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