The prismatic coefficient Cp is the ratio of the volume of displacement to the volume of a prism having the same length as the ship and a cross-section equal to the midship section area:

Cp = Volume of displacement/(L x Am)

where L is the length and Am the midship section area.

Derivation: The block coefficient Cb = Volume/(L x B x d), where B is the beam and d the draught. The midship section coefficient Cm = Am/(B x d). Therefore:

Cb/Cm = [Volume/(L x B x d)] / [Am/(B x d)] = Volume/(L x Am) = Cp.

Hence Cp = Cb/Cm. The prismatic coefficient indicates the fullness of the ends of the ship: a high Cp means full ends, a low Cp means fine ends. It is used in resistance calculations.

The half-ordinates of a waterplane at 15 m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0 m. There are 9 ordinates, so the waterplane length is 8 x 15 = 120 m.

(i) TPC.

Area A = 2 x (h/3)[y0 + y8 + 4(y1+y3+y5+y7) + 2(y2+y4+y6)]

= 2 x (15/3)[1 + 0 + 4(7+11+10.5+4) + 2(10.5+11+8)]

= 10[1 + 4x32.5 + 2x29.5] = 10[1 + 130 + 59] = 10 x 190 = 1900 m2.

TPC (salt) = A x 1.025/100 = 1900 x 1.025/100 = 19.48 t/cm.

(ii) Distance of the centre of flotation from midships.

First moment of area about the after perpendicular:

M = 2 x (h/3) x sum of (weighted y x distance). Using station distances 0,15,30,...,120 m with Simpson weights (1,4,2,4,2,4,2,4,1):

weighted sum = 1x0 + 4x(7x15) + 2x(10.5x30) + 4x(11x45) + 2x(11x60) + 4x(10.5x75) + 2x(8x90) + 4x(4x105) + 1x0

= 0 + 420 + 630 + 1980 + 1320 + 3150 + 1440 + 1680 = 10620.

M = 2 x 5 x 10620 = 106,200 m3.

Distance of centroid from AP = M/A = 106,200/1900 = 55.89 m. Midships is at 60 m from AP, so the centre of flotation is 60 - 55.89 = 4.11 m aft of midships.

(iii) Second moment of area about a transverse axis through the centre of flotation.

Second moment about AP: I_AP = 2 x (h/3) x sum of (weighted y x distance^2):

weighted sum = 1x0 + 4x(7x225) + 2x(10.5x900) + 4x(11x2025) + 2x(11x3600) + 4x(10.5x5625) + 2x(8x8100) + 4x(4x11025) + 1x0

= 0 + 6300 + 18900 + 89100 + 79200 + 236250 + 129600 + 176400 = 735,750.

I_AP = 2 x 5 x 735,750 = 7,357,500 m4.

Transfer to the centre of flotation: I_CF = I_AP - A x (55.89)^2 = 7,357,500 - 1900 x 3123.7 = 7,357,500 - 5,935,000 = 1,422,500 m4.

Answer: TPC = 19.48 t/cm; LCF = 4.11 m aft of midships; I about the transverse axis through CF = about 1.42 x 10^6 m4.