STABILITY BASIC FORMULAS

S SOME BASIC FORMULAS

Area of Waterplane = L x B x CW . L = Length of vessel . B = Breadth of vessel ...CW = Co-efficient of Waterplane

Volume of Displacement = L x B x d x CB . d = depth of vessel .CB= Block co-efficientVolume (V) = L x B x d

Displacement (W) = L x B x d x R.D ... R.D = Relative density of water

TRANSVERSE STABILITYRectangular WaterplanesBM =I . where I =LB3 V 12 . V = Volume of vessel

Depth of centre of buoyancy below water line: =1(d+V ) 3 2 ALONGITUDINAL STABILITYa)Rectangular WaterplanesIL =L3B and BML =IL. 12 V

b)Box ShapesBML =L2 12d

LISTGG1(Horizontal) =w x d W .d = distance moved horizontal .w = weight .W = Final Displacement

GG1(Vertical) =w x d W .d = distance moved vertical .w = weight .W = Final DisplacementTAN =GG1 GM .GM = Metacentric height

TAN =Listing Moments W x GM .GM = Use Fluid GM . W = Final Displacement

------------o------------

DRY DOCKING

P =trim x MCTC l ...l = Distance of COF from where vessel touches blocks first ...P = Upward force acts on ship where block first touches

P =COT x MCTC l ..In case of declivity of Dock

Virtual loss of GM =P x KM WVirtual loss of GM =P x KM (WP) if Pforce is very small

After taking the blocks (F & A):P = Change in TMD ( cms ) x TPC orP = Reduction in water level x TPC .TMD = True mean draft

Change in Draft (rise) (cms) = Palways subtract from draft TPC

TMD = Draft Aft (LCF x Trim ) LBP .subtract if vessel is by the stern .add if vessel is by the head

DRY DOCKINGHYDROSTATIC TABLES ANDVESSEL A TYPE PROBLEMS

Proceed as follows :

1.Findmean draftfrom the present given drafts.2.From this mean draft, look in tables forLCF3.Using that LCF, calculateTMD4.From the TMD, look in tables and findMCTC, LCF and DISPLACEMENT5.Calculate nowP-Force6.For Displacement (W) at Critical Instant, findW-P7.From this new (W), look in tables forKMT8.Now findVirtual loss of GMand usenew KMTbutold Displacement (W)9.Find nowinitial GM, using thenew KMT10.Apply Virtual loss of GM in it and find theEFFECTIVE GM.

------------o------------FREE SURFACE EFFECT / MOMENT

FSE =l.b3.R.D 12W

FSM =l.b3.R.D 12 .R.D = Density of liquid in tankFSE =FSM W

Corrected FSM = Tabulated FSM xActual R.D Assumed R.D

New FSM = Original FSM x1 n2

.n = number of tanks which are subdivided

------------o------------

DYNAMICAL STABILITY

Dynamical Stability = W x Area under the curve

STATICAL STABILITY

Statical Stability = W x GZ

KN CURVES

GZ = KN KG.SIN

INCLINING EXPERIMENT

GM =w x dxLength of Plumbline W Deflection

------------o------------

RIGHTING MOMENT

SMALL ANGLES OF HEEL (UPTO 10OHEEL):

GZ = GM x SIN

LARGE ANGLES OF HEEL (WALL SIDED FORMULA):

GZ = SIN( GM +1.BM.TAN2) 2WIND HEELING MOMENT:

Total Wind heeling moment =F.A.d 1000GZ (at angle of heel) =F.A.d 1000W

.d = Distance of centre of buoyancy to centre of windage area.F = Steady wind force of 48.5 kg/m2

------------o------------

SIMPSONS RULES

SIMPSONS FIRST RULE:

Area =hx ( a + 4b + 2c + 4d + 2e + 4f + g ) 3

Remember : 1 4 1

SIMPSONS SECOND RULE:

1 3 3 1

h h h

Area =3x h x sum of products 8

Remember : 1 3 3 1

SIMPSONS THIRD RULE:5 8 1

h h

Area =hx ( 5a + 8b c ) 12

Remember : 5 8 1

NB:Divide the value of h (in degrees) by 57.3 while calculating the area.

NB:In the 3rd rule of Simpson, we are only looking for a particular piece between the area i.e., from one co-ordinate to other and this is mainly used by surveyors for calculating sludge in bunker tank etc. Also for knowing the full area, we use Simpsons first rule.

GM CONDITIONS

GM AT LOLL:

GM =2(Initial GM) COS .answer will be ive but write +ive sign

WHEN GM IS NEGATIVE:

WHEN GM IS NIL:

------------o------------

TURNING CIRCLE

TAN(Heel) =v2BG gGMr .v = velocity of ship(m/s) .r = radius of turning circle .g = Acceleration due to Gravity

(9.81 m/s) .T = Period of Rolls (seconds) .K = Radius of Gyration

.= 3.142857143 (constant) .I = Weight Moment of Inertia about Rolling axis (tonne - metres2)Hence we get,

Actual New Draft = [ Initial draft +BTan] Cos 2------------o------------AIR DRAFT

CALCULATING LENGTH OF THE IMMAGINARY MASTWHICH IS EXACTLY ABOVE THE CF:

Correction to Aft Mast

=Dist. of center mast from Aft Mast x Diff. of ht between masts Dist. between the two masts

.subtract this value from the ht of Aft mastor

Correction to Fwd Mast

=Dist. of center mast from Fwd Mast x Diff. of ht between masts Dist. between the two masts

.add this value from the ht of Fwd mast

FOR FINDING DRAFT FWD AND AFT

Trim between masts=Trim of vesselDistance between masts LBP

..(from this, calculate trim of vessel and proceed as follows)

Trim Effect Aft =la xTrim L

Trim Effect Fwd =lfxTrim L

GRAIN

Weight of Grain =Volume S.F

Weight of H.M =Volumetric H.M S.F

Approx. Angle of heel =Total H.Mx 12o Max.H.M

. Max.H.M can be found in the Tables of Maximum permissible Grain heeling moment against W and KG

GG1(o) =w x d W .w = weight of Grain liable to shift while rolling .d = horizontal distance of Grain shift

o =Total volumetric H.M (in m4) S.F x W

40= GG1(o) x 0.80 .80% ofo (GG1)

NB:If value for cargo is given for centroid then follow as normalbut if value given for Kg of cargo then,

Multiply H.M value for fully filled compartment by 1.06 andMultiply H.M value for partially filled compartment by 1.12

TRIMHYDROSTATIC TABLES ANDVESSEL A TYPE PROBLEMS

Proceed as follows :

1.Findmean draftfrom the present given drafts.2.From this mean draft, look in tables forLCF3.Using that LCF, calculateTMD4.From the TMD, look in tables and findMCTC, LCB and DISPLACEMENT5.Calculate nowINITIAL LCG6.Now CalculateFINAL W and FINAL LCG by MOMENTS7.With this FINAL W, go in tables and look findTMD, LCB, LCF and MCTC8.CalculateTRIM9.After this calculateTRIM EFFECTS ( F & A )10.Now apply this TRIM EFFECT tofind FINAL DRAFTS.

------------o------------

TRIM

Trimming Moment = w x d ( d = distance from COF )

Area of Waterplane = L x B x Cw

Volume of Displacement = L x B x D x CB

TPCsw =1.025A 100

FWA =W . 40 TPC

DWA =FWA (1.025 R.D) 0.025

MCTC =WGML 100L

TPCDW=R.D x TPCSW 1.025

MCTCDW=R.D x MCTCSW 1.025

Displacement(DW) =RD x Displacement(sw) 1.025

Sinkage (cms) =w . TPC

COT =Trimming Moments MCTC

COD Aft =lax COT L

COD Fwd = COT COD Aft

WHEN THE VESSEL IS EVEN KEELLCG = LCB

FOR A BOXED SHAPED VESSEL

BM =B2 12d

KB =draft 2

FOR A BOX SHAPED VESSEL WHEN DISPLACEMENT CONSTANT

New Draft =Old DensityOld Draft New Density

FOR A SHIP SHAPED VESSEL WHEN DRAFT CONSTANT

New Displacement =New DensityOld Displacement Old Density

TO KEEP THE AFT DRAFT CONSTANT

d =L x MCTC la x TPC .keeping the aft draft constant

d =L x MCTC lf x TPC .keeping the fwd draft constant

d = Distance from the CFla= Distance from the APlf= Distance from the FP

TO PRODUCE A REQUIRED TRIM

Change in Draft (cms) = (l. xw x d )w . L MCTC TPC(ive for Draft Aft)( + ive for Draft Fwd)( lafor aft and lffor fwd)

Trim (cms) =W (LCB LCG) MCTC

(Values for LCB, LCG and MCTC should be final)COT WITH CHANGE OF DENSITY

COT =W(RD1 RD2)(LCF LCB) RD1x MCTC2

LCGINITIAL= LCB(Trim (cms) x MCTC) W .( ive for stern trim ) .( + ive for head trim )

TRIM EFFECT AFT =la x Trim L

TRIM EFFECT FWD =lfx Trim L

------------o------------

BILGING

WHEN HEIGHT OF COMPARTMENT IS GIVEN AND ABOVE WATER LEVELCALCULATE SINKAGE BY RECOVERABLE BUOYANCY METHOD:

Sinkage =Buoyancy still to be recover L x B

Buoyancy still to be recover = Lost buoyancy Recoverable BuoyancyVolume of Lost Buoyancy = l x b x draft

Recoverable Buoyancy = ( L l ) x B x ( Depth Draft )

To find the Final Draft, add the Sinkage to Tanks height

WHEN IN QUESTION PERMEABILITY OF THE CARGO IS GIVENCALCULATE THE EFFECTIVE LENGTH OF THE TANK:

Permeability () =Broken Stowage Stowage Factor

Broken Stowage = Actual Stowage Solid Stowage

Solid Stowage =1 . R.D of liquid in tank

Effective Length = Tanks lengthORIGINAL x Permeability ()

NBAfter calculating Effective length always use this length for tanks length.

------------o------------BILGING

MIDSHIP COMPARTMENT

NON WATER TIGHTWATER TIGHT

Sinkage =v . A a If NON WATER TIGHT

Sinkage =v . A If WATER TIGHT

BM =LB3 12V If WATER TIGHT

BM =(L l)B3 12V If NON WATER TIGHT

BILGING

SIDE COMPARTMENT

PLAN VIEW OF A SHIP

Sinkage =v . A a If NON WATER TIGHT

Sinkage =v . A If WATER TIGHT

TAN =BB1 GM . = List

BB1 =a x d Final A .d = Distance from center of tank to ships center line .Final A = A aBM =IOZ V

IOZ= IABAd2.d =B+ BB1 2 .A = AaIAB =LB3lb333BILGING

END COMPARTMENT

AFT COMP. BILGEDFWD COMP. BILGEDNON WATER TIGHTNON WATER TIGHT

Sinkage =v . A a If NON WATER TIGHT

Sinkage =v . A If WATER TIGHT

If KG is not given, then GML = BML

BM =L3B 12V If WATER TIGHT

BM =(L l)3B 12V If NON WATER TIGHT

COT =w x d MCTC .w = l x b x dft x R.D .d =L..(Non water tight case) 2 .d = tanks center to CF ..(Water tight case)

MCTC =WGML 100L

COD Aft =lax COT L .la = (Ll) + tanks length 2(For measuring the CF from AP) ..(Non water tight case) . la =L 2(CF hasnt changed and is amidships) ..(Water tight case)

When Fwd compartment is bilged (and non water tight), then just use .la = (Ll) 2(Again for measuring the CF from AP) ..(Non water tight case)

IN CASE OF WATER TIGHT COMPARTMENT BELOW WATER LINE AND BELOW THE TANK THERE IS AN EMPTY COMPARTMENT

a)Deal as normal water tight caseb)Use volume of the tank only which is filled with water but not the portion beneath it.c)But for KB of tank, use from K to center of tank

NBIN WATER TIGHT CASEBM remains the same before and afterKB is different before and after bilging KB1is half of Original Draft KB2is found by moments

IN NON WATER TIGHT CASEBM is different before and after bilging BM1isLB3andBM2is(Ll)B3 12V 12V

KB is different before and after bilging KB1is half of Initial Draft KB2is half of New Draft

PLEASE NOTE THE FOLLOWING CONDITIONS

WATER TIGHT CASENON WATER TIGHT CASECalculate:a) Sinkage by non w/t method b) KB2by Moments

NB:In all cases of WATER TIGHT COMPARTMENT, calculate KB by the MOMENTS METHOD& use New Draft in calculating this KB when calculating volume.

------------o------------TANKER CALCULATIONS

TOTAL OBSERVED VOLUME (T.O.V.):The Total Observed Volume of all Petroleum Liquids and Free Water at observed temperature.

GROSS OBSERVED VOLUME (G.O.V.):The Total Volume of all Petroleum Liquids, excluding Free Water at observed temperature.

G.O.V. = T.O.V. - Vfw (at observed temperature)

GROSS STANDARD VOLUME (G.S.V.):The Total Volume of all Petroleum Liquids, excluding Free Water, corrected by appropriate Volume Correction Factor for the observed temperature and API Gravity 60 F, Relative Density 60 F / 60 F or Density 15 C.

G.S.V. = G.O.V. X V.C.F.

FREE WATER (Vfw):The volume of water present in a tank which is not in suspension in the contained liquid at observed temperature.

ONBOARD QUANTITY (O.B.Q):Quantity of water, oil, slops, residue, sludge or sediment, remaining in the tanks prior to loading.

TOTAL CALCULATED VOLUME (T.C.V.):It is the Gross Standard Volume plus Free Water.

T.C.V. = G.S.V. + Vfw

TOTAL RECEIVED VOLUME (T.R.V.):Is equal to the Total calculated Volume minus O.B.Q.Weight Correction Factor (W.C.F.) is applied to this Volume to obtain Weight in Metric Tons or Long Tons.Shore Gross B/L figure is to be compared with this figure.Whenever Free Water is found in Cargo:

T.R.V. = T.C.V. - O.B.Q.

LOADED OIL WEIGHT:Is equal to the Gross Standard Volume minus O.B.Q.Weight Correction Factor is applied to this Volume to obtain weight in Metric Tons or Long Tons.Shore Gross B/L figure is to be compared with this figure when no Free Water is found in Cargo.

LOADED OIL WEIGHT = (G.S.V. - O.B.Q.) W.C.F.

VESSELS EXPERIENCE FACTOR (V.E.F.):Is equal to the Total of Gross B/L figures divided by the Total of Ships figures over the last 10 voyages.

For the purpose of calculating V.E.F., 10 TO 20 Voyages may be taken. However all voyages must qualify. A minimum of 5 qualified voyages is needed for some level of V.E.F.

The defination of a qualified voyage is one that meets the following criteria:Any voyage that is within +/- 0.0030 of then average ratio of all voyages listed. (eg. If the average listed is 1.00105, then all voyages within the range 0.99805 through 1.00405 would qualify)Excludes all voyage prior to any structural modification which affected the vessels cargo capacity.Excludes load or discharge data where shore measurements were not available.

This Factor is not to be applied to ships figure for assessing Ship / Shore difference.

The Factor may be applied to Ships figure to obtain an approximate B/L figure, only as a counter check where:

SHIPS FIGURE X V.E.F = APPROXIMATE B/L FIGURE.

TABLES, VOLUME AND DENSITY:API = AMERICAN PETROLEUM INDEX

ASTM = AMERICAN STANDARD OF TESTING MATERIALS

Previously there were 3 versions of ASTM Tables:USA Version - Giving API at 60 FUK Version - Giving Specific Gravity at 60 / 60 F (Ratio of density of Oil at 60 F to density of Water at 60 F = Specific Gravity)Metric Version - Giving Density at 15 C (eg. 0.865 kg/m3)

IN PRACTICE:Volume at observed temperature is calculated by taking ullages.

Density at 15 C is given by Shore authorities.

Now Volume at 15 C = Volume at observed temperature X V.C.F.

And Volume at 15 C X Density at 15 C = Weight at 15 C (IN VACUUM)

But we want Weight IN AIR, therefore we apply the W.C.F.

W.C.F. = DENSITY AT 15 C - 0.0011

NOTES:Now all 3 Versions are combined together and made into total 14 Volumes which contain all calculations regarding CRUDE, LUBE OIL, DIESEL OIL and all kinds of fluids / liquids.A particular ship may have selected Volumes only, for the trade on which she is being run.

HYDROMETER - TO MEASEURE DENSITY OF WATERPICNOMETER - TO MEASURE DENSITY OF ANY LIQUID OTHER THAN WATER

SOLVED NUMERICALS:

1.Volume at observed temperature = 10000 m3Density at 15 C. Use observed temperature = VCFVolume at 15 C = 9000 m3 (Volume at observed temperature X VCF)Therefore Weight in Vacuum = 9000 X 0.8 = 7200Density at 15 C - 0.0011 = WCF0.8- 0.0011 = 0.7989 Therefore W = Volume at 15 C X 0.7989

OBQ = ON Board QuantityROB = Remaining On Board

TOV = Vo + Vfw at observed temperatureGOV = TOV - VfwGSV = GOV x VCFTCV = GSV + VfwTRV = TCV - obq

TRU x WCF is the figure used to compare B/L figure.(GSV OBQ) x WCF = Weight of Oil Loaded

API at 60 F =141.5 - 131.5 SG 60 / 60 F

API 10 is for fresh water.Higher the API, the lighter the product

1m3 = 6.28981 barrels

6A A - stands for Crude Oil6B B - stands for Product Oil

2.On commencement of discharge of No. 3 tank at 1324, EK draft of 9.00 m, ullage of 0.20 m with waterdip 15 cm.On completion of bulk discharge at 1800, sounding of 3 was 20 cm.The tank is box shaped with dimensions L = 14, B = 12, D = 10, density at 15 C = 0.8937Average Cargo temperature = 26.0 C. Find the rate of discharge.

TOV = 14 x 12 x 9.8 = 1646.4 m3Vfw = 14 x 12 x 0.15 = 25.2 m3GOV = 1646.4 25.2 = 1621.2 m3GSV = GOV x VCF = 1608.026 m3

By Interpolation: 0.8937 0.890 0.89526 C 0.9918 0.9919therefore VCF = 0.991874

WCF = 0.8937 - 0.0011 = 0.8926

Therefore Weight of oil on arrival = 1608.026 x 0.8926 = 1435.324 TonsVolume ROB (0.2 x 14 x 12) x VCF x WCF = 29.747 Tons

Therefore Rate of Discharge in MT =(1435.324 - 29.747) TIME

WEDGE FORMULA:It is applicable to Center Tanks only and when the ship is upright and trimmed.

Refer to Figure 1.Let Breadth of the tank = b, trim = 0, dist. of ullage port from aft b/head = d, height of the tank = h.

In triangle DGB, angle DGB = 90DB = DG Cosec 0 [ DG = Sounding = Pm = Pmiddle ] = Pm Cosec 0EC = Dist of ullage port from aft b/head = dNow, BE = EC + BD CDTherefore BE = (d + Pm Cosec 0 h Tan 0) . (A)

BE is < / = AE [Wedge is formed]EF = BE Tan 0 = (d + Pm Cosec 0 - h Tan 0) x Tan 0Therefore Volume of Wedge = [1/2 x BE x EF] x bVwedge = [1/2 x BE x P] x b (B)

Now, Trim = 0Therefore Tan 0 =Trim =T.(1) Length L

Also from Figure 1Tan 0 =EF (2) BE

Comparing (1) and (2)T =EFL BE

Therefore BE =EF x L =P x L(3) T T

Putting value of BE (3) in (B),

Vwedge = x(P x L)x P x b T

Vwedge =L x b x P2 Where P = EF = Ht. Of liquid at aft b/head 2T T = TrimSee Figure 2. below

IN CASE WHEN TRAPEZIUM IS FORMED INSTEAD OF WEDGE

Refer to Figure 3 as above.Corrected Sounding (P) = {Obs. Sounding (P) + [d (h T/L)] x T/L}

Area of Trapezium = [(Pmax + z) + z] x l = [2P Pmax] x l12

Where P = Corrected sounding at aft b/head P = Sounding observed from ullage port d = Dist of ullage port from aft b/head h = Height of the tank T = Trim, L = Length of the ship, l = Length of the tank, b = Breadth of the tank

Pmax Max EF due to wedge

Pmax = T/L x l .(1)

Pcorr. sdg = P + [ d (h x T/L) ] x T/L .(2)

If (1) Pcorr. sdg, therefore a wed ge is formed

P = BE Tan 0 = [ d + Pm Cosec 0 h Tan 0 ] x Tan 0.

(Tan 0 = 2/150. Therefore 0 = 0.1458. Cosec 0 = 81.404 )

P = [ 1.0 + 0.2Cosec 0 10.5 x 2/150] x 2/150 = [ 1.0 + 0.2 x 81.404 10.5 x 2/ 150 ] x 2/150 = 0.22854

Volume of wedge =L x b x P2 =150 x 8 x (0.228544 x 0.228544) 2T 2 x 2

Therefore Volume of wedge = 15.6697 m3

3.l x b x h = 40 x 20 x 20 m3 Ullage of oil = 1.24 m Trim = 3 m Height of ullage point = 1.10 m LBP = 200 m Depth of free space above oil = 0.14 m d = 1.6 m Depth of oil = P = 19.86 m Water dip = 21.1 20.94 = 0.61 m

Pmax = T/L x l = 3/20 x 40 = 0.60 m

Pcorr.sdg = { [P] + [ (d h x T/L) ] x T/L } = [ 19.86 + ( 1.6 20 x 3 / 200) x 3 / 200 ] = 19.8795

Since Pmax < Pcorr.sdg, a trapezium is formed.

Volume of trapezium = (19.2795 19.8795) x 40 x 20 = 15663.8 m3 2Now considering the water dip,Pmax = EF /AE = T/LPmax = 3/200 x 40 = 0.60 m

Present P = ( d + Pm Cosec 0 h Tan 0) x Tan 0 = ( 1.6 + 0.16 x 66.901 20 x 0.015 ) x 0.015 = 0.180 m

Since Pmax > P a wedge is formed

Volume of wedge EBF = L x b x P2/ 2T = 200 x 20 x 0.18 x 0.18/ 2 x 3 = 21.6 m3

TOV = 15663.8 m3Vfw = 21.6 m3GOV = 15642.2 m3VCF = x 0.980784GSV = 15341.619 m3WCF = x 0.8135

Therefore amount of oil in Tonnes = 12480.407 MT

Email ThisBlogThis!Share to TwitterShare to FacebookShare to Pinterest