# Imo Weather Criteria For Pontoons / Barges

Discussion in 'Stability' started by QUICKNAVAL, May 14, 2013.

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### QUICKNAVALJunior Member

Pls. help. Im making a stability calculation for a certain deck cargo barge, is there a need to include the weather criterion on the stability booklet?

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### TANSLSenior Member

IMO code of the ship intact stability for pontoons, chapter 4.7.3.2, only speak to study a uniformly distributed wind load of 540 Pa

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### jehardimanSenior Member

There is also the load line assignment to consider. You cannot load her down below her load line for the season and waters.

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### QUICKNAVALJunior Member

What does it mean the static angle of heel due to a uniformly distributed wind load of 0.54kpa should not exceed an angle corresponding to half the freeboard for the relevantloading condition?
Does tan theta = (FB/2)/(B/2) ?

Then i found this on an old stability booklet:

Part(HULL)
LPA=159.6m2
HCP=1.098m
Arm=2.593m
Pressure=0.514kpa
Moment=22.79t.m.

In the table above,,,im confuse of one thing this value "HCP". What does it stands for.

THANK YOU FOR SHARING YOUR KNOWLEDGE....

Last edited: May 14, 2013
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### QUICKNAVALJunior Member

You're right Mr. Jehardiman. What is just my concern is the heeling moment derivation on the relevant loading condition. (provided the mean draft should not exceed the assign freeboard).

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### jehardimanSenior Member

Basicly, when loaded down to the load line the amount of heel caused by the static wind load cannot cause the loss of 1/2 the freeboard...i.e. the distance between the load line and freeboard line (generally the deck edge).

LPA stands for the laterial projected area...the amount of area exposed to the wind
HCP stands for the height of the center of the projected area generally above the waterline.
The arm is the distance from the CG of the barge to the HCP.
pressure and moment are given an calculated respectively.

In the case of a deck cargo, rasing the projected area increases the moment. Since the freeboard is fixed by the load line, you are effectively limited in how large and how high the deck cargo can be.

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### QUICKNAVALJunior Member

Can i consider the area of the coaming at deck (height = 1.55m) and the lateral area of cargo which exceeds the coaming's height?

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### TANSLSenior Member

You have to con sider all exposed áreas, coamings, structures, cargo, ... any exposed área above the waterline.

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### jehardimanSenior Member

Concur, everything above the waterline.

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### QUICKNAVALJunior Member

Does the lever arm from VCG of the barge to HCP? or half of the draft plus the HCP?

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### TANSLSenior Member

I do not understand your question.
The righting lever is GZ.
The heeling arm is heeling moment (wind) divided by displacement.
Both should be equal in the equilibrium point.

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### philSweetSenior Member

Tansl, he wants to know how to score the heeling moment.

I'm just part of the peanut gallery here. But I would take a first guess at the heeling arm by adding half the freeboard to the windage force and vc of windage force calcs, and add 1/2 the freeboard to the draft for vc of resistance calc.

In other words, load it, heel it, and use the high side for windage force and center calcs, and the low side for figuring the center of resistance. The resulting moment should be less than the righting moment. Depending on the type of cargo, less than full loads might need to be looked at also.

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### QUICKNAVALJunior Member

Does the lever arm from VCG of the barge to HCP? or half of the draft plus the HCP? (Sorry i wrote the wrong term,,,,what i mean is the distance "z", instead of lever arm)

Wind Heeling lever (lw1) = (P*A*z)/(1000*9.81*Displacement)

where:
P = wind pressure of 0.504 kpa
A = windage area above w.l.
z = ?

Is "z", the distance from VCG of the barge to HCP (z=VCG+-(d+HCP))? Or it is distance half of the draft of the relevant loading condition plus the HCP (z=0.5d+HCP)?

Last edited: May 14, 2013
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### TANSLSenior Member

"z" is the distance between the centers of gravity of the submerged area and the total area above the waterline.
The attached figure, although it corresponds to a sailboat, can clarify the issue.
z = hlp + hce

#### Attached Files:

• ###### Z dist.jpg
File size:
18 KB
Views:
637

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### QUICKNAVALJunior Member

Thank you....

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