View Full Version : How to Calculate Vessel Capacity
4th February 2010, 20:01
How to calculate the max iron ore cargo , a ship can carry?
Hello everyone ,
if u are given the data of a ships gross tonnage , net tonnage and suez canal net tonnage , how can u find out the maximum iron ore carrying capacity of tht ship .
Now i cant find the ships design draft deadweight because no other data like the length or breadth is given ...
I hope someone sheds a lil light on this pls, thnk you
5th February 2010, 0:13
The ships net tonnage should be the tare weight of the vessel without fuel and the Suez Canal tonnage the draft of the vessel with the reduced fuel loading for safe passage I would think.
And the gross tonnage would dependent upon the allowable ships net tonnage plus what ever tonnage would creat the maximum tonnage for the Suez Canal in that case to avoid lightering and loss of ballast.
I thought anyway.
5th February 2010, 13:01
The data “ships gross tonnage , net tonnage and Suez canal net tonnage” are volume based.
Gross tonnage is based on the ship’s volume from keel to funnel.
Net tonnage is based on the available cargo hold volume.
Suez canal net tonnage is similar to net tonnage with some changes.
These “tonnages” are used to calculate passing fees and harbor fees.
What you are looking for is a relationship between these volumes and the respective deadweight.
This relationship is very much depending on the type of ship.
A 100% ore carrier can have little net tonnage (small holds) and still carry a lot of ore, due to the high bulk density of the ore.
However, if you know how much ore you want to ship, you also know approximately the deadweight of the ship.
Knowing the deadweight of the ship it is possible to determine the ship’s size in length, width, depth and draft , because there exists a relationship between deadweight and ship dimensions.
Using the ship’s register books, it is possible to estimate the respective tonnages.
If you select a series of one type of ship in the registers, you can try to formulate a certain relationship between tonnage and deadweight as a regression line yourself.
All the best
5th February 2010, 19:58
hmmm so it means , say the stowage factor is 0.40 cubic mtr per tonne , stow factor being the volume of space per ton and the Net tonnage is 48,018. hence dividing the net tonnage by 0.40 we get 120045 tons which is the cargo carrying capacity of this capesize bulk carrier , jee tht was easy thnks
5th February 2010, 20:49
The tonnage is calculated with a formula:
V is cargo hold volume.
Conclusion: NT is not the same as the total hold volume.
In your example, you use the total NT (which is not the same as the hold volume) for storing the ore.
I never saw a ship fully loaded until the hatch coaming with ore.
Doing that will sink the ship.
Assume the cargo to be grain with a stowage factor of 1.25, then your ship could only carry 38415 tons.
To do these type of calculations, you have to know for which density the ship is designed.
No. it is not easy.
6th February 2010, 2:46
solving this equation, NT= (0.2+0.02*logV)*V
V is cargo hold volume.
fr NT-48,018 we get the cargo hold volume V=157,968 cubic mtrs
I refrred to the book on cargo work by david house , which states tht usually when loadin bulk carriers with iron ore the holds are only one fourth full ( in my case the ship is strengthened fr heavy cargoes with alternate holds 1-3-4-5-7-9 loaded ) assuming tht we load the cargo holds to one fourth of their capacity ( 157968/4= 39492 ) dividing this volume of space 39492 cubic mtrs by the stow factor 0.40 we get 98, 730 tonnes .
i hope this is the correct carrying capacity of my ship , im sorry , just a little confused and thanks fr bothering to reply , i really appreciate it .
6th February 2010, 9:41
If the ship, you have in mind, follows the assumption of ¼ filled holds, the calculation seems to be correct.
Repeating your calculation for light grain (Stowage factor 1.4 m3/ton) resulted in a hold filling factor of 0.875, which seems OK.
However, I wonder why you know exactly the NT of the ship and not the deadweight.
You even know that the ship has, at least, 9 holds.
Your ship of approx. 98730 dwt has the following approx. dimensions:
Length = 250 m
Width = 38.5 m
Depth = 21.13 m
Draft = 15.56 m
I did your (inventive) calculation for the Berge Stahl:
and calculated the filling factor of the 5 holds, which turned out to be 0.724.
This is much higher than your 0.4.
The conclusion is that the intended trade for which the ship is designed (ore between Brazil and Rotterdam) is of great influence on this factor.
Be careful in applying parameter figures for situations where they might not be meant for.
Have a nice day
7th March 2010, 2:50
its me again , a quick question . If you could direct me to some sites withe current iron ore freight rate trend ?
basically its abt getting into a contract fr a year to supply around 1.7 million tonnes on iron ore to turkey frm aus and brasil to turkey at 20$ per tonne and 15$ per tonne . I want check with the latest trends if this is a good deal or not
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