Practical Ship Hydrodynamics

Numerical example for BEM 259
This boundary condition must be fulfilled at any time. The steady terms give
the steady body-surface condition as mentioned above. Because only terms of
first order are left, we can exchange Ex and Ex at our convenience. Using some
vector identities we derive:
Enr O ⊲1⊳ C O Eu[⊲Enr⊳r ⊲0⊳
iωeEn] C O E˛[En ðr ⊲0⊳ CEx ð ⊲⊲Enr⊳r ⊲0⊳
iωeEn⊳]D 0
where all derivatives of potentials can be taken with respect to the inertial
system.
With the abbreviation Em D ⊲Enr⊳r ⊲0⊳ the boundary condition at S⊲Ex⊳ D 0
becomes:
Enr O ⊲1⊳ C O Eu⊲ Em iωeEn⊳ C O E˛⊲Ex ð ⊲ Em iωeEn⊳ CEn ðr ⊲0⊳ ⊳ D 0
The Kutta condition requires that at the trailing edge the pressures are equal
on both sides. This is automatically fulfilled for the symmetric contributions
(for monohulls). Then only the antisymmetric pressures have to vanish:
⊲ i t Cr ⊲0⊳ r O i ⊳ D 0
This yields on points at the trailing edge:
iωe O i Cr ⊲0⊳ r O i D 0
Diffraction and radiation problems for unit amplitude motions are solved independently
as described in the next section. After the potential O i ⊲i D 1 ...8⊳
have been determined, only the motions ui remain as unknowns.
The forces EF and moments EM acting on the body result from the body’s
weight and from integrating the pressure over the instantaneous wetted surface
S. The body’s weight EG is:
EG Df0, 0, mgg T
m is the body’s mass. (In addition, a propulsive force counteracts the resistance.
This force could be included in a similar fashion as the weight. However,
resistance and propulsive force are assumed to be negligibly small compared
to the other forces.)
EF and EM are expressed in the inertial system (En is the inward unit normal
vector):
�
EF D ⊲p⊲Ex⊳ p0⊳En⊲Ex⊳ dS C EG
S
�
EM D ⊲p⊲Ex⊳ p0⊳⊲Ex ðEn⊲Ex⊳⊳dS CExg ð EG
S
Exg is the centre of gravity. The pressure is given by Bernoulli’s equation:
p⊲Ex⊳ p0 D ⊲ 1
2⊲r total 2 1
⊲Ex⊳⊳ 2V2 C gz C total
t ⊲Ex⊳⊳