The centre of flotation (F) is at the geometric centre of the water plane area. The centre of buoyancy (B) is at the geometric centre of the underwater volume. The centre of gravity is at G. Their longitudinal positions are seldom the same. Usually F is abaft of B at any given draught. But B & G are vertically in one line. If B & G are not in one vertical line, a ship will trim till B & G are in one vertical line. A ship The builder provides 'the relevant information for B & F in curve or tabular form.

CHANGE OF TRIM DUE TO CHANGE OF DENSITY

For a box shape vessel on even keel B, G & F are vertically in line. So when it passes from SW to FW, it merely sinks more & B changes, G stays same, F moves up but stays in same vertical line in a new horizontal plane i.e new WPA.

The sketch shows a ship on an even keel, in salt water, on waterline WL. ln this condition B is vertically below G. On passing into FW (or water less density than SW) the ship will (first) sink bodily to W1L1. This new waterline can be assumed to have the centre of flotation as F1. The added layer of buoyancy due to the sinkage (as shown) will have its centroid at 'b' . The addition of a layer of buoyancy at 'b' causes B to move to some position up and towards 'b' to a point B1.

FBB1

GF1

d

W L

L1W 1

b

W 2

L2

Because B1 is out of line with G a trimming couple exits, which causes the ship to trim till 'B1' is again vertically below 'G'. This new waterline is represented by W2L2.

TO CALCULATE THE FINAL DRAFTS(i) Find the bodily sinkage : this will be the FWA (= W/TPC) if the

ship is passing from SW to FW. In other cases it will be the dock allowance or a proportion of the FWA.

(ii) Calculate the weight of this layer of buoyancy acquired at ‘b’. This is (sinkage x TPC) tonnes. The value of TPC must be for the fresh water (or dock water).

(iii) Find the trimming moment = (weight of layer x horizontal distance B to b) tonne.metres. (Note: Distance B b = (LCB ~ LCF)

(iv) Find the change of trim = moment/MCTC (v) Proportion the change of trim to find Tf and Ta according to the

position of LCF.

NOTE : The above method avoids the need to calculate the position B1 but if B1 is known, or determined, the trimming moment can be given by ( W x B B1 horizontally or W x GB1). The shift BB1 can be found by taking moments,

about B. Thus moment of added layer = moment of total displacement x distance BB1 Therefore volume of layer x bB = total volume displacement x B B1or BB1 = volume of layer x bB1

total volume of displacement ( and bB is of course (LCB ~ LCF)).The volume of the layer is always the difference between the volumes displaced in two different densities. In the greater (original) density d1 the volume is (V/d1).In the lesser (final) density d2 the volume is (V/d2 ) and the volume of the layer = ( W/d2 - W/d1) m3 = W ( d1 - d2 ) m3

d1 x d2Sinkage/Rise = FWA or DWA depending on density of second liquid.

COT cms = W ( RD1 - RD2) (LCF -LCB) MCTC2 + by stern - by head