12262013, 01:22 AM
  Senior Member   Join Date: Feb 2013
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Location: india   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 tm
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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athvas

12262013, 06:54 AM
  arch. nav.   Join Date: Dec 2003
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Location: Filia pulchra Lubecć   athvas,
have a look at this thread. It deals with exactly the same kind of problem/task as yours.
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12262013, 10:03 PM
 Senior Member   Join Date: Oct 2010
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Location: Australia   How about emptying the water tank ? Needs no software ! 
12262013, 10:46 PM
 Naval Architect   Join Date: Oct 2008
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Location: Japan   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? 
12262013, 11:22 PM
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Location: india   Quote:
Originally Posted by Ad Hoc 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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athvas

12262013, 11:33 PM
 Naval Architect   Join Date: Oct 2008
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Location: Japan   Quote:
Originally Posted by athvas 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. 
12272013, 12:07 AM
  Senior Member   Join Date: Feb 2013
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Location: india   Quote:
Originally Posted by Ad Hoc 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 Tm; XB = 12.636 m; TPC= 2.7 t
Trimming moment = 464.81 (12.636 12.804) = 78.088 tm
Change in trim = 78.088 / 6.779 = 11.519 cm fwd
Trim for the desired condition = 0.4180.115 = 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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athvas

12272013, 01:10 AM
 Naval Architect   Join Date: Oct 2008
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Location: Japan   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. 
12272013, 01:30 AM
  Senior Member   Join Date: Feb 2013
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Location: india   Quote:
Originally Posted by Ad Hoc 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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athvas

12272013, 02:39 AM
 Naval Architect   Join Date: Oct 2008
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Location: Japan   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! 
12272013, 03:21 AM
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Location: Australia   I have a piece of chalk I can lend to mark the waterline fore and aft, before and after the water is pumped out. 
12272013, 07:49 PM
 Naval Architect   Join Date: Oct 2008
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Location: Japan   athvas
Have you been able to get the same values yet? 
12282013, 01:27 AM
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Location: india   Quote:
Originally Posted by Ad Hoc 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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12282013, 02:29 AM
 Naval Architect   Join Date: Oct 2008
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Location: Japan   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 = 6.0143.127 = 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. 
12282013, 02:45 AM
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Location: india   Quote:
Originally Posted by Ad Hoc 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 = 6.0143.127 = 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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