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Formulae

Form Coefficients

WPA = Waterplane area L = Length overall B = maximum breadth d = maximum depth

CwC_w

= coefficient of waterplane area

CmC_m

= coefficient of midship area

WPA=L×B×Cw WPA = L\times B \times C_w AreaAmidships=B×d×Cm Area Amidships = B \times d\times C_m

when displacement is constant ( box shaped vessels)

NewDisplacementOldDisplacement=NewDensityOldDensity\frac {New Displacement}{Old Displacement} = \frac{New Density}{Old Density}TPCSW=WPA97.56TPC_{SW}= \frac {WPA} {97.56} FWA=W4timesTPCSWFWA = \frac {W} { 4 times TPC_{SW}}

change in draft

=FWA×(1025−ρDW)25 = \frac {FWA \times (1025-\rho_{DW}) }{25}

Trim

MCTC=W×GML100×LMCTC = \frac {W \times GM_L}{100 \times L}GML=KML−KGGM_L = KM_L-KG

KML=KB+BMLKM_L = KB + BM_L

BML=ILL∇BM_L = \frac{I_{LL}}{\nabla}

Change in Trim ΔT\Delta_T

ΔT=TrimmingmomentMCTC=W×(LCBfoap−LCGfoap)MCTC\Delta_T = \frac{Trimming moment}{MCTC}= \frac{ W \times (LCB_{foap}-LCG_{foap})}{MCTC}

Change of draft Aft δaft=1L×ΔT\delta_{aft} = \frac{1}{L} \times \Delta_T

distance ’d’ to keep aft draft constant MCTC×LTPC×l\frac{MCTC \times L}{TPC \times l}

Simpson’s Rules

Common Interval = h

  • Area=13×h×Σ1Area = \frac{1}{3} \times h \times \Sigma_1 Multipliers (1,4,1)
  • Area=3h8×Σ1 Area = \frac {3h}{8} \times \Sigma_1 Multipliers (1,3,3,1)
  • Area=h12(5a+8b−c)Area = \frac{h}{12}(5a+8b-c)

Second Moment of Inertia (I)

Essentially moment of an Area about an axis . Here the formulas are given for its own centroidal axis.

rectangleB×LB \times L

For rectangular waterplane Area

ICG=L×B312 I_{CG} = \frac { L \times B^3}{12}BM=IVBM = \frac{I}{V}

triangle base =‘B’, height =‘H’

ICG=B×H336I_{CG}= \frac{B\times H^3}{36}

Circle of Radius ‘R’

ICG=π×R44I_{CG} = \frac{\pi \times R^4}{4}

Polar moment is (for rudder stock problem) J=π×R42J = \frac{\pi \times R^4}{2}

Prism base ‘B’,‘C’ and height ‘H’

ICG=H336×B2+4BC+C2B+CI_{CG}= \frac {H^3}{36} \times \frac{B^2+4BC+C^2}{B+C}

for the Sectional Modulus Problem:

where b is the horizontal length and h can be height or can be thickness

ICG=bh312I_{CG} = \frac {bh^3}{12}

General Formula (Inclined Surface)

hCP=hˉ+IGsin⁡2θAhˉh_{CP} = \bar{h} + \frac{I_G \sin^2\theta}{A\bar{h}}

For vertical surfaces θ=90°\theta = 90°: hCP=hˉ+IGAhˉ\quad h_{CP} = \bar{h} + \dfrac{I_G}{A\bar{h}}


Shape Properties & Center of Pressure

Shape Area AA Centroid Depth hˉ\bar{h}* IGI_G (about centroid) hCPh_{CP} (Vertical Surface)
Rectangle width bb, height hh bhbh h2\frac{h}{2} (top at surface) bh312\frac{bh^3}{12} 2h3\frac{2h}{3}
Triangle (base bb, height hh, apex up) 12bh\frac{1}{2}bh 2h3\frac{2h}{3} (top/apex at surface) bh336\frac{bh^3}{36} 3h4\frac{3h}{4}
Circle (radius RR) πR2\pi R^2 RR (top at surface) πR44\frac{\pi R^4}{4} 5R4\frac{5R}{4}
Trapezoid (bases BB bottom, CC top, height hh) B+C2h\frac{B+C}{2}h h3⋅B+2CB+C\frac{h}{3}\cdot\frac{B+2C}{B+C} (top at surface) h336⋅B2+4BC+C2B+C\frac{h^3}{36}\cdot\frac{B^2+4BC+C^2}{B+C} hˉ+h218hˉ⋅B2+4BC+C2(B+C)2\bar{h} + \frac{h^2}{18\bar{h}}\cdot\frac{B^2+4BC+C^2}{(B+C)^2}
Semicircle (radius RR, flat side up) πR22\frac{\pi R^2}{2} 4R3π\frac{4R}{3\pi} (diameter at surface) 0.1098R40.1098R^4 hˉ+0.1098R4πR22⋅hˉ\bar{h} + \frac{0.1098R^4}{\frac{\pi R^2}{2}\cdot\bar{h}}

*Measured from the free surface to the shape’s centroid. For submerged cases, add the depth of the top edge.


Quick Exam Shortcuts

  • Rectangle (top at surface): hCP=23hh_{CP} = \frac{2}{3}h (measured from top)
  • Triangle (apex at surface): hCP=34hh_{CP} = \frac{3}{4}h (measured from apex)
  • Circle (top at surface): hCP=54Rh_{CP} = \frac{5}{4}R (measured from top)
  • Universal Check: hCPh_{CP} is always below the centroid ( hˉ \bar{h}) for vertical/inclined surfaces.

Bilging Connection (from your previous point)

  • Sinkage: Lost VolumeIntact Waterplane Area\dfrac{\text{Lost Volume}}{\text{Intact Waterplane Area}}
  • New GMTGM_T: Use IT,intactI_{T,\text{intact}} (about centerline) →BMT=IT∇\rightarrow BM_T = \frac{I_T}{\nabla}
  • New GMLGM_L: Use IL,intactI_{L,\text{intact}} (about midships) →BML=IL∇\rightarrow BM_L = \frac{I_L}{\nabla}

Pro Tip: If the surface is inclined, remember the sin⁡2θ\sin^2\theta factor. If the tank wall is curved, find the horizontal/vertical force components first; the center of pressure formula above applies only to plane surfaces.


Simpson’s Rules for Moment of Inertia

Area = A, y is athwartship, x is longitudinal, da is an elemental strip of the area.

1st moment of Area about X axis:

2nd moment of inertia about X axis: For a small area dA the moment of inertia about its centroid is

Therefore, when taking moments about X Axis:

Therefore, for Simpson’s Rule 1: Where ‘h’ is the common interval along X axis and ‘y0, y1, y2, y3, y4’ are y values along the X axis .

1st Moment of Area about Y Axis:

2nd Moment of Area about Y Axis: (ignoring first term as it is too small)

For  (stations ) and common interval h,

Now, if Y is Longitudinal and X is athwartship then switch x for y. Therefore, for TRANSVERSE items like Centre of Floatation use For Longitudinal items use