Chapter 4. Engineering Wave Properties
4.1 Introduction
etc.
energy, wave
field, pressure ,
kinematics elevation Wave
Surface
potential Velocity
⎭→
⎬⎫
Wave transformation (Shoaling, Refraction, Diffraction) Wave breaking
4.2 Wave Kinematics for Progressive Waves
Particle velocity
) cosh sin(
) ( cosh
2 kx t
kh z h k g
H σ
φ =− σ + −
) sinh cos(
) ( cosh 2
) cosh cos(
) ( cosh 2
t kh kx
z h k H
t kh kx
z h k gk
H u x
σ σ
σ σ φ
+ −
=
+ −
∂ =
−∂
=
) sinh sin(
) ( sinh 2
) cosh sin(
) ( sinh 2
t kh kx
z h k H
t kh kx
z h k gk H w z
σ σ
σ σ φ
+ −
=
+ −
∂ =
−∂
=
where we used the dispersion relationship, . Note that and are out of phase.
kh gk tanh
2 =
σ u w
90°
Acceleration
) sinh sin(
) ( cosh 2
2 kx t
kh z h k H
t u dt
du = σ + −σ
∂
≅∂
) sinh cos(
) ( sinh 2
2 kx t
kh z h k H
t w dt
dw =− σ + −σ
∂
≅∂
At t=0
At x=0
Particle displacements
Consider a water particle moving around the mean position (x1,z1):
z dt u x
z u x u
dt z
x u t z x
z x z
∫
x∫
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡ +
∂ + ∂
∂ + ∂
=
+ +
=
L
) , ( )
, ( 1
1 1 1 1
1
1 1 1
1
) , (
) ,
( ) , , (
ξ ς
ξ ς ς
⎟
⎠
⎜ ⎞
⎝
⎛ T
H ⎟
⎠
⎜ ⎞
⎝
⎛ L
T H H/
) sinh sin(
) ( cosh 2
) , (
1 1
1 1
t kh kx
z h k H
dt z x u
σ + −
−
=
≅
∫
Similarly,
) sinh cos(
) ( sinh 2
) ,
( ) , , (
1 1
1 1 1
1
t kh kx
z h k H
dt z
x w t z x
σ ξ
ς ξ
+ −
=
+ +
=
∫
Let
) cos(
) ( ) , , (
) sin(
) ( ) , , (
1 1
1 1
1 1
1 1
t kx z
B t z x
t kx z
A t z x
σ ξ
σ ς
−
=
−
−
=
in which
kh z h H k
z B
kh z h k z H
A
sinh ) ( sinh ) 2
(
sinh ) ( cosh ) 2
(
1 1
1 1
= +
= +
Then
1 ) (
cos ) (
sin2 1 2 1
2 2
=
− +
−
⎟ =
⎠
⎜ ⎞
⎝ +⎛
⎟⎠
⎜ ⎞
⎝
⎛ kx t kx t
B
A ξ σ σ
ς
which is the equation of an ellipse. Note that A is always greater than B because . Also note that
x x sinh
cosh > B=H/2 at z1 =0, that is, the water particle on the free surface moves vertically between crest and trough.
In shallow water,
) 1 (
2 f z1
kh
A= H ≠ but constant over the depth
⎟⎠
⎜ ⎞
⎝⎛ + + =
= h
z H kh
z h k
B H 1 1 1
2 ) (
2 varies from H/2 at surface to at bottom 0
In deep water,
1 1
2 2
kz kh
kh kz
H e e
e e
A= H =
A H e
e e e
B H kh kz
kh kz
=
=
= 1 1
2
2 : Circle with diameter decreasing exponentially with depth
4.3 Pressure Field under a Progressive Wave
Consider the Bernoulli equation:
( )
( )2
1 2 2
t t C w
u p gz
∂ =
−∂ + +
+ φ
ρ
Neglecting small nonlinear terms,
0
0
∂ =
−∂
⎟⎟ ≅
⎠
⎜⎜ ⎞
⎝
⎛
∂
−∂ +
⎟⎟ =
⎠
⎜⎜ ⎞
⎝
⎛
∂
−∂ +
= z=
z
z g t
gz t p gz t
p φ η φ
ρ φ
ρ η
) cosh cos(
) ( cosh
2 kx t
kh z h g k
gz H gz t
p φ σ
ρ −
+ +
−
∂ = + ∂
−
=
) (
) cosh cos(
) ( cosh 2
z K g gz
t kh kx
z h k g H
gz p
η p
ρ ρ
σ ρ
ρ +
−
=
+ − +
−
=
∴
in which
kh z h z k
Kp
cosh ) ( ) cosh
( = +
is the pressure response factor, which varies from 1/coshkh at z=−h to 1 at . Note that
=0 z 1
) (z ≤
Kp . In the equation for pressure, the first term represents the hydrostatic pressure, and the second term dynamic pressure due to waves. The hydrostatic pressure increases linearly with depth, but the dynamic pressure decreases exponentially with depth.
Above the mean water level, i.e., in the range of 0<z1<η,
( ) ( )
) (
) 0 ( ) (
1 1
1 0 0 0 1 1
z g
gz g
K g z gz
z K
g gz
z z p p z p
p z p z
z
−
=
−
=
+ +
∂ − + ∂ +
−
=
∂ + + ∂
=
=
=
=
η ρ
ρ η ρ
η ρ ρ η
ρ
ρ L
L
That is, the pressure is hydrostatic for 0< z<η.
Wave measurement using a pressure transducer
) ) (
( gK z
z p K g p
p d p
d =ρ η ⇒ η= ρ
If the pressure transducer is located at the bottom (z=−h),
) gK ( h
p
p d
= − η ρ where
⎩⎨
≅⎧
=
− 0 for deep water (short-period wave) wave) period - (long water shallow for
1 cosh
) 1
( h kh
Kp
Dynamic pressure due to short-period wave is hardly recorded by a pressure transducer located at the bottom. It is difficult to distinguish real signal and noise of the gauge.
Therefore, pressure should be measured near the water surface when water depth is large.
4.4 Wave Kinematics for Standing Waves
t kh kx
z h k H g
t kh kx
z h k H g
t kx t
kx t
kx t
kh kx z h k H g
t kh kx
z h k H g
t kh kx
z h k H g
s p p
p p
σ σ σ σ
σ σ
σ σ σ
σ σ σ σ
φ
sin cosh cos
) ( cosh 2
sin cosh cos
) ( cosh
) sin cos cos
sin sin
cos cos
sin cosh (
) ( cosh 2
) cosh sin(
) ( cosh ) 2
cosh sin(
) ( cosh 2
= +
= +
+ +
+ + −
=
+ + +
+ −
−
=
t kh kx
z h H k
t kh kx
z h k gk
H u x
s s
σ σ
σ σ φ
sin sinh sin
) ( cosh 2
sin cosh sin
) ( cosh 2
= +
= +
∂
−∂
=
t kh kx
z h H k
w φz sσ σ
sin sinh cos
) ( sinh 2
− +
∂ =
−∂
=
t Hs kx σ
η cos cos
= 2
At node, coskx=0, sinkx=1, w=0, ⎟π
⎠
⎜ ⎞
⎝⎛ +
= 2
n 1 kx
At antinode, coskx=1, sinkx=0, u=0, kx=nπ
At vertical wall,
=0
∂
−∂
= x
u φ
⇒ Antinode at the wall
Local accelerations:
t kh kx
z h H k
t
u sσ sin cosσ
sinh ) ( cosh 2
2 +
∂ =
∂ ;
t u
∂
∂ is max at node and zero at antinode.
t kh kx
z h H k
t
w s σ cos cosσ
sinh ) ( sinh 2
2 +
−
∂ =
∂ ;
t w
∂
∂ is max at antinode and zero at node.
Particle displacements:
t z
x A
t kh kx
z h k H
dt z x u dt z
x u z x
s
σ
σ ξ
ς ς
cos ) , (
cos sinh sin
) ( cosh 2
) , ( )
, ( ) , (
1 1
1 1
1 1 1
1 1
1
−
=
− +
=
≅ + +
=
∫ ∫
t z
x B
t kh kx
z h H k
z
x s
σ
σ ξ
cos ) , (
cos sinh cos
) ( sinh ) 2
, (
1 1
1 1
1 1
=
= +
ς ς ξ
ξ ⎟
⎠
⎜ ⎞
⎝
−⎛
=
⇒
−
= A
B A
B : Straight line with slope ⎟
⎠
⎜ ⎞
⎝
−⎛ A B
4.5 Pressure Field under a Standing Wave
Consider the Bernoulli equation:
( )
( )2
1 2 2
t t C w
u p gz
∂ =
−∂ + +
+ φ
ρ
By the same way as for progressive waves,
) (
cos cosh cos
) ( cosh 2
z K g gz
t kh kx
z h H k
g gz gz t p
p s
η ρ ρ
σ ρ
ρ ρ φ ρ
+
−
=
+ +
−
=
∂ + ∂
−
=
Again
η η
ρ − < <
= 1 1
1) ( ) for 0
(z g z z
p
Wave force acting on a vertical wall
∫ ∫
∫
−
⎥⎦ +
⎢⎣ ⎤
⎡ +
+
=
=
= −
w w
η
-h w w
h w
z)dz ρg(η
kh dz z) ρgη k(h
gz -
dz z p F
0 0
cosh cosh
wall at the )
( ρ
η
η η
The first integral is a first-order term, and the second integral is a second-order term.
However, note that this F is neither first-order nor second-order. For a complete second order solution, the nonlinear terms (i.e., terms) should be included from the beginning.
u2
To the first order,
2
0 2
0
2 tanh 1
cosh ) ( sinh 2
1
cosh cosh
gh k kh
g
kh z h k k
gz g
kh dz z) ρgη k(h
gz - F
w
h w
-h w
ρ ρ η
ρ η ρ
ρ
+
=
⎥⎦⎤
⎢⎣⎡ +
+
−
=
⎥⎦⎤
⎢⎣⎡ +
+
=
−
∫
The first term is the wave force due to dynamic pressure, while the second term is the hydrostatic pressure force.
Since
Hs t
w σ
η cos
= 2 we have
2 2
2 max
2 1 2
2 tanh 1 2
gh C
H
gh kh
k H F g
s s
ρ ρ ρ ρ
+
=
+
=
4.6 Partial Standing Waves
Partial reflection from beaches or breakwaters: where = incident wave height, and = reflected wave height. The wave potential is
r
i H
H > Hi
Hr
) cosh sin(
) ( cosh ) 2
cosh sin(
) ( cosh
2 σ ε
σ σ
φ σ + + +
+ + −
−
= kx t
kh z h k H g
t kh kx
z h k
Hi g r
where ε = phase lag.
) 2 cos(
) 2 cos(
1
0
ε σ φ σ
η = − + + +
∂
= ∂
=
t H kx
t H kx
t g
r i
z
Since , there is no true node. Instead, we have wave envelopes within which the surface elevation varies (see Fig. 4.7 of textbook). In other words, the upper and lower envelopes are the upper and lower boundaries of surface elevation, i.e.
r
i H
H ≠
ηmax and η . Therefore, min ∂η/∂t=0 on envelopes.
Separation of incident and reflected waves
t x F t x I
t H kx
H kx t H kx
H kx
t kx
t H kx
t kx t
H kx
r i
r i
r i
σ σ
σ ε
σ ε
σ ε σ
ε σ
σ η
sin ) ( cos
) (
sin ) 2 sin(
2 sin cos
) 2 cos(
2 cos
) sin ) sin(
cos ) (cos(
) 2 sin sin cos
2 (cos
+
=
⎭⎬
⎫
⎩⎨
⎧ − +
⎭ +
⎬⎫
⎩⎨
⎧ + +
=
+
− +
+ +
=
To find the envelopes,
t x F t x
t σI σ σ σ
η =0=− ( )sin + ( )cos
∂
∂
t x F t x
I( )sinσ = ( )cosσ
) (
) ) (
tan( I x
x t m = F σ
2 2
2 2
2 2
2 2
2 max 2
) ( ) (
) ( ) (
) ( ) (
) ( ) (
) ) (
( ) ( ) (
) ) (
(
x F x I
x F x I
x F x I
x F x I
x x F
x F F x I
x x I
I
+
=
+
= +
+ +
= + η
2 2
2 2
2 min 2
) ( ) (
) ( ) (
) ) (
( ) ( ) (
) ) (
(
x F x I
x F x I
x x F
x F F x I
x x I
I
+
−
=
+ + −
+
= − η
) 2 2 cos(
2 2
) 2 sin(
2 sin )
2 cos(
2 cos
2 2
2 2
min max
ε
ε ε
η
+
⎟ +
⎠
⎜ ⎞
⎝ +⎛
⎟⎠
⎜ ⎞
⎝
± ⎛
=
⎭⎬
⎫
⎩⎨
⎧ − +
⎭ +
⎬⎫
⎩⎨
⎧ + +
±
=
H kx H H
H
H kx H kx
H kx H kx
r i r
i
i r i r
1 ) 2
cos( kx+ε =− cos(2kx+ε)=1 π
ε (2 1)
2kx+ = n− 2kx+ε =2nπ
(
i r)
r i r
i
r i r
i
H H
H H H
H
H H H
H
−
=
− +
=
⎟ −
⎠
⎜ ⎞
⎝ +⎛
⎟⎠
⎜ ⎞
⎝
= ⎛
2 1 2 2 1
2 2
2
2 2
2 2
ηmax
(
i r)
r i r
i
r i r
i
H H
H H H
H
H H H
H
+
=
+ +
=
⎟ +
⎠
⎜ ⎞
⎝ +⎛
⎟⎠
⎜ ⎞
⎝
= ⎛
2 1 2 2 1
2 2
2
2 2
2 2
ηmax
Distance between x1 and x2 = x2 −x1
k n k
n
2 ) 1 2 ( 2
2 π −ε − − π −ε
=
k 2
= π
4
= L same as perfect reflection
( )
=(
Hi −Hr)
2 1
max node
η
( )
=(
Hi +Hr)
2 1
antinode
ηmax
( ) ( )
(
ηmaxmax)
antinodeantinode(
ηmaxmax)
nodenodeη η
−
=
+
=
∴
r i
H H
To measure the wave envelopes, we have to measure waves as slowly moving a wave gauge along the wave tank (see Fig. 4.9 of textbook). This technique for separation of incident and reflected waves is applicable only to regular waves. For irregular waves,
use multiple fixed gauge method (Suh et al. 2001, Coastal Engineering, 43, 149-159).
4.7 Wave Energy
Total energy (E) = potential energy (Ep) due to displacement of free surface + kinetic energy (Ek) due to water particle movement
Potential energy
Potential energy per unit surface area:
( )
2 2
2 2
2 2
2
16 1 2
1
2 1 2
1 2 2 1
) 2 (
1
gH gh
dx dx
h L L h
g
dx h
L h g
dx g h
E L
L x x L
x x L
x x L x p x
ρ ρ
η ρ η
η ρ η
ρ η
+
=
⎭⎬
⎫
⎩⎨
⎧ + +
=
+ +
=
+
=
∫
∫
∫
∫
+ +
+ +
The first term is the potential energy in still water, while the second term is the potential energy due to wave. Therefore, the potential energy of only wave is
2
16 1 gH Ep = ρ
Another view:
L g H H
g HL 2
2 16 16 energy 2
Potential π ρ
ρ π × × =
=
∴ For unit area, 2
16 1 gH Ep = ρ
Kinetic energy
( )
{
[ ]
2
2 0
2 2
2 2
2 2 2
16 1
) (
2 cos ) ( 2 2 cosh 1 cosh
1 2
2
)}
( sin ) ( sinh
) (
cos ) ( cosh cosh
1 2
2
2 1
gH
dzdx t kx z
h kh k
gk H L
dzdx t kx z
h k
t kx z
h kh k
gk H L
dzdx w L u
E
L x
x h
L x
x h
L x
x h
k
ρ σ σ ρ
σ σ σ
ρ
ρ
η η
=
− +
⎟ +
⎠
⎜ ⎞
⎝
≅ ⎛
− +
+
−
⎟ +
⎠
⎜ ⎞
⎝
= ⎛
+
=
∫ ∫
∫ ∫
∫ ∫
+
− +
− +
−
Now, the total energy per unit surface area is
⎟⎠
⎜ ⎞
⎝
= ⎛ ⋅
= +
= 2 2 2 2
m Joule m or
m N 2
1 8
1 gH ga
E E
E p k ρ ρ
Energy flux
Energy flux, or rate of work done by wave, is
∫
−= η
h pdudz F
Average energy flux over one wave period is
g T t
t h
T t
t h d
EC ECn
kh kh gH k
dzdt t kh kx
z h k gk
H kh
z h g k
T
udzdt T p
F
=
=
⎥⎦
⎢ ⎤
⎣
⎡ ⎟
⎠
⎜ ⎞
⎝⎛ +
=
⎭⎬
⎫
⎩⎨
⎧ + −
⎭⎬
⎫
⎩⎨
⎧ +
=
=
∫ ∫
∫ ∫
+
− +
−
2 sinh 1 2 2 1 8
1
) cosh cos(
) ( cosh 2
cosh ) ( cosh 1
1
2 σ ρ
σ σ η
η ρ
η
where Cg = group velocity at which wave energy is transmitted.
dk nC d Cg = = σ
where σ2 =gk tanhkh is used for the last step.
In shallow water, n→ ,1 ∴ Cg =C In deep water,
, 2 2
1 0
0
C C
n→ ∴ g =
So far, we considered the dynamic view of the group velocity.
Kinematic view of group velocity
Consider two waves with slightly different frequencies and wave numbers, which gives a wave group.
) 2 cos(
)
2 cos( 1 1 2 2
2 1
t x H k
t x
H k σ σ
η η η
− +
−
= +
=
where
, 2 2
, 2 2
2 2
1 1
k k k
k k k
+ Δ Δ =
+
=
−Δ Δ =
−
= σ σ σ
σ σ σ
such that
1 ,
1 Δ <<
<<
Δ
k k σ
σ
Now, using the relationship
cos 2 cos 2
2 cos
cos x y x y
y
x+ = + −
we have
( ) ( ) ( ) ( )
⎥⎦
⎢ ⎤
⎣
⎡ ⎟
⎠
⎜ ⎞
⎝
⎛ Δ
−Δ Δ
−
=
⎟⎠
⎜ ⎞
⎝
⎛− Δ + Δ
−
=
⎥⎦⎤
⎢⎣⎡ − − −
⎥⎦⎤
⎢⎣⎡ + − +
=
k t x k t
kx H
t kx
t kx H
t x
k k t
x k k H
σ σ
σ σ
σ σ σ
σ η
2 cos 1 ) cos(
2 1 2
cos 1 ) cos(
2 1 2
cos 1 2
1 2
cos 1 1 2 1 2 1 2 1 2
In the last term of the above expression, the first part represents the individual wave propagating at the speed of C=σ /k, while the second part represents the modulation or envelope propagating at the speed of Cg =Δσ /Δk.
The speed of propagation of envelope or wave group is
Cg k Δ
≡Δσ
This is why we call Cg “group velocity”.
4.8 Wave Transformation Due to Depth Change
In this section, we deal with wave shoaling, refraction, and breaking.
Conservation of waves equation
{
( , ,)}
Re AeiS xyt η=
where
(
k x σt) (k x k y t)
S≡scalar phasefunction = ⋅ − = x + y −σ
∂t
∂ / and ∇ are independent operators so that the change of order does not make any difference. Thus
0 )
( ⎟=
⎠
⎜ ⎞
⎝
⎛
∂
∇ ∂
−
∂ ∇
∂
t S S
t
or
=0
∇
∂ +
∂ σ
t k
This is the equation for conservation of waves, which states that the temporal variation of wave number k should be balanced by the spatial variation of frequency σ .
Consider 1-D case:
Number of waves coming in at section 1 per unit time =
π σ 1 = 2
T .
x ⎟dx
⎠
⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎠
⎜ ⎞
⎝
=⎛
⎟⎠
⎜ ⎞
⎝
⎛
π σ π
σ π
σ
2 2
2 2 1
2
1 2
2
2 ⎟
⎠
⎜ ⎞
⎝
−⎛
⎟⎠
⎜ ⎞
⎝
=⎛
⎟⎠
⎜ ⎞
⎝
⎛
∂
− ∂
π σ π
σ π
σ dx x
Number of waves in dx = k dx L
dx π
= 2 . The rate at which the number of waves changes in dx = ⎟
⎠
⎜ ⎞
⎝
⎛
∂
∂ k dx t 2π .
x dx k dx
t ⎟
⎠
⎜ ⎞
⎝
⎛
∂
− ∂
⎟=
⎠
⎜ ⎞
⎝
⎛
∂
∴ ∂
π σ
π 2
2
=0
∂ + ∂
∂
∂ x t
k σ
Conservation of waves equation in 1-D
For a steady state,
0
0 =
∂
⇒ ∂
∂ =
∂
x t
k σ
that is, σ (or T) is constant along x in a steady state.
Refraction
0 ) (∇ =
×
∇
=
×
∇ k S by vector identity (curl of gradient = 0)
ˆ 0 0
ˆ ˆ ˆ
⎟⎟ =
⎠
⎜⎜ ⎞
⎝
⎛
∂
−∂
∂
= ∂
∂
∂
∂
∂
∂
= ∂
×
∇ k
y k x k k
k
z y x
k j i
k y x
y x
) 0 cos ( ) sin
( =
∂
−∂
∂
∂
y k x
k θ θ
Simple example: straight and parallel contours
constant sin
0 ⇒ =
∂ =
∂ k θ
y
Dividing by σ ,
constant sin =
C θ
or
0
sin 0
sin
C C
θ θ
= Snell’s law
Since C<C0, θ < θ0
Variable bathymetry:
) 0 cos ( ) sin
( =
∂
−∂
∂
∂
y k x
k θ θ
Since k =k(x,y) and θ =θ(x,y),
0 cos
sin sin
cos =
∂
− ∂
∂ + ∂
∂ + ∂
∂
∂
y k k y
x k
k θ θx θ θ θ θ
This is a 1st-order nonlinear partial differential equation for θ, because is given by for given
k kh
gk tanh
2 =
σ σ and . This equation was solved numerically by Noda et al. (1974).
h
A simpler way by Perlin and Dean (1983), Proc. Coastal Structures ’83, ASCE 988-999:
Defining
θ θ cos )
, (
sin )
, (
k y x B
k y x A
=
=
we have
x A y B
∂
=∂
∂
∂ ← solved numerically
2 2
2 B k
A + = ⇒ θ is calculated.
Wave height H is calculated by the similar way using the conservation of energy equation. Also see Dalrymple (1988). JWPCOE, ASCE, 114(4): 423-435, which includes wave-current interaction and nonlinearity.
Conservation of energy
Assume a steady state and no energy input or loss.
Assume wave energy is not transmitted across wave rays. Then
( )
=0∂
∂ bECg
s
In Cartesian coordinates,
(
cos) (
sin)
=0∂ + ∂
∂
∂ g θ ECg θ
EC y x
or
( )
=0⋅
∇ ECg Conservation of energy equation
For straight and parallel contours, ∂ y/∂ =0. Therefore,
(
cos)
=0∂
∂ ECg θ x
( )
ECg 1b1=( )
ECg 2b2 2 2 2 2 1 12
1 8
1 8
1ρgH Cg b = ρgH Cg b
2 1
2 1 1 2 1
2 1 1
2 cos
cos θ θ
g g
g g
C H C b b C H C
H = =
If position 1 is in deep water,
r s g
g
K K C H
H C b b C H C
H 0
2 0
2 0 0 2 0
2 0 0
2 cos
cos 2
2 = =
= θ
θ
where Ks = shoaling coefficient, and Kr = refraction coefficient.
Ray tracing method
1) Graphical method: Use Snell’s law over two adjacent bottom contours.
L sin ,
, sin sin
sin
3 3
3 2
2 2
2 2 1
1 θ θ θ θ θ
θ = → = →
C C
C C
Draw → refraction diagram
⎭⎬
⎫
⎩⎨
⎧
crests s orthogonal
Measure b ⇒ Calculate
b
Kr = b0 ⇒ H =H0KrKs
2) Numerical method Munk and Arthur (1952)
Griswold (1963), Journal of Geophysical Research, 68(6): 1715-1723.
Let = travel time along a ray of a wave moving with speed . Let the wave is located at at time . Then
t C(x,y)
[
x(t),y(t)]
t θCcos
dx =dt (1) θ
Csin
dy =dt (2)
? ) , (x y = θ
θ
θ sin
cos dn
ds
dx= =− ds
n C C dn
d dCdt
d ∂
− ∂
=
−
=
≅ 1
tan θ θ
θ
θ cos
sin dn ds
dy = =
Cdt ds=
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂
− ∂
∂
= ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
−
⎟⎟=
⎠
⎜⎜ ⎞
⎝
⎛
∂
∂
∂ +∂
∂
∂
∂
− ∂
∂ =
− ∂
∂ =
∂
y C x
C C
y C x
C C
n y y C n x x C C n C C s
θ θ
θ θ θ
cos 1 sin
cos 1 sin
1 1
y C x
C C s
t s s
t ∂
− ∂
∂
= ∂
∂
= ∂
∂
∂
∂
=∂
∂
∂θ θ θ θ θ
cos
sin (3)
Solve (1), (2), and (3) to find a ray using a numerical method (ray by ray).
Let β =b/ b0 = ray separation factor such that
2 / 0 = −1
= β
b Kr b
β can be found analytically (Munk and Arthur, 1952)
θ θ sin
cos n
s
x= −
θ
θ cos
sin n s
y= +
By chain rule,
y x
s y y s x x
s ∂
+ ∂
∂
= ∂
∂
∂
∂ + ∂
∂
∂
∂
= ∂
∂
∂ cosθ sinθ (4.112b)
y x
n y y n x x
n ∂
+ ∂
∂
− ∂
∂ =
∂
∂ + ∂
∂
∂
∂
= ∂
∂
∂ sinθ cosθ (4.112c)
n C C n k k x k k
y k k
y x
s ∂
− ∂
∂ =
= ∂
∂
− ∂
∂
= ∂
∂ + ∂
∂
= ∂
∂
∂ 1 1
1sin 1cos
sin
cosθ θ θ θ θ θ
θ (4.113)
θ θ θ θ
sin sin ,
cos cos
, C
dt ds dt C dy
dt ds dt C dx dt
ds = = = = =
At the point A,
ds d db nb
d =
∂
=∂θ θ
θ ,
n s b
b ∂
=∂
∂
∂ θ
1
n s b
b ∂
=∂
∂
∂β θ
β
0 0
1
n s ∂
=∂
∂
∂β θ
β
1 (4.120b)
Now
2
2 2
2
2 2
2 2
2 2 2
2 2
2
2 2 2
2 2 2
2 2 2
2 2
2
2 2 2
2 2 2
1 1
cos sin sin
sin cos
sin
cos cos
sin cos
sin cos
cos sin cos
cos sin
cos
sin cos
sin cos
sin sin
cos sin
sin cos
sin cos
cos sin
cos sin
sin cos
⎟⎠
⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
= ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
= ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
= ∂
∂
− ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
∂ + ∂
∂
∂
∂
− ∂
∂
∂
− ∂
∂
∂
∂ + ∂
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
∂ + ∂
∂
∂
∂
− ∂
∂
∂
− ∂
∂
∂
∂
− ∂
∂
− ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
= ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
⎟⎟ ∂
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
=
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
∂
− ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
∂
∂
= ∂
∂
∂
∂
− ∂
∂
∂
∂
∂
s n
C C
n s
y x
y y
y x x
y
y x y
x x
x
y y
y x x
y
y x y
x x
x
y x
y x
y x
y x
y x
s y x
n n s s n
β β θ θ
θ θ
θ θ θ θ
θ θ θ θ
θ θ θ
θ θ θ
θ θ θ θ
θ θ θ
θ
θ θ θ θ
θ θ θ θ
θ θ θ
θ θ θ
θ θ θ θ
θ θ θ
θ
θ θ θ θ
θ θ
θ θ θ θ
θ θ
θ θ θ θ
θ θ θ θ
θ θ
On the other hand, directly from (4.113) and (4.120b),
2 2 2
2 2 2 2
2
1 1
1 1
1 1
s s
n C C n
C C
s s
n C C n n s s n
∂
− ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
= ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂
∂
∂
− ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
− ∂
∂
= ∂
∂
∂
∂
− ∂
∂
∂
∂
∂
β β β
β β β θ
θ
Comparing the two results,
1 0 1
2 2 2
2 =
∂ + ∂
∂
∂
s n
C C
β β 1 0
2 2 2
2 =
∂ + ∂
∂
∂ β β
n C C s
where
2 2 2 2
2 2 2
2 2 2 2
2 2 2
2 2 2 2
2 2 2
2 2 2 2
2 2 2
2 2 2 2
2
2 2
2 2 2
2 2
2 2
cos cos
sin 2 1 sin
sin cos
2
cos cos
sin 2 sin
2
cos cos
sin 2 sin
sin cos
cos cos
sin 2 cos
sin sin
sin cos
sin cos
cos cos
sin cos
sin cos
cos sin sin
sin cos
sin
cos sin
cos
cos sin
sin sin
cos
cos sin
sin cos
cos sin
y C y
x C x
C s
y C x
C
y C y
x C x
C n
s C
y C y
x C x
C y
C x
C n
n s C
y C y
x C y
x y
C
x C y
x x
C n
s C
y C y
C y y
x C x
C y
y x
C y
C x x
C x
C x n
s C
y C x
C y
y C x
C x
n y C x
C
n y
C y
C n n
x C x
C n
y C x
C n
n C
∂ + ∂
∂
∂
− ∂
∂ + ∂
∂
⎟⎟ ∂
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
=
∂ + ∂
∂
∂
− ∂
∂ + ∂
∂
∂
∂
− ∂
=
∂ + ∂
∂
∂
− ∂
∂ + ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
∂
∂
− ∂
∂
∂
∂
−∂
=
∂ + ∂
∂
∂
− ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂
− ∂
∂
∂
∂ + ∂
∂ + ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂
− ∂
∂
∂
∂ + ∂
∂
∂
∂
−∂
=
∂ + ∂
∂
∂
∂
− ∂
∂
∂
− ∂
∂
∂
∂
− ∂
∂
∂
− ∂
∂
∂
∂ + ∂
∂ + ∂
∂
∂
∂ + ∂
∂
∂
∂
−∂
=
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
∂ + ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
∂
− ∂
∂
⎟⎟∂
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
=
∂
∂
− ∂
∂
∂
∂
− ∂
∂
∂
− ∂
∂
∂
∂
− ∂
=
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
∂
= ∂
∂
∂
θ θ
θ β θ
θ β θ
θ θ
θ θ θ
θ θ
θ θ
θ θ θ
θ
θ θ
θ θ θ θ
θ θ
θ θ θ θ
θ θ θ
θ θ θ θ θ
θ θ θ
θ θ θ
θ θ θ
θ θ θ
θ θ
θ
θ θ
θ θ θ
θ
θ θ θ θ θ
θ
θ θ
Now, we have an ODE for β(s):
2 0
2β + β + β =
ds q pd ds d
where
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
− ∂
= y
C x
C s C
p 2 cosθ sinθ )
(
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
∂
− ∂
∂
= 1 sin2 ∂22 2sin cos 2 cos2 22 )
( y
C y
x C x
C s C
q θ θ θ θ
Initial condition:
β =1, =0 dt dβ
at t=0 in deep water
which means that no refraction occurs, i.e. Kr =1, in deep water
A ray tracing method gives θ and H along rays not at grid points. So, it is difficult to use the results from ray tracing method as input data for numerical models (e.g.
sediment transport model) using finite difference method. We need interpolation.
Furthermore, infinite wave heights are calculated at the ray-crossing points. To resolve these problems, finite-difference refraction models have been developed, e.g. Noda et al.
(1974), Perlin and Dean (1983), Dalrymple (1988).
Wave breaking
(1) Finite, constant depth Miche (1944):
⎟⎠
⎜ ⎞
⎝
= ⎛
⎟⎠
⎜ ⎞
⎝
⎛
L h L
H
b
π tanh 2 142 . 0
As h/L→0, H /h=0.89 at breaking.
But, Nelson (1983, 6th Australian Conf. on Coastal and Ocean Eng.) found based on experimental results that
55 .
≤0
⎟⎠
⎜ ⎞
⎝
⎛ h b
H on a constant depth
(2) Shoaling waves in shallow water
where hb = breaking depth, and Hb = breaker height.
⎟⎟⎠
⎜⎜ ⎞
⎝
= ⎛ , 2 gT m H h
H b
b
b κ as in Fig. 12.7 of textbook,
where m = beach slope.
Surf similarity parameter based on (Battjes, 1974, Proc. 14th Coastal Eng. Conf., ASCE, 466-480):
Hb
/ L0
H m
b b = ξ
Breaker types depending on ξb:
Spilling ξb ≤0.4 smaller beach slope and steeper wave Plunging 0.4≤ξb ≤2
Collapsing 2≤ξb ≤3.5
Surging 3.5≤ξb larger beach slope and milder wave
For other breaker height formulas, see Rattanapitikon and Shibayama (2000) Coastal Eng. Journal, 42(4): 389-406.
Analyzing many laboratory results, Weggel (1972) proposed
) 2
( )
( gT
m H a m
b − b
κ =
where
(
e m)
m
a( )=43.81.0− −19
(
1.0 19.5)
156 . 1 )
(m = +e− m − b
On a beach with straight and parallel contours,
r s g
K K C H
H C
H 0 0 0 0
cos cos
2 =
= θ
θ
For breaking wave in shallow water, Cg ≅ ghb and θb ≅0. Therefore
2 / 1
0 0
0 cos
2 ⎟⎟
⎠
⎞
⎜⎜
⎝
≅ ⎛
=κ θ
b b
b gh
H C h H
0 2 0
0 2
2 cos
2 θ
κ
b
b gh
H C h =
2 1 02 0 cos 0
2 2 / 1 2 /
5 θ
κ
C H hb = g
5 / 2 0 0 2 0 5 / 4 5 /
1 2
cos
1 ⎟⎟⎠
⎜⎜ ⎞
⎝
= ⎛ θ
κ
C H hb g
5 / 2 0 0 2 0 5 / 1
2
cos ⎟⎟⎠
⎜⎜ ⎞
⎝
⎟⎟ ⎛
⎠
⎜⎜ ⎞
⎝
=⎛
=κ κ H C θ
h g Hb b
For a plane beach with h=mx, where x = distance from shoreline,
5 / 2 0 0 2 0 5 / 4 5 /
1 2
cos
1 ⎟⎟⎠
⎜⎜ ⎞
⎝
= ⎛ θ
κ
C H mxb g
5 / 2 0 0 2 0 5 / 4 5 /
1 2
cos
1 ⎟⎟⎠
⎜⎜ ⎞
⎝
= ⎛ θ
κ
C H xb mg
where xb = distance from shoreline to breaker point.
4.9 Wave Diffraction
High energy region →(Energy transfer)→ Low energy region
Examples:
Diffraction behind a breakwater: Penney and Price (1952) Philosophical Transaction of Royal Society of London, A244: 236-253.
BBC, DFSBC, and KFSBC are same as before. Far offshore from breakwater,
) cosh sin(
) ( cosh
2 ky t
kh z h k g
H σ
φ =− σ + − as y→−∞
Assume
{
F x y Z z e iσt}
φ =Im ( , ) ( ) −
where
) ( cosh )
(z k h z
Z = +
Substituting into the Laplace equation,
2 0
2 2
2 2 2
2 ⎟⎟=
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂
= ∂
∇ −
z F Z y
ZF x
e iσt ZF φ
2 0
2 2 2
2 + =
∂ + ∂
∂
∂ Fk Z
y Z F x Z F
2 0
2 2 2
2 + =
∂ + ∂
∂
∴ ∂ k F
y F x
F Helmholtz equation for F(x,y)
The solution is given by Eq. (4.135) of textbook.
4.10 Combined Refraction-Diffraction
Assume locally
t
e i
kh z h y k
x
F σ
φ = + −
cosh ) ( )cosh , (
Berkoff (1972), Proc. 13th International Conference on Coastal Eng., 471-490:
(
∇)
+ 2 =0⋅
∇ F
C F C
CCg h g
h σ mild-slope equation
The mild-slope equation is an elliptic equation, which gives a boundary value problem in horizontal space and can be solved when all the boundary conditions are specified.
Radder (1979, Journal of Fluid Mechanics, 95: 159-176) proposed the parabolic approximation, which gives an initial value problem so that the parabolic equation can be solved by marching toward shore, with given offshore boundary condition.
Also, Copeland (1985, Coastal Engineering, 9: 125-149) proposed a hyperbolic equation, and Massel (1993, Coastal Engineering, 19: 97-126) proposed an extended refraction-diffraction equation, which includes higher-order bottom effect terms compared with the mild-slope equation.