*To establish the relation between GM _{L} and change of trim (t_{rc}). To prove *MCTC =

In above fig a weight w is shifted aft (longitudinally) by d metres to cause a trimming moment = (w.d) t.m. This causes the shift of COG =

The couple caused due to forces at B and G_{1} will trim the vessel aft so as to bring COB vertically below G_{1} as the vessel settles at new waterline W_{1}L_{1}.

The vertical through B_{1} meets the original vertical through B at M_{L}.

Triangle GG_{1}M_{L} is a right angled triangle. This triangle is similar to the right angled triangle formed by sides t_{rc }and LBP. It can be said that,

Tan θ =

Also tan θ =

If the change of trim due to shift of weight is 1cm, then w x d is the trimming moment to change the trim by 1cm.

Tan θ =

But tan θ is also equal to

Equating (1) and (2) we get

Since w x d is MCTC, we have MCTC =

The change of draft at any point along the ships length is due to:

1. Bodily rise or bodily sinkage,

2. Change of trim. Let us consider a case where a certain weight is loaded at center of flotation. The drafts allover will change uniformly and the changes of drafts are *purely due to bodily sinkage*.

Let us now consider a case where a certain weight is shifted longitudinally on board a ship. The drafts will change all over the vessel except at center of flotation. The changes in drafts at different points along the ships length are *purely due to change of trim alone*.

Whenever loading or discharging operations are conducted on board a ship, the changes in drafts forward and aft are due to both the reasons stated above. In the numerical to follow we will apply these changes in two steps.

**Ex. A box shaped vessel 110m x 16m x 6m floats in salt water on an even keel draft of 5m. Find the new drafts if a weight of 80 tonnes already on board is shifted longitudinally forward through a distance of 40 metres. (ship’s KG = 3.5 m)***Sol ^{n}: *

**Ex. A ship 116 m long is floating at draft 6.5m forward and 7.8 m aft. The centre of flotation is 2.5m aft of amidship, TPC 18, MCTC 196 tm. Find the new drafts if a weight of 250 tonnes is loaded in a position 44m forward of midship & 300t is loaded 20m aft of midship.***Sol ^{n}:*

LBP = 116m

Bodily sinkage =

Forward trimming moment = 250 x (44+2.5) = 11625 tm

Aft trimming moment = 300 x (20-2.5) = 5250 tm

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