If we use (Trim/LBP)=(GG1/GML) then could we write GML=GG1/(Trim/LBP) = GG1*LBP/Trim? With this relation and if we know GG1 provided a center of mass equation, could you write Moment needed to change Trim by 1 unit (MCTC) = W * (GG1 * LBP/Trim) / * LBP This is assuming the trim is 1/100th of LBP
If B shifts from B to B1 how come when we move cargo from forward to aft the AP to B distance before shift from h. tables and then again AP to B1 are exactly the same since LCB distance is same for same displacement? In your image distance from AP to G1 is larger then AP to B1, that makes trim lever forward not aft? How do you calculate BB1 longitudinal shift? You have seen the problem your self when you drag the green arow to B1, if you placed it correctly it would be left from G1 and that would make ship trim forward. ???
I understood that, if we load above G , then G will not move longitudinally, and therefore no trimming moments will take place, and trim will remain the same. and if we load above C.F, which is not vertically above G will do the same, i.e the trim will not change. now the question is, Is loading above G will not affect the trim as the same effect as loading above C.F? thanks.
If the ship is box shaped where COF and COG will lie in the same line , it will not make any difference .However for normal ship shape COG and COF are different, hence loading at COF will not result in trim but loading at COG will result in some trim. FYI COF is the geometrical center of water plane area where ship trims and COG is the point where total weight of ship is assumed to be acting vertically downward.
