IMPORTANT SUMMARY:

The stresses caused by lunar tides in the Earth's atmosphere and oceans should be a maximum when the tidal forces of the Moon acting upon the Earth change by the largest amount in either strength or direction over a relative short period in time.

There are two such conditions where this takes place:

When a new/full moon occurs at or near the times when the lunar line-of-apse (see figure 1) is pointing towards the Sun. This is true because a new/full moon at closest perigee produces the strongest lunar tidal forces upon the Earth while a new/full moon at apogee produces the weakest lunar tidal forces upon the Earth (see figure 2). Thus, when the new/full moon takes place very close to perigee it will be followed roughly 14 days later by a full/new moon not far from apogee. Similarly, when the new/full moon takes place very close to apogee it will be followed roughly 14 days later by a full/new moon not far from perigee.

Figure 1.

Figure 2.

We are looking for the times when the stresses caused by lunar tides in the Earth's atmosphere and oceans will be a maximum because the direction of the tidal forces of the Moon acting upon the Earth are changing by the largest amount over a relative short period in time.

When a new/full moon occurs at or near the time when the lunar line-of-nodes is at a right angle to the Earth-Sun line (see figure 3 and 4). This is true because a new/full moon at a major lunar standstill produces the largest changes in the meridional (north-south) lunar tidal forces upon the Earth (see figure 4).

Hence, when the new/full moon takes place very close to a time when the lunar line-of nodes is at a right angle to the Earth-Sun line (i.e. the new/full moon is at a major lunar standstill) it will be located at a declination of roughly 28 degrees in one hemisphere, followed roughly 14 days later by a full/new moon located at a declination of roughly 28 degrees in the other hemisphere.

Figure 3.

Figure 4.

a) the rate of change in the stresses caused by lunar tides in the Earth's atmosphere and oceans, as a result of a change in the

b) the rate of change in the stresses caused by lunar tides in the Earth's atmosphere and oceans, as a result of a change in the

Now if the period of the rate of change in stresses caused by the change in strength of the lunar tides (i.e. 10.14686 tropical years)

Interestingly, the time period of the positive side-lobe is almost exactly the same as that of the Quasi-Biennial Oscillation (QBO). The QBO is a quasi-periodic oscillation in the equatorial stratospheric zonal winds that has an average period of oscillation of 28 months, although it can vary between 24 and 30 months (Giorgetta and Doege 2004). Of even more interest is the 1.589(9) tropical year negative side-lobe period, which just happens to be synodic period of Venus and the Earth = 583.92063 days = 1.5987 years, to within an error of ~ 1.8 hours).

**strength**of the lunar tidal forces, should reach a maximum every**0.563714**tropical years (= 205.89223 days = 0.5 FMCs) and**10.14686**topical years (= 9.0 FMC's). [Note: the longer time period is the more precise alignment of the two and FMC = Full Moon Cycles]b) the rate of change in the stresses caused by lunar tides in the Earth's atmosphere and oceans, as a result of a change in the

**direction**of the lunar tidal forces, should reach a maximum every**1.89803**tropical years (= 2.0 Draconic year).**amplitude modulates**the period for the rate of changes in stresses caused by the change in direction of the lunar tides (i.e. 1.89803 tropical years), you would expect that the 1.89803 year tidal forcing term would split into two spectral peaks i.e. a positive and a negative side-lobe, such that:**Positive side-lobe****[10.1469 x 1.89803] / [10.1469 – 1.89803] = 2.334(7) tropical yrs = 28.0 months**

**Negative side-lobe**

**[10.1469 x 1.89803] / [10.1469 + 1.89803] = 1.598(9) tropical yrs**

**REWRITTEN BLOG POST - UPDATED: 05/10/2015**

**1. The rate of change of the Moon's tidal stresses upon the Earth's atmosphere and oceans - When is it maximized?**

The stresses caused by lunar tides in the Earth's atmosphere and oceans should be a maximum when the tidal forces of the Moon acting upon the Earth change by the largest amount in either strength or direction over a relative short period in time.

There are two such conditions where this takes place:

**a) Changes in lunar tidal strength.**When a new/full moon occurs at or near the times when the lunar line-of-apse (see figure 1) is pointing towards the Sun. This is true because a new/full moon at closest perigee produces the strongest lunar tidal forces upon the Earth while a new/full moon at apogee produces the weakest lunar tidal forces upon the Earth (see figure 2). Thus, when the new/full moon takes place very close to perigee it will be followed roughly 14 days later by a full/new moon not far from apogee. Similarly, when the new/full moon takes place very close to apogee it will be followed roughly 14 days later by a full/new moon not far from perigee.

Figure 1.

A time when a new moon exerts maximal tidal forces upon

the Earth because of its distance from Earth (i..e it is at closest

perigee). Approximately 14 days later, a full moon will exert

minimal tidal forces upon the Earth because of its distance (i.e.

it is at apogee).

**b) Changes in lunar tidal direction**We are looking for the times when the stresses caused by lunar tides in the Earth's atmosphere and oceans will be a maximum because the direction of the tidal forces of the Moon acting upon the Earth are changing by the largest amount over a relative short period in time.

When a new/full moon occurs at or near the time when the lunar line-of-nodes is at a right angle to the Earth-Sun line (see figure 3 and 4). This is true because a new/full moon at a major lunar standstill produces the largest changes in the meridional (north-south) lunar tidal forces upon the Earth (see figure 4).

Hence, when the new/full moon takes place very close to a time when the lunar line-of nodes is at a right angle to the Earth-Sun line (i.e. the new/full moon is at a major lunar standstill) it will be located at a declination of roughly 28 degrees in one hemisphere, followed roughly 14 days later by a full/new moon located at a declination of roughly 28 degrees in the other hemisphere.

Figure 3.

Figure 4.

This file is available under Creative Commons CC0 1.0 Universal Public Domain Dedication

*Hence, the maximal rate of change in the strength of the lunar tidal forces acting upon the Earth should vary with a period set by**the minimum time required for the lunar line-of-apse to realign with the Sun, at the same time as a new/full moon is taking place.*

*In addition, the maximum rate of change in the direction of the lunar tidal forces acting upon the Earth should vary with a period set by the minimum time required for the perpendicular to the the lunar line-of-nodes to re-align with the Earth-Sun line, at the same time as a new/full moon is taking place.***2. Minimum Realignment Times for the Lunar Line-of-Apse**

**Starting out with a new moon at perigee with the Perigee end of lunar line-of-nodes pointing at the Sun (see figure 5).**

Figure 5

The time required for the Perigean end of the lunar line-of-apse to point back at the Sun close to New Moon is (see figure 6):

**1.0 Full Moon Cycle = 1.0 FMC = 411.78445 days = 1.1274 tropical years (1)**

The Moon almost returns to being New at perigee after 1.0 FMC because 14 Synodic months = 413.42824 days and 15 anomalistic months = 413.31825 days.

[NB: There is a slight miss-match of 0.110 days between these two alignments which means that the new Moon occurs ~ 1.3 degrees away from actual lunar perigee.]

Figure 6.

This means that minimum time required for the lunar line-of-apse to realign with the Sun when there is new/full moon is (see figure 7):

**0.5 FMC = 205.89222(5) days = 0.56371(4) tropical years (2)**

The Moon almost returns to being New at apogee after 0.5 FMC because 7 Synodic months = 206.714122 days and 7.5 Anomalistic months = 206.659125 days. [NB: producing an error of ~ 0.8 days between 7 Synodic months and the length of 0.5 FMC.]

Figure 7.

The next shortest time required for the lunar line-of-apse to re-align with the Sun when there is a new/full moon (in this case a Full Moon) is at:

**9.0 FMC = 3,706.06005 days = 10.1468(6) tropical years (3)**

The Moon almost returns to Full Moon at apogee after this period because 125.5 synodic months = 3706.08891 days and 134.5 anomalistic months = 3706.08698 days. Note that there is a mismatch of only 0.029 days between 125.5 Synodic months and 9.0 FMC which means that this realignment period is about four times more precise than that at 0.5 FMC realignment period.

**3. Minimum Realignment Times for the Perpendicular to the Lunar Line-of-Nodes**

**Starting out with a New Moon at a major lunar standstill (i.e. the line-of-nodes is at right angles to the Earth-Sun direction).**

**Figure 8.**

**The lunar line-of-nodes return to being perpendicular to the Earth-Sun direction after 0.5 Draconic Year (DY), where:**

**0.5 DY = 173.310038 days = 0.47450(7) tropical years (4)**

**[NOTE: 1.0 Draconic Year (DY) = 346.620076 days = 0.94901(4) tropical years]**

Figure 9,

Unfortunately, 173.310038 days is 5.869 Synodic months and 6.369 Draconic months and so the Moon phase is not new or full.

It turns out that the minimum time required for the Moon to return to a major standstill (i.e. the time required for the perpendicular to the lunar line-of-nodes to realign with the Sun), when there is a new/full moon is:

**2.0 DY = 4 x 0.5 DY = 693.24015(2) days = 1.8980(3) tropical years (5)**

The Moon almost returns to being Full after 2.0 DY because 23.5 Synodic months = 693.96884 days and 25.5 Draconic months = 693.91161 days. [NB: producing an error of ~ 0.73 days between the 23.5 Synodic months and 2.0 DY.]

The 1.89803 tropical year period is part of the 19.0 year Metonic Cycle with the Moon's phase returning to new moon at a node after 3.79606 tropical yrs = 1387.481264 days (since 47 Synodic months = 1387.937678 days and 51 Draconic months = 1387.82322 days). With this cycle, the synodic lunar cycle realigns with the seasonal calendar every 4 tropical years, 4 + 4 = 8 tropical years, 4 + 4 + 3 = 11 tropical years, 4 + 4 + 3 + 4 = 15 tropical years and 4 + 4 + 3 + 4 + 4 = 19.0 tropical years. to give and average spacing of roughly (4 + 4 + 3 + 4 + 4)/5 = 3.8 years.

**3. Discussion**

The above analysis tells us that:

a) the rate of change in the stresses caused by lunar tides in the Earth's atmosphere and oceans, as a result of a change in the

is the more precise alignment of the two.]

b) the rate of change in the stresses caused by lunar tides in the Earth's atmosphere and oceans, as a result of a change in the

Now if the period of the rate of change in stresses caused by the change in strength of the lunar tides (i.e. 10.14686 tropical years) amplitude modulates the period of the rate of changes in stresses caused by the change in direction of the lunar tides (i.e. 1.89803 tropical years), you would expect that the 1.89803 year tidal forcing term would split into a positive and a negative side-lobe, such that:

Positive side-lobe

[10.1469 x 1.89803] / [10.1469 – 1.89803] = 2.334(7) tropical yrs = 28.0 months

Negative side-lobe

[10.1469 x 1.89803] / [10.1469 + 1.89803] = 1.598(9) tropical yrs

Interestingly, the time period of the positive side-lobe is almost exactly the same as that of the Quasi-Biennial Oscillation (QBO). The QBO is a quasi-periodic oscillation in the equatorial stratospheric zonal winds that has an average period of oscillation of 28 months, although it can vary between 24 and 30 months (Giorgetta and Doege 2004).

a) the rate of change in the stresses caused by lunar tides in the Earth's atmosphere and oceans, as a result of a change in the

**strength**of the lunar tidal forces, should reach a maximum every 0.563714 tropical years (= 0.5 FMCs) and 10.14686 topical years (= 9.0 FMC's). [Note: the longer time periodis the more precise alignment of the two.]

b) the rate of change in the stresses caused by lunar tides in the Earth's atmosphere and oceans, as a result of a change in the

**direction**of the lunar tidal forces, should reach a maximum every 1.89803 tropical years (= 2.0 DYs).Positive side-lobe

[10.1469 x 1.89803] / [10.1469 – 1.89803] = 2.334(7) tropical yrs = 28.0 months

Negative side-lobe

[10.1469 x 1.89803] / [10.1469 + 1.89803] = 1.598(9) tropical yrs

Even more interesting, is the 1.589(9) tropical year negative side-lobe period, which just happens to be synodic period of Venus and the Earth = 583.92063 days = 1.5987 tropical years, to within an error of ~ 1.8 hours).

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[NB: The lunar and planetary periods used in this post are:

Synodic month = 29.5305889 days

Anomalistic month = 27.55455 days

Draconic month = 27.21222 days

Tropical Year = 365.242189 days

Sidereal orbital period of the Earth = 365.256363 days

Sidereal orbital period of Venus = 224.70069 days

a = 29.53059 days

ReplyDeleteb = 27.55455

c = 27.21222

y = 365.25

9*a*b/(a-b)/y*2*a*c/(a-c)/y/(9*a*b/(a-b)/y-2*a*c/(a-c)/y)

simplifies to:

18 a b c / y / (9ab -2 ac -7 bc)

this does give 2.3347, but the choice of multiples of 9 and 2 is

I am not sure whether or not you are making fun of me here Paul??

ReplyDeleteI have just chosen the two shortest physical times for the precession of the line-apse (10.1446 years) and the line-nodes (1.8980 years) and asked myself how long do they take to realign with the Synodic cycle. This beat is based upon straight forward physical principles.

I am sure that you are aware that what you have proposed is an arbitrary combination of numbers unless you can justify the multiples 9 and 2. Indeed you have noted that in your post. So, I will assume that your inquiry is genuine.

I am sorry, I have made a mistake in my post. I have not used the time for the shortest alignment between the apse and the synodic period. What I have used is the shortest alignment between that of syzygy of the Earth, Moon and Sun (i.e. the time of spring tides) and the synodic cycle. My Bad!

ReplyDeleteJust trying to follow your recipe. What I presented were the only multiples (2 and 9) that would give the 2.335 result.

ReplyDeletePaul,

ReplyDeleteThanks for the clarification. It makes you wounder if the 2 and 9 have some physical significance.

My post on October 1st 10:39 PM (above) was made in haste as I was rushing out the door. Consequently, it should not to be take seriously. I have added an addendum to the above post to clarify what it should have said.

Sorry for the confusion.

Paul,

ReplyDeleteYou posted the following:

a = 29.53059 days

b = 27.55455

c = 27.21222

y = 365.25

9*a*b/(a-b)/y*2*a*c/(a-c)/y/(9*a*b/(a-b)/y-2*a*c/(a-c)/y)

[Note I have used: a = 29.5305889 days; b = 27.55455 days; c = 27.21222 days;

and y = 365.242189 days]

It turns out that a*b/(a-b)/y is just the Full Moon Cycle (FMC) in years (1.1274 trop yrs).

The FMC is the time required for the line-of-apse to realign with the Sun.

While a*c/(a-c)/y is just the Draconic Year (DY) (0.94901 trop yrs),

so what you have written is just:

9*FMC*2*DY/(9*FMC-2DY)

where 9* FMC = 3706.06005 days = 10.146856 trop yrs

is the shortest time in which the lunar line-of-apse precisely re-aligns with the Sun at New or Full Moon.

[Note: this because 125.5 Synodic months = 3706.08890 days and 134.5 Anomalistic months = 3706.086975 days]

and

2*DY = 693.2415(2) days

is the shortest time in which the lunar line-of-nodes precisely re-aligns with the Sun at New or Full Moon.

[Note: this because 23.5 Synodic months = 693.96884 days and 25.5 Draconic months = 693.91161 days]

Hence, there is actually a physical reason for choosing this beat period.

Sorry, forgot to add - The Draconic year [=346.620076 days] is the time for the lunar line- of-apse to realign with the Sun.

ReplyDeleteIan, I think you are spot on. What I earlier wrote was a Mathematica factored expression, which is why it looked so concise.

ReplyDeleteAlso have a multiple linear regression algorithm fit for the QBO and what it can do is test sinusoidal factors and find out which values give the highest correlation coefficient. So if I tune around the 2.33 and 1.6 values, I get the following chart.

http://imageshack.com/a/img633/2783/V7QffL.png

You predicted 2.334(7) and the fit finds a peak at 2.334.

You predicted 1.598(9) and the fit finds a peak a hair under 1.599.

This is likely not coincidence, and instead is the starting point for explaining the behavior of QBO.

Thank you for your further investigations. I would not have noticed this phenomenon if it was not for your comments above. It helped me realize that it might have something to do with the minimum period between the peaks in the rate of lunar tidal stresses upon the Earth as a result of changes in strength and direction of the lunar tidal forces.

ReplyDeleteI do not have the mathematical skills and ability to model these changes, that I will have to leave up to someone else. However, I hope that this blog post will get others thinking and investigating. First to check if what I am saying makes any sense [There is always a possibility that I have missed something important in the conceptualization and the physics], and second to expand on this initial idea.

WHT, your discoveries and investigations have been truly ground breaking and I cannot wait to see your paper on QBO and ENSO published so that I can use my limited mathematical skills to further look into the details.

I am still trying understand the full physical implications of the results of this post and I know that you have a task ahead of you to collect together all of the details of your investigations about the QBO and ENSO for their inclusion in your CSALT model.

Best Wishes

Ian, the question I have is why is the 2.33 factor much stronger than the 1.6 factor? It is a faster frequency so perhaps it gets more quickly damped.

ReplyDelete2.33 is also close to the Draconic aliasing (2.368) and also double the Chandler wobble (1.185) so it may have more of a reinforcing effect.

Any other insight?

WHT,

ReplyDeleteThe side-lobes produced by amplitude modulation of a carrier signal should be of roughly equal intensity - so can't help you here.

The QBO lengths cluster into two groups of roughly equal number around ~ 26.0 months (2.17 yrs) and ~ 30.0 months (2.5 years) with the average holding steady at ~ 28.0 months, so I would expect the spectral region in the data to have a number of close spectral peaks spread between about 2.2 and 2.5 years.

It is just possible that 2.335 year peak may be a(n unknown) constant integer multiple in strength of the 1.589(9) year peak e.g. 2 x, 3x, 4x, etc. if the line strengths of the spectral peak cluster are super-harmonics of the 18.6 year period and sub-harmonics of the Chandler wobble - but this is along shot.