Longitudinal Stability

7/28/2019 Longitudinal Stability

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Longitudinal Stability(MAR Rev 03/01/01) 1

LONGITUDINAL STABILITY (1)

Trim

Is the difference in cms between the forward and aft draughts,as measured at the forwardand aft perpendiculars respectively.

The ship above has draughts F 2.20 m A 2.68 m.

The trim of the ship is: 2.68 -

2.20

0.48 m by the stern;48 cms by the stern.

The ship above has draughts F 2.70 m A 2.32 m.The trim of the ship is: 2.70 -

2.32

0.38 m by the head;38 cms by the head.


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Longitudinal centre of gravity (LCG)Is the position of the ships centre of gravity relative to the

length of the ship. Termed GL in diagrams.

Longitudinal centre of buoyancy (LCB)Is the position of the ships centre of buoyancy relative to the

length of the ship. Is termed BL in diagrams.

Longitudinal Metacentre

Is the point of intersection of the lines of action of buoyancy

force acting through the LCB when the ship is in the initialeven keel condition and subsequently trimmed conditions. Is

termed ML in diagrams.

Longitudinal Metacentric height (GML)

Is the vertical difference between the centre of gravity and the

longitudinal metacentre. Termed GML

in diagrams.

Consider the ship shown.


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M

L

w

FBL1

B

L

G

L

GL1

d

)(

Consider a ship initially on even keel. A weight already onboard is moved aft through d metres. This causes GL to move

to GL1.

GLGL1 = w dW


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Rearranging this gives: GLGL1 W = w d

Trimming moment = GLGL1 W = w d

The ship trims until both LCB and LCG are in the same verticalagain.

Change of trim (COT)

Is the difference in cmsbetween the trim in the initial condition

and the trim in the final condition.

SAQA ship has the following initial draughts:

F 6.00 m A 5.86 m

and final draughts: F 5.66 m A 6.20 mafter cargo on board is shifted.Calculate the change of trim that has occurred.


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Answer

Initial draughts:F 6.00 m A 5.86 m: Trim = 0.14 m by HEAD

Final draughts:F 5.66 m A 6.20 m: Trim = 0.54 m by STERN

Therefore: Change of trim = 0.68 m by STERN

= 68 cms by STERN

Moment to change trim by one centimetre (MCTC)This is the trimming moment (w d) required to change the

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

hydrostatic particulars and used to determine the change in trim

when cargo is either shifted, loaded or discharged.

COT (cms) = w d = Trimming momentMCTC MCTC

SAQ

A weight of 150 tonnes is moved aft by a distance of 20 m. Ifthe MCTC for the current draught is 250 t-m determine thefinal trim of the ship if the initial trim was 0.20 m by thestern.


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Answer

COT (cms) = w d = 150 20 = 12 cmsMCTC 250

COT = 0.120 m

Initial trim: 0.200 m by STERNCOT: 0.120 m further by the STERNFINAL TRIM 0.320 m by STERN

Longitudinal centre of flotation (LCF or F)

Is the geometric centre of the ships water-plane area at a

particular draught and is the point about which the ship willtrim. Its position will change with draught.

F

)(


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The position of the LCF determines how the change of trim

(COT) will be apportioned between the forward and aft

draughts.

Ship with LCF amidships

AP FP

F

Ta

Tf

)(

If LCF amidships then: Ta = Tf = COT

2

where: Ta = change of draught aft due to trim; and

Tf = change of draught forward due to trim.

SAQA ship floats at draughts F 6.50 m and A 6.80 m. Determinethe final draughts if 25 tonnes is moved 45 m forward giventhat MCTC is 112.5 t-m and the LCF is amidships.


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AnswerCOT = w d = 25 45 =10 cms = 0.100 m

MCTC 112.5

Ta = Tf = 0.100 = +/- 0.050 m2

Weight is moved forward so the ship will trim by the HEAD.

Initial draughts F 6.500 A 6.800Trim + 0.050 - 0.050

FINAL F 6.550 m A 6.750 m


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AP FP

F

Ta

Tf

Ship with LCF not amidships

In this case the change of trim (COT) will have to be

apportioned to the forward and aft draughts according to theposition of the LCF within the ships length.

)(LBP

a f

If the similar triangles are considered then:

a = f and Ta + Tf = COT

Ta Tf

Therefore: Ta = a COT and Tf = f COTLBP LBP

SAQA ship has initial draughts F 10.25 m and A 10.15 m. A weight

of 95 tonnes is moved aft through a distance of 42 m.Calculate the final draughts given that LBP is 100 m, LCF is48 m foap and MCTC is 285 t-m.


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AnswerCOT = w d = 95 42 = 14 cms

MCTC 285Ta = 48 14 = 6.7 cms = 0.067 m

100Tf = 52 14 = 7.3 cms = 0.073 m

100Weight is moved aft so the ship will trim by the STERN.

Initial draughts F 10.250 A 10.150

Trim - 0. 073 + 0.067FINAL F 10.177 m A 10.217 m


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SAQ

A ship 100 m in length floats at draughts F 7.00 m and A 6.80m. Calculate the final draughts if 150 t is loaded 20 m foap

given that TPC is 15 and MCTC is 150 t-m and LCF is 45 mfoap.

TIP

Always draw a sketch to help you picture what is happening!

The effect of loading and discharging weights

The effect ofbodily sinkage/risemust be taken into account:

Sinkage/Rise cms = w

TPC

The following procedure should be followed:

1. Load/discharge the weight from the LCF, calculating the

sinkage/rise using the TPC value given.

2. Calculate the COT by moving the weight from the LCF

position to its actual loaded/discharged position.3. Find Ta/Tf by apportioning the COT according to the

position of the LCF.

4. Apply both the sinkage/rise and Ta/Tf to the initial

draughts to determine the final draughts.


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F

Answer

Sinkage = w = 150 = 10 cms = 0.100 mTPC 15

COT = w d = 150 (45 - 20) = 25 cmsMCTC 150

Ta = 45 25 = 11.25 cms = 0.113 m100

Tf = 55 25 = 13.75 cms = 0.137 m100

Weight is loaded aft of the LCF so the ship will trim by theSTERN.

Initial draughts F 7.000 A 6.800Sinkage + 0.100 + 0.100

7.100 6.900Trim - 0.137 + 0.113FINAL F 6.963 m A 7.013 m

AP FP

)(

45 m

20 m

150 t


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Loading/discharging multiple weights

A tabular approach needs to be adopted where moments are

taken about the LCF.

Consider the following example:

A ship 120 m in length floats at draughts F 6.24 m and A 6.36m. LCF is 54 m foap, TPC 14.2 and MCTC 116 t-m.

The following cargo is worked:

Load 120 t lcg 10.0 m foap;Load 68 t lcg 86 m foap;Discharge 36 t lcg 22 m foap;Discharge 48 t lcg 60 m foap.

Calculate the final draughts.


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TRIM - MULTIPLE WEIGHTS

Enter data

LBP = 120 m

LCF = 54 m foapTPC 14.2

MCTC 116 t-m

Weight (t) Weight (t) Dist from LCF HEAD STERN

120 120 44 5280

68 68 32 2176

-36 36 32 1152

-48 48 6 288

NET 104 3328 5568

2240

Using: Sinkage = w

TPC

Sinkage 7.3 cms Using: COT = Trimming moment

= 0.073 m MCTC

COT = 19.3 cms

Apportion the COT using: Ta = a x COT

LBP

Tf = f x COT

LBP

Ta = 8.7 cms = 0.087 m

Tf = 10.6 cms = 0.106 m

Find final draughts

Fwd Aft

Initial 6.240 6.360

Sinkage 0.073 0.073

6.313 6.433

Trim -0.106 0.087

FINAL DRAUGHTS 6.207 6.520


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Most trim problems are straight forward provided that you

understand the information that is being given and can recognise

the formula to which it belongs.

Sinkage/Rise cms = wTPC

COTcms = w d = Trimming moment

MCTC MCTC

Apportion COT to forward and aft draughts using:

Ta = a COT and Tf = f COT

LBP LBP

NOTE

In practice, the mean value of TPC must be used to determinethe sinkage/rise of the ship. Similarly, the mean values ofMCTC and LCF must be used when calculating the change oftrim. The change of trim is then apportioned to the final

waterline using the final LCF . If a hydrostatic particularstable is not given, then it has to be assumed that the values of

TPC, MCTC and LCF position do not significantly change i.e.they remain constant for the range of draughts concerned.

Longitudinal Stability

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