Draft of a oil tanker

Hi first timer here!

Im doing a project and I wanted to know how to calculate the draft of a tanker boat.
My tanker is 240 meters long, 32 meters beam, 20 meters high and 60000 DWT. I dont know any other parameter. Could you help me?
Can I even have an aproximated draft? Thank you for help!!

 

the DWT is only what is carried. We need to know the total displacement

 

For an approximate draft, the Panamax and Aframax limits might be relevent.

google turner this up: https://www.shipowners.dk/services/...on-analysis-for-tankers-and-bulk-carriers.pdf

 

I'm sorry I don´t have any more info about the tanker . How do you find displacement?

Thank you so much! How did you find this article?

Do you know if one can asume ballast load as 40% of DWT as a good aproximation? Is there any other aproximate method??

 

The displacement is the sum of all the weights, including cargo, of a ship.

 

That's not right at all. For example lightship displacement does not include the ship's cargo and several other loads.

 

Thats strange, I didn't think Gonzo said 'lightweight' displacement ! Hey wait a minute, he DIDNT say lightweight displacement.

Just feeling contrarian again there Tansl ?

 

rwatson, you are quite right, Gonzo has never mentioned the word lightship. It amazes me see how well you can read in English. But besides knowing English, in the present case you should know theory of the ship and speak properly. When discussing the displacement of a vessel must be indicated in what condition is the boat. Displacement, without more, does not make sense, you must say what the loading condition is the boat or what draft refers to.
Sorry I can not express myself in English as well as you. In return, I can express in terms of ship's theory much better than you, because I have clear concepts, although poorly expressed.
Thank you, as always you chasten me, for the noble attempt but this time you are not right.
Cheers.

 

I think you will find tha its Draft is roughly half its Beam, unless its restricted for some reason, such as the water depths at its likely ports.

I am, of course, referring to its loaded draft.

The reason I'm so confident about this number is that, with it, the ship has the minimum Whetted Area for its shape and diplacement.

 

Thank you for your responses, but lets calm down
Can you refer me to any guides on displacement?

@sharpii2: Thank you, I'll read about minimum Whetted Area to understant you better

Also:
Do you know if one can asume ballast load as 40% of DWT as a good aproximation? Is there any other aproximate method?

 

The definition of displacement is the weight of the displaced liquid (water), which is the same as the weight of the ship. Are you trying to find the displacement in a particular condition or setting?

 

There are several types of tankers. For its dimensions, your project looks like the "Panamax" type. Search with Google ships of this type.
The only rule as per the ballast volume refers says the boat, in the worst condition, must carry sufficient for the propeller is completely submerged.
To give a figure, orientation, for 60,000 dwt you should think of a ballast of around 36,000 m3.

 

Thank you.

About ballast load. I think i've got some good references here: AQIS. 1993b. Ballast Water Management, it says: "modern ships on transoceanic
voyages may carry 25-35% of their deadweight tonnage as ballast water.", I have some other refs so I think I'll use that :?:

About displacement: So a very rough aproximation for displacement could be area by draft by water density? in my case it would be: 240m long x 32m beam x 12m draft x 1000kg/m3. Is that right?

 

35% ballast seems little, but that should be calculated, as I said earlier, so that the ship in ballast condition have a fully submerged propeller.
As for the displacement (for a draft "T"), you must have into account the block coefficient (Cb), which in this type of boat can be close to 0.95. That is, displacement (m3) = Cb * L * B * T (usually L and B are the length and the breadth of waterplane to the draft T, all in meters)).
The Cb expresses the difference between the actual volume of the hull and what would be a parallelepiped of dimensions L, B and T.

 

[ About displacement: So a very rough aproximation for displacement could be area by draft by water density? in my case it would be: 240m long x 32m beam x 12m draft x 1000kg/m. Is that right? (GUST) ]

And multiply by the Cb (block coefficient) which would be about 0.85