# Question 4.2: The tidal constituents for four harbours are given in the fo......

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
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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.

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