and by some means is made to float at the waterline W'L', the centre of buoyancy must shift, owing to the changed shape of the displacement, from B to B' say. Then the original vertical through B and the new vertical through B' will intersect in the point M, which is the longitudinal metacentre. This point is precisely analogous to the transverse metacentre, the difference consisting in the direction of the inclination. The distance between the C.G. and the longitudinal metacentre is the longitudinal metacentric height.

The point M is determined with reference to the C.B., and the distance BM is given by the equation—

where Io is the moment of inertia of the waterplane about a transverse axis through its centre of gravity (this C.G. is termed the centre of flotation), and V is the volume of displacement.

The calculation for Io is somewhat complicated,* but it may be approximately written

Ion'. L3. B (n' being a coefficient)

also Vk. L. B.D ( being the coefficient of fineness)

where L is the length of ship between perpendiculars in feet D is the mean draught in feet

b is a coefficient which does not vary much from 0.075.† This approximate formula shows the great influence of the length in determining the position of the longitudinal metacentre.

Longitudinal shift of Weights already on Board. The trim of a ship is the difference between the forward and after draughts. Thus H.M.S. Pelorus is designed, under normal load conditions, to float at a draught of 12 ft. forward and 15 ft. aft, giving a trim of 3 ft. by the stern.

Change of trim is the sum of the changes of draught forward

* See the author's "Theoretical Naval Architecture."

In a vessel with full waterplane n' will be large, but k will also be large, the ship being of full form. If n' is small, k also will be small. So that the quotient

and aft. Thus if Pelorus, when floating at 12 ft. forward, 15 ft. aft, has certain weights shifted on board resulting in a draught of 12 ft. 10 in. forward and 14 ft. 2 in. aft, she is said to have changed trim 10 in. + 10 in. = 20 in. Change of trim can be produced by the fore-and-aft shift of weights already on board, this being analogous to the inclining experiment, in which heeling is caused by shifting weights in a transverse direction.

In Fig. 180, let w be a weight on the deck at A when the vessel is floating at the waterline WL. Now suppose w is moved forward through a distance d. G, the centre of gravity of the ship, will in consequence move parallel to the shift of w to G', such that GG' = (w ×d). Under these circumstances the vessel

W

cannot float at the waterline WL, as in the second sketch in Fig. 180, because the C.G. and the C.B. are not in the same vertical line. The ship must adjust herself to the line W'L', as shown, so that G', the new centre of gravity, and B', the new centre of buoyancy, are in the same vertical. Then the line through B'G' intersects the original vertical through BG, in M, the longitudinal metacentre. If is the small angle of inclination—

Having thus two values of tan 0 we can equate, so that

Transposing we have the moment causing the change of trim, viz.—

W x GM

w x d =

× change of trim in inches

12 x L

L= length between draught marks

In ships of small ratio of length to breadth, like battle-ships up to Royal Sovereign, the longitudinal GM was approximately equal to the length. In these ships, therefore, the moment to change trim 1 in. is very nearly one-twelfth the displacement in tons. This does not hold so well for more modern ships, and the following give good approximations to this moment to change trim 1 in. :

where L = length between perpendiculars in feet

B = breadth of ship in feet

The length and breadth do not vary much for considerable changes of draught; the above formulæ, therefore, show that the moment to change the trim 1 in. varies very little for the range of draughts at which a vessel is likely to float.

The following examples will illustrate the use of the above in determining change of trim :

EXAMPLE.-A vessel is Alvatiny at a draught of 12 ft. 1 in. forward and 14 ft. 10 in. aft. Determine the draughts forward and aft after shifting 5 tons from forward to aft through 210 ft. The moment to change trim 1 in. is 295 foot tons.

=

In this case w = 5 tons, d 1050 change of trim is therefore 295 increase of draught aft of 1 So that

210 ft., so that w x d = 1050 foot tons. The

= 3 in. This change of trim will cause an in., and a decrease of draught forward of 13 in.

draught forward = 12 ft. 1 in.

13 in.

=

11 ft. 11 in.

draught aft = 14 ft. 10 in. + 13 in. = 14 ft. 11 in.

EXAMPLE.-Determine approximately the shift of 50 tons on board the above ship necessary to bring her to an even keel.

The mean draught is

There is therefore necessary a change of trim of

(13 ft. 5 in. 12 ft. 1 in.) + (14 ft. 10 in.

13 ft. 5 in.) = 33 in.

The necessary moment from aft forward is therefore 295 x 33 = 9735 foot

tons, so that 50 tons would need to be shifted in order to bring the ship to an even keel.

9735
50

=

195 ft., say, from aft forward

Effect on the Trim due to adding a Weight of Moderate Amount. If a ship sinks from a waterline WL to a parallel waterline W'L', the added buoyancy of the layer will act at the C.G. of the layer. If the sinkage is small this point will very nearly be in the same section as the centre of flotation. If, therefore, we wish to add a weight to a ship so that the trim shall not be changed, the added weight must be in the same section as the added buoyancy, and if the added weight be not large, it must, therefore, be placed in the same section as the centre of flotation. The centre of flotation in ships of the Navy is usually abaft amidships, on the average about one twenty-fifth the length.

If a weight is placed on board anywhere else both increase of draught and change of trim must occur. We imagine the weight is added first at the centre of flotation, by which we can see how much she will sink to a parallel waterline, and then we shift the weight to the position given, and determine the change of trim. The following will illustrate the method of dealing with the addition of weights on board a ship.

EXAMPLE.-A ship is floating at a draught of 20 ft. forward and 22 ft. aft, when the following weights are placed on board in the positions given, viz.—

What will be the new draught, the moment to change trim 1 in. being 800 foot tons and the tons per inch 35.

The total weight added is 155 tons, and if placed at the centre of flotation, the increased draught is 15 4 in.

=

The weights to be moved forward give a moment of (20 × 100) + (45 × 80) 5600 foot tons, and the weights to be moved aft give a moment of (60 × 50) +(30 × 10) = 3300 foot tons. The forward moment is thus in excess by 2300 foot tons, and this will cause a change of trim of 200 = 3 in. nearly, and the new draught forward is 20 ft. 4 in. + 11⁄2 in. draught aft is 22 ft. 4 in. - 11 in.

=

22 ft. 3 in.

=

20 ft. 6 in., and the new

Strictly speaking, the change of trim ought not to be divided equally forward and aft. It should be less than half (about 0·46) aft, and rather more than half (about 0:54) forward. This will be understood by reference to Fig. 180. Since F, the centre of flotation, is abaft amidships, WW' is not equal to LL', but rather less. Unless the change of trim is very considerable, however, the error involved in taking the half forward and aft is small.

The following example will illustrate how to deal with the problem of bringing a vessel to such a draught as will allow her to pass over a place like a bar at the mouth of a river. It is assumed also that no information is available, except the dimensions of the ship, so that the approximations already given have to be employed.

EXAMPLE.-A cruiser 300 ft. × 361 ft. is floating at a draught of 12 ft. forward and 15 ft. aft. It is desired to bring her to an even keel at a draught of 12 ft., in order to pass over a bar at the mouth of a river. Determine approximately how this could be accomplished.

The mean draught is 13 ft. 6 in., so that the ship would need lightening 300 × 36.5 18 in. The approximate tons per inch (page 174) would be = 18-25 600 tons, so that 18 x 18-25 = 330 tons, say, must be removed. Suppose this is done so that the vessel lightens to a draught of 10 ft. 6 in. forward, 13 ft. 6 in. aft.

=

=

The approximate moment to change trim 1 in. is 11800 × (300) × 36·5 = 300 foot tons, say. As a change of trim of 36 ins. is necessary to bring the vessel to an even keel, we require a moment of 300 × 36 10,800 foot tons. The 330 tons must therefore be removed 1988 33 ft., say, abaft the centre of flotation to give the necessary moment. Taking the centre of flotation as the length abaft amidships, or 12 ft., the 330 tons would have to be removed so that its C.G. was about 45 ft. abaft amidships to give the required draught of 12 ft. forward and 12 ft. aft.

(It may be stated that, taking the correct data for the above ship, the answer would be 350 tons, taken out 421 ft. abaft amidships, so that the approximation is a very good one.)

Passage through the Suez Canal.-In order to pass through the Suez Canal it is necessary that the maximum draught should not exceed 26 ft. 3 in. In the deep draught battle-ships like Majestic class which may have to pass through the canal, a special set of instructions is drawn up, showing how the ship may most readily be brought to the required mean draught. The following is a specimen set of such instructions (the maximum draught allowable has since then been increased to 26 ft. 3 in., instead of 25 ft. 7 in.).

1 If, as would probably be the case, the bar is in fresh water, allowance must be made for the bodily sinkage in going from salt to fresh water, see p. 177.