Need assistance on boat question

[Aegis]

Hello, I am a novice in the marine field and I need advice on doing my tutorials.

THe question is:

a ship approaches a seawater canal with drafts 28'0" foreward and 29'6" aft. the max permitted draft in the canal lock is 29'0". The master proposes to take on seawater ballast in deep tank 1 in order to enter the lock.

at 28'9" draft the hydrostatics are: TPI = 69.4 tones
MTI = 1870 ft tons
LCF = F = 280.5 ft
A = L - F = 528-280.5 = 247.5ft
Deep tank 1 has a capacity of 137.4 tons of seawater and its LCG is 40.3ft aft of the forward perpendicular.

1)how many tons of seawater will be needed?
2) will the final forward draft be satisfactory?

Thank you for your help

[alan white]

I think the only way to get an accurate answer is to get an actual ship and try it out on a windless day, see what happens,

[Aegis]

the correct answer is 131 tons of seawater
the forward draft will be 28'11" and that is satisfactory. This is suppose to be a purely theoretical question. I am seeking for advice on how to approach such problems regarding trim and ballasting

[alan white]

Sorry Aegis. I have nothing to do and it's too hot to sleep.

[Ad Hoc]

Not sure if your "F" is fwd or aft of FP or AP?, this is also assuming that the LCF remains unaltered, for the purposes of the calculation.

This is not a one line calculation, since there are 2 variables.

1) sinkage
2) trimming moment.

So, firstly, assume a certain mass of water being added, say 10, 20, 40 ton and so on.

This mass divided by the TPI, gives the parallel sinkage, to both draft values.

Secondly, the distance form the LCG of the SW Tank to the LCF is the lever. This lever multiplied by the mass you have added is the trimming moment. So, divide this moment by the MTI, will give you how much trim, over the Lpp, total. (Using "similar triangles" over the distance from AP to F and F to FP)

You can then summate the final drafts, by adding the parallel sinkage, to both drafts values, then take the total trim calculated , via the MTI, and add/subtract from the aft, fwd drafts as calculated.

You can do this just twice (in reality - assume it is linear), and then just interpolate between the 2 masses, to give you your answer. Then do a final check calculation, just to confirm.

[Aegis]

yes F = FP and A = AP

However, when additional seawater is taken on does it alter the position of the center of gravity of the ship?? If it does does, will this have any effect on the trim of the vessel??
Anyway if I were to assume a certain mass of seawater as a ballast? Wouldn't I have to try multiple values before I can arrive at the correct answer??
I was thinking I should approach the question by trying to reduce the aft draft first reducing the draft by 7".

Then use the draft to be reduced from the aft to calculate the trimming moment needed to cause the trim with the MTI.
Next with the trimming moment needed, I can deduce the weight ot seawater needed by dividing with the lever distance from the LCF to the LCG of the tank.
And finally get the amount of parallel sinkage by dividing the weight of seawater by the TPI.

Thank you for your efforts and advice

[Ad Hoc]

All depends upon the mass being added relative to the displacement of the vessel. For small mass additions, the above is a rough and ready acceptable method. But when adding large masses, more information, such as LCB, LCG et al, are required.

In your "thinking", you haven't accounted for the sinkage owing to the mass being added, the TPI.

[Aegis]

So should I account for the parallel sinkage due to the weight being added first
or should I consider the adjustment to the aft draft mark first??

Coz I wanted to account for the parallel sinkage after I have deduced the weight of seawater needed to be taken in to reduce the aft draft

[Ad Hoc]

You do the parallel sinkage first. Since you are adding a mass to the boat, so it will sink wherever you place it!

[Aegis]

Ok then...THank you for all your suggestions

[Fanie]

Ad Hoc, sounds very interesting. Any chance you could explain how that is done (without the abbrev) ?

[Ad Hoc]

Fanie

If you're referring to the original question, then the rough and ready reply is in my post #5.
Referring to post #8...is no different, as the methodology for post #5.

Basically whenever you place a weight/mass on board a boat it can be categories into 2 main effects, minor or major.
1) minor, relates to post #5. that is minor weights being added. One assumes the mass being added as a percentage is very small compared to the displacement of the boat. As such the boats hydrostatic properties, such as MCT (moment to chnage trim), TPC (tonnes per centimetre) unlike above, these are in metric not imperial, LCB (longitudinal centre of buoyancy) etc do not chnage. This then makes the calculation very quick and simple. As per post #5.

2) This is when the mass is sufficient to cause a chnage in the LCG (longitudinal centre of gravity) and hence this in itself creates a trimming moment. Therefore more data on the boat, from the hydrostatics is required. But, the principal method of calculation remains the same....calculate the sinkage and then calculate the trimming moment.

To calculate the sinkage is just => mass/TPC = sinkage in centimetres

The trim, is trimming moment (mass x distance)/ MCT = total trim over the LWL.

You then have trim/length as a ratio or "similar triangle". This triangle of height and length as a ratio, is the same wherever the triangle. Hence, the triangle of height or "X"/ distance from AP to LCF (centre of flotation) is the same. So you just cross multiply to get "X". X being the rise or fall in draft owing to the trim.

Therefore the new draft at, say AP, is the original, +sinkage + or - the chnage in draft owing to trim = final draft.

Is that what you mean Fanie??

[Fanie]

Hi Ad Hoc,
Yes, thank you, I get the idea.

The master proposes to take on seawater ballast in deep tank 1 in order to enter the lock.

So this is done to 'balance' the weight inside the boat so it will 'tilt' to the required depth prior to entering the channel ?

[Ad Hoc]

Fanie

Correct, or more to the point, one must calculate it to prove it is possible first. Better to make a calculation first than running the ship aground!

Since the draft aft = 29'6" and fwd = 28'

Since the canal depth is 29'...obviously the stern will hit the bottom....so, how much SW can be added, to trim the boat, so that the aft draft is reduced to allow the clearance. But in doing so, the fwd draft will increase, but also, by how much and will this then hit the canal bottom?

[Fanie]

In the old days they threw stuff off the boats to make them lighter... now they're taking on water

Ships is a bit out of my league, but it is interesting to know. Can't give you any points, I'm beginning to think Jeff is terrorising me