Solution Naval Numerical 9
Numerical
9: A ship 120m long floats has draughts of 5.50m forward and 5.80m aft;
MCT1 cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the
new draughts which a mass of 110 tonne is added 24m aft of midships.
Solution:Given L=120m
draught fwd =5.5m
draught aft =5.8m
MCT1cm = 80t-m
TPC = 13
LCF = 2.5m
m = 110 t
d= 24m
bodily sinkage= 110/13 =8.46cm=0.085m
Chnage in Trim = 110(24+2.5)/80 =36.44cm by stern.
Draught change fwd/aft }{MCT1cm} "=\pm \frac{\frac{Change in trim}{L}\left ( \frac{L}{2}\pm LCF \right )}{MCT1cm}")
Change fwd }{80} "=-\frac{\frac{36.44}{120}\left ( \frac{120}{2}- 2.5 \right )}{80}")
= -17.46cm = -0.1746m
Change Aft }{80} "=\frac{\frac{36.44}{120}\left ( \frac{120}{2}+ 2.5 \right )}{80}")
= 18.98cm = -0.1898m
New draught fwd = 5.5+0.085-0.1746 =5.4104m
New draught aft = 5.8+0.085+0.1898 = 6.0748m
new draught is wrong...... need to add 0.1898....also instead of 5.5 its 5.8 answer is 5.8+0.085+0.1898=6.075
ReplyDeleteDear Unknown
DeleteThankyou for pointing out the error.
Suggested correction is made.
Kindly keep posting.
While calculating changes in draft ,why to devide it again by MCTC ? Simply no need ....
ReplyDeleteJust use changes in draft=(t/L)* (L/2 +_ LCF)