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Question 4.7: This example demonstrates how a cotidal chart can be used to......

This example demonstrates how a cotidal chart can be used to reveal information about the depth of the seabed. Use the phase information for M_{2} in the cotidal chart below to estimate the speed of the tidal wave along the coast of Finnmark and Kola. You may assume that 1° longitude at 70° latitude is 38 km.

From your answer and assuming the tide travels as a shallow water wave, find the corresponding water depth. What is the feature at ‘A’?

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Remember we have been asked to estimate, not calculate in fine detail.

The distance along the Finnmark to Kola coast is about 20° longitude. At 70° latitude this is approximately (38 km/degree)×20° = 760 km.

From the chart, note that the phase of the tide changes approximately from 360° (or 0°) to 150° between Finnmark and Kola. Now, 150° is 150/360 of one M_{2} period (=12.42 hours). Thus, this change in phase corresponds to a time interval of (150/360) × 12.42 ≈ 6 hours. The tide wave speed is therefore approximately 760/6 km/hour or 125 km/hr.

Assuming the tide wave propagates as a shallow water wave, its speed, c, =√(gh), where h is the water depth. Knowing c and g we can estimate h, the water depth. Now, making h the subject of the wave speed equation and converting the tide speed into units of metres per second, we have

h = (125,000/(60 × 60))²/g ≈ 1205.6/9.81 ≈ 120 m.

Thus, the depth of the sea along the coast is approximately 120 m. The feature at ‘A’ is an amphidromic point.


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