater th€ moment of inertia, the greater the resistance to rotation

What is moment of inertia andupon what properties ofbodies doesit depend? Consider the see-sawshown in Fig. 39 (a), consisting ofawooden plank, lB, balanced acrossa block, CD. If it were struck adownward blow at l, it would startto oscillate up and down. Supposethat the wooden plank were nowreplaced by an iron one of exactlythe same dimensions. Strike thisiron "plank" a blow of exactly thesame force as before and it will alsooscillate, but much less than did thelighter wooden one. This showsthat moment of inertia depends onweieht.

Suppose that we now replace the pla* by a shorter one of equal weight, asshown in Fig. 39 (b). Ifthis plank is struck a downward blow with the same force asbefore, it will oscillate quite violently. Since the weight is the same, the increase inthe rate ofoscillation must be due to the decrease in length, so we can conclude that

{.

{A

FtG.39

B".;7j

I

J


moment of inertia will also depend on length. Unfornrnately, it is not proportionalto actual length, but to the square of what we may term the "effective length" oftheplank. This effective length is called 'ladius of gyration ' and it can be shown thatwhere / is the actual length:-

P.adius ofevration = ffi

MomeDt of inertia : weight x (radius of ryration)2.

Thus, where I is the moment of inertia, m is the weight of the plank, and / theactual l€ngth:-

If the plank is ofunit weight per unit length (i.e., if it ha4 say, a weight ofonekilogramme per metre of length), zr and / will be numerically the same (i.e., a 5metre plank would weigh 5 kilogrammes, so that rz and / would both be 5).

For our purpose, we can assume this and substitute / for rz in the above formula,which becomes:-

One factor which we have not so far considered is that momeDt of inertiaalso depends on the position of the axis about which the body is assumed to rotate.Throughout the above, we have only considered the plank of a see-saw rotatingabout ie cenhe-line and the above formula only holds good for such rotation. In thecase ofrotation about other axes. the formula will be modifed.

Moment of lDcrtia of r Body About its Centre,Line.--{ur discoveriesabout the mom€nt of inertia ofplanks also extend to bodies ofany size. The integral

calculus is involved in most cases, but roughlyspeaking, the method of calculating the mom€nt ofinertia about the axis ofrotation is to divide the bodyinto strips, which may be likened to planks. The totalmoment of inertia of all the planks will give that ofthe body. For instance, the moment of inertia of thebody shown in Fig. 40 about its centre-line,,4B couldbe found by dividing it into a number of rhin "planks"as shown. If the moment of inertia of each ' plank"were then found. the sum ofall ofthese would be themoment of inertia of the whole bodv.

r=+

| =!-

FORCESANDMOMENTS 35

Moment of Inertla of r Wrterphne rbout its Centre'Line.-If the bodyshown in Fig. 40 were infinitely thin, it would resemble a ship's waterplane and it isobvious that the moment of inertia of such a waterplane could roughly be found as

described above. For ship shapes, this involves the useofthe calculus, but for box shapes, having a rectangularwaterplane, the calculation is simpler. Consider sucha waterplane, PQRS having the centre-line CD. Letl be its length and D its breadth. If we consider thiswaterplane to be composed of a number of infinitelythin planks parallel to QR, the momant of inertia willbe equal to the sum of the moments of inertia of all theplanks. The -moment ofinertia ofeach plank about CDwill be:-fr-

Since the planks, placed side by side, extend over the length I we could say thatif there were one plank for each unit of length, there would be / planks. The totalmoment of inertia of / similar planks would be / times that of one plank, so where 1is the moment of inertia ofthe area, / its length and D its breadth:-

So, for a rectangular waterplane:

Equilibrium.-This may roughly be defined as the state of balance of a body.There are three states of equilibrium-stable, neutral and unstable- and the bodymay be in either ofthese, according to the relative positions of the centre of gravityand the point of support.

A ball resting on a surface can be used to demonstrate the three states ofeouilibrium.

Fig. 42 shows a ball balanced on a ooncave surface. If the ball moves to theside it will continue to roll off the surface. It is said to be unstable or in unstableeouilibrium

Fig. 43 shows a ball balanced on a convex surface. If the ball moves to the sideit will roll back. It is said to be stable or in stable eouilibrium

r - lb-

MERCHANT SHIP STABILMY

(c I

36

Fig. zl4 shows a ball on a flat surface. If the ball moves to the side it neithert€nds to roll away or roll back. It is said to be neutal or in neutral equilib'rium

Stable means tbat if moved it will tend to move back.

Unstable means that if moved it will move awayNeuhal means that if moved it has no t€ndency to move back or move away.

All the above are general cas€s, and apply to the equilibrium of any body. Thespecial cases of ships will be considered later.

CHAPTER4

DENSITY, DEADWEIGHT AND DRAFTWe have already seen that the volume displaced by a floating body varies

inversely as the density of the water in which it floats. This means that ifa ship'sdisplacement remains the same, her draft will increase if she enters water of lessdensity; or will decrease if she enters water of greater density.

We also know that a ship which loads cargo (i.e. increases her deadweight) willincrease her draft, whilst a ship which discharges cargo will decrease her draft, ifthedensity ofthe water does not change.

The Effect ofDensity on Draft ofBox Shapes,-In the case ofbox shapes, thevolume displaced is equal to the product oflength, breadth and draft; so we can say:-

New volume displaced Old densitvbiacffitEd;a- = N"*den#t

Length x Breadth x New Draft _ Old densityLength x Breadth x Old Draft New density

New Draft _ Old densityOld Draft New density

Effect of Density on Draft of Ship Shapes.-These also increase their draftwhen the density ofthe water decreases and vice versa, but in this case the change isnot in proportion and its calculation is more complicated. We overcome this difficultyby giving each ship a "Fresh Water Allowance" when her loadlines are assigned.This allowance is the amount by which the ship's draft will change when going fromsalt water to fresh water or vice versa when at sumrner load line displacement.

The ordinary loadlines show the draft at which a ship can safely remain at sea.In the smooth water ofa harbour or river, it would be quite safe to load her a littlebelow these marks, provided that she rises to them when or before she reaches theopen sea. A ship loading in a harbour of fresh water could submerge her load linesby the amount ofher fresh-water allowance, since she would rise to her proper loadline on reaching salt water.

Ships often load in dock water which is brackish-that is, which has a densityof more than 1000 and less than 1025 kg/m3. In this case, the amount by which the

38 MERCHANTSHIPSTABILITY

salt water load-lines can be submerged is found from the freshwater allowance bysimple proportion. This amount is often called the "Dock Water Allowance",

Let x equal the amount by which the load-line can be submerged in water ofdensity, d. Let Fa equal the fresh-water allowance.

A change of density from I 000 to I 025 kglm3 will produce a change of draft ofFZI millimetres.

A change of density from d to 1025 will produce a change of draft of .rmillimetes.

We can assume, for practical purposes, that the change of draft is proportionalto the change ofdensity. Hence:-

x - 1025-6FWA 1025-1000

FWA(r025-6\, = __-

25-

Note that where, as in this case, we are dealing with change ofdraft and changeof density, these are directly proportional. In box shapes, when we are dealing withactual draft and density, they are inversely proportional.

Example l.-A ship floats at a draft of 5.42 m in water of density l'007 t/m3. Ifher fresh-water allowance is 242 mm, what would be her draft on passing into saltwater?

Dock water allowanc e 1"1= FwA(9zs - a)

_242 (102s-1007)25

= 174 mm, or 0. 17 m (Rise)Old draft =5.42 mNewdraft=ll! m

Exanple L-A ship floats at a draft of 6.83 m in water of density 1.022 tlms.What would be her draft in water of density I .0 I 0 fin3, if her fresh water allowanceis 156 mm?

F7,{ x change of densig _ 156 (1022-1010)25 25

= 75 mm, or 0. 08 m (sinkage)Old draft = 6.83 m

Newdraff = q:2!m

G mf*#yr$"r":#, :rlimlft sto be ,4, and one centimetre to represent the changeof draft, we can see how the above is arrived at.

DENSITYDEADWEIGHTANDDRAFT 39

Tonnes per Centimetre Immersion-"T.P.C."-A ship must always displaceh o*l weight of water. If a weight is added to her, it will cause her to sink until* dkplaces an extra layer of water of equal weight. The tonnes per centimetreLasion is the weight which must be added to cause the ship to sink one centimetre

-'-rhat

is, to increase her mean draft by one centimetre.

h is safe to assume that, for all practical purposes, two waterplanes which areG centimetre apart will have the same areas and also that the ship's sides will beEid b€tween them. The volume of the layer between such water-planes can thusbosidered as having the area of one waterplane and as being one centimetre thick.|lb ctnrimetre is one-hundredth of a metre so, ifl is the area of one waterplane inqre metres, the volume of the layer will be:-

A100

cubic metres

FtG.45

h order to submerge the above layer, it is necessary to add to the ship a*ilr equal to the weight of water which the layer will displace. Since the layer. a certimetre thick, this weight is that which is necessary to sink the ship one

-'nete bodily, or the "Tonnes per Centimetre Immersion".

Let d be the density ofthe water in which the ship floats.

Weight of the layer = volume x density = $xdIUU

so. r.p.c. = 4l 1' 100

Eunple l.-A ship floats at a waterplane of area 1520 m2. What is her T.P.C.h raer of density 1'020 t/mr.

1.p.6, ._ 5Xl _ 1x02011520 = 15.5 tonnes100 100

Erontple 2.-A ship is 120 m long, 18 m beam and the coefficient of finenessdb s'aterplane is 0'788. Find her T.P.C. in salt water.

f .p.C.= I 025x,4 - | '025xl20Xl8x0 788 = 17.4 tonnes100 100

F',1:*$,;1I.',c

fi;f

:


To Adjust T.P.C. for Density.-If, as is usual, we are given a ship's T.P.C. insalt water, we can adjust it for water ofother densities by direct proportion:-

r.p.L* d" ity q = a

T.P.C. i" salt *"t"r - 1.025

T.P.C. in density 6 = T.P.C. in salt water x l. &,

Exarnple.-Aship has a T.P.C. of 18.35 tonnes in salt water. What would it bein waGr of density 1.010?

r.P.c. indensity 1.010=18-35x1*i3 = 18.08 tonnes.

Loading to a Given Lordline.-To fnd out how much to load in order to floatat a given loadline 61 1666[ing salt wa!er:-

(a) Find the ship's present mean draft or freeboard. If she has a list, thefreeboards on the Port and Starboard sides will be different: if so. take themean of the two.

(b) Calculate the dock water allowance and apply this to the requiredsalt water draft or freeboard. This will give the allowable draft or freeboardto which the ship can be loaded in the dock water.

(c) The difference benneet (a) and (b), above, will be the allowablesinkage in the dock water.

(d) Adjust the T.P.C. for the density ofthe dock water.

(e) The allowable sinkage, multiplied by the adjusted T.P.C. will be theamount to load to bring the ship to her appropriate load line on reachingsalt water.

A If the ship will use fuel, stores, etc., after leaving her berth, but beforereaching salt water, this will reduce her draft to less than that allowable. Tocompensate for this, extra cargo, equal to the weight of fuel and stores so used,may be loaded before sailing.

Example l.-A ship is loading in an upriver port, where the density of thewater is l'006 t/m3. Her present freeboards are 1832 mn on the Port side and1978 mm on the Starboard side. Her statutory summer freeboard is 1856 mm;Fresh water allowance is 148 mm; and her T.P.C. is 18.62 t. On the voyagedownriver, she is expected to use 24 tonnes of fuel and 5 tonnes of stores andfresh water. Find how much more cargo she can load to be at her summer loadline in salt water.

DENSITY, DEADWEIGHT AND DRAFT

present mean free6sn16 = 1E32j.1928 = 1905 mm

Dock water allowance = f;x

la8 = I l2 mm

Summer freeboard 1856 mmAllowable fieeboard (in dock water) 1744 mmPresent 1905 mmAllowable sinkage (in dock water) _Il run (16.1cm)

. 1.no6T.P.C. in dock water = f f ix lS

62 = 18'21 t .

Total to load = 16. l x 18.27

Cargo to replace fuel, etc., used

Cargo to load

7 500 m

0.105 m or 10.5 cm

4l

294 tonnes

29 tonnes

323 tonnes

Example 2.-A ship anchors off a Port, in salt water, with the upper edge ofher Tropical load line 3'0 cm above water. Her Tropical draft is 7.562 m; T.P.C. is22'54 t.; F.W.A. is 153 mm. She is to discharge some cargo into lighters in order toenter a dock, where the density of the water is 1.013 t/m3 and the depth on the docksill is 7'70 m. Find the minimum amount ofcargo to discharge in order to cross thedock sill with a clearance of20 cm, assuming that the vessel remains on an even keelthroushout.

Present draft = 7 .562-0.03 = 7.532 m (in salt water)Required draff in dock water =7.70-0.20=7 50m

Dock water allowance = .!a x | 53 = 73 mm 0.073 m

SW draft + DW allowance 7.532 + 0.073 = 7.605 mRequired DW draft

Required rise in DW

Dock water T.P.C.= 22.54x1.013x1.025 =22.28

Cargo to discharge=10.5x22.28= 234 tonnes

CHAPTER 5

CENTRE OF GRAVITY OF SHIPSCentre of Gravity of e Ship-"G-.-Is the point through which all the weight

ofthe ship is considered to act vertically downwards.

A ship may be regarded as a hollow shell, inside which weights may be adddremoved, or shifted about. The position of the centre of gra.vity will change withevery condition of loading and must be calculated each time that the ship's stabilityis to be found. The transverse and longitudinal positions are always consideredseparately, as in the case of any other body (see Chapter 3). As far as the Fansverseposition is concemed, G is usually assumed to be on the centre-line; since if it werenot so the ship would list. Longitudinally, it may be forward of, or abaft midlengthand is considered accordingly.

*[email protected]\e vertical heigbt of the centre of gravity above the keel is usuallycalled "KG". This is due to the fact that, in stability diagrams, K is usually taken todenote the keel and G the c€nte ofgravity.

"Light .KG.'-The height of G above the keel in the ligbt ship, before anycargo, storcs or fuel are placed on board, is calculated by Naval Architects. It is givento the seaman in the ship's stability information.

Before a ship is built, the KG is estimated, usually by comparison with someexisting ship of similar size and lines, although in some unusual cases it is actuallycalculated approximately. The KG of the completed ship, when light, can be foundby means ofthe "lnclining Experiment", which will be described later.

Shift of sG'.-The cenfe of gravity of a ship obeys the same laws as that ofany other body. Let us summarise the conclusions which we drew in Chapt€r 3 withregard to this matter.

G moves directly towards the centre of gravity of any weight added to theship, directly away from the centre of gravity of any weight taken away from theship and parallel to the shift of the cenae of gravity of any weight moved from oneplace to another.

CENTREOFGRAVITYOFSHIPS 43

The distance t}rough which G will move can be found from the formula:-

GGr=#

Where GG, : the shift of G.

w = the weight shifted

d: (for weights removed or addedFthe distance from G to the centreof gravity of the weight,

(for weights shiftedFthe distance through which the centre ofgravity ofthe weight is moved.

gr: (for added weightsfthe displacement after the weight has beenadded,

(for weights removedfthe displacement after the weight has beenremoved,

(for weights shiftedfthe displacement of the ship.

Example l.-Find the shift of the centre of gravity of a ship of 7000 tonnesdisplacement, ifa weight of50 tonnes is shifted for a distance of80 metres

oo,=#=1#=0'57 metres

Example 2.-A ship has a displacement of 3200 tonnes. What would be theshift ofher centre of gravity ifa weight of200 tonnes is added at a distance of60metres ftom the original position of her centre of gravity?

(Where Z equals the displacernent after adding the 200 tonnes)

CG' =w\d =2W'. |90 =3'53 metres' W J4UU

Example 3.-Aweight of48 tonnes is removed from a lighter, the centre ofgravityofthis weight being 2'0 metres above tlre keel. What will be the new KG of the lighterif its original displacement and KG were 690 tonnes and 6'2 metres, respectively?

d =6.2-2.0=4.2 metres

7 = Displacement after weight has been removed= 690 - 48 = 642 tornes

GG, =re\d = 48)! '2 = 0.31 metres-- , w 642G moves vertically upwards, away from the weight removed, so:

New KG = Old KG + Gq = 6.2+ 0.31 = 6. 51 metres


"f,G- foreny Condition of Loeding.-The naval architects who build a ship,give the seaman her light KG and displacement. The KG for any other condition ofloading must be found by inspection ofthe hydrostatic data supplied.

When weights are added to the ship, G will move upwards or downwardsaccording to whether the centre ofgravity ofthe weight is above or below that oftheship. (Note that we are here only considering the shift ofG in the vertical direction.)

In this case, we could use the method just described to find successive shiftsof G due to each weight: but this would be laborious and errors could easily creepinto the calculations. A better and simpler method of finding the new r(G is to takemoments about a horizontal line through the keel, known as the "base line".

Now, moment is weight multiplied by length of lever. In this case, the lenglhof lever will be the distance from the base line to the centre ofgravity of the weight(sometimes written as Kg).

So the moment of each weight : w x rKg

And final KG = Total moment : total weight.

The method now becomes:-(a) The ship's original displacement and iKG are multiplied together to giveher original moment.(b) Each weight, added or removed, is multiplied by its height abovethe base line, to give its moment, Added weights and moments are added to thoseof the ship. Weights removed and their moments ar€ subtracted.(c) The totalmoment, dividedbythetotalweight, willgivethenewKGofthe ship.

Example l.-A ship arrives in port with a displacement of 4250 tonnes and l(Gof 5'96 metres. She then loads 520 tonnes at 6'3 m above the keel; 1250 tonnes at4'2 m above the keel; 810 tonnes at I l'6 m above the keel. She also discharges 605tonnes from 2'4 m above the keel. What will then be her KG?

Ship LoadedWeight Heisht Moment4250520t250810

s.966'34'211.6

2.4

25330327652sO9396

6830- 605

43252- t452

6225 41800

Discharged

Totals

1J"on K6 =.rglal-lqoment. = art8rT = o. zr ."t ".


Example 2.-Aship has a light displacement of 4559 t and light iKG of 6.62 m.Find her rKG and displacement when she has the following weights on board:-

Cargo 730 tonnes at 5.1 metres above the keel.480 tonnes at 9.3 metres above the keel.

1470 tonnes at 4.7 metres above the keel.860 tonnes at 8.5 metres above the keel.

Fuel 355 tonnes at 0.6 metres above the keel.Stores 60 tonnes at 12.4 metres above the keel.F. Water 80 tonnes at 8'8 metres above the keel.

Weight Height Moment4559 6.62 30181730480

t470860

) .1

9.3

6)

. t tz5

446469097310

J)) 0.6 21360 12.4 74480 8.8 704

8594 54248

New displacement = 8594 tonnes.

N"* KG = r-IE#ffi = ffi = o . : r metres

Nort.-The KGs which we have found above are often called the "Solid KG'.Ifthe ship has free liquid on board, it may be necessary to apply a correction to this,in order to find the "Fluid i(G', as we shall see in Chapter 9.

Deadweight Moment.-This term is sometimes used to describe the sum of themoments about the base line of all the items, which make up the ship's deadweight(i.e. cargo, fuel, stores, etc.). It does not include the moment of the light ship. Sowe can say that:-

Total moment : Moment of light ship + Deadweight moment.

For instance, in tr'Janp le 2, above:-

Moment of light ship = 36131 y-

Deadweight moment = 24067 tlm

Total moment = 5 4248 t/m

Light ShipCargo

FuelStores

F. WaterTotals


Real and Vlrtu.l Centre of Gnvity.-If a weight is free to move about,the position of its cenhe of gravity is moving. If the weight is moving in an arc,then the position of its ce,ntre ofgravity can be considered to be at the centre ofthearc instead of being at its real centre of gravity. The position where its cenfe ofgravity can be considered to act is called the "Virtual Centre of Gravity" and forall stability calculations, the centre of gravity ofthe weight it considered to be atthis virtual point.

An exarnple of this can be found in the case ofa weight, which is being liftedby a derrick. Suppose the weight to be one centimetre off the bottom of the hold"The real cente of gravity has scarcely moved-cerlainly not enougb to affect thecente of gravity of the ship appreciably. If the weight is now free to swing in anarc. the cente of which is at the head of the denick. so that G will move as if thecentre of gravity of the weight were actually at the head of the derrick and notdown in the hold. t ih

Fig. 46 shows a weight being lifted bya derrick. When the weight is resting on thebottom of the hol4 its centre of gravity and itseffect on the ship is at g. As soon as the weighthas been lifted clear of the bottom of the hol4so that it is free to swing, its virtual ceotre ofgravity appears at the head ofthe denick, ft. Theeffect on the ship is now as if the weight wereactually moved from g to lr. The virtual centreof gravity will remain at i, whatever the actualposition of the weight (i.e. whether itbe at & g2or gr) provided that the weight is free to swing.

FtG.46

A similar effect occun in "slack" tanks, which contain liquid which is free tomove. At small angles of heel the cente of gravity of the liquid moves in an arc.This is called "Free Surface Effect" and causes a virtual centre ofgravity to appear atthe centre ofthe arc, above the actual centre of gravity ofthe liquid in the tank. Thiseffect is discussed more fully in Chapter 9.

A somewhat similar effect is also found where chilled meat is hung from beamsin a hold and is more or less free to move.

Exanple l.-A ship displaces 6200 t and has a i(G of 6'12 m when a lift of 80 tis resting on the inner bottom, at 2'50 m above the keel. What would be the ship's iKGwhen the lift was being hoisted by a derrick, the head ofwhich is 2l m above the keel.

Head of denick is 21 - 2'5 = l8'5 m above c.g. of lift. So c.g. of the lift risesl8'5 m when lifting.


GG - \vld = 80x1*& 5 =0.24 metres (upwardslw 6200

Old KG = 6. 50 metres

New KG = q:4 metres

Example 2.-A ship displaces 7400 t and her centre of gravity is on the centreline at 6'50 m above the keel. A locomotive, weighing 100 t, is to be lifted from thequay by means ofa derrick. The head ofthe denick, when lifting, will be 25 m abovethe keel and 12 m to starboard ofthe centre line. Find the position ofG whilst thelocomotive is beine lifted. To find the new r(G:-

Loco

Weight Height Moment7400 6.50 48100

100 25.00 25007500 50600

NewI(G=#=6x75 metres

To find the horizontal shift of G:-

oo, = w\d = tg=x^12 = 0 . I 6 metres to starboard of the centrelinew 7500

The Elfect of Tanks on *G-.-When a tank is filled with water or oil, weightis added to the ship and G will move directly towards the centre of gravity of thetank. When a tank is emptied, the reverse happens and G moves directly away fromthe centre of gravity of the tank. The distance it moves will be the same as for anyother weight added or taken away and can be found by the same formula, viz.:-

F]G.47A

One or two examples will illustrate this more fully.Ifa forward double-bottom tank, having its centre of gravity atl were filled, G

would move forward and domward to G, If a double-bottom tank, having its centre

Ship

GG. = !z-dW

"t:,,'.*.- ----


of gravity at,8 were filled, G would move vertically downward to q. If a tlnk aft, atC, were filled, G would move directly toward C, downward and aft.

In the case ofa deep tanlq the one shown nFig,47 (a) is abaft G and has itscenhe of gravity (D) level with it, so in this case, G will move horizontally aft toG' directly towards D. The relative positions of the centre of gravity ofa deep tankand that of the ship vary considerably with different ships and different conditionsof loading. So the above case is only an example and not an invariable rule. Fillinga deep tank may have a very different effect from that shown here and the only nrlethat can be laid down is that G will move directly toward the centre ofgravity oftheliquid in the tank.

When tanks are emptied, G will move in exactly the opposite direction fromthe above in each case. That is, if ank I were emptied, it would cause G to moveupwards and aft; tank B would cause it to move vertically upward; tank C, upwardand forward; and tank D, directly forward.

Now consider the transverse effect offilling the same tanks. As long as the tanksare symmetrical about tle centre-line of theship, their centres of gravity will be on thatline. Thus, so far as the transverse stability ofthe ship is concemed, G is merely assumedto move vertically upwards or downward.For instance, if we filled or emptied thewhole rqnk here, G would move down to G.or up to q.

If the tank weie not symmetrical about the ceotre-line of the ship, its c€ntre ofgravity would be to one side ofthat line, so that G would move sideways as well asup or down. For instance, ifwe only filled one side ofa double-bottom tank, with itscenE€ of gravity at g then G would move in the direction shown to Gr.

When it is necessary to calculate the effect offilling or emptying tanks they aretreated as ordinary weights added or removed. The vertical shift of G is found bytaking the weight and KG of the ship, weight and i(G of the liquid in the tank andthen adding or subtracting the moments (see "iKG for any condition of loading").

It must be remembered that, in all tbe above cases, the tanks are considered asbeing completely filled or emptied. Ifthey are not, the free surface ofthe liquid maycause a virtual cenhe of gravity, which will produce some very different effects, asdescribed in Chapter 9.

G.t

Giui\i't \

FIG.47B

CHAPTER 6

CENTRES OF BUOYANCY AND FLOTATIONCentre of Buoyancy-"8".-This is the geometrical centre of the underwater

part of the ship. That is, it is the centre ofm-:--i+olsplaceo.

Tllforce qf-Quoyaqcy is considered to act vertically up-rvard through the centreof buoyggfr, with a force equal to the weight of the displaced water. Since, by thii-I76-of Archimedes. a floating body displaces its own weight of water, this forcemust be equal to the weight ofthe ship.

The transverse and longitudinal positions ofB are always considered separately. Inthe transverse direction, B will always be on the centreline as long as the ship is upright,but will move out to one side ofthis line when she heels. Longitudinally, B may be a littleforward ofor abaft the midlength. (The method offinding B is given in Chapter 3)

(KR'.-This is the vertical height of the centre of buoyancy above the keel.For box'shaped ships, floating upright, KB is always equal to one-half of the draft;this is obvious if we consider B as the centre of gravity of the underwater body. Forship shapes, KB is about 0'52 ofthe draft and can be found by seamen fiom the ship'sstability curves or scales or tables.

Centre of Flotation - *F".-This is the point about which the ship heelsand trims. Consider a shipwhich is heeled to a smallangle, as shown in Fig. 48. Letthe plane STMN be the originalwaterplane and the planeSTTTPQ be the new waterplane.The wedge SS,FNpE hasemerged fiom the water,whilst the wedge FTTTPMEhas become immersed. Theseare known as the "emergedwedge" and "immersed wedge".respectively.


By the Law ofArchimedes, a ship must displace her own weight of water at alltimes if she is to remain afloat. Thus, the volume which she displaces when heeledmust be the same as that which she displaced when upright; so that the volumesof the immersed wedge and of the emerged wedge must be equal. When the sidesof the ship are parallel, the line forming the apex of each wedge must divide eachwaterplane into exactly equal areas. For instance, in the figure, the line EF mustbe such that ttre area SIfEF is equal to the arca TMEF and lhe area S,QEF is equalto the area ?,PEF. This will hold good whether the ship swings longitudinally ortransversely, or in any other direction. It is obvious that all such 'tentrelines" mustcut each other at one point-{he geometrical centre of each waterplane; or, in otherwords, the centre of gravity of the waterplane.

ln box-shaped ships, the centes ofgnvity ofthe upright and heeled waterplanesmust coincide, unless the deck-edge becomes submergd or the bilge emerges fromthe water, In the case of ship-shapes this is not strictly true, but for small angles ofheel or trim it can be taken as correct for all practical purposes. This gives us a newdefinition for the centre of flotation, namely that the centre offiotation is the centreof gravity ofa shipb waterplane.

The transverse position of the centre of flotation is always at the c€ntreline ofthe waterplane; that is, the intersection of the waterplane and the centreline of theship. Longitudinally, it is in the waterplane and at the mid-length for box shapes; butmay be a little abaft or forward of the mid-length in ship shapes. Chapter 3 showshow it may be found.

FtG.49

Shift of "B'.-The centreof buoyancy is the cenhe ofgravity of the water which hasbeen displaced by a ship. It obeysthe same laws as any other centreof gravity. Fig. 49 representsthe ship shown in Fig. 48, as itwould be seen transversely. ,9,lrepresents the original waterplaneand S, T, the new waterplane whenthe vessel is heeled. S,S.F is theemerged wedge and 77,F the

centres of gravity at g and g' respectively. I isthe position of the centre of buoyancy before the ship heeled. The effect of heelingthe ship is the same as if we took the emerged wedge away and placed it in theposition ofthe immersed wedge. Since the centre ofgravity ofa body moves parallelto the shift ofthe cenhe of gravity ofany weight shifted, B must move out to B, in adirection parallel to the line gg,

immersed wedge, which have theu

CENTRXS OF BUOYANCYAND FLOTATION 5I

The distance which .B will shift can be found by taking moments, in the sameway as for the shift of any centre of gravity. Let Wbe the displacement of the shipand w the weight of water in either wedge.

Now, for the centre of gravity, l{ x GG, = y'v /

So, for the centre ofbuoyancy, W x BB, = wx gg,

The weight of any body equals its volume in cubic metres multiplied by itsdensity. So if I/ represents the volume of displacement of the ship; v, the volumeof either wedge; and d the density ofthe water; the above formula can be modified,thus:-

Il x BBt = wx gg,

oo _wxggtDDt_-- f r -

RR, =vx6 x ggl-- t Vx6

vY ooBBr=::;et

All the above will also be true if the ship changes trim.Example.-A ship displaces 28,800 cubic metres and is heeled so that the

volume of the immersed wedge is 1550 cubic metres. The distance between thecentres ofgravity ofthe immersed and emerged wedges is found to be 9 metes. Findthe shift of B.

84 =ry-=5#=0.484 metres

Horizontal and Vertical Components of the Shift of 6B'.-For moreadvanced problems in stability, it is sometimes necessary to know how far B will

FtG.50shift in either the horizontal orthe vertical direction only, whenthe ship heels. Fig. 50 shows thesame ship as in the last example,heeled and with B moved out to

!l-- 8.. In the new vertical direction.: B' h". moved downwards foithe distance RB, In the newhorizontal direction, it hasmoved out for the distance BR.These two components of 88,can be found as follows:-


Drop perpendiculars from g and g, on to the new waGrplaue S,I,. Let theseperpendiculars be gh ard grh,

For the horizontal shift, BrR:-The shift ofg in the horizontal direction is &i,.If we take moments in this direction about any vertical line, the moment of thewedges must equal that of the ship. Thus, using volume in lieu of weight, as in thelast example.

V x BR=vx hl\

^^ vxhhon=__V_

For the vertical shift, B,R:-j has risen for a vertical distance ofgft above thehorizontal, S,I,. g, has fallen for a vertical distance of Srir below the horizontal.Thus, the total vertical shift of g is (gh + grhr). If we take moments about anyhorizontal line, we shall see that:-

YxBrR=v(gh+ grhr)

4R=v(sh!s\)

CHAPTER 7

THE RIGHTING LEVERAND METACENTREEquilibriun of Ships.-We have seen in Chapter 3 that a body's state of

equilibrium determines whethet when it is tilted, it will right itself, remain as it is,or tilt more. Seamen are, naturally, very much concemed as to whether their shipswill remain upright and so the study of equilibrium forms an important part ofshipstability.

In the normal ship, the centre of gavity is always higher than the centreof buoyancy; that is, KG is greater than KB. The force of gravity acts verticallydownwards through the former and the force of buoyancy vertically upwardsthrough the latter. As we have already seen, these two forces must be equal. It hasbeen shown in Chapter 3 that the equilibrium of a tilted body depends on the relativepositions of the centre of gravity and the point of support. Unless one is verticallyover the other, the body will try to tum in one direction or the other. This will holdgood for ships, if we substitute "centre ofbuoyancy" for "point of support"; thus fora ship to remain at rest, G must be vertically over B.

When a ship is upright andcorrectly loaded, ,B and G will bothbe on the centreline, as sbown inFig. 51. The forces of gravity andbuoyancy will be equal and oppositeand there will be no tendency for theship to move from the upright. Assoon as any other transverse forceacts on her, however, she will heel.The centre of buoyancy will moveout towards the low side and the

results ofthis movement will decide whether the shio is in stable. neutral or unstableequilibrium.

In the next six sections we shall consider Figs. 52 (a), (b) and (c), which show aship inclined by some extemal force such as wave action. In each case, B has movedout to Br but since no weights have been moved on board, G remains in its original

- - - - - - t

EQUIUERIUT,t

EQUILIER'UM

tz/ . \

r ! l. . . . - . - - . - :

€OU'UERIUM

THE RIGHTING LEVERAND METACENTRE 55

position on the centre line. The force of gravity now acts venically downwardsthrough G, in the direction Q and the force of buoyancy vertically upwards throughthe new centre of buoyancy, in the direction 8,.r. These two forces form a "couple",and, in the case ofFigs. 52 (a) and (c), are trying to tum the ship in one direction orthe other.

The Righting Lever- "GZ".-In Figs. 52 (a) and (c), a horizontal lne, GZ,has been drawn perpendicular to B,x. This perpendicular distance betrveen G and thedirection ofthe action of the force ofbuoyancy is called the "righting leveC', on theends of which the forces ofgravity and buoyancy act to produce a tuming movement.It can be seen that when this lever is on the immersed side ofthe ship, she will try toright herself(Fig. 52a); when it is on the other side, she will try to heel firther over(Fig. 52c); when it does not exist, there will be no tuming effect (Fig. 52b).

The Metacentre-'M'.-In Fig. 52, the point at which the force of buoyancy,acting in the direction B,x, cuts the centreline ofthe ship, has been rnarked M. Forsmall angles ofheel, up to about l0 or l5 degrees, the shift ofthe centre ofbuoyancyis an arc of a circle and Mis the centre of the arc and for these small angles may beregarded as a fixed point. This point is called "the metacentre". Mmay be consideredas the point at which a vertical line upward through 8,.r when the ship is heeled asmall amount cuts the centreline

For larger angles of heel, I moves out more quickly and this causes Mto moveand it can no longer be regarded as a fixed point. A point called the "pro-metacentre"is sometimes considered to exist in this case. This point changes its position withevery change of the angle ofheel and may not even be on the centreline, so it is notused for ordinary stability calculations.

Metacentric Height-'GM"-This_ rg !r9_ disglce -betwee[jhe- a9[8.9 .-of

-Cruuj][3n.1]E 9999"$re.Jt he: a definite relationship with.GZ and, since it iseFrer to find than GZ, C_4 ,_r:nSl.lrlp9* for calculating stability at small angles of

_!99!.-It carp,o1b9 9s94 for ?lg!9!_gfggri!gg!_l! degees, since the metacentre thenhas moved. GMis termed "positive" if G is below Mand "negative" if G is above M.

Notr.-TGGi. :'-nel!frt oFthe metacentre" is sometimes used in stability todenote the height of M above the keel; that is, KM. It should not be confused withmetacentric height, or GM.

Stable Equilibrium.-A ship is said to be in stable equilibrium when, if shewere inclined by some extemal force, she would try to retum to the upright. If weconsider Fig. 52 (a), we shall see that this condition will exist when:-

L For small angles of heel, the ship has a positive GM.

2. For any angle ofheel, the righting lever, GZ, is on the low side ofthe ship.

56 MERC}IANT SHIP STABILITY

Neutrel Equilibrium.-A ship which, if heeled by some extemal force, wouldhave no tendency, either to retum to the upright, or to heel further over, is said to bein neutral equilibrium. We can see from Fig. 52 (b) that this will occur when G andMcoincide, so that GM and GZ equal zero.

Unstable Equilibrium.-We say that a ship is in unstable equilibrium when, ifinclined by some extemal force, she would try to heel still further. Fig. 52 (c) showsthat this will occur when:-

For small angles ofheel, the ship has a negative GM.For any angle of heel, the righting lever, GZ, is on the high side ofthe ship.It should be pointed out that a ship in this condition would not necessarily

capsize. As she heels further over, the centre ofbuoyancy will usually move fudheroutwards and may become vertically under the centre of gravity at some larger angleofheel. She will then have developed conditional equilibrium at *rat angle.

Conditional stability means that if she heels further she becomes stable, if sheheels less she becomes unstable.

Longltudinal Metacentric Height-DGM"'.-We have so far only consideredtransverse stability in this chapter. A ship also has a longitudinal metacentre andmetacentric height, which obey the same rules as the transverse ones, althoughthe longitudinal metacentre is not in the same position. To distinguish them, thelongitudinal metacentre and metacentric height are usually denoted by M, andGMr. We shall consider them later, as they are used in calculating trim and for otherpurposes.

The longitudinal righting lever is not considered in stability, although itobviously exists. The angles of inclination in this direction are so small that GM,,which is more easily found, can be used for all necessary calculations.

CHAPTER 8

TRANSVERSE STATICAL STABILITYMoment of Statical Stability.-This is the rnoment which will try to retum,a,

tUS t" 31"l19!ge5.lp 1s-hggtr{Jt is termed "positive" if it tends to right theship and "negative" if it tries to cause her to heel still further over.

We have seen in Chapter 3 that moment is equal to force multiplied by thelength ofthe lever. In the case ofstatical stability, the lever is GZ and the force actingon this lever is equal to the weight (i.e. displacement) of the ship. So, if I/ is thedisplacement and GZ the righting lever:-

Moment of Static al Stabllity = W x GZ .

For any given condition of loading, when the displacement is constant, a ship'smoment of statical stability will increase or decrease, or be positive or negative, withGZ. So we can say that, whilst the measure of statical stability is ,t/ x GZ, the lengthand direction of GZ alone is an indication of the ship's statical stability at any angleofheel.

Relation Between (GM" and (G2".

Fig. 53 represents the GZM triangle from Fig. 52 (a). The angleZ is a right angle, so that, if0 is the angle ofheel:-

-SZ - = si11gGMGZ = GM xsir0

This means that for angles of heel of less l5o we can useGM as the indication of statical stability, instead of GZ, This is anadvantage, since the former is more easily found, but it must beremembered that at large angles Mmoves, so that we must use GZfound from GZ curves for large angles.

Initial Stability and Range of Stability.-Initial stability means the value ofGMwhen the ship is upright. It determines whether the ship will be "stiffor "tender"and if she is likely to develop a list during a voyage. It gives no real indication as tohow the ship will behave at angles ofheel more than l5o.


Range of stability is the angular range over which a ship will have positivestatical stability. It is important because it indicates the theoretical angle to which theship could heel before she would capsize.

A ship's initial stability does not necessarily indicate what her range of stabilityis likely to be, or vice versa. The two have little to connect them and a ship with alarge initial stability may have either a large or small range ofstability. It is also quitepossible for a ship to have negative initial stability, yet to become stable at a smallangle ofheel and thereafter to be able to heel to quite a large angle before she capsizes.

Calculrtion of r Ship's Initiel Strbility.-When a ship is built, the navalarchitects calculate her displacemen! deadweight and the height of the ceotre ofbuoyancy above the keel (r<B). They also find the distance of the netacentre abovethe centre of buoyancy (BM) arlld, by adding this to the KB, obtain the height of themetacentre (1i11).

Once the ship is nearly completd the "IncliningExperiment" is performedtofndthe metacentric height (Gn{ ofthe ship in the light condition. This is subtracted &omthe light KMto give the height of the centre of gravity above the keel (the light KG).

KM is tabulated in the 'Deadweight Scale", or given in the form of graphscalled "Curves of Stability" or in Hydrostatic Data tables. The righting levets forvarious angles of heel and for assumed KG's are also calculated and added to thestability information; usually in the form of i(N Curves or Tables. Care is takento see that the range of stability is adequate to ensure the safety of tbe ship at anyreasonable angle ofheel if she is properly loaded.

This completes the naval architects'part of the lvork. Armed with the aboveinformation, the seaman can calculate the KG ofhis ship at any stage ofloading, andcan thus find her rKM and GM, her righting levers at various angles of heel and herapproximate range of stability.

Crlculation of (BM" forShip Shapes.-Let a ship be heeledby some extemal force, as shown inFig. 54. ,S? is the original waterlineand S, I, is the new one. The originalcenhe of buoyancy, B, has movedout to ,r. g and g, are the centresof gravity of the emerged andimmersed wedges, SCS' Md TCTIrespectively. The Metacentre is atthe intersection of the centrelinewith the vertical line through.B,.

TRANSVERSESTATICALSTABILITY 59

The proof ofthe following formula is outside the scope of this book, but it canbe shown that:-

Where .I = Moment of inertia of the waterplane

I/ = Ship's volume of displacementIBM =+

Example.-T\e inclining experiment is performed on a ship and her GMis foundD be l'90 metres. Her displacement in salt water is 3200 tonnes and the moment ofirrtia of her waterplane was 17,070. If her KB is 1.62 mehes, what is her KG?

Volume of displacement = ,3?09= = 3122 ml--- ' -" ' 1.025 - ' - - " '

BM =+=W=5.47 metresBM =5.47 menes

KB=l.62metres

KM =7{l9 mettes

GM=l.90metres

Kc = l:19 metres

Celculation of BM for Box Shrpes.-A box-shaped vessel has a rectangularraerplane. It was shown, in Chapter 3, that the moment of inertia of a rectangleiout is centre line is

-Etz

where / is the length and 6 the breadth of the waterplane.

We can thus substitute this for / in the last formula:

BM=i

BM=!+lzr

When d equals the draft, Y = lxbxd

so. BM=---lt-tzx lxbxd

RM= b212d

It must be remembered that the abave formula holds good for box shapes only.


Example.-Abox-shap€d vessel is 60 metres long, l0 metres broad and floatsat a draft of 3'0 metres. Find her.BMand height of the metacentre.

t , =*=*=2.78 metres

i(B=* draft = 1.50 metres

BM =2.78 menes

KM =!.28 nrctxesApproximrte Formula for gBM'.-A close approximation for the BM of a

ship shape, which is sometimes useful, can be found by the following formula:-Where D = the shio's breadth.

d = her mean draft.

4 = a coemcient.

BM=4cl

a is about 0'07 in very fine ships and about 0.083 in very full-formed ships.Its average value for merchant ships is about 0.075.

The Inclining Experiment-This is performed to find the ship's light GMandhence her light r(G. It consists of shifting weights transversely across the deck of aship when the latter is free to heel. The angle of heel is measured by the shift ofaplumb-bob along a batten.

Certain conditions are necessary for this experimen! if it is to give good results,viz:.-

(a) Mooring lines must be slack and the ship must be clear of the whari so thatshe may heel freely.(b) The water must be smooth and there should be little or no wind. If there isany wind, the ship should be head-on or stem-on to it.(c) There must be no free surface of water in the ship. The bilges must be dryand boilers and tanks dry or pressed up.(d) All moveable weights must be properly secured.(e) All persons should be ashore, except the men actually engaged in theexperiment who should stand on the centreline when all readings are taken.

0 The ship must be upright at the beginning ofthe experiment.When this experiment is performed in practice, four weights are generally used,

two on each side of the ship. These are shifted altemaGly, fust one and then both,across the deck. Ttvo or three plumbJines are used and all weights and plumbJines

TRANSVERSESTATICALSTABILITY 6I

rt identical in order that they may provide a reliable check on each other. For theppose of proving the formula, etc., the effect of shifting one weight and of usinge plumb-line only is considered.

In the figure, Dtrepresents a batten,fi xed horizontally acrossthe ship and having thepoint, F, at which it cutsthe centre line, markedon it. CI is a plumb-line, suspended at Cand free to move acrossthe batten when theship heels. The weight,w, is shifted across thedeck to w, through adistance of d metres. Gthen moves out to Grand the ship heels until

i has moved to B, vertically under both G, and M. The plumbJine moves out acrosst batten for the distance FZ.

Let 0 be the angle ofheel and let W'be the ship's displacement. Consider theiiff of G, as described in Chapter 4:-

oor=vConsider the triangle FCL:-

Tbe angle F is a right angle and the angle C is equal to 0.

so, ff=ta1 6

Consider the triangle MGG r:- w has been shifted across the deck at rightrgles to the centreline. Since G moves parallel to the shift ofthe weight, the angleG must be a right angle. The angle Mis equal to 0.

^ GG,5o,

bi: = Ian u

GGt=G1t4rtn t

sbstioting for GG, and for tan 0, from formulae (l) and (2), above, this gives us:-

GM =vzc\cFWFL

(1)

(2)


Example.-Aweight of 15 tonnes is moved horizontally across the deck of aship for a distance of I 1.0 metres. The ship heels so that a pendulum, suspended E.0metres above a horizontal batten, moves out along the batten for a distance of 244millimetes. The ship's displacernent is 3150 tonnes and her 1(Mis 6.12 metres. FindtheGMandKG.

GM = #d-x?;= 5#"Hl = r' 72 metesKG = KM - GM =6.12- l '72 =4'40 mebes

Strticd Strbllity rt Smdl Angler of Heel.-The metacentre is consideredto exist for angles of heel up toabout l5o.

For such angles, the ship'smoment of satical stability canbe found as follows:-

0 = angle ofheel

GZ = the righting leverI/ = the ship's displacementWe have aheady seen tha!

since the ngle GZM is a rightangle,

GZ= GM x sin 0

Also that, moment of statical stability = W x GZFrom the above it can be seen that for small angles ofheel -

Moment of statical stability =If xGZ

Moment of statical stability = ,'z 1 62' "

.io 6

Example.-A shtp of 6240 tonnes displacement has a GMof 0.67 mete. Findher moment of statical stability at an angle ofheel of9o.

Moment of statical stabilitv = 7 x G,ll x sin 0=6240x0.67xsin 9 '= 654 tonne-metres

cGZ' by the iWall-Sided Formuh ".-This formula is relies on the sides ofthe ship being vertical. It gives better results than GZ = GM tang.

It is used to calcul*e GZ atlarye angles ofheel until the deck edge becomesimmersed. Its main use is for investigating the stability of a ship when GM is


very small or zero or negative. The positions of G, B and Mmust be known. Itstates that:-

GZ = sin Q {GM + ttv xtan2 el

It is based on the assumption that the ship is "Wall'Sided": that is, that hersides are vedical at all points along the waterplane, when she is upright. Althoughthis assumption is not strictly true, the formula will give reasonably accurate resultsfor all angles ofheel in ships of normal shape, provided that the deck edge has notbecome submerged.

Loll, or List.-A ship may develop a list for one of two reasons:-(a) If the centre of gravity is out ofthe centreJine ofthe ship.

(b) If the ship has a negative GM.

These conditions are usually caused by faulty loading of the cargo and aregenerally avoidable if the weights in the ship are properly distributed. They are notnecessarily dangerous, provided that the ship has an adequate range ofstability, but areobviously bad seamanship. In either case the ship will heel over until she is in neutralequilibrium; that is, until 3 has moved out sufficiently to come vertically under G.

The first condition can occur in either stiff or tender ships and the list willalways be towards that side ofthe centre-line to which G has moved. In the secondcondition the list may be to either side and may, under the influence of extemalforces, change from one side to the other. It may also increase or decrease if weightsare taken away from, or added to the ship.

Angle of Heel due to "G" being out of the Centre-line.-If a weight is shiftedor loaded so that G comes out of the ship's centreJine, the ship must heel until Gcomes vertically over.B. This is precisely what happens in the "inclining experiment",as we have already seen. In that case, the heel was caused by weights shifted. Now let

us consider how weights may be loaded soas to cause or to correct heel.

Case l.-Consider a ship which isupright to begin with, but in which a weightof w tonnes is then loaded at a distance ofd metres off the centre line, as shown inFig. 57(a). If the ship's centre of gravitywas originally at G^, it will now moveupwards and outwards to G, QG will bethe vertical component of this shift andGG, the horizontal component. The ship'snew metacentric height will be Gi[FIG.57A


The moment trying to heel the ship is v x dt.m, which e{uals I/ x GG, t.m.

The ship will now heel until, at some angle 0, the centre of buoyancy hasmoved out to come vertically under G, The effect of buoyancy will then exactlycounteract the effect of gravity, so that:-

Moment of buoyancy = heeling moment=wxd =WxGG.

But in the triangle GG,M : GG, = GM xtan 0

So; w x /= If x GM\ tan 0 (Where W tndGM arethe new displacement andnew metacentric height, respectively.)

If several weights are loaded or discharged, we must take their combinedmoments in lieu of w x d in the above formula.

Case 2.-Now take the case of aship which is already heeled becauseof uneven disfibution of weightson board, as shown in Fig. 57(b).Let us see how to bring her upright.The ship's centre of gravity must besomewhere offher centre line, say atG, . So the heel must be caused by theship's weight, W', actrng verticallydownwards through G, Let GG, bethe perpendicular distance of G,from the ship's centre line: then thedownward force through Gr at rightangles to GG, will be 7 x cos 0.

So, the moment heeling the ship = lll , GGr, cos 0.

Now, to bring the ship upright, let us load a weight of lr tonnes at a distance ofd metres from the ship's cente line, so as to balance the heeling moment.

The moment to right the ship = ]rx dxcoso

And, wxdxcos0--W xGG xcosO

Cos 0 cancels out, so:-

But,

So,

wxd =I l xGG,

GG,=614' t^ '

wxd =WxGMxtang

Which is the same formula as in the last case, except that *" ".s

hsle using theoriginal (heeled) Wa dGM.

j.ij-"dt?.1

, t


Note that, in the figure, the weight, w, is shown at the same height as G. Thisis done for simplicity: but in fact, the height at which the weight is loaded does notmatter in this case, provided that it does not cause the ship to become unstable. Inother words, we could load the same weight anywhere on the line ry, in the figure,to bring the ship upright.

General Rules.--:l\e following will apply to both the above cases.(a) ex 6l=f f lx GMxtangi

(b) Always use the heeled W rnd GM in the abovel

(c) If a nunber of weigbts are loaded or discharged, use their combinedmoments forw x d.

(d) When a ship, which is heeled, is to be brought upright, the height atwhich the weight is loaded is immaterial, provided that it does notrender the ship unstablo.

Example l.-A ship, which is upright, has a GM of 2.64 m and displacementof 3200 t. A weight of 160 t is then loaded into the wing of a tween deck, so that itis 4'25 m above G and 6'36 m offthe shio's centre line. Find the anele to which theship will heel

In this case, the heeled W' and G Mwill be those after the weight has been loaded.

c"G = w:d = l6Ql4.,?tr = 0. 20 m

New GM = Old GM -GoG =2.64 -0.20 =2.44 m

wxd =W xGMxtart 0

l60x 6.36 = 3360 x2.44xrang

Tan e= l!9936 =o.r24l336|-Jx2\44

0=7.1'

Example 2.-A ship displaces 5700 t and is listed 6' to starboard. Her rKG is6'24 m andher KM is 7'01 m. Calculate the weight of cargo to load into the port wingofa hold, at a distance of l2'0 m offthe centre line, in order to bring the ship upright.

In this case, the heeled W ud GMwill be the original ones.

wxd =lryxGMxtanO

wx12.0 = 5700x0.77 xtan 6"

re= 5700x0;J7l<0.1051 = 38.4 tonnes' ' 12.0


Example 3.-A ship displaces 8820 t, has a KG of 6.73 m and is upright. Shethen loads:-

540 tonnes at 3.0 m above the keel and on the centeline

300 tonnes at 9'0 m above the keel and 2'5 m to port ofcenteline

200 tonnes at 5.0 m above the keel and 6.8 m to port of centreline

280 tonnes at 7'5 m above the keel and 4.5 m to starboard ofcentreline

If the KrWis then 7.81 m, fnd the angle of heel.

Weight Height Vert. Moment Dist. Hor. Moment88205403002002E0

6'7t3.09.05.07.5

593591620270010002100

002.56.84.5

00

750 (to port)1360 (to port)1260 (to sta6oard)

10140 66779 E50 (to port)

r..r"*,(c=ff#=6.5e mNew GM = KM - KG=7'81-6.59=1.22m

Heeling moment (fot wx =VxGMxt^nO

850 = 10140x 1.22x tan 0

rane= rffizz =o'0687Fleel = 3'9' to port

Example 4.-A ship displaces 64701 tonnes, has a KG of 5.97 m, KM of 7.62m and is heeled 5" to starboard" 400 tonnes of cargo are to be loaded into the wingsof a tween deck at distances of 5.0 m off the centre line to port and 2.0 m off thecente line to staxboard. Find the weight to load in each wing in order that the shipmay then be upright.

Load rry tonnes into the port wing: (400-w) tonnes into the starboard wing.Moment to port = 5]r t/m Moment to starboard : 7(400 - w) t/m

The ship is heeled to starboard, so we require the greater moment in the portwing in order to bring her upright.


So, moment required to right ship = 5w -7 ( 4OO - w) tlm

Righting moment =WxGMx tan 0

5w -7 (4OO - w\ = 6470x l. 65x tan 5'

5w - 2800 + 7 w = 6470x l. 65x0. 0875

l2w-2800=934'l

w =934'l.t-2800 =3t I tonnest2

So we must load 311 tonnes in the port wing and 89 tonnes in the starboardwing.

Angle of Loll due to a Negative "GM".-A ship may develop a list throughhaving a negative GM. In such a case, she will heel until she is in neutral equilibrium:

that is, until B has moved out toa position where it is verticallyunder G.

The figure represents aship in the above condition.fy,,6 things are obvious:-

l. That GMmust benegative in the first instance

2. That, after the shiphas heeled and I is verticallyunder G, there can be no GZ.

We have seen. from thewall-sided formula. that:-

GZ =sin' (GM +rBM xt?tr2 e)

But it has just been shown that GZ = O, so: -

o=sin o (GM ++BM xt^r] o)

For two quantities, when multiplied together, to be equal to O, one at least ofthem must be O. This means that either.

(1.)s in0:O.

1

or (2;)GM+ ); BMxtar20:O


If sin 0 equals O, then the angle 0 must also equal O, and the ship would beupright. This is obviously impossible, since the ship has a lisl so we can concludethat:-

cu+B{ tan'?e=o

ffr^;e=-e,,ntS=+#

,^rO=.@-. \BM

Rernember GMis negative and two negatives make a positive.

Example.-Ashtp has a displacement of 4500 tonnes and a GMof 0.24 metres.A weight of 270 tonnes is then loaded on deck, so that its centre of gravity is 5.0metres above that of the ship and the BMis then found to be 4.40 metes. Find thenew GM and the angle of loll, if any.

New displacement 4500 + 270 = 4770 tonnes

GG,=#{=AH# = 0.28 metresNew GM =0.24-0.28=-0.04 metres

To find the angle of loll: -

angle of loll = fo"

Note that a GMof-O.M m causes in this case a loll of nearly 8'If the GM is 0.2 the ship will probably capsize.

CHAPTER 9

FREE ST]RFACE EFFECTThe Elfect of Free Surface of Liquids.-lf a tank is completely filled with

liquid, the latter becomes, in effect, a solid mass. It can be treated in exactly the sameway as any other weight in the ship; that is, its weight can be regarded as being at itsactual centre of gravity.

In a tank which is only partly filled, the surface ofthe liquid is free to move andpossesses inertia. The moment of inertia of this free surface about its own centre-line causes its centre of gravity to appear to be at some height above it known asthe virtual centre of gavity. (he centre of the arc of the movement of the liquid.)The effect on the ship's stability will then be as ifa weight, equal to the weight ofthe liquid in the tank, were raised from its position in the tank to the position ofthevirtual centre of gravity.

Fig. 59 shows a ship which is heeled and whichhas free water in a doublebottom tank.

st is the originalsurface of the water and gits original centre of gravity.When the ship heels, itssurface becomes srtr and itscentre of gravity shifts outto 8r'8r

m is the new verticalthrough 9,. The water nowbehaves almost as if it werea weight, free to swing andhanging from m, which is its

virtual centre of gravity. This means that although the actual centre of gravity of thewater is at g org, its efrect on the ship's stability is as if it were placed at m. The centreof gravity ofthe ship if the free surface effect were ignored (i.e. as if the water werea solid weight and not free to move) would be at G and would be called the "SolidCentre of Gravity". When the free surface is taken into account, the ship's cenhe ofgravity would be at G, and would be called the "Fluid Cenhe of Gravity".


When the GMof a ship is being calculate4 the fluid KG and GMshould alwaysbe taken as the tnre values if free surface of liquid exis6.

We have already seen thal for a ship, when I is the moment of inertia of thewaterplane and Zthe volume of displacement:-

BM=+If we consider Fig. 59, it is apparent that scs, and tctr are equivalent to the

immersed and emerged wedges, respectively, ofa ship: that the shift ofg, out to g,is parallel to the shift of the centre of gravity of these wedges: that m occun at theintersection ofthe centeline and the vertical line through g,

This is obviously similar to what happ€ns when B moves out to produce M inthe case ofa ship.

So if i is the moment of inertia of the free surface and v the volume of water inthe tank:-

cn=+ ( l )

The effect ofthe virtual centre ofgravity is the same as would be that ofa solidweight placed at m: that is, as if the weight of the water in the tank were shifted fromgIam.

So, if Wis the displacement of the ship and w the weight of water in the tark:-

wxemLi(it=i:- (Shift of ..G'-{haprcr 5)

Let v be the volume of water in the tank, Zthe volume of displacement of theship, d, the density of the liquid in the tank and d the density of the water in whichthe ship floats. Then:-

w=vx4

I l =Y x6

So, OOr='"!n#If we now substitute for grr, from fonnula (l):-

vx6, x iuu, =zl5rl

ec,=lxl'vd

(Note Iz" d = Displacement) nv =iff

FREESURFACEEFFECT 7I

Example.-A ship displaces 9750 t and has a solid GMof 0.96 m when floatingin water of density 1.020 t/m3. Free surface exists in a tank which contains water ofdensity 1.005 t/m3. The moment of inertia of this free surface is 1640. What is thefluidGMl

to""of GM J#

= 0.17 m

Solid GM 0.96 m

Fl,tidGM 9:lL n

The Effect of Rectangular Free Surfaces.-The free surface in a tank is oftena rectangular area. In this case, the calculation of its effect is much simplified. If/ isthe length and b the breadth of such a surface, its moment of inertia will be found bythe formula:-

. - rbr x6,

'12

u<'t=fr

Since ca _tb ' xol-- I l2W

Free Surface Effect when Tanks are Filled or Emptied.-When liquid isrun into, or removed from a tank so as to cause free surface, it will affect the ship'sstability in two ways:-(t) The solid rKG and displacement will be changed by the weight added or

removed.

(b) There will be a rise of G caused by the free surface. This will have to beadded to the new solid KG to give the fluid KG.

Example .-A ship displaces 5 I 54 t, floats in sea water and has a KG of 6. I 0 mwhen a double bottom tank is empty. The tank is rectangular, 15 m long, 12 m wideand 1'3 m deep. What would be the ship's GMafter fresh water has been run into thetank to a depth of 0'80 m, if the KMis then 7'45 m.

Added volume = 15 x 12 x 0.8 = 144 m3 (144 t of fresh water).

Weieht Heieht Moment51541M

6.100.40

3143958

s298 3t497

ShipWater


SotidKG=ry.=5.9s m5294

11'l x6lRise of G due to free surface = 'l

12W

= ,!sx=t?='= = o +r ml2x 5298Solid KG = 5.95 m

Fluid KG = 6 J6 m

KM =7.45 m

Fluid GM = I .09 m

Free Surface in Divided Tanks.-Figure 60 shows a tank having a rectangular

free surface and divided at the centreline. Let / be the length and b the breadth of thetank. The breadth ofthe free surface on either side ofthe tank will thus belD .

/ l \ r

Rise of G due to free surface oo oo" ,id" = 4?Al2v .=

"r"TGG, is due to the free surface on two sides however, so:-

cct=zxfix(

=2\ tb j8 t2V

GG, =!x!-E-4 t2I/

If we compare this with the formula found in the last section, we shall see thatit is one quarter ofthe latter. So, in the case ofa rectangular free surface, dividing thetank at the centre-line will decrease the rise ofG to one quarter ofwhat it would be inan undivided tank.. In the case of surfaces other than rectangular ones, t}e decreasemav not be the same. but it will be considerable.

FREESURFACEEFFECT 73

It can be proved, in the same way, that dividing the tank into three woulddecrease the rise ofG due to free surface to one-ninth ofthat for an undivided tank.Dividing the tank into four parts will decrease the free surface effect to one-sixteenthofthe above.

Iffree surface exists in more than one tank, this will cause a total decrease inthe ship's GMequal to the sum of the effects for each tank.

Example 1.-A ship is floating in salt water and displaces 6000 t, has a solidl(G of 5.810 m, and KM of 6.424 m. Free surface exists in a rectangular bunker tank,3'0 m long and 12'0 m wide, divided at the centre line and containing oil of density0'890 Vm3. What is the shin's fluid Gi14

FI

E

Free surface effec t=ccr=lbjr#t xf,

_ 3xl2r x0.gg0x=12x6000 | 4

=0.016m

Sol id l (G = 5.810 m

Fluidr(G=5.826m

KM =6-424 m

Ftuid cM = ql9! m

Example 2.-A ship, floating in salt water, displaces 8440 t, has KG of 6'21 matrtd KM of 7.21 m, when a double bottom tank is full of fresh water. The tank isrectangular, l8 m long,20 m wide, 1'5 m deep and is divided into 4 equal tanks byfore and aft watertight divisions. Find the ship's GM after 360 t of fresh water havebeen pumped out ofthe tank, leaving it slack?

Depth of water removed = #-L

= 1.00 m

c.g. of water removed = 0 . 5 m from top of tank

Ks of water removed = I .5 - 0.5 = 1.00 m

Weight Height Moment8440360

6.211.0

52412360

8080 52052

Sordrc=ff i =6.442n

The surface is divided into 4 parts

ShipWater

!

IhE


th3 x6, IFree surface effect = :----l- x t:

l2w 16

_ 18x203 x I12x8080 l6

=0'09m

Solid KG = 6 .,14 m

Fluid KG = 6 J3 m

KM =7.21m

Fluid GM = 0 .68 m

Free Surface Moments.--One of the requirements of the Load Line Rules isthat a ship shall be provided with a staternent showing the loss of GM due to freesurface in each tank. The Departrnent of Trade now recommend that this be givenas "Free Surface Moments", ass"ming that the liquid in the tank has a density ofl'000 t/m3.

Where i is the moment of inertia of the free surface (allowing for longitudinalsubdivisions):-

Free surface moment = t(t = l)

Free Surface Effect if liquid has density 6tiv j

tossof GM =T

Example.-A ship has a solid GM of l'02 m an,d displaces 7600 t when a tankis partly filled with oil of relative density 0'E50. If the free surface moment of thistank is 1250 m'. find the fluid Grtd

Loss of GM=F.S.r"r.xfi =125s)<8-94 =0'14 m

Sol id GM=l '02 m

Fluid GM = 0. 88 m

So,

I

I

t

CHAPTER IO

TRANSYERSE STATICAL STABILITY IN PRACTICEFactors Alfecting Steticel Stability.-Statical stability is governed principally

b,*-(a) The position ofthe ship's centre of gravity.

ft) The underwater shape of the hull, upright and as the ship rolls.

The position of the centre of gra.vity depends on the loading ofthe cargo andder weights in the ship. It affects the statical stability, because it is one of theEctors which determine the length ofthe righting lever, GZ.

The form ofthe ship decides the shape ofthe emerged and immersed wedgesrten the vessel heels. These in their tum will determine the position and shift ofthecam of buoyancy and hence the length of GZ.

An example will best show the efrect of the above. Let us consider a graph$owing a ship's moment of statical stability at various angles of heel.

FtG.61

a

F

IN

x)


Curve I is for a vessel 160 mehes long, 20 metres beam, 8 metres draft,3 metres freeboard and having a KG of 7.00 metres. The maximum righting momentfor this ship is about I1,600 tonne-metres and occurs at 23' of heel. Her range ofsrabiliry is 58'.

Curve B shows the effect of adding 2 metres of freeboard to the above ship,if all other details remain the same. The two curves run together at first, but curveB continues to rise to a maximum of about 27,500 tonne-metres at 48o ofheel. Therange of stability has increased to 81o.

Curve C shows the effect of adding 2 metres of beam to the original ship incurve ,{. The maximum stabilify has increased to 25,000 tonne-metres, but it onlyoccurs at about 25o heel. The range of stability has increased to 68'.

The effect of raising the centre of gravity of ship A by 0.5 metre is shown incurve D. The maximum righting moment is now nearly 7000 tonne-metres and therange is 39o- a considerable reduction in each case.

Curve E shows the effect ofraising G, in ship,4, by 1.20 metres, so as to giveher anega/rive G M of 0'03 metres. The negative stability causes her to loll to an angleofabout 7", but thereafter she develops positive stability and has a range to 23o.

Curve F is an example of what would happen if ship A had G raised by 1.20metres (as in curve C), but at the same time had the freeboard increased by 2 metres.In this case she will still loll to 7"; but thereafter she will have a range to about 55o,because of the increased freeboard.

Let us tabulate these results:-

ShipGM

meres

Maximum stabiliWRange ofstabiliryTonne-metres

Angle ofheel

A t . t7 l I,600 23" 58"B t 'r7 27,500 43" 81.C 2.4 24,700 25" 66"D 0.67 6,800 22" 390L -{.03 1.100 17" 23"

-o.03 8,800 39' 55"

(Only ship B complies with the Loadline Rules regarding minimum stabilityrequirements)

From the above we can draw the following conclusions:-(a) Increase of freeboard does not affect initial stability, but increases rangeof stability.

ll

rl

tnIy

TRANSVERSE STATICAL STABILITY IN PRACTICE 77

Increase of beam increases initial stabiliw. but has very little effecton range.

(c) Raising the centre of gravity decreases both initial stability and range.

(d) A ship which has negative initial stability will not necessarily capsize,but may become stable at some small angle of heel and may, thereafteq havea reasonable range of stability before she will capsize, provided that she hassufficient freeboard.

It must be remembered that the curves shown are for one particular case and areintended as a demonstration only. In practice, the average merchant ship often has alarger range ofstability than that shown, but the conclusions that we have drawn willhold good in almost all cases.

Placing of Weights.-The naval architects who design a ship, make sure thatshe will be reasonably safe if she is properly loaded, as regards both her staticalstability and her range of stability. They can, however, onlp!-r the po_sition of thecentre of gravity for the ship whel qle is in her light condilion. Its position duringaiii aftijr the loading ofcargo will deplnd on the distribution oftne weiEEGlthich isthe duty o_f the ship's officers. It haS already been sein ihai both ihe statical stabiiityand the range -of slability depend partly on the position of the centre of gravity,so those who load the ship must always remember that the final responsibility ison them.

It is not always possible to load ships exactly as we would wish, since we donot control the kind of cargo we receive, or the order in which it comes alongside.Thus, we sometimes have to "make the best of a bad job"; but even in the worstcases we can do quite a lot to control the stability of our ships by the judiciousdistribution ofweights. Ifthe seaman loads his ship so that she has a reasonably largemetacentric height, he need not worry unduly about the range of stability, since thenaval architects can be relied on to do their part of the work faithfully. In practice,the average merchant ship, when properly loaded and with a sufficient metacentricheight, usually has a range ofat least sixty to seventy degrees. Many still have a largerighting leverevel_at ninety degrees of heel.

-{i1qle of thumb";hethod sometimes used at sea, is to place about one third of. - - . - - i .+- - - . :

. - - . t - - -the weight in fhE-trveeri decks and two{hirds in the holds. This is a reasonably saferule in irost cases, but it must be remembered that all ships have their peculiaritiesand what is good for the average ship is not necessarily good for every one. The onlytruly reliable method is that ofcalculating the metacentric height.

Stiff and Tender Ships.-A stiff ship is one which has too much metacentricheight (Glr'). A tender ship is one iith-n-ot eirough mefacEniri-height. These terms

(b)

P'rele

lnhe

ivegle

I-20tres.( io

rbility

range


are relative: a ship does not suddenly become either stiffor tender at a given GM, butchanges gradually from one condition to another.

| .- i , i". A good metacentric height for a fully:loe4Bdlt$*StShip is^usqa]Ujg1$n| ' one half and one mg.Ep. A ship with a GMof less than this yiill-go_!.nglly.bel9nder.

-...--..--aa* It is difficult to say just when a ship becomes stiff. A GMwhich would renderone ship too stiff might be quite comfortable in another: also, generally speaking,much larger GMs are considered reasonable in modem ships than would have beenregarded as permissible some thirty years ago. It is probably fair to say, however,that a loaded ship with a GMof over one metre has a tendency to stiffiress: whilst ifher GMis much greater than this she will probably be too stiff. As we shall see later,a ship in the light condition normally has a large GM often as much as from two tofour metres.

Stiff Ships.-tf a ship is too stiff, she will have an exce_ssi_v:Jigb[Iryg:g!,and will tend to right herself violently when inclined. HEr period of roll may berather srnall and she will roll heavily and quickly in a seaway. This will cause her tobe uncomfortable at sea and there is a risk that she may strain herself, or may causeher cargo to shift or to be <lamaged. Such a condition is not usually dangerous, butshould be avoided whenever possible, for obvious reasons.

One is sometimes asked if it is advisable to pump out double bottom tanks'ina stiff ship. In port, this would be perfectly safe and good practice, subject to theship being left with sufficient ballast for seaworthiness, since tJre metacentric heightwould thus be decreased. Whilst the tank is being pumped out, free surface effectwould cause the centre of gavity to rise somewhat above its final position, but thisshould do no harm in the circumstances.

It would probably be safe to work tanks at sea in the same way, but it is notusually considered good practice to do so, unless absolutely necessary; because ofthe risk of structural damage to the tank due to free water sloshing about.

Tender Ships.-A tender ship will have a small righting moment and acomparaqygly,-l_o1g period of roll. She will have an easy motion in a seaway andmay be quite safe.|rdvidcifthifner GM and freeboard are sufficient to give her anadequate range of stability.

This does not mean that it is good practice for a ship to be in a very tender state:on the contrary, such a condition should be avoided as much as undue stiftress. Itis important to remember that the consumption of fuel and stores during a voyageusually causes the ship's centre of gravity to rise, so that she will probably arrive inport with a smaller GMthan that with which she set out. If the ship is tender to beginwith, this may cause her to become more so and she may even develop a negativeGMbefore she finishes her voyage.

ul

TRANSVERSE STATICAL STABILITYIN PRACTICE 79

If a ship should become tender, the solution is to fill double-bottom tanksin order to lower the centre of gravity. Whilst tanks are being filled, free surfaceeffect will cause G to rise slightly and the ship to become more tender, but thiswill disappear when the tank becomes full. In modem ships the double-bottoms areusually well divided up by the cenneline and intercostal girders, which reduce fieesurface. It is always advisable to fill tanks which have a watertight sub-division ifmore GMis required at sea.

Unstable Ships.-In British law, Masters and Officers must make sure that theirships are be stable at all times at sea. (Minimum GM0.l5 m) The fact that a ship wasstable at the outset ofa voyage is not accepted as extenuation, should she suffer lossor damage through becoming unstable during that voyage, unless the circumstancesare extraordinary and unforeseeable. Apart from this, good seamanship demands thata ship should be so loaded as to ensure that the GMwill be sufficient at all times.Ships do sometimes become unstable, however, and if this happens, every effortmust be made to rectif matters. The necessary steps are simple and obvious to thosewho understand stability, but unforhmately the wrong thing has sometimes beendone occasionally with serious results.

Let us consider the causes of instability in practice:-

(") Weights may be placed too high in the ship when cargo is being loaded andthus cause a negative GMat the outset ofthe voyage.

@ Fuel and stores consumed at sea are often taken from a position low downin the ship. This may cause G to rise sufficiently, to give her a negative GM.

@ When a deck cargo is carried, it may soak up water during thevoyage and thus increase its weight. This may also cause G to rise sufficiently togive the ship a negative GM.

(d) Water or oil will be taken from double-bottom tanks at sea. There will bea rise of G due to the removal of weight from the bottom and a further rise due tofree surface effect in the slack tanks. This is particularly liable to happen in shipscarrying oil fuel in the double-bottoms. The free surface effect is then unavoidable,But by using small tanks and emptying one pair before using the next tanks, theFree Surface Effect should not be too excessive.. Tanks can always be filled withballast as they become empty.

It is obvious from the above that the ultimate cause of instability is a negativeGM and that the only cure is to lower the ship's centre of gravity. This must bedone very carefully, since it is always a difficult operation and it can actuallybe danserous.

Er,lo

not;of

ln

t

lerE,e1l

er,r i f

be

: lo

!se

Dut

nntheght[ecthis

daiaran

ragefe ln

Rgrnfive

Brc:s. It


Weights such as bunkers, stores and cargo must, ifpossible, be trimmed downand/or double-bottom tanks filled, whilst in extreme cases it may become necessaryto jettison cargo. The best general rule is to add weight as low as possible or toremove it fom the highest possible point.

The most common and practical method of curing instability in a ship is to filla double-bottom tank or tanks, but this may be dangerous if it is not done properly.Tanks which are sub-divided at the centreline should always be filled first, inorder to minimise the effect of free surface. One tank should be filled at a time,commencing with the low side and when this is about two.thirds full, it will be safeto start running up the high side. Free surface effect and the added weight on the lowside will probably cause the ship to increase her list at first, but as the tank fills, shewill gradually come upright. The high side ofa tank should never be filled first, eventhough it may evenh:ally achieve the desired result. There are two reasons againstthis: that G will not be lowered so quickly as by filling the low side first: that atsome time the added weight on the high side will cause the ship to change her list,suddenly and violently, from one side to the other.

It is extremely dangerous and worse than useless to pump out double'bottomtanks in an attempt to correct loll. It might seem at first sigbt that if we pump out atqnk on the low side of the ship, the removal of weight from that side would allowher to right herself. We must remember, however, that an unstable ship lists becauseher centre ofgravity is too high and ifwe remove weight from the bottom ofthe ship,we shall only cause G to rise still higher. This rise ofG will probably be aggravatedby free surface efect. When sufficient weight has been removed from the low sideof the tank, the ship will give a sudden "heave" and develop an even greater list tothe other side: or she may even capsize.

A ship rarely becomes unstable when all the double-bottom tanks are full, butif this does occur, it is obvious that they should on no account be pumped out. If theship is still dangerously unstable in such a case, after all possible cargo, fuel andstores have been shifted downwards, the only resort is to jettison cargo. When this isdone, the cargo should first be taken from the high side of the ship and levelled offlater. The reasons for this are the same as those for fillins double-bottom tanks onthe low side first.

Ships in Ballast.-When a ship has to make a voyage with no cargo on board,it is usually advisable to carry a certain amount ofballast. This makes the ship moreseaworthy generally and immerses the propeller more deeply, thus increasing itsefficiency and decreasing vibration. Modern ships use water ballast carried in tanksfor this purpose.

I

dsiln

d,retsts

TRANSVERSE STATICAL STABILITYIN PRACTICE 8I

A ship which is light usually has a large GM and is often excessively stiff.When water ballast is loaded into such ships it is important not to increase the GMfirther. Peak and Deep tanks will minimise the stiffiress and keep the GM down.

It can be seen from the hydrostatic curves in the back ofthis book that, near thelight draft, Mfalls quickly as draft and displacement increase. This is all to the good,since it tends to reduce the GMif we load water ballast. But if we load the ballast indouble-bottom tanks alone it will cause G to fall considerablv. so that the nett resultis usually an increase in GM.

In order to avoid this, most ships have deep tanks, which can carry a largeamount ofwater ballast higher up in the ship. Loading water into these will not lowerthe ship's centre of gravity appreciably. In this case, Mwill fall much more than willG and the nett result is usually a decrease in GM-

To illustrate this, consider a ship which, when light, has a draft of 3'20 m,displacement of 4986 t, ,(G of 6'50 m, KM of 10.14 m: and,hence a GM of 3.64 m.

Now consider what will happen if we load 1500 t of water ballast: (a) in doublebottom tanks, with their centres of gravity at 0.60 m above the keel: (6) in a deeptank, with its centre of glavity at 5.00 m above the keel. Suppose that, in each case,the new draft has become 4'00 m and the new,(Mis 9.02 m.

To find the new GM:

For (a) For (b)Weight Height Moment Heisht Moment4,9861,500

6.500.60

32,409900

6'505.00

32,4097,500

6,486 33,309 39,909

NewKG=

KM=GM=

11lnq6486

- "14m

9-02 m3.88 m

39909_6.15t116486

9.02m2'87 m

In this case, the water ballast, when loaded in the double bottom, will increasethe GMby 24 cm: when loaded in the deep tank, it will decrease the GMby 77 cn.

The Effect of "Winging-out" Weights. "Winging-out" means placingweights well out from the centre-line towards the sides ofa ship. Most seamen knowthat a ship so loaded is steadier in a seaway than one in which the heaviest weightsare concentrated at the centre-line, all other things being equal.

A ship's period ofroll depends largely on her moment ofinertia. We have seen,in Chapter 3 , that the greater the moment of inertia of a see-saw, the less quickly willit swing. Similarly with a ship; if her moment of inertia is increased, her period of


roll will also become greater. If the weights in the ship are winged well out, they willcause her to have a greater radius of glration than she would have if they were nearthe centreJine. This will increase her moment of inertia and period ofroll, so that shewill be steadier in a seaway. This should always be done whenever possible.

It must be remembered, to avoid confusion, that we are here considering themoment of inertia of the ship herself; not that ofthe waterplane, as we did when wewere finding 8M.

Deck Cargoes.-Ships carrying heavy deck cargoes are always liable tobecome unstable, since the additional weight is placed high in the ship. Ifthe cargo isofa type which is likely to soak up water during the voyage, the consequent increaseof weight on deck may cause G to rise sufrciently to make the vessel unstable. Whensuch a cargo is being loaded, therefore, a sufficient margin of safety must be allowedfor this eventuality.

Timber Deck Cargoes.-The remarks made in the last paragraph also applyto the particular case of a timber deck cargo. When such a cargo is properly secured,however, it becomes in effect an addition to the ship's hull and thus increases thefreeboard. We have seen that an increase in freeboard will increase the range ofstability so that a ship carrying a timber deck cargo may be perfectly safe, eventhough she is tender. In Fig. 61, curves E'andFshow that such a ship may even havea small list on account ofthe extra weight on deck, and yet have quite a large rangeof stability. This would be bad seamanship, but not necessarily dangerous as far asstability is concemed.

The advantage of increased range of stability can obviously only be gainedif the deck cargo is efficiently secured so as to fomr a solid block with theship's hull. It is worth noting that the regulations with regard to deck cargoesof timber carried on ordinary ships lay down that such cargo must be compactlylashed, stowed and secured and that it must not render the vessel unstable duringthe voyage.

The Load Line Rules require that an allowance equal to 15 per cent of theweight ofa timber deck cargo shall be made for water soaked up by the timber.

Ships marked with lumber loadlines are allowed to load more deeply whencarrying a full timber deck cargo, than at other times. Since the additional weightwill normally be on the deck in such cases, it is important that the stability of theseships should be even more carefully considered. Three points from the regulationswith regard to this are worth noting particularly:-

(a) The double-bottoms must have adequate longitudinal sub-divisions. Thisis a precaution against undue fiee surface effect when the tanks are partly filled, toprevent the ship from becoming unduly tender.

TRANSVERSE STATICAL STABILITY IN PRACTICE 83

(b) The timber must be stowed solid to a certain minimum height. Thisensures sufficient freeboard to give an adequate range of stability, if the shipbecomes very tender. It also means that if she were to lose the deck cargo, shewould rise approximately to her ordinary loadlines.

@ The lashings have to conform to very stringent rules, which ensure that thedeck cargo forms a solid mass with the ship.

Free Liquid in Tanks.-The importance of longitudinal subdivisions in tankshas been referred to several times. A study of Chapter 9 will show that the smaller thearea offree surface in a tank, the less will be the rise of the ship's centre of gravitydue to such surface; also that a decrease in its breadth will have a much greater effectthan a decrease in length. Hence, the best way of minimising the effect is to use atank which has as many longitudinal subdivisions as possible. The modem cellulardouble-bottom tank has, at least, a watertight centre girder so that the free surface isdivided into at least two parts. Slack double-bottom tanks should always be avoidedif possible but they should not cause instability if the number ofsuch slack tanks iskept to a minimum, unless the ship is initially very tender.

The amount of liquid in a tank will not appreciably affect the position of thevirtual centre of gravity due to free surface, unless it changes the shape of thatsurface. The weight of the liquid does, however, affect the final position of theship's centre of gravity for two reasons. In the first place, it will have an influenceon the original position ofG. Secondly; it will change the volume of displacementofthe ship and will thus cause a slight change in the position ofM In theory onecentimetre of water in a double-bottom tank would cause the centre of gravity oftheship to rise much higher than, say, one metre of water: the free surface effect wouldbe the same in each case, but in the second case, the original centre of gravity wouldbe lowet on account ofthe extra weight in the bottom of the ship. This would holdgood in practice as long as the ship were perfectly upright, but as soon as she heeledslightly, the water would run down into one comer. If the tank were nearly empty, ornearly full, this would cause a considerable decrease in the free surface.

Free surface is considered not to exist if the tanks are less than 2olo full or morethan 98% tull.

There is always a large free surface effect when deep tanks are being filled.This is not normally dangerous, since, in the average ship, such tanks are only filledwhen she is light and, therefore, comparatively stiff. Some modem ships carry liquidcargoes and/or bunkers in deep tanks and peak tanks, however, and may only have asmall metacentric height when such tanks are filled. In this case, free surface effectbecomes important and must be considered carefully.


Free liquid in tanks, as distinct from pure free surface effect, is not usuallyconsidered, because it has peculiar and apparently unpredictable effects on therolling ofships. There is no doubt, however, that the period of surge ofthe liquid issometimes the same as the ship's period ofroll and when this happens, it increasesthe rolling.

Apart from any question of stability, it must be remembered that slack tanks arealways bad from a structural point of view. Free liquid exerts a considerable liftingeffect on the tank tops and may cause considerable damage to them.

It can be seen from the above that fiee liquid in tanks should be avoidedwhenever possible, even when the free swface effect is not dangerous. When it doesoccur, one should keep a sense ofproportion and neither underestimate nor overrateits possibilities.

Free Surface Effect in Oil Tankers.-This effect presents a special problemin the case of oil tankers, since, when tanks are "fu11", a certain amount of space(or "ullage") must be left between the surface of the oil and the tank top to allowfor expansion of the cargo due to changes of temperature. The usual methods ofminimising free surface effect, in this case, are shown diagrammatically in Figs. 62(a) and 62 (b).

(a) Frc.62 ro,

Fig. 62 (a) shows an arrangement which may be used in small vessels. Alongitudinal bulkhead, B, is fitted at the centre line and an "expansion trunk",4extends upwards above the freeboard deck. When the tanks are full, the free surfaceis confined to the expansion trunk and is there subdivided by the bulkhead.

Fig. 62 (b) shows an arrangement which is often adopted in modem tankers.There is no expansion trunk, but two longitudinal bulkheads B are fitted, one at eachquarter line and also a washplate l/, or a non-watertight bulkhead, at the centre-line.When the tank is partly full, the free surface is divided into three nearly equal parts.

CHAPTER I I

DYNAMICAL STABILITYDefinition,-Dynamical stability is the amount of work done in inclining a

ship to a given angle ofheel.

Work.-Suppose that we wish to push a weight across the deck of a ship.The weight will resist our efforts to move it on account of inertia, friction with thedeck, etc., and we shall have to exert force in order to start it moving. If we thenstop pushing, the fiiction between the deck and the weight will soon cause thelatter to stop moving, so we must continue to push until it is in the desired position.The greater the weight, the harder we must push and the greater the distance, thelonger we must push. In other words, we must do work and the amount of workdone depends on the distance we have to move the weight and the amount of forcewe have to exert in order to move it. Thus, work done is equal to the force exerted,multiplied by the distance over which it is exerted.

Dynamical Stability.-tonsider a ship which is being heeled by some extemalforce. As soon as she heels to a small angle, her moment ofstatical stability will tryto force her back to the upright. In order to heel her further, sufficient force must beexerted to overcome this statical stability and must continue to be exerted for as longas the ship continues to heel. We can liken this case to that ofthe weight mentionedin the last paragraph and say that the work done to heel the ship to any given angleis equal to all the force exerted, over all the distance through which the ship hasheeled. This is obviously only another way ofexpressing the definition ofdynamicalstability, which is given above.

Dynamical Strbility from a Curve of Strtical Stability.-Fig. 63 shows acurve of statical stability, in which the moment of statical stability (trx GQ is plottedagainst the angle ofheel. The statical stability at any angle is found by the perpendicular

FtG.63 distance from the base line to the curveat that angle. For instance, the momentof statical stability at 30' of heel isfound by drawing the perpendicularline,,48, and then the horizontal one,BD. The required moment is then CD-in this case about 13,000 tonne-metres.

t5,OOO

to,ooo


We have just said that dynamical stability is equal to all the force exerted overall the distance through which the ship has heeled. This can be taken to mean the sumofall the moments of statical stability, for every small angle ofheel, up to the givenangle. If we consider Fig. 6l again, we shall see that the sum ofall the moments ofstatical stability up to 30' of heel will be equal to the shaded area ABC. Similarlyfor other angles. This means that the dynamical stability at any angle is equal tothe area under the curve of statical stability up to that angle. For this purpose, thevertical distances to the curve are always measured in terms of statical stability andthe length along the base line in terms of circular measure (or "radians"). A radian isequal to 57.3', so that the lenglh along the base line becomes:-

Angle ofheel, in degrees57.3

Thus, given a scale or curve of righting levers or moments, we can useSimpson's Rules to find the Dynamical Stability.

Strictly speaking, we should always use moments of sto,nc l stabllity (V x Gato obtain dynamical stability, which should be measured in 'tonne metre radians".For purposes of comparison, however, we often use righting levers (GZ), to givea quantity called "mehe radians". The latter can be converted to actual dynamicalstability by multiplying it by the ship's displacemefi (n, as illustrated in thefollowing example.

Example.-lalculate the dynamical stability at 40' heel, for a ship whichdisplaces 6000 tonnes and which has the following righting levers:-

Heel 10" 20. 30' 40"GZ 0.18 0.41 0.67 0.85

Common interval (fi) = l0' = ,$

radians

Ordinate Multiplier Product0.00 I 0'000.18 4 0.720.41 2 0.820.67 4 2.680'85 I 0.85

5.07

Area=lxs.07

= 0. 2949 metre-radians

Dynamical Stability = 0.2949x displacement= 0 . 2949 x 6000 = I 770 tonne metre radians

DYNAMICAL STABILITY 87

The word radians is used here to indicate that the angles used to calculatethe dynamical stability is radians and not degrees. Some administrations measuredynamical stability in tonne.metre.degrces.

Calculation of Dynamical Stability.-Dynamical stability can, if desired, befound by direct calculation.

When a ship heels, the centres of gravity and buoyancy separate vertically.G moves slightly upwards with the ship, whilst B moves outward and doumwards,because its shift is parallel to the shift of the centres of gravity of the immersed andemerged wedges. The upward shift of G is resisted by gravity and the weight of theship, which acts vertically downwards through it. The downward shift ofB is resistedby the upward thrust of the force of buoyancy, which is also equal to the weight ofthe ship. Thus, when a ship heels, -B and G are forced apart against a resistance equalto the ship's weight, or displacement.

Theworkdone, ordpamical stability, isthus equal to the displacement, multipliedby the increased vertical separation of the centres of gravity and buoyancy.

Consider Fig. 64. When the ship was upright, the vertical distance between thecentres ofbuoyancy and gravity was BG. With the ship heeled to the angle 0, -B hasmoved out to 8,, and the vertical distance has become B,Z So the increase ofverticalseparation ofB and G, due to the heel of the ship is:-

(B.tZ - BG)

If l/ is the ship's displacement, this means that:-

Dynamical Stability at the angle 0 = (BrZ - BG) x Displacement


Notes.-Dynamical stability is important in ship stability for two reasons.Since it is the measure of the work that must be done to heel a ship:-(a) It is a big factor in deciding how a ship will roll; in this case the waves are

doing the work.

(b) It determines the ability of a sailing ship to stand-up under sail; thepressure ofthe wind on the sails supplying the work in this case. This is notof much interest to the average merchant seaman today, but is important toyachtsmen and others who have to deal with sailing craft.

It must always be remembered that anyfting which reduces statical stabilitywill also reduce dynamical stability. This is yet another count against our old enemy,the free surface of liquids.

olls.

i are

thei notnt to

nlirysmy,

CHAPTER 12

LONGITUDINAL STABILITYRecapitulation.-Before we begin, it will be as well to refresh our memories

on certain matters which were discussed in previous chapters.The centre of gavity (G) and centre of buoyancy (B) will not necessarily be

amidships, but may be forward or aft of midlengh. They will however be in thesame vertical line.

The centre offlotation (F) is the cenhe ofgravity ofthe ship's waterplane and isthe point about which the ship heels and trims. In box shapes, it is always amidships:in ship shapes it may be a little abaft or forward of the centres of buoyancy andgravity. Its longitudinal position changes with change of draft and sometimes withchange of trim also. It is sometimes called the "tipping centre".

The longitudinal metacentre (Mr) is a different point from the transversemetacentre, although it is found in a similar way and obeys similar laws.

The longitudinal metacentric height (GMr) is always very large, often well overone hundred metres. If the position ofG is not known, 8M, can often be used insteadof GMr,the enor thus caused being negligible in practice.

Draft is the depth ofthe bottom ofthe keel below the waterline. It is marked andmeasured forward and aft at the stem (Forward Perpendicular) and the rudder post(After Perpendicular), respectively.

Mean draft is the mean ofthe drafts fore and aft: that is, the draft at midlengthin way of the ship's loadlines, midway between the stem and the rudder post.

14"- 6."n = Draft forwar-d + Draft aft

Tonnes per centimetre immersion (T.P.C.) is the weight required to cause theship to rise or to sink bodily by I centimehe.

Longitudinal Metacentric Height-"GM"". The calculations to findthis arevery similar to those used for transverse metacentri c heighl. BMris first calculated,whilst KB and KG will have already been found for the transverse calculations, sothat GMrcanbe found by addition and subhaction.

89


For instance, suppose that a ship's BM. is found by calculation to be 130'4metres, that KB is 3'3 metres and KG is 6'8 metres. Then:-

BMr = l30.4metresKB = 3.3 metres

KM t = 133 ", ^"O""KG : 5.8metres

GM" : 12i7_:2menes

The Calculation of "BM"" for all Shapes.-Figure 65 shows a ship whichhas tipped longitudinally through a small angle, 0. .8 has moved out to 8r and thelongitudinal metacentre, M, is the intersection of the vedicals through B and 8,.

Compare this figure with Fig. 54, Chapter 8, which shows the transversemetacentre appearing as the ship heels. In each case there is an emerged and animmersed wedge; a shift ofB to B, parallel to the shift ofthe centres ofgravity ofthewedges; and a heel or trim to the angle 0. The transverse centre of flotation, aboutwhich the ship heels, is on the centre line, at C, The longitudinal centre of flotation,about which she trims, may or may not be on the longitudinal centreJine-in Fig.65 it is shown at 4 abaft that centreline. Its position is at the centre of gravity of thewaterplane area. The only other diference is one of perspective, so that where weconsidered breadth before, we now have length, and vice versa.

It can be seen from the above that we could show by means ofa proof similarto that for transvers e B M, thal:-

Where -1, is the moment of inertia of the waterplane, longitudinally, about tllecentre of flotation, and Z the ship's volume of displacement; then for all shapes,including ship shapes,

BM , =I+

LONGITUDINAL STABILITY 9I

The Calculation of BM" for Box Shapes.-In the case of box shapes, thecentre of flotation is on the longitudinal centreline of the ship. If we consider themoment of inertia ofa rectangle about its centreJine, it follows that ifl is the ship'slength, b her breadth and d the draft at which she floats:-

, _ bl''L- 12

but BM. =ILL- y

t 'r=#V = lxbxd, so the above becomes:-

tM,=a#nBM L =

ndJL (for box shapes only)

Example.-Find the BMrand GMrof a box'shaped vessel, 120 metres long, 15metres beam, which floats at a draft of 4'00 metres, the l(G being 5.00 metres

BM, = ruL=ffi=toometres

KB +&&KMt

KG

GMt

Trim.-This is the longitudinal equivalent of heel, but whereas the latter ismeasured in angle, trim is measured by the difference ofthe drafts fore and aft.

Ifthe draft forward is the greater ofthe two, the ship is said to be "trimmed bythe head". Ifthe draft aft is the greater, she is said to be "trirnmed by the stem". Ifthedrafts are the same, fore and aft, she is said to be "on an even keel".

In many ways, the calculation ofa ship's trim is simpler than that for heel, butthere is one complication that we do not meet with in transverse calculations. A shipheels and trims about her centre of flotation and when she is upright, the transversepositions of the centres of gravity, buoyancy and flotation are all vertically overone another, on the centreline. Longitudinally, none of them will necessarily beamidships; and furtheq although B and G must be in the same vertical line, the centreof flotation is very rarely directly over them. We shall shortly consider the effect ofthis on trim and sinkage due to added weights.

SO

but

= 2 metres

= 302 tn"n"t: 5 metres: 297 metres

F

mleIrtn,

c.bve

ar

hecs,

92 MERCHANT SHIP STABILIry

Change of Draft due to Change of Trim.-When a ship changes her trim,she can be considered to increase her draft at one end and to decrease it at the otler.The sum ofthe changes at both ends is the change of trim assuming that there is noincrease of draft due to added weights.

The change of draft due to change of trim will depend on the position of thecentre of flotation. When this is at midlength, the ship will increase her draft at oneend by exactly half the change of trim and will decrease it by a like amount at theother end. The mean draft will not change.

When the centre offlotation is not amidships, the draft will change more at oneend than at the other, because the ship will be tipping about a point which is not midwaybetween the ends. In this case, the change ofdraft can be found by a simple proportion.

Figures 66 (a) nd (b) show what happens in the above cases. F represents

FtG.66the centre of ffotation and CZ the centreline of the ship, whilst SI and S,t are theold and new waterlines, respectively. In each case the change of trim is,LS, + ZZ,.

In Fig. 66 (a/, F is amidships It can be seen that SF equals Zf, S,F equals {4whilst the angle 0 is the same on either side of .E The riangles ,S^t,F and TT,F are,therefore, equal in all respects, so:-

SSt =rrl

But SSr + fq = the change oftrim

So, SSt = 2; = half the change of trim'

This means that when the centre offlotation is amidships the change ofdraft ateither end is equal to halfthe change of trim.

Fig. 66 (b) shows what happens when F is not amidships. In the triangles ,SS,and 77, the angles 0 are equal and the angles at S and Imay be considered to be rightangles, so that the triangles are similar but not equal. One ofthe properties of similartriangles is that the sides are proportional to each other, so:-

SS,:77,:: SF: 7FNow, SS, + 77, : the change of trim (call this t)and ,SF' + 7F : the length of the ship on the waterline (call this I)

ss, =f xr

ITt=?xt

So,

and

LONGITUDINAL STABILITY 93

Example l.-A box-shaped vessel has its centre of flotation amidships andfloats at drafts of 5.00 metres forward and 5.16 metres aft. A weight is shiftedforward and causes a change oftrim of l2 centimehes. Find the new drafts.

Since the centre offlotation is on the centre-line, the draft at each end will changeby halfthe change of trim. The draft forward increases and that aft decreases, so:-

Old drafts Forward 5 00m Aft5.16mHalfchange oftr im +0.06 m -0.06 mNew drafts Forward l.lQ! m Aft l:!! m

Example 2.-A ship is 120 metres long and her centre of flotation is 3 metresabaft amidships. She floats at drafts of 7.45 metres forward and 7.33 metres aft. Findthe new drafts ifa weight is shifted aft so as to change the trim by 56 centimetres.

The centre offlotation is 63 metres from forward and 57 metres fiom aft.

Change ofdraft forward =-63 ;.56=29 grn

Change of draft aft =ffixs6=27 cmOriginal drafts F. 7 .45 m A. 7 .33 m M. 7. 39 m

Change due to trim 4.29 m 0 .27 m

New drafts F. Z:l! . A. Z:!9 . M. 7.!! -

Note that in the above case, although no weight has been added, the mean drafthas decreased by one centimetre. The effect ofthis is discussed later.

At first sight this might appear impossible, since the ship's displacement curvesgive only one mean draft for a given displacement. The displacements given in thecurves are calculated for the ship on even keel. When the ship changes her trim, shetips about the centre of flotation: if this is not amidships, where the mean draft ismeasured, the mean draft must change as the ship trims.

Fig. 67 (a) shows the shipfloating at her designed trim, at thewaterline SZ, C is the point amidshipsat which the mean draft is taken:F the centre of flotation, abafi C.

In Fig. 67 (b), a weight has beenshifted to cause the ship to trim by thestem. She has tipped about F to thenew waterline S, I. which cuts thecentre-line at C, the new mean draft.

st F

(o)


The mean draft has decreased from C to C, although the displacement hasnot changed.

Fig. 67 (c) shows the sameship trimmed by the head. The newwaterline is now ,9rl and the meandraft has increased to C^. for the samedisolacement.

We can convert mean draft to draft at the centre offlotation, if necessary, by asimple calculation:-

Fig. 68 shows a ship trimmed by the stem and floating at the waterline.9,L S7,is the equivalent even-keel waterline. The even-keel drafts are at C and F, whilst themean draft is at C, The ship's trim is equal to S,Sr + 77, The line S,.r is drawn parallelto S{so that,S,.r is equal to the ship's length (I.).

Let the distance ofI' from amidships (FC) be d metres.

r; = SSr + 77i = ship's trim.

The triangles S,.rI nd FCC, are similar triangles.

CC, =FCTx S,.x

CC, =dTrim L

so, cct=dxTl im

Ifthe designed trim is not for even-keel drafts, we can modifr this to:-

,.. Amount out ofdesigned Trim

If Fis on the same side of amidships as the end which is "down", the draft atF will be greater than the mean draft. If F is on the opposite side of amidships to theend which is "down", the draff at F will be less than the mean draft.

The draft at F is known as the true mean draft


Exampla-Aship is 120 m long and floats at drafts of6'76 m forward and 7'48 maft. What is the equivalent even-keel draft, ifF is 3 2 m abaft amidships?

u.unarun=tfrlA=7.12m

f nm=l .48- 6.16 = 72 cm (by the stem)

Difference of draft s (C) = l:#f= l 9 cm

Even-keel draft = mean draft + CC,=7 .12+0 02=7 .14 m

Displacement out of the Designed Trim.-We can see from the above that aship's displacement may vary slightly for a given mean draft, according to her trim.The displacement scale or curves will only give the correct displacement for the evenkeel condition which in this condition is the draft at F. This means that if we use themean draft to obtain displacement from the scale or curve, there may be a small errorinthe displacement so found. In most cases, this would not matter; but if it is nec€ssaryto obtain the displacement with greater accuracy, we can do so in one of two ways:-

(a) Convert the mean draft to draft at the centre of flotation, as above. Thentake out the displacement for the latter from the scale or curves-(b) Take out the displacement for the mean draft from the scale or curve.Then apply a correction, called a "layer conection", to that displacement. Themethod of finding this conection is as follows:-

FtG.69

In Fig. 69, SZ represents the even keel waterline for the draft at F or C. r/ isthe even keel waterline for the mean draft, at C,. The shaded area between the twowaterlines is the layer between them.

Weight of such a layer = T.P.C.xits depth in cms

So, layer correction = T.P.C.x CCr in cms

But CC' =7YJriJn

So. layer correction = T P C x-dxTrim

This correction must be added to the displacement tak€n for the mean draft, iftheendnearerto the centre offlotation is "down": or subtracted ifthe other end is "down".


Example.-A ship is 130 mehes long, her T.P.C. is l3'0 tonnes and her centreof flotation is 2'0 metres abaft amidships. She floats at drafts of 6'82 metres forwardand 7'18 metres aft. Her designed trim is at even keel. Ifthe displacement for a meandraft of 7'00 metres is given in the scales as 8016 tonnes, find the true displacement.

The mean draft is 7'00 metres.The ship is trimmed 36 cm by the stem. Since the desigaed trim is an even keel,

this means that she is 36 cm out ofdesigned trim.I-^ffqf.'.^_ T.C.P.xDistance ofF from midlength x Amount out of designed TrimLUiTEL-uun =

= l3X?i<36= 7 tonnes130

Displacement from scale = 8016 tonnesTrue displacement from = q94 tonnes

Moment to Chrnge Ttim by I Centimetre-"M.C.T.1c."- If we add,remove, or shift a weight forward or aft, the ship will change her trim. The greaterthe weight, or the greater the distance it is shifted or placed away from the tippingcentre, the greater the change of trim. Moment is weight, or force, multiplied by thelenglh of the lever on which it acts. tn this case the weight is that adde4 removed,or shifted; the length ofthe lever is the distance through which the weigbt has beenmoved, or the distance between the centre ofgravity ofthe weight added or removedand the centre of flotation. If we add or remove a weight ofw tonnes at a distanceofr metres from the centre offlotation, or shift it for a distance ofJ metres, then themoment changing the trin is l"J tonne-metres. The moment lo change trim by onecentimetre is the moment which will change a ship's trim by exactly one centimetre.Its amount will vary in the same ship with change of draft.

Fig. 70 represents a ship in which a weight, w, has been shifted forward to w,through a distance of d metres. B and G have moved to 8, and G, respectively. .SI

E

dnL


is the original waterplane and Sl', the new waterplane, the ship trimming about thecentre of flotation, F.

Let 0 be the angle between the original and new waterplanes, I/ the ship'sdisplacement and / her length.

Let us do a little simple trigonometry:-In the triangle GG, M,,the angle G isequal to 90" and the angle M. to 0. So:-

GG, =GM, xtalJ.g ( r )And yet more trigonometry:-In the triangle F77,, the angle Iis equal to 90"

and the angle Fto 0. In the triangle FS.l, the angle S is equal to 90'and the angle Fto 0. So:-

4= FTx tan O

SS, = f'S x tan 0

Adding these, we get:-7'{ + '9'S, =P7t1"te+FSxtan0

= (Ff + F,t) x tan 0

But (77i + SSr) equals the change of trim, whilst (Ff + F,S) is equal to the ship'slength, so:-

Change of trim = / x tan 0

If the trim is to change one centimetre-i.e., one-hundredth of a metre:-

/xtane=fi

tan0=-J.I uux length

The change of trim is due entirely to the shifting of the weight, so we cansay that:-

Moment changing trim = l'x dBut, wxd =WxGG,

So, Moment chan ging tim = llt x GG,Substituting for GG, from formula (l), we get:-

Moment changing trim = llt xGM L xtar:' 0Substituting for tanO from formula (2), this gives:-

Moment to chang etrim lcm=ll xGM tx1|-WYGML

, .e. M.C.T. lC. = ; ; ;l u0/

m

ld,ET

nghed"3trrdEEhe)tene.

(2)

l r isr


If GM, is not known, we can substitute,BM. for it in the above fonnula withoutcausing any appreciable enor.

Example.-Aship is 120 metres long, has a displacernent of 36l g tonnes and aGM" of 180 metres. Find her M.C.T.1C.

u.c.r.tc. =w ;,f;#, = *l%# = ro. 27 tonne_metres

. The Effect of Shifting a Weight.-We have already seen that for a ship to bein transverse equilibrium, her centre of gravity must be vertically over her centre ofbuoyancy. This applies equally in the longitudinal direction. If a weight is shiftedforward or aft, the centre of gravity will move parallel to it and will no longer bevertically over the centre ofbuoyancy, unless the ship changes her trim.

I II

\ , r l "GB

'-Gr

FI v

t t l<

This can be seen fiom Fig. 7l (a). The weight, w, has here been moved forwardt9 w, and G has moved to G,. The force of gravity nolv acts vertically downwardsthrough G,. in-the direction xG, whilst the force of buoyancy acts vertically upwardthrougb B, in tbe directionyB. The effect ofthis is to force the ship's head downwards,so that she starts to change her trim about the centre offlotation, F.

._- Ifwe now refer to Fig. 7l (b), we can see that as the ship trims, the wedge SFS,will emerge from the water and t}re w edge TFTrwill become immersed. _B will movi


out parallel to the shift of the centres of gravity of the wedges (g and g,), until itreaches 8, vertically under G,. The ship is then in equilibrium and floats at the newwaterline S,I,.

The moment changing the trim is the weight, in tonnes, multiplied by thedistance, in metres, through which it has been shifted. Since M.C.T.IC. is themoment to change the trim by one centimeffe, we can say that:-

Moment changing trimunangeol Inm=

MCT:fti-So, where w is the weight, in tons, and dthe distance through which it is shifted

forward or aft,Moment changing trim = w x d

Change of trim. in centimetres = M##.

The change ofdraft due to this can be found from the change of trim, as shownin the last section. If the weight is shifted aft, the after draft will increase and theforward draft decrease; if the weight is shifted forward, the reverse will happen.lf the position of the centre of flotation is not known, we usually assume it to beamidships since the error thus caused, if any, is generally small.

Example.-Find the change of trim and the new drafts in a ship of 125 meneslength, when a weight of 100 tonnes is shifted aft for a distance of60 metres. Theoriginal drafts were 4'80 metres forward and 4.76 metres aft. The centre offlotationis 2'5 metres abaft amidships and the M.C.T.lC. is 120 tonne-metres.

. Moment changing rrim _ 100 x 60uhange oI l t t=-mc.T.rc. lzo

=50cm

F. is 65 m from forward and 60m from aft.

Chanee of draft fory316 = $x50 = 26 cm- lz5

Change of draft aft = #

= rO "

Original drafts

Change due to trim

New drafts

F.4.80 m A.4.76m M. 4.78 m

- 0.26 m +0.24m

F.4.54m A.5.00m M.4-77 m

The Effect of Adding a Weight rt the Centre of Flotation.-Suppose that,as in Fig. 72 (a), a ship floats at a waterline, ,97, and that a weight, w is then addeddirectly over the centre of flotation, F. Let B and G be vertically under F. When the

IOO MERCHANT SHIP STABILITY

weight is added, G will move vertically upwards, towards. the centre ofgravity oftheweight. The ship will sink to the new waterline S,I, so that the weight of the addedlayer, .S.9,I,1", is equal to that ofthe added weight. The new centre of flotation is atF, but if the weight is not too large, it will be approxinately over the old centre offlotation. I will move vertically upwards towards the centre of gravity of the addedlayer, which is somewhere between F and F, In this case, both I and G have merelymoved upwards and are still in the same vertical line. Consequently, there will be noreason for the ship to change her trim and she will merely sink bodily, increasing herdraft by the same amount at each end.

Now consider what will happen when B and G are not vertically under thecentre offlotation, as shown in Fig. 72 (b). G will move directly towards the centre ofgravity of the weight and I towards the centre of gravity ofthe added layer, SS,{7;so they both move aft and upwards. Their upward movement will not affect the trim,so we need only consider the fore and aft movement. Thus, we can consider G asmoving to G, and B to 8,. The horizontal distance between the centre of gravity ofthe ship and that of the added weight, (d), is the s.rme as that between the centre ofbuoyancy ofthe ship and the centre of gravity of the added layer. The shift ofB andG can be found in the same way as in transverse stability, that is:-

When 7 is the weight of the ship; w, the weight added; V the volume ofdisplacement of the ship; and v, the volume ofthe layer:-

GG,=wfrd

!.-{ r:

FtG.72

BB,=ry

LONGITUDINAL STABILITY IOI

But the weight ofthe layer must be equal to the added weight, so:-

Therefore

Original drafts

Bodily sinkage

New drafts

B4=

Gq=

F.4.42 m A.4.64 m

+ 0.05 m +0.05 m

F.4.47 m A.4.69 m

wxdW

BBt

Thus B and G will move aft for the same distance and will remain in the samevertical line, so thatthe shipwill again merely sinkbodily andwill not change hertrim.

In either ofthe cases mentioned above, the sinkage ofthe ship will be:-

Weight addedT.P.C.

The vessel will increase her draft at each end by the amount of the bodilysinkage

Example l.-A ship has a T.P.C. of 25'2 tonnes and her drafts are 4.42 metresforward and 4.60 mehes aft. What will be her new drafts ifa weight of 126 tonnes isadded directly over the centre offlotation?

aoditv sintaee = !$f49 = #= s "

Example 2.-Find the new drafts in a ship which has a T.P.C. of 7.0, if 70tonnes of cargo are discharged from a point directly below the centre of flotation.The original drafts were 3'05 metres forward and 3'30 metres aft.

Bodily rise = w"'+T :4d"d = f = ro

".Original drafts F. 3 . 05 m A. 3 . 30 m

Bodi ly r ise -0.10 m -0.10 m

New drafts F. 1 .9! -

A. 3_!9 -

Moderate Weights Loaded off the Centre of Flotation.-When a weight isloaded at any point which is not in the same vertical line as the centre of flotation,the ship will change both her draft and trim. If the weight is of moderate amountthe mean draft will only be altered by a few centimetres so that the T.P.C. and theM.C.T.lC. will be nearly the same for both the old and the new drafts. In such a case,we can calculate the new draft and trim as follows:-

IO2 MERCHANT SHIP STABILITY

(a) First, assume the weight to be added at the centre of flotation and calculatethe sinkage by the formula:-

Bodilv sinkaee =

-ft.@ Next, assume the weight to be shifted from the centre offlotation, forward oraft, to its new position. Calculate the change oftrim caused by this by the formula:-

Chaneeof trim=;ffi.

(c) Calculate the change of draft at either end, due to the above changeof trim, as described in "Change of draft due to change of trim".(d) Add the results of fal to the &aft at each end; then apply the results of (c.).This will give the ship's new drafts after the weight has been added.(e) When weights are discharged, we follow the same procedwe as for weightsloaded. Remember, however, that they will cause a bodily rise, instead ofsinkage:also be careful about which way they will cause the trim to change.(f) If a number of weights are loaded or discharged, use the total effects ofall of them. The bodily rise or sinkage will equal the nett weight, divided by theT.P.C. The moment changing trim will be the algebraic sum of all the moments.The following Erample 3 illustrates this.

Example 1.-A weight of 200 tonnes is loaded into a ship at a distance ofl5 metres abaft amidships. The ship's length is 120 metres, T.P.C. is 10.0, M.C.T.lC.is 96 and the centre of flotation is 3.0 metres abaft amidships. If the original draftswere 5'85 metres forward and 6.00 metres aft, find the new drafts.

\Veiohr 4.ldadBodily sinkage =

ff =

ffi = Zo c.

Distance of added weight from F = I 5 -3 =12 mMo€$oglu t.il! = 2oo=\t2 = 25 cmLnatrge oI mm =

M.c.T.lc. 96

Change of draft due to trim: po*-6=Sx25=l3cm

Original drafts F. 5'85 mBodily sinkage +0.20 m

6 {5- m

eft =ffx2s= 12 cmA.6.00m

+0. 20 m

O.ZO.-0.13 m +0.12 m

F. 5.92 m A. 6.32 mNew drafts

LONGITUDINAL STABILIry 103

Example 2.-A vessel floats at drafts of 5.12 m forward and 4.69 m aft. HerT.PC. is 14.4; M.C.T.tC. is 102; and F is amidships. What will be the new draftsafter 98 tonnes of water has been pumped out ofthe fore peak tank, 52 m forwardof amidships?

Boditvrise=ffi=f|=2"rll

chanee oftrim=;ffi. =9!f =so ".

^ _ Since F is amidships, the drafts fore and aft will each change by halfthe changeof trim.

(a) Loaded

(b) Loaded

(c) Discharged

Find the new drafts.

F. 5.12 m A. 4.69 m-0.07 m -0.07m5os. q42^

-0.25 m +0.25 m

F.4.80 m A. 4.87 m

120 tonnes, 50 mehes forward ofl'.70 tonnes, 20 metres abaft F.90 tonnes, 30 metres forward ofF.

Original draftsBodily sinkage

j change oftrim

New drafts

^ Exaryde 3.-A ship is 140 metres long, has a T.p.C. of 20 and an M.C.T.IC.

of 120, whilst the centre offlotation is 3'0 metres abaft amidships. The draft is 7.10mefes forward and 7.25 metres aft. The following cargo is therrworked:_

Weight Distance fiom F Moment120 tonnes70 tonnes90 tonnes

(+)(+)(-)

50 m. (Forward)20 m. (Aft)30 m. (Forward)

6000 by head1400 by stem2700 by stem

100 tonnes (+) | 900 by head

Bodity sinkage - Nett wgigllll loaded = ioEb

=, ",

change of trim= I&!9{ {318in8Jnl! = #

= ,u "-

Forward=ftx16=8 cm

Aft=ffix16=8 crn

Change of draft for trim:



F.7'10m 4.7'25m+0'05m +0.05 mZ.tS. Z :0.

+0.08 m -0.08 mF.7.23 m 4.7.22 mNew drafts

Large Weights Added OIf the Centre of Flotation.-When large weightsare added to a ship, the methods described in the last section must be modified.This is necessary because, as the ship changes her draft, there may be considerablechanges in the position ofthe centre offlotation and in the amounts ofthe tonnes percentimetre immersion and moment to change trim by one centimetre. If the weight isIlot too large (say, not more than 8 o/o or 10% ofthe ship's deadweight), the followingmethod will give results which are accurate enough for most practical purposes. Forlarger weights, or where greater accuracy is required, it is better to use momentsabout the after perpendicular, as described later in this chapter.

(a) Find the approximate bodily sinkage, as shown in (a) above, using theoriginal torures per centimetre immersion. Apply this sinkage to the old draft tofind an approximate new mean draft and find the tonnes per centimetre immersionfor this new draft. Find the mean of the two tonnes per centimetre immersion byadding them together and dividing by two. Use this mean T.P.C. to find the truesinkage and new mean draft, using the formula:-

r*".Altematively, ifa sumciently accurate displacement curve or scale is available,

this may be used to find the new draft, as shown inthe following Example 2.(b) Find the position ofthe centre of flotation for this new mean draft, usingthe ship's stability curves. Also the moment to change trim by one centimetre atthis draft. Use these to find the change oftrim by the formula:-

cban I\'l

' T

se or tnm = Mtiicl

(c) Calculate the change of draft due to the change of trim, using the newposition of the centre offlotation.

(d) Find the new drafts fore and aft by applying sinkage and changes, as in thelast section.

Example l.-An oil tanker, which is light, is found to be trimmed too much bythe stem. In order to change the trim and also to increase the draft, a tank which hasits centre of gravity 70 metres abaft the stem is filled with I100 tonnes of water.

LONGITUDINAL STABILITY

The following details are found from the stability scales:-Draft T.P.C. M.C.T.IC. Centre of Flotation

4.00 m4'50 m5'00 m

20.821.221.7

160165170

2'9 m abaft amidships3'0 m abaft amidships3'l m abaft amidships

Ifthe original drafts were 2 80 metres forward and 5 20 metres aft, find the newdrafts. The ship is 200 metres long.

The original mean draft of 4'00 m gives a T.PC. of20'8.

Approximate sinkur" = y+*e+94 = -L!-Qg = 53 6,1

Approximate new mean draft = 4' 00 + 0' 53 = 4' 53 mT.P.C. for draft of 4 00m = 20 8T.P.C . fior draft of 4 . 53m = 2l . 2

vean T.p.C. = Z1_!

Corrected .inkage = I9E[149ed = I !-Q0 = 52 qrn

Corrected new mean draft = 4 . 00 + 0 .52 = 4.52 m

At this draft, the M.C.T.l C. is 165 and the centre of flotation is 3'0 metres abaftamidships.

The tank will be 33 metres forward ofthe centre offlotation.Moment changing trim _ ll00x33 _ ̂ ^^L nange oI mm -

Mcjlc:- = --i6- = ttu

"-

Change of draft for trim: Forward = !-Q1x220 = I l3 cm^z|Jll

Aft = lf^ x 220 = l07 cmzUU

F.2 80m A.2.20 m+0.52 m +0.52 m

:. :Z -

SIZ^+ 1.13 m -1.07 m

F.4.45 m A. 4.65 m

Example 2.-The ship for which the hydrostatic particulars are given in theback ofthis book, is 140 m long and floats at drafts of5'56 m forward and 5'60 m aft.She then loads 600 tonnes of cargo at 35 m forward of F; 1800 tonnes at 2 m forwardofI; and 1400 tonnes at 3l m abaft F. Find her new drafts.

Present mean draft is 5'58 m

105


Change due to trimNew drafts


The hydrostatic particulars show that F is amidships at this draft, so the draft atf'will also be 5.58 m. The particulars also show that the displacement at this draftis 9702 tonnes.

Weieht Distance from F Moment600,

1800,1400 t

3s m (F)2m(F)

3l m(A)

21000 by head3600 by head

43400 by stem3800 r 18800 by stern

New displacement = 9702 + 3800 = 13502 tonnesNew draft at F for 13502 t displacement = 7.32 m

M.C.T.lC. at draft of 7 '32 m = 196 .9 t/m

_{68 rmfromaft

[ .7 m abaft midships

But the ship was originally trimmed 4 cm by stem

So, the new trim = 95 + 4 = 99 cm by stem

Difference of draft for trim: Forward = ffix99

= 5l cm

chance of trim = ;ffi; = 13*93 = * cm (by stem)

Aft = 98, -3 x99 = 48 cml4u

F.7 .32 m A.7 .32 m

-0.51m + 0.48 m

F. 6.81m A. 7.80 m

New draft at F

Difference for trim

New drafts

Special Trim or Draft.-It is often desirable to obtain a certain trim whenthe ship is loaded. Generally, we do not require the trim to exact limits, but merely"a little by the stem". In this case, it is usually sufficient to load weights as desiredduring the earlier stages ofloading. Later on, we can get tle trim roughly as we wantit by using a combination of experience and common sense. Finally, near the endof the loading, we can trim our ship exactly as we want her by'Juggling" with theM.C.T.lC. and the amount of weight to come on board. Sometimes, as when a shiphas to cross a bar, we require a certain maximum draft, or wish to keep the after draftconstant. This can be done quite easily.

LONGITUDINAL STABILITY IO7

Loading Weights to Obtain a Desired Trim.-First, find the difference of thedrafts forward and aft, thus finding the existing trim. The difference between this andthe desired trim is the change oftrim required.

Let , be the existing trim and /, the desired trim. Let w be the added weight andd its distance from the centre offlotation.

Change of trim required = I - t,

Change of trim _ M.C.T.M.C.T.1C.

M.C.T. _ r _,M.C.T.IC. 1

M.C.T. = M.C.T.1 C.x(t - t)

From which:-

,_ M.C.T. lC.( t - r r )w

M.C.T. lC.( , - r , ),=------d -

Example 1.-A ship has an M.C.T.lC. of 150 tonne-metres and floats at draftsof 6'00 metres forward and 7.00 metres aft. How much water must be run into a tank,the centre ofgravity ofwhich is 50 metres forward ofthe centre offlotation, to bringthe ship to a trim of 20 centimetres by the stem?

Present trim (r) = 100 cm bY stem

Required trim (t' ) = 20 cm bY stem(, - 4) = 80 cm by head

Let w be the required weight

Then: wxd =M.C.T.lC.(t - t,)

50w = 150x 80

w = E9I& = 240 tonnes.JU

Example 2.-Aship, which is completing loading, has drafts of 7.58 m forwardurdT'72 m aft.Het M.C.T.1C. is I l8 t/m. 360 tonnes ofcargo remain to be loaded andthis is to be distributed between No.l hold (45 m forward ofI) and No.4 hold (25 mabaft F). How much cargo should there be loaded in each hold in order to finish witha trim of. 50 cm by the stem?


Present trim =7 -72-7-58=l4cm by stern

Required trim = 50cm by stem

Required change oftrim

into No.l; (3r

Moment of No.l = 45w t/m by head.

Moment of No.4 = 25w (360 - w) t/m by stem.

To increase the trim bythe stem, the momentofNo.4 mustbe greater than that ofNo.1.So' "":;ff;#),"-oo;'"'""'"'*"'Change of trim = M€J.

M.C.T. lC.

36= 9000-70wi l8

36x I 18 = 9000- 70w

70w = 9000-36x1l8

- = 9000 -_3=6 x I | 8 = 68 tonnes in No. I

1l)

360 - 68 = 292 tonnes in No 4

Position to Load Weight so rs not to Change the Draft Aft.-When a weightis loaded, there is usually a bodily sinkage ofthe ship and also a change of trim. Thebodily sinkage tries to increase the draft aft: whilst if the weight is loaded forwardof the centre of flotation, the change of trim tries to decrease the after draft. It ispossible by balancing these effects against each other, to keep the after draft constantwhen a weight is loaded.

Let w be the added weight and d its distance forward ofthe centre offlotation.Let Z be the length ofthe ship and / the distance ofthe centre offlotation from aft.

Bodily sinkage = ft

This is equivalent to an increase ofdraft aft.

Total change of tri* = M#j: I .fL

Change of draft aft. due to trim =;ffixf

This is eouivalent to a decrease of draft aft.

= 36cm by stem

Load w tonnes

LONGITUDINAL STABILITY IO9

For the after draft to remain constant, the increase due to the sinkage must beequal to the decrease due to the change of trim. That is:-

Bodily sinkage = Change of draft aft, due to trim.

wxd * l_ wM.C.T.IC. l , T.P.C.

wxd _ wxLM.C.T.1C. T.P.C.x/

-r _ wx l , x M.C.T. lC." -

,>.T-PC,

-r- M.C.T. lC.xZ"- TPc,

Note that the after draft will remain constant if the weight is loaded in theposition so found, inespective ofthe amount ofweight added or discharged, as longas the MCTC and the centre offlotation do not change.

Example.-A ship which is loading cargo has drafts of 6.50 metres forwardand 7 00 metres aft, when there are a further 200 tonnes ofcargo to come on board.Where must this cargo be placed in order to maintain this draft of 7.00 metres aft?The ship is 140 metres long, her T.P.C. is 20, M.C.T.lC. is 135 and the centre offlotation is 3.0 metres abaft amidships.

Distance ofF from aft = 70 - 3 = 67 m.

,_ M.C.T. lC.xL _ l35xl40 _, . r | _ c^_.o = *tEt;;: m torward of F

Load cargo 14.l metres forward of F; or ll.l metres forward of amidships.Loading a Weight to Produce a Desired Drrft Aft.-This may be achieved

by a little modification of the above. When a weight is loaded, the bodily sinkagecauses an increase of draft aft, whilst the change of trim may cause an increase ordecrease, according to whether the weight is loaded forward ofor abaft the centre offlotation. So:-

To obtain a decrease of draft, aft:-Change ofdraft aft = change due to trim - bodily sinkage.

To obtain an increase of draft, aft:-Change of draft'aft : Bodily sinkage + change due to him.

In the latter case, ifthe bodily sinkage is less than the change required, the signwill be + and the weight loaded abaft the cenhe offlotation. If it is geater, the signwill be - and the weight loaded forward ofthe centre offlotation.

l l0 MERCHANT SHIP STABILITY

Example.-Aship has a draft, aft, of 5'14 m. She is 120 m long, has a T.P.C. ofl5'1, M.C.T.lC. of 102, and F is 2'0 m forward of amidships. Where must 90 tonnesof cargo be loaded in order that the ship may sail with a draft, aft, of 5'00 m?

Required decrease of draft aft = 5.14 - 5.00 = 14 cm

BodilYsinkaee=1fa. = i?ai

= o.-

Required rise aft due to trim 2! "

To causes the ship to rise aft, we must load the weight forward ofF

Rise aft due to tri- =,#i+.";

)o_90d v 62to2" t20

6 = 20xL02x120 = 43.9 m (forward oftr')

Maximum Weight to Lood for a Given Draft -We may wish to load asmuch cargo as possible, but to sail with a given maximum draft in order to cross abar or a dock sill. This can be done by bringing the ship on to an even keel at therequired draft.

Example.-A ship has drafts of 6'72 m forward and 6'94 m aft. Her T.P.C. is21'7, M.C.T.IC. is 183, and F is 1.5 m abaft amidships. In order to cross a bar, herdraft on sailing must not exceed 7'00 m. What is the maximum weight of cargo thatshe can load, and where must it be placed?

Bring the ship to an even keel at 7'00 m draft.

Present mean draft 6.83 m

Maximum oermissible draft 7 .00 m

Maximum sinkage 0. 17 m ( l7 cm)

weight to load Sinkage t r.pErzJr.T=369r

Present trim = 6. 94 - 6. 72 = 22 cmby sterrr

To bring the ship to an even keel, the cargo must be loaded forward ofF so asto change the trim by 22 cm by the head.

Chanee of trim =

"ffi,"" -

369d"- lnd = 22al=83 =rc'9m forward of F

J69

LONGITUDINAL STABILIry III

The Use of Moments about the After Perpendicular.-This is an altemativemethod offinding moment to change him when weights are loaded or discharged. Itsimplifies the working, in that the moments of all weights loaded are added, whilstthe moments ofall weights discharged are subtracted: so that it is not necessary toconsider the effect on the trim ofeach separate weight as we go along.

To use the method, we must be able to find:-(a) the distance of the centre of buoyancy from the after perpendicularat any draft;

(b) The distance of each parcel of cargo from the after perpendicular.

The Stability Information Booklet, as recomrnended by the Deparhnent ofTrade, gives the above information.

IA/P I

F- - - - - - - d - - - - r- - - 5] tt

lEgw4F IG

6,L- D

FtG.73

Fig. 73 illustrates the method. Before any weights are loaded, B and G rnust bein the same vertical line in order that the ship may be in equilibrium, fore and aft.Let the ship's displacement be I/ tonnes and let B and G be at a distance ofr metresfrom the after perpendicular.

The moment of gravity about the A.P. = ,/.r tonne-metres.

Now let a weight ofw tonnes be loaded at a distance ofD metres from the afterperpendicular. This will cause the ship to sink bodily to some new mean draft. It willalso cause G to move to G, so that the new displacement (I/,) will now act verticallydown through G, The ships' moment of gravity about the after perpendicular willnow be W , multiplied by the distance of G, from the after perpendicular.

Ifthe weight loaded is large, the bodily sinkage may cause B to move a little say,to some point Br We must, therefore, use the stability information to find the distanceof,B, from the after perpendicular. The new displacement l/, will now act verticallyupward through 8,. so the moment of buoyancy about the after perpendicular willnow be I/, multiplied by the distance ofB, from the after perpendicular.

t

II2 MERCTIANT SHIP STABILITY

We now have the moment of gravity acting downwards t}rough G, and themoment ofbuoyancy acting upwards through,B, The difference between the two willbe the moment trying to change the ship's trim. That is: -

Moment to change trim = Moment of gravity - Moment of buoyancy.Ifthe moment of gavity is the greater, the ship will now trim by the head. Ifthe

moment of buoyancy is the greater, the ship will trim by the stem.To use this method, therefore, we must:-

(a) If necessary, corect the original mean draft to give the draft at F. Fromthe stability information, find the distance of .B (and hence of G) from the afterperpendicular and also the ship's displacement for this draft.(b) Draw up a table ofweights and their moments about the after perpendicular.including those ofthe ship. Moments of weights loaded must be added, those ofweights discharged must be subtracted. This will give the ship's new displacementand moment of gravity.(c) Find the bodily rise or sinkage and, ffom this, the new draft at F.(d) Use the stability information to find the new distance ofB from the afterperpendicular for the new draft. This, multiplied by the new displacement, willgive the new moment ofbuoyancy.(e) The difference between the new moments of gnvity and buoyancy will bethe moment to change the trim. If the moment of gravity is the greater, the ship * illtrim by the head: ifthe moment ofbuoyancy is the greater, she will trim by the stern-A We can now use the methods described previously to find the new trim anddrafts fore and aft.

Example l.-A ship has a displacement of 8600 tonnes at a mean draft of5 01 m on even keel. B is 71.90 m from the after perpendicular and the T.p.C_is 20'60 tonnes. She then loads 600 t at 40 m from the after perpendicular.300 t at ll0 m, and 4OO t at 70 m. She also discharges 500 t from 50 mfrom the after perpendicular, and 400 t from 60 m. The stability informationthen shows that at a mean draft of 5.20 m,8 is 71.80 m fiom the afterperpendicular and the T.P.C. is still 20.60 t. Find the moment changing the trin

ShipLoadedLoadedLoaded

DischargedDischarged

Weight d from NP Moment about A./P8600

+ 600+ 300+ 400- 500- 400

71.9040

l l0705060

618340+ 24000+ 33000+ 28000- 25000-24000

9000 654340

LONGITUDINAL STABILIry

Added weight = !000 - 8600 = 400 tonnes

BodilV sin-kaee =ffi = 19.4 cmNew mean draft 5 '01+ 0.19 = 5.20 m

Moment of buoyancy = Distance of B from A,1P x new l/= 7l .80x9000 t .m

Moment changing trim = moment ofgravity - moment ofbuoyancy654340 - 646200

8140 t/m by the head.

Example 2.--:f1rc hydrostatic particulars given in the back ofthis book are for avessel of 140 m long. She is floating at drafts of6.14 m forward and 6.26 m aft. Shethen loads 800 t at 52 m from the after perpendicular, 700 t at 74 m; and discharges550tfrom80m. Find the new drafts?

Old mean draft : 6.20 mAt this draft, F is 0.67 m abaft amidshipsCorrection to draft at F:0.67 x 0.12 + 140:0.0006 mSo, corrected draft at F = 6-20 mFordraftof6'20m, the displacement is I t027 tandBis 7l.56mandMCTC is 177.5The ship has a trim of4 cm8-G=TrimXMCTC+Wwhich means that G-B:4 x 177-5 = 11027 = 0.06 m aft ofBPosition of G = 71.56 - 0.06 : 71.50 foao

Weight dfromA.IP Moment about A,{Pt to27+ 800+ 700- ))u

71.5052

80

788430+ 41600+ 51800- 44000

11977 837830

New position ofG is 837830 = 11977 :69.953

From the hydrostatic particulars, with new displacement of I 1977 ronnes:-Draft = 6.635 MCTC: 185.4 LCB=7r.372 LCF = 68.9G-B = 7 l'372-69.953 : 1.419Trim = B-G x W - MCTC =11977 x 1.419 + 185.4 = 91.7 cm (92)tud tr im effect = 92 x 7l-1 +140 =47 6.64-0.47:6.17aft t r im effect =92x68.9=l4O =45 6.64+0.45=7.09Answer Final draft = 6'17 m fivd and 7.09 m aft

l t3

omfter

thewill

ithe

ilar,;ofEnt

fterrill

lbeIillltn.

md

lofLc.lar,m

ionfrerim.

ShipLoadedLoaded

Discharged

CHAPTER 13

STABILITY CURVES AND SCALESWhen a ship is built, the Naval Architects calculate certain data affecting

her stability and set it out in the form of tables, curves and scales. Some of thisinformation must be supplied to ships, as described in Chapter 1 5 . Meanwhile, let usconsider the basic data and their uses.

Hydrostatic Curves.-In the back of this book will be found a hydrostatic datatable for a ship and a Deadweight Scale for another ship. Curves are also supplied toships, which are a graphical representation ofthe information contained in the tablesand scales. Each ship has its own unique set of this information. As we have alreadymentioned, such curves vary considerably in detail, but are fundamentally the same,so that anyone who can understand those given here should have no difficulty intaking off information from any others he may encounter.

A scale of mean drafts of the ship runs vertically up the left-hand side of thesheet, lines being drawn horizontally across the plan at every half-metre. The mainbody ofthe plan consists ofa number ofcurves, each showing the amount or positionof any one item for any mean draft on the scale. Along the top or bottom edges arescales from which the amounts ofthe various items can be read off.

To obtain information, find the ship\ mean draft on the left-hand scale anddraw a line horizontally across the plan from this, until it cuts the curve required; oruse dividers to measure up to this point from the nearest horizontal line. From thepoint thus found, drop a perpendicular line to the appropriate scale on the bottom ofthe plan, or again use dividers, and read offthe required information.

It will be noticed that the curves have their descriptions written on them in full,whilst, to save space, abbreviations have been used for naming the scales. Sometimesabbreviations are used along the curves also, but this should not cause any difficulty,since they are standardised and should be known by anyone studying stability. Thoseused in the curves given here are as follows:-

Height ofthe centre ofbuoyancy above the keel ..............................K8Height ofthe transverse metacentre above the keel......-.-.....-..-.--.--.KMHeight of the longitudinal metacentre above the keel..... . .. . . .. ..KM,Moment to change trim by one centimetre............................ M.C.T.lC.Tonnes per centimetre immersion.... .................T.P.C.Centre of buoyancy from A.P........ ........ B. ffom A.P.Centre of flotation from A.P ............. ...... F. from A.P.Displacement ...... ."Displacement"

Always be careful that you take of information from the proper scale.tt4

Lgtisus

STABILITYCURVESANDSCALES II5

Use of the Hydrostatic Curves.-The uses of the information, which we canobtain fiom the curves, are fairly obvious to anyone who has read through this book.Let us consider an example ofthe information that it is possible to obtain.

Suppose that we wish to find all possible stability information from the curvesgiven in the back of this book, assuming that the ship is floating at a mean draft of5.40 metres. We draw a horizontal line across the plan at this draft and can then takeoff the following information from the scales at the foot:-

Displacement...... .......9300 tormesK8.. . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 '93 metresKM... . . . . . . . . . . . . . . . . . . . . . . . . . . 8 '18 metresKMr.. . . . . . . . . . . . . . . . . . . . . . . . . . . .247 metresM.C.T.1C............ 162 tonne-metresT.P.C.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20.9Centre of flotation from A.P............. ...... 70'2 metresCentre of buoyancy fromA.P.......... ....... 7l'9 metres

The Deadweight Scale,-This scale is familiar to most ship's officers and isanother method of giving certain stability information, which they are most likelyto need. A typical scale will be found in the back of this book, with the hydrostaticcurves, and is made out for the same ship as the latter. To obtain information from thescale, draw a horizontal line, or lay a ruler, across the scale at the appropriate draft,then read-offthe figures shown against this.

For example, to obtain information from the scale for a draft of 6'40 m, lay aruler across it at that draft and you will find the following:-

Deadweight in salt water................ ........6900 tonnesDisplacement in salt water.............. ...... 11450 tonnesT.P.C. in sal t water. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 l '8M.C.T.lC. in salt water......... ......... 181 tonne-metresDeadweight in fresh wa1er............... .......6600 tonnesDisplacement in fresh water. ............. ... 1l150 tonnesT.P.C. in fresh wa1er........................ .........21'3 tonnes

Hydrostatic Particulars.-Sometimes, instead of hydrostatic cuwes, similarinformation is set out in tabular form, as "Hydrostatic Particulars". This is themethod recommended for the Stability Information Booklet which is described inChapter I 5 . Its main advantage is that it will usually give more accurate informationthan that which we could obtain from the hydrostatic curves or the deadweightscale. An example of such a table, in the recommended form, is given at the backofthis book.

[aloES

dyle,in

hednonlt€

Ddorheof

il1,leS

ry,tse


To obtain information from the table for a certain draft, take it out for thenearest draft or drafts and then interpolate, as necessary, according to the accuracyrequired. For instance, suppose that ri/e wish to find information for a draft of 6'34m: we take out the information for drafts of 6'20 m and 6'40 m and then interpolatebetween these for the required draft. This will give:-

Displacement ...... .....11330 tonnesT.P.C.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21.75M.C.T.1C............ 180'09 tonne-metresICB from the After Perpendicular...... .. 71'50 metresICF from the After Perpendicular...... .. 69'20 metresZCB above base ........ 3'45 metresKM... . . . . . . . . . . . . . . . . . . . . . - . . . .8 '03 metresKMr.. . . . . . . . . . . . . . . . . . . . . . . . .226'2 mef ies

Curves of Statical Stability.-These are graphs, which show the value ofGZ for a parlricuJar ship with a particular KG (or Gtr4) at a particular displacementfor all angles of heel. The curves sometimes show righting moments (W x G4, orsometimes show righting lewrs (GQ, according to the purpose for which they areintended.

There is a theorem which says that we can find the shape of a GZ curve at itsorigin (i.e., 0') by drawing a perpendicular at l radian (57'3') and on this, markingoff the GMfrom the GZ scale. A straight line from the point so found to the point oforigin will form a tangent to the curve at the origin.

Figure 74 shows a single curve, on which the righting levers have been plotted.It has been constructed by calculatingthe GZ's d vaious angles ofheel and plotting

5

?

I

FlG.74

these as a graph. The dotted linesillustrate the method offinding the shapeof the curve at its origin, referred to inthe theorem above: the assumed GMhasbeen marked ofr on the perpendicular at57.3' and the line fiom here to the pointof origin forms the tangent to the curveat the origin.

A single GZ curve enables us to find the range of stability, the angle of vanishingstability; the angle and amount of maximum GZ; the approximate GM for which thecurve has been drawn. The latter is found by drawing a tangent to the curve at thepoint of origin and then measuring offthe GMat the point where this tangent meetsthe perpendicular at I radian.(57'3')

>l Gl't

STABILITY CURVES AND SCALES

If we wish for a comprehensive picture of the stability of a ship w'e mayconstruct a series of curves of statical stabiliry, using the same iKG for all, but adifferent displacement for each. A t)?ical set of such curves, drawn for one ship, isshown in Figure 75, above.

Cross-Curves. -These areanother common method ofdrawing graphs of rightinglevers. They are different fromthe ordinary curves of staticalstability, however, because eachcurve shows the righting leversfor one angle of heel only, butat different displacements.The principle on which theyare constructed is showndiagrammatically in Figure 76;in which the curves of staticalstability are represented asstanding up, one behind theothel at equal distances apart.Points 15" apart have been

tt7


o{,JtrFlrJ:

uJ ot . I o>t irY

oo:(uJon-E

o

t!

:.

9LEO

c., ln

N9(9 t l

trFL-

za

Fttc)Er l jF-gEo9

bL

STABILITYCURVESANDSCALES 1I9

marked on each curve and each "set" ofpoints has beenjoined by a fair curve, whichis the cross curve for that angle of heel. It can be seen from this that their name isderived from the fact that these curves. in effect. cut across the curves of staticalstability.

Figure 77 shows a typical set ofcross curves, which has been derived from thecurves of statical stability given in Figure 75.Tl;'e GZ canbe found from these, forany angle for which a curve is given, by measuring vertically upwards to the curve,at the displacement chosen, and then reading offthe GZ from the scale on the left-hand side. For example, the GZ for 30" of heel at 7000 tonnes displacement wouldbe 0'55 metres.

The Effect of Height of 33G". -If the actual height of G is different from theassumed height used in constructing the curves, the GZs obtained from the curveswill be incorrect. For instance,in Fig. 78, let G 6e the assumedposition of the ship's centre ofgravity and let G, be its actualposition. The righting lever givenby the curves will be GZ, but thetrue righting lever willbe GrZr.

If Gx is a vertical lineparallel to ZZt, theni-

GZ = xZ,

GrZr= GZ* Grx

But, in the triangle GG rx:-G,x = GG, x $i1 $

Thus, in this case, True GZ: GZ from atrves * GG, x sfu $.

On the other hand, if G were the true position and G, the assumed position ofthe ship's centre of gravity:-

GZ: GrZ - Grx And True GZ: GZ ftom ctwes - GGr x Sin eSo, if the actual KG is different from that used in the curves, we must apply

a correction of GG. x Sin 0 to the GZ found from the curves. This correction mustbe added if the true KG is less than the assumed KG: or subtracted ifthe true iKG isgreater than the assumed KG.

For convenience, a table of corrections is often placed on a set ofcross curves,as in the right-hand top corner ofFig.77. This is actually a table of natural sines;which gives us Sin 0, for use as above.

xz

c7

n

AII.IISVIS dIHS INVHJUIIW

STABILITYCURVESAND SCALES I2I

Example.-Ihe cross curyes show a GZ of 0.73 m at 45o heel and displacement7000 tonnes, for an assumed KG of 7.00 m. If the actual KG is 7.54 m, what is theturc GZ at 45"?

Correction = GG xSin 0 = 0.54xSin 45" = 0.38 m

Since the acf.nl KG is greater than the assumed r(G, this correction must besubtracted from the GZ found from the curves.

So, True GZ:0.73 - 0.38 = 0.35 metres.

KN Curves.-Formerly the assumed height of G was taken to be somewherenear the typical height offc. The corrections were then added or subtracted to thecross curve value of GZ. Modem practice is to assume the height ofKG as NIL. Inthis case the correction is always subtracted because KG is always more than NIL. Ifwe refer to Fig. 77 (a), we can see that, in this case, Ki/ would be the righting lever.If, however, G were the tw position of the ship's centre of gravity, then:-

TrueGZ= KN - Ky= r(ir'-KGx sin 0

A typical set of KM curves is shown in Fig. 78.

The advantage of this method is that the correction is always r(G x sin 0 and isalways subtracted from the Kly' found from the curves. This reduces the possibiliwof error.

Example. :flte curves show a KN of 3.98 m at 30" of heel, for a displacementof9000 tonnes. Ifthe ship's KG is 6.84 m, what is the GZ at 30"?

-:il:-:;:i:l*The Metacentric Diagram.-This diagram is another method of showing the

heights of the metacentre and centre of buoyancy for various mean drafts. A scaleofmean drafts is drawn up and a line is drawn across the paper at an angle of45. toit-this is called "the 45' line". A curve ofcentres ofbuoyancy is drawn in below thisline, so that the distance from the curve to any point on the line is equal to the distanceofthe centre ofbuoyancy below the water line at that point. The curve ofmetacentresis drawn in a similar way above or below the 45. line. This is best seen fromFig. 79. To find the positions of B and M for my particular draft, draw a horizontalline through that draft. Then, through the poitt where this cuts the 45" line, drawa vertical line to meet the .B and M curves. The scale readings opposite the points


thus found give the positions of the centre ofbuoyancy and the metacentre above the keel,whilst the distance between the points is equalto the BM.

For example, n Fig. 79, to find theKB. KM and BM. at a draft of 5'50 metres.Draw the horizontal line, AB, thrbugh the5'50 metre mark and then the vertical line.CBD. C then represents the height of thecentre of buoyancy, D that of the metacentreabove the keel and CD, the ship's BMWe can see from the scale that KB equals 3'5metres, KM equals 6'8 metres and BM is 3'3metres.

These curves are still to be found insome ship's stability information, but modempractice is to obtain this information fromHydrostatic Data tables as shown at the back

this book.

FtG.79

Es.E

B.

E

]e{.5.3

)fI,rl

mmm*

CHAPTER 14

BILGING OF COMPARTMENTSThe Effect of Bilging a Compartment.-When a hold or compartment is

bilged (i.e., holed, so that it becomes flooded), a number of things can happen.(a) The ship will increase her mean draft in order to compensate for thebuoyancy which she has lost, since she must displace her own weight of water inorder to float. Ifan empty hold is bilged, it will cease to displace any water and sothe ship must sink until the remaining, intact part ofher has made up this loss anddisplaces a weight of water equal to the weight ofthe ship. Ifthe hold has cargo init, such cargo will continue to displace a certain amount ofwater, so that the bilgedcompartrnent only loses a part of its displacement. The amount of displacementthen lost, expressed as a percentage ofthat which would have been lost had thehold been empry is called the Permeability ofthe hold.

(b) Ifthe centre of gravity ofthe compartment is in the same vertical line asthe centre of buoyancy of the ship, the latter will merely sink bodily to a newwaterline. Ifthese two points are not in the same vertical line, B will shift forwardor aft, as the case may be. As the bilging is the cause ofloss ofbuoyancy only andnot actual addition ofweight to the ship, G will not move, so the ship must changeher trim in order to bring I back into the same vertical line as G.

(c) Note the difference between this case and that ofweights added, removed,or shifted. In the latter case, G moves as well as B, so that the relative positionsof the weight and the centre of flotation govem whether the ship will change hertrim. In this case, where G does not shift, the change of trim, if any, is governedby the relative positions ofthe bilged comparfinent and the centre of buoyancy.

(d) If the compartment is divided longitudinally, the ship may list on accountofthe lost buoyancy being out of the transverse centre-line.

Permeability.-is the ratio between the space available for water and the totalspace in the comparfinent.

For instance, suppose that a compartment has a volume of5000 cubic metres.This would be the volume available for water if the empty compartment was bilged.If this compartrnent was filled with cargo, the solid pads of that cargo would take

t23


up space which would be otherwise available for water, so that less water would

be able to enter the compartment if it was bilged. Ifthe actual solid material in the

compartment occupied 3000 cubic metres, then only 2'000 cubic metres of space

would be available for water.

rn this case, the per-.ubilir), = !!399+:llffi3ly3!91

=?94=0 400.or40oo5UUU

For certain types of cargo, it is possible to calculate the permeability if the

stowage factor and relative density are known:-

Relative density is the ratio between the weight ofsolid material and the weight

ofan equal volume of fresh water.

Space occupied by I tonne fiesh water = | ml

Space occupied by I tonne solid cargo =Relative density

For each tonne of cargo, as stowed, the space available for water = Space

occupied by I tonne as stowed (Stowage factor) - Space occupied by I tonne of

solid cargo.= Stowage factor - (1+ Relative density)

sEsgig'liblg f"l.g!:! (per ronne)Permeabiliry =:r Total space

Sto*ug" f-to.-(l*n"l@Permeabrlrty =@

Example.-F ind the permeability of a cargo which has a relative density of 1'80

and stowaqe factor of l'20.

r".,''"utititr = ry\Sg = 0 537, or 53'7Yo

Bilging of an Empty Compartment Amidships.-Fig. 80 represents a vessel

in which a comparftnent amidships has been bilged. The vessel is shown as boxshaped, for the sake of simplicity, but the

effect will be the same for ship shapes.

Suppose that the shiP floated

originally at the waterline SI and that

an empty compartment ABCD is bilged.

The buoyancy of this compartment is

m'

FtG.80

ilec

BILGING OF COMPARTMENTS 125

now lost to the ship. Call this lost buoyancy u The ship must continue to displaceher own weight of water and so must displace the same volume as before. In order todo this, she will sink to the new waterline S,I, so that the total volume of the layersn and n, which have become immersed, must be equal to that ofthe lost buoyancy.This means that:-

m+n:v

If the total areas of the original and the new waterplanes are the same, thevolumes of rz and n are together equal to the area of intact waterplane multiplied bythe increase of draft (,S,S, or ffr). So, ifl is the total area of waterplane, d the areawithin the comoartment and Xthe increase of draft in metres:-

Tnan=(A-a)X

m+ n=vy=(A-a)X

-- A-a

Example.-Abox-shaped lighter is 50 metres long, 8 metres wide and floats ata mean draft of 2.00 metres. Find the sinkage if an empty compartment amidships,20 metres long, is bilged.

Lost buoYancY (v) = 20x 8x 2'0 ml

Original waterplane area (,4) = 50x8 m2

ginksss=---lL ==-2q-E! - =1.33 metresA-a 50x8-20x8 -

Bilging of rn Amidships Compartment, with Cargo,-If there is cargo in thebilged compartment, it will still continue to displace a certain amount of water andwe shall not lose the whole of the buoyancy in the compartment. We shall also gain alittle buoyancy by submerging more cargo between tlte old and new waterplanes.

In this case, ifp be the permeability:-

(a) The lost buoyancy in the compartrnent rvill now become up cubic metres.

(b) In Fig. 80, consider the space between the two waterplanes, ,SZ and ,S,{and within the compartment IBCD. As the ship sinks to her new waterplane, thecargo in this space will become submerged and will thus add to the buoyancyeained.

hr

But,

so,

DE

Df

t0

selbxhe

tsdhatFd.l is


The volume of this space - aXcubic metres.

So, space available for water = apX cubic metres.Buoyancy gained in the compartment = volume space available for water

=aX-apX

= X(a _ ap)And, total buoyancy gained, = m + n + X(a - ap)

Lost buoyancy = BuoYancY gained

vp=m+n+X(a-ap)

But (flom previous sectiort) m+n=(A- aV

So, vp=( l -a)X + x(a-ap)

vp=x(A-a)+(a-ap)

., vp

^ =7:tp

Example.-Abox-shaped vessel is 60 metres long, l0 metres wide and floatsat a draft of 4'00 metres fore and aft. A compartment amidships is 15 menes longand contains cargo of permeability 40o/o. Find the new drafts if this compartmentis bilged.

Volume of compartment (v) = l5x10x4 ml

Original waterplane area (l) = gQy lQ 62

Area in bilged compartment = l5x l0 m2Permeability 40% = 0.40

si*ase = rh; =aii6*ii&ta = o 44 metresNew Draft = 4 . 00-+ 0. 44 = 4. 44 metres

Bilging of an Empty Compartment, not Amidships.-In this case, there willbe both a bodily sinkage and a change oftrim. These may be calculated as follows:-

(a) The bodily sinkage is found in the same way as for an amidshipscompaxtment.(b) The method of finding change of trim is illustrated by Fig. 81. Here,it can be seen that, since there is a loss of buoyancy at one end of theship, B will move away from that end to a new position 8,. The forces ofbuoyancyand gravity then form a couple, forcing the bilged end of the ship downwards.

BILGING OF COMPARTMENTS

t_- _ _ --,f

{

' fco1--'i"

'T

r,-- - - - -d---7

127

FtG.81

As the ship changes her trim, the centre of buoyancy will move back towards Guntil it again comes vertically under G.

It can be seen from the diagam that the couple causing the ship to change her trimconsists of the forces of gravity and buoyancy acting on the lever 88, .

So, Moment changing trim = Il x BB,

Moment changing trimt hange ot tnm =

MrTjai-lry xBBl

(c) To find BB,:-

It has been shown, in Chapter 5, that BB, =!J39t and that.B obeys the samelaws as any other centre ofgravity.

So ifD is the distance ofthe centre of gravity ofthe bilged compartment from B,then:-

BBr=Y

Note.-In the case of an end compartment in a box shaped vessel, BB, willbe equal to half the length of the bilged compartment. This will only apply to thisparticular case.

(d) If the new drafts are required, we must calculate the change of draft dueto trim, as shown in Chapter 11. Remember, in this case, that the vessel will trimabout the new centre offlotation, which will be at the centre ofgravity ofthe neqintact waterplane.

Example.-A box-shaped vessel, 100 metres long and 20 metres beam,floats at drafts of 4'00 metres fore and aft. Find the new drafts after an empty endcompartment, l0 metres long, is bilged forward.

Bodily sinkage = ---!- - , . .10I20I4 -- = 0.44 mA-a l00x 20 - lOx 20

-t

TS

nglnt

iillr-tips

lle'

'theDcyds.

M.C.T.1C.


To calculate the M.C.T.lC. we must first find W and BMr.Ilt = LxBxDxl.025

:100x 20x4x1.025:8200 tonnes (D

t r ,=+=#=ff i=tsz^ ( i i )r,t.c.r:c=wii{, =ffif,#=tro r, (iii)

I48, =j length of compartment = 5.0 m (i")

F is at half length of intact water plane : 45 m ffom aft.

change of tri6 = Z I n4- = s2991 5 = r:t ..M.C.T. lC 124

Change of draft for trim: Forward =ffix::f = 182 cm

Original drafts

Bodily sinkage

Change for trim

New drafts

Aft = '41

x 331 = 149 cmIUU

F.4.00 m A.4.00 m

0.44 m 0.44 m

Ar 4q m 4.441t1+ 1.82 m -1.49 m

F.6.26m A.2.95 m

Notes:- (i) The displacement remains unchanged, so we can use, here, the originallength and draft.

(ii) 1 is for the intact part of the waterplane, so if we wish to convert I/l/ toL2ll2D we must here use the length of the intact waterplane and thenew draft.

(iii) Since we do not know GM. we must use BM, here,(iv)This only applies to an empty end compartrnent in a box-shaped vessel.

The Effect of a Watertight Flat, at or below the Waterline. -If the bilgedcompartment has a watertight flat at or below the waterline, the sinkage and trim(if any) will be found in the same way as before. From Fig. 82, however, we can seethat two modifications will have to be made in the calculations:-

(a) The volume oflost buoyancy (u/ will be that below the watertight flat.(b) The whole waterplane will remain intact, so that the buoyancy gained asthe vessel sinks will be lXcubic metres (instead of (A - a) X1.

BILGING OF COMPARTMENTS

FtG.82

Thus, for an empty compartment: X =;

Or, if the compartment contains carCo: X =)

(c) BB, even for a box shape, must be found by the formula:-

BB, =y (for an empty compartment)

tnv )

or BB, =:!la ( if the compartmenl contains cargo).

(d) Since the waterplane remains intact, F will not move appreciably whenthe compartment is bilged; so we can assume that the ship will still trim aboutthe same point as before. In the case ofa box shape, this means that the ship willcontinue to trim about amidships.

Example.-Abox-shaped vessel, 120 metres long and 20 metres beam, floatsat drafts of 4.00 metres fore and aft. An end comparftnent, 25 metres long, has awatertight flat 3.00 metres above the keel and contains cargo of permeab llily 60%.Find the new drafts ifthis compartment is bilged.

Sinkase = ?

= 25x-Z0l!32!0'60 = 0. 38 m

W =l20x20x4x1. 025 = 9840 tonnes

BML=h=t?|Yf,4oorn

BBt =y?z ! = 25x2gL3,r!U.5t 47 . s = 4. 45 m

M.c.r:c=wif#' = ti33i r3o =146 tonne-metres

chanse of trim = offi = %ffi:!tr - nt "

t29

T

I

I

vli?t4.a,.;i.

I3O MERCHANT SHIP STABILITY

Since the waterplane remains intacl F will remain amidships and the change ofdraft at either end will be halfthe change of fim.

Original drafts F. 4.00 m A. 4. 00 m

Bodilysinkage 1.* - 1.*.

4.38 m 4.38m

Halfchange of trim +0.89 m -9:E9 m

New drafts F.s:n ^

A. 119 -

Compartments with Cargo, not Amidships.-Unless such compartmentshave a watertight flat at or below the waterplane, accurate calculations becomevery complicated because of the submersion of further cargo as the ship trims. Suchcalculations are outside the scope ofthis book.

of

ES

neDh

CHAPTER 15

STABILITYAND THE LOAD LINE RULESMinimum stability requirements are laid down in the Load Line Rules and

ships must be provided with prescribed stability information for the use of Mastersand Deck Ofrcers.

Stability Requirements. -The ship's stability must be sufrcient for thefreeboard assigned to her and her light GM ascertained by means of an incliningexperiment. Also:-

(a) The initial GM is to be not less than 0.15 metres for ships loaded to.ordinary load lines. Ships loaded to timber load lines may, however, havean initial GMof not less than 0.05 metres.

ft) The maximum righting lever (GZ) is to occur at an angle ofheel ofnot lessthan 30o and must be at least 0'20 metres.

(c) The area under the cuwe ofrighting levers must not be less than:-0'055 metre-radians up to 30o ofheel.0'09 metre-radians up to 40o of heel: or up to the angle at which non-weathertight openings become submerged, if this is less than 40".0'03 metre-radians between the above.

(d) Certain ships may be assigned less freeboard than others, provided thatthey meet certain additional requirements.

Requirements for Special llpes of Ships.-For freeboard purposes, shipsare divided into two basic types: T)?e A, which are ships intended to carry onlyliquid cargoes in bulk: and Type B, which includes all other ships. If, however, aType B ship has steel hatch covers, improved protection for the crew, better freeingarrangements and special subdivision against flooding, she may be designated Type8-60, or 8-100.

Ships of Types A, 8-60 and 8-100 are allowed to have less freeboard than thebasic Type B ship. To qualifu for this they must be able to withstand the floodingof one or two compartrnents, according to their length and type: whilst after suchflooding, they must meet the following requirements:-

(a) The new waterline must be below any opening through which the shipcould become flooded.

@ Any heel due to unsymmetrical flooding (i.e. excess weight of water on


one side ofthe ship), should not be more than 15". This may be extended to 17' ifno part ofthe deck is then immersed.

G) The ship must have a positive GM, when she is upright, of at least 0'05metre.(d) The range of positive stability must be at least 20". (for example, if thevessel heels to, say, l2' after flooding, her angle ofvanishing stability must be notless than 32').ld The maximum GZ must be at least 0'l metre.

Information to be Supplied to Ships.-Full details of this may be found in theLoad Line Rules. The following is a summary of the requirements:-

(a) A plan of the ship to show the capacity and Kg of each space: weightand Kg of passengers and crew; weight, disposition and Kg of any anticipatedhomogeneous deck cargo.

ft) The light displacement and KG; also the weight, disposition and Kg ofpermanent ballast, if any.

@ Curves or scales to show displacement, deadweight, KM, T.P.C. andM.C.T.1C.(d) A statement of the free surface effect in each tank.

@ Cross curves, stating the assumed KG.(l) Statements and diagrams to show displacement, disposition and weightsof cargo, etc., drafts, trim information, KG, KM, G, free surface corrections andcurves of statical stability when the ship is:-

(i) Light.

(ii) In ballast condition,

(iii) Loaded with homogeneous cargo,

(iv) In service loaded conditions.

@ Written instructions conceming any special procedure necessary tomaintain adequate stability throughout the voyage.(h) Vessels more than 150 metres in length are to be provided with informationrelating to shear forces and bending moments and the maximum values allowed.

The Stability Inforrnation Booklet.-There is no statutory requirement as tohow the specified stability information is to be set out, and this has varied from shipto ship. It would be an advantage to ship's Masters and Officers if a standardisedmethod were used in all ships. To this end, the Department of Trade have producedtheir own recommended form of Stability Information Booklet.

In order to illustrate the Department's recommendations, the main informationfor an imaginary ship is set-out in the suggested form in this chapter and in the scales

STABILITYAND THE LOAD LINE RULES I33

in the back of this book. Some parts have here been abbreviated, or merely described,in order to save space: but it is hoped that this will be sufficient to give the reader aclear idea of the main contents ofthe booklet, which are as follows:-

(") General particulars of the ship (name, official number, dimensions,tonnage, etc.).(b) Plans ofthe ship, showing cargo, tank, store spaces, etc.

@ Special notes regarding the stability and loading ofthe ship: both in generaland as applied to that particular vessel.(d) Hydrostatic particulars for the ship in salt water. (See the example given inthe back ofthis book.)(e) Capacities and centres ofgravity ofcargo spaces, storerooms, crew spaces,etc. (See Plate in this chapter.)(fl Capacities, centres of gravity and free surface moments ofoil and watertanks. (Also in this chapter.)

@ Notes on the use offree surface moments. (Here described in Chapter 9.)(h) Special information required if the ship is designed to carry containers:including a container stowage plan and a statement indicating the position of thecentre of gravity of each container.(i) Cross curves of stability (Kl{ curves) and an example showing their use.(Described here in Chapter 13.)(j) A deadweight scale. (See the example given in the back ofthis book.)(k) Condition sheets, giving aplanand details ofweights on board, informationon stability on deparhue or arrival, and a curve of statical stability: all for at leasteach of the following conditions:-

(i) The light ship.

(ii) Ballast conditions on (a) deparhre and (b) arivil,(iii) The ship loaded to the summer load line with homogeneous cargo on:

(a) departure and (b) arival,

(iv) The ship loaded to the summer load line in at least one service loadedcondition on (a) departre and (b) arival.

For the above purposes, it is assumed that:-

For each "Arrival Condition", all fuel. fresh water and consumable stores havebeen reduced to l0% oftheir original amounts.

Inthe"departurecondition", fueltankswhichare "full"ofoil aretakenas 98% full.An abbreviated example of the ship in condition 3 (a,) is given in this chapter.


CAPAG|NES AND CENTRES OF GRAVITY OF CARGO SPACES,STOREROOMS AND CREW ANO EFFECTS

(A.8.- AtovE s !E) (t.P - AfrER PeRPstorouuR,

CAR GO

REFRICERATED

STOREROOI.IS

CREW, STORES ANO EFFECTS

coliPAmt{ErTL0cAn0(FRAItExunEEns)

clP ctT|Es c€t{rREs 0F cRlvrTtBAL€ BiAIN VTPTI6AL

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AfT D€EP TAT'X 9J - tO5 toaf l |06 1.76 77.2

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CENTRE5 OF CRAVITYv€pTrcAL

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BAtE0 P cfTY

CENTRES OF GRAVITYVERTICIL

A.A.LONGTTUOIruL

rROH A.PcuSrc hETn€sFORCCIsTLE STORT t f1- ta1 3+ 12,7? t35.2EORE PtI(

'fME174 - ts+ 5l to.15 154.9

ND$ P SfORE 6A-17 50 I.90 16, A

moP saoaE A-E 60 lO. O.l - t .s

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T0r{xEscEt{TREs 0F GRAVrt

TONNESOENTRES Of CRAV]'Y

VERTIG^IA.B

ton6|Tuo||r/Lfnon A. e

VERNCALA-4.

tot|6tluo||{^|.fno A.P.

sfoREa eq 9.96 7t.+ 2 a .46 7l.+

crtv a ftFfcts l+ I t .87 65. t l4 r l .9 o 53. I

STABILIryAND TIIE LOAD LINE RULES

CAPAGITIES, CENTRES OF GRAVITY AND FREE SURFACEHOMENTS OF OIL AND WATER TANKS

(4.8. - ABo!E BtsE) (A.e - Art€R PaRPE OTCUL^R)

OIL TU€L TAI{XS

135

ENGINE TANKS

ir[

FRE8H, FEED AND BALLAgT WATER TANKS

CARGO.OIL TANKS

CO4PARTrltNTtoclTt0x(rRinEllu|.|9:RS)

CAPA6ITIES CENTRES OFGRAVITY tnerres)

FFTESURFACEHO ENTlT 56.| 0

DOT FUtr- tsr FULLcuEro

! ITTRESCUBICEtR€S 'ONNE5AT s.G.t.O

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MERCIIANT SHIP STABILITY

L0i0n{; cout

DIIY CAR6O

FRrcfliArfD C RCo

cAR60 orL

orl FutL l=1:-:-:.

FR€SH VATER

WATER BALLAST

C0N0|110N No.3A H0n06€xEoU5 toAo|t{G coNor oN - DEPARTURECAR60 oF SToVAEE FAcTOR 1.507 '..rr-

ITEII OFoEA0rtr6[T WEIG}II v.c.6. VERTI

nonEriT l .c.G. LON6 !f10r.r6Nl SAII-ING STAIE

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F.D'A I( 7t9 4.81 5i |80 46. I 6t9Oa K. M. 8. 0 '?5

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*- llnx trtt FPEE slRFlo€ €FrEor

1I

fl

STABILITYAND T}IE LOAD LINE RULES 137

The Use of Maximum Deadweight Moments.-This is a simplified methodof making sure that a ship's GMis large enough to enable her to comply with, atleast, the minimum requirements set out in the Load Line Rules. It is intended foruse in small ships, for which it is particularly suitable.

The method is, basically, an adaptation of that described in Chapter 5 forfinding a ship's,(G by taking moments about the base line. In that chapter, it hasbeen shown that:-

(a) Deadweight Moment is the sum of the moments about the base line of allweights on board the ship (i.e. cargo, fuel stores, etc.; which are the componentsof deadweight).

(b) The total moment about the base line equals the moment of the light shipplus the deadweight moment.

(r) a6=Ftq1_!qerns+ulsplacemenl

_ Moment of light ship + Deadweight moment,v

Now, the moment of the light ship is equal to the light displacement multipliedby the light fG, which is a fixed quantity for a given ship. The displacement dependson the ship's draft and will be known. So we can use the above formula to calculatethat deadweight moment which will give the ship a desired KG.

But, GM:KM-KGKMdepends on the ship's draft and will also be known

Moment of light ship + Deadweight momentSo. Glrl = Kirl -

W

This means that we can calculate that deadweight moment which will give us adesired GMfor a ship at given draft.

In practice, the Naval Architects first find the least GM which, at a givendraft, will satisry the requirements of the Load Line Rules. They then calculate thedeadweight moment which will give this GM: this will be the maximum allowabledeadweight moment for the chosen draft (maximum because the greater the moment,the greater the KG and the less the Gi11).

Example.-A small vessel has a light displacement of 737 tonnes and a lightKG of 3'300 m. At 3 00 m draft, her displacement is 1259 tonnes and her KM is3.845 m. It is calculated that, at this draft, her GMmust be at least 0'238 m in orderto comply with the Load Line Rules. Find her maximum allowable deadweightmoment at 3'00 m draft.

-1I

--1

Ir lil. lI. l;l;t;l;til


Required minimum GM 0.238 n

KM 3.845 m

Max. allowable KG 3_:607 m

Maximum total moment =W xKG =1259x3 607 = 4541 tlm

Moment of light ship = 737 * 3 rOO =2432tlm

Maximum permissible deadweight m6ment = !p t/m

The above is repeated for a series of drafts between the light and loadwaterlines and, from this, a scale or graph is drawn up to show the maximumpermissible deadweight moment for each draft. The seaman is given a copy ofthis scale and/or graph: also a form on which is shown a profile of the ship andthe heights of the centres of gravity of the various compartments. He enters on thisform, the amount and of each item on board and multiplies them together to findits deadweight moment. The sum of these moments will be the actual deadweightmoment ofthe ship. The seaman also extracts from the scale or graph, the maximumpermissible deadweight moment for his ship's draft or displacement. As long as theactual deadweight moment is less than the maximum permissible moment, the ship

will have a suffrcient GM

An example of the above is shown in the following two diagrams. The firstof these shows the maximum permissible deadweight moments for an imaginarysmall ship. The second shows a completed form for the same ship, when loaded:

indicating that at a displacement of 1861 tonnes, the ship has an actual deadweightmoment of 3702 tonne-metres. The maximum permissible deadweight moment forthe ship's displacement is then extracted from the first diagram (in this case it is 4l4lVm): this is entered at the bottom ofthe second form. This then shows that, in this

case, the ship has sufficient GM, since the actual moment is less than the maximumpermissible moment.

"Simplffied Stability Information".-This may be provided as an addition to

the basic data and sample loading conditions required by the Rules. This informationmay be presented in one of three ways, provided that it is accompanied by clearguidance notes for its use:-

(a) A maximum deadweight diagram or table.

(b) A diagram or table showing maximum permissible KGs.

(c) A diagram or table showing minimum permissible GMs.

The method of setting out the diagrams or tables for KGs or GMs would be

basically similar to those shown here for deadweight moments.

STABILIryAND THE LOAD

SIMPL IED STABILITY

LINE RULES

INFORMAT1ON

139

s3=3

1800

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OEADIdEIGHT HOI'ENT$ IN TONNE -fiErRES

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SIMPLIFIED STABILITY INTORMATION

HETCHTS -'h'| - ?.67^.5 - O.Oa'n.9 * 0'55re.

:6 'OJr.

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- 5 'O?n.

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ltont T rusr 0TDEAOVEIGHT

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VIEIGHT(rn)

HEIGHT ABO\E KEEL TOCENTRE OF WEIGHT.'Ii

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(lnco *r f0RE li{tr 437 J. r? r365i 860 rt{ Itr t${I) 597 3.56 2t37CIR6O O |)EGK

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STABILITYAND TTIE LOAD LINE RULES 141

Another method by which simplified stability information might be given is bymeans of a maximum KG table. Part of such a table is illustrated below

Displacementt

Max. KGm

19500 7.85r9000 7.9318500 8.0218000 8.1017500 8.1717000 8.2016500 8.1916000 8.1815500 8.1815000 8.1914500 8.2014000 8.2213500 8.2413000 8.28

The table shows that if the displacement is 15600 tonnes, then the KG must notexceed 8'18 m.

As long as the r(G is less 8'18 m, the ship will comply witl all requirements oftlre loadline rules regarding GM, GZ values and areas under the GZ cvve.

CHAPTER 16

MISCELLANEOUS MATTERSDrydocking.-When a ship is drydocked, her support has to be transferred

from the water to the keel blocks and shores. She may be considered safe whilst

she is waterbome, or once the shores have been set up, but there is a danger that

she may become unstable during the intervening period, which is often termed the

"critical period '.

Whilst the dock is being pumped out, the ship at first sinks bodily as the water

level falls, but as soon as she touches the keel blocks she stops sinking, and the water

falls around her. She thus loses displacement so that weight, equal to the amount

of the lost displacement, is transferred to the blocks. As far as the ship's stability is

concemed, this weight is equivalent to a force acting vertically upwards at the keel

and it will decrease the metacentric height. The latter must, sooner or later, become

negative and if this were to happen before the shores were properly set up, the ship

might capsize in the dock. It is thus ofthe utmost importance to keep full control of

the ship during the critical period and to get the shores set up as soon as possible. To

assist in this, it is usual to have the ship trimmed a little by the stem when she enters

the dock, so that the heel ofthe stem post is the first part to touch the blocks.

FtG.83

Fig. 83 illustrates what happens in the above case. As soon as the ship's stem

touches the blocks the upward force, .P comes into existence. This force is small at

first, but gradually increases as the waterlevel falls and the ship's bow comes down.

The advantage ofthis is that the decrease in metacentric height, caused by the force

.f is more gradual than it would be if the ship suddenly sat flat on the blocks fore and

aft, so that we have more control over the ship's stability. Also, although the shores

cannot be set up before the ship comes down flat on the blocks, we can start to put-

t42

D

I

L

E

d

sF

MISCELLANEOUSMAITERS 143

in the after shores loosely as soon as the stem touches. By the time that the ship isright down on the blocks a large number of shores are already in place, so that therernainder can be put in and all set up with the minimum ofdelay. This decreases therisk ofthe ship falling over in the dock.

It is important to have the ship upright when she enters a drydock. Ifshe werenot. this could be due to one of two causes: a negative metacentric height, or theweights on board not being symmetrical about the centreJine. In the first case, theship would be certain to fall over as soon as her keel touched the blocks. In thesecond, she might fall over at some time during the critical period on account oftheercess of weight on one side.

Before the ship is floated again, it is very important to check any weights whichmay have been shifted whilst she is in the dock; otherwise we may have a similareffect to the above whilst the dock is being filled. In this respect, do not forget tonake sure that boilers have not been filled or emptied, or to check-up on any weightsshifted in the engine-room. All tanks must be sounded as soon as the ship is on theblocks and they must be sounded again just before flooding the dock to make surethey are in the same state regarding weights contained therein.

The procedure of dry docking is, briefly, as follows. As soon as the ship entersthe dock she usually comes under the control ofthe dock master or shipwright, whomanoeuwes her into the position he requires. The dock gates are then closed andpumping-out commences. When the ship's stem is nearly on the blocks, pumpingis stopped whilst the ship is aligned so that her centreline is exactly over them.Pumping is then resumed slowly until the stem touches the blocks, when the aftershores are put-in loosely. As the ship settles down, more shores are put-in, workingfrom aft forward, and as soon as the keel comes flat on the blocks any remainingshores are put in place and all are set-up as quickly as possible. The heads of shoresshould always be placed on frames and not between them, in order to eliminate therisk of denting the ship's plating. Once the shores have been set-up, pumping iscontinued quickly until the dock is dry.

The following formula will give the ship's metacentric height at any timeduring the process of drydocking:-

P = the force acting upwards through the keel

KM = The KMcorresponding to the actual displacement - P

W = The ship's displacement entering the dock (the real displacement)

Virtual rise of G (or virnral loss of K.[4 = ryM


The force P is the difference between the displacement ofthe ship on enteringthe dock and her displacement at the time for which we wish to calculate her G,llAfter the ship has come ffat on the blocks, this calculation is quite simple, since thetwo displacements will be those for the respective mean drafts: that is:-

P: displacement at original draft - displacement at new draft.

It is more difficult to find P during the critical period, after the ship's stem hastouched the blocks, but before she comes flat on them. The most dangerous partof this period, and hence the one with which we are most concemed, occurs at theinstant before the ship takes the blocks fore and aft. For this instant, p can be foundapproximately by the following formuta:-

Where / = the trim, in centimetres, on entering the dock.

/ = the distance between the after block and the centre of flotation.

o- M.C.T. lC.xrI

Example.-A ship entered a drydock with drafts of 3.00 metres forward and3'50 metres aft. At this draft, her displacement was 4650 tonnes; KG, 7.4 metres;KM 9'2 mefiesi M.C.T.1C. 148 tome-metres: whilst the centre of flotation was 72metres from aft. Find:-

(1) The GMat the instant before the ship came flat on the blocks.

(2) T"he GM when the water had fallen so that the draft read 2.65 metres. herdisplacement was 3650 tormes, and KMwas 10.0 m.

(l) Trim entering dock : 50 cm by stem

p = M.C.T.lC.x trim =.!4%S = tol ro*".I

Vimral r iseofG= Px.{M =l03l2 '2 = 0.20 mw 46so

KG 7.40 m

fuse 0.20 m

Virtual KG 760- m

KM 9.20 m

Yirfial GM LlQ m Answer

MISCELLANEOUSMATTERS U5

(2) P = Old displacement _ displacement corresponding to nertr draft= 4650 - 3650 = 1000 tonnes.1(Mcorresponding to new draft: 10.0 mVirtual rise of G = = 2.15 m

KG 7.40mRise 2.15 mVirhral t(G -9T5 mKM 10.00 mVirfial GM 0.45 m Answer

^_ "^9_11"odtog:-*:n a ship nrns ashore, her metacentric height will decreaseor become negative as in drvdocking, but the exact effect of this on her stability isalmost unpredictable. It wil-l vary according to the nature of the ground, how theship is placed on the bottom, what damag" ,f," hu, .*t"i*d anitrre nature and stateof the lides. In practice, we can only aftimp, t" c;;;ll;;;at again as soon aspossible, if it appears safe to do so.

The Effect of Density on Stability._When there is a change in the density ofthe water in which a ship floats, she will "h-g"

h;; d*;.-el inliase oraensity wiltgenerally have the following effects on her stabilitv:_(a) When the ship is near..her light draft, the centre of buoyancy will belowered and the metacentre will probibly rise. ffr", ,fr".. JU be an increase ofB M and probably of GM also.(!) When the ship is near her

_load draft, the cenrre of buoyancy wil fall and*:.T""1.1I" will probably do likewise. rh"* ;i,l;;"ti;rbe tittle change inBM but GM may decrease.(:)

.The trim may change. Whether the ship trims by the head or rhe sterndepends on the rerative positions ofB and F and'"" *i"rrt'"r,i" o"nsity increasesor decreases.

(d) It should be remembered that all the above effects are small, but in somecases they may be critical and cannot be ignored.The Elfect of Density on the Draft of Ships._In Chapter 4, we considered

consider it in relation to ship shapes. In Fig. g4, letthe ship be first floating at the waterline SI in waterof density ,. Then let her pass into water of a lesserdensity, d, and sink to the new water_line g Z..

FtG.84

146 MERCIIANT SHIP STABILITY

Let X be the increase of draft, in centimetres_ that is, the distance betweenthe two water-lines. Let Wbe the ship's displacement: since the ship must alwaysdisplace her own weight of water, this will remain the same for both drafts.

WeightvOtume =;_Density

So, volume below the water-line ST =4o

And volume below the water-line.i,l! = {dl

The difference between these two volumes must be equal to the volume ofthelayer.9S,{4 so:-

Volumeof the taVerSS,f,f ={-$

w6 -1r6,&t

w.^ ^,= 6dr

(o -dr ) ( l )

The weight ofa layer such as SS,I,Z must be equal to its depth in centimetresmultiplied by the tonnes per centimetre immersion.

Weight of the layer,y,sr?i? = XxT.p.C.

Volume of the layer SSr4Z = xxt.P.C.5Q)

If we now combine formulae (1) and (2) above, we can see thar:_

n*o=6,u_u,,

xxr.p.c.=ff(6-4).. w(6 -5,1"

= a,"r-p.fDerivation of the Fresh Water Allowance._The F.WA, is the amount, in

millimetres, by which a ship would change her draft on passini from salt to freshwater, or vice-versa, when floating at, or very near, her Summer Loadline.

- Let the displacement and T.pC, at suruner draft be / and Z respectively. A

change of density from salt to fresh water will give d as 1025 tglm, and a, as iOOOkg/mr. Substituting these in the above formula sives:_

MISCELLANEOUS MATTERS 147

change of draft (xl = 4$ffif9Q "rn

But F.W.A. is expressed in millimetres, so we must multiply the result by 10, toconvert it from centimetres.

r'w n' = d(tl?s;iooo), r0,"-r.w.e.=1| mm

.. R9r:*9 Buoyancy.-is the watertight volume of the hull between the waterline and the freeboard deck. It amounts, approximately to the difference between theactual displacement and that which the ship would have if she were submerged toher freeboard deck.We can calculate the reserve. bloyancy for any floating body by finding thedifference between the total watertight volume ofthe body an? the vorume of waterwhich it displaces.

. .Continuous Watertight Longitudinat Bulkheads,_These give greatlongitudinal strength to a ship and also reduce free surface effect when liquids arecarried in bulk. They have one serious disadvantage, however, in that if the ship isholed on one side and the bulkhead remains intaci tie compartrnent could becomeflooded on one side only. This would give the ship a list, which may be dangerous ifthe compartment is large.

^ . In,ordinary cargo ships, having large holds, there would be considerable riskof-the ship capsizing in the above circumstances. There is normany no free surfaceeffect to be reduced in the holds and the bulkheads have the additional disadvantage

that they interfere with the handling of cargo. consequently continuous longitudinalbulkheads are not fitted in ordinary cargo ships, since the disadvantages outweighthe gain in longitudinal strength.

_ , .In the case of oil tankers, carrying bulk liquid cargoes, some form of longitudinalsubdivision is necessary to minimise fiee surfaceeffect. Intertbrence with thestowage of cargo does not have to be considered and great longitudinal strengthis required. In such ships, the advantages of continuous Tongitudinal bulkheads areobvious and one or two are always fitted. The danger of th" u1s."t,,

""p.izing in theevent of her being bilged is overcome by restricting the length of hei tanksl eho,in the event ofa tank on one side becoming floodedl the con-esponding tank on theother side could be filled quickly to counterbalance this.

. . Non.Continuous Longitudinal Bulkheads._These are often fitted in ordinaryships, as they have a number of structual advantages. Since they are not continuous

throughout any hold, they do not affect the ship's itabilitv.


Bulkhead Subdivision and Sheer.-The subdivision of a ship intocompartrnents by means of transverse bulkheads is a great factor in determining hersafety if she is holed- It is not generally realised by seamen that sheer also plays animportant part in this if the ship is holed forward or abaft the centre ofthe flotation.

In 1912, a committee was set up to investigate the spacing ofbulkheads and thesuggestions which were made in their report are now compulsory for passenger ships.It was not possible to apply them to cargo ships also and the bulkheads in the latterare usually more widely spaced than would be allowed in passenger vessels. Thecommittee introduced the "Margin Line" and the ..Curve of Floodable Lengths',.

The Margin Line is an imaginary line, 76 millimetres below the bulkhead deck.It is assumed that a ship which was sunk to this line would still be navigable in fineweather.

The Curve of Floodable Lengths is a graph from which can be found thefloodable lengths for any part ofthe ship, i.e. that length ofthe ship which, ifflooded,would cause her to sink to her margin line. When this is calculated, allowance ismade for an assumed average permeability in each ofthe various comparhnents. Thelength allowed for any comparhnent is found by multiplying the floodable length bya factor that depends on the length ofthe ship and the proportion ofcargo, engine andpassenger spaces below the margin line.

Sheer is the upward rise ofthe ship's deck from amidships towards the bow andstern. Ifa compartment becomes bilged at one end ofthe ship so that she changesher trim, the sheer helps to prevent the margin line from becoming submerged at thatend. It thus increases the floodable lengths forward and aft.

Pressure on Bulkheads.-When a compartment is flooded, the water pressureon the end bulkheads is greatest at the bottom and decreases to nothing at the waterlevel. The greatest support is needed at a point somewhere between the bottom andthe surface level, such that the total pressure above it is equal to the total pressurebelow it. This point is known as the "Centre of Pressure" and its position is asfollows:-

(a) For rectangular bulkheads, at the two{hirds depth ofthe bulkhead belowthe waterlevel.(b) For triangular bulkheads (apex downwards), at the half-depth of thebulkhead below the water-level.(c) In the case of ship shapes, the majority of the bulkheads are nearlyrectangular and need shoring most strongly at a point about one-third of the heightof the bulkhead measured from the bottom. Some of the bulkheads are, however,ofan intermediate shape and in this case the centre ofpressure falls between thehalf-depth and the one-third depth.

MISCELLANEOUSMAITERS 149

Three formulae may be useful here:-Where ft is the depth below the surfaceof any point on a bulkhead in metres; 11, the depth below the surface of the centre ofgravity of the underwater part of the bulkhead in metres; l, the underwater area ofthe bulkhead in square metres; IZ the density ofthe water in tonneVm3:-

Pressure at any gi ven point = Wh (tonnes/m2 )

Average pressure on the bulkhead = I/11 (tonnes/m2)

Total pressure on the bulkhead = lZIll (tonnes)

Note that A, which is sometimes called the "Centroid", must not be confusedwith the centre of pressure. For instance, in a rectangular bulkhead, the centre ofpressure is at one-third from the bottom, whilst the centroid is at the halfheight fromthe bottom.

II

I

I

riB

3

ItII

CHAPTER 17

ROLLINGThe Forrnation of Waves.-Waves are produced by friction between the wind

and the sea surface. The wind blows, to a greater or less degree, in gusts and alsoappears to blow somewhat obliquely down on to the sea surface. The effect of thisls to cause slight depressions in that surface in some places, with correspondingelevations elsewhere so that "ripples" are formed. The wind will now act directl|on these ripples, and if it blows long enough and strongly enough it will tum theminto waves. It appears that, within certain limits, the siie of the-waves will dependlargely on the force of the wind and on the distance from the point at which thewaves originated.

The. Trochoidal Theory.-This theory is generally used to explain theconstruction ofwaves and also certain phenomena connected with them. It is rathertoo complicated for us to consider fully here, so we will merely extract two pointsfrom it.

l*I

FIG.85 B

(a) The shapes ofwaves are approximately the same as a..trochoid", which isthe curve traced out by a point inside a circle, rolling along a straight line. This isshown in Fig. 85. Suppose that a wheel, with centre C, bi rolled along the levelsurface AB, from left to right. Apoint, x, on the wheel would trace out the trochoid-r7x. The shape ofa wave approximates to a troichoid as shown except that thewave would be figure 85 upside down. The crests are sharper than the troughs(b) The water in a wave is not considered to have any appreciable horizontalmotion; that is, it does not travel along with the wave in a horizontat direction.Each particle of water moves in a circular orbit, forward at the crest, backward inthe trough, upwards in ffont of the wave and downward behind it. This producesa progressive "heaping-up" ofthe water, which causes the wave_outline to travelalong, although the water itselfis not doing so. For our purposes, we can considerwaves as comparatively shallow, vertical movements of water.

150

R

J

s

ROLLING 15I

The T?ue Period of Waves.-This is the interval between the passage of anytwo consecutrve wave crests at a stationary point. Ifa ship were stopped and had nomovement, the period of the waves would be the interval between the time she wason one wave-crest and the time she was on the next. In theory, it is often assumedthat a series ofwaves all have the same period; in practice, this very rarely occursand successive groups of waves often have slightly different periods. The periodusually increases with the size ofwaves, but rarely exceeds ten or twelve seconds.

The Apparent Period of Waves.-When a ship is moving through the watetthe period ofthe waves may appear to those on board to be greater or less than thetrue period. A ship which is steaming head-on into a sea will be moving to meet eachsuccessive wave, which will thus reach her more quickly and will appear to have ashorter period than it actually has. A ship which has the sea aft, on the other hand,is moving away from the waves, so that these will take longer to catch up with herand will appear to have a longer period. When the sea is exactly abeam of the ship,her motion will have no effect and the apparent period will be the same as the trueperiod. The apparent period will thus depend on the ship's speed through the waterand on her course relative to the direction ofthe waves.

The apparent period is important, because it is the one which is actually felt bythe ship and which thus affects her rolling.

The Period of a Ship.-This is the time taken by a ship to roll from one sideto the other and back again. When the period is exactly the same for every roll,the rolling is termed "isochronous". It is often assumed that isochronous rollingoccurs in every ship for any angle ofroll, but this is not correct. We may accept thefollowing general rules:-

(a) Different ships have different periods ofroll.

(b) The same ship will have a different period for different conditions ofloading.

(c) The same ship will have a longer period when she is tender than when sheis stiff.

"Winging out" weights will increase the period, all other things being

(e) Rolling is isochronous for small angles ofroll, up to about ten degrees, butthe period increases slightly for larger angles.

Synchronisrn.-This is said to occur when the ship's period of roll is the sameas the apparent period ofthe waves. When it occurs, the waves give the ship a ..push"

each time she rolls, in the direction in which she is rolling, causing her to roll moreand more heavily. In theory it would continue until she capsized, but this does nothappen in practice because of certain resistances, which we shall consider later.

(d)equal.

btdI

ilL

D

5

d

152 MERCI{ANT SHIP STABILITY

Unresisted Rolling.-Let us assume for a moment that there are no forces inexistence to damp a ship's roll and that she merely rolls under the influence of thewaves and of her own period. Let us also assume that the period ofthe waves is thesame throughout and that the ship's period is isochronous for all angles ofroll.

Ifthe ship's period is much less than that ofthe waves, she will always take up aposition at right angles to the lvave-slope and will not roll to either greater or smallerangles than this. In other words, she will behave in exactly the same way as a raftwould do.

If synchronism occurs, the ship will roll through increasingly greater anglesuntil she capsizes.

When the ship's period is much greater than that of the waves, she will rolleasily and never to large angles. The waves would set her rolling, but would soonbecome out of time with the roll and would thus cause her to steady-down again.

Normally the ship's rolling period is longer than the wave period. When headinginto a sea the apparent wave period is less than the wave period and much less thanthe ship's rolling period. Synchronism is not likely to happen.

When sailing with the sea on the quartet apparent waves period is more than thewave period and becomes closer to the ship's rolling period. It is in such conditionsthat synchronism occurs.

The obvious remedy is to alter couse so as to bring the waves onto the bow ifthis is possible.

In practice, we find that the ship's period increases with the angle of roll andalso that we rarely meet a long series ofwaves ofexactly the same period. So, even ifthe ship herself offered no resistance to rolling, synchronism would seldom exist forlong. A normal ship might roll heavily, but she would be unlikely to capsize.

Resistances to Rolling,-We all know that it is possible to set a boat rolling byleaning from side to side in time with her period. If we first start her rolling and thensit upright and without moving, the roll will gradually decrease and will finally dieout altogether.

A ship can be set rolling in still water, in a similar way, by shifting weights orbodies of men from side to side. Similarly also, if we stop moving them, the ship'sroll will die out. For this to happen, the boat or ship must be setting up resistance toits own roll; otherwise it would merely continue to roll to the same angle and with thesame period. These resistances are usually considered to be as follows:-

(a) A rolling ship creates waves and these require a considerable amount ofwork to produce them. This work is provided by the ship and would other wisehave formed part of the forces producing the rolling. Instead, it now passes awaywith the waves and is lost. This wave-formation is one of tJre chief resistances tobe considered.

ROLLING I53

@ Friction between the water and the hull ofthe ship sets up a slight resistance.The effect is very slight in a ship having a much rounded bottom. In a ship whichhas a full bilge, or particularly bilge keels, it is much greater but is never veryconsiderable.(c) There is a certain amount ofresistance between the hull and the air, but itseffect is negligible.(d) Properly fitted bilge keels have a g.eat damping effect on rolling. Theirexact effects are complicated and will be considered more fully in the next section.

It can be seen that since the above resistances are capable of eliminating theship's roll in still water, they will also resist the forces causing the ship to rolfin aseaway. They will not eliminate rolling, but they can damp it considerably.

The Elfects of Bilge Keels.-These may be said to have three main effects inresisting rolling, namely:-

(a) They offer a certain amount of direct resistance to the wate! but thiseffect is comparatively weak.(b) They cause the ship's period ofroll to increase slightly.(c) They set up eddy currents and pressute under water.

The latter effect is by no means simple and may be considered to produce anumber of subsidiary effects:-

l. The wave formation due to rolling, described in subdivision /a) of the lastsection, is considerably increased.

2. The water pressure on the hull is increased on that side ofeach bilse keeltowards which the ship is rolling. This pressure acts at right angles to the hull-and itsdirection is such that it forms a resistance to the roll ofthe ship.

3. Water is unable to run around the hull in an unintemrpted streamline.This will also reduce rolling, since any upsetting of streamline effects will causeeddy currents and resistance to motion.

Bilge keels have a greater effect when the ship is moving than when she isstationary, and the greater the speed the geater the effect. This is often considered tobe due to the fact that as the ship moves ahead she is passing out ofthe water whichhas been disturbed by her rolling. A certain amount of the bilge keel is thus workingin

'ndisturbed water and the effect of this part is increased accordingly. The fastei

the ship moves, the more ofthe keel is in such water and the greater thi anti-rollingeffect. There may also be some "planing" effect.

_ Cures for Heavy Rolling.-When, for any reason, a ship is found to be rollingheavily, the proper cure is to alter course and./or speed. This will alterthe apparent, perioiofthe waves and destroy synchronism, which is nearly always the cause oisuch rolling.

It is usually unwise, and sometimes dangerous, to attempt a cure by workingwater ballast in these circumstances.

CHAPTER 18

SUMMARY

Abbreviations

d................ Waterplane area in bilged comparment.

A................ Area of waterplane.

A/P ............ The after perpendicular.

0................ Breadth of the ship.

8................ Centre of buoyancy.

D ............... Distance (of shift of weights, etc.)

D............... Depth, or draft.

.F'................ Centre of flotation.

FWA .......... Fresh water allowance.

I ................ Centre of gravity of weight or wedge.

G............... Centre of gravity of ship.

GM............ Transverse metacentric height.

GM"........... Longitudinal metacentric height.

G2............. Righting lever.

lr . . . . . . . . . . . . . . . . Common interval (Simpson's rules, etc.).

hh,........................Hoi.ontaldisplacement of the centres of gravity of wedges.

i................. Moment of inedia of free surface.

1................. Transverse moment of inertia of a waterplane.

I L . . Longitudinal moment of inertia of a waterplane.K................ Denotes the keel.

K8............. Height of the centre ofbuoyancy above the keel.

t54

ABBREVIAIIONS

KG............. Height of tlte centre of gravity above the keel.

Kg.............. Height ofthe centre of gravity ofa weight above the base line.

l(M............ Height of the transverse metacentre above the keel.

KMr...........Heig}i of the longitudinal metacentre above the keel.

Kir'............. The righting level assuming G to be at the keel.

/ . . . . . . . . . . . . . . . . . Length.

2................ Length.

M............... Transverse metacentre.

M".............. Longitudinal metacentre.

M.C.T.lC.' Moment to change trim one centimetre.

p ................ Permeability of a comparanent.

/ . . . . . . . . . . . . . . . . . Tr im.

7.. . . . . . . . . . . . . . . T.P.C.

T.P.C.......... Tons per centimetre immersion.

v................. Volume of immersed or emerged wedge, or volume of buoyancy.

V................ Volume of displacemant of a ship.

w................ Weight shifted, added, etc.

W............... Displacement of a ship.

0 ................ Angle ofheel.

d ................ Density.

./................ Displacement at summer draft.

155

FORMULAE

Angle of Heel, or Loll.

Due to weight out of centreline-tan 0=#,

Dueto negative GM-tano= P'

Areas of Common Figures.

Circle

Rectangle

Square

Trapezoid

Triangle

Areas of Waterplanes, etc.

Five-eight rule Area

Simpson's first rule Area

Simpson's second rule Area

B-Shift of B.

BB, ='x-!!''v

${s,+tr-r)

! $* +r*zu* lw+2x+4y+ 2)

] h (t + 3u + 3v + 2w + 3x + 3y + z)

Area

Area

Area

AIea

Area

Area

Area

n2

ab

d

i@+b\bhTcDxsin 0

2

s(s-a)(s-D)(s-c)

156

FORMULAE

Bilging.

Sinkage due to empty compartment = 7!;

Sinkage due to comparEnent with carg "

= U" *

Sinkage, if W/T flat exists in comp ar;lrnefi = +

Wx BBILnange ol tnm =

Mci.lc

BB, = v\-d'v

BM.

t57

For all shapes

For box shapes only

Approximate, for ship shapes

Bt{,

For all shapes

For box shapes

Circumference of a Circle.

Density and Draft.For box shapes

For ship shapes

For ship shapes

Fresh water allowance

Draft.

New dmlt _ Old densityOld draft New densityrG025 - d)

2>

w(6 -6)6r xT.P.C.

A* millimetres4tt

M.C.T. lC.xZT.P.C.x/

d x amount out of designed trim

tM: +

"*: hBM= T

tr"= +tr"= h

=2r

Sinkage

Sinkage

Loading to keep constant aft D =

Mean draft - draft at F

I58 MERCHANT SHIP STABILITY

Drydocking.

NewGM=otdGM-ryt

Where P = Old displacement - New displacement.

p= M.C.llC.xr (approx.)

Free Surface of Liquids.fuse of G, due to free surface of any shap e GG, =

f

Rise of G, due to rectangular free surface cc, = !!1r"6) or lQ.l!j,

' lzyx6"-- t2W

G - Shift of G.GG. =vz-d-w

GM.By Inclining Experiment

GZ,At small angles

Permeability.

Permeability =

cc,=ffxff

GZ: GM x sin0

By the Wall Sided Formula GZ =sinQ(GM +;BMxtan g)

Moment to Change Trim by I Centimetre,

For all shapes

Layer Correction.

-'.C. x d x Amount out of trimDisplacement of layer = ffi

Moment of Inertia.Transversely, for rectangular waterplanes I = +

lz-

Longitudinally, for rectangular waterplanes I =+' t2

M.c.r.\c.=w#yL

Stowage factor - (l - Relative density)Stowase factor

FORMIJLAE

Pressure.At depth DOn area A, (where D is centroid of area)

Sinkage.

Due to bilging an empty compartrnent

Surface Areas.

Box shape

Cube

Cylinder '

Sphere

Tons per Centimetre Imrnersion.In water ofdensity d

In salt water

Tiim.

Arer = 2(al+bl+ab)

Aret : 6d

Area : 2tn (r + I)

Area : hd

r.PC.=%#T.p.C.- 1.0_2_5;l

159

dD tonneVml:dlD tonnes

X=-J-A-a

Due to bilging a comparhnent with cargo * = f *

Due to added weights BodilV sinkage = 1*Ja.

Staticd Strbility.

At small angles Moment of statical stability = W x GM x sn9At any angle Moment of statical slabiliq = ry x 67

Moment changing trimunange oI rnm =- Mcjlfu:-

Loading to produce desired trim M.C.T. = M.C.T.1C. = (l - a )

Volumes.

Box shape

Cube

Cylinder

Hollow round section

Volume : a6l

Volume : ri

Volume = zrll

Volume = r(R + r) (R - r)


Volumcc (cont )Hollow sphene

Sphere

Wedge, or prism

Wetted Surfrce.

4n(R3 -rt)J

4n13J

AI

Volume

Volume

Volume

Arer = L{1.7d +CrxB}

DEF'INITIONSAngle of Vanishing Stability.-The angle at which a ship's stability becomes 0:

numerically the same as t}e range of stability except in ships with and angle ofloll.

Centre ofBuoyancy.-The geometrical centre ofthe underwaterpart ofthe ship.Centre of trlotation.-The point about which the ship heels and trims. The

centre of gravity ofthe waterplane.Centre of Grrvity.-The centre of all the weight in a body. The point about

which the body would balance.

Deadweight -The weight of all cargo, stores, bunkers, etc., in a ship.Deadweight Moment.-The total moment about the base line of all the

components of deadweight (cargo, stores, fuel, etc.).Displacement -The actual weight of the ship and all aboard her at any time.Draft.-The depth of the bottom of the keel below the surface of the water.

Measured forward and aft.

Draft at F.-The ship's draft, measured at the position ofthe centre offlotation.Dlnamicel Stability.-The amount of work done in inclining a ship to any

given angle ofheel.Equilibrium.-The state of balance of a body.Fluid KG or GM.-The ship's l(G or GM after allowance has been made for

the effect of free surface of liquids.Force.-Any push or pull exerted on a body.Freeboard.-The distance from the deck line to the water.

Fresh Water Allowance.-The amount by which the ship would increase herdraft on passing from salt to fresh water at summer loadline displacement.

Ileight of the Metacentre.-The height of the metacentre above the keelInertia.-The resistance of a body to motion or to change of motion.Initial Stability.-The GMwhen the heel is zero.

Isochronous Rolling.-The name given to the rolling of a ship when tlteperiod of each roll is exactly the same.

KG.-The height of the centre of gravity above the keel.l6 l


KN.-The righting lever which the ship would have if G were placed at the keel.

Law of Archimedes.-A body immersed in a liquid appears to suffer a loss inweight equal to the weight of liquid which it displaces From this law we concludethat a floating body displaces its own weight of water.

Light Displacement.-The displacement of a ship when she is floating at herdesigned light draft. The weight of the hull, machinery spare parts and water inthe boilers.

Loaded Displacement.-The displacement of a ship when she is floating ather designed summer draft. The light displacement plus the deadweight.

Longitudinal Metacentric Height.-The height of the longitudinal metacentreabove the centre of gravity.

Mean Draft.-The mean of the ship's drafts fore and aft.Metacentre.-The point at which the vertical line through the centre of

buoyancy, at a small angle ofheel, cuts the ship's centreJine. It is only considered toexist for angles ofheel ofup to about l5o.

Metacentric Height.-The height of the transverse metacentre above thecentre of gravity.

Moment.-The attempt of a force to tum a body. It is usually measured by theproduct of the force and the length of lever.

Moment to Change Trim by One Centimetre.-The moment to change theship's trim by one centimetre.

Moment of Statical Stability.-The moment which will try to retum a ship tothe upright when she is inclined.

Period of a Ship.-The time taken by a ship to roll from one side to the otherand back again.

Period of Waves.-The interval between the passages of any two consecutivewave crests.

Prismatic Bodies. -The term "Prismatic" is used in stability to indicate abody which has a constant cross-section throughout its lenglh. For example in thecase ofa box-shaped vessel which is on an even keel fore and aft, but heeled, theimmersed and emerged wedges are prismatic.

The volume of a prismatic body is the area of its cross-section multiplied byits length.

Range of Stability.-The angular range over which a ship will have positivestatical stability. The angle to which the ship could heel before she would tend to

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DEFINTTIONS r63

capsize. It is measured from zero heel or tlte angle of loll to the angle of vanishingstability.

Reserve Buoyancy.-The volume of a ship's hull between the waterplane andthe freeboard deck.

Righting Lever.-The perpendicular distance between the centre of gravityand the vertical line through the centre ofbuoyancy. The lever on the ends of whichthe weigbt ofthe ship acts to retum her to the upright when she is heeled.

Solid KG or GM.-The ship's i(G or Gild neglecting the effect of free surfaceof liquids.

Stilf Ship. -A ship which has a large moment of statical stability. One havinga large metacentric height or righting lever.

Synchronism.-Said to occur when the ship's period of roll is the same as thatofthe waves.

Tender Ship. -A ship which has a small moment of statical stability. Onehaving a snall metacentric height, or righting lever.

Tonnages.-A number relating to the volume of a ship, and approximatelyequal to Volume = 2'83 m3.

Tonnes per Centlmetre Immersion. -The weight which must be added to aship in order to cause her to sink one centimetre bodily.

Thim.-The difference of the drafts forward and aft. The longitudinalequivalent ofheel.

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PROBLEMS

Increase of Pressure with Depth

A flat plate, 5 mehes long and 2 metres wide, is placed horizontally at a depthof 12 metres below the sea surface. Find the pressure and the total thrust onthe plate.

A sealed box is made ofa metal which is capable of withstanding a pressure of500 grammes per square centimetre. To what depth, in salt water, can the box besunk before it will collapse?

Find the total thrust, in tonnes, on a keel plate which has an area of 82 m2 andwhich is 9.0 m below the sea surface.

Find the average pressure, in kilogrammes per square centimetre, on the body ofa diver who is working in salt water at a depth of 17 metres.

A tank is being tested and is pressed-up with salt water until it overflows fromthe air pipe. If the top of the pipe is 8.6 metres above the tank+op, find thepressure on the tank-top in kg/cm2.

A double bottom tank is 1.6 metres deep and contains oil of relative density0'895. If a sounding gives a reading of 4.8 metres, what is the pressure on thetank top in kg/cm'??

Answers-

l. 123 tonnes; 123 tonnes.

2. 4.878 m.

3. 756.5 tonnes.

4. 1.742kg.

5. 0'882 kglcm3.

6. 0.286k!cm2.

Floating Bodies and Density

7 . A log weighs 4 tonnes and floats in water of density 1.024 tonnes/m3. Whatvolume of water does it disolace?

A floating body displaces 21.45 cm3 of water of density 1.010 tonnes/m3. Findits weight in grammes.

164

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PROBLEMS 165

9. A ship displaces 2941 m3 of water of density 1022 kgn . Find the density of thewater in which she would displace a volume of 2945 m3.

10. A log is l0 x I x 0.7 metres and weighs 4.08 tonnes. How much of it would beabove the surface in water of 1.020 grlcm3?

I l. A block of hardwood, 4.0 m x 0.8 m x 0.3 m, weighs 985 kilogrammes. Will itfloat or sink if it is placed in water ofrelative density 1.026?

Answers-

7. 3'906 m3.

8. 2l'66 gr.

9. 1.0206.

10. 3.0 m3.

11. Block has same densiw as the water-in theory it neither floats norsinks.

Loadlines

12. A ship is 120 metres long. At her summer draft of 7.626 metres, her freeboard is2186 mm, her displacement is 10720 tonnes and her T.P.C. is 16 58 t. Find thedistances between her load lines; also draw the lines.

13. Find the distances between the load lines ofa vessel which is 60 metres long.At her summer draft of 4'053 metres, her displacement is 1861 tonnes and herT.PC. is 4'67 t.

14. A ship, which is marked with lumber load lines is 135 metres long and hersummer draft is 8.042 m. Find the distances between: (a) the L9 and ZZ lines;(b) the LS and LW lines; (c) Where would the LI(NA line be placed?

Answers-

12. ,S to ? and S to Il': 159 mm.,S to F and T to TF: 162 mm.

13. Sto Zand,Sto IV':84 mm.S to I' and Z to 7F: 1 00 mm.I( to IINA: 50 mm.

14. I.9 to Z?.1 168 mm.LS to LW: 223 mm.No LWA if Z is more than 100 m.

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166 MERCI{ANT SHIP STABILITY

Areas

15. Find the area ofa square whose sides are ll cm long.

16. A square has an area of 43 mr. Find the length of its sides.

17. What is the area of a rectangle which has sides of 2'6 cm and I1.5 cm?

18. A box-shaped ship is 172 m long and 27 m wide. Find the area of its water planewhen it is upright and on an even keel.

I 9. What would be the area of the waterplane of the ship in the last question, if shewere on an even keel but had a list of20"?

20. A hiangle has a vertical height of 6.2 cm and a base of 8.8 cm. Find its area.

21. Find the area of a triangle having two of its sides 3.8 cm and 9.5 cm long, theangle between them being 36'.

22. A triangular plate has sides of 10.7 m, 16.5 m, and24.0m. What is its area?

23. Find the area and circumference ofa circle ofradius 2.62 metrEs.

24. A circle has a circumference of 32.8 cm. Find its area.

25. Acircle has an arca of 57.6 crfr. Find its circumference.

Answers-

15. l2l crd.

16. 6.36 m.

17 . 29.9 cm2.

18. 4644 m2.

19. 4942.1 m2

20. 27.28 cm?.

21. 10.613 cm,

22. 74.5 cmz.

23. 21.57 m'?; 16.47 m.

24. 85.58 cmr.

25. 26.91 cm.

PROBLEMS 167

Surface Areas and Volumes

26. Find the surface area and volume of a cube which has edges of 3 cm length.

27. A crbe is made ofa metal which has a density of 7.290 gr/cm3.Its edges are2'4 cm long. Find its weight.

28. Find tlte underwater volume and displacanent, in seawater, of a box-shapedvessel of 120 m long, 17 m beam and 3.00 m draft.

29. Abox-shaped lighter is 25.00 m long, 6.00 m wide and has a light draft of0.60 m.How many tonnes of cargo must she load, when floating in salt water, in orderto increase her draft to 2.00 m?

30. Find the weight of a log of wood, l0 metres long, and of cross-section 90centimetres by 40 centimetres, which has a relative density of0.750.

31. A box-shaped lighter is 35 m long, 8 m beam and weighs 420 tonnes. Find herdraft in salt water.

32. A box-shaped lighter is 25'0 m long, 6.5 m wide and floats at a draft of 80 cm insalt water. What is her weight?

33. Find the surface area of a rectangular tank of 9.0 x 3.0 x 1.5 metres.

34. Find the volurne ofa prismatic-shaped lighter, if the area of each end is 35 rr?and its length is 50 m.

35. Find the volume and surface area ofa sphere ofradius 2.1 cm.

36. A sphere of radius 2l centimetres weighs 200 kilogrammes in air. What will beits apparent weight when immersed in sea water?

37. A hollow sphere has an intemal radius of 3 cm and an extemal radius of 6 cm.Find the volume of material in it.

38. Find the weight ofa hollow sphere ofintemal radius 6 cm, external radius l0 cmand relative density 2.700.

39. Find the volume and surface area of a cylinder 80 cm long and having a radiusof 14 cm.

40. A cylindrical boiler has a diameter of 3.0 m, a length of 4.5 m and weighs 80tonnes. Would it be possible, by sealing the openings in it, to float it ashore in anout-of-the-\ray port?

41. Find the weight, in tonnes, ofa hollow mast which is l0 m long and 50 cmoutside diameter, made of steel 20 mm thick. Assume that the steel weiehs 8.00tonnes/mr.


Answers-

26. 54 cm2;27 cm3.

27. 100.8 gr.

28. 6120 mr; 6273 tonnes.

29. 215.25 tonnes.

30. 2700kg.

31. 146.3 cm.

32. 133.25 tonnes.

33. 90 m,.

34. 1750 m3.

35. 38 81 cmr; 55.u14 cm:.

36. 160.2kg.

37. 792 cm3.

38. 8.869 kg.

39. 49267 cm3;8271cm2.

40. Yes. Weight ofboiler is 30 tonnes.Weight ofequal volume ofsalt water is 32.6 tonnes.

41. 2.413 tonnes.

Simpson's Rules, Etc,

Use Simpson's First and Second Rules to find the area of part of a water planewhich has the following ordinates, spaced 6.0 metres apart:-2'4, 3'7, 5 3, 6.6, 7.6, 8.1, 8.2 metres.

Use Simpson's First Rule to find the area of part ofa waterplane, which has acommon interval of 3.0 metres and ordinates of:l '3, 3 '5, 4 '8, 5.4, 5.6, 5.0,3.9 metres.

Find the area of a waterplane which has the following half-ordinates and acommon interval of 8 metres:-0,3 0,4.7, 6.2,7.3,7-8,7.3,6.9, 5.3,3.1, 0 metres.

Find the area of a waterplane which has the following ordinates and a commoninterval of6 metres:-0, 5'4, 8'2,9 0, 8.6, 7.6,5.4, 2 1, 0 metres.

42.

43.

44.

45.

PROBLEMS 16946. Part of a waterplane has the following half-ordinates and a common interval of+ u menes:-

l l ,2 2,3.7, 5.3,6.9,8.2,8.7, 9.0, 8.9, 8.8 metres. Find i ts area.47. A prismatic-shaped tank is 15 metres long. Its end is divided into ordinates of5'0' 4 5' 3'8 and2'0 metres, spaced 2.0 metres apart. Find the area ofthe end andthe volume of the tank.

48. Find the a^rea of a waterplane which has the following ordinates and a commontnrervai ot J metres._0.2,3.6,6.7,8.9, 10.7, t2.0, 12.7, 12.0, 11.4,8.8, 5.0, 0.3 metres.

zl9. Find the area ofa bulkhead which has the following ordinates, spaced 2.0 metresapan:-3.0,4.7,7.4,9.5, I l . l , 12.5, 13.6,14.4 mehes.

50. Three ordinates are spaced 12 metres apart and have lengths of 7.4, 11.7 and17.5 metres. Find the areas between the first and seconi; also between thesecond and third ordinates respectively. How does the total area so foundcompare with that found by Simpson,s First Rule?51. Find the areas between two ordinates l0 metres apad, which have lengths of10.3 and 15.0 metres ifthe next ordinate has a length oi 19.6 metres.52. Find the coefficient of fineness ofa waterplane which has the tbllowing half_ordinates, spaced l0 metres aoan:_

O-5.4 0,7 2,9. t . 10.0. 9.S, 8. t , 5. t , 0 metres.The grealest breadth of the waterplane is 20 metres.

53' The midship section of a boat is 3'0 metres deep and is divided horizontally intothe following equally spaced ordinates:_7.5, 7 2, 6.8, 6.0, 5.3, 3.3,0.5 mehes.Find the coefficient offineness ofthe midship section.

54. A ship is divided up into a number ofwaterplanes, spaced 0.6 metres apart andhaving the following areas:-KeelWaterplane

120A- 1092B- 1242L _ IJ)J

D- 1439E - 1499r'- 1548G 1593

square metressquare metessquare metressquare metressquare metressquare metressquare metressquare metres

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I7O MERC}IANT SHIP STABILITY

D is the light waterplane and G the load waterplane. Find:-

(a) the light displacement;

@) the loaded displacement;

(c/ the block coefrcient offineness, assuming the length and breadth ofthe

ship on waterplane G to be 130 metres and 17 metres, respectively.

55. The midship section ofa ship has the following ordinates, spaced 3 metres apart,

below her waterplane:-

16 4, 16'0,15'5, l5'0, l4'8, 14'0, l2'8, l0'9, 7'9, l'0 metres.

Find the area of the midship section. Also, if the length of the ship is 130

metres and her displacement is 12,095 torures, find her prismatic coefrcient of

fineness.

56. A waterplane has ordinates, spaced 9 metres apart of:-

0'6, 8'4, 14'5, 17.2,17'5, 17.0, 15'9 and l2'5 metres.

There is an intermediate ordinate, midway between the first two, of 5'1 metres.

Find the area ofthe waterplane.

57. Find the area of a waterplane which has ordinates, 7.0 metres apart of:-

12.0,11.7,10.9, 9.5, 6'8, 1.8 metres.

There is an intermediate ordinate of 5'4 metres between the last two.

58. A waterplane has ordinates, from forward to aft, of:-

0'2, 8'0, 12'3, l2'8, 13'1, l0'2,7'5,4'0, 0'6 metres, spaced l0 metres apart.

There are two intermediate ordinates: one, midway between the two forward

ones, is 4'2 m long; the other, midway between the two after ones, is 2'0 m long.

Find the area ofthe waterplane.

59. A vessel has waterplanes, 1 metre apart, from the keel upwards, of areas:-

152, 402, 625, 807 ,950 and 1032 square metres.

There is also an intermediate ordinate,50 centimetres above the keel, of26l

square metres. Find the vessel's displacement.

PROBLEMS

Answers-

42. First nrle, 220.0 nf. Second rule,220.3 m2.

43. 81.6 m,.

44. 838.4 n .

45. 281.6 m,.

46. 463.8 mz.

47. 23.9 m2:358.9 m3.

48.464.6t l l2.

49. 135' l d.

50. 113.1 m,; 173.7 m2; the same;286.8 d.

51. 126.6 m2.

52. 0.68.

53. 0.73.

54. Light displacement, 2835.4 tonnes. Load displacement, 5642.9tonnes. Block coefficient, 0.5930.

55. 116.3 m3:0.780.

56. 886.8 m'?.

s7. 327.8fr.58. 694.7 m2.

59. 3467 tonnes.

Forces, Moments, Etc.A weight of 3 tonnes is placed on a beam, so that its centre of gravity is25 metres from the end. What is the moment about this end?Abar is pivoted in the middle and a man pushes on it, in a clockwise direction,with a force of 25 kilogrammes at a distance of 300 centimetres from the middle.Another man, on the other side, also pushes in a clockwise direction with a forceof 20 kilogrammes. If the second man is 250 centimetres from the middle of thebar, find the moment about its centre.

If, in the above question, the second man tumed and pushed with the same forcein an anticlockwise direction, what would be the moment?

17l

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! 60.

61.

62.

172 MERCIIANT SHIP STABILIry

63. A man presses down on the longer end ofa bar with a force of 50 kilogrammes.The bar is 7 metres long and is supported at a distance of2 metres from its end.How much weight can the man lift on the shorter end?

64. Four men, working at a capstan, each push on the bars at a distance of3.0 metresfrorn the centre. If they push with forces of 20, 25, 30 and 35 kilogrammes,respectively, find the moment to turn the capstan.

65. Ifa rope is wound around the capstan in the last question, and the radius fromthe centre ofthe capstan to the centre ofthe rope is 40 centimetres, find the pullon the rope.

66. A wire, wound around a capstan, supports a weight of0.50 tonnes. The diameterofthe capstan barrel is 60 cm. How much force must be exerted by each of fourmen, each at a distance of220 cm from the centre, in order to lift the weight?

67. Two weights are placed on a beam: one of2 tonnes at a distance of 10 metresfrom one end; the other of3 tonnes at a distance of7 metres from the same end.What is the moment about the end of the beam?

68. A weight of 150 kilogrammes is placed on a see-saw at a distance of 400 cmfrom the middle. What weight must be placed on the other side, at a distance of300 cm from the middle, in order to balance the seesaw?

Answers-

60. 75 tonne/metres.

61. 12500 kg/cm.

62.2500k/cm.

63. 125 kg.

64. 330 kglmetres.

6s. 82s kg.

34.r kg.

4l tonne/metres.

200 kg

66.

67.

68.

tnes.€nd.

PROBLEMS 173

Centre of Gravity (General)

A plank weighs 62 kilogrammes. What will be the shift of its centre of gravitybe if a weight of 3 8 kilogrammes is placed on it at a distance of 400 cm from itsoriginal centre of gravity?

A weight of 0.40 tonnes is added to a body which weighs 2.00 tonnes, at adistance of 120 cm from its centre of gravity. Find the shift ofG.

A body, weighing 32 tonnes, has a weight of 8 tonnes removed from it. Whatis the shift of G, if the centre of gravity of the weight removed was 5.00 metresfrom that ofthe body?

A beam carries a weight of 4.8 tonnes at a distance of 6.0 metres from one endand the centre of gravity of the whole mass is 20 metres from that end. If theweight ofthe beam alone is 8.4 tonnes, find the distance ofthe centre of gfavityofthe beam from the end mentioned.

A loaded boat weighs 6 tonnes. Ifa weight of300 kilogrammes is shifted fromone end to the other, through a distance of20 metres, find the shift of G.

A boat weighs 1.512 tonnes and when a man, weighing 70 kg, sits in the bottom,the centre ofgravity is 80 cm above the keel. What will be the new height ofthecentre of gravity if the man stands up so as to raise his weight by 160 cm?

A see-saw has a number of weights placed on one end at a distance of 150 cmfrom the fulcrum. The centre of gravity of the whole mass is 30 cm from thefirlcrum and the total weight is 270 kilogrammes. A number of the weights arethen moved for a distance of300 cm, along the see-saw to the other end, so thatit balances. Find the amount ofweight shifted.

A table{op has a number of weights on it and the whole mass weighs 275kilogrammes. How far would it be necessary to shift a weight of 25 kilogrammesin order to cause the centre of gravity ofthe mass to shift for one centimetre?

Find the position of the centre of gravity of the waterplane in Question zl4.

Find the position of the centre of gravity of the waterplane in Question 45.

A waterplane has the following ordinates spaced l0 metres apart, fiom forwardto aft:-0'6, 8'4, 14'9, 18'6,20'0, 19.3, 16.1, 10.7,2.5 metres.Find the position of its centre of gravity relative to the mid-ordinate.

A waterplane ha< ordinates, spaced 20 metres apart, from forward to aft, of:-0 '2,2 '8,6 '9,9.6,10.3, 10.7, 9.8, 7.7, 3.2 metres.There is also an appendage of20 square metres, with its centre of gravity 2.0metres abaft the aftermost ordinate. Find the position of its centre of gra.vity.

pull

69.

70.

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76.

77.78.

79.


8l . A waterplane is 60 metres long and has ordinates of:-0'4,6'6, l0'1, l4'3, I l'8, 8.2 and 3.5 metres, from forward to aft.There is also an appendage of 12.4 square metres, with its centre of gravity32'2 metres abaft the mid-ordinate. Find the position ofthe centre of gravity.

82. Find the height of ,B in a ship which has waterplanes 1.0 metre apart, from thekeel upwards, of:-ll0, 1071, 1203,1331, 1412, 1486,1525 square mehes.

83. A small vessel, 48 metres long, has the following underwater cross-sections,from forward to aft:-0'9, ll'7 , 17'3,24.2,29.5,26.0,20.4, 12.8, 3.4 square metres.Find the longitudinal position ofB.

84. A ship has the following underwater cross-sections, 12 metres apart, fromforward to aft:-4,31, 59,78, 80, 82, 81,79, 63,45,20 square metres.There is also an appendage of 384 cubic metres, with its centre of gravity70 metres abaft the mid-ordinate. Find the position ofB.

85. A yacht has waterplanes, 20 cm apart, fiom the base line (a horizontal linethrough the top ofthe keel amidships) to the waterplane as follows:-6 '3,9 '7,11 9,13.6, 14.5 square metres.In addition, the ballast keel forms an appendage of 0.75 cubic metres, with itscentre of gravity 80 centimetres below the base line. Find the position of Brelative to the base line.

86. A waterplane has ordinates, 8 metres apart, from forward to aft, of:-0'2, 4'8, 8 6, 12'6, 13'4, 12'2,9.7 , 4.0 mefiesThere is also an intermediate ordinate of 2.1 m midway between the two foremostordinates. Find the area of the waterplane and the position of its centre ofgravity.

87. A waterplane, 120 metres long, is divided into eight equal parts, giving half-ordinates, from forward to aft of:-0 '0,5 4,9 '0, 10'7, l l '0, 10.8,9.5,6.8, 1.2 metres.There are also two intermediate half-ordinates: one of 3.8 metres, between theforemost two; one of 5'0 metres between the aftermost two. Find the area of thewaterplane and the position of its centre of gravity.

88. A ship, floating at her light draft, has waterplanes I metr€ apart, from the keelupwards, of:-270,2748,3020, and 3231 squar€ metres.There is also an intermediate waterplane, 0.5 metre above the keel, of 2228square metres. Find the ship's displacement and KB at her light draft.

PROBLEMS

Answers-

69. 152 cm.

70. 20 cm.

71. 166.7 cm.

72. 28 m.

73. I m.

74. 83.7 cm.

7s. 27 kg.

76. l l cm.

77. 0'55 m from mid-ordinate.

78. 2.46 m fiom mid-ordinare.

79. 1.28 m abaft mid-ordinate.


81. 2.21m abaft mid-ordinate.

82. 3.36 m above keel.

83. 0.59 rn abaft mid-ordinate.


85. 35 cm above base line.

86. 511.7 nP1'31 .3 m from forward.

87. 987'5 m,; 1.62 m abaft mid-ordinate.

88. 8208 tonnes; 1.67 m above keel.

Moment of InertiaA rectangular surface is 12 metres long and 5 metres wide. Find its moment ofinerfia about (a) the transverse centreJine; (6) the longitudinal centre_line.A box-shaped lighter is 120 metes long and 20 metres beam. Find the momentsof inertia of its waterplane about its longitudinal and transverse centre_lines.Find the moments of inertia of a rectangle, 6 metres long and 2 metres wide,about its centre-lines.

175

89.

90.

91.

176

Answers-

89. (a) 1251,

90. 80,000;

91. 4;36.


(b) 720.

2,880,000.

The Effect of Density on Draft

92. Abox-shaped barge floats at a draft of2.55 m in water ofdensiq/ 1.004 tonnes/m3.What would be her draft in water of density 1.020 tonnes/m3?

93. A ship has a fresh water allowance of 175 mm. To what depth could shesubmerge her load-line when loading in dock water ofdensity l0l1 kg/m3?

94. A ship has a fresh water allowance of 185 mm and a summer draft of 8.52 m.Find the draft to which she may load in river water of density 1.007 tonnes/mr.

95. A box-shaped vessel has a draft of 4.90 m in salt water. On entering a dock, herdraft becomes 5.00 m. Find the density ofthe dock water.

96. A ship has a fresh water allowance of 175 mm. By how much will she changeher draft if she passes from water of density 1004 kg/m3 to water of density1021 kg/mr?

97. A box-shaped barge has a draft of I . 12 m in salt water, when she is fully loaded.To what draft could she load in water of 1.015 tonnes/m3?

98. A box-shaped lighter draws 1.95 m in water of density 1004 kgim3. Find herdraft in water ofdensity 1020 kg/mr.

99. To what depth can a ship submerge her loadline in dock water ofdensity l0l2kg/m3, if her fresh water allowance is 150 mm?

100.A ship enters port with a salt water draft of7.18 m. Ifher fresh water allowanceis 122 mm, what will be her draft in a dock where the relative density of thewater is 1'009 tonnes/m3?

101.4 box-shaped vessel has a draft of 3.31 m in salt water Find her draft in waterof density 1.000 tonnes/mr.

102.A box-shaped lighter draws 4.10 m in water of relative density 1.010 tonnes/m3.What will be her draft in sea water?

103.By how much can a ship submerge her loadline in water ofdensity 101I kg/m3,if her fresh water allowance is 202 mm?

PROBLEMS

Ansu)ers-

92. 2'51 m.

93. 98 mm.

94. 8.65 m.

95. l'005 tonnes/m3.

96. ll9 mm.

97. 1 '13 m.

98. l'92m.

99. 78 mm.

100. 7'26 m.

101. 3 '39 m.

102. 4'M m.

103. 113 mm.

T.P.C.

l04.Find the tonnes per centimetre immersion of a box-shaped vessel, 70 metreslong and l0 metres beam.

105.A ship is 150 metres long, l6 metres beam and floats at a draft of5'00 m. If thecoefficient of fneness of the waterplane at that draft is 0'763, find her T.P.C.

106.A box-shaped lighter is 25 metres long, 6 metres wide and floats at a draft of1 .10 metres fore and aft. What is her T.P.C. and what will be her new draft after30 tonnes ofpig-iron have been spread evenly over the bottom?

1 07. At a given draft, a ship of I 20 metres length and I 5 metres beam has a coefficientoffineness ofthe waterplane of0'770. Find her T.P.C. at this draft.

108.A vessel's waterplane has a common interval of 6'0 metres and ordinates of:-0, 4'4,7.4,9.7, 10.3, 10'0, 9'8, 5'5, 0'6 metres. Find her T.P.C.

109.A ship's waterplane has ordinates of:-0.5,3'8,5'9, 6'5,6'8,6'6,5'7,4'4, and 2'5 metres, spaced 12 metres apart.Find her T.P.C. in fresh water.

110.A waterplane has an area of 1960 square metres. Find the T.P.C. in waterof density 1.012 t/m3.

177

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I ll.Avessel has a T.P.C, of 18.50 t in salt water. What would the T.P.C. be in waterof density 1.010 t/m3?

112.What would be the effect of loading a weight of 140 tonnes, in a ship which hasa T.P.C. of l0'0 t?

113.A ship floats at a mean draft of 6.25 metres. If her T.P.C. is 12.0 t, what wouldbe her new mean draft after 300 tonnes of cargo have been loaded?

114.Find the bodily rise of a ship which has a T.P.C. of 7.0 t, if a weight of120 to les is discharged.

115.A ship is loading in dock water of density 1.012 t/m3 and has a mean draft of7 16 metres. Her T.P.C, in salt water is 17.82 t. Find the new mean draft after shehas loaded a further 264 tonnEs.

l16.,{ ship has a mean draft of4.38 metres in water of relative density 1.015 t/m3.Her TP.C. in salt water is 15.20 t. What will be her new draft after she hasdischarged 360 tonnes of water ballast?

Answers-

104.7.2t .

105. 18.8 t .

106. 1 '5 t ; l '3 m.

107. 14.2 t.

108.3.571.

109. 10.08 t.

110. 19.841.

111. 18.23 t .

112. Ship sinks 14 cm bodily.

113. 6.50 m.

I14. l7 cm.

115. 7.31 m.

116. 4'14 m.

PROBLEMS

Loading to a Given Loadline117.A vessel has a summer draft of 6.730 mehes, a F.W.A. of 152 mm and T.p.C. of

21.86 t. She is loading in a dock, where the density of the water is 1.013 t/m3and her present mean draft is 6.64 metres. How much more cargo can she loadin order to be at her summer load line on reaching salt water?

I l8.A^ship is loading in an upriver port, where the relative density ofthe water is1.008 and her present mean draft is 6.9g metres. Her summer draft is 7.165metres, F.W.A. is l 16 mm, and her T.p.C. is 17.91 t. Calculate how much morecargo she can load in order to float at her summer load line on reaching saltwater, if she expects to use 32 tonnes of fuel and stores on the way downriver

119.A ship has a summer freeboard of 1764 mm, T.pC. of 19.54 t, and F.W.A. of135_mm. She is loading in dock water of density 1.015 Vm3 and her presentfreeboards are 1792 mrn on the port side and tAiO mm on the starboard side.How much more cargo can she road in order to float at her summer load line onentering salt water?

120.4 vessel arrives offa port, in salt water, with an even_keel draft of 7.91 metres.Her F.W.A. is 162 mm and her T.p.C. is 23.04 t. She is to discharge cargo intolighters in order to enter a dock, where the densify ofthe water ls 1.0t6 Vmr.The depth on the dock silt is 8'10 metres, and she is to cross it with a clearanceof30 cm. Assuming that the ship remains on an even keel, find the least amountof cargo to discharge into tlle lighters.

121.A ship is loading in dock water ofrelative density 1.020. The upper edge ofhersummer load line is lever with the water on the port side: whilsi the loi,er edgeof her summer load line is 6.5 cm above water on the starboard side. Her F.W.l.is 154 mm and her T.p.C. is 22.9g t. How much more can she load in order tofloat at her summer load line in salt water?

122.,4' ship has a summer draft of 8.094 m, her T.p.C. is 23.25 t and her displacementat summer draft is 15250 t. She is loading in dock water of relative density 1.010and her present mean draft is 8'22 metres. How much more can she load in orderto float at her tropical load line on reaching salt water?

123.4 vessel is loading at a berth in river water of density 1.009 t/m3. Her summerfreeboard is 1948 mm, T.pC, is 23.0g t, and F.W.A, is 164 mm. Her presentfreeboards are 2077 mm on the port side and 2053 mm on the starboard side.O-n

-completion of loading, she is to proceed to the river mouth, using 1 2 tonnes

of fuel and stores on the way, and is then to load,24} tonnes of iargo fromlighters

_before proceeding to sea. How much more catgo can she load in theupriver berth, in order to be at her summer load line in sa-it water on leavins thenver mouth?

179

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I8O MERCHANT SHIP STABILITY

124.4 wall-sided ship has a waterplane area of 2175 square metres and displaces12800 tonnes when floating at her summer draft of 7.200 metres. She is floatingin water ofrelative density 1.015 and the lower edge ofher T.F. load line is 3 cmabove water on the port side, whilst the lower edge of the summer load line islevel with the water on the starboard side. How much cargo can she load in orderto float at her Tropical load line in salt water?

125..4 vessel arrives at a port at a river mouth in water of density 1.022 t!m3, floatingat an even-keel draft of 8.06 metres (Freeboard, 2.47 metres). She is to dischargeas much cargo as possible and then to proceed to an upriver port, where therelative density of the water is 1.012 and where there is a bridge under whichshe must pass. The bridge is 23.00 mehes above water level; the truck of theship's mast is 19.40 metres above the freeboard deck, and must have a clearanceof 1.0 metres for passing under the bridge. F.W.A. is 125 mm and the T.p.C. is21.04 t in salt water. Ifthe vessel will bum 9% tonnes of fuel on the way upriverand assuming the ship to remain on an even keel throughout, find the maximumamount of cargo that she can discharge at the river mouth.

Answers-

117. 352 t.

118. 497 t.

119. 182 t .

120. 387 t.

l2l. 174 t.t t ) 1) \ t

123.276t.

t24. 223 t.125. 368 t.

Shift of G in Ships126.A ship has a displacement of2000 tonnes. Find the shift ofher centre of gravity

ifa weight of 100 tonnes is shifted l2 metres across a hold.127.A weight of 500 tonnes is loaded into a ship so that its centre of gravity is

l0 metres ffom that of the ship. Find the shift of G if the ship's originaldisplacement was 3000 torures.

128.A ship and her cargo displace 7200 tonnes. What will be the shift ofthe centreofgravity ifa weight of80 tonnes is removed from a point 100 metres from theoriginal centre of gavity ofthe ship?

PROBLEMS

l29.Find the effect ofadding a weight of80 tonnes at a distance of 120 metres fromG in a ship, the original displacement of which was 7600 tonnes.

l30.What will be the shift ofG in a ship of3600 tonnes displacement, ifa weight of40 tonnes is moved 16 metres across the deck?

131.A ship has a displacernent of 11000 tonnes. Calculate the shift of G ifa weightof 1000 tonnes is removed from a point 60 metres from the original centreof gravity.

132.A ship has a displacement of 3600 tonnes and a KG of 3.12 metres. A weight of60 tonnes is raised from a hold into a 'tween deck, through a vertical distance of7'2 metres. Find the new KG.

133.A weight of 250 tonnes is loaded into a ship at a height of 6.6 metres aboveher centre of gravity. If her original iKG was 4.75 metres and her newdisplacunent, after the weight has been loaded, is 2200 tonnes, find the new KG.What would the iKG have been if the weight had been loaded at a distance of 6.6mekes below the ship's centre ofgravity, instead of above it?

134. A ship has a displacanent of2550 tonnes and a KG of 7.40 metres. What wouldbe the KG after 950 tonnes of cargo had been loaded, with its centre of gravity2'50 metres above the keel?

l35.Find the shift ofG and the new KG ifa weight of40 tonnes is discharged from apoint 1.20 metres below the centre ofgravity ofa lighter? The lighter's originaldisplacement and KG were 680 tonnes and 3.00 metres respectively.

136.A ship's displacernent is 2800 tonnes and her KG is 4.15 metres. What will bethe new i(G if a weight of 35 tonnes is lowered vertically downwards into thehold for a distance of 12.0 metres?

137.A weight of 9 tonnes is lifted from a hold by means of a derrick, the headof which is 20 metres above the original position ofthe centre of gravity ofthe weight. If the ship's displacement is 2700 tonnes, what will be the shiftof G?

138.A ship has a rKG of 5.00 metres and a displacement of 3000 tonnes. A weightof 20 tonnes is lifted from the lower hold and placed on deck by means of aderrick, the head ofwhich is 25 metres above the keel. The centre of gravity ofthe weight was 1.0 metres above the keel when in the hold and 11.5 metres abovethe keel when on deck. Find:-

(a) +he KG when the weight is hanging on the derrick;

(b) the KG when the weight has been landed on deck.

181

182 MERCIIANT SHIP STABILITY

l39.Fifteen tonnes is lifted by a derrick, the head of which is 15 metres above theoriginal position ofthe weight, in a ship of 1250 tonnes displacement and KG3'10 metres. What will be the new KG when the weight has been lifted through(a) 2 metres; (b) 12 metres?

140.A ship displaces 2415 tonnes and has a KG of 4.50 metres. A weight of35 tonnes is lifted from the shore by a derrick, the head of which is 20 metresabove the keel and is placed in the hold, the final position of its centre of gravitybeing 3 metres above the keel. Find the KG (a) when the weight is hanging onthe derrick, and (6) when the weight has been landed in the hold.

141.A double-bottom tank, when full, has its centre of gravity at a height of60 cmabove the keel and can hold 380 tonnes of water. The KG of the ship is 9.40 mand her displacement is 3700 tonnes when the tank is empty. What will be herrKG when the tank is filled?

142.Atank holds 252 tonnes ofwater and its centre of gravity is 44 metres from thatofthe ship. Ifthe ship's displacement is 3024 tonnes when the tank is full, whatwill be the shift of G caused by pumping it out?

143.150 tonnes ofoil are transferred from a fore peak tank to an after peak tank, thedistance between their centres of gravity being 130 metres. Find the shift of Gdue to this, ifthe ship's displacement is 7500 tonnes.

Answers-

126. 0 60 m.

127. 140m.

128. l'12 m.

129. l'20 m towards the c.g. ofweight.

130. 0 '18 m.

l3 l . 6 '00 m.

132. 3'24 m.

133. 5 '50 m; 4 '00 m.

134. 6'07 m.

135. 7'50 cm; 292'5 cm.

136. 4'00 m.

137. 6 6 cm upwards.

138. (a) 5'16 m. (b) s'07 m.

139. (a) 3'28 m.. (b/ The same.

PROBLEMS

t40. (a) 4.72 m. (b) 4.48 m.14l . 8.58 m.

142. 4.00 m.

143. 2.60 m.

Stores: 98 tonnes from 9.5 m above the keel.Fresh water: 87 tonnes fiom 10.0 m above the keel.

What will be her r(G on arrival at her port of destination?l47.The original displacement of a ship was 42g5 tonnes and her 1(G was

6.00 metres. Find her new KG after she has loaded the following weights:_800 tonnes at 3.6 metres above the keel.440 tonnes at 7.0 metres above the keel.I l0 tonnes at 5.8 metres above the keel.630 tonnes at 3.0 metres above the keel.

148.4 ship has a KG of6 5 metres and a displacement of6020 tonnes. Find her r(Gafter she has loaded and discharged the following weights:_

Loaded: 500 tonnes at 2.5 metres above the lieel.850 tonnes at 5.0 metres above the keel.220 tonnes at 9.4 metres above the keel.

Discharged: 300 tonnes from 5.5 metres above the keel.700 tonnes from 2.6 metres above the keel.

183

KG144..{ ship displaces 2730 tonnes and has a KG of 6.00 metres. She then loads the

following weights:-540 tonnes at 5.0 metres above the keel.370 tonnes at 8.5 metres above the keel.I l0 tonnes at 10.4 metres above the keel.850 tonnes at 4.6 metres above the keel. Find her new rKG.

145.4 loaded lighter displaces 856 tonnes and has a KG of 1.50 metres. Find the newKG after the following weights have been discharsed:_

160 tonnes from 2.5 metres above the keel.40 tonnes from 3.7 metres above the keel.395 tonnes from 1.2 metres above the keel.

146.A ship leaves port with a displacement of9060 tonnes and a r(G of 5.20 metres.During the voyage she consumes the followine:_

Oil fuel: 260 tonnes from 0.g m abJve the keet.320 tonnes from 0.7 m above the keel.

\1


l49.Find the new l(G of a lighter which has loaded and discharged the followingweights:-

Discharged: 140 tonnes from 2.5 metres above the keel.270 tonnes from 1.4 metres above the keel

Loaded: 215 tonnes at 1.0 metres above the keel.

The original displacement and KG were 646 tonnes and 2.00 metres.150.,{ ship arrives in port with a KG of 6.80 metres and a displacement of

6080 tonnes. Whilst in port, she discharges and loads the following cargo:-Discharged: 1250 tonnes from 5.0 metres above the keel.

675 tonnes from 3.5 metres above the keel420 tonnes from 7.2 metres above the keel.30 tonnes from 0.7 metres above the keel.

Loaded: 980 tonnes at 3.2 metres above the keel.550 tonnes at 6.5 metres above the keel.700 tonnes at 0.6 metres above the keel.70 tonnes at I 1.0 metres above the keel.

She then sails on a voyage during which she burns 840 tonnes of oil from0'6 metres above the keel and uses 60 tonnes of water from l l.0 metres abovethe keel. Find the KGs at the beginning and end ofthe voyage.

15l.The light displacement of a ship is 2875 tonnes. She loads 390 tonnes at 7.0metres above the keel and 710 tonnes at 2.5 metres above the keel. Ifher KGwas then 5.20 metres, what was the light r<G?

152.A ship displaces 7425 tonnes and has aKG of 6.30 metres. She then loads 670 tof cargo at 6.0 metres above the keel and 840 t at 3.0 metres. How much morecan she load at 7.0 metres above the keel in order to finish with a r(G of 6.0Cmetres?

153.A ship displaces 9500 tonnes and has a KG of 5.84 metres. She then loads 550 tof cargo at 4.2 metres above the keel and 720 t at 6.l metres. She has a firther500 t to load. At what height must this be loaded if the ship is to sail with a KGof 5'70 metres?

154.A vessel displaces 4750 t and has a i(G of 7.05 metres. She then loads 920 t ofcargo at 4.5 metres above the keel and 630 t at 7.0 metres; she also discharges350 t from 8.5 metres above the keel. How much fuel oil can she load into adouble bottom tank, at an estimated height of 0.50 metre above the keel, tofinish with a KG of 6.30 metres; allowing for an estimated rise of G of 0.02metres caused by free surface ofthe oil?

PROBLEMS

155.A ship which is completing loading has a KG of 7'23 metres and displaces14600 tonnes. On passage to her next port, she is expected to use 360 t of fueloil from 0.70 metres above the keel; 40 t of fiesh water fiom 9.40 metres abovethe keel; and 20 tonnes ofstores from 8.8 metres above the keel. Before sailing,she is to load cargo into a 'tween deck at a height of 9.50 metres above the keel.How much can she load in the 'tween deck in order to arrive at her next portwith a KG of 7.45 metres?

156.4 ship which is completing loading has a KG of 6.82 metres and displaces11250 t. Before sailing, she has to load deck cargo at an estimated height of10.5 metres above the keel. On the voyage she is expected to use 180 t of fuelfrom 5.6 metres above the keel; 30 t of fresh water from 8.8 metres above thekeel; and 10 t of stores from 8.2 metres above the keel. How much deck cargocan she load in order to arrive at her next port with aKG of6.95 metres, allowingfor a rise of G of 0.05 metre due to fiee surface appearing on voyage?

157.A vessel displaces 9740 t and has a l(G of 6.06 metres. A 50-tonne lift is to betaken on board by means ofa derrick, the head ofwhich will be 25 metres abovethe keel when lifting. To prevent excessive heel when lifting, the iKG ofthe shipmust not exceed 6'00 metres when the lift is hanging from the derrick. Find theleast amount of cargo to be loaded into a lower hold, at a height of 3.5 metresabove the keel, to satisry this condition.

158.A ship which displaces 7925 t and,has a KG of 5.42 metres, has to load a deckcargo oftimber at a height of 12.00 metres above the keel. On voyage to the nextport she expects to use 135 t of fuel and water from the double bottom, from aheight of 0.6 metres above the keel; causing a rise ofG of 0.04 m, due to freesurface effect. Calculate how much timber she can load, allowing 15% of theweight ofwood for absorption of water on voyage, to arrive at her destinationwith a l(G of 6.10 metres.

Answers-

144. 5.93 m.

145. 1.00 m.

146. 5.41 m.

147 . 5'46 m.

148. 6'53 m.

149. l'73 m.

150. 6.26 m;7.13m.

185


151. 5.62 m.

152.292t.

153. 4.11 m.

154.294t.

155. 433 t.

156. 204 t.

157. 613 t .

158. 635 t.

BM

159.A box-shaped ship is 120 metres long, 18 metres beam, and floats at a draft of5'00 metres. What isher BM!

160. Find the EMof a box-shaped lighter which has a beam of 6.00 metres and floatsat a draft of 2.00 metres.

16l.What is the height ofthe metacentre (Krl1) in a box-shaped vessel, of 10.00 metresbeam, when floating at a draft of 5.00 metres?

162.Fndthe KM ofa box-shaped lighter which has a beam of7.00 metres and floatsat a draft of l'40 metres.

163.A ship displaces 3860 tonnes and the mometrt of inertia of her waterplane is23350. Find her rKM if the iKB is 3'50 metres.

164.What is the BM of a ship of 4160 tormes, if the moment of inertia of herwaterplane is 32470?

l65.Find the GMof a box-shaped lighter,20 mehes long and 6 metres wide, whichhas a draft of 2.40 metres and a KG of 1.70 metres.

Answers-

159. 3.75 m.

160. 1.50 m.

161. 4.17 m.

162. 3.62 m.

163. 9.70 m.

164. 8.00 m.

165. 0'75 m.

PROBLEMS

The Inclining Experiment166.A weight of25 tonnes is shifted transversely for a distance of l0 metres across

the deck of a ship. A plumbJine, which is suspended 4.00 metres above ahorizontal batten moves out for a distance of 0:30 metres along the batten.Ifthe ship's displacement is 4950 tonnes, what is her GM assuming that she wasupright at the beginning?

l67.When the inclining experiment is performed on a ship of 2304 tonnesdisplacement, a weight of 15 tonnes is moved for 12 metres across the deck.The plumbJine is 8.00 metres long and moves out 0.43 metres when the shipheels. Find the Gll

168.A ship which has just been completed has a light .KM of 10.30 metres and adisplacement of37g0 tonnes. A weight of 12 tonnes is moved across the deckfor a distance of I I metres; when a plumbJine, suspended g .00 metres above thebatten, moves out 8 centimetres. Find the ship,s light KG.

l69.Find the,(G of a ship which has a r(Mof g.l5 metres and displaces 2400 tonnes.When the inclining experiment was performed, a weighi of l0 tonnes wasshifted 16 metres across the deck and caused a plumb_iine, 5 metres long, tomove out 18 centimetres.

l70.In an inclining experiment, a weight of 12.50 tonnes was moved l0 metresacross the deck and caused a ptumbJine, 12 metres long, to move out32 centimetres. A double-bottom tank in the ship was firll of water, whichweighed 450 tonnes and had its centre of gravity 0.90 metres above thekeel; otherwise the ship would have been in the light condition. If the ship,sdisplacement at the time of the experiment was 37i0 tonnes and her r(M was9.00 metres, fnd:-(a) T\e KG at the time of the exDerimenr.(b) ThetightKc.

Answers-

166. 0.6i3 m.

167. 1.453 m.

168. 6.81 m.

169. 6.30 m.

r70. (a) 8.00 m; (b) 8.97 m.

187

188 MERCHANT SH,IP STABILITY

Moment of Statical Stability17l.Find the moment of statical stability of a ship of 3165 tonnes displacement and

GM0 80 metres, when she is heeled to an angle of 12o.

172.A ship of 1068 tonnes displacement has a GMof l'20 metres. Find her momentof statical stability at an angle ofheel of6'.

173.What is the moment of statical stability of a ship which displaces 6752 tonnesand has a righting lever of 0'45 metres?

174.A ship of5124 tonnes displacement has the following righting levers:

Angle ofheel: l0' 20" 30" 40" 50" 60" 70"

Conesponding GZ 0'12 0'33 0'48 0 52 0 39 0'18 {'09 metre

Draw a curve of moments of statical stability and find fiom this:-

(a) The moment of statical stability at 24o ofheel.

(b) The maximum moment of statical stability and the angle at whichthis occurs.

@ The range of stability.

175.4 ship of 7200 tonnes displacement has a KB of 4'00 metres and a KG of6'50 metres. At an angle of heel of 23' the volumes of the immersed andemerged wedges are each 1200 cubic mehes and the horizontal shift of theircentres of gravity is 7'00 metres. Find the length of the righting lever and themoment of statical stability at this angle ofheel.

176.A ship which is heeled to an angle of57" has immersed and emerged wedgesof 2500 cubic metres each, with their centres of gravity 12'00 metres apart.The ship displaces 12500 cubic metres of salt water, has a KG of 7'10 metresand a KB of 4'30 metres. What is the moment of statical stabilitv and is the shipin stable equilibrium?

l77.Abox-shaped ship is 120 metres long, l8 metres wide and floats at a draft of4'00 m.Assuming that the deck edge does not submerge, nor the bilge emerge, find thevolumes ofthe wedges and the shift oftheir centres of gravity at an angle ofheelof25o. Thence find the righting lever at 25' ofheel, if the KG is 8'30 metres.

Answersl7l . 526'7 t/m.

172. 134'6 tlm.

173.30384Vm.

174. 20390 t/m:37":26900 t/m;67'.

175. 0 '218 m; 1569'6 Vm.

176. - 0'03 m; Ship is unstable.

177 . 0'66 m.

PROBLEMS

Angle of Heel

178.A weight of 50 tomes is shifted tmnsversely across the deck of a ship for adistance of 12 metres. The ship's displacement was 4350 tonnes and her GMwas 0.40 metres. If the ship was upright before the weight was shifted, find theangle to which she will heel.

179.A ship of 4800 tonnes displacement has a list of 8o, due to unequal loadingof weights. If her GM is 0'30 metres, find how much weight must be shiftedtransversely across a 'tween deck, for a distance of 12 metres, in order to bringthe ship upright.

180.A weight of 120 tonnes is loaded into a 'tween deck so that it is 3 80 metres fromthe ship's centeJine and at a vertical height of 6'00 metres above her centreof gravity. Before the weight was loaded, the ship was upright, had a GM of O'70metes and a displacement of7080 tonnes. What will be the effect of the weight?

181.A ship has a KG of 3'8 metres and displacement of5750 tonnes and is listed 12"to port. 250 tonnes of cargo are to be loaded into the port and starboard wingsof a 'tween deck at a height of 7'50 metres above the keel and a distance of8.00 metres on either side of the centreline. If the r(M is 4'60 metres, how muchweight must be placed in each wing to finish loading with the ship upright?

182.80 tonnes of grain shifts in a hold, 10 metres horizontally and 3 metresdownwards. Before this happened, the ship was upright, had a GM of l'35metres and a displacement of2320 tonnes. Find the angle ofheel caused by theshift of the grain.

183.A ship displaces 11600 t, has KG of 6'10 metres, KM of 6'95 metres and isheeled 5o to starboard. 350 t ofcargo are to be loaded into the wings of a'tweendeck at distances of 5'00 metres to port and 7 00 metres to starboard of thecentre line. How much ofthe cargo must there be loaded into each wing in orderto finish with the ship upright?

184.A ship displaces 14500 t, has KG of 7'lO metres, KM of 8 05 metres, and isheeled 4'to starboard. 500 t of cargo are to be loaded into a 'tween deck at aheight of 10'00 mehes above the keel: ofthis, 300 t are to be loaded in the squareof the hatch, whilst the remainder is to be distributed between the wings, at6'00 metres to port and 8'00 metres to starboard ofthe centre line. How much mustbe placed in each wing if the vessel is to be upright after it has been loaded?

185.A vessel is heeled 7" to starboard and has a KG of 6'02 metres, rKM of 6'41metres, and displaces 8800 t. She then loads 75 t cargo at 5'60 metres above thekeel and 5'00 metres to starboard of the centre line; 100 t at 4 20 metres abovethe keel and 6 50 metres to port ofthe centre line; and 90 t amidships and at 3'80metres above the keel. What will be her final heel, if any?

189


186.A ship displaces 12720 t, has KG of 6.90 meies, KM of 7.50 metres and isupright when a 60-tonne locomotive is stowed on deck, at 5.80 metres tostarboard of the centre line and I1.50 metres above the keel. The locomotive isto be discharged over the port side ofthe ship by means ofa derrick, the head ofwhich will be 26 metres above the keel and 13 mehes offthe centre line whenplumbing the quay. Find the maximum heel, the final heel and the final Gi4

187.Two heavy lifts, each of40 t, are to be lifted from the quay by a derrick and areto be placed on the ship's deck at 12.00 m above the keel and 6'00 m on eitherside ofthe centre line. The first lift is to be landed on the offshore side of thedeck and the second lift on the onshore side. The head ofthe derrick will be 22 mabove the keel and will plumb a point 11.50 m from the centre line when liftingfrom the quay. Before lifting, the ship was upright, displaced 9600 t, hadKG of6 85 m and iKM of 7'46 m. Find: (a) the maximum heel; (b) the heel when thefirst lift has been landed on deck and the second is being lifted from the quay.

188.A vessel displaces 12420 t and has a KM of7'84 mehes. She is to take on board alift of 80 t, using a derrick, the head of which will be 12.50 metres offthe centreline and 2l m above the keel when lifting. What must be the ship's maximumi(G before lifting, in order that the heel may not exceed 5' when lifting?

189.Find the angle to which a ship will loll, if her GMis -{.06 metres and her BMis3'60 metres.

190.A ship displaces 2040 tonnes and has a GMof 0.10 metre. 120 tonnes of cargoare then loaded on deck at a vertical height of 4.00 metres above the ship's centreof gravity, when the BM was found to be 3 .20 metres. What will happen?

l91.To what angle will a ship loll if she has a BM of 6.00 metres and a GM of-O'20 metres?

Answers-178. 19".

179. 16.9 t.

180. 6 '3 'heel .

181. 64t(P);186t(S)

182. 13".

183.278 1(P);72t(S).

184. 183 t (P); 17 t (S).

185. 2'l' to Starboard.

186. 9'5' (P); 2'5" (P); 0'622 m.

PROBLEMS

187. 5 '0" (s) ;2.s" (S).

188. 6 '835 m.

189. 10.4".

190. Ship lolls 15.3".

l9 l . 14.5 '

Free Surface Effect192.4 seaman calculates his ship's GMas 0.68 metres and her displacement as 4320

tonnes, when floating in salt water. He has forgoften to allow for free surface ofsalt water in a rectangular double-boftom tank, 15 metres long and 12 metreswide, with no subdivisions in it. What is the ship's true GMwhen upright?

l93.Ifthe tank in the last question had contained oil ofrelative density 0.875, whatwould then have been the Gld2

194.A ship of 6000 tonnes displacement has a KG of 3.45 metres and a KM of3 72 metres, neglecting the effect of free surface. Free surface exists in anundivided rectangular tall/r, 12 metres long and 10 metres wide, which is partlyfilled with sea water. Find the tnre GMwhen the ship is upright.

195.100 tonnes of water are run into a rectangular tank, l0 metres long and 12metres wide, in a ship of5300 tonnes displacement: when the tank is found tobe about three-quarters full. Ifthe centre of gravity ofthis water is 0.50 metresabove the keel and the ship's original KG was 5.00 metres, find the new KG.

196.A box-shaped lighter is 30 metres long and 8 metres wide and floats at a draftof l'00 metre: whilst its i(G is 0.80 metre. If 15 centimetres of water is thenallowed to run into the bottom, so that the new KMbecomes 5.20 metres, whatwill be the new Gllfl

197.A ship has a displacement of4880 tonnes and a r(G of6.00 metres when all herdouble-bottom tanks are full of salt water. What would be the KG if 80 tonnesof water were pumped out of a rectangular tank, leaving it slack: The tank being7'00 metres long and 16.00 metres wide, whilst the centre ofgravity ofthe waterremoved was 5'00 metres below G.

198..4 rectangular deep tank, l0 metres long, 12 metres wide and 5 metres deep isdivided at the centreline. When the tank is full, the ship has a displacement of6080 tonnes and a KG of6'00 metres. Ifthe tark is then pumped out until thereare 2'00 metres of water left in it, what will be the new KG of the ship, assumingthat the tank extends risht down to the keel?

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199.A ship displaces 10400 t, has a KG of 6.45 metres, a KM of 7.52 metres and

floats upright in salt water when the starboard side of a double-bottom tank is

full ofsalt water and the port side of the tank is empty. The tank is rectangular,

35 metres long, 16 metres wide and 1.60 metres deep; with one side gider on

each side and a watertight centre girder. To what angle will the ship heel if

exactly one halfofthe ballast is transfened from the starboard to the port side

ofthe tank?

200.A rectangular tank is 24 metres long, 15 metres wide and has a fore and aft

division at its centre line. What is its free surface moment?

2Ol.Calculate the free surface moment of a rectangular double bottom tank which

is 16 metres long, 18 metres wide, and has a centre girder and one side girder

on each side.

202.A ship displaces 9850 t and has a solid GMof l'08 metres when a tank is partly

filled with oil ofrelative density 0.910. Ifthe tank has a free surface moment of

1220, what is the ship's fluid G,l4z

203.A vessel has a KG of 5'982 metres, a KM of 6'493 metres and displaces 7486 t.

200 tons of salt water are then run into a tank, leaving it slack. The centre of

gavity of the water is then 0'65 metres above the keel and the tank has a free

surface moment of 588. Find the ship's new fluid GM

204.A ship displaced 10540 t, had a KG of 5.421 metres and a KMof 5.873 metres.

She then discharged 150 t of salt water from l'20 metres above the keel from

a tank which had a free surface moment of 395: and 100 t of salt water from

1'30 metres above the keel from a tank which had a free surface moment of457.

Find the new fluid Gi4

205.A ship loads 300 t ofoil of relative density 0'895 at 0.85 metres above tlre keel

in a tank which has a free surface moment of 512 She also loads 250 t of salt

water at 0'70 metres above the keel in a tank which has a free surface moment

of3 86. Both tanks are then slack. The original displacement and KG were 8720 t,

and 5'01 m respectively, whilst the final KMis 5'641 m. Find the new GM.

PROBLEMS r93Answers-

192.

r93.194.

195.

196.

197.l9E.

199.

200.

201.

202.203.

2M.

205.

0'17 m.0.24m.

0'10 m.

5'19 m.

-{'15 m-6.53 m.6'22m.

9.9".1688.

486.

0'967 m.0'572 m.

0'265 m.

0'790 m.

Dynamical Stability206.Find the dynamical stability at 40' heel ofa ship which displaces 6826 tonnes

and has the following GZ;-

Heel l0 ' 20" 30" 400GZ 0.1 l5 0.228 o33l 0.429

metres

207.Find the dynamical stability at 45. heel of a ship which hs a displacement of5000 tonnes and GZs of:-

208.Find the area undet a crrve ofrighting lwers up to 40o heel, given:- ,

Heel 10" 20" 300 40"GZ 0.145 0.248 0.2& 0.234

Answers-206. 1068 t/m.207. l3l2t1m.

208. 0.138 mehe-radians.

Heel 15" 300 450GZ o'205 0.482 0:611


BM"2O9.Abox-shaped ship has a length of 120 metres and floats at a draft ofS 00 metres.

What is her longitudinal BMr?

210.Find the BMLand, GMrof abox-shaped lighter, 30 metres long, l0 metres wideand floating at a draft of 2.00 metres, if herKG is 1.50 metres.

211. A box-shaped lighter is 40 metres long and floats at a draft of 1.20 metres foreand aft. Find her BM, and, KM".

Answers-

209. 150 m.

210. 37'5 m; 37'0 m.

2l l . l l l '1 m; 1l l '7 m

Change ofDraft due to Change ofTrim2l2.Find the new drafts ifa weight is shifted aft in a ship, sufficiently to change the

him by 0'26 metres. The centre offlotation is amidships and the original draftswere 6'45 metres forward and 6'48 metres aft.

2l3.Thecentre offlotation ofaship is4.00 metres abaftamidships. Weights are shiftedaft so as to change the trim by 0.56 metres. If the ship's length is 140 metresand her original drafts were 5.92 m forward and 6.08 m aft, find the new draftsfore and aft and also the original and new mean drafts.

214.,{ ship is 120 metres long and floats on an even keel at drafts of 5.36 metresfore and aft. Weights are shifted aft so as to change the trim by 0.48 metres.Find the new drafts, fore, aft and mean, if the ship's centre of flotation is5 00 metres abaft amid ships

215. An oil tanker is 260 metres long and floats at drafts of 5. 12 m forward and 6.88 maft. The centre of flotation is 4.00 metoes forward of amidships. Oil is thenshifted from an after tank to a forward tank so as to change the trim by1'20 mehes. Find the new drafts.

Answers-

212. F.6.32 m; A. 6.61 m.

213. F.5'62 m; A. 6'34 m; Original Mean 6.00 m; New Mean 5.98 m.

214. F. 5 '10 m; A. 5 '58 m; M. 5.34 m.

215. F.5'70 m; A. 6 '26 m.

PROBLEMS

Draft and Displacement Out of the Designed Tiim216.A ship which is 150 metres long, floats at draffs of 5'16 metres forward and

6'32 metres aft. If her designed trim is for an even keel and F is 4'00 metresabaft amidships, find the draft at fl

217.A vessel is 110 metres long, F is 3'00 metres forward of amidships andher designed trim is an even keel. If her drafts are 3 98 metres forward and4.86 metres aft, what is the draft at F?

218.A ship is 140 metres long, has a T.P.C. of 20 and the centre of flotation isl'50 metres abaft amidships. Find the layer correction when the drafts are 6'30metres forward and 6'70 metres aft.

219.A ship is 133 metres long, has a T.P.C. of 19.0 and her centre offlotation is 3.00metres abaft amidships. She floats at drafts of 6'12 m forward and 6'54 m aft.The displacement for the designed trim (even keel) at a draft of 6.33 metres is8243 tonnes. Find the layer correction and the true displacement.

220.Aship, 160 metres long, floats at drafts of7'95 metres forward and 8'59 metres aft.The designed draft is an even keel: the T.P.C. is 25'00 and ttre centre offlotation is 2'5 metres abaft amidships. The displacement is given in the scaleas 10942 tonnes for a mean draft of 8'27 metres. Find the draft at F and the truedisplacement.

221.A ship, is 90 metres long, her T.P.C. is ll'0 and her centre of flotation is 2'00metres abaft amidships. The designed trim is 15 centimetres by the stern.The drafts are 4.15 m forward and 3 85 m aft. The designed displacement for adraft of4'00 m is 3283 tonnes. Find the draft at F and the true disolacement.

Answers-216. 5'77 m.

217. 4'40 m.

218. 8.6 t .

219. 18t;'8261t.

220. 8'28 m; 10967 t.

221. 3'98 m;3264 r.

M.C.T.1C.222.A ship is 140 metres long and displaces 4340 tonnes, her i(G is 5.10 metres and

her KMr 132'20 metres. Find her M.C.T. I C.

223.Find the M.C.T.IC. of a ship, 120 metres long and displacing 3600 tonnes,which has a GM, of 150'00 metres.

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195

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224.4 box-shaped vessel is 80 metres long, 15 metres wide and floats at a draft of5'00 metres in sea water. i(G is not known. Find her M.C.T.lC.

225.4 ship is 192 metres long, has a KG of6.80 metes, a KM of200.20 metres anddisplaces 9200 tonnes. Find her M.C.T.1C.

Answers-222. 39 4 tlm.

223. 45.0 tJm.

224. 82'O tlm.

225. 92'7 tJm.

Change of Trim due to Shifting Weights226.A ship floats at drafts of 6.84 metres forward and 7.14 metres aft. Her M.C.T.lC

is 105 tonne/metres and her centre offlotation is amidships. Find the change oftrim and the new drafts ifa weight of42 tonnes is shiftedlorward for a disLnceof 60 metres.

227.Find the change of trim and the new drafts fore and aft, if 120 tonnes of oilis transferred from the forepeak to the afterpeak tank in a ship which has anM.C.T.lC. of 144 tonne/metres. The distance between the cenfes of gravity ofthe tanks is 60 metres, the ship's centre offlotation is amidships, and the originaldrafts were 4.82 metres forward and 4.58 metres aft.

228.What will be the change of trim and the new drafts fore and aft if a weightof90 tonnes is shifted aft for a distance of 100 metres in a ship which has anM.C.T.1C. of 180? The length of the ship is 150 metres, her centre of flotation is2'00 metres abaft amidships and her original drafts were 6.00 metres forward and6.10 metres aft.

229..4 ship is 160 metres long and her centre offlotation is 4 metres abaft amidshios.Her drafts are 6.58 metres forward and 8'08 metres aft, whilst her M.C.T.iC.is 180 t/m. Find the change of trim and the new drafts if 300 tonnes of oilis transferred from No.4 double-bottom tank to No. I double-bottom tank: adistance of72 metres.

230..4 ship has drafts of 4.10 metres forward and 5.50 metres aft. Her M.C.T.lC. is200. Find how much oil to transfer fiom No. 5 D.B. tank to No. 2 D.B. tank.through a distance of 100 metres, to bring the ship to an even keel.

231..4 ship has an M.C.T.IC. of 210 and floats at drafts of 6.20 metres forwardand 7.60 metres aft. Find how much weight must be moved forward, through adistance of60 metres, to bring her to a trim of50 centimetres by the stem.

Answers-

226. 24 cm;'

227. 50 cml'

228. 5O cm1,

229. 120 cml'

PROBLEMS

F. 6'96 m; A. 7'02 m.

F. 4 57 m; A.4'83 m.

F. 5.74 m:, A. 6'34 m.

F. 7'21 m; A. 7'51 m.

197

230. 280 tonnes.

231. 315 tonnes.

Moderate Weights Loaded Off the Centre of Flotation

232.A ship is 150 metres long, has a T.P.C. of 12'5 and an M'C.T.IC. of 120.Her drafts are 4'76 metres forward and 5'40 mehes aft. 250 tonnes of cargo areloaded at a distance of50 metres abaft the stem. Find the new drafts, assumingthat the centre of flotation is amidships.

233.The centre of flotation of a ship is amidships, her length is 120 metres, herT.P.C. is 15 and her M.C.T.IC. is 100. A weight of 240 tonnes is dischargedfrom a position 40 metres abaft the stem. Ifher original drafts were 5'10 metresforward and 5'20 metres aft, what will be her new drafts?

234.The following particulars are known about a ship: Length 136 metres; T.P.C.20;M.C.T.lC. 120; centre of flotation 70 metres abaft the stem; drafts 6'50 mforward, 6'60 m aft. Find her new drafts after 80 tonnes of cargo have beenloaded at a distance of40 metres from aft.

235.A ship has drafts of4'72 metres forward and 5'64 metres aft. In order to bringher more nearly to an even keel, 240 tonnes of water are run into No. I doublebottom tank, the centre of gpvity ofwhich is 20 metres abaft the stem. The shipis 160 metres long, has aT.P.C. of 22, an M.C.T.IC. of 172 and the centre offlotation is 2'00 metres abaft amidships. Find the new drafts.

236.Find the new drafts after 120 tonnes of cargo have been discharged from a pointwhich is 15 metres abaft amidships. The ship is 130 mehes long, has a T.P.C.of 15. and M.C.T.1C. of 110, whilst her centre of flotation is 3'00 metres abaftamidships. The original drafts were 7 00 m forward and 7'40 m aft.

237.A ship which is 130 metres long has a T.P.C. of 20 and M.C.T.IC. of 125'Her centre of flotation is amidships and her drafts are 6'20 metres forward and6.50 metres aft. What would be the new drafts if 140 tonnes of cargo are loadedat a distance of30 metres abaft the stem and 56 tonnes are loaded at a distanceof 100 metres abaft the stem?


238.A ship floats at drafts of 4.30 m forward and 4.80 m aft. Her length is l0gmetres, T.P.C. 16, M.C.T.lC. I18, whilst her centre of flotation is 2.00 metresabaft amidships. Find the new drafts after the following weights have beenloaded and discharged:-

Loaded: 55 tonnes at 40 metres abaft the stem.Loaded: 100 tonnes at 70 metres abaft the stem.Discharged: 3 tonnes from 86 metres abaft the stem.

239.Avessel is 90 metres long and her centre offlotation is 2 metres abaft amidshios.Her T.P.C. is 80 and her M.C.T.lC. is 48. 60 tonnes of cargo is discharged fiom22 metres forward of the centre of flotation; whilst 40 tonnes is loaded at 13metres abaft the centre of flotation. If the original drafts were 3.g5 metresforward and 3.79 metresaft, find the new drafts.

240.A ship is 140 metres long and floats at drafts of 6.38 metres forward and 7.06metres aft. Her T.P.C. is 22.1, M.C.T.lC. is 186 and,F is 2.00 metres forward ofamidships. She next takes in 310 t ofwater ballast at 6.00 m abaft amidshios andthen sails for her next port. On the voyage, she uses 370 t ofoil from 15 metresabaft amidships; 25 t ofstores from 5l m forward of amidships; and 4g t offreshwater ffom 3 metres abaft amidships. Find her draft on arrival.

Answers-232. F. 5.22 m; A. 5.34 m.233. F. 4.70 m; A. 5.28 m.234. F. 6.45 m; A. 6-72 m.235. F.5 28 m; A. 5.33 m.236. F. 6-99 m: A.7.26 m.237. F. 6.42 m; A. 6.48 m.238. F.4.42 m; A. 4.78 m.239. F. 3.62 m; A.3.94 m.240. F. 6 39 m; A. 6.92 m.

Large Weights Loaded Off the Centre of Flotation241.A ship floats at drafts of 4.90 menes forward and 5.10 metres aft. 600 tonnes of

oil is then loaded into a deep tank, the centre of gravity of which is 3.00 metresforward of amidships. Find the new drafts, if the following information is foundfrom the deadweight scales:-

Draft 5.00 m. T.P.C. 21.0. M.C.T.tC. 148.Draft 5.50 m. T.P.C. 21.8. M.C.T.lC. 154.Centre of flotation amidshins.

PROBLEMS 199

242.The following information is given in the ship's deadweight scale:-Draft 6'00 m. T.P.C. 16 1. M.C.T.lC. 108.Draft 6'50 m. T.P.C. 16'6. M.C.T.1C. 113.Draft 7'00 m. T.P.C. 17.0. M.C.TlC. 117.Centre offlotation, 3'00 m abaft amidships

The ship is 126 metres long and floats at drafts of 6'10 m forward and aft. Findthe new drafts ifthe following cargo is loaded:-

450 tonnes at 40 metres abaft the stem500 tonnes at 110 metres abaft the stem

243.The ship, for which the hydrostatic particulars are given inthebackofthisbook,is 140 metres long. She is floating at drafts of5'16 metres forward and 5'24 maft. Use the curves or scales to find the new drafts after the ship has loaded:-

In No. I hold; 190 tonnes; c.g. 52 metres forward of amidships.In No. 2 hold; 260 tonnes; c.g. 28 metres forward of amidships.In No. 3 hold; 180 tonnes; c.g. 5 metres forward of amidships.In No. 4 hold; 380 tonnes; c.g. 32 metres abaft amidships.In No. 5 hold; 250 tonnes; c.g. 46 metres abaft amidships.

244. The same ship as in the last question had drafts of2'58 m forward and 4'72 m aft-To bring her to a better trim, the forward deep tanks (c.g. 15'20 m forward ofamidships) were then filled with I I 50 t of water. Find her new drafts.

245.The same ship, as above, when loading cargo, had drafts of 6'23 metres forwardand 6'69 metres aft. She then loaded:-

480 tonnes at 48 metres forward of amidships.710 tonnes at 24 metres forward of amidships.630 tonnes at 32 metres abaft amidships.370 tonnes at 43 metres abaft amidships.

This completed her loading and she then sailed for her next port. On the voyageshe used 210 t ofoil from 2 metres forward of amidships, 30 t of fresh water from8 metres abaft amidships and 10 t of stores from 12 metres abaft amidships.What were her drafts on arrival?

Ans'r9ers-241. F. 5'24 m; A. 5'32 m.

242. F. 5.74 m:4.7'88 m.

243. F. 5'58 m; A. 6'02 m.

244. F. 3'72 m; A. 4'71 m.

245. F.7'25 m; A. 7'41 m.

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Loading a Weight to Produce a Desired Trim

246.4 ship, which is completing her loading, has 120 tonnes of cargo to come onboard. Her drafts are 7.00 m forward and,7.82 m aft. Her M.C.T.lC. is 125.Where must the cargo be loaded in order that the ship may sail with a trim of50 centimetres by the stem?

247.Findthe weight of water which must be run into a double-bottom tank in orderto bring the ship on to an even keel. The centre of $avity ofthe tank is 25 metresabaft the stem. The ship is 140 metres long, has an M.C.T.IC. of 144 and hercentre of flotation is 5'00 metres abaft amidships. Her present drafts are 5.70metres forward and 6.60 metres aft.

248.A ship has been in collision and her fore peak is flooded, causing her to trim 0.60 mby the head. It is desired to bring her to a trim of 0.20 m by the stem. The afterpeak tank, which is empty, can take 240 torures ofwater and its centre ofgravityis 70 metres abaft the ship's centre of flotation. If the M.C.T.lC. is 168, will itbe possible to bring the ship to the desired trim by running up this tank and, ifso, what weight of water must be taken in?

249.4 ship floats on an even keel and has an M.C.T.IC. of 116. A total of 400tonnes ofcargo are to be loaded into No. I hold (60 menes forward of F.) andinto No. 4 hold (20 metres abaft F.). How much cargo must be loaded into eachhold in order that the ship may finish loading with a trim of 50 centimetres bythe stem?

250.A ship floats at drafts of 5.60 metres forward and 7.00 metres aft. Her M.C.T.l C.is 140. 280 tonnes of water is then pumped out ofNo. 4 double-bottom tank,which has its centre of gravity at 15 metres abaft the centre of flotation.Calculate how much oil would have to be transferred from No. 6 double-bottomtank (55 metres abaft F.) to No. I double-bottom tank (65 metres forward of F.)to bring the vessel to an even keel.

Answers

246.33.3mforwardofF.

247. 259 tornes.

248. Yes. 192 tonnes is required.

249. 27 5 tornes into No 1; 372.5 tonnes into No.4.

250. 128 tonnes.

PROBLEMS

Loading for Required Draft Aft

251.A ship is 138 metres long, has an M.C.T.IC. of 132, a T.PC. of 18, and hercentre offlotation is 3 metres abaft amidships. How far forward ofthe centre offlotation must a weight be loaded if the after draft is to remain constant?

252.How farr abafr the stem must a weight be loaded if the draft aft is not to changein a ship 140 metres long? The centre of flotation is 2 metres abaft amid ships,the T.P.C. is 22, and the M.C.T.lC. is 160.

253.The centre of flotation of a ship is amidships and her lenglh is 120 metres. HerT.P.C. is 16 and her M.C.T.1C. is 115. Where, with relation to amidships, musta weight of 140 tonnes be loaded, if the draft aft is not to change?

254.A ship is 120 metres long and has drafts of 5'83 metres forward and 5'49 metresaft. Her T.P.C. is 16'4, M.C.T.1C. is 115 and F is amidships. How much cargomust she load into an after hold, the centre ofgravity ofwhich is 48 metres abaftamidships, in order to increase the draft, aft, to 6 00 metres?

255.An oil tanker, when light, floats at drafts of2'62 metres forward and 5 24 metresaft. Her length is 200 metres, T.P.C. is 38, M.C.T.lC. is 190, whilst F is 5'00metres forward of amidships. On sailing, she has to cross a bar, on which thedepth is 5'00 metres, with a clearance of 50 cm. Find the least amount of waterballast to load into a trimming tank, the centre of gravity of which is 95 metresforward of amidships. Find, also, the final draft forward.

256.A ship is 140 metres long and has drafts of 7'54 metres forward and 7'68metres aft. Her T.P.C. is 23 0. M.C.T.lC. is 207, whilst F is 2 00 metres abaftamidships. Her fore peak tank is full of water ballast, with its centre of gravity64 metres forward of amidships. Calculate the amount ofthis water ballast to bedischarged in order to bring the ship to a draft, aft, of 8'00 mehes. What wouldthen be the draft forward?

257.A ship has 400 tonnes of cargo to load and her present drafts are 6'16 metresforward and 6'92 metres aft. She is 126 metres long, herT.P.C. is 20'0, M.C.T' lC.is 158, and F is l'00 metres abaft amidships. How far from amidships should thecargo be loaded in order to bring the after draft to 7'00 metres?

Answers-251. 15'3 m.

252. 57 m.

253. 14.4 m forward.

254. 189 t.

201


255. 333 t;3'46 m.

256. 287 t; 6'94 m.

257. 8'6 m forward.

Weight to Load for a Given Draft258.A ship has drafts of 5'86 metres forward and 6.12 metres aft. Her M.C.T.lC. is

114 and F is 2'00 metres abaft amidships. She has 220 tonnes of cargo to loadinto two holds, one at 56 metres forward of amidships and the other at 19 metesabaft amidships. How much should be loaded into each hold in order to bringthe ship to a trim of l0 cm by the stem?

259.A ship arrives offa port with drafts of 7.12 metres forward and 7.88 metres aff-Her T.P.C. is 19'5, M.C.T.lC. is 110 and her centre of flotation is amidshipsIn order to enter the port, she has to reduce her draft to not more thatr7'40 metres. Find the minimum amount of cargo which she must dischargsinto lighters from two holds, one 32 metres forward of amidships and the other54 metres abaft amidships.

260.A vessel has a T.PC, of 16.3, a M.C.T.1C. of ll5, whilst.lc is 2.00 metres abaftamidships. Her present drafts are 6'18 m forward and 6.40 metres aft. In orderto cross a bar, her maximum draft on sailing must not exceed 6.50 metres.Find the maximum amount of cargo which she can load into each of two holdsone 44 metres forward of amidships and the other 35 metres abaft amidships.

261 . A ship which is loading cargo has to load 240 tonnes into No. 3 hold, at 12 merresforward of amidships; then to distribute as much cargo as possible betweerNo. I hold (49 metres forward of amidships) and No.4 hold (32 metres abafiamidships). On sailing, she has to cross a bar, on which the depth is 6.00 metes.with a clearance of 20 cm. What is the maximum amount of cargo to be loadedinto each hatch, if her present drafts are 5.37 metres forward and 5.61 metcsaft? The ship's T.P.C, is 20.1, her M.C.T.lC. is 166, whilstFis amidships.

262.1n a Tropical Zone, a ship which has a summer draft. of 7.94 metres, arrivsin port with salt water drafts of 7.98 metres forward and 8.16 metres afi-Her T.P.C. is 23.2, M.C.T.lC. is 208, F' is 3.00 metres abaft amidships and herFresh Water Allowance is 164 mm. She then has to cross a dock sill, where thcrelative density of the water is 1.010 and where her mean draft must not exce€d8'00 metres. How much cargo must she discharge, before entering the doclcfrom each oftwo holds, one of which is 18 metres forward ofamidships, and thcother 36 metres abaft amidships?

PROBLEMS

Answers-258. Forward, 7 4 t; Aft, 146 t.

259. Forward, 25 t; Aft,170 t.

260. Forward, 175 t;4ft, 167 t.

261. No. l , 165 t ; No.4,218 t .

262. Forwafi, 171 t: 4ft,223 t.

The Use of Moments About the After Perpendicular

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203

263.A ship displaces 9870 t and B is 58'25 metres from the after perpendicularShe loads 750 t at 22'00 metres from the after perpendicular. IfB is then 58'37metres fiom the after perpendicular, what is the moment changing trim?

264.4 vessel displaces 5260 t and her B is 59'72 metres from the A,/P. She then loads250 t at 94 metres from the A/P and 320 t aI35 metres fiom the A,{P: she alsodischarges 180 t from l6 metres from the A,/P. If B is then 59 74 metres from theA,{P, find the moment changing trim.

265.The following is an exfiact from the ship's hydrostatic information:-Draft Displacement -B from the A/P

6'0tl m6'20m6'4O rn

8147 t8461 tR79), t

5E'558 m58'516 m5R.477 m

Her present drafts are 6'08 metres forward and 6 04 metres aft. She then loads260 t at 84 metres from the A/P, 430 t at 37 rnetres from the A-IP, and discharges180 t ffom 54 metres from the A,rP. What is the moment changing trim?

266.A vessel is 120 metres long and floats at drafts of 5'28 metres forward and6'14 metres aft. At this draft she displaces 7620 t, her T.P.C is l5'94, B is 58'76metres from the A,/P. whilst F is 58.02 metres from the A,tP. She then loads920 t of cargo at 67'80 metres from the A.rP. The T.P.C is then 16'18, M.C.T.lC.is 114.0, B is 58.67 metres from the A./P and I'is 57.90 metres from the A./P.Find the new drafts.

267.A ship is 164 metres long and has drafts of 8.02 metres forward and 8.10 metresaft. At this draft her displacement is 18050 t, T.P.C. is 27.80, B is 82.14 metresfrom the A,/P, whilst F is 78'94 metres from the A/P. She then:-

Loads 8 I 0 t at 69 metres from the A./P.Discharges 650 t from 1 l8 metres from the A,iP.Discharges 430 t from 64 metres from the A,?.Discharges 720 t from 56 metres from the A,{P.

At the new drafts, the T.P.C. is27'76,M.C.T.1C. is 248, B is 82'32 metres fromthe A,? and F is 78'99 metres from the A,/P. Find the new drafts.

l

204 MERCIIANT SHIP STABILIry

268.The hydrostatic information for a ship which is 122 metres long, shows:Draft Displacement M.C.T.lC. B from A,/P FfromA/P

6'00 m6'2Om640m6.60 m

8128 t8409 t8702 t9007 t

r12.4113.6114.7115.7

60.091 m60'068 m60'M5 m60'021 m

59'258 m59'169 m59'079 m58'988 m

Her present drafts are 5.47 m forward and 6'59 m aft. She then loads 520 t atI 13 m from the A/P, 330 t at 98 m from the A,/P, and 410 t at 17 m from the A/P.She also discharges 490 t from 85 m from the A/P. What are her new drafts?

269.A vessel which is 150 metres long, floats at drafts of 4'28 metres forward and4'24 metres aft. At this draft she displaces 7520 t and the centre of buoyancy is77'30 metres forward ofthe after perpendicular. She then loads:-

No. t hold: 290 t at 114 metres from the A./P.No. 2 hold; 670 t at95 metres from the A,/P.No. 3 hold: 930 t at 82 metres from the A/P.No. 4 hold: 780 t at 4l metres from the A./P.No. 5 hold: 5 l0 t at 24 metres from the A./P.Oil fuel: 400 t at 85 metres fiom the A/P.

At the ship's new displacement, the draft at F is 6.01 metres. The M.C.T.1C. is194'2 tJm, B is 7 6.43 metres ftom the A,{P, whilst F is 74'56 metres from the A/P.Find her new drafts.

270.The ship for which the hydrostatic particulars are given in the back ofthis bookis 140 metres long and floats at drafts of 2.74 metres forward and 3.60 metresaft. She then loads 1 1 00 t of water ballast into a deep tank, at 77-20 metres fromthe A/P, and sails on a voyage. During the voyage she used 380 t of fuel from80'10 metres from the A,/P, 100 t of fresh water from 50.50 metres from theA,?, and 20 t of stores from 71.40 metres from the A,/P. What were her drafts onarrival at her next port?

Answers-263. 28461 t/m by the stem.264 . 8416 tlmby the head.265. 1277 tlmby the stem.266. F. 6.26 m: A. 6.32 m.267. F.7'44 m; A.7'94 m.268. F. 6.48 m; A. 6.66 m.269. F. 5.62 m: A. 6.40 m.270. F.3.23 m; A. 3.75 m.

)

iat/P.

ndl is

PROBLEMS 205

Hydrostatic Curves and ScalesThe curves and scales given in the back of this book are for a vessel of 140

metres long; light draft 2.967 metres; summer load draft 8.094 mehes; and light KGof6'623 metres. Use them, as appropriate, to solve the following questions.

2T l.Extract all possible information from:(i) the hydrostatic particulars;(ii) the deadweight scale;(iii) the hydrostatic curves,

for each ofthe following drafts in salt water:(a) 4 20 metes;(D/ 5.50 metres;(c) 6.34 metres;(d 7'49 metes.

272.Find the displacement and deadweight at the summer draft.

273.What is the ship's light Gn4

274.Find the ship's fresh water allowance.

275.The ship is loading in a dock, where the relative density of the water is 1.012,and her present drafts are 8.04 mehes forward and 8.12 rnetres aft. How muchmore cargo can she load in order to be at her summer load line in salt water ifshe expects to use 42 t of fuel, water and stores on the way from the dock to theopen sea?

276.The vessel is loading in a tropical zone, in water of relative density 1.015.The lower edge of the tropical load line is 74 mm. above water on the port side,whilst the upper edge of the summer load line is level with the water on thestarboard side. How much more cargo can she load in order to be at her tropicalload-line in salt water?

277.The ship anchors offa port with drafts of 8.12 metres fore and aft, in water ofdensity 1.026 Vm3. On the way to her berth, she has to cross a dock sill on whichthe depth of water is 8.20 metres and the density is 1.018 t/m3. Find the leastamount of cargo which the ship must discharge into lighters, before proceedingto her berth, in order to cross the dock sill with a clearance of 15 cm, assumingthat she remains on an even keel throughout.

278.The ship has a,KG of 6.61 metres and drafts of 7.13 metres forward and 7.85metres aft. She then loads 720 t at 6.20 metres above the keel; 615 t at 9.50metres above the keel; and 430 t of deck cargo at I 1.40 metres above the keel.Find the new Gil4

, tslP.

okGS

mmhe


279.The vessel is empty of cargo and has only the following weights on board:-Fuel 630 t at 0'60 metres above the keel.Fresh water 95 t at 0'70 metres above the keel.Stores 60 t at 10'00 metres above the keel.

She then completely fills a deep tank with 1102 t of water ballast at 4'84 metresabove the keel, leaving no free surface effect. Find her GMs before and afterfilling the tank. How do you account for the change in G,142

280.The ship, which has a deck cargo of660 t of timber at 12'00 m above the keel,cornmences a voyage with drafts of 6'72 metres forward and 6'88 metres aft andKG of 7'67 metres. During the voyage she uses 480 t of fuel from 0'60 metresabove the keel, 50 t offresh water from 0'70 m above the keel, and 20 t of storesfrom 7'00 metres above the keel; whilst the timber increases its weight by 15%through absorption of water. Find her GM on arrival and the angle of loll.

281 . The ship sails with drafts of 6'83 metres forward and 6'91 mehes aft. KG of 6'77metres, and all fuel and fresh water tanks full (no free surface effect). During thevoyage she expects to use:-

8 t of stores from 10'40 metres above the keel.30 t offresh water from 9'30 metres above the keel, leaving the tank slack.(Free surface moment 104 t/m).210 t ofoil fuel from 9'30 metres above the keel, leaving the tank empty.300 t ofoil fuel from 0'80 metres above the keel, leaving the tank slack(free surface moment 378 t/m and density ofoil 0'950 Vm3).

Estimate the ship's GMon arrival at her next port.

282.When a deep tank is 100% fulI of water of relative density l'020, the ship hasdrafts of 4'04 metres forward and 4'20 metres aft, whilst her GMis 3'08 metres.920 t of water, with its centre of gravity at 5'47 metres above the keel is thenpumped out of the tank, leaving it slack. If the free surface moment of the tankis 1278 t/m, find the ship's new G,4l

283.The ship has drafts of 7'18 metres forward and 7'50 metres aft. It is desiredto load 200 tonnes of cargo, so as to maintain the after draft at 7'50 metres.Where should the cargo be placed and what will be the final draft forward?

284.The vessel is loading cargo and has drafts of 6'72 metres forward and 6'88metres aft. She is to load cargo into No. t hold (115 m from the A/P) and No. 4hold (40 m from the A/P). How much cargo must be loaded into each hold inorder to finish with the ship on an even keel with drafts of7'00 m fore and aft?

285.The ship has drafts of 4'93 metres forward and 5'57 metres aft, whilst her i(G is7.28 mehes. She then loads:-

PROBLEMS 2O7

480 t at 123 metres from the A/? and at 8'80 metres above the keel.

720 r at l0l metres tom the A/P and at 5'70 metres above the keel.

560 t at 78 metes from the A/? and at 6'30 meires above the keel.

610 t at 36 metes from the A.t? and at 7'10 metres above tle keel.

370 t at 23 mehes fiom the A./P and at 8'50 metres above the ke€I.

She then sails and on voyage uses the following firel, water and stores:-

425 t from 80 metes from the A/? and 0'70 metres above the keel.56 t from 121 mehes from the AiP and 0'80 mefes above the keel.19 t from 7l metres from the A/? and 9'00 mehes above the keel'

Find her drafts and solid GMon arrival.

Answers-

i

I

271.DisplacementDeadweightT.P.C.M.C.T.1C.LCBfromNPLCFftonNPKBKMW"

272. 15279t;10720t.273. 3'777 m.274. 164 mn.

275.269t.276. 461t.277. 284.t.278. l'19m.279. t'75m;t'18m.2E0. -0'02;5Y2".281. 0'99 m.282. 3'32m.

(") @ @ (d)6893 9534 ll33l 138902334 4975 6772 9331t9.g 20-99 2l-75 22.74140.1 l&.2 180.1 199.572.33 71.85 71.50 70.977l-17 70.06 69.20 6E022.27 2.99 3.45 4'098.66 8.16 8.03 7.99286-9 244.2 226.2 205.1

283. 86.08 m from NP;7'36 m.

284. No. l, 210 t; No. 4, 236 t.

285. F. 6'19 m,A. 6'41 m;0'53 m.


Stabitity Curves286.Use the cross curves given in Fig. 77 to find the righting levers for:-

(a) Heel 30' , Displacement 9000 t, KG 7.00 metres.(b) Heel60o, Displacement 6700 t, KG 7.50 metres.(c) Heel 45" , Displacement 7800 t, KG 6.75 metres.(d) }Ieel l5o, Displacement 5600 t, KG 7.20 metres.

287. From the cross curves in Fig. 77, find the righting levers for a KG of 7.00 metresand displacement of 5000 t. Use these to draw a curve of statical stability andfrom this 6nd the range of stability, amount and angle ofmaximum stability andthe approximate GM

288.Use the cross curves (Fig. 77) to find the GZs for a displacement of 8500 tand l(G of 6.50 metres. Draw a cuwe of statical stability and find the angle ofvanishing stability and the approximate Gr'll

289.Use the r(M curves given in Fig. 78 to find the righting levers for:-(a/ Heel 30", Displacement 6000 t, KG 6.00 metres.

@) Heel 45", Displacement 7900 t, KG 7.25 metres.(c/ Heel 15', Displacement 9800 t, iKG 6.43 metres.(d) Heel75", Displacement 8500 t,l(G 6.92 metres.

290.Use the KN curves given in Fig. 78 to find the righting levers for a displacementof7600 t and a i(G of6'78 metres. From these, draw a curve ofstatical stability,and find the range of stability and the amount and angle ofmaximum stability.

291.A ship has a l(G of 6.56 metres and a KM of 6.45 metres. Her KNs are asfollows:-

Heel 10" 15' 30' 45" 60'KN (m) 0.56 1.13 1.72 3.43 4.80 5.63

Find the righting levers, draw a curve of statical stability and find the range ofstability, the approximate angle of loll and the amount and angle of maximumstability.

Answers-286. (a) 0.37 m; (b)0.r8 m; (c) 0.79 m; (d)0.26 m.287. 85"; 1.02 m at 44' ;l-02 m.288. 85': 0.77 m.289. (a) 1'r7 m; (D)0'41 m; (c) 0.28 m; (d)0.0s m.290. 85'; 0.79 mat45".291. 57":. 12: 0.19 m at 38".

PROBLEMS

The Metacentric Diagram

292.Construct a metacentric diagram for drafts ofbetween 3'00 and 8'00 metres forthe ship for which the hydrostatic information is given in the back ofthis bookFrom this, find KB, KM ufi BM for dralls of (a),4 50 m and (b) 6'25 m.

293.Calculate the KM and BM of a box-shaped lighter, 30 metres long and6 metres beam, for every half-metre of draft from l'00 to 4 00 metres. Constructa metacentric diagram and from this find the KB, KM and BM for drafts of(a) 2'60 metres and (D) 3'40 metres.

Answers-292. fu\ KB 2'44 m r(M 8 50 m BM 6'06 n.

(b) KB 3'40 m KM8'04m BM4'64m.

293. (a\ KB 1'30m KM2'46 m BMI'16m.6\ KB l'70m KM2'59 m BM0'89m.

Bitging

294.Find the permeability ofthe following cargoes:-

(") Stowage factor 2'60; Relative density l'12

(b) Stowage factor 0'40; Relative density 8'00.(c) Storvage factor l'50; Relative density 1'75.

295.A box-shaped lighter, 30 metres long and 8 metres wide, floats at drafts of1'00 metres fore and aft. It is divided into three equal comparhnents by two

transverse bulkheads. Find the new drafts if the centre comparhnent, which is

empty, is holed below the water line.

296.What would have been the draft, in the last question, if the compartment had

been filled with cargo of permeability 40%?

297.4 box-shaped vessel is 75 metres long, 12 metres beam and floats at a draft of

6'20 metres fore and aft. A compartment amidships is 15 metres long and has apermeability of 60o10. Find the new drafts if this compartrnent is bilged.

298.A box-shaped vessel is 60 metres long, 12 metres beam and 5'75 metres deep.

She floats on an even keel at a draft of 4 80 metres. What will happen if anempty compartment amidships, 12 metres long, is bilged?

299.A ship is 120 metres long, l8 m beam and floats at a mean draft of6'00 m. The

coefficient of fineness of the waterplane is 0 750. A rectangular compartmentamidships is 15 m long, extends for the full width and depth ofthe ship, and has

a permeability of 60%. Find the sinkage if this compartment is bilged.

209

2t0 MERCHANT SHIP STABILITY

300.A box-shaped lighter, 30 metres long and 8 metres beam, floats at drafts of1'20 metres fore and aft. Find the sinkage and the new drafts if an emptycomparhnent, right forward and 3.00 metres long, is bilged.

301.A box-shaped vessel is 80 metres long, 15 metres beam and floats at drafts of3'00 metres fore and aft. Find the sinkage and new drafts if an emptycompartrnent, right forward and 12 metres long, is bilged.

302.Aship, 120 metres long, floats at drafts of4.50 metres fore and aft. The waterplanearea is 1400 square metres, the displacement is 5800 tonnes, M.C.T.lC. is 96,whilst.B is l'50 metres abaft amidships. At this draft, the forepeak, which isempty, has a volume of 50 cubic metres below water, a waterplane area of 25square metres, whilst its centre of gravity is 3'5 metres abaft the stem. Find thesinkage and change of trim if the forepeak is bilged.

303.A box shape, ll0 metres long and 12 metres beam, floats on an even keel at adraft of 5.00 metres. An empty comparanent, 10 metres long, has its centre ofgravity 30 metres forward ofamidships. Find the new drafts ifthis compartmantis bilged. M.C.T.lC. is 102.4 tonne/metres.

304.A box-shaped vessel, 72 metres long and 7 metres beam, floats at drafts of4.00metres fore and aft. An empty compartnent, right forward, is 6 metres longand has a watertight flat 3.00 metres above the keel. Find the new drafts if thiscompartment is bilged below the flat.

305.,{ box-shaped lighter is 30 metres long and 7 metres beam. She floats at a draftof 1'50 metres fore and aft. An end comparhnent, 5 metres long, has a watertightflat l'50 metres above the keel and has a permeability of 45%. Find the newdrafts ifthis compartrnent is bilged.

306,A ship is 120 metres long and floats on an even keel at a dmft of 6.00 metres.Her displacement is 8000 tonnes; M.C.T.lC. is 110; the waterplane area is 1680square metres; I and F are both 2.00 metes abaft amidships. The after peak has acapacity of240 tonnes ofsalt water and its centre ofgravity is 53 metes abaft.EFind the new &afts ifthe after peak is bilged.

Answers-294. (a) 66%; (b) 680/o; (c) 62%.

295. 1.50 m.

296. 1.15 n.

297.6.74m.

298. Vessel sinks.

299. 0.67 m.

PROBLEMS

3fi). {'13 m; F. 1'87 m; A 0.88 m.301. 0'53 m; F. 6'06 m; A. 1.66 m.

302. 3.6 cm; 3l cm.303. F.6'45 qA.4'65 m.304. F.4'93 m;A. 3'57 m.

305. F. 1'E9 m;A. l'33 m.

306. F.5'54m:A.6'70m.

Drydocking307.A ship enters a drydock with drafo of 3.00 metes forward and 3.50 mefies

aft. Her displaoement is 3000 tonnes. KG, 7.3O; GM 2.N; M.C.T.IC., 88.The ccnte of flotation is 55 metres from aft. What will be the ship's GMat theinstant of settling on tbe blocts, fore and aft?

308.In the case of the ship in the last question, what would be her GMwhen she wasflat on the blocks and the water-level had fallcn to 2.80 mehes, her displacementthen being 25fi) tonnes?

309.The ship is timmd 60 c€ntimehes by the stern when she eirters a dockHet displacement is 43fi) tonnes; KG,6.n m; KM,8.4O m; M.C.T.lC., 100;whilst the centre of flotation is 70 metres from aft. Find the GM at the instantbefore the ship comes flat on tie blocks fore and aft.

310.A box-shaped vessel is 100 metres long 12 metes beam and floats at drafo of2.210 metres forc 8nd aft. Her KG is 4.90 metres. Find her new GMwhen she isflat on the blocks in a drydock and the water-level hss fall€n so that the draft is2'00 metes fore and aft.

Ansuers-307. l'75m.

308. 0'45 m.

309. l'33 m.

310. O'27 m.

2ll

A sacrAbbreviations t<^

PAGE

f{f*9,er9"-r"orsmvity _ -'- _tll ff'#""jrtr"tJjilf;ff"n""r" : -*_.?:Archimedes, Law 4,162 Circles E7Arca-- l , 156Co€ffc ientof f i t reness___----17Areasofplanef igures_--gcoupl€___23Areas of shipshapes 9 Cross curvesareas ofwalrpunes - 9, 156 cures for heavy roltitrg - - : : - ittli;A-tDe ships - - - _ _ l3l Curyes ofstatical stability _ _ _ - - 116

INDEX

Depth with pressureDimensioas of ShipsDisplacemetrtDisplacament out of designated him

Efect ofaddioS weight at F

Deadweight - - 5,35, 16lDead\f,eight momed 45,16lDcadwcight scalevw.*wsrEx, s,_.uE ll5Deck cargo€s _ _ - _ _R,82

DefinitionsDeDsity

l6 l

BBale measurement 5

80153

Bilging - ]-23. tsl

Bilging a partially full midship conpartrnenr 125Bilging an empty eod compartrneni - _ t26Bilging an empty nidship compartDent _ 124Bifgingwitbawatertightflat _ _ _ _ l2EBlock coefrcient t7

157

D

BallastBilge keels

BM

'15 t <?

- -2

5, 161- -95BM by approximate formula - - _ _ _60

BM for box shapes _ - _ _59BM for shipshap€s _ _ _ -58BML - - - _ _ _ 90. t57BML for Box sbaped sbips - _ _ _ _91B-type ships ---__ l3 lBulkhead suMivision 148

?.! - - - - 6,3s, ts7, t6 lDnftatF - _ _ _ _ _ 16lDrydocking - - - - l42,t51Dynamic srability _ g5,87, 16lDynamic stability from GZ curves _ _ -85

E

ccapaci typlansrr /EtrectofdeDsityondraf t - - - -37,145cereoriuoyancy - - ;r.iil "t$1:flf,:'gll'tabiritv

-^- - - t4s;#;ffifft- - - -

-_ -- 4e't:: llectoffltag-tanl*oncenteorgravity -47

Cedrc of sravirw _

-: : Effect of KG oD saatical stability curves - I I 9cetrrcorsravity

?1,1? #;;;;il;ruii"Lilt:ff"_ i;;ccnteorgravityorabody 24,27 4;;ffiff;1,il;weishtsontrim l0l:3X::*:E::l:1,1**- - - -2s Etr*,"i"h*;;;;;;;;"_. _ - _e8cenre ofgravity ofa v.aterplane _ _ _ -28 Ed tbril--" -

":':":"--'_ ,r,rr.,il

214

Height of metacentrcHyfuostatic data curves - - -:Hydrostatic data scales - - - - - -Hydrostatic particulars

F PAGE

F49Factors affecting statical stability - - - -jsFloat ingbodies-----4FluidGM - - - - - - 16lFluid KG - - - - - - 16lForce - - 1,19, 16lFormulae - - - - - - 156Free liquid in Oil tankeN - - - - - -84Free liquid in taDks - - - -83Free surface effect - - - -69Fr€e surface emptying t""ts - - - - -ilFree surface filling taDks - - - - - -'llFreesurfaceindividedtanks - - - - - i2Frce surface moments - - -74Free surface ofliquids - 158, 16lFrse suface ofrectatrsular tarks - - - - 7 l

6, 16lFreeboardFresh watcr allowance - - 37,38,146,16l

GG 16lG out ofcenheline - - - -63GM 55,57,58, 158GML - - 56,89Orain measwemetrtGross toDnageGroudingGz--- - - - 55,57, 158GZ curves - - - - - -76

H156161lt4114l l5

IInclining experime - - -60Inertia - - - - - - 33, 16lInitial stability - - 57, 58, 16lIsochronous rolling 162

L PAGELayer conection 158Length--- lLight displacemeot -

INDEX

s, t62LighrKG - - - - - - -42

LoadDisplacement----5t62

List

Loaded displacement - -Loading for a constant draft affLoading for a desired draftLoading for a desired trim -Loading !o a givel loadline

Metacente - -Metacentric diagamMetacentric height

Moment combination -Moment of inertia - -Moment of statical stability

63,64,65,66

108106

- - 106, 107- - - -40

Loading to produce a desired draft aft - 109Loadline rules l3lLoadline rules requirements - - - - 131Loadlines - - - - - - - 6Loll - 63,68, 156Loll due to rcgative GM - - - - - -67Longitudinal metacentric height - 56,89,162Longitudinal stability - - -89Longitudinal watertight bulkheads 147

M

5

145

MMaximum weight to load for a given draftMCTIC---__96,Mean draft

55l l0158162

Heel

Metric system - - - - - IMidship coefficient - - - - 16Moment - 1,19,20,162Moment about after perpendicular, trim - lll

- 53,55,162t2l

- _ - _2133,34,35, 158- - 57,162

Moment properties - - - -21

KBKGKN

49t62t62

K

Moment to chanee tim I cm 96, t62

NNeft tonnage------5Neutral equilibrium 54,56

PPeriod ofa ship - - - l62,t ' lPeriod ofwave 162Period ofwaves, appateot - - - - - l5lKN curyes - - - - l20,l2l

INDEX 2t5

PAGE

- r1254 55

PACE

Period ofwaves, true l5lPermeability - - - - 123, 158Positioning weights - - - -77Pressure - l, 159PressureaDddepth----2Presswe on bulkheads 148Prismatic Mies lAPrismatic coefficient - - -17Prismatic wedgcs - - - -25

R

Radius ofgyration - - - -33Rangc ofstability - - 5'1,163Relative deDsity - - - - - 2Removing weighb, cent€ ofgravity - - -26Reserve buoyaacy - - 147,163Resistanc€ to rolling 152Resultant forc€s - - - - -19Righting lwer - -

Stability infomation supryrlied to shipsStable equilibrium

Rolling

sSheer-------148ShiffofB - - - - - 50, 156Shift ofB horizontal - - -51Shift ofB vedical - - - -51Shif tofc ---- 42,158Shifting weights, centre ofgravity - - -26Ship scctions - - - - - - 9Simplified stability

infomation 137, 138, 139, l,!0, l4lSimpson's mles - - - - - 10Sirnpson's rules for intemediate odimtes 15,31Simpson's rules. Appendages

Statical stability 159Stltical stability at small angles ofhecl - - 62Stiffships - - - 77,78,163Surface are{s 8, 159Syncbronism - - - - l5l, 163

TTender ships - - - 77,18,163Timbq dcck cargo€s - - -82Timberloadl ines----7Toonage - 4,163Tonn€s per centimebe inunersion 163TPC - - - - - - 39, t59TPC, etrect ofdensity - - -40Transverse statical stability 57,75Trim - - - - - 91, 159, 163

53,55. 163 Trochoidat theory 150150

U(JMS toDnage - - - - - 5Unresisted rolling 152Unstable equilibrium - 54,56Unstable ship's - - - - -79

Si*age 159SolidGM - - - - - - 163SolidKG - - - - - - 163Sounding pipes -----2Stability Informatior 134Stabilityinformationbooklct - - - - 132

Usc of hydrostatic dat& curves l l5

vVanishing stability 16lVidual CetrE€ ofgravity - - - - - -46Volumes - - - - -1,8, 159Volumes ofship ehapes ----

wWall sided fonnula - - - -AWaterplane coefrcient - - -17Weiebt --- lWetted surface arca - - 18, 160Wingirg out wcight - - - -81Wor* - - - - - - - -85

l5

HYDROSTATIC CURVES & SCALES

I{YDROSTATIC PARTICULARS(in Salt Water)

DRAFT

Metres

DISPLAC.EflE1{T

Tonnes

T.P.C.l.TonneSPer cm.

lmmolsion

[.c.T.1C.lSoment

to ChangeTdm

ono cm.

L.C.B.Fwd. of

A.P.

Metres

L.C.F.Fwd. of

A.P.

Metres

v.c.B.AB.

Metres

Kil.fi.)Tramver3eIetac€ntro

A.B.

Metres

K.r. (1.)Longl.

t$etacenrgAB.

Metres

2.40 JCCU 17.38 109.3 72.79 72.50 1.29 11.62 427-4.60 3901 17.68 112.6 72.76 72.36 1.40 11.16 402-8.80 4257 17.96 1160 72.72 72.22 1.51 10.73 382.1

3.00 4619 18.21 119.4 72.64 72.07 1.62 10.33 363.5.20 4986 18-47 122-8 IZ05 71.93 1.73 9.94 346.9.40 5358 '18'72 126.2 72.57 71.78 1.U 9.58 332.8.60 5735 18.97 129.6 72.51 71.63 1.94 9.26 319.3.80 6't17 19.20 133.1 72.45 71.48 2-05 9.01 306.9

4.00 6502 19.42 136.6 72.39 71.33 2.16 8.82 ZYO'O

-20 6893 19.64 140.1 72.33 71.'17 2.27 8.66 286.9.40 7296 19.86 143-7 72.26 7't.o1 2.38 6.CC 278.2.60 7696 20.o7 147.3 72.'19 70.8s 2.49 8.45 270.3.80 8107 20-28 150.9 72.12 70.68 2.60 8.36 263.5

5.00 8496 20.50 154.6 72.05 70.51 2.71 8.29 257.5.20 8908 20.69 158.4 71.97 70.33 282 6'ZJ 251.9.40 9323 20.89 't62.2 71.89 70.15 2.93 8.18 246.6.60 9744 21.09 1661 71.81 69.96 3.04 8.13 24'1.7.80 10167 21'27 170.0 71.73 69.76 3.15 8.09 237.1

6.00 10596 21.44 173.8 71.65 69.56 326 8.06 232.9.20 11027 21.63 177.5 71.56 69.35 3.37 8.04 225.9.40 11461 21.80 18't.2 71.48 69.14 3.49 8-02 225.O.60 11899 2'1 98 184.8 71.39 68.92 360 8.00 221.2

.80 12UO 22-15 188.3 71.29 68.71 3.71 7.qq 2'17.4

7.00 12786 22 32 19't.7 71.20 68'50 3.82 7.99 213.7.20 13234 22-49 1950 71.11 68.30 3.93 7.99 210.1.40 13686 22.66 198.1 71-O1 68.11 4.M 799 206.6.60 't4140 22.83 201.2 70.91 67.92 4.15 8.00 203 2

'80 14599 23.00 204 2 70.82 67.73 4.26 8.01 200.0

8.00 15060 23.17 207-1 70.72 67 54 4.37 8.02 196.9.20 15525 23.U 210.0 70.62 67.36 4.48 8.03 194.1.40 '15994 23.51 212.8 70.52 67.19 4.59 8.0s 191.2.60 16467 23.68 2't5.5 70.41 67.03 4.71 8.07 188.4

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Book-Merchant Ship Stability

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