First Class Naval Arc

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A.DUG
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First Class Naval Arc

Postby A.DUG » Fri Jun 07, 2013 9:09 am

A rectangular Block of wood of square cross section of side Sand length L greater then S is to float in fresh water with its horizontal axis parallel to the waterline, the sides are vertical and in neutral equilibrium. Calculate the relative density of wood.


Now how am i supposed to approach this question and give a number when no numbers are given to me

Mass= VOL X Density

RD= Density of wood \ density of water


I have no mass, no definite vol, the only variable i have to work with is the 1000kg\m3 density of fresh water.
Ive been looking for similar problems in the Derrett Stability book with no success.

Any help would be greatly appreciated and i will most definatly have more questions as i go through the questions.
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Re: First Class Naval Arc

Postby A.DUG » Fri Jun 07, 2013 12:36 pm

A fuel tank bulkhead is made in the shape of a trapezoid 13 m wide at the top, 10 m wide at the bottom and 7.5 m deep. When the tank is filled with fuel of 5 m, calculate: 1- the load on the bulkhead 2 the position of the center of pressure relative to the surface of the fuel
note: R.D. of fuel = 0.8

INA (rectangle) = BD3 12

INA (triangle) = BD3 36

For the Load :

Area of the 2 triangles in the trapezoid = 2 X BH\2 --> the 2 and \2 cancel ou so BH = 1.5 X 5m (depth of liquid) = 7,5m2
Area of the rectangle (center of trapezoid) = BH = 10m X 5m (depth of liquid again) = 50m2

Pressure = Depth of liquid X density of liquid = 5m X 0.8 = 39.54 T\m2

Thrust = Pressure X Area = 39,54 X (50m2+7.5m2) = 2256.3 T

That part im pretty sure of now for the center of pressure thats where im lost. Can anyone guide me to a solution?
It must involve the two moments of inertia listed otherwise why would they list them?
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Re: First Class Naval Arc

Postby A.DUG » Sat Jun 08, 2013 5:17 pm

How do i even start this?!?!?!?!


C24. For a ship 135 m in length, 16.0 m breadth the values of tonnes per centimeter immersion (TPC) in water of density 1025 kg/m3 are as follows:

Draught 1.2 1.8 2.4 3.0 3.6 4.2 4.8
TPC 14.6 14.83 15.1 15.36 15.54 15.7 15.82

The displacement of the ship below the 1.2 m draught is 1200 tonnes. If at a draught of 4.8 m the position of the longitudinal center of buoyancy below the metacenter (BML) is 140 m and the second moment of area of the waterplane about midships is 935000 m4, calculate:


a) the distance of the longitudinal center of flotation (LCF) from midships

b) the block coefficient
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Big Pete
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Re: First Class Naval Arc

Postby Big Pete » Wed Jun 12, 2013 1:29 pm

Hi Dug,

Welcome to the wonderful world of Exams!!


Ref Q1)
You have not been given any numerical values so you can not work out a numerical answer. They say that the horizontal axis will be parrallel to the waterline, if they said what the offset was you could work out a numerical value, but they dont!

All you can do is to combine and transpose your formuales so that you have them in the Form R.D. = ????

Ref Q2

The pressure on the Bulkhead varies uniformly from zero at the top of the liquid to Rho.g.h at the base of the bulkhead so you could use our old friend Calulus or something similar to calculate the total load on the triangles. To do it by simple arithmetic or square counting or something similar would be very hard.
The total load on the rectangle relatively is easy, because it is symetrical relative to the pressure:- area x average pressure will give you the total force.

You then have to calculate the centres of pressure of the rectangle and the two triangles, which is where the moments of inertia come in.

You then take moments about the base of the Bulkhead, multiplying the force on each Bulkhead by the height of of the centre of pressure above the base. Take the totsl of these moments and divide by the total force or load, that will gibe you an overall centre of pressure.

Ref Q3

Dont know how to calculate part a) from the information give, but it is about 28 years since I wrote my last Naval Architecture paper!! Like the first question you could just rejig the formulae so that it is in the form: -
"The distance of LCF from midships = ???

The Bloc coefficient is the ratio between a soild rectangular Block of the same Length beam and draught as the ship and the actual measure volume (or calculated from the displacement so that should be very easy for you if you have already written your Class 2 .

Best of Luck.

BP
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Re: First Class Naval Arc

Postby A.DUG » Wed Jun 12, 2013 3:20 pm

too late i passed it

thanks anyways though
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carbob
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Re: First Class Naval Arc

Postby carbob » Wed Jul 03, 2013 4:12 pm

If I remember correctly, the mass of the volume of water displaced is equal to the mass of the immersed object, n'est pas?

Volume of water displaced = S x S/2 x L, which = S(squared)L /2 meters cubed. Mass of body = S(squared)L/2 x 1000 kg/m3.

If Mass = Volume x Density, it follows that Density = Mass / Volume.

Therefore, Density = S(squared)L/2 divided by S(squared)L, which is the same as S(squared)L/2 X 1/S(squared)L.

When you cancel out the S(squared)L from the top and bottom, you're left with 1/2, which means to me that the density of the body is half that of water, RD = 0.5.

Sorry if this looks confusing, looks much easier to understand in numeric form on paper. The fact you're not given any numbers just means you don't need them. They're not looking for exact value, just the RD.

Does this seem right to anybody, or am I right out to lunch??

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Big Pete
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Re: First Class Naval Arc

Postby Big Pete » Fri Jul 05, 2013 1:23 pm

You still cant give an answer from the information given.

The breadth and height are both S (square cross section) and the length is L, so the volume of the wood is S squared L.
If you were told that the draft was S/2 then you would be correct, but we are not given the draught, so we can not calculate the volume of water displaced.

If you say the draft is D then volume of water displaced is: S x D x L

If you divide the volume of water displaced by the total volume of the Block, the Ls cancel out and one s from each equation leaving Relative Density = D/S, but we are not given the draught.
The rubric says that "its horizontal axis parallel to the waterline" i.e. that there is no Trim, it does not say that the Horizontal axis is at the waterline.

BP
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carbob
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Re: First Class Naval Arc

Postby carbob » Fri Jul 05, 2013 5:30 pm

You may well be right, Big Pete. I took 'floating at neutral equilibrium' to mean floating at 'arf and 'arf. Perhaps that is wrong, who knows, but it is how I would have approached it had I seen in during my exam.

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Big Pete
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Re: First Class Naval Arc

Postby Big Pete » Sat Jul 06, 2013 9:05 am

Hi Carbob,
I can sort of see where you are coming from, but to me the expression: -

'floating at neutral equilibrium'

Implies that this is a static situation, rather than a dynamic situation. i.e. the block has not just been dropped in the water and is bouncing up and down until it reached the equilibrium position.

Sometimes answering the questions correctly takes very close reading and an excellent grasp of Technical English.

BP
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Re: First Class Naval Arc

Postby JollyJack » Sun Jul 07, 2013 11:38 am

Best advice I can give is "RTQ", Read The Question. Sometimes, an "F" is thrown in there too.

Next bit of advice, "KISS" Principle, keep it simple, stupid. Examiner's brains are not baffled by bullshit, they are far better at it than you'll ever be, that's why they are what they are :)

For 1st and 2nd Class EKs, explain it as if you would to a cadet, or a brand new Junior first-tripper. Make it understandable, buzz words and jargon indicate you don't have a clue what you're talking about
Discourage incest, ban country "music".


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