Simpson's Second and Third Rules

In the previous lesson we explored Simpson's First Rule for calculating the area of a water-plane.
You will recall that Simpson's First Rule can be used when there are an even number of area sections and hence an odd number of ordinates.

Simpson's rule calculations can also be carried out when a water-plane is divided into an odd number of sections (has an even number of ordinates). One way to approach this situation is to utilize Simpson's Second Rule.

Simpson's Second Rule

Simpson's Second Rule can be applied when there are an even number of ordinates, but only when a second condition is also satisfied. When a water-plane is subdivided using an even number of ordinates, Simpson's Second Rule can be applied, if and only if, the number of ordinates, less one, is a multiple of 3.

second Condition for simpson's second rule:

Number of ordinates - 1 = Multiple of 3

For example, we know that both 4 and 6 are even numbers but only 4 satisfies the condition for Simpson's Second Rule since:
$\begin{array}{l} 4 - 1 = 3 a n d 3 i s a m u l t i p l e o f 3 \\ 6 - 1 = 5 a n d 5 i s n o t a m u l t i p l e o f 3 \end{array}$ This means that you need to have at least 4 ordinates to apply Simpson's Second Rule. The Second Rule states that the area under the curve (which satisfies all the necessary conditions for this rule) is given by:

A r e a = \frac{3 h}{8} (y_{1} + 3 y_{2} + 3 y_{3} + y_{4})

This rule is also called the three-eighths rule. The Simpson's multipliers in this case are 1,3,3,1 and if we follow the same process of applying the rule more than once in a problem as we did in the last lesson, the multipliers become: 1, 3, 3, 2, 3, 3, 2, 3, 3, 2....... 3, 3, 1

Example 1: Find the area of the following shape using Simpson's Rule:

Solution:

There are 4 (even) half-ordinates given so we cannot apply Simpson's First Rule to solve this equation. Now let us check if we can apply Simpson's Second Rule:

Even number of ordinates
$4 - 1 = 3 (m u l t i p l e o f 3)$

Hence we can use the Second Rule to evaluate this area.

Value of h:

h = \frac{30 m}{3} = 10 m

Complete the table:

Half-ordinates (1)	Simpson's Multiplier (2)	Area Function (3)=(1)x(2)
2.5	1	2.5
3.5	3	10.5
4.5	3	13.5
5.0	1	5.0
$(T o t a l) Σ_{2}$		31.5

Calculate the area:

A r e a = \frac{3 h}{8} \times Σ_{2} = \frac{3}{8} \times 10 m \times 31.5 m = 118.125 m^{2} = 118.13 m^{2}

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Example 2: The lengths of the half-ordinates of a 90 m long water-plane commencing from forward are as follows: 0, 2.0, 3.6, 5.0, 5.5, 6.0, 4.8, 3.4, 2.0, 0.5 Find the area of the water-plane.

Solution:

We are given 10 half-ordinates, 10 is even and 10 - 1 = 9 is a multiple of 3; therefore we can use Simpson's Second Rule to calculate this water-plane area.
Find the value of "h" :
$h = \frac{90 m}{9} = 10 m$

Complete the table:

Half-ordinates (1)	Simpson's Multiplier (2)	Area Function (3)=(1)x(2)
0	1	0
2	3	6
3.6	3	10.8
5	2	10
5.5	3	16.5
6	3	18
4.8	2	9.6
3.4	3	10.2
2	3	6
0.5	1	0.5
$(T o t a l) Σ_{2}$		87.6

Table 5.2

Calculate the area: $Area of the water-plane = 2 \times \frac{3 h}{8} \times Σ_{1} = 2 \times \frac{10 m}{8} \times 87.6 m = 657 m^{2}$

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Appendages

We have discussed Simpson's First and Second Rules which can be used to calculate the area of a ship's water-plane. Many water-plane areas are not defined by a single curve however, but can be divided into two or more separate water-plane areas, each with its own shape. In such cases we can calculate each section of the water-plane separately, using one or more of Simpson's Rules. Often, the main area of the water-plane is calculated separatedly from the bow or stern ends of the ship, which are referred to as an appendages.

Example 3: A ship's water-plane area has been divided into common intervals of lenth 9 m, with the following half-ordinates commencing from forward: 0, 1.0, 2.0, 3.0, 2.0 metres respectively Aft of the last ordinate is an appendage of 25 m². Find the total area of the water-plane.

Solution:

The total water-plane area can be looked upon as the sum of two areas: Area 1 and Area 2 (area of appendage)

We are given 5 half-ordinates so we will use Simpson's First Rule to find the value of Area 1.
Area 1:
Complete the table :

Half-ordinates (1)	Simpson's Multiplier (2)	Area Function (3)=(1)*(2)
0	1	0
1	4	4
2	2	4
3	4	12
2	1	2
$(T o t a l) Σ_{1}$		22

Calculate the area:

A r e a 1 = 2 \times \frac{h}{3} \times Σ_{1} = 2 \times \frac{9}{3} \times 22 = 132 m^{2}

$A r e a 2 = 25 m^{2}$ $T o t a l a r e a = A r e a 1 + A r e a 2 = 132 + 25 = 157 m^{2}$

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Simpson's Third Rule

Simpson's Third Rule helps us to evaluate the area included by two consecutive ordinates when an external ordinate is also known. According to this rule: $A r e a = \frac{h}{12} (5 y_{1} + 8 y_{2} - y_{3})$

where,

h = common interval
y₁ = the first ordinate
y₂ = the middle ordinate
y₃ = the external ordinate

It is very important to understand which ordinates represent y1, and y3 as y2 is always the middle one. y1 is one of the ordinates that includes the area of interest. The following images show the appropriate half-ordinates while calculating the area of the shaded strip:

Example 4: Find the area of a steel plate of the following shape: The area of section A is known to be 25 cm².

Solution:

The area of section A is given to be 25 cm², therefore all we need to calculate is the area of section B which we can find using Simpson's Third Rule. We can then add the areas of sections A and B to determine the total area of the steel plate.

Clearly, $y_{1} = 7 c m, y_{2} = 6 c m a n d y_{3} = 5 c m . A l s o h = 4 c m$ Hence area of section B will be:

$A r e a = \frac{h}{12} (5 y_{1} + 8 y_{2} - y_{3}) = \frac{4}{12} (5 \times 7 + 8 \times 6 - 5) = 26 c m^{2}$ $And the total area = 25 c m^{2} + 26 c m^{2} = 51 c m^{2}$

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Example 5: The lengths of the half-ordinates of a 60 m long water-plane commencing from forward are as follows:
0, 4.2, 5.8, 6.0, 3.5, 0.2 metres respectively Find the area of the water-plane.

Solution:

We are given 6 half-ordinates and 6 is even. Therefore, we cannot apply Simpson's First Rule. Also,6 - 1 = 5 which is not a multiple of 3; therefore we cannot use Simpson's Second Rule either to calculate this water-plane area.
So in this case we are going to use both rules to find the area of two parts of the water-plane and add the results to find the total area.
Now, if we divide the water-plane into two sections such that the first section includes the first two area strips and the second section includes the rest, then we have two areas:

Find the value of h:

h = \frac{60 m}{5} = 12 m

AREA 1:Number of half-ordinates = 3, 3 is odd. Therefore we can apply Simpson's First Rule.
Complete the table:

Half-ordinates (1)	Simpson's Multiplier (2)	Area Function (3)=(1)*(2)
0	1	0
4.2	4	16.8
5.8	1	5.8
$(T o t a l) Σ_{1}$		22.6

Calculate the area:

Area of the water-plane = \frac{h}{3} \times Σ_{1} = \frac{12 m}{3} \times 22.6 m = 90.4 m^{2}

AREA 2: Number of half-ordinates = 4, 4 is even and 4 — 1 = 3 is a multiple of 3. So we will use Simpson's Second Rule.
Complete the table:

Half-ordinates (1)	Simpson's Multiplier (2)	Area Function (3)=(1)*(2)
5.8	1	5.8
6.0	3	18.0
3.5	3	10.5
0.2	1	0.2
$(T o t a l) Σ_{2}$		34.5

Calculate the area:

Area of the water-plane = \frac{3}{8} \times h \times Σ_{2} = \frac{3}{8} \times 12 m \times 34.5 m = 155.25 m^{2}

Total area = 2 \times (90.4 + 155.25) = 491.3 m^{2}

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Simpson's Second and Third Rules

Simpson's Second Rule

second Condition for simpson's second rule:

Example 1: Find the area of the following shape using Simpson's Rule:

Example 2: The lengths of the half-ordinates of a 90 m long water-plane commencing from forward are as follows: 0, 2.0, 3.6, 5.0, 5.5, 6.0, 4.8, 3.4, 2.0, 0.5 Find the area of the water-plane.

Appendages

Example 3: A ship's water-plane area has been divided into common intervals of lenth 9 m, with the following half-ordinates commencing from forward: 0, 1.0, 2.0, 3.0, 2.0 metres respectively Aft of the last ordinate is an appendage of 25 m2. Find the total area of the water-plane.

Simpson's Third Rule

Example 4: Find the area of a steel plate of the following shape: The area of section A is known to be 25 cm2.

Example 5: The lengths of the half-ordinates of a 60 m long water-plane commencing from forward are as follows: 0, 4.2, 5.8, 6.0, 3.5, 0.2 metres respectively Find the area of the water-plane.

Example 3: A ship's water-plane area has been divided into common intervals of lenth 9 m, with the following half-ordinates commencing from forward: 0, 1.0, 2.0, 3.0, 2.0 metres respectively Aft of the last ordinate is an appendage of 25 m². Find the total area of the water-plane.

Example 4: Find the area of a steel plate of the following shape: The area of section A is known to be 25 cm².

Example 5: The lengths of the half-ordinates of a 60 m long water-plane commencing from forward are as follows:
0, 4.2, 5.8, 6.0, 3.5, 0.2 metres respectively Find the area of the water-plane.