MARPOL Consolidated Edition 2017
Annex /: Regulations for the prevention of pollution by oilRegulation 238 The probability Ps of breaching a compartment from side damage shal l be calculated as follows:.1 Ps = PsL. Psv . Psr.2where:P 5 L = 1 - Psr - Psa= probability the damage will extend into the longitudinal zonebounded by Xa and x,;Psv = 1 - Psu - Psi= probability the damage will extend into the vertical zone boundedby Zi and Zu; andPsr = 1 - Psy = probability the damage will extend transversely beyond the boundarydefined by y.Psa, Psr, Psi, Psu and Psy shal l be determined by linear interpolation from the tables of probabil itiesfor side damage provided in paragraph 8.3 of this regulation, where:Psa = the probability the damage w ill lie entirely aft of location ~ ;Psr = the probability the damage will lie entirely forward of location { ;Ps 1= the probability the damage will lie entirely below the tank;Psu = the probability the damage will lie entirely above the tank; andPsy = the probability the damage will lie entirely outboard of the tank.Compartment boundaries Xa, x,, Z 1, Zu and y shall be developed as follows:= the longitudinal distance from the aft terminal of L to the aftmost point on thecompartment being considered, in metres;= the longitudinal distance from the aft terminal of L to the foremost point on thecompartment being considered, in metres;= the vertical distance from the moulded baseline to the lowest point on the compartmentbeing considered, in metres;= the vertical distance from the moulded baseline to the highest point on thecompartment being considered, in metres. Zu is not to be taken greater than 0 5 ; andy = the minimum horizontal distance measured at right angles to the centreline betweenthe compartment under consideration and the side shell, in metres;*• For symmetrical tank arrangements, damages are considered for one side of the ship only, in which case all "y" dimensions are tobe measured from that same side. For asymmetrical arrangements, refer to Explanatory Notes on matters related to the accidental oiloutflow performance (resolution MEPC.122(52), as amended).84 MARPOL CONSOLIDATED EDITION 2017
Chapter 4 - Requirements for the cargo area of oil tankersRegulation 23.3 Tables of probabilities for side damagex.TPsax,T0.00 0.000 0.000.05 0.023 0.050.10 0.068 0.100.15 0.117 0.150.20 0.1 67 0.200.25 0.217 0.250.30 0.267 0.300.35 0.317 0.350.40 0.367 0.400.45 0.417 0.450.50 0.467 0.500.55 0.517 0.550.60 0.567 0.600.65 0.617 0.650.70 0.667 0.700.75 0.717 0.750.80 0.767 0.800.85 0.817 0.850.90 0.867 0.900.95 0.917 0.951.00 0.967 1.00Psr0.9670.9170.8670.8170.7670.7170.6670.6170.5670.5170.4670.4170.3670.3170.2670.2170.1670.1170.0680.0230.000z,Os Psi0.00 0.0000.05 0.0000.10 0.0010.15 0.0030.20 0.0070.25 0.0130.30 0.0210.35 0.0340.40 0.0550.45 0.0850.50 0.1230.55 0.1720.60 0.2260.65 0.2850.70 0.3470.75 0.4130.80 0.4820.85 0.5530.90 0.6260.95 0.7001.00 0.775z.Ds Psu0.00 0.9680.05 0.9520.10 0.9310.15 0.9050.20 0.8730.25 0.8360.30 0.7890.35 0.7330.40 0.6700.45 0.5990.50 0.5250.55 0.4520.60 0.3830.65 0.3170.70 0.2550.75 0.1970.80 0.1430.85 0.0920.90 0.0460.95 0.0131.00 0.000Psy shall be calculated as follows:Psy = (24.96 - 19:~6y)(JJPsy = 0.749 + (5 44.4(lsPsy = 0.888 + 0.56(ls - 0.1)for X ~ 0.05s0.05) )(J50.05) for 0.05 < ls < 0.1for ls 2:: 0.1Psy shall not be taken greater than 1.9 The probability P 6of breaching a compartment from bottom damage shall be calculated as follows:.1 Ps = PaL Par Pavw here:Pal = 1 - Par - PaaPar = 1 - Pap - PasPav = l - Paz =probability the damage w ill extend into the longitudinal zonebounded by Xa and Xr;probability the damage will extend into the transverse zonebounded by YP and Y 5; andprobability the damage w ill extend vertically above the boundarydefined by z .. 2 Paa, Par, Pap, Pas, and P 82 shall be determined by linear interpolation from the tables of probabilitiesfor bottom damage provided in paragraph 9.3 of this regulation, where:Paa = the probability the damage w ill lie entirely aft of location ~;MARPOL CONSOLIDATED EDITION 2017 85
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Chapter 4 - Requirements for the cargo area of oil tankers
Regulation 23
.3 Tables of probabilities for side damage
x.
T
Psa
x,
T
0.00 0.000 0.00
0.05 0.023 0.05
0.10 0.068 0.10
0.15 0.117 0.15
0.20 0.1 67 0.20
0.25 0.217 0.25
0.30 0.267 0.30
0.35 0.317 0.35
0.40 0.367 0.40
0.45 0.417 0.45
0.50 0.467 0.50
0.55 0.517 0.55
0.60 0.567 0.60
0.65 0.617 0.65
0.70 0.667 0.70
0.75 0.717 0.75
0.80 0.767 0.80
0.85 0.817 0.85
0.90 0.867 0.90
0.95 0.917 0.95
1.00 0.967 1.00
Psr
0.967
0.917
0.867
0.817
0.767
0.717
0.667
0.617
0.567
0.517
0.467
0.417
0.367
0.317
0.267
0.217
0.167
0.117
0.068
0.023
0.000
z,
Os Psi
0.00 0.000
0.05 0.000
0.10 0.001
0.15 0.003
0.20 0.007
0.25 0.013
0.30 0.021
0.35 0.034
0.40 0.055
0.45 0.085
0.50 0.123
0.55 0.172
0.60 0.226
0.65 0.285
0.70 0.347
0.75 0.413
0.80 0.482
0.85 0.553
0.90 0.626
0.95 0.700
1.00 0.775
z.
Ds Psu
0.00 0.968
0.05 0.952
0.10 0.931
0.15 0.905
0.20 0.873
0.25 0.836
0.30 0.789
0.35 0.733
0.40 0.670
0.45 0.599
0.50 0.525
0.55 0.452
0.60 0.383
0.65 0.317
0.70 0.255
0.75 0.197
0.80 0.143
0.85 0.092
0.90 0.046
0.95 0.013
1.00 0.000
Psy shall be calculated as follows:
Psy = (24.96 - 19:~6y)(JJ
Psy = 0.749 + (5 44.4(ls
Psy = 0.888 + 0.56(ls - 0.1)
for X ~ 0.05
s
0.05) )(J
5
0.05) for 0.05 < ls < 0.1
for ls 2:: 0.1
Psy shall not be taken greater than 1.
9 The probability P 6
of breaching a compartment from bottom damage shall be calculated as follows:
.1 Ps = PaL Par Pav
w here:
Pal = 1 - Par - Paa
Par = 1 - Pap - Pas
Pav = l - Paz =
probability the damage w ill extend into the longitudinal zone
bounded by Xa and Xr;
probability the damage will extend into the transverse zone
bounded by YP and Y 5
; and
probability the damage w ill extend vertically above the boundary
defined by z .
. 2 Paa, Par, Pap, Pas, and P 82 shall be determined by linear interpolation from the tables of probabilities
for bottom damage provided in paragraph 9.3 of this regulation, where:
Paa = the probability the damage w ill lie entirely aft of location ~;
MARPOL CONSOLIDATED EDITION 2017 85