Ship Stability Trim

r5

Trim

l. Trim TRtu may be consideredas the longitudinal equivalent of list, but instead of being measuredin degreesit is measuredby the differencebetweenthe drafts forward and aft. Consider a ship to be floating at rest in still water and on an even keel as shown in figure 84.

w

--

d - - +

_l=lw

G.F GI

w

B

\

)

w Fig, 84

The centre of gravity (G) and the centre of buoyancy (B) will be in the same vertical line and the ship will be displacingher own weight of water.

r27

t28

TRIM

Now let a weight "w", alreadyon board, be shifted aft through a distance"d", as shown in figure 84. This causesthe centre of gavity of the ship to shift from G to G, , parallel to the shift of the centre of gravity of the weight shifted, so that:-

GG,:HX d or WX GGt= w X d A trimmingmomentof W X GG, is therebyproduced. B u t W X G C T= w X d Thetrimmingmoment: w X d

w

M

--

G,

B \

w Fig. 85

TRIM

t2g

The ship will now trim until the centresof gravity and buoyancy are againin fhe same vertical line, as shown in figure 85. When trimmed, the wedge of buoyancy LFLr emergesand the wedge WFWr is immersed. Since the ship, when trimmed, must displacethe sameweight of water aswhen on an evenkeel. the volume of the immersedwedgemust be equal to the volume of the emerged wedge and F, the point about which the ship trims, is the centre of gravity of the waterplane area. The point F is called the "Centre of Flotation" or "Tipping Centre". A vesselwith a rectangularwater-planehas its centre of flotation on the centre line amidshipsbut, on a ship, it may be a little forward or abaft amidships,dependingupon the shapeof the water-plane. In trim problems, unlessstatedotherwise.it is to be assumedthat the centreof flotation is situated amidships. Trimming moments are taken about the centre of flotation sincethis is the point about which rotation takesplace. The longitudinal metacentre(Ml) is the point of intersection oetweenthe verticalsthrough the longitudinal positions of the centresof buoyancy. The vertical distancebetweenthe centre of gravity and the longitudinalmetacentre (GMr-)is called the longitudinalmetacentricheight. BMlis the height of the longitudinal metacentre above the centre of buoyancy and is found for any shape of vesselby the formula: BMr whereIt

and V

= IT V = the longitudinal secondmoment of the water-planeabout the centreof flotation, : the vessel'svolume of displacement.

The derivation of this formula is similar to that for finding the transverseB.M. (page97). For a rectangularwater-planearea: BL3 , 11 =12 where L and B

= the length of the water-plane, = the breadth of the water-plane.

Thus, for a vesselhavinga rectangularwater-plane:

BMr = ?*

Jl|.

rRrM

t30 For a box-shapedvessel: Ir BMr ='-! V

- BL'

r2v BL3

IzXLXBXd BMI :* where L and d

tzo

= the length of the vessel, :

the draft of the vessel.

-

IL

For a triangularprism: BMr

V

:@ BMr :*

BL3

12

od

It should be noted that the distanceBG is small when compared with BMp or GMr and, for this reason,BM1 ftay, without appreciableerror, be substituted for GM1 in the formula for finding MCT I cm.

2. The Moment to Change Trim one centimetre (MCT I cm. or MCTC) The MCT I cm., or MCTC, is the moment required to changetrim by I cm., and may be calculatedby using tfre formula:

McTtcm. - \GMr l00L where W GMr L

= the vessel'sdisplacementin tonnes = the longitudinal metacentricheight in metres,and : the vessel'slength in metres.

The derivation of this formula is as follows:Consider a strip floating on an even keel as shown in figure 86 (a). The strip is in equilibrium.

r3l

TRIM

I

w

I

rut

I

G \

B

L )

w Fig. 86 (a)

Now shift the weight 'w' forward through a distanceof 'd'metres. The strip'scentre of gravity will shift from G to G, , causinga trimming moment of W X GGr, as shownin figure 86 (b).

Fig. 86 (b)

The ship will trim to bring the centresof buoyancy and gravity into the same vertical line as shown in figure 86 (c). The ship is alain in equilibrium. Irt the ship's length be L metresand let the tipping centre (F) be I metres from aft. The longitudinal metacentre(Ml) is the point of intersectionbetweenthe verticals through the centre of buoyancy when on an even keel and when trimmed.

TRIM

r32

w

vL

H

,___ | HI

w_j

--

e

|

'/J t

G

L

Gl

AF; v Fig. 86 (c) .1r\WXd GGr

and GG, :

Tan0 : butTan0

GML Tan 0

wXd

WXGML

:L

(See figure 87(b) ) kt the changeof trim due to shifting the weightbe one centimetre.Then w X d is the moment to changetrim one centimetre. .'. Tan 0 but Tan 0 Tan 0 or

MC T I c m . wxGML

and MCT I cm.

I l00L wX d WX GML MCT I cm. WXGML I t00L WX GML l 00L

TRIM

I33

3. To find the change of draft forward and aft due to change of trim When a ship changestrim it will obviouslycausea changein the drafts forward and aft. One of thesewill be increasedand the other decreased.A formula must now be found which will give the changein drafts due to changeof trim. the Considera ship floating upright as shownin figure 87 (a). F, represents position of the centre of flotation which is i metresfrom aft. The ship'slength is L metresand a weight 'w' is on deck forward.

.w

I

w I

\

I

) Fig. 87 (a)

'd'metres. The ship will trim I-,etthis weight now be shiftedaft a distanceof 't'cms. by the sternasshownin figure 87 (b). about F1 and changethe trim

- - x TCi r

Fig.87(b)

Wt C is a line drawn parallelto the keel. 'A' representsthe new draft aft and 'F' the new draft forward. The trim is therefore equal to A - F and, sincethe original trim was zeto, this must also be equalto the changeof trim. [€t 'x' representthe changeof draft aft due to the changeof trim and let 'y' representthe changeforward. In the trianglesWW Fr and W, Ia C, using the property of similar triangles:

t34

TRIM

x cm.

= t cm. tr:

,m-

/m-Ll cm. orxcm. .'. Change of draft aft in centimetres : I t Change of trim in centimetres L where / = the distanceof centre of flotation from aft in metres,and L : the ship'slength in metres. It will alsobe noticed that x * y : t Changeof draft F in cms. : Changeof trim - Changeof draft A.

4. The effect of shifting werghtsalreadyon board ExampleI A ship l26m.long is floating at drafts of 5.5m. F and 6.5m. A. The centre of flotation is 3 m. aft of amidships.MCT I cm. = 240 tonnes-m.Dsplacement= 6,000 tonnes. Find the new drafts if a weight of 120 tonnesalreadyon board is sttifted forward a distanceof 45 metres.

Trimmingmoment

^, F.. Lnange or rnm

:wXd -

l20X 45

:

5400 tonnes-m.by the head

= Trimminemoment ffi 5400 240 :

Change of draft aft

22.5 cm. by the head

: /

x Ctrarge of trim

-_4 x 22.s 126 : change of draft forward : : :

l 0 ' 7 c ms . change of trim - change of draft aft 2 2 . 5 - 1 0 . 7c m . I1.8 cms.

{

135'

TRIM

Original drafts Changedue to trim Ans.

5'500m. F + 0.118 m.

6'500m. A - 0 ' 1 0 7m .

5.618 m. F

6.393m. A

New drafts

Example 2 A box-shapedvessel90 m. X l0 m. X 6 m. floats in salt water on an evenkeel at 3 m. draft F and A. Find the new drafts if a weight of 64 tonnesalreadyon board is shifted a distanceof 40 metresaft. BMr

-

lj

r2d 90x 90 12x 3 BMr = 225m.

Fig. 88

$1/=

\{:

cUL-T00r Mcr l cm. - wI

L X B X d X I.025 90x10x3x1.035 2767'5tonnes

Since BG is small compared with GM1 :-

Mcrlcm.

^ wXBMt 100L. 2767.5X 225 : -JboFo.-

MCT I cm. Changeof trim

:

wXd MCT I cm 64x 40 69.19

Changeof trim

: 37 cm.by the stern

Changeof draft aft

- +I X Chanseof trim

Changeof draft aft

-- l8'5 cm.

L

- +x37cm.

69'19tonnes-m.

136

TRIM

Changeof draft forward

= l8'5 cm.

Originaldrafts 3.000m. F Change duetrim - 0'185m. Ans. New drafts

2 ' 8 1 5m .

3'000m. A + 0.185 m. 3 ' 1 8 5m .

5. The effect of loading and/or discharging weights When a weight is loaded at the centre of flotation it will produceno trimming moment, but the ship's drafts will increaseuniformly so that the ship displaces an extra weight of water equalto the weightloaded.If the weight is now shifted forward or aft away from the centre of flotation, it will causea changeof trim. From this it can be seenthat when a weight is loaded away from the centre of flotation, it will causeboth a bodily sinkageand a changeof trim. Similarly, when a weight is being discharged,if the weight is first shifted to the centreof flotation it will produce a changeof trim, and if it is then discharged from the centre of flotation the ship will rise bodily. Thus, both a changeof trim and bodily rise must be consideredwhen a weight is being dischargedaway from the centre of flotation. Example I A ship 90 m. long is floating at drafts 4.5 m. F and 5.0 m. A. The centre of flotation is 1.5 m. aft of amidships. TPC l0 tonnes. MCT I cm. 120 tonnes-m. Find the new drafts if a total weight of 450 tonnesis loaded in a position 14 m. forward of amidships.

Fig. 89

TRIM

137

w TPC

Bodily sinkage

450 t0 Bodily sinkage

= 45 cm.

Changeof trim

= -Trim M c r moment lr.m_ 4 5 0x r 5 . 5 t20

Changeof trim

_ 5 8 ' 1 2 c m. by the head

Changeof draft aft

- : x Chaneeof trim L

I

4?.5 = -ff x s8.r2

Changeof draft aft

:

28.09 cm.

Changeof draft forward :

Changeof trim - Changeof draft aft = 5 8 ' 1 2 - 2 8 . 09

Changeof draft forward :

30.03 cms.

Original drafts 4.500 m. F Bodily sinkagea 0'450 m.

4'950m. Change due trim + 0.300m. Ans.

Newdrafts

5.250m. F

5'000 m. A + 0.450 m.

t-** - 0'281 m. 5'169m. A

Note. In the event of more than one weight being loaded or discharged,the net weight loaded or dischargedis used to find the net bodily increaseor decrease in draft, and the resultant trimming moment is used to find the changeof trim. Also, when the net weight loaded or dischargedis large, it may be necessary to use the TPC and MCT 1 cm. at the original draft to find the approximatenew drafts, and then rework the problem using the TPC and MCT t cm. for the mean of the old and the new drafts to find a more accurateresult. Exomple 2 A box-shapedvessel40 m. X 6 m. X 3 m. is floating in salt water on an even keel at 2 m. draft F and A. Find the new drafts if a weight of 35 tonnesis discharged from a position6 m. from forward. MCT I cm.: 8.4 tonnes-m.

138

TRIM

1.0?sA TPC =

100 1 . 0 2xs 4 0 x 6

100 : 2'46 tonnes TPC 90 w

Bodily rise

rpc 35 2.46 l4'2 cms. ulXdMCTI cm.

Bodily rise Changeof trim

35X t+ 8.4 Changeof trim

:

58'3 cm. by the stern

Changeof draft aft

:

- X Chaneeof trim

I

L

- + x 5 8 . 3c m s . Changeof draft aft

:

Changeof draft forward

:

Changeof draft forward

= 29'15 cms.

29.15cms.

Changeof trim - Changeof draft aft _ 5 8 ' 3 - 2 9 ' 15

Original drafts 2.000 m. F - 0'140 m. Bodily rise Changedue trim Ans. New drafts

1 ' 8 6 0m . - 0'290m. l ' 5 7 0m . F

2'000m. A - 0 . 1 4 0m . 1 ' 8 6 0m . + 0.290m. 2'150m.A

Example 3 A ship 100 m. long arrivesin port with drafts 3 m. F and 4'3 m. A. TPC l0 tonnes. MCT I cm. 120 tonnes-m. The centre of flotation is 3 m. aft of amidships. If 80 tonnes of cargo is loaded in a position 24 m. forward of amidshipsand 40 tonnes of cargois dischargedfrom 12 m. aft of amidships,what are the new drafts?

TRIM

139

50m ! l

24m

-

l-r?t

\

-I

L

Bott

Fig.9l

Cargoloaded Cargodischarged Net loaded

80 tonnes 40 tonnes 40 tonnes

Bodily sinkage

: :

#a 40 l0

Bodilysinkage :4cms. To find the changeof trim take momentsabout the centreof flotation.

Moment to changetrim by

Weight Distancefrom C.F.

80 40

head

27 9

stern

2t60 360

2s20

change oftrim

:

InHfHlt _ ?!4_

Change of trim

t20 : 2l cms.by thehead

Change of draft aft

: -

Change of draftaft changeof draft forward

I

x Changeof trim

4\xzl 100

- 9.87cms. - changeof trim - changeof draft aft

TRIM

140

= 2l - 9'8'l = I I ' 1 3c m s . Changeof draft forward 4 ' 3 0 0m . A . 3'000 m' F . Originaldrafts 0'040 m. m' 0'040 BodilYsinkage 4340 n. 3 ' 0 4 0m. - 0'099 m. * m' I I 0'l Changedue trim 4 ' 2 4 1 m .A . 3 ' 1 5 1m. F . A ns . New dr a fts Example 4 MCT I cm' A ship of 6,000 tonnesdisplacementhas drafts 7 m. F and 8 m. A' tonnesof 500 amidships. is 100 tonnes-m., TPC 20 tonnes, centreof flotation holds: from eachof the following four cargois discharged No. I hold, centreof gravity40 m. forward of amidships ') " " " " 25m. No.2hold. " ,t tt )' )' aft " 20 m. No.3 hold, ,, , t ') ') ') " 50m. No.4 hold, The following bunkersare alsoloaded: 150tonnesat l2 m. forwardof amidships 50 tonnesat 15 m. aft of amidshiPs Find the new drafts forward and aft.

Fig.92

Total cargodischarged Total bunkersloaded Net weight discharged

2000 tonnes 200 " I 800 tonnes

Bodily rise :

w

tPc 1800 20

Bodilyrise : 90 cms.

l4l

TRIM

Moment to changetrim by

Iileight Distancefrom C, F.

500 s00 s00 s00 150 50

stern

head

40 25 20 50 t2 l5

20000 I 2500

10000 2s000 I 800

750 36800 332s0

33250

35s0

Resultantmoment 3550 tonnes-m.by the head Changeof trim

moment = Trim MCT I .*

= Changeof trim

3550

loo

= 35'5 cms.by the head

Sincecentre of flotation is amidships, : Changeof draft forward Changeof draft aft :

I changeof trim 1 7 . 7 5c m s .

Originaldrafts Bodily rise Chang ed u e tri m Ans.

New draft,

7.000 m. F. 0.900m. 6 . 1 0 0* + 0 ' 1 8 0m . OZSOtn f .

g.000 m. A. - 0.900m. 7.100.. -- 0.180m. e CO rn. n.

.I-RIM

142

Example 5 A ship arrivesin port trimmed 25 cms. by the stern. The centreof flotation is amidships. MCT I cm. 100 tonnes-m.A total of 3,800 tonnesof cargois to be dischargedfrom 4 holds, and 360 tonnes of bunkers loaded in No. 4 double bottom tank. 1,200tonnesof the cargois to be dischargedfrom No. 2 hold and 600 tonnesfrom No. 3 hold. Find the amount to be dischargedfrom Nos. I and 4 holds if the ship is to completeon an evenkeel. Centreof gravity of No. I hold is 50 m. forward of the centreof flotation "No.2 " "30m. )t

,'

)'

)t

,t

,,

"

"

))

"No.3

"

"20m.abaft

t,

tt

tt

tt 45

No.4

t'

')

tt

t'

m.

"

"No.4DBtankis5m."

(29oo-x)t 6oot

*:i'li

tt

"

"

"

)t

t)

tt

')

"

"

(x)t

l2OOt

discharged ",." I ffi^ IT Nos.I &4 2000tonnes ":-ilJ'ii from No. I hold tonnesof cargobe discharged

'x' bt "(2000-x) "

"

"

"

"

)' No.4 "

Takemomentsaboutthecentreof flotation. Weight Distancefrom C, F.

x l 200 600 2000-x 360

50 30 20 45 5

Moment to changetrim by head

stern

50x 36000 l 2000 9000H5x 1800 102000-45x

37800+50x

T R IM

143

: 25 cms.by the stern -0

Original trim Requiredtrim Changeof trim required

:

25 crrr. by the head

Trimmingmoment required

:

Changeof trim X MCT I cm.

:25X Trimming moment required : Resultantmoment

:

100

2500 tonnes-m.by the head Moment to changetrim by head - MCT by stern

2500 - (102,000- 45x) - (37,800+ 50x) : :

102,000- 45x - 37,800- 50x 64,200- 95x

: 61.700 . or 95x x : 649.5 tonnes and2000-x : 1 3 5 0 ' 5to n n es

Ans. Dscharge649'5 tonnesfrom No.l hold and 1350.5tonnes from No.4 hold.

6.

Using trim to find the position of the centre of flotation

Example A ship arrives in port floating at drafts of 4.50 m. A and 3:g0 m. F. The following cargo is then loaded: 100 tonnesin a position 24 m. aft of amidships 30 ') " " " 30 m. forward of amidships 60

t)

t'

tt

tt

l5m.

tt

tt

"

The drafts are then found to be 5.10 m. A and 4.40m. F. Find the position of the longitudinal centre of flotation relativeto amidships.

Fig.94

144

TRIM

O r iginaldr af ts 4 .5 0 m.A 3 .8 0 m. F g i v e0 .7 0m. tri m by the stern. New dr af t s5' 1 0 m . A 4 .4 0 m . F g i v e0 .7 0 m . tri m by the stern. Thereforetherehasbeenno changein trim, or, The moment to changetrim by the head : The moment to change trim by the stern' L,etthe centreof flotation be L'metres aft of amidships. Then,

100(2a- x) 2,400 - 100x

_ 30 (30 + x) + 60 (15 + x) :

190x : x

:

900 + 30x * 900 + 60x 600 3'16

Ans. Centreof flotationis 3'16 metresaft of amidships. Note. In this type of questionit is usualto assumethat the centreof flotation is aft of amidships,but this may not be the case. Had it been assumedthat the centre of flotation was aft of amidshipswhen in actualfact it wasforward, then the answerobtainedwould havebeenminus.

7. Loading a weight to keep the after draft constant When a ship is being loaded it is usually the aim of those in chargeof the operationto completeloading with the ship trimmed by the stern. Shouldthe ship's draft on sailingbe restrictedby the depth of water over a dock-sillor by the depth of water in a channel,then the ship will be loadedin sucha manneras to producethis draft aft and be trimmed by the stern. Assumenow that a ship loaded in this way is ready to sail. It is then found that the ship has to load an extra weight. The weightmust be loadedin sucha position that the draft aft is not increasedand also that the maximum trim is maintained. lf the weight is loadedat the centreof flotation, the ship'sdrafts will increase uniformly and the draft aft will increaseby a number of centimetresequal to w/TPC. The draft aft must now be decreasedby this amount. Now let the weight be shifted through a distanceof 'd'metres forward. The ship will change trim by the head, causing a reduction in the draft aft by a number of centimetresequalto llL X Changeof trim. Therefore, if the samedraft is to be maintainedaft, the abovetwo quantities must be equal.

TRIM

i.e. I

L

^ '-x chanse ' of t r l m

But,

Change of trim

t45

w

=TPC

:

##.

f'#fu :#. d =Tffif*

or whered : L : / -

the distanceforward of the centre of flotation to load a weight to keep the draft aft constant, the ship'slength, and the distanceof the centre of flotation from aft.

Example A box-shapedvessel60 m. long, 10 m. beam, and 6 m. deep,is floating in salt water at drafts 4 m. F and 4.4 m. A. Find how far forward of amidshipsa weight of 30 tonnes must be loaded if the draft aft is to remain at 4.4 m.

'.025A

TPC::ib0=

_ 1 . 0 2 x5 6 0 x t 0 100 TPC = 6'15 tonnes W:LXBXdXI'025tonnes = 6 0 X l 0 X 4 . 2X l ' 0 2 5 W = 2583tonnes BMr:L

r2

r2d

- 60X60 1 2x 4 . 2 BMr = U metres 7 wX M C T l c m . = - i * f -B M I _

2583X 500 100x 60x 7

TRllvl

t46

MCTI cm. :

L X M C T1 c m . /XTPC

d: : d:

12314tonnes-metres.

60 Yry

4 30 l0 metres

x __l

6.15

Ans. Load the weight l0 metresforward of amidships. 8. Loading a weight to produce a required draft Example A ship 150 metreslong arrivesat the mouth of a river with drafts 5.5 m. F and 6'3 m. A. MCT I cm. 200 tonnes-m. TPC 15 tonnes. Centreof flotation is l'5 m. aft of amidships. The ship has then to proceedup the river where the maximum draft permissibleis 6'2 m. It is decidedthat SW ballastwill be run into the forepeaktank to reducethe draft aft to 6'2m. If the centreof gravity of the forepeak tank is 60 metresforward of the centre of flotation, find the minimum amount of water which must be run in and also find the final draft forward.

:ffi

-----

w \f

60m

FF

|

\ w Fig.95 (a) Load

'w'

tonnes at the centre of flotation

Bodily sinkage =

w TPC w t*' 15

New draft aft

:

o'J m. +

Requireddraft aft

:

6.2 m.

Reduction required = 0.1 m. *

w ra. 15

15

t-'

,l

TRIM (b ) Shift

'w'

tonnes from the cente of flotation

Changeof trim

=

t47 to the forepeak tank

wX d

Mcr I c;. 60w 200

Changeof trim

Changeof draft aft due to trim

: 3w cms.by thehead l0 I

- + X Chaneeof trim L

=

Changeof draft aft dueto trim But changeof draft requiredaft 0'417w

7 3 ' 5. . 3 w 1 5 0" 1 0

-x

- 0 ' 1 4 7w c m s . = (' 1 0* l l . ') . * r . 15

: l0++ l)

2'205w = 1 5 0* w 1.205w = 1 5 0 w = 124'5tonnes Thereforeby loading124.5tonnesin the forepeaktank the draft aft will be reducedto 6.2 metres.

(c) To find the new draft forward Bodily sinkage -

w TPC

:

wl 5

Bodily sinkage : 8'3 cms. wXd Changeof trim : MCT t c*. __ 124'5X 60 200 Changeof trim = 37.35cms.by the head

148

TRtM

Changeof draft aft due trim

:

Change of draftaft duetrim : Changeof draft forward due trim

Changeof draft forward due trim Originaldrafts 5.500m. F Bodily sinkage* 0.080 m.

tatt * Changedue t r i m New drafts

* 0 ' 1 9 0m . 5.770m. F

:L

ffi

X Changeof trim

X 37.35

: l8'3 cms. - Changeof trim - Changeof draft aft : 3 7 ' 3 5 - 1 8 ' 3c m . :

19'05 cms. 6.300m. A + 0.080 m.

aa* t* - 0 .180m. 6.200m. A

Ans. Load 124'5 tonnesin forepeaktank. Final draft forward is 5'770 metres.

9. Usingchangeof trim to find the longitudinal metacentricheight (GMr). It was shown in Section 3 of this chapterthat, when a weight is shifted longitudinallywithin a ship, it will causea changeof trim. It will now be shown how this effect may be usedto determinethe longitudinalmetacentricheight. Considerfigure 96 (a) which representsa ship of length "L" at the waterline, floating upright on an even keel with a weight on deck forward. The centreof gravity is at G, the centre of buoyancy at B, and the longitudinalmetacentreat M1. The longitudinalmetacentricheight is thereforeGMs.

IRIM

t49

w M1

G

W

L

B

\

w Fig. 96 (a)

w

ML

-tG,

W

G B

\ w

Fig. 96 (b)

L

150

TRIM

w

ML

o"

-1' \e' A

Gr

\,

B,

B

I

L,

.LI

-f'I 'F

\ tv Fig. 96 (c)

Now let the weight be shifted aft horizontally asshown in figure 96 (b). The ship's centre of gravity will also shift horizontally, from G to Gt, producinga trimming moment of W X GG, by the stern. The ship will now trim to bring G1 under M1 as shown in figure 96 (c). In figure 96 (c) Wt Lt represents the new waterline,F the new draft forward, and A the new draft aft. It was shown in figure 87 (b) and by the associated notes, that F - A is equai to the new trim (t) and since the ship was originally on an evenkeel, then 't' must alsobe equal to the changeof trim. If the angle betweenthe new and old verticalsis equal to 0, then the angle between the new and old horizontalsmust also be equal to 0 (the anglebetween two straightlines being equal to the anglebetweentheir normals). It wiil also be seen in figure 96 (c) that the trianglesGG1M1 and CDE are similar triangles.

.

GMr G Gt

or

GMI

: L

r

:

I

:XGG, t'

(All measurements are in metres).

TRIM

l5t

Example I When a weight is shifted aft in a ship 120 metreslong, it causesthe ship'scentre of gravityto move 0'2 metreshorizontallyand the trim to changeby 0.15 metres. Find the longitudinalmetacentricheight. GMI G G,

GMr_

:- L t

: fx cc, t20 x 0-2 0.15

Ans. GMr-

1 6 0m e tre s

Example 2 A ship 150 metres long has a displacementof 7,200 tonnes, and is floating upright on an evenkeel. Whena weight of 60 tonnes,alreadyon board,is shifted 24 metres forward, the trim is changedby 0'15 metres. Find the longitudinal metacentricheight.

GMr- :

I

GGt

t

GMr-

G G' ,t X "

I

wXd Wt

*

60x 24 7200 Ans. GMr_

L

150 0.l s

200 metres.

EXERCISE15 l. A ship of 8,500 tonnesdisplacement has TPC 10 tonnes, MCT I cm = 100 tonnes.m. and the centre of flotation is amidships. She is completingloading under coal tips. Nos.2 and 3 holds are full, but spaceis availablein No. I hold (centreof gravity 50 m. forward of amidships),and in No. 4 hold (centre of gravity 45 m. aft of amidships). The present drafts are 6'5 m. F and 7 m. A, and the load draft is 7.1 m. Findhow much cargois to be loaded in each of the end holds so as to put the ship down to the load draft and complete loadingon an evenkeel.