19. Trim #1

A.A. B. Dinariyana

Jurusan Teknik Sistem PerkapalanFakultas Teknologi Kelautan ITS Surabaya

2013

Trim may be considered as the longitudinal equivalent of list.

Trim is also known as `longitudinal stability'.

Instead of trim being measured in degrees it is measured as the difference between the drafts forward and aft. If difference is zero then the ship is on even keel.

If forward draft is greater than aft draft, the vessel is trimming by the bow. If aft draft is greater than the forward draft, the vessel is trimming by the stern.

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Consider a ship to be floating at rest in still water and on an even keel as shown in Figure

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

So W = b.

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Now let a weight `w', already on board, be shifted aft through a distance `d', as shown in Figure.

This causes the center of gravity of the ship to shift from G to G1, parallel to the shift of the center of gravity of the weight shifted, so that:

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The ship will now trim until the centers of gravity and buoyancy are again in the same vertical line.

When trimmed, the wedge of buoyancy LFL1 emerges and the wedge WFW1 is immersed.

the volume of the immersed wedge must be equal to the volume of the emerged wedge

F, the point about which the ship trims, is the center of gravity of the water-plane area. The point F is called the `center of Flotation' or `Tipping center'.

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A vessel with a rectangular water-plane has its center of flotation on the center line amidships.

On a ship, the center of flotation may be a little forward or abaft amidships, depending upon the shape of the water-plane.

Trimming moments are taken about the center of flotation since this is the point about which rotation takes place.

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The longitudinal metacenter (ML) is the point

of intersection between the verticals through

the longitudinal positions of the centers of

buoyancy.

The vertical distance between the center of

gravity and the longitudinal metacenter

(GML) is called the longitudinal metacentric

height.

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BML is the height of the longitudinal metacenter above the center of buoyancy and is found for any shape of vessel by the formula:

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910

The MCT 1 cm, or MCTC, is the moment

required to change trim by 1 cm, and may be

calculated by using the formula:

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Consider a ship floating on an even keel as shown in Figure (a). The ship is in equilibrium.

Now shift the weight `w' forward through a distance of `d' meters.

The ship's center of gravity will shift from G to G1, causing a trimming moment of W x GG1, as shown in Figure (b).

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(a)

(b)

The ship will trim to bring the centers of buoyancy and gravity into the same vertical line as shown in Figure (c). The ship is again in equilibrium.

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Let the ship's length be L meters and let the tipping center (F) be l meters from aft.

The longitudinal metacenter (ML) is the point of intersection between the verticals through the center of buoyancy when on an even keel and when trimmed.

(c)

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When a ship changes trim it will obviously

cause a change in the drafts forward and aft.

One of these will be increased and the other

decreased.

A formula must now be found which will give

the change in drafts due to change of trim.

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Consider a ship floating upright as shown in Figure below.

F1 represents the position of the center of flotation which is l meters from aft. The ships length is L meters and a weight `w' is on deck forward.

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Let this weight now be shifted aft a distance of `d' meters. The ship will trim about F1 and change the trim `t' cms by the stern as shown in Figure.

W1C 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 zero, this must also be equal to the change of trim.

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Let `x' represent the change of draft aft due to the change of trim and let `y' represent the change forward.

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EXAMPLE#1

A ship 126m long is floating at drafts of 5.5m F and 6.5m A. The center of flotation is 3m aft of amidships. MCT 1 cm . 240 tonnes m. Displacement = 6000 tonnes. Find the new drafts if a weight of 120 tonnes already on board is shifted forward a distance of 45 meters.

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EXAMPLE#2

A box-shaped vessel 90m x 10m x 6m floats in salt water on an even keel at 3m draft F and A. Find the new drafts if a weight of 64 tonnes already on board is shifted a distance of 40 meters aft.

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Ship Stability for Masters and Mates, Fourth

Edition, Revised, D.R. Derrett, B-H Newnes,

1990

Basic Ship Theory, Fourth Edition, Vol. 1,

Chapter 1 to 9, Hydrostatics and Strength,

K.J. Rawson and E.C. Tupper, LST 1994

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