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’?
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.
