The tidal constituents for four harbours are given in the following table. Classify the tidal regime at each harbour using the tidal ratio. Estimate the maximum tide level at each harbour. Calculate the length of the spring-neap cycle at harbours A and D. The mean water level relative to the local datum, Z_{0}, is also given. Note that this can be positive or negative and, strictly speaking, its value is a magnitude rather than an amplitude.

Constituent |
Period (hours) |
Harbour A Amplitude (cms) |
Harbour B Amplitude (cms) |
Harbour C Amplitude (cms) |
Harbour D Amplitude (cms) |

M_{2} | 12.42 | 233 | 53 | 22 | 3 |

S_{2} | 12.00 | 68 | 14 | 7 | 4 |

K_{1} | 23.93 | 15 | 35 | 32 | 70 |

O_{1} | 25.83 | 17 | 26 | 26 | 68 |

Z_{0} | – | 0 | 50 | 0 | −10 |

Step-by-Step

Learn more on how do we answer questions.

We calculate F from the amplitudes of the constituents:

For Harbour A: F = (15 + 17)/(233 + 68) = 0.11, so the tide is semidiurnal.

For Harbour B: F = 0.91, so the tide is mixed but predominantly semidiurnal.

For Harbour C: F = 2.00, so the tide is mixed but predominantly diurnal.

For Harbour D: F = 19.71, so the tide is diurnal.

Note that an estimate of the maximum tide level may be obtained by adding the amplitudes of the tidal harmonics to the mean water level. This corresponds to the case when all of the harmonics are exactly in phase with each other. The constituent denoted by Z_{0} is a fixed correction term that can be used to adjust the tide level to a particular reference datum. The maximum tide levels are therefore 233 + 68 + 15 + 17 + 0 = 333 cms at Harbour A; 53 + 14 + 35 + 26 + 50 = 178 cms at Harbour B; 22 + 7 + 32 + 26 = 87 cms at Harbour C; and 3 + 4 + 70 + 68 − 10 = 135 cms at Harbour D.

Note also that the spring-neap cycle is of a slightly different period for semidiurnal and diurnal tidal forms. The period can be determined by calculating the time taken for the slower constituent to fall exactly one whole cycle behind its companion. Thus, for the semidiurnal case, Harbour A, we have:

Period difference between M_{2} and S_{2} = 0.42 hours

Number of S_{2} cycles required for S_{2} to lag one M_{2} cycle = 12.4/0.42

= 29.52 cycles

=354.2 hours (12.00 × 29.52)

=14.8 days

The corresponding spring-neap cycle period for diurnal tides, Harbour D, is 13.7 days.

Question: 4.1

We may estimate the magnitude of the effect of the...

Question: 4.6

We expand the Coriolis parameter in a Taylor serie...

Question: 4.4

a. From the tidal range information we can determi...

Question: 4.5

Following the steps outlined in the section above:...

Question: 4.8

In the first case we may use the 1-dimensional ver...

Question: 4.9

The governing equation is
\frac{\partial²\e...

Question: 4.11

We assume the wave is a shallow water wave so that...

Question: 4.3

We calculate F from the amplitudes of the constitu...

Question: 4.7

Remember we have been asked to estimate, not calcu...