Reduction of the Centre of Gravity

[Mads]

Hi, I'm trying to figure out the equivalent change of the Centre of Gravity when an amount of liquid is reduced. This could be a fuel oil tank on a ship.

In other words, if I e.g. reduce the content what would be the equivalent vertical change of Centre of Gravity on a ship with a total displacement (weight) of x tonne?

Example: On a ship with a deplacement of 6000 tonne the fuel oil (947 kg/m3) tank is reduced 50%. Tank measurements: 18 m long, 8 m wide and 2.2 m deep (see attached illustration).

Hope I've made myself understandable and that some of you can help me

Thanks in advance!

 

[Mads]

I've thought of using the formula:

GG2 = q*d/disp

GG2 = change of centre of gravity
q = mass moved
d = distance moved
disp = displacement of the ship

But cannot figure out what would be the right q and d since it is not quit a relocation of mass?

 

[rumborak]

Isn't the center of gravity always half of the height of the filling level in this setup? Doesn't seem you would need a formula here.

 

[Mads]

Isn't the center of gravity always half of the height of the filling level in this setup? Doesn't seem you would need a formula here.

Yes, if it was of the tank. But I want the new centre of gravity for the ship...

 

[rumborak]

You are lacking sufficient information for calculating that. Without the knowledge of the weight of the ship itself you can't know what percentage of the total mass is the fuel.

 

[sophiecentaur]

It's in three stages. You just find the CM and mass of the ship plus empty tank, and the CM and mass of the remaining fuel and do a third calculation to find the overall CM position by combining the two.

 

[Mads]

You are lacking sufficient information for calculating that. Without the knowledge of the weight of the ship itself you can't know what percentage of the total mass is the fuel.

The weight of the ship is 6000 tonne as mentioned in the text.

 

[sophiecentaur]

You still need to know where the tank is, relative to the CM of the ship.
Ahh- from the diagram are you assuming that the oil is filling the bilge of the ship?

 

[Mads]

It's in three stages. You just find the CM and mass of the ship plus empty tank, and the CM and mass of the remaining fuel and do a third calculation to find the overall CM position by combining the two.

You still need to know where the tank is, relative to the CM of the ship.
Ahh- from the diagram are you assuming that the oil is filling the bilge of the ship?

Yes, the tank is placed at the very bottom of the hull

 

[Mads]

It's in three stages. You just find the CM and mass of the ship plus empty tank, and the CM and mass of the remaining fuel and do a third calculation to find the overall CM position by combining the two.

I'm not sure what CM stands for... I'm used to danish notaions :) Can you maybe exlpain, please?

 

[Mads]

In the attached pdf it is assignment 4.4 where I have some trouble finding the reduction of the GMt (centre of gravity to meta center height)

 

[sophiecentaur]

Sorry. Centre of Mass. CM is a Physicists pedantic name for Centre of Gravity.
You would still need to know the depth of the hull and the shape so you can work out where the CM of the empty ship is. (Open top? and very thin roof to tank?)

 

[Mads]

Sorry. Centre of Mass. CM is a Physicists pedantic name for Centre of Gravity.
You would still need to know the depth of the hull and the shape so you can work out where the CM of the empty ship is. (Open top? and very thin roof to tank?)

Thanks. I have just uploaded the assignment as an pdf. The only information I have is the hydrostatic data... Don't know if you can see it?

 

[sophiecentaur]

In the attached pdf it is assignment 4.4 where I have some trouble finding the reduction of the GMt (centre of gravity to meta center height)

OMG
Not trivial then. But the part you are asking about is straightforward, I think. Point is that you can work out the various masses and CM positions and then work out overall CM position.

Relative to any origin
Total mass times distance of overall CM = mass of one times CM distance + mass of the other times CM distance + + +
It's basically the principle of moments.

 

[Mads]

OMG
Not trivial then. But the part you are asking about is straightforward, I think. Point is that you can work out the various masses and CM positions and then work out overall CM position.

Relative to any origin
Total mass times distance of overall CM = mass of one times CM distance + mass of the other times CM distance + + +
It's basically the principle of moments.

Yeah, that was what I was afraid of. I have uploaded my solution here if it could have anyones interest. Thank you very much sophiecentaur!