<p>The age of the tide (a.k.a, <a href="http://co-ops.nos.noaa.gov/publications/glossary2.pdf" rel="nofollow noreferrer">age of phase inequality</a>, <a href="https://en.wikipedia.org/wiki/Lunitidal_interval" rel="nofollow noreferrer">lunitidal interval</a>) 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.</p>

<p>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. </p>

<p>There are two different mechanisms to consider: resonance and friction.</p>

<p>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.</p>

<p>The inclusion of friction is necessary to explain the observed time lags. <a href="http://dx.doi.org/10.1111/j.1365-246X.1941.tb00439.x" rel="nofollow noreferrer">Proudman (1941)</a> 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. <a href="http://dx.doi.org/10.1016/0011-7471(71)90073-8" rel="nofollow noreferrer">Garrett &amp; Munk (1971)</a> and <a href="http://dx.doi.org/10.1016/0011-7471(73)90006-5" rel="nofollow noreferrer">Webb (1973)</a> 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.</p>

<p>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.</p>

CC BY-SA 4.0
2024-09-29T08:12:13.500Z
In order to measure the age of the tide at a location, it is necessary to know the time of spring tide at that location.  How is this defined?  If it is the time of the highest high tide, which tide is chosen when (as often happens) there is more than one tide of the same height?  Or is it defined as the maximum of the envelope of the tidal curve at that location, which might occur at between high tides?
Steve Lenartowicz
CC BY-SA 4.0
2024-10-01T12:05:13.123Z
The tidal spring-neap cycle is defined as a combination the M2 and S2 constituents: The M2 and S2 tidal constituents have a slightly different period resulting in a beating (some time they add up: spring; sometimes they oppose: neap) of the two signals resulting in spring-neap tide. No need to measure at a specific time, as a measurement would include other constituents and other effects such as surge. The measurements are taken over a long enough time (e.g., 29 days) and then harmonic analysis is used to extract the magnitude and phase of each of the constituents
arkaia
CC BY-SA 3.0
2016-11-08T02:46:09.373Z
2016-11-08T02:46:09.373Z
arkaia
<p>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.</p>

<p>What causes this delay?</p>

<ocean><oceanography><tides>
Why don't spring tides occur at full/new moon (age of tide)?
arkaia
# Why don't spring tides occur at full/new moon (age of tide)?

(Asked by: username_0 on 2016-10-26)

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?

**Tags:** ocean, oceanography, tides


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## 1 answers

### Answer by username_0 on 2016-11-08. Score: 3 | Accepted answer

The age of the tide (a.k.a, [age of phase inequality](http://co-ops.nos.noaa.gov/publications/glossary2.pdf), [lunitidal interval](https://en.wikipedia.org/wiki/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)](http://dx.doi.org/10.1111/j.1365-246X.1941.tb00439.x) 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)](http://dx.doi.org/10.1016/0011-7471(71)90073-8) and [Webb (1973)](http://dx.doi.org/10.1016/0011-7471(73)90006-5) 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.

> **Comments:**
> **username_1**: In order to measure the age of the tide at a location, it is necessary to know the time of spring tide at that location.  How is this defined?  If it is the time of the highest high tide, which tide is chosen when (as often happens) there is more than one tide of the same height?  Or is it defined as the maximum of the envelope of the tidal curve at that location, which might occur at between high tides?  
> **username_0**: The tidal spring-neap cycle is defined as a combination the M2 and S2 constituents: The M2 and S2 tidal constituents have a slightly different period resulting in a beating (some time they add up: spring; sometimes they oppose: neap) of the two signals resulting in spring-neap tide. No need to measure at a specific time, as a measurement would include other constituents and other effects such as surge. The measurements are taken over a long enough time (e.g., 29 days) and then harmonic analysis is used to extract the magnitude and phase of each of the constituents  

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