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Long term wave statistics
Distributions of
significant wave heights
peak periodsmean wave directions
etc
Extreme value analysisindividual wave height statistics
within a given sea state
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Sources of data
Measurements (often proprietary)
Buoys
Remote sensing systems
Wave staffs
Hindcast (Norwegian Metorological Institute)
Computation of winds and waves in the past based
on weather obseservations.
Specific sites
Every 6 hours (typically)
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Table 3. Long
term wave
statistics table
Above:
Significant wave
height Hm02
Below:
Average period
Tm02.
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Figure 32.
Scatter plot ofHmo andTp.
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Joint occurrence table ofHm0and Tm02.
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Joint occurrence table for wave height and wave direction.
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The graph shows the actual
variation over a winter season for
Hm0, Tp
and the wave direction at Tp, qp.
The solid lines are
measurements and the dotted
line are daily numericalpredictions carried out by a
numerical wave model run by the
Norwegian Meteorological
Institute.
The purpose of long termstatistics is to extract and
compress the information in such
recordings in the best possible
way!
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Joint occurrence table ofHm0and wind speed (WS).
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Table 7.
Significant wave
height statistics
table for Weibull
plotting.
ClassInterva[
m]
Upper limit
[m]
ni ni Fe,j (h) = Pr(Hm0
< h)
1 0.0 - 0.49 0.49 29 29 0.00800
2 0.5 - 0.99 0.99 158 187 0.05150
3 1.0 - 1.49 1.49 830 1017 0.28055
4 1.5 - 1.99 1.99 746 1763 0.48634
5 2.0 - 2.49 2.49 626 2389 0.65903
6 2.5 - 2.99 2.99 415 2804 0.77352
7 3.0 - 3.49 3.49 298 3102 0.85572
8 3.5 - 3.99 3.99 189 3291 0.90786
9 4.0 - 4.49 4.49 152 3443 0.94979
10 4.5 - 4.99 4.99 57 3500 0.96552
11 5.0 - 5.49 5.49 44 3544 0.97766
12 5.5 - 5.99 5.99 36 3580 0.98759
13 6.0 - 6.49 6.49 26 3606 0.99476
14 6.5 - 6.99 6.99 11 3617 0.99779
15 7.0 - 7.49 7.49 1 3618 0.99807
16 7.5 - 7.99 7.99 3 3621 0.99890
17
8.0 - 8.49 8.49 0 3621 0.9989018 8.5 - 8.99 8.99 2 3623 0.99945
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0
0
( )
1 expc
F h
h H
H H
: ln( ln(1 ))Y axis F
0: ln( )X axis H H
Weibull fitting of Hm0
l ( l (1 ))
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0
0
( )
1 expc
F h
h H
H H
: ln( ln(1 ))Y axis F
0: ln( )X axis H H
Weibull fitting of Hm0
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Pr(Hm0 < h)
0.00800
0.05150
0.28055
0.48634
0.65903
0.77352
0.85572
0.90786
0.94979
0.96552
0.97766
0.98759
0.99476
0.997790.99807
0.99890
0.99890
0.99945
Upper
limit [m]
0.49
0.99
1.49
1.99
2.49
2.99
3.49
3.99
4.49
4.99
5.49
5.99
6.49
6.997.49
7.99
8.49
8.99
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Pr(Hm0 < h)
0.00800
0.05150
0.28055
0.48634
0.65903
0.77352
0.85572
0.90786
0.94979
0.96552
0.97766
0.98759
0.99476
0.997790.99807
0.99890
0.99890
0.99945
Upper
limit [m]
0.49
0.99
1.49
1.99
2.49
2.99
3.49
3.99
4.49
4.99
5.49
5.99
6.49
6.997.49
7.99
8.49
8.99
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00 00
Pr ( ) 1 exp ,mc
h HH h F h h HH H
0 0 0 0
1E 1
mH m cH H H H
0 0
2 22 2
0 0
2 1E 1 1
m mH m H cH H H
( ) ( )0
3 3 3
3 0 0
3 1 2 1E 1 3 1 1 2 1
m H cH H Hm m
g g g g
= - = - G + - G + G + + G +
Weibull distribution .
Estimation ofparameters by fitting the
statistical moments of
1., 2. and 3. order to
observed sets of Hm0
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The return period
0,1 ( )pp m RR
F H
0,
0,
Prob(exceeding the design value
only once in years)
1 ( )
p
p
m R
p
m Rp
H
R
F HR
is the average time betweeneach observation used
to establish the probability distributuion
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= 6hrs (H0 =0.8 m)
R50 = 50yr
0,50
6( ) 1
50 365 24
1 0.0000134
0.999986
mF H
0,50( )
mF H
Hmo,50yr H0 =10.8 m
Hmo,50yr= 11.6 m
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The encounter probability
0,1 ( )pp
m R
RF H
( , ; ) is the probability
that the design level associated with a return period
will occur during a period of years
p
p
E Y R
R
Y
1 1
( , ; ) 1 1 1 1/p p p
YY
E Y R R R
Percent chance of exceeding the return period design level during different time periodsY.
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Distribution of maxima
1) What is the largest wave experienced for a given sea sate?
2) How do we use the answer of 1) when the sea state varies?
.
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max 1 2Pr Pr , , ... , NX x x X x X x
1 2Pr Pr ...Pr N
N XX X X x X x F x
2
0
1 exp 2 ,Hm
hF h
H
max
2
max
0
( ) Pr 1 exp 2
N
H
m
hF h H h
H
max 00,57
ln / 2 8ln
mE H H N N
Distribution of maxima
Question 1) What is the largest wave experienced for a given sea sate?
Consider the stochastic variableXandNindependent outcomes ofX:X1,...,XN.
LetXmaxbe largest ofX1,...,XN,
The statement thatXmax x is equivalent to thatX1x,X2x,...,XN x.
By the assumption of independence:
whereFXis the cumulative distribution function ofX.
We recall that the wave heights in a sea state follow the Rayleigh distr.
the highest wave expected in a given sea sate
02/ ( is the duration of the sea state)mN A T A
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Distribution of maxima
Question 2) How do we use the answer of 1) when the sea state varies??
Consider now a sea state "1" and a sea state "2".
Exactly as before
Pr(Hmax < h during both "1" and "2")
= Pr(Hmax < h during "1")Pr(Hmax < h during "2")
Ai =NiTm02
It is obvious how this generalises as a product involving several different sea states:
1 02 2 021 2
1 2
/ /2 2
0 0
1 exp 2 1 exp 2
m mA T A T
m m
h h
H H
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Distribution of maxima
:
max
2
max
0( | ) Pr 1 exp ( 2
ij
i
N
H ij
m
h
F h A H h H
i
j
A= 7 years = 7 365 24 3600 s =220752000 s
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Distribution of maxima
:
max
2
max
0( | ) Pr 1 exp ( 2
ij
i
N
H ij
m
h
F h A H h H
Long term probability
for individual wave heights
based on several years of
measured (H,T ) and Hm0 atan offshore site in Norwegian
waters.
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Exercise 12.1
Find the expected maximum wave at a point in
the sea ifHm0 has been 1 m and Tm02 = 6 s
from the creation of the earth ( 4109 years).
max 0
8
8
0,57ln / 2
8ln
0,571 ln 6.7 10 / 2 3 m
8ln 6.7 10
mE H H N
N
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Exercise 12.2
Two sea states, (Hm0 = 4 m, Tm02 = 10 s) and(Hm0 = 8 m, Tm02 = 12 s), have both lasted for 12
hours. Determine the expected maximum wave
for the heaviest sea state and show that the
probability that a larger wave should haveoccurred during the first sea state is vanishingly
small.
max 00,57
ln / 2 8lnmE H H N N
max
2
max
0
( ) Pr 1 exp 2
H
m
hF h H h
H