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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.

e..; z.oo C.h 8

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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.