Question 1 )
a) State the formula for the parallel axis theorem :
b) In your own words state the meaning of each term with in the formula
c) State in your own words how the water plane area is derived.
The water plane area is an irregular shape and can therefore be found by using Simpson’s rules. There are three basic rules:
1)
2)
3) Area of segment=
(Additionally is we wanted to find the water plane area from the area of segment given by the Simpson’s rules we would need to multiply through by two.)
Where:
y= half the breadth of the ship at the given points h= the distance between the half ordinates. For example if the ship was 120meters in length and has 7 ordinates then h= 120/(7-1) i.e.20m
(The third rule is not commonly used when finding the WPA area as it is only used to find the area between two ordinates when three are known.)
The water plane area can also be described in a calculus format as:
To best describe how the waterplane area is derived I shall show an example:
A ship is 120 meters long at the waterline and has equidistantly spaced half-ordinates commencing from forward as follows:
0, 3.5 , 6 , 8 , 7.5, 4.5 , 0.2 meters respectively
Find the area of the water-plane:
½ ordinates
Simpsons multipliers
Area function
0
1
0
3.5
4
14
6
2
12
8
4
32
7.5
2
15
4.5
4
18
0.2
1
0.2
Sum =91.2
d) Explain how the transverse first moment of area is derived.
As suggested in the name the formula takes in to consideration the moment created by the area and is as follows:
If we look at a box shaped vessel then the second moment of area can be described as:
As you can see from the diagram
Also I similar situation is presented should we want to look at a circular vessel e.g. a buoy
e) Explain how the transverse first moment of area is used to derive a position of stability. The first moment of area is used to find the centre of flotation (COF) an important stability term. Labelled as LCF (lateral centre of flotation) below:
The LCF is a certain distance from the amidships position (shown in the diagram) this can be either aft. or fwd. depending on whether the answer calculated is positive or negative when subtracted from half the length of the vessel.
1) The formula can be rearranged so that
2) The area can be found by integrating an elementary strip of half the water plane area:
3)
4) By substituting 2 and 3 into one:
5) Now finally if we want to find the CF we must subtract from half the length between the perpendiculars (Lpp)
6) This can be more plainly described as:
Again I will use an example to illustrate my point. For ease of comparison I will use the same example of A ship is 120 meters long at the waterline and has equidistantly spaced half-ordinates commencing from forward as follows:
0, 3.5 , 6 , 8 , 7.5, 4.5 , 0.2 meters respectively
Find the area of the centre of flotation :
½ ordinates
Distance from forward perpendicular
Moment
Simpsons multipliers
Product of SM and moment
0
20x0=0
0
1
0
3.5
20x1=20
70
4
280
6
20x2=40
240
2
480
8
20x3=60
480
4
1920
7.5
20x4=80
600
2
1200
4.5
20x5=100
450
4
1800
0.2
20x6=120
24
1
24
Sum =5704
f) Explain how the second moment of area is derived:
As mention before the first moment of area is area multiplied by distance. The second moment of area however is multiplied by two times the distance to the ss axis referred to previously as. This is why it is termed the second moment.
So if we take the first moment of area and multiply this by y again we will create the second moment of area:
So if I integrate this now the second moment of area will be equal to
So let’s look at the rectangle example form earlier and
