r5 Trim

l. Trim

TRtu may be considered as the longitudinal equivalent of list, but insteadof being measured in degrees it is measured by the difference between the draftsforward and aft.

Consider a ship to be floating at rest in still water and on an even keel asshown in figure 84.

Fig, 84

The centre of gravity (G) and the centre of buoyancy (B) will be in the samevertical line and the ship will be displacing her own weight of water.

w

w

d- - - - + _l=lw

GIG.F

\ B )

w

r27

t28 TRIM

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

X dGG,:H

or WX GGt= w X d

A trimming moment of W X GG, is thereby produced.

B u t W X G C T = w X d

The trimming moment : w X d

w

M

--G,

\

B

wFig. 85

TRIM t2g

The ship will now trim until the centres of gravity and buoyancy are again in

fhe same vertical line, as shown in figure 85. When trimmed, the wedge of

buoyancy LFLr emerges and the wedge WFWr is immersed. Since the ship,

when trimmed, must displace the same weight of water as when on an even keel.

the volume of the immersed wedge must 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 vessel with a rectangular water-plane has its centre of flotation on the

centre line amidships but, on a ship, it may be a little forward or abaft

amidships, depending upon the shape of the water-plane. In trim problems,unless stated otherwise. it is to be assumed that the centre of flotation is situated

amidships.Trimming moments are taken about the centre of flotation since this is the

point about which rotation takes place.

The longitudinal metacentre (Ml) is the point of intersection oetween the

verticals through the longitudinal positions of the centres of buoyancy. The

vertical distance between the centre of gravity and the longitudinal metacentre(GMr-) is called the longitudinal metacentric height.

BMlis the height of the longitudinal metacentre above the centre of

buoyancy and is found for any shape of vessel by the formula:

where It = the longitudinal second momentof the water-plane about thecentre of flotation,

and V : the vessel's volume ofdisplacement.

The derivation of this formula is similar to that for finding the transverse B.M.

(page 97).

For a rectangular water-plane area:

, BL31 1 = 1 2

where L = the length of the water-plane,

and B = the breadth of the water-plane.

Thus, for a vessel having a rectangular water-plane:

BMr = ?*

BMr = IT

V

Jl|.

t30 rRrM

For a box-shaped vessel:

I rB M r = ' - !

V

- BL'r2v

BL3I z X L X B X d

B M I : *t z o

where L = the length of the vessel,

and d : the draft of the vessel.

For a triangular prism:

BL3:@

1 2B M r : *

od

It should be noted that the distance BG is small when compared with BMp orGMr and, for this reason, BM1 ftay, without appreciable error, be substitutedfor GM1 in the formula for finding MCT I cm.

2. The Moment to Change Trim one centimetre (MCT I cm. orMCTC)

The MCT I cm., or MCTC, is the moment required to change trim by I cm.,and may be calculated by using tfre formula:

McTtcm. - \GMrl00L

where W = the vessel's displacement in tonnesGMr = the longitudinal metacentric height in metres, and

L : the vessel's length 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 stripis in equilibrium.

BMr - IL

V

TRIM

Fig. 86 (a)

Now shift the weight 'w' forward through a distance of 'd'metres. Thestrip's centre of gravity will shift from G to G, , causing a trimming moment ofW X GGr, as shown in figure 86 (b).

Fig. 86 (b)

The ship will trim to bring the centres of buoyancy and gravity into the samevertical line as shown in figure 86 (c). The ship is alain in equilibrium.

Irt the ship's length be L metres and let the tipping centre (F) be I metresfrom aft.

The longitudinal metacentre (Ml) is the point of intersection between theverticals through the centre of buoyancy when on an even keel and whentrimmed.

r 3 l

I Irut

Lw IG

B\ )

w

w

v

H

L e

'/J tw_j, _ _ _ || --H I G G l L

AF;

v

r32 TRIM

Il00L

w X d

Fig. 86 (c)

. 1 r \ W X dGGr and GG, : GML Tan 0

T a n 0 : w X dW X G M L

b u t T a n 0 : L

(See figure 87(b) )

kt the change of trim due to shifting the weight be one centimetre. Then w X dis the moment to change trim one centimetre.

.'. Tan 0

but Tan 0

Tan 0

MCT I cm.o r w x G M L

and MCT I cm.

W X G M L

MCT I cm.W X G M L

It00L

W X G M L

l00L

T R I M I 3 3

3. To find the change of draft forward and aft due to change of trim

When a ship changes trim it will obviously cause a change in the drafts forwardand aft. One of these will be increased and the other decreased. A formula mustnow be found which will give the change in drafts due to change of trim.

Consider a ship floating upright as shown in figure 87 (a). F, represents theposition of the centre of flotation which is i metres from aft. The ship's lengthis L metres and a weight 'w' is on deck forward.

I . w

wI

\ I)

F ig . 87 (a )

I-,et this weight now be shifted aft a distance of 'd'metres. The ship will trim

about F1 and change the trim 't'cms. by the stern as shown in figure 87 (b).

Wt C is a line drawn parallel to the keel.'A' represents the new draft aft and 'F' the new draft forward. The trim is

therefore equal to A - F and, since the original trim was zeto, this must also beequal to the change of trim.

[€t 'x' represent the change of draft aft due to the change of trim and let 'y'

represent the change forward.In the triangles WW Fr and W, Ia C, using the property of similar triangles:

C r- -xT i

Fig.87 (b)

t34 TRIM

x cm. t cm.,m-

= tr:

orxcm. - /m-Ll cm.

.'. Change of draft aft in centimetres : I t Change of trim in centimetresL

where / = the distance of centre of flotation from aft in metres, andL : the ship's length in metres.

It will also be noticed that x * y : t

Change of draft F in cms. : Change of trim - Change of draft A.

4. The effect of shifting werghts already on board

Example IA ship l26m.long is floating at drafts of 5.5m. F and 6.5m. A. The centre offlotation 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 tonnes already on board issttifted forward a distance of 45 metres.

T r i m m i n g m o m e n t : w X d

- l 2 0 X 4 5

: 5400 tonnes-m. by the head

^ , F . . T r i m m i n e m o m e n tLnange or rnm =

ffi

5400240

: 22.5 cm. by the head

Change of draft aft : /

x Ctrarge of trim

-_ 4 x 22.s126

: l0 '7 cms.

change of draft forward : change of trim - change of draft aft {: 22 .5 - 10 .7 cm.

: I 1 . 8 c m s .

Original draftsChange due to trim

Ans. New drafts

TRIM

5'500 m. F+ 0 . 1 1 8 m .

5 .618 m. F

6'500 m. A- 0 '107 m.

6.393 m. A

135'

Example 2A box-shaped vessel 90 m. X l0 m. X 6 m. floats in salt water on an even keelat 3 m. draft F and A. Find the new drafts if a weight of 64 tonnes already onboard is shifted a distance of 40 metres aft.

BMr - ljr2d

90x 90

$1/ =

\ { :

Fig. 88

L X B X d X I .025

90x10x3x1 .035

2767'5 tonnes

12x 3

BMr = 225 m.

Mcr l cm. - wI cUL-- -T00r

Since BG is small compared withGM1:-

M c r l c m . ^ w X B M t100 L.

2767.5 X 225: -JboFo.-

MCT I cm. : 69'19 tonnes-m.

w X dMCT I cm

64x 40

Change of trim

Change of trim

Change of draft aft

Change of draft aft

69.19: 37 cm. by the stern

I- + X Chanse of trimL

- +x37cm.-- l8'5 cm.

1 3 6 TRIM

Change of draft forward = l8'5 cm.

Original drafts 3.000 m. FChange due trim - 0'185 m.

3'000 m. A+ 0 . 1 8 5 m .

Ans. New drafts 2 ' 8 1 5 m . 3 ' 1 8 5 m .

5. The effect of loading and/or discharging weights

When a weight is loaded at the centre of flotation it will produce no trimmingmoment, but the ship's drafts will increase uniformly so that the ship displacesan extra weight of water equal to the weight loaded. If the weight is now shiftedforward or aft away from the centre of flotation, it will cause a change of trim.From this it can be seen that when a weight is loaded away from the centre offlotation, it will cause both a bodily sinkage and a change of trim.

Similarly, when a weight is being discharged, if the weight is first shifted tothe centre of flotation it will produce a change of trim, and if it is then dischargedfrom the centre of flotation the ship will rise bodily. Thus, both a change oftrim and bodily rise must be considered when a weight is being discharged awayfrom the centre of flotation.

Example IA ship 90 m. long is floating at drafts 4.5 m. F and 5.0 m. A. The centre offlotation 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 tonnes is loaded in a position 14 m.forward of amidships.

Fig. 89

Bodily sinkage

TRIM

wTPC

= 45 cm.

Trim moment=-Mc r l r .m -

_ 450 x r5 .5t20

_ 58'12 cm. by the head

I- : x Chanee of trim

L

4 ? . 5= -ff x s8.r2

Change of draft aft : 28.09 cm.

Change of draft forward : Change of trim - Change of draft aft

= 58'12 - 28.09

Change of draft forward : 30.03 cms.

Original drafts 4.500 m. F 5'000 m. ABodily sinkage a 0'450 m. + 0.450 m.

t-**- 0 '281 m.

5'169 m. A

137

450t0

Ans.

Bodily sinkage

Change of trim

Change of trim

Change of draft aft

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

New drafts 5.250 m. F

Note. In the event of more than one weight being loaded or discharged, the netweight loaded or discharged is used to find the net bodily increase or decreasein draft, and the resultant trimming moment is used to find the change of trim.

Also, when the net weight loaded or discharged is large, it may be necessaryto use the TPC and MCT 1 cm. at the original draft to find the approximate newdrafts, and then rework the problem using the TPC and MCT t cm. for the meanof the old and the new drafts to find a more accurate result.

Exomple 2A box-shaped vessel 40 m. X 6 m. X 3 m. is floating in salt water on an even keelat 2 m. draft F and A. Find the new drafts if a weight of 35 tonnes is dischargedfrom a position 6 m. from forward. MCT I cm.: 8.4 tonnes-m.

1 3 8 TRIM

wrpc3 52.46

l4'2 cms.ulXd-MCT I cm.

3 5 X t +8.4

1.0?sA100

1 .02s x 40x 6

TPC =

100

TPC : 2'46 tonnes

90

Bodily rise

Bodily rise

Change of trim

Change of trim

Change of draft aft

: 58'3 cm. by the stern

I: - X Chanee of trim

L

Change due trim

Ans. New drafts l ' 570 m. F

cms.

2'000 m. A- 0.140 m.

1 '860 m.+ 0.290 m.

2'150m. A

- +x 58.3 cms.Change of draft aft : 29.15 cms.

Change of draft forward : Change of trim - Change of draft aft

_ 58'3 - 29'15

Change of draft forward = 29'15

Original drafts 2.000 m. FBodily rise - 0'140 m.

1'860 m.- 0'290 m.

Example 3A ship 100 m. long arrives in port with drafts 3 m. F and 4'3 m. A. TPCl0 tonnes. MCT I cm. 120 tonnes-m. The centre of flotation is 3 m. aft ofamidships. If 80 tonnes of cargo is loaded in a position 24 m. forward ofamidships and 40 tonnes of cargo is discharged from 12 m. aft of amidships, whatare the new drafts?

TRIM

50m

Cargo loaded 80 tonnes Bodily sinkage : #aCargo discharged 40 tonnes

Net loaded 40 tonnes : 40l 0

Bod i lys inkage :4cms.

To find the change of trim take moments about the centre of flotation.

Weight Distance fromC.F.

Moment to change trim by

head stern

8040

279

2t60360

2s20

1 3 9

change of trim : InHfHlt

_ ?!4_t20

Change of trim : 2l cms. by the head

Change of draft aft : I

x Change of trim

- 4 \ x z l100

Change of draft aft - 9.87 cms.

change of draft forward - change of trim - change of draft aft

! l

24m - I- L

\ l-r?t Bott

F i g . 9 l

Example 4A ship of 6,000 tonnes displacement has drafts 7 m. F and 8 m. A' MCT I cm'

100 tonnes-m., TPC 20 tonnes, centre of flotation is amidships. 500 tonnes of

cargo is discharged from each of the following four holds:

No. I hold, centre of gravity 40 m. forward of amidships

N o . 2 h o l d . " ' ) " 2 5 m . " " "

1 4 0

Change of draft forward =

Original drafts 3'000 m'

BodilY sinkage 0'040 m'3'040 m.

Change due trim * 0'l I I m'

Ans. New draf ts 3 '151 m.

TRIM

= 2l - 9'8'l

)' " 20 m. aft' ) ' ) 5 0 m . "

F ig . 92

2000 tonnes200 "

I 800 tonnes

I I ' 1 3 c m s .

F . 4 '300 m. A .0'040 m.4340 n.

- 0'099 m.F . 4 ' 2 4 1 m . A .

No. 3 hold, ) '

No.4 ho ld , ' )

t t , t

, t , ,

Bodily rise :

180020

Bodily rise : 90 cms.

The following bunkers are also loaded:

150 tonnes at l2 m. forward of amidships

50 tonnes at 15 m. aft of amidshiPs

Find the new drafts forward and aft.

Total cargo dischargedTotal bunkers loaded

Net weight discharged

w

tPc

T R I M

Iileight Distance fromC, F.

Moment to change trim by

head stern

500s00s00s0015050

40252050t 2l 5

100002s000I 800

20000I 2500

750

36800332s0

33250

35s0

Resultant moment 3550 tonnes-m. by the head

Trim momentChange of trim

= MCT I .*

3 5 5 0= loo

Change of trim = 35'5 cms. by the head

Since centre of flotation is amidships,Change of draft aft : Change of draft forward

: I change of trim

17.75 cms.

Original drafts 7.000 m. F. g.000 m. A.Bodily rise 0.900 m. - 0.900 m.

6 . 1 0 0 * 7 . 1 0 0 . .Change due tr im + 0 '180 m. -- 0.180 m.

Ans. New draft, OZSO tn f . e CO rn. n.

l 4 l

142 . I -RIM

Example 5A ship arrives in port trimmed 25 cms. by the stern. The centre of flotation is

amidships. MCT I cm. 100 tonnes-m. A total of 3,800 tonnes of cargo is to be

discharged from 4 holds, and 360 tonnes of bunkers loaded in No. 4 double

bottom tank. 1,200 tonnes of the cargo is to be discharged from No. 2 hold and

600 tonnes from No. 3 hold. Find the amount to be discharged from Nos. I and

4 holds if the ship is to complete on an even keel.

Centre of gravity of No. I hold is 50 m." N o . 2 " " 3 0 m .

) t , ' ) ' " N o . 3 " " 2 0 m . a b a f t t ' ' ) " " ") t , t , , t ,

N o . 4 t t t t

4 5 m . t t t t t ' ) t t ) t t

" " ) ) " N o . 4 D B t a n k i s 5 m . " " t t " ' ) "

l2OOt (x)t

*:i'li I T

discharged ":-ilJ'ii I ffi^ ",."Nos. I &4 2000tonnes

bt 'x' tonnes of cargo be discharged from No. I hold"(2000-x) " " " " " ) ' No.4 "

Take moments about the centre of flotation.

forward of the centre of flotation

(29oo-x)t 6oot

Weight Distance fromC, F.

Moment to change trim by

head stern

xl 200600

2000-x360

50302045

5

l 20009000H5x

50x36000

1800

102000-45x 37800+50x

Original trimRequired trim

Change of trim required

Trimming moment required

Trimming moment required

Resultant moment

2500

. or 95x

x

and 2000-x

TRIM 143

: 25 cms. by the stern- 0

: 25 crrr. by the head

: Change of trim X MCT I cm.: 2 5 X 1 0 0

: 2500 tonnes-m. by the head: Moment to change trim by head

- MCT by stern- (102,000 - 45x) - (37,800 + 50x)

: 102,000 - 45x - 37,800 - 50x

: 64,200 - 95x

: 61 .700

: 649.5 tonnes

: 1350'5 tonnes

Ans. Dscharge 649'5 tonnes from No.l hold and 1350.5 tonnes from No.4 hold.

6. Using trim to find the position of the centre of flotation

ExampleA ship arrives in port floating at drafts of 4.50 m. A and 3:g0 m. F. Thefollowing cargo is then loaded:

100 tonnes in a position 24 m. aft of amidships30 ') " " " 30 m. forward of amidships6 0

t ) t ' t t t t l 5 m .

t t t t "

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

F i g . 9 4

144 TRIM

Original draf ts4.50m. A 3.80m. F give 0.70m. t r im by the stern.New draf ts 5 '10m. A 4.40m. F give 0.70 m. t r im by the stern.Therefore there has been no change in trim, or,The moment to change trim by the head : The moment to change

L,et the centre of flotation be L'metres aft of amidships. trim by the stern'

Then, 100(2a - x) _ 30 (30 + x) + 60 (15 + x)

2,400 - 100x : 900 + 30x * 900 + 60x

190x : 600

x : 3 ' 1 6

Ans. Centre of f lotation is 3'16 metres aft of amidships.

Note. In this type of question it is usual to assume that the centre of flotation isaft of amidships, but this may not be the case. Had it been assumed that thecentre of flotation was aft of amidships when in actual fact it was forward, thenthe answer obtained would have been minus.

7. Loading a weight to keep the after draft constant

When a ship is being loaded it is usually the aim of those in charge of theoperation to complete loading with the ship trimmed by the stern. Should theship's draft on sailing be restricted by the depth of water over a dock-sill or bythe depth of water in a channel, then the ship will be loaded in such a manner asto produce this draft aft and be trimmed by the stern.

Assume now that a ship loaded in this way is ready to sail. It is then foundthat the ship has to load an extra weight. The weight must be loaded in such aposition that the draft aft is not increased and also that the maximum trim ismaintained.

lf the weight is loaded at the centre of flotation, the ship's drafts will increaseuniformly and the draft aft will increase by a number of centimetres equal tow/TPC. The draft aft must now be decreased by this amount.

Now let the weight be shifted through a distance of 'd'metres forward. Theship will change trim by the head, causing a reduction in the draft aft by anumber of centimetres equal to llL X Change of trim.

Therefore, if the same draft is to be maintained aft, the above two quantitiesmust be equal.

TRIM t45

i .e. I x chanse of ̂ ' -- wL ' t r l m = T P C

But, Change of trim : ##.

f'#fu :#.or d =Tffif*

where d : the distance forward of the centre of flotation to load a weight tokeep the draft aft constant,

L : the ship's length, and/ - the distance of the centre of flotation from aft.

ExampleA box-shaped vessel 60 m. long, 10 m. beam, and 6 m. deep, is floating in saltwater at drafts 4 m. F and 4.4 m. A. Find how far forward of amidships aweight of 30 tonnes must be loaded if the draft aft is to remain at 4.4 m.

' . 0 2 5 ATPC: : i b0=

_ 1 .025 x 60 x t0100

TPC = 6'15 tonnes

W : L X B X d X I ' 0 2 5 t o n n e s

= 60 X l0 X 4 .2 X l ' 025

W = 2583 tonnesr 2

B M r : Lr2d

- 6 0 X 6 01 2 x 4 . 2

BMr = U metres7

w X B M IMCTlcm. = - i * f -

_ 2583 X 500100 x 60x 7

t46

MCT I cm. :

d :

:

d :

TRllvl

12314 tonnes-metres.

L X M C T 1 c m ./ X T P C

60 Yry x __ l3 0 4 6 . 1 5

l0 metres

Ans. Load the weight l0 metres forward of amidships.

8. Loading a weight to produce a required draft

ExampleA ship 150 metres long arrives at the mouth of a river with drafts 5.5 m. F and6'3 m. A. MCT I cm. 200 tonnes-m. TPC 15 tonnes. Centre of flotation isl'5 m. aft of amidships. The ship has then to proceed up the river where themaximum draft permissible is 6'2 m. It is decided that SW ballast will be runinto the forepeak tank to reduce the draft aft to 6'2m. If the centre of gravityof the forepeak tank is 60 metres forward of the centre of flotation, find theminimum amount of water which must be run in and also find the final draftforward.

60m | ,lw: f f i

\f -----

FF\

wFig.95

(a) Load 'w' tonnes at the centre of flotation

Bodily sinkage =

New draft aft w: o 'J m. + 15

ra .

Required draft aft : 6.2 m.

1 5 t - '

wTPC

w15

t * '

Reduction required = 0.1 m. *

TRIM

(b ) Shift 'w' tonnes from the cente of flotation to the forepeak tank

t47

Change of trim

Change of trim

Change of draft aft due to trim

Change of draft aft due to trim

But change of draft required aft

0 '417w

2'205w

1.205w

w

Therefore by loading 124.5 tonnesreduced to 6.2 metres.

(c) To find the new draft forward

Bodily sinkage -

:

Bodily sinkage :

Change of trim :

__

Change of trim =

w X d= Mcr I c;.60w200

3w: l0 cms. by the head

I- + X Chanee of trim

L

73'5 . . 3w= - x

1 5 0 " 1 0- 0 '147 w cms.= (10 * l l . ) . * r .'

1 5 '

: l0++l )

= 1 5 0 * w= 1 5 0

= 124'5 tonnes

in the forepeak tank the draft aft will be

wTPC

wl 5

8'3 cms.

w X dMCT t c*.

124'5 X 60200

37.35 cms. by the head

1 4 8 T R t M

Change of draft aft due trim : : X Change of trimL

Change of draft aft due trim : ffi X 37.35

: l8 '3 cms.Change of draft forward due trim - Change of trim

- Change of draft aft

: 37 '35 - 18 '3 cm.

Change of draft forward due trim : 19'05 cms.

Original drafts 5.500 m. F 6.300 m. ABodily sinkage * 0.080 m. + 0.080 m.

tatt * aa* t*Change due tr im * 0 '190 m. - 0.180 m.

New drafts 5.770 m. F 6.200 m. A

Ans. Load 124'5 tonnes in forepeak tank.Final draft forward is 5'770 metres.

9. Using change of trim to find the longitudinal metacentric height(GMr).

It was shown in Section 3 of this chapter that, when a weight is shiftedlongitudinally within a ship, it will cause a change of trim. It will now be shownhow this effect may be used to determine the longitudinal metacentric height.

Consider figure 96 (a) which represents a ship of length "L" at the waterline,floating upright on an even keel with a weight on deck forward. The centre ofgravity is at G, the centre of buoyancy at B, and the longitudinal metacentre atM1. The longitudinal metacentric height is therefore GMs.

I R I M

w

t49

M 1

LW G

\ B

w

Fig. 96 (a)

w

ML

-t-G LW G ,

\B

wFig. 96 (b)

w

M L

o"

I

Gr \, L ,

A\

\e' B, B

tv

1 5 0 TRIM

Fig. 96 (c)

Now let the weight be shifted aft horizontally as shown in figure 96 (b). Theship's centre of gravity will also shift horizontally, from G to Gt, producing atrimming 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 associatednotes, that F - A is equai to the new trim (t) and since the ship was originallyon an even keel, then 't' must also be equal to the change of trim.

If the angle between the new and old verticals is equal to 0, then the anglebetween the new and old horizontals must also be equal to 0 (the angle betweentwo straight lines being equal to the angle between their normals).

It wiil also be seen in figure 96 (c) that the triangles GG1M1 and CDE aresimilar triangles.

-1'.LI-f'I

' F

. GMr :GGt

or GMI :

L

rI: X G G ,t '

(All measurements are in metres).

TRIM

Example IWhen a weight is shifted aft in a ship 120 metres long, it causes the ship's centreof gravity to move 0'2 metres horizontally and the trim to change by 0.15 metres.Find the longitudinal metacentric height.

GMI L: -GG, t

GMr_

l 5 t

: fx cc,

Example 2A ship 150 metres long has a displacement of 7,200 tonnes, and is floatingupright on an even keel. When a weight of 60 tonnes, already on board, is shifted24 metres forward, the trim is changed by 0'15 metres. Find the longitudinalmetacentric height.

GMr-

t20 x 0-20 .15

160 metresAns. GMr-

Ans. GMr_

GMr- :GGt

It

IG G , X "' t

w X d * LW t

6 0 x 2 4 1 5 07200 0.l s

200 metres.

EXERCISE 15

l. A ship of 8,500 tonnes displacement has TPC 10 tonnes, MCT I cm = 100 tonnes.m.and the centre of flotation is amidships. She is completing loading under coal tips. Nos.2and 3 holds are full, but space is available in No. I hold (centre of gravity 50 m. forward ofamidships), and in No. 4 hold (centre of gravity 45 m. aft of amidships). The presentdrafts are 6'5 m. F and 7 m. A, and the load draft is 7.1 m. Findhow much cargo is to beloaded in each of the end holds so as to put the ship down to the load draft and completeloading on an even keel.