Irr WavesLongterm

7/28/2019 Irr WavesLongterm

1/24

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


2/24

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)


3/24

Table 3. Long

term wave

statistics table

Above:

Significant wave

height Hm02

Below:

Average period

Tm02.


4/24

Figure 32.

Scatter plot ofHmo andTp.


5/24

Joint occurrence table ofHm0and Tm02.


6/24

Joint occurrence table for wave height and wave direction.


7/24

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!


8/24

Joint occurrence table ofHm0and wind speed (WS).


9/24

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


10/24

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 ))


11/24

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


12/24

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


13/24

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


14/24

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


15/24

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


16/24

= 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


17/24

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.


18/24

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?

.


19/24

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


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


20/24


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


21/24


:

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


22/24


:

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.


23/24

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


24/24

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

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