1h longitudunal trim lcb lcg method

7/24/2019 1h Longitudunal Trim Lcb Lcg Method

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NOTE :

As of 2005 and the new MCA syllabus Longitudinal Trim

uestions will ha!e to be done by the LC"#LC$ method% The

old method of sol!ing trim uestions using LC& will not be

allowed e'(e)t in e'(e)tional (ir(umstan(es% One has to

remember that ta*ing moments about the LC& is only (orre(t

when the LC& has not (hanged substantially% As a shi) rises

due to dis(harged weight or sin*s further due to added weight+

the LC& (hanges% ,nless the (hange is !ery small the

LC"#LC$ method must be used% Now what is small will !ary

from shi) to shi)%

-f the hydrostati( data has been )ro!ided+ then LC"#LC$method must be used% Otherwise LC& method will ha!e to be

used%

Longitudinal centre of gravity (LCG represented by GL )

Is the position of the ships centre of gravity relative to the

length (longitudinally) of the ship taken from the A.P. Symbol

!in diagrams. It is the point "here all #f can said to be acting

do"n"ard through.

!$%I&''%A! &I*

'SI% (!+, - !+) *&/$


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0

)(

Longitudinal centre of buoyancy (LCB represented by BL )

Is the volumetric centre (or point) of the ships under "ater

volume "here all the buoyancy force ,f can be said to act

up"ards through and is e2pressed relative to the A.P. of the

ship. Symbol ,! in diagrams

Longitudinal centre of Flotation (LCF represented by F )

It is the point about "hich the ship pivots. Is the geometric

centre of the ships "ater3plane area at a particular draught and

is the )oint about whi(h the shi) will trim. Its position "ill

change "ith draught.

It is the point through "hich a line dra"n transversely bisects

the "ater plane into t"o e4ual halves. It is e2pressed relative

to distance from A.P.


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5

F

BL

GL

)(

Wf

w

+onsider the ship sho"n.

,y la"s of 6lotation #f 7 ,f

In the above diagram 8

Since 6 is the pivot point if "e take moments about 6 then

Clockwise moments = Wf * d (distance Wf is acting from F).

Counter Clockwise moments = Bf * d (distance Bf is acting from F).

Since #f 7,f by la"s of flotation and 9d

the distance "here #f : ,f are acting are vertically above

each other i.e. same distance from 96 (also from A.P.)

so Counter (lo(*wise moments . Clo(*wise moments%

In this case the resultant moments about centre of flotation 96

is ;ero. So the vessel "ill be at ven


4/18

?

w

FBL

GLGL1

d

)(

%o" let us take the same vessel again on even keel i.e.

#fis vertically inline "ith ,f . A "eight 9" "hich is

already on board is shifted aft through a distance 9d. &he

moment caused by this shift is 7 " 2 d

/o"ever "e kno" from previous that "hen a "eight is

shifted /7 " 2 d@# so can say

!

!12 # 7 " 2 d

Since the vessel "as on even keel i.e. !"as above ,!.%o"

because !has moved to !1 moments about 96 due to #f

and ,fare not the same.

+lock"ise moments about 6 7 #f d1

+ounter +lock"ise *oments about 6 7 ,f d0

Wf

d1

d2

Since #f 7 ,f

&rimming moments about 6 7 #f d1B,f d07 (#f d1) B (#f d0)

7 #f (d1- d0) but distance d1-d0 7 distance (!+-!+,)

Tus we can say C!"!T! # W $ (LCG%LCB)

&CTC


5/18

C

We 'now Triing oent # w d ( only sift of weigt)

Triing oent # (LCB%LCG) Wf

&he ship trims until both !+, and !+ are in the same verticalline again.

+.$.&. 7 # (!+b-!+g)

*+&+

Cange of tri (C!"!T!)

Is the difference in (msbet"een the trim in the initial condition

and the trim in the final condition provided the displacement of

the vessel has not changed.

/A

A shi) has the following initial draughts:

& 1%00 m A 5%1 m

and final draughts: & 5%11 m A 1%20 m

after (argo on board is shifted%

Cal(ulate the (hange of trim that has o((urred%

/ee ne't )age for solution%


6/18

D

L"*G+T,-+*.L /T.B+L+T0 (1)

Tri

Is the difference (in (ms3bet"een the for"ard and aft draughts

as measured at theforward andaft )er)endi(ularsrespectively.

&he ship above has draughts 6 0.0E m A 0.DF m.

&he trim of the ship isG 0.DF

3 0.0E

E.?F m by the stern8

4 (ms by the stern.

&he ship above has draughts 6 0.HE m A 0.50 m.&he trim of the ship isG 0.HE

3 0.50

E.5F m by the head8

(ms by the head.


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H

C"T (cs) # w d # Triing oent

&CTC &CTC

# W $ GG

&CTC

# W $ (LCB % LCG)

&CTC

Answer

-nitial draughts:

& 1%00 m A 5%1 m: Trim . 0%64 m by 7EA8

&inal draughts:

& 5%11 m A 1%20 m: Trim . 0%54 m by /TE9N

Therefore: Change of trim . 0%1 m by /TE9N

. 1 (ms by /TE9N

Assuming no weights loaded or dis(harged%

&oent to cange tri by one centietre (&CTC)

&his is the trimming moment (" d) re4uired to change the

ships trim by e2actly 1 cm. It is tabulated in the ships

hydrostatic particulars and used to determine the change in trim

"ithout change of draft.


8/18

F

&he position of the !+6 determines ho" the change of trim

(+$&) "ill be apportioned bet"een the for"ard and aft

draughts.

/ip wit LCF aidsips

. F

F

Ta

Tf

)(

If !+6 amidships thenG Ta # Tf # C"T

2

"hereG &a 7 change of draught aft due to trim8 and

&f 7 change of draught for"ard due to trim.

/A

A shi) floats at draughts & 1%50 m and A 1%0 m% 8etermine

the final draughts if 25 tonnes is mo!ed 45 m forward gi!en

that MCTC is 662%5 tm and the LC& is amidshi)s%


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Answer

COT . w ; d . 25 ; 45 .60 (ms . 0%600 m

MCTC 662%5

Ta . Tf . 0%600 . eight is mo!ed forward so the shi) will trim by the 7EA8%

-nitial draughts & 1%500 A 1%00

Trim < 0%050 0%050

&-NAL & 1%550 m A 1%?50 m


10/18

1E

. F

F

Ta

Tf

/ip wit LCF not aidsips

In this case the change of trim (+$&) "ill have to be

a))ortioned to the for"ard and aft draughts according to the

position of the !+6 "ithin the ships length.

)(LB

a f

If the sketch above triangles are similar thus G

a 7 f and &a K &f 7 +$&

&a &f

&hereforeG Ta # a C"T and Tf # f C"T LB LB

Were a # distance of CF (LCF) fro .!!

f # distance of CF fro F!! ( L!B!! 3 LCF )

/A

A shi) has initial draughts & 60%25 m and A 60%65 m% A weight

of @5 tonnes is mo!ed aft through a distan(e of 42 m%

Cal(ulate the final draughts gi!en that L" is 600 m+ LC& is

4 m foa) and MCTC is 25 tm%

/ee ne't )age for solution%


11/18

11

Answer

COT . w ; d . @5 ; 42 . 64 (ms

MCTC 25

Ta . 4 ; 64 . 1%? (ms . 0%01? m

600

Tf . 52 ; 64 . ?% (ms . 0%0? m

600

>eight is mo!ed aft so the shi) will trim by the /TE9N%

-nitial draughts & 60%250 A 60%650

Trim 0% 0? < 0%01?

&-NAL & 60%6?? m A 60%26? m


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Calculating Final -raugt by LCG%LCB etod

&he follo"ing procedure should be follo"edG

1. +alculate the A.*..0. 'sing A* enter hydrostatics find !+6.

5. 'sing !+6 apply correction to raught Aft and obtain &.*..

?. 'sing &* note isplacement !+, *+&+ !+6 interpolating if

necessary.

C. 'sing +.$.&. 7 # 2 (!+,-!+)@*+&+

calculate position of Initial !+.

D. 'sing !+ obtained from above calculate moments about A.P.similar to calculating


13/18

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Loading4discarging single weigt LCG%LCB etod

Lessel on even keel find !+ and subse4uent draft : trim

+onsider the follo"ing e2ampleG

A shi) 640 m in length floats at draughts & 5%10 m A 5%10 m

A weight of 220 tonnes is loaded 50 m &OA%

Cal(ulate the final draft and trimD

%ote G Solve this 4uestion using the hydrostatic sheet provided

on ne2t slide.


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1?

HYDROSTATIC PARTICULARS EXTRACTS

-5.FT -+/L! -+/L! TC TC &CTC &CTC 6&t 6B LCB LCF

t t t t t7 t7 foap foap

/W FW /W FW /W FW

5- 1!829 5- 1!888 5- 1!829 5- 1!888 5- 1!829 5- 1!888

:!88 1;9:< 1;228 2=!1= 22!9: 1>;!< 1>8!1 >!=; =!!=9 =!9> :8!8> 1!; 1::!8 >!=< =!9= :8!12 < 1=9;> 22!?2 22!=< 1:?!? 1:9!9 >!=: =!;> :8!1< != 1:;!8 >!=> =!;= :8!28 1:2!9 >!=? =!=> :8!2; !;1 =!== :8!2> !88

22!=8 21!?9 21!;1 19>!> 19;!? >!?> 2!9> :8!:? 9: ?2 9 >:


15/18

1C


16/18

1D

Loading4discarging ultiple weigts using (LCB%LCG)

A tabular approach needs to be adopted "here moments are

taken about the A.P..

+onsider the follo"ing e2ampleG

A shi) 640 m in length floats at drafts & 1%24 m and A 1%1 m%

The following (argo is wor*ed:

Load 620 t l(g 60%0 m foa)Load ? t l(g 1 m foa)

8is(harge 1 t l(g 22 m foa)

8is(harge 4 t l(g 10 m foa)%

Cal(ulate the final draughts%

,se hydrostati( tables on )re!ious )age for this uestion%


17/18

1H


18/18

1F

*ost trim problems are straight for"ard provided that you

understand the information that is being given and can

recognise the formula to "hich it belongs.

C"Tcs # W $ (LCB%LCG)

&CTC

Apportion +$& to for"ard and aft draughts usingG

Ta # a C"T and Tf # f C"T

LB LB

*"TA

In the (!+,-!+) method the &.*.. for the appropriate

isplacement is calculated and all other relevant factors such as

!+6 *+&+ !+, are obtained for the calculated &.*..

interpolating as necessary. If a 9hydrostatic particularstable is

not given then it has to be assumed that the values of &P+*+&+ and !+6 position do not significantly change i.e. they

remain constant for the range of draughts concerned.

1h longitudunal trim lcb lcg method

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