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1.8-3 On the same sheet of graph paper plot the following sine waves. Or, you may program these waves
into a spreadsheet and plot them. Assume all waves are in phase and have zero amplitude at time zero.
(a) Wave amplitude 10 ft, wave period 6 seconds. Amplitude(t) = 10 ft · sin(1.0472 radsec · t)
(b) Wave amplitude 6 ft, wave period 4 seconds. Amplitude(t) = 6 ft · sin(1.5708 radsec · t)
(c) Wave amplitude 4 ft, wave period 3 seconds. Amplitude(t) = 4 ft · sin(2.0944 radsec · t)
(d) Using the superposition theorem, plot the wave that would be created by the combining the three
waves, assuming all waves are traveling in the same direction. What is the maximum height of
the resulting wave?
Superposed wave has a peak amplitude of 18.12 ft.
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2.8-5 In Chapter 7 we saw that ship resistance was a force that countered the thrust produced by the ship’s
propellers. In other words, two opposing forces act in the direction of surge. However, surge motion is
not a simple harmonic motion for ships. Explain why.
This is simply because the magnitude of the ship’s resistance is a (complicated) function of velocity,
not a function of position (like for heave) or angle (like roll or pitch). This is to say that while the
resistance is a force in the opposing direction of the motion, it is not a restorative (conservative) force
in the way that gravity is.
3.8-6 An object is moving with simple harmonic motion.
(a) For the motion to be maximized, what relationship must exist between the object’s motion and
the forcing function?
z(t) =F
k
1
1−(
ωωnat
)2
The timing of the object motion and the forcing function (freqeuncies) must match up (i.e. the
frequency of the forcing function and the natural response frequency of the object must be close
in magnitude).
(b) What would happen if the magnitude of the forcing function were doubled?
The amplitude of the object response should also be doubled.
(c) What would happen if the frequency of the forcing function were doubled?
If ω > ωnat, the amplitude of the response of the object would be reduced. If ω < ωnat
2 , the
amplitude of the response of the object would be increased.
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4.8-8 How does wave height affect the response of a ship?
The wave height is a measure of wave amplitude and, therefore, the amplitude of the forcing function
on the ship. As wave height ↑, the amplitude of ship response ↑.
Note: wave energy is proportional to wave height squared(Ewave ∝ Hwave
2).
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5.8-11 A ship has the following rigid body motion and structural response frequencies:
ωheave = 0.42rad/s ωlong bend = 0.50rad/s
ωpitch = 0.53rad/s ωtorsion = 0.41rad/s
ωroll = 0.50rad/s
The ship is currently traveling at 12 knots directly into a sea system rated at Sea State 7 using the
NATO classification table.
(a) Using Table 8.1, what is the modal wave frequency associated with this system?
From the table, SS 7 → TSS 7 = 15.0 sec within a range of periods of 11.8− 18.5 sec.
ωwave =2π
TSS 7=
2π
15.0 sec= 0.4189
rad
sec
(b) Comment on the motion being experienced by the ship.
ωe = ωw −(ωw)
2 · Vs · cos(µ)
g= 0.4189
rad
sec−(0.4189 rad
sec
)2 · 12 kts · 1.688fts
kt · cos(180◦)
32.17 ftsec2
ωe = 0.5294rad
sec
This excitation frequency is very close to the natural frequencies of both pitch and roll accelerations
and torsional bending on the ship structure. The ship is expected to have large pitch motions and
very torsional stresses (which my exceed the capabilty of the crew or the hull structure).
(c) The ship is about to alter course by 45◦. Comment on the feasibility of this course change.
Examine the encounter frequency at this course relative the the waves:
ωe = 0.4189rad
sec−(0.4189 rad
sec
)2 · 12 kts · 1.688fts
kt · cos(135◦)
32.17 ftsec2
ωe = 0.4970rad
sec
This course is also extremely undesirable as the ship will be excited in roll and (to a lesser extent
pitch) and and will experience very high longitudinal bending stresses.
6.8-12 A warship is underway on a course of 090◦ T at a speed of 14 kts. The ship is 465 ft in length, a
displacement of 8,000 LT , and its center of gravity is located 19 ft above the keel.
The ship has the following natural frequencies of rigid body motion and structural response:
ωheave = 1.17 radsec ωlong bend = 1.05 rad
sec
ωpitch = 1.07 radsec ωtorsion = 1.20 rad
sec
ωroll = 0.69 radsec
The ship is steaming in waves coming from the west. Waves are 450 feet in length, 8 feet in height,
and are at a period of 9.4 seconds.
(a) Calculate the ship’s encounter frequency and comment on the ship’s response at this frequency.
ωwave =2π
T=
2π
9.4 sec= 0.6684
rad
sec
ωe = ωw −(ωw)
2 · Vs · cos(µ)
g= 0.6684
rad
sec−(0.6684 rad
sec
)2 · 14 kts · 1.688fts
kt · cos(0◦)
32.17 ftsec2
ωe = 0.3402rad
sec
Excitation from the waves is coming at a frequency far from any of the ship’s natural frequencies
→ no significant concerns.
(b) The ship turns to a course of 270◦ T. Determine the ship’s encounter frequency on this new course
and comment on the ship’s response.
ωe = ωw −(ωw)
2 · Vs · cos(µ)
g= 0.6684
rad
sec−(0.6684 rad
sec
)2 · 14 kts · 1.688fts
kt · cos(180◦)
32.17 ftsec2
ωe = 0.9966rad
sec
This encounter frequency is very close to the pitch and heave natural frequencies and the structural
resonance for longitudinal bending which is undesirable.
(c) In order to conduct flight operations, the ship turns into the wind on a course of 325◦ T. Calculate
the new encounter frequency and comment on the motion of the ship. Flight operations require
a steady deck . . . is this a good course for flight operations?
ωe = ωw −(ωw)
2 · Vs · cos(µ)
g= 0.6684
rad
sec−(0.6684 rad
sec
)2 · 14 kts · 1.688fts
kt · cos(125◦)
32.17 ftsec2
ωe = 0.8567rad
sec
The ship is angled to the seas, causing some roll, but the excitation frequency is fairly far from
the natural frequencies of roll and heave, and reasonably far from pitch. This should be a decent
heading for flight operations (provided 8 ft seas don’t frustrate operations).
(d) The Navigator realizes that the ship is 4 hours ahead of schedule and recommends slowing the
ship to 6 kts. Is this a wise choice with flight operations in progress?
ωe = ωw −(ωw)
2 · Vs · cos(µ)
g= 0.6684
rad
sec−(0.6684 rad
sec
)2 · 6 kts · 1.688 fts
kt · cos(125◦)
32.17 ftsec2
ωe = 0.7491rad
sec
This speed brings the encounter frequency much closer to the natural frequency in roll which will
produce large amplitudes in roll which will seriously degrade flight ops.
(e) As the ship burns fuel, its center of gravity rises. What is happening to the ship’s metacentric
height, stability, and roll response as fuel is burned?
KG ↑⇒ GMT ↓⇒ overall stability (R.O.S., stiffness, GZmax, etc.) will be lower. With ω =√
kM
where kroll ∝ GMT , kroll ↓⇒ ωnat ↓ and roll amplitude ↑.
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7.8-15 A small frigate 340 ft in length is pursuing a target at a speed of 25 kts on a course of 045◦ T. Seas,
350 ft in length, are from the southwest at a height of 8 feet and a period of 8.27 seconds. The ship
has the following natural frequencies of rigid body and structural response:
ωheave = 1.03 radsec ωlong bend = 0.75 rad
sec
ωpitch = 1.21 radsec ωtorsion = 1.12 rad
sec
ωroll = 0.67 radsec
(a) Determine the frigate’s frequency of encounter with the waves and comment on its rigid body and
structural response.
ωwave =2π
T=
2π
8.27 sec= 0.7598
rad
sec
ωe = ωw −(ωw)
2 · Vs · cos(µ)
g= 0.7598
rad
sec−(0.7598 rad
sec
)2 · 25 kts · 1.688fts
kt · cos(0◦)
32.17 ftsec2
ωe = 0.002557rad
sec
The ship speed is nearly matching the wave propagation so there is no significant ship motion is
induced.
(b) On its current course and speed, how are the waves affecting the power required to achieve a
speed of 25 knots?
As Lwave ≈ Lship, there is the potential for having large resistance (or very low resistance)
depending on the location of the hull and the wave causes wave constructive interference (high
resistance) or destructive interference. This is like attempting to power the hull at speeds above
the Hull Speed (trying to climb out of the trough of its own wake).
(c) Is the current course of 045◦ T a good course for the ship’s structure? Explain why or why not.
Lwave ≈ Lship, there is the potential for maximal hogging or sagging loading of the hull structure.
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8.8-17 Describe one active and one passive anti-roll device commonly used on ships. Why are there no
similar anti-heave or anti-pitch devices?
Passive: Bilge Keels (or Stabilization Tanks/Weights) serve to dampen roll motion by increasing viscous
drag associated with angular velocity about the x axis.
Active: Fin Stabilizers (or active Stabilization Tanks/Weights or gyros) provide counter moments to
the roll by generating forces at each of two lifting surfaces in directions opposing the roll.
As long, slender vessels are efficient to power but are more susceptable to excitation in roll than pitch
and heave (due to larger damping and k-m ratios), compensation devices for these motions in ships
have been viewed as a much lower priority.