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Forced 5 motion result in vertical motion of ship 5 and unbalanced force in 3-direction in a ship:3 = 5 2 .

Integrating along the hull: 5 = 2 5 = 35 5 35 =

2

3. As shown above, 35 = 53 is true.

c)

1. Eigenperiod in roll is:

4 = 2 4 + 4444

44 = 42 = 42 44 = 0.3 44 44 =

4 = 2 42 (1 + 0.3)= 2 1.3 4

2

= 2 1.3 5.89.81 2 = 9.4[ ] The period of the wave 0 is the same as 4

2. The new encounter frequency:

= 0 +02

=29.4

+29.4

2

9.8128.28 0.5144 cos 135 [1] = 0.2[ 1]

=

2

0.2 = 31.4[ ]

3. The new roll period is the same as

4. The wavelength is =2

= 02

2= 138[ ]

d)

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1. The vessel is heeled < 10 0 by a known moment = = . . , and this heelmoment is equal to the restoring moment = 4 . The only unknown is , whichcan easily be found by setting = . Then we have 44 = .By using + = + , the only unknown term is KG. KB and BM is known from the

hydrostatic curves as the shape of the hull is known.

The contribution from the weights of equipment which should not be included in the lightweight shipmust be subtracted:

= + = 1 2. The vessel is heeled and then released. The oscillation is measured and the plotted result should look like figure below. We find 4 = from the figure.

By neglecting damping and using 4 = 2 4 + 4444 , ( 44 + 44 ) can be found.3. Damping can be found by using figure above. By measuring the decay in displacement from oscillationto oscillation, the damping can be determined. 40 is given and 41 is measured. With 4 1 known, we canuse the general expression for the rotation and by logarithmic decrement find 44 (with T, ( 44 + 44 ) and

44 given.)

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Solution of problem 2:

a)

The roll damping consists of two main contributions 44 4 + 44 4 , where one part is due to thegenerated outgoing waves(potential theory) and the other part due to friction and viscous effects alsononlinear terms.

1.

Skin friction, bilge keel, active stabilization fins2.

Active and passive roll damping tanks

3.

If one have active/passive roll damping tanks, these could be used. There is also a possibility of adding bilge keel, but the hull may need strengthening.

b)

1. The tank reduces the due to the free surface effect2.

Check for shallow, finite or deep water.

B is the breadth of the tank, h is the water filling depth in the tank(B = 10, h=1).

Deep water: 2 = > 12 2 = 1210 = 12012 So finite water depth must be assumed.

3.

finite water dispersion relation: 2 = tanh( )

= 4

tanh(

)

=2

=2

=2

2=

4.

As high up on the vessel as possible, because this will give a long arm from the center of rotation of the vessel to the forces created in the tank.

c)

1.

Wave period (deep water)

0 = 2 = 2 1009.81 = 8.0 2.

Wave encounter period: = 0 +

02

cos( ) = 2 + 2 cos( ) = 2100 9.81 + 2100 cos(0) 15 0.5144 1 = 1.27 1 = 4.9 [ ]

3.

Wave encounter period is set to 0: = 0 = 0 +02

cos( )

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= 0 cos( ) = 9.81 2100 9.81 cos(180)= 12.5 = 24.3 [ ]

4.

The ship and the wave moves at the same speed. This may lead to loss of stability or control, andhave been the cause of some accidents. (parametric roll)

d)

1.

Eigenperiod in heave with no damping: 3 = 2 + 3333

3 = 2 + 0.8 = 2 1.8

3 = 4.66 [ ]

= 33 = 0.8 33 =

2.

The stiffness coefficients:

33 = 35 = 53 = 0 (Because of symmetry in the bow/stern.)44 = = 44 = = = +

= 1.5 +

112

3

1 = 3.3 [ ] 55 = = = +

= 1.5 +

112

3

1 = 70.0 [ ]