Change of draft at any point due loading or longitudinal shift of weight

To establish the relation between GML and change of trim (trc). To prove MCTC = \dfrac{\Delta XGM_{L}}{100L}

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 = \dfrac{w\times d}{\Delta }=GG_{1}

The couple caused due to forces at B and G1 will trim the vessel aft so as to bring COB vertically below G1 as the vessel settles at new waterline W1L1.
The vertical through B1 meets the original vertical through B at ML.
Triangle GG1ML is a right angled triangle. This triangle is similar to the right angled triangle formed by sides trc and LBP. It can be said that,
Tan θ  = \dfrac{t_{rc}}{LBP}=\dfrac{GG_{1}}{GM_{L}}
Also tan θ = \dfrac{GG_{1}}{GM_{L}}=\dfrac{\omega \times d}{\Delta }=\dfrac{1}{GM_{L}}

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 = \dfrac{\dfrac{1}{100}}{L}=\dfrac{1}{100L} ……………………1
But tan θ is also equal to \dfrac{\omega \times d}{\Delta }=\dfrac{1}{GM_{L}} …………………..2
Equating (1) and (2) we get \dfrac{1}{100L}=\dfrac{\omega \times d}{\Delta \times GM_{L}}
Since w x d is MCTC, we have MCTC = \dfrac{\Delta \times GM_{L}}{100L}

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)

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.
LBP = 116m
LCF=\left( \dfrac{116}{2}-2.5\right) =55.5m

Bodily sinkage = \dfrac{250+300}{\left( 100\times 18\right) }=0.306m

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

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

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