Ship Motions Ship motions in a seaway are very complicated but can be broken down into 6-degrees of
freedom motions relative to 3 mutually perpendicular axes passing through the center ofgravity (CG) of ship as shown below.
Translational motions:1. Surge: Longitudinal disturbance fore & aft along shipβs track superimposed on the
shipβs forward velocity.2. Sway: Lateral disturbance along the π¦ βaxis as port & starboard drift.3. Heave: Vertical disturbance caused by the imbalance between the weight of the ship
and the instantenous changes in the buoyant force resulting from wave action.
Rotational motions:1. Roll: Transverse oscillatory rotation about shipβs transverse axis.2. Pitch: Longitudinal oscillatory rotation about shipβs transverse axis.3. Yaw: Rotation about shipβs vertical axis.
Motion of a Buoy in Waves
We shall consider the simplest of these motions for thesimplest possible geometry; namely, heaving motion of abuoy of circular cross-section in waves.
Motion of a Buoy in Waves Newtonβs second law:
π π§ = πΉπ
where the right-hand side is the summation of all vertical forces acting on thebuoy, m is the mass, and π§ = π2π§ ππ‘2 is the instantaneous acceleration.
Supposing that the buoy is subjected to only a steady external force it will floatat an equilibrium position with a hydrostatic or buoyancy force of
π βπ§ + π
where π§ is the vertical displacement of the buoy from its still water levelpozition, π is the surface elevation, and π = πππ΄π€, where π΄π€ is the water planearea. The expression represents the total change in buoyancy from the initialcalm-water condition as the result of both change in water level and verticaldisplacement, neglecting the pressure attenuation with depth. The attenuation
correc tion would require that π be multiplied by the factor πβπ(π§+π), where πis the wave number, and π is the draft. If π is large in relation to π§, then πβππ.
Motion of Buoy in Waves When regular waves are present they generate an exciting force which causes
heaving motion. Both π§ and π are harmonic but not necessarily in phase,π§ = π§0 cos ππ‘ + π , π = π cos ππ‘
where π§0 and π are amplitudes of heave and wave motions, respectively, π iscircular frequency 2π ππ€, ππ€ wave period, and π is the phase angle by whichheaving motion lags the wave.
Hydrodynamic Forces: One component of hydrodynamic force can be relatedto relative vertical acceleration between buoy and fluid, and therefore it is180π out of phase, or opposite to, the buoyancy force,
π β π§ + π
where π is the so-called βadded massβ or βhydrodynamic massβ.
The component of hydrodynamic force which is 90πout of phase with both therelative acceleration and the buoyancy is the damping force:
π β π§ + π
where π is the damping coefficient. Here, the damping force is assumed linear.The approximate attenuation factor πβππ applies to all π-related terms.
Motion of a Buoy in Waves The equation for dynamic equilibrium of the buoy at any instant, based on
Newtonβs law, can now be stated as follows, without depth attenuation effect
π π§ = πΉ = π β π§ + π + π β π§ + π + π βπ§ + π
Hydrodynamic Hydrostatic
After rearranging,
π+ π π§ + π π§ + ππ§ = π π + π π + ππ
The right-hand side is customarily referred to as the exciting force, whichrepresents the force exerted by the waves on the buoy when it is restrainedfrom vertical motion. Using the previously defined π = π cos ππ‘ ,
π π + π π + ππ = π π β ππ2 cos ππ‘ β ππ sin ππ‘ = πΉ1 cos ππ‘ + πΉ2 sin ππ‘
where πΉ1 = π π β ππ2 and πΉ2 = βπππ. The exiciting force can then be writtenπΉ = πΉ1 cos ππ‘ + πΉ2 sin ππ‘ = πΉ0 cos ππ‘ β π
where πΉ0 = πΉ12 + πΉ2
2 1/2= π π β ππ2 2 + ππ 2 1/2 , π = tanβ1 βππ
πβππ2.
Here, π is the phase angle by which the force πΉ lags the wave elevation π. Wemay now consider the depth attenuation factor as follows.
Motion of a Buoy in Waves The equation of motion for the buoy in waves
π+ π π§ + π π§ + ππ§ = πΉ0 cos ππ‘ β π πβππ
Substituting π§ = π§π cos ππ‘ + π results in
π§π = πΉ0 π β ππ2 β ππ2 2 + π2π2 β1/2, π = tanβ1 πππβ π+π π2
where π is the phase angle by which the heaving motion lags the force. Toobtain the total phase lag angle π between the motion and the wave we musttake into account the phase angle π, hence π = βπ + π. Using πΉ0 the heavingmotion amplitude π§π is written in the final form of
π§π = ππ β ππ2 2 + π2π2
π β ππ2 β ππ2 2 + π2π2
1/2
πβππ
andπ§ = π§π cos ππ‘ β π + π
In these equations π is known and π is a simple geometrical quantity. Thecoefficients π and π can be determined by experiment.
Motion of a Buoy in WavesConsider the free (unforced) motion:
π + π π§ + π π§ + ππ§ = 0
Letting π§ = π§0πππΌπ‘ results in
β π + π πΌ2 + πππΌ + π = 0
which in turn gives for πΌ
πΌ1,2 =βππ Β± ππ 2 β 4 β π + π π
β2(π + π)
πΌ1,2 = ππ
2(π + π)β
π
(π + π)β
π2
4(π + π)2
Defining π = π/ π(π + π) 1/2 as the dimensionless damping parameter
π =π
2(π + π)ππ = ππ 1 β
π 2
4
1/2
ππ =π
π + π
1/2
where π is the damping coefficient, ππthe undamped natural frequency and ππ is the damped frequency of the system.
Motion of a Buoy in WavesThe solution of the damped unforced system is then
π§ = π§0πβππ‘ cosπππ‘
This solution means that the amplitude gradually decreases with time because of damping, as expressed by the factor πβππ‘.
Motion of a Buoy in WavesAs shown in the figure we may designate the successive amplitudes by π§0, π§1, π§2, ... Then considering the first complete swing, we can obtain the expression for π§ when π‘ = ππ β ππ (for small damping); thus
π§2 = π§0πβπππ
Hence π§2/π§0 = πβπππ
or β ln π§2/π§0 = πππ =ππ
2
π
π+π
This is referred to as the logarithmic decrement. Substituting ππ = 2π(π+π
π)1/2
β ln π§2/π§0 =ππ + π
π
12 π
π + π= ππ
If the damping is linear, π§4/π§2 = π§2/π§0 etc., and the samme result will be obtained for each successive circle. Hence, knowing π, π, and ππ, we can solve for the damping coefficient, π. The complete solution to the equation for the buoy in waves can thus be obtained. π , which is used in the forced solution, can also be computed from the above equation.
Motion of a Buoy in WavesNow turning back to the forced (due to waves) solution
π§ = π§π cos ππ‘ β π + π
π§π = ππ β ππ2 2 + π2π2
π β ππ2 β ππ2 2 + π2π2
1/2
πβππ
Let us define a nondimensional quantity known as the tuning factor
Ξ =π
ππ= π
π+ π
π
1/2
and rearrange π§π as
π§π = π
1 βππ2
π
2
+π2π2
π2
1 β(π + π)π2
π
2
+π2
π2π2
1/2
πβππ
The above amplitude may be expressed in terms of the nondimensional tuning factor Ξ and the nondimensional damping parameter π .
Motion of a Buoy in WavesDefining a dimesionless ratio π called the magnification factor, representing the ratio of buoy motion to wave motion at draft π as
π =
1 βΞ2
3
2
+ π 2Ξ2
1 β Ξ2 2 + π 2Ξ2
1/2
We can then write the solution as π§ π‘ = πππβππ cos ππ‘ β π + π .
Unresisted Rolling in Still Water Unresisted rolling in still water: One of the important motions
is rolling. Capsizing of ships mostly occurs in rolling motion in waves. Here, we first consider the rolling in still water without any external forcing. The equation of motion for undamped roll motion is given by
πΌπ2π
ππ‘2+π = 0
where πΌ is the mass moment of inertia of the ship about a longitudinal axis through the center of gravity, π is the righting moment, and π is the angle of inclination of the ship from the vertical. Letting
πΌ =Ξ
ππ2
where Ξ is the displacement, π the gravitational acceleration and π is the radius of gyration of mass of ship about a longitudinal axis through the center of gravity. For small angles of inclination
π = Ξ β πΊπ = Ξ β πΊπ β sinπ = Ξ β πΊπ β π
Unresisted Rolling in Still WaterSubstituting these values we have
π2π
ππ‘2+ππΊπ
π2π = 0
The above equation is the equation for simple harmonic motion having period
ππ =2ππ
ππΊπ=
2π
πΊπthe latter expression being valid for SI units.
Example: A 10,000 ton-ship has πΊπ = 0.9 π and ππ = 15 π . Determine the rolling period after moving 1000 tons symetrically away from a mean distance of 3 m to a mean distance of 6 m.
Solution: Calculate the radius of gyration of mass π =ππ πΊπ
2=
15 0.9
2= 7.1 π
then πΌ = Ξπ2/π = 10000 β 7.12 = 504100 π‘ππ β π2. The altered mass moment of inertia πΌβ² = 504100 + 1000 62 β 32 = 531100 π‘ππ β π2. The new radius of
gyration of mass πβ² =531100
10000= 7.29 π and the rolling period ππ
β² =2β7.29
0.9=
15.4 π .
Unresisted Rolling in Still WaterSolving the differential equation
π2π
ππ‘2+ππΊπ
π2π = 0
gives
π π‘ =πππ΄ππ‘
ππ
2πsin
2ππ‘
ππ+ ππ΄ cos
2ππ‘
ππ
π π‘ =πππ΄ππ‘
1
ππsinπππ‘ + ππ΄ cosπππ‘
where at π‘ = 0, π = ππ΄ and ππ/ππ‘ = πππ΄/ππ‘ are the initial angle of roll and the initial angular velocity of roll while ππ = 2π/ππ is the circular frequency. If
we assign as initial conditions that ππ΄ = 0 and ππ/ππ‘ = πππ΄/ππ‘ when π‘ = 0,
π π‘ =πππ΄ππ‘
1
ππsinπππ‘
whereas if, when π‘ = 0, the inclination is equal to ππ΄ and πππ΄/ππ‘ = 0π π‘ = ππ΄ cosπππ‘
Unresisted Rolling Among Waves Unresisted rolling among waves: To a first approximation the
wave-disturbing moment is proportional to the wave slope tanπΌπ = 2ππππΏπ€For small angles πΌπ in radians may be substituded for tanπΌπ hence
πΌπ2π
ππ‘2+ ΞπΊππ = ΞπΊπ 2πππ/πΏπ€ sinππ€π‘
where ππ is the wave amplitude, πΏπ€ the wavelength, and ππ€ the wave frequency. The above equation may be re-written in the following form
π2π
ππ‘2+ ππ
2 = ππ2 πΌπ sinππ€π‘
where the right-hand side is the exciting moment. Solving the above differential equation with initial conditions that ππ΄ = 0 and πππ΄/ππ‘ = 0 gives
π π‘ =πΌπ
1 βππ
2
ππ€2
sinππ€π‘ βππ
ππ€sinπππ‘
Note that when ππ€ = ππ the equation reduces to 0/0, which must be evaluated for the limit to get π π‘ = πΌπ/2 sinππ€π‘ β ππ€π‘ cosππ€π‘ .
Resisted Rolling in Still Water Resisted rolling in still water: If the resistance to rolling is
π΄ ππ/ππ‘ ,
πΌπ2π
ππ‘2+ π΄
ππ
ππ‘+ ΞπΊππ = 0
The solution of the above equation yields the following results
π΄ =ΞπππΊππΎ1
π2, ππ
β² =ππ
1 βπΎ1
2
π2
1/2
where πΎ1 is a coefficient related to π΄ by the above relationship and is less than
unity; therefore πΎ12 /π2 is less than 0.1 and the denominator of ππ
β² is less
than unity. Consequently, the period of resisted rolling differs from the period of unresisted rolling only by a small amount. For instance, let πΎ1 = 0.1 then
1 βπΎ1
2
π2
1/2
= 0.999 = 0.999, ππβ² =
ππ
0.999= 1.0005ππ