3
$\begingroup$

Tides observations everywhere show that spring tides occur not at full or new moon, at the time of maximum driving forces, but typically a day or two later.

What causes this delay?

$\endgroup$

1 Answer 1

3
$\begingroup$

The age of the tide (a.k.a, age of phase inequality, lunitidal interval) at a location is a term that denotes the interval between the time of a Full or New Moon (when the equilibrium semi-diurnal tide force is maximum), and the time of the local spring tide. In most cases, there is a delay between maximum forces and occurrence of maximum elevation.

The tidal forces generate waves of very long wavelength that travel all around the ocean with a velocity determined by the shallow water progressive wave velocity equation, $c=\sqrt(gH)$, where $H$ is water depth. The delay between the moon phases and their effect on the tide is the time that it takes for the wave to travel around the ocean.

There are two different mechanisms to consider: resonance and friction.

The fact that the age of the tide is analogous to the increase of the phase lag of a forced, linearly damped harmonic oscillator as the frequency increases was known since the XIX century. The damped oscillator analogy does not explain the phase magnitudes in the real ocean.

The inclusion of friction is necessary to explain the observed time lags. Proudman (1941) showed that by including frictional effects on coastal areas, a positive age of tide appear in the tidal solution. If there is some resonant frequency in an ocean basin close to the semi-diurnal frequencies ($M_2$, $S_2$...), friction enhances the age of the tide resulting in higher delay values. Garrett & Munk (1971) and Webb (1973) provided a methodology for the calculation and estimates around the oceans. The longest delays (longer than 60 hours) were estimated in areas with the largest continental shelf, as they represent areas of strong tidal dissipation: the North Sea, the Ross Sea, Hudson Strait, the shelf off Argentina, the Bering Sea, and the Sea of Okhotsk.

The age of tide can be obtained from the harmonic constants by dividing the difference in phases of the $S_2$ and $M_2$ constituents by frequency separation between both constituents: $ Age\_of\_tide = {Phase\_of\_M_2 - Phase\_of\_S_2 \over 1.0163 }$, where the phases are measured in degrees and the age of tide is in hours.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.