How to use Simpsons Rule Maths for Ship Stability

Question

I did a Civil Engineering course some years ago and from my textbook I had this question. As I am interested in this I have been trying to solve this, but unfortunately I haven't been able to find a good solution yet. I would be appreciative if someone can discuss my solution and why I have not got the answer from textbook of 0.74 m.

The half ordinates of the contour of the load water plane of a ship of 2,800 metric tons displacement, measured at intervals of 6 m, are 0.30, 3.33, 5.07, 6.18, 6.75, 6.90, 6.75, 6.18, 5.61, 4.47, 0.60 m respectively. If the centre of gravity is 1.87 m above the centre of buoyancy, calculate the transverse metacentric height. Sea water has a density of 1,025 kg/m^3.

Also, just wondering how you would best Explain the relation between metacentre and curve of buoyancy as vessel heals over.

My Calculations so far

The Tranverse Metacentric Height measures the stability across the Width of the Ship

We can abbreviate Tranverse Metacentric Height = GM Transverse

We will use Simpsons Rule to calculate GM Transverse, because the shape is irregular and not a rectangle or circle.

If you want to understand in more detail how Simpsons Rule works visit https://www.youtube.com/watch?v=7MoRzPObRf0

Simpsons Multiplier, which is a constant will be used during the calculation and we can abbreviate this as follows

Simpsons Multiplier = SM

If you want to understand in more detail how Simpsons Multipliers work visit http://www.snamelearning.ord/learn/assignments/00000010/00000108/simpsonsrulenotes.doc or even better https://www.youtube.com/watch?v=vpfy3sGw8tI

We will use the following in a table to calculate what we need to know

Function of Area = F (A) = Half Ordinate * SM

Function of First Moment of Area = F (M) = Half Ordinate SM Lever

Function of Second Moment of Area (about Longtitudinal Axis for Longtitudinal Stability) = F (I) Long = Half Ordinate SM Lever * Lever

Function of Second Moment of Area (about Transverse Axis for Stability across Ship Width) = F (I) Trans = SM * Cube Half Ordinate

If you want to understand more on why we need to calculate this visit https://www.usna.edu/NAOE/_files/documents/Courses/EN400/02.02%20Chapter%202.pdf

This is found from a simpsons rule ship stability calculations google search

We have been given half ordinates, which are the half widths of the ship at 6 m spacings.

Now we can Draw up a table as below to calculate what we need to know:

Half  Ordinate SM F (A) Lever F (M) Lever F (I) Long Cube Half Ordinate F(I) Trans

0.3           1    0.3     5       1.5    5     7.5            0.027         0.027
3.33          4   13.32    4     53.28    4   213.12          36.926037       147.704148
5.07          2   10.14    3      30.42   3    91.26         130.323843       260.647686  
6.18          4   24.72    2      49.44   2    98.88         236.029032       944.116128  
6.75          2   13.5      1      13.5   1    13.5          307.546875       615.09375  
6.9           4   27.6      0        0    0     0            328.509         1314.036  
6.75          2   13.5     -1     -13.5  -1    13.5          307.546875       615.09375  
6.18          4   24.72   -2     -49.44  -2    98.88         236.029032       944.116128  
5.61          2   11.22   -3     -33.66  -3   100.98         176.558481       353.116962  
4.47          4   17.88   -4     -71.52  -4   286.08          89.314623       357.258492  
0.60          1    0.6     -5       -3   -5    15              0.216            0.216  

Sums 157.5 -22.98 938.7 5551.426044

We also need the sums for each of the functions and then use simpsons rule to calculate what we need

Sea Water density = 1,025 kg/m^3

Ship Volume Displacement = V = 2,800 * 1,000 / 1,025 = 2731.707317 m^3

Using summations from the table and applying simpsons rule https://en.wikipedia.org/wiki/Simpson%27s_rules_(ship_stability)):

Using Simpsons Rule

As we have used half ordinate, we can firstly work out the the half area of the waterplane.

Area of Half of Waterplane = ( 1 / 3 ) spacing between Half Ordinates Total Sum of F (I) Trans

Area of Half of Waterplane = ( 1 / 3 ) 6 157.5

so Total Waterplane area = 2 (1/3) 6 * 157.5 = 630 m^2

Distance of Centre of Flotation aft of the mid-ordinate = Spacing between Half Ordinates * Sum F (M) / Sum F (A)

Distance of Centre of Flotation aft of the mid-ordinate = 6 * 22.98/157.5 = 0.875428571 m

Transverse I has to be doubled (* 2) because we have only calculated F (I) trans from the half ordinates

Using Simpsons Rule

Transverse I (for Half Waterplane Area) = ( 1 / 3 ) ( 1 / 3 ) Spacing between Half Ordinates * Sum F (I) Trans

Transverse I = 2 (1/3) (1/3) 6 5551.426 = 7401.901392 m^4

BM Transverse = I (Trans) / V= 7401.901392 / 2731.707317 = 2.709624617 m

Transverse Metacentric Height is the distance from the Centre of Gravity to the metacentre

and this is GM Transverse

GM Transverse = BM Transverse - BG

We know from the question that

Centre of Gravity is 1.87 m above Centre of Buoyancy, so BG = 1.87 m

GM Transverse = 2.709624617 - 1.87 = 0.839624617 m This is what I have calculated

Answer from Textbook is 0.74 m

So with textbook Answer

BM must equal 1.87 + 0.74 = 2.61 m so Transverse I would equal 7129.75527 m^4

My question is, is the textbook answer correct or my 0.84 m what I have calculated, is this the correct answer? I hope this makes it clear what I am asking in this question