13.42 Spring 2005
13.42 Design Principles for Ocean Vehicles
Prof. A.H. Techet
Spring 2005
Froude Krylov Excitation Force
1. Radiation and Diffraction Potentials
The total potential is a linear superposition of the incident, diffraction, and radiation potentials,
φ = (φI + φD + φR ) eiωt .
(1)
The radiation potential is comprised of six components due to the motions in the six directions,
φ j where j = 1, 2, 3, 4, 5, 6 . Each function φ j is the potential resulting from a unit motion in j th
direction for a body floating in a quiescent fluid. The resulting body boundary condition follows
from lecture 15:
∂φ j
= iω n j ; ( j = 1, 2, 3)
(2)
= iω (r × n) j −3 ; ( j = 4, 5, 6)
(3)
∂n
∂φ j
∂n
r = (x, y, z )
(4)
n = n j ( j = 1, 2, 3) = (nx , ny , nz )
(5)
In order to meet all the boundary conditions we must have waves that radiate away from the body.
Thus φ j ∝ e ∓ ikx as x → ± ∞ .
For the diffraction problem we know that the derivative of the total potential (here the incident
potential plus the diffraction potential without consideration of the radiation potential) normal to
the body surface is zero on the body:
version 1.0
updated 3/29/2005
∂φT
∂n
= 0 on S B , where φT = φI + φD .
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13.42 Spring 2005
∂φI
∂φ
= − D ; on S B
∂n
∂n
(6)
We have so far talked primarily about the incident potential. The formulation of the incident
potential is straight forward from the boundary value problem (BVP) setup in lecture 15. There
exist several viable forms of this potential function each are essentially a phase shifted version of
another. The diffraction potential can also be found in the same fashion using the BVP for the
diffraction potential with the appropriate boundary condition on the body. This potential can be
approximated for a long wave condition. This long wave approximation assumes that the incident
wavelength is very long compared to the body diameter and thus the induced velocity field from
the incident waves on the structure can be assumed constant over the body and approximated by
the following equation:
φD

∂φ
∂φ
i  ∂φI
φ1 + I φ 2 + I φ3 

∂y
∂z 
ω  ∂x
(7)
Further explanation of this approximation can be found in Newman (p. 301).
Ultimately, if we assume the body to be sufficiently small as not to affect the pressure field due to
an incident wave, then we can diffraction effects can be completely ignored. This assumption
comes from the Froude-Krylov hypothesis and assures a resulting excitation force equivalent to
the froude-krylov force:
F FK (t ) = − ρ ∫ ∫
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∂φI
n dS
∂t
(8)
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13.42 Spring 2005
2. Vertical Froude-Krylov Force on a Single Hull Vessel
z
x
T
B
Deep water incident wave potential is:
φI =
aω kz
e Re i ei (ωt −kx) 
k
(9)
The force in the vertical direction is found from the incident potential using eq. 8 along the bottom
of the vessel. Here the normal in the z-direction, nz , is negative: nz = −1, so the force per unit
length in the z-direction is
iaω −kT i (ωt −kx ) 
 B/ 2
−ρ iω
FzFK = Re  ∫
e e
dx 
− B/ 2
k




ρ oω 2


k2
= Re 
(10)


a e −kT eiωt  e −ikB/ 2 − eikB/ 2  
(11)


 ω2

= Re 2 ρ 2 a e −kT eiωt sin(kB/2) 
k


Recall that sin z =
eiz − e − iz
2i
(12)
.
Using the vertical velocity we can rewrite the force in terms of the velocity. version 1.0
updated 3/29/2005
w(t ) = Re  aω ekz i ei (ωt − kx ) 
(13)
w(t ) = Re  −aω 2 e kz ei (ωt −kx) 
(14)
w( x = 0, z = 0,t) = Re  a ω 2 eiωt 
(15)
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Now we can write the force in the vertical direction as a function of the vertical (heave)
acceleration,
 2ρ

Fz = Re  2 e −kT sin(kB/ 2) w(0, 0,t )  .
k

(16)
Let’s look at the case where ω → 0 the wavenumber, k = ω 2 /g → 0 , also goes to zero and the
following simplifications can be made:
e kt
1 − kT
sin(kB/2)
(17)
kB/2
(18)
to yield a simplified heave force.


ω2


k2
Re  2 ρ
FzFK


a (1 − kT ) (kB/ 2) eiωt 
(19)



 ω 2  iωt 

Re  ρ g aB 1−
Te 


g 



(20)
If we look at the case where ω → 0 and consider the heave restoring coefficient, C33 = ρ g B ,
and the free surface elevation, η ( x, t ) = Re  a ei (ωt −kx)  we can rewrite this force as
Re {C33 η (x = 0, t)}
FzFK
(21)
3. Horizontal Froude-Krylov Force on a Single Hull Vessel
The horizontal force on the vessel above can be found in a similar fashion to the vertical force.
Fx = ∫ ∫
SB

= Re  ρ i ω


iω a
k
∫
ρ
∂φI
nx dS
∂t
(22)

0
−T
e kz dz  ei (ωt − kB/ 2) − ei (ωt +kB/ 2)  
(23)


 aω 2 

− kT  iω t
= Re i ρ
2 sin( kB/ 2) 
1 − e
e


k


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As frequency approaches zero similar simplifications can be made like above for the vertical
force:
 aω 2

Re i ρ
( KT ) eiωt 2 kB/ 2 
k


Fx (t )
(25)
u (t ) = Re  aω ekz ei (ωt − kx ) 
(26)
u (t ) = Re i a ω 2 e kz ei (ωt − kx ) 
(27)
Fx (t )
Re { ρTB u ( x = 0, z = 0, t )}
(28)
Where ρTB = ρ∀ , and ∀ is the vessel volume such that we are left with the surge force
ρ∀ u
(29)
C33η + ρ∀w
(30)
Fx
Fz
4. Multi Hulled Vessel
z
-B/2
B/2
T
T
b
x
b
Again, let’s make a few basic assumptions: ( b/λ << 1 ), ( B /λ ∼ 1 ), ( a < b ), and ( b ∼ T ).
Let’s look at the force in the x-direction:
FxFK
ρ bT u ( x = − B/ 2, z = 0, t ) + ρ bT u ( x = B/ 2, z = 0, t )
(31)
η ( x, t ) = a cos(ωt − kx)
(32)
u ( x, z, t ) = − a ω 2 e kz sin(ωt − kx)
(33)
(34)
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13.42 Spring 2005
FxFK
ρ bT (−a ω 2 ) {sin(ωt + kB / 2) + sin(ωt − kB/ 2)}
(35)
−2 ρ b T (a ω 2 ) cos(kB/ 2) sin(ωt )
(36)
Note that when kB/ 2 = π/2 (or B = λ/2 ) then FxFK (t ) = 0 .
4.1. Multi Hulled Vessel with additional pontoon
z
-B/2
B/2
T
T
x
c
b
b
Use the same assumptions from above to find the x-force adjusted for the additional pontoon
between the two hulls.
FxFK
− 2 ρ b T (a ω 2 ) cos(kB/2) sin(ωt)
+c p ( x = −B / 2 + b / 2, z = 0, t)
− c p ( x = B / 2 − b / 2, z = 0, t)
The last two terms are the adjustment to the force for the addition of the pontoon, δ FxFK (t ) .
Pressure is found from the incident potential: p(x, z, t ) = ρ g a e kz cos(ωt − kx) .
k
2


δ FxFK = −2 ρ g a sin(ωt) sin  (B − b) 
(37)
For B >> b using g = ω 2 /k we get a force:
FxFK (t )
version 1.0
updated 3/29/2005
−2 ρ a ω 2 sin(ωt ) {bT cos(kB/ 2) + δ/ 2 sin(kB/ 2)}
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