I have a Bachelor’s degree in Civil Engineering and recently completed a Master’s degree in Marine Civil Engineering from the University of Liverpool. In the course of studying for both my Bachelor’s and Master’s degrees, I have had the opportunity to write numerous academic papers on a wide range of subjects, ranging from science to management. For 12 months, I worked as a consultant engineer in the construction of roads, dams, irrigation and drainage schemes for developing countries around the world. I am planning to do a Ph.D. in Engineering but at the moment I am engaged in a graduate training scheme. I have always loved Engineering activities.

The use of the harmonic analysis method for analysing tidal levels.

INTRODUCTION
Equilibrium Theory of the tides was the first attempt to describe tidal phenomena. Significant tide-producing forces on earth are caused by the relative configuration and motions of the earth-moon-sun system. Harmonic analysis is another way by which tides can be analysed. Since the tide is composed of various constituents that interact at the same time with the river flow, the resulting records of water levels in the river will show a rather complicated pattern. Harmonic analysis of the hourly water level of a station could be used to record individual interaction of each constituent of the tide in each month. The characteristics of the individual constituents of the tide can then be correlated with the river flow.  There are nine constituents of the tide but the first four are the main ones and are analysed, based on their characteristics which are:

Table 1: Major tidal constituents

No.

SYMBOL

CONSTITUTION

PERIOD, T (HOURS)

FREQUENCY, ω (RAD/HOUR)

1

M2

Lunar Semi-diurnal

12.42

0.5059

2

S2

Solar Semi-diurnal

12.00

0.5236

3

K1

Luni-Solar diurnal

23.93

0.2626

4

O1

Lunar declinational diurnal

25.82

0.2433

5

N2

Larger Lunar elliptic Semi-diurnal

12.66

0.4963

6

M4

First over-tide of M2

6.21

1.0118

7

S4

First over-tide of S2

6.00

1.0472

8

MS4

Compound Luni-solar

6.10

1.0300

9

M6

Second over-tide of M2

4.14

1.5177

The harmonic analysis of tides is developed from the following two parts: the resultant tide at any location is composed of a finite number of constituents, each with its own periodicity, phase angle and amplitude and also the constituents are each simple harmonics in time and are mutually independent. i.e.
ɳ=ao+james barton clip image002 James Barton………………………..1
Where, ɳ = tidal level, ao= amplitude at mean water level, ai= amplitude at i-constituent, james barton clip image004 James Barton = phase angle at i- constituent, t= time.

AIMS
The exercise is designed to illustrate how the measured water levels from an estuary can be analysed to produce harmonic tidal constants, how the tidal levels can be predicted using harmonic tidal constants and how to use dominate constants (M2, S2, K1 & O1) to identify a tidal type that can affect transport and engineering operations in an estuary.

HARMONIC ANALYSIS THEORY

Harmonic Analysis is the summation of the amplitude at water level and all the harmonic functions (M2, S2, K2, O1, N2, M4, S4, MS4, M6).
i.e.
ɳ=ao+james barton clip image002 0000 James Barton

Where ωi =james barton clip image004 0000 James Barton and φi= phase angle

n= number of samples
i= 1……9

Where ao can be calculated from
ɳ=ao+james barton clip image002 0001 James Barton
We have
james barton clip image006 James Barton= james barton clip image008 James Bartondt + james barton clip image010 James Barton [james barton clip image012 James Bartont-james barton clip image014 James Barton)dt]
As
james barton clip image010 0000 James Barton [james barton clip image012 0000 James Bartont-james barton clip image014 0000 James Barton)dt] =0

So,
james barton clip image016 James Barton = james barton clip image018 James Barton……………………………………………………….2

While ai (amplitude of constituents) can be decided by the following derivation:

                   ɳ=ao+james barton clip image002 0002 James Barton (Multiply through by sin(james barton clip image020 James Bartont))
ɳ sin(james barton clip image020 0000 James Bartont) = james barton clip image016 0000 James Barton sin(james barton clip image020 0001 James Bartont) + james barton clip image002 0003 James Barton sin(james barton clip image020 0002 James Bartont)

Integrate and take the limit of the above

james barton clip image022 James Bartondt=james barton clip image024 James Bartondt+james barton clip image026 James Barton=james barton clip image028 James Barton+james barton clip image030 James Barton]james barton clip image032 James Bartondt
=james barton clip image034 James Barton james barton clip image010 0001 James Barton james barton clip image036 James Bartondt + james barton clip image038 James Bartondt
Using the features gives,

james barton clip image010 0002 James Barton james barton clip image040 James Bartondt = james barton clip image042 James Barton  james barton clip image044 James Barton

Similarly we have:

james barton clip image010 0003 James Barton james barton clip image046 James Bartondt = james barton clip image048 James Barton  james barton clip image044 0000 James Barton
Or for james barton clip image050 James Barton  we have,

= james barton clip image052 James Barton

james barton clip image054 James Barton can be decided as
james barton clip image054 0000 James Barton= james barton clip image056 James Barton………………………..….3

Or simply,
james barton clip image058 James Barton= 2 james barton clip image060 James Barton

Phase angle φi is derived from:

 james barton clip image052 0000 James Barton
We have,
james barton clip image014 0001 James Barton= james barton clip image062 James Barton( james barton clip image064 James Barton )……………………………………………………4
james barton clip image014 0001 James Barton= james barton clip image062 James Barton (james barton clip image066 James Barton)

The tide is classified using the classification criteria given below:

F factor: F= james barton clip image068 James Barton……………………………………………………………………5
where the symbols of the constituents indicate their respective amplitudes.

Table 2: Classification of tides

F

Category

< 0.25

Semidiurnal

0.25-3

Mixed (Mainly Semidiurnal 0.25 < F < 1.5)

Mixed (Mainly Diurnal 1.5 < F < 3)

>3

Diurnal

RESULTS AND DISCUSSION

SOLUTION TO QUESTION 1

Based on the given template, reconstructing the water level changes relative to Chart Datum over a period of at least half a year at quarter-hourly intervals using the harmonic tidal data from Liverpool (Gladstone Dock) was done by calculating for the harmonic functions of each constituent (M2, S2, K1, O1, N2, M4, S4, MS4 and M6) of the template. A sample of the results is shown in Table 2.

Theory
To obtain the values of A1, B1, A2, B2, A3, B3, A4, B4, A5, B5, A6, B6, A7, B7, A8, B8, A9, B9, manually, we would first need to calculate the individual component for each, using equation 6 and equation 7, after which, the sum of each column divided by the total number of samples (n) will give the corresponding value for each of the terms. To do this we need the parameters ωi for each of the tidal constituents which can be found in Table 1, above, over a range of time t

                          ηsin (ωit)..……………………………………….………………….6
ηcos (ωit)………………….…………………………………………7
Ai =james barton clip image002 0004 James Barton)..……………………………………………..8
Bi =james barton clip image004 0001 James Barton)..……………………………………………..9
Where, ɳ = tidal water level at a particular time (m)
ωi = angular frequency at a particular time (rad/hr)
t= time (hrs)
i= 1-9
n= number of samples considered
Substituting the parameters into the equations above we obtain the same values for A1…B9 as shown in Table 3 below.

SOLUTION TO QUESTION 2

Carrying out harmonic analysis on the data in Table 3, we can find out the harmonic constants based on the formulae given below (the summation of the observed data over a range of samples and then divided by the sum by the total number of samples being considered).
i.e.
From Table 3 to calculate the constants a0 , james barton clip image006 0000 James Barton andjames barton clip image008 0000 James Barton:
For i =range of 1 – 9

 a0 = james barton clip image010 0004 James Barton
where n= number of samples being considered (18801)

Therefore,

 a0 =james barton clip image012 0001 James Barton)/(18815-15+1)
a0 = james barton clip image014 0003 James Barton
a0 =5.30525

while

         james barton clip image006 0001 James Barton = james barton clip image006 0001 James Barton= james barton clip image016 0001 James Barton
or simply,
james barton clip image006 0001 James Barton= 2 james barton clip image020 0003 James Barton
by substituting into the above equation for when i=1, 2, 3……….9, we have
for i=1
james barton clip image006 0001 James Barton= 2 james barton clip image024 0000 James Barton
where A1=-1.130 and B1=0.977 (from Table 3)
james barton clip image006 0001 James Barton= 2 james barton clip image026 0000 James Barton =2.9880

Similarly, all other constants for each constituent are calculated using the same method as above.

Likewise for the phase angle from i=1, 2, 3,……..9 going by the formulae

For i=1
james barton clip image008 0000 James Barton= james barton clip image062 James Barton (james barton clip image032 0000 James Barton)
where A1=-1.130 and B1=0.977

james barton clip image008 0000 James Barton= james barton clip image062 James Barton james barton clip image034 0000 James Barton
james barton clip image008 0000 James Barton=-0.85788
In the same way, the rest of the phase angle for each constituent is calculated. See Table 3 for the full results of all the constituents.

Table 3: The Harmonic Constants (a0, ai  and φi)

Port:

P234

 

 

 

Consti

M2

S2

K1

O1

N2

M4

M6

S4

MS4

Site:

Liverpool

(Gladstone

Dock)

 

i =

1

2

3

4

5

6

7

8

9

Latitude:

53.4497

 

 

 

ωi =

0.5059

0.5236

0.2626

0.2433

0.4963

1.0118

1.5177

1.0472

1.03

Longitude:

-3.0181

 

 

 

ai =

2.9882

1.0218

0.1221

0.1372

0.6168763

0.2341694

0.0505411

0.0221983

0.104331

Start

Date:

01JAN2005-00.00.00

 

 

φi =

-0.8578

-0.7361

-0.8295

1.5594

-0.729238

0.0460049

-0.633921

0.1507001

1.106581

End

Date:

31DEC2005-23.45.00

 

 

n =

18801

 

a0 =

5.3053

 

 

 

 

 

Contributor:

Proudman

Oceanographic

Laboratory

 

F=

0.064664

(Semidiurnal)

 

 

 

 

 

 

 

Datum

information:

The

data

refer to

Admiralty

(ACD)

Chart

 

Datum

 

 

 

 

 

Parameter

code:

ASLVBG02

=

Sea level

Bubbler

(second

tide

sensor)

gauge

 

 

 

 

 

The calculated constants in summary

M2

S2

K1

O1

N2

M4

M6

S4

MS4

ai =

2.9882

1.0218

0.1221

0.1372

0.6168763

0.2341694

0.0505411

0.0221983

0.104331

φi =

-0.8578

-0.7361

-0.8295

1.5594

-0.729238

0.0460049

-0.633921

0.1507001

1.106581

n =

18801

 

a0 =

5.3053

 

 

 

 

 

SOLUTION TO QUESTION 3

Comparing the predicted and the observed tidal water level, Figure 1 shows the graph of the observed and predicted tidal water level against time.
The observed and predicted values are taken from Table 3, and t = 0, 0.25, 0.5…371.25 hours.

james barton clip image002 0005 James Barton

Figure 1: Plot of comparison between observed water level and predicted water level vs. time

Figure 1 above gives the graph of both observed and predicted tidal levels against time (every quarter of an hour). It can be seen that the predicted values obtained were all positive; there were no negative values. Also from the graph it can be seen that the predicted value has a very close relationship with the measured value. If more constituents are used in the calculation, accuracy is improved. (In this case, all nine constituents were used)

SOLUTION TO QUESTION 4

Plotting of the individual constituents (M2, S2, K1, & O1), as shown in Figure 2 below, is done by calculating the equations below. The summary is given in Table 4:
η = a0 +a1 cos(ω1t – φ1) + a2 cos(ω2t – φ2) + a3 cos(ω3t – φ3) + a4 cos(ω4t – φ4)……..10
M2 = a1 cos(ω1t – φ1)……………………………………………………………………………………….11
Ss = a2 cos(ω2t – φ2)……………………………………………………………………12
K1 = a3 cos(ω3t – φ3)…………………………………………………………………..13
O1 = a4 cos(ω4t – φ4)…………………………………………………………………..14
The coefficient tidal values are taken from Table 4, and t = 0, 0.25, 0.5…146.25 hours.
Table 4: The Values for the Four Main Constituents

Time(hours)

M2

S2

K1

O1

 

 

 

 

 

0

1.954719

0.75721

0.082444

0.001565

0.25

1.654006

0.661187

0.076358

0.009903

0.5

1.326871

0.55385

0.069944

0.018203

0.75

0.978539

0.437037

0.063229

0.026437

1

0.614576

0.312747

0.05624

0.034572

1.25

0.240795

0.183105

0.04901

0.04258

1.5

-0.13683

0.05033

0.041569

0.05043

1.75

-0.51227

-0.08331

0.033948

0.058094

2

-0.87953

-0.21552

0.026181

0.065543

2.25

-1.23274

-0.34404

0.018301

0.072749

2.5

-1.56626

-0.46668

0.010343

0.079687

2.75

-1.87475

-0.58133

0.00234

0.086329

3

-2.1533

-0.68603

-0.00567

0.092653

3.25

-2.39745

-0.779

-0.01366

0.098633

3.5

-2.6033

-0.85864

-0.02159

0.104249

3.75

-2.76756

-0.92358

-0.02943

0.10948

4

-2.88761

-0.97273

-0.03714

0.114305

4.25

-2.96154

-1.00523

-0.04469

0.118708

4.5

-2.98815

-1.02052

-0.05205

0.122671

4.75

-2.96703

-1.01836

-0.05918

0.126181

5

-2.89852

-0.99878

-0.06606

0.129224

5.25

-2.7837

-0.9621

-0.07265

0.131789

5.5

-2.62441

-0.90896

-0.07893

0.133867

5.75

-2.4232

-0.84027

-0.08487

0.13545

6

-2.18328

-0.7572

-0.09045

0.136532

6.25

-1.90848

-0.66118

-0.09563

0.137109

6.5

-1.60319

-0.55384

-0.10041

0.137178

6.75

-1.2723

-0.43703

-0.10475

0.136741

7

-0.92108

-0.31274

-0.10864

0.135797

138.5

-0.7076

-0.55371

0.107173

0.102999

138.75

-1.06816

-0.43688

0.110779

0.097298

139

-1.41166

-0.31258

0.113908

0.091237

139.25

-1.7326

-0.18293

0.116545

0.084839

139.5

-2.02587

-0.05016

0.118681

0.078127

139.75

-2.28677

0.083478

0.120305

0.071125

140

-2.51114

0.215686

0.121411

0.063861

140.25

-2.6954

0.344202

0.121994

0.056361

140.5

-2.8366

0.46683

0.122052

0.048652

140.75

-2.93249

0.58147

0.121583

0.040763

141

-2.98153

0.68616

0.120591

0.032724

141.25

-2.98294

0.779111

0.119079

0.024563

141.5

-2.93671

0.85873

0.117054

0.016311

141.75

-2.84355

0.923656

0.114525

0.008

142

-2.70498

0.972778

0.111502

-0.00034

142.25

-2.52319

1.005256

0.107999

-0.00868

142.5

-2.3011

1.020533

0.104031

-0.01699

142.75

-2.04224

1.018349

0.099614

-0.02523

143

-1.75077

0.99874

0.094769

-0.03339

143.25

-1.43132

0.962042

0.089515

-0.04142

143.5

-1.08901

0.908884

0.083875

-0.04929

143.75

-0.72931

0.840174

0.077874

-0.05698

144

-0.35795

0.757089

0.071537

-0.06447

144.25

0.019126

0.661049

0.064893

-0.07171

144.5

0.395895

0.553699

0.057968

-0.07869

144.75

0.76634

0.436875

0.050794

-0.08537

145

1.124543

0.312575

0.043401

-0.09175

145.25

1.464782

0.182928

0.035821

-0.09778

145.5

1.781622

0.05015

0.028087

-0.10345

145.75

2.070001

-0.08349

0.020232

-0.10874

146

2.325312

-0.21569

0.012289

-0.11362

146.25

2.543477

-0.34421

0.004294

-0.11809

The plotting produced is a cos function.

james barton clip image002 0006 James Barton

Figure 2: Plot of the individual constituents (M2, S2, K1 & O1) vs. time

SOLUTION TO QUESTION 5

In order to decide what type of tide it is at this location we use the formulae given below:

F factor: F= james barton clip image002 0007 James Barton = james barton clip image004 0002 James Barton

Where K1= a3 = 0.1221
O1= a4 = 0.1372
M2= a1 = 2.9882
S2= a2 = 1.0218

Substituting into the equation above we have:

F= james barton clip image006 0003 James Barton

F= 0.064663.

From Table 2 above, the value of F falls in the category of “SEMIDIURNAL” (0.064< F< 0.25).
Therefore the type of tide at this location is the SEMIDIURNAL TIDE.

SOLUTION TO QUESTION 6

  1. The harmonic analysis method is a very good method for predicting tidal level
  2. It can enable us to predict future values of the tidal level since we know that tidal levels in estuaries don’t change often. i.e. they have a similar pattern over a period of time
  3. The results that are obtained by this method will be more accurate if more constituents are considered
  4. Taking a regular and consistant reading/ measurement we can also reduce the error in the analysis, because if the reading is not done in a regular way then the result would be wrong and inaccurate.
  5. This method is more straightforward and easier to use.
  6. It allows a larger data sample in the analysis

CONCLUSION: The Harmonic analysis is a better method for analysing tidal levels. Further, it gives the option of using more than the four main constituents, which gives it an edge over other methods. In short, it is better, faster and easier to use.