# How to calculate displacement with simpson's rule

Discussion in 'Stability' started by Adeyele, Aug 1, 2012.

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Dear all, am a student and i was given an assignmnt to find displacemnt with simpsons rule without common interval. the draught are 0.762M Forwd and 1.905M Aft.

The segments are 0, 3.04, 4.9, 5.02, 4.62, 4.21, 3.8, 3.38, 2.93, 1.79 and 0 metre-sqs

pls can any1 hlp me on how to find d displacemnt.

Regards
Mike

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### DCockeySenior Member

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You still need a common interval for the 5 8 -1 rule. Which the OP says he doesn't have.

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### DCockeySenior Member

Any suggestions for what he could use?

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### DCockeySenior Member

What are the distances between the stations?

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There is Tchebycheff's rule, failing that, just treat each as a trapezoid and work each individually.

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### TeddyDiverGollywobbler

OP's numbers look like something isn't right or is missing..

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### PARYacht Designer/Builder

Durand's rule still requires uniform spacing doesn't it? I'd agree he'll have to treat each section with the trapezoidal rule.

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Yes:
http://mathworld.wolfram.com/DurandsRule.html

He could use Guass' Rules, but that is very long and tedious. I think trapezoidal rule for each segment is quicker and easier.

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### gonzoSenior Member

Simpson's rule is based on equal intervals. I agree that the trapezoidal rule is easier.

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### ldigasSenior Member

Mike, please, it would be very useful if you could tell what were your exact input data, for (I at least) cannot make sense of what is the above you've been given. What are the 0, 3.04, 4.9 ... values? Metre-sqs is meters squared I presume?

Anyway, as others have already noted for Simpson's rule (either the most popular one using a quadratic curve) or some of the less popular variants, you need to have equidistant spacing of points. That means the distance between the points can change, but you need AT LEAST three equidistant points.

For example, this series could work
0 1.5 3 6 9 10 11
Because the distance in the first few points is 1.5, then 3 then 1. But you need to have at least three points to put the quadratic curve through.

That being said, Simpson's rule is merely a way to integrate an area beneath the curve. You could fit a curve through your points, interpolate a new set of equidistant points and then use Simpson's rule to calculate the area beneath it.

All of this, however, is rather moot since we would need to first determine what is the given set of data.

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Dear all, sorry for late reply; I av a problm wit my connection.

Actly i saw an example in d ship stability for masters and mates txtbk. Dey use first rule SM to multiply d areas, and divide d sum of area by Product sum. dis givs displacemnt.

Many thanks.
Mike

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### ldigasSenior Member

Still, please, if you could present the assignment as it was given. It would help to clarify the problem at hand. And not everyone has the copy of the book in question.

Willing to help, by all means, but the given explanation doesn't make much sense (could be a language issue).

Displacement is usually calculated in one of three ways, which are analogous ... either by integrating the areas below the design waterline of the buttocks, or the sections, or by integration of waterline areas up to the design waterline.

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My Assignment

Dear all,

Please kindly checK the information below; for the assignment i was talking about.

Type of vessel is Single Screw Dry Cargo Coaster
Length Overall 31.85 Metres
Moulded Depth Amidships 2.87 Metres
Full Load Draught (Amidships to underside of keel) 2.53 Metres
Displacement to Full Load Draught (Amidships to underside of keel) 423.5 Tonnes

QUESTION 1
When an inclining experiment was performed on the vessel the draught forward was noted as 0.762 Metres (frame 53) and aft 1.905 Metres (frame 3).
The waterline length is 29.50 Metres and for the purposes of this assignment has been divided into ten equal segments with the half-area at each section, from aft, being as follows: -
0 0.00 M2
1 3.04 M2
2 4.90 M2
3 5.02 M2
4 4.62 M2
5 4.21 M2
6 3.80 M2
7 3.38 M2
8 2.93 M2
9 1.79 M2
10 0.00 M2
Using Simpsonâ€™s rules calculate the displacement in salt water including the position of the longitudinal centre of buoyancy.
The total weight used for the inclining experiment was 3.60 Tonnes, positioned on the side decks port and starboard, (half each side), 18.22 Metres forward of the transom and 3.25 Metres above the keel, the shift was 5.200 Metres transversely. The forward pendulum length was 3.040 Metres and the aft pendulum 2.545 Metres. What effect does the removal of the weights have on both the vertical and longitudinal centre of gravity?

Question 2

Below are the pendulum deflections from the inclining experiment: -
Experiment
Weight Moved
Forward Pendulum Deflection
Aft Pendulum Deflection
1
0.400 Tonnes starboard to port
35.00 mm
23.00 mm
2
1.200 Tonnes starboard to port
75.00 mm
65.00 mm
3
0.800 Tonnes port to starboard
55.00 mm
45.00 mm
4
0.400 Tonnes port to starboard
22.00 mm
20.00 mm
5
0.400 Tonnes port to starboard
29.00 mm
21.00 mm
6
0.800 Tonnes port to starboard
51.00 mm
45.00 mm
7
0.800 Tonnes starboard to port
51.00 mm
43.00 mm
8
0.800 Tonnes starboard to port
53.00 mm
44.00 mm
9
0.800 Tonnes starboard to port
50.00 mm
43.00 mm

Using the answers to Question 1, the pendulum deflections and the hydrostatic data below calculate the GM and VCG of the vessel in the inclined condition.
Draught Amidships (Metres)
0.914
1.067
1.219
1.372
1.524
1.676
1.829
Displacement (Tonnes)
128.2
153.0
178.2
204.0
230.2
257.0
284.3
Block Coefficient
0.504
0.535
0.562
0.580
0.592
0.602
0.610
LCB from Transom (Metres)
14.833
15.048
15.198
15.302
15.370
15.411
15.430
LCF from Transom (Metres)
16.178
16.143
16.070
15.962
15.831
15.691
15.539
KB above Base (Metres)
0.501
0.579
0.658
0.737
0.8183
0.900
0.983
BMt (Metres)
3.905
3.382
2.982
2.669
2.418
2.213
2.042
KMt above Base (Metres)
4.355
3.908
3.590
3.362
3.196
3.076
2.991
BML (Metres)
64.033
56.545
51.169
47.253
44.282
41.874
39.959
KML above Base (Metres)
64.535
57.124
51.827
47.991
45.100
42.775
40.941
TPc (Tonnes/cm)
1.607
1.641
1.673
1.707
1.742
1.775
1.810
MTc (Tonne. Metres)
2.609
2.755
2.912
3.087
3.274
3.467
3.671
In the full load condition the vessel can carry 280 Tonnes of homogeneous cargo having an LCG of 16.55 Metres from the transom and 1.94 Metres above the keel. Calculate the draught, trim, position of the longitudinal and vertical centres of gravity for the loaded vessel? Using the KN data below prepare a GZ curve for the loaded vessel?

I will really appreciate any help on how to solve this questions.

Regards
Mike

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