Gross Tonage

Does any one know of a way to calculate the hull volume with out having a linesplan?

 

You have to know a few basics, but here is a great resource:

http://www.uscg.mil/hq/msc/docs/CG-5397_Rev_6-04_v1_0b.pdf

and

http://www.qualitymaritime.info/Press/Guide_Tonnage_Simplified_0.pdf

 

Sure. If you know the weight of the hull and add weights to bring it down to a waterline, divide the total weight (hull weight + added weights) in pounds by 62.4 for fresh water and 64 for salt water to get the volume in cubic feet. multiply by 0.03 to get cubic meters.

If you are doing this in kilograms, 1 cubic meter of fresh water weighs 1000 kg. So you simply divide the number of kilograms (hull wt + Added wt) by 1000 for fresh and 1027 for salt water.

This of course means you already have the boat in the water.

 

You can make a gross estimate by multiplying the main dimensions and CB (CB can be estimated according to the ship type). But that is only a gross estimate...
Rodrigo

 

Gross tonnage is a legal measurement. You need the line plans or measure the cargo area.

 

Quite a strange question coming from a naval architect.

Daniel

 

These comments brings up the point wether you are after the submerged volume of the hull or the enclosed volume of the hull. Both volumes can be estimated from the general dimensions as RAaujo stated, but Gonzo is correct that the actual Gross Tonnage is requirements based and cannot be accurately calculated without plans and arrangements because of what is and is not counted in the volume.

Submerged volume on the other hand can be accurately determined wothout a lines plan if other drawings like the D&O's, Bonjeans Curves, or the Stability Book is avaiable.

 

Hey guys many thanks for your comments. I am new to the forum and I must say that it is very useful. I should have given you more details at the very beginning.

In short I have a situation in which an old work boat has had some of it's superstructure removed and the new GT needs to be calculated. The only drawing I have in hand is a GA.

What I was trying to figure out is that if I know the original GT, can the new GT be calculated by using the volume of the superstructure removed which I can obtain form the GA.

Using GT=K1*V and the block coef.(Cb) formula. If Cb is assumed the total volume of the vessel can be estimated. Using the estimate of the total volume, K1 can be found and by deducting the the volume removed from the total volume, the new GT can be calculated.

Has anyone had similar situation? Is this a impossible task?

 

What rule are you calculating for?

 

No rule as such.

 

Hi,
As you probably already know it, the calculation of ship's Gross Tonnage is defined in the Annex1, Regulation 3 of the International Convention on Tonnage Measurement of Ships (1969). You can find the text of the Convention in this webpage, for example: http://www.admiraltylawguide.com/conven/tonnage1969.html

Now, if you know the previous GT (let me call it GT1), then you can calculate the associated volume by resolving the following system of equations:
Eq.1) GT1 = K1 V1
Eq.2) K1 = 0.2 + 0.02 Log10 (V1)

An iterative process is necessary to resolve this system, you can proceed like this:

Step 1: V1a = GT1 / (0.2 + 0.02 Log10 V1)
Step 2: V1b = V1 + 0.5 (V1a - V1)
Step 3: Difference = V1b - V1
Step 4: V1 = V1b
Step 4: Go to Step 1 until Difference is sufficiently close to 0.

I am enclosing here an xls spreadsheet with implemented procedure which converges after 15-16 iterations. It will give you the moulded hull volume V1 from the known GT1.

Now that you know the V1, you can simply subtract the removed volume to obtain the moulded hull volume of the modified ship, V2.
Calculating the new Gross Tonnage (GT2) is then straightforward from equations 1 and 2 given above.
Hope that I have understood well your question and that this is what you needed.

Cheers!

 

Where does the 0.5 in step 2 come form?

 

To keep the iteration steps at adequate level I think..

 

That's correct, TeddyDiver. It is called Fixed Point Functional Iteration (More about it here, for example: http://pathfinder.scar.utoronto.ca/~dyer/csca57/book_P/node34.html).
In this case, step 1 "predicts" the next value of V1, while step 2 "corrects" the prediction by halving (hence 0.5) the difference between the previous-step V1 and the predicted next-step V1. 0.5 is a value which gives a pretty fast convergence. Should there have been problems with convergence, I would have used a smaller multiplier, say 0.2 or 0.1.

 

I assume that since I have two equations and two unkowns this iteration method results in a fairly accurate result no?

Daiquiri thats for the spreadsheet it worked fine!