94.NA March 2020 Q.8(b)

 The half-breadth s of waterplane of a ship of 120m length and 15m breadth are given below.

Station	0	1	2	3	4	5	6	7	8
Half- Breadth	1.6	2.8	5.5	6.4	7.3	6.2	4.2	2.0	0

Calculate i.Waterplane area ii.TPC in salt water iii.Cw iv.LCF from midship

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Given

L = 120m
B = 15m

To find

i.Waterplane area
ii.TPC in salt water
iii.Cw
iv.LCF from midship

Solution

The lever is taken about midship as 0 Since the centroid lies near midship.

½ Ordinates	SM	Product	Lever	Product
1.6	1	1.6	+4	+6.4
2.8	4	11.2	+3	+33.6
5.5	2	11	+2	+22
6.4	4	25.6	+1	+25.6
7.3	2	14.6	0	Total (A) = +87.6
6.2	4	24.8	-1	-24.8
4.2	2	8.4	-2	-16.8
2.0	4	8	-3	-24
0	1	0	-4	0
		Total (C) = 105.2		Total (B) = -65.6

Since there are 9 ordinates there will be 8 spaces
Common Interval h = 120 / 8 = 15

i. Waterplane area = 1/3 x h x Total (C)
   Since half ordinates are given.
   Therefore water plane area = 2/3 x h x Total (C)

                                               = 2/3 x 15 x 105.2

                                         Aw = 1052 m2

ii. TPC in Sw = Aw/100 x Density of SW

                      = 1052/100 x 1.025

              TPC =  10.78

iii. Cw = Aw / L x B

           = 1052 / 120 x 15

     Cw = 0.584 

iv. LCF from midship = (h x (Total (A) + Total (B) / Total (C))

                                  = (15 x (87.6 - 65.6) / 105.2)

                                  = 330 / 105.2
                                  = 3.136 m

    The Answer is positive.Therefore ordinate is aft of midship.

     LCF from midship = 3.136 m aft