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  #1  
Old 12-26-2013, 01:22 AM
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athvas athvas is offline
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Calculation of draft and trim

I have the following loading condition.

Loading condition - 50% condition
Displacement - 482.65 T
LCG - 12.636 m from AP
VCG - 3.107 m
Draft at FP -2.623 m
Draft at Mid - 2.832 m
Draft at AP - 3.041 m
Trim 0.418 m aft (bet Perpendiculars)
MCT 1cm = 6.779 t-m
TPC = 2.7 t
LBP = 28.5 m

I need to remove the following fresh water tank from the above condition

Fresh water = 17.84 T
LCG = 8.236 m from Ap
VCG = 0.760

After removing the fresh water of 17.84 Tons from the above conditon what would be the change in draft and trim. how to calculate manualy. Can you guide to solve this without using software.
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  #2  
Old 12-26-2013, 06:54 AM
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Olav Olav is offline
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athvas,

have a look at this thread. It deals with exactly the same kind of problem/task as yours.
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  #3  
Old 12-26-2013, 10:03 PM
Mr Efficiency Mr Efficiency is offline
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How about emptying the water tank ? Needs no software !
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  #4  
Old 12-26-2013, 10:46 PM
Ad Hoc Ad Hoc is offline
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It is straight forward.

You know the TPC...so, you can address the parallel rise/sinkage of the vessel. You know the MCT 1cm, so you can calculate the change in trim from the trimming moment. The only assumption you need to make is that the LCF is at amidships, since the vessel shall trik about the LCF, which value is not given.

Thus you can establishe the levers, you know the weight..there is your moment. The change of trim is also from said moment and you can see how much the vessel changes draft by a parallel sinkage/rise from the removal.

What aspect of this do you need assistance with? This is very basic naval architecture calculations even for a student by drawing out the location of the tanks, the weights involved and the levers from each knowing the vessels geometry. Thus what is the objective/purpose of your enquiry?
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  #5  
Old 12-26-2013, 11:22 PM
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athvas athvas is offline
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Quote:
Originally Posted by Ad Hoc View Post
It is straight forward.

You know the TPC...so, you can address the parallel rise/sinkage of the vessel. You know the MCT 1cm, so you can calculate the change in trim from the trimming moment. The only assumption you need to make is that the LCF is at amidships, since the vessel shall trik about the LCF, which value is not given.

Thus you can establishe the levers, you know the weight..there is your moment. The change of trim is also from said moment and you can see how much the vessel changes draft by a parallel sinkage/rise from the removal.

What aspect of this do you need assistance with? This is very basic naval architecture calculations even for a student by drawing out the location of the tanks, the weights involved and the levers from each knowing the vessels geometry. Thus what is the objective/purpose of your enquiry?
i calculated manually and also cross checked in software but i found some variations in the draft in manual calculation and the software values. so i posted this...
thanks for your reply
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  #6  
Old 12-26-2013, 11:33 PM
Ad Hoc Ad Hoc is offline
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Quote:
Originally Posted by athvas View Post
so i posted this...
Well, you haven't provided your manual calculations of your own nor the software calculated values, thus how is anyone able to assist you?
Since whatever I or any one else calculates, if it is different from yours...back to sq1. Thus please post your values so any one can inspect them for their accuracy.
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  #7  
Old 12-27-2013, 12:07 AM
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Quote:
Originally Posted by Ad Hoc View Post
Well, you haven't provided your manual calculations of your own nor the software calculated values, thus how is anyone able to assist you?
Since whatever I or any one else calculates, if it is different from yours...back to sq1. Thus please post your values so any one can inspect them for their accuracy.
Manual Calculation
For standard condition,
Trim = 0.418 m by aft; Mean draught = 2.832 m
MCT 1cm = 6.779 T-m; XB = 12.636 m; TPC= 2.7 t
Trimming moment = 464.81 (12.636 -12.804) = -78.088 t-m
Change in trim = 78.088 / 6.779 = 11.519 cm fwd
Trim for the desired condition = 0.418-0.115 = 0.302 m aft
Parallel emersion = 17.84/2.7 = 6.607 cm = 0.066 m

Base line draft at FP 2.832 - 0.066 - 0.302/2 = 2.615 m
Base line draft at AP 2.832 - 0.066 + 0.302/2 = 2.917 m
Mean draft = 2.766 m
Trim = 0.302 m aft

Software Values
Draft at FP = 2.608 m
Draft at AP = 2.943 m
Mean draft = 2.776 m
Trim = 0.335 m aft
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  #8  
Old 12-27-2013, 01:10 AM
Ad Hoc Ad Hoc is offline
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The problem you have is that the softare has more data than you are using for hand calculations.

Since, the trim occurs around the centre of flotation, LCF. This is not given, so one can only assume the LCF is as per midships, ie LBP/2.

Additionally, the removal of the weight will mean the LCG moves. But since the LCG and LCB must remain in line to be in equilibrium, where is the LCB??...Since there is now an additional trimming moment from the difference between the new LCG and the LCB of the vessel in this new condition. The LCB is not given....the software does not need this data, it can calculate the location!

Hence the software has this additional data, since the software has the 3D geometry of the hull. And as such has recalculated the location of the LCB in the new condition, which provides the additional trimming moment. This is information/data which is not included. Which is probably where you are getting the differences.
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  #9  
Old 12-27-2013, 01:30 AM
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athvas athvas is offline
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Quote:
Originally Posted by Ad Hoc View Post
The problem you have is that the softare has more data than you are using for hand calculations.

Since, the trim occurs around the centre of flotation, LCF. This is not given, so one can only assume the LCF is as per midships, ie LBP/2.

Additionally, the removal of the weight will mean the LCG moves. But since the LCG and LCB must remain in line to be in equilibrium, where is the LCB??...Since there is now an additional trimming moment from the difference between the new LCG and the LCB of the vessel in this new condition. The LCB is not given....the software does not need this data, it can calculate the location!

Hence the software has this additional data, since the software has the 3D geometry of the hull. And as such has recalculated the location of the LCB in the new condition, which provides the additional trimming moment. This is information/data which is not included. Which is probably where you are getting the differences.
The initial values of LCB and LCF before removing fresh water contents reads
LCB = 12.616 m
LCF = 11.127 m
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  #10  
Old 12-27-2013, 02:39 AM
Ad Hoc Ad Hoc is offline
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I get different values too!!

Reason is that there is no allowance in the change for the LCG change. If I ignore that, i get the same values as the software...weird!
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  #11  
Old 12-27-2013, 03:21 AM
Mr Efficiency Mr Efficiency is offline
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I have a piece of chalk I can lend to mark the waterline fore and aft, before and after the water is pumped out.
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  #12  
Old 12-27-2013, 07:49 PM
Ad Hoc Ad Hoc is offline
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athvas

Have you been able to get the same values yet?
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  #13  
Old 12-28-2013, 01:27 AM
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athvas athvas is offline
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Quote:
Originally Posted by Ad Hoc View Post
athvas

Have you been able to get the same values yet?
No iam not, i am trying to get the values atleast closer to the results not exactly.
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  #14  
Old 12-28-2013, 02:29 AM
Ad Hoc Ad Hoc is offline
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Well, I always use midships as my datum.

So, the LCF @ 11.127 = 3.123m aft of midships

And the tank @ 8.236m fwd of AP = 6.014m aft of midships.

So, the distance from the tank to the LCF = 6.014-3.127 = 2.887m

Trim = 2.887x17.84/6.779 = 7.59 call it 8cm

The parallel rise/sinkage = 17.84/2.7 = 6.6cm

The LCF is 11.127m fwd of AP or 3.123m aft of midships.

Thus the trim and hence change of draft at the AP=> 8.0/28.5 = X/11.127, thus X = 3.12cm reduction since aft of LCF.

Where X = the trim change at AP centred @ the LCF.

So draft aft = 3.041 – 0.066 – 0.031 = 2.944m your software gives you 2.943m...good enough owing to rounding errors.
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  #15  
Old 12-28-2013, 02:45 AM
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athvas athvas is offline
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Quote:
Originally Posted by Ad Hoc View Post
Well, I always use midships as my datum.

So, the LCF @ 11.127 = 3.123m aft of midships

And the tank @ 8.236m fwd of AP = 6.014m aft of midships.

So, the distance from the tank to the LCF = 6.014-3.127 = 2.887m

Trim = 2.887x17.84/6.779 = 7.59 call it 8cm

The parallel rise/sinkage = 17.84/2.7 = 6.6cm

The LCF is 11.127m fwd of AP or 3.123m aft of midships.

Thus the trim and hence change of draft at the AP=> 8.0/28.5 = X/11.127, thus X = 3.12cm reduction since aft of LCF.

Where X = the trim change at AP centred @ the LCF.

So draft aft = 3.041 – 0.066 – 0.031 = 2.944m your software gives you 2.943m...good enough owing to rounding errors.
Yes, thanks AdHoc.... Your results are more closer to the software values than my calc. thanks for spending your time on this.
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