## Ian Wilson: Help Needed to Solve an Interesting Lunar Puzzle

Posted: April 29, 2017 by tallbloke in Astronomy, Astrophysics, Celestial Mechanics, moon, solar system dynamics

Astrophysicist Ian Wilson has emailed me to ask for a brainstorming session at the talkshop to assist him. Ian writes:

“I was wondering if you or your colleagues (e.g. oldbrew) could help me work out the solution to the following lunar puzzle”

The Conundrum

The diagram below shows the Perigee of the lunar orbit pointing at the Sun at 0.0 days. In addition, the diagram shows the Perigee of the lunar orbit once again pointing at the Sun after one Full Moon Cycle (FMC) = 411.78443025 days. It takes more than 1.0 sidereal year (= 365.256363004 days) for the Perigee to realign with the Sun because of the slow pro-grade (clockwise) precession of the lunar line-of-apse once every 8.85023717 sidereal years.

1.0 FMC falls short of 15 anomalistic months (= 413.31824817 days) by 1.53381792 days (= 1.5117449198O). During these 1.5117449198 days the Perigee end of the lunar line-of-apse rotates by 0.17081406in a prograde direction, producing an overall movement of the line-of-apse (red line) of 1.34093086O (= 1.5117449198O – 0.17081406O) with respect to the Earth-Sun line (blue line).

if we let:

DT  =  (15 anomalistic months — FMC) = (413.31824817 — 411.78443025) days

= 1.53381792 days

S = the angular revolution (in degrees) of the Earth about the Sun over DT days.

= 1.5117449198 degrees

L = the orbital precession (in degrees) of the lunar line-of-apse over DT days.

= 0.1708140574 degrees

then

D = S — L = angle between the lunar line-of-apse and the Earth-Sun line after DT days.

= 1.3409308624 degrees

we find that if we take the incremental angle between the lunar line-of-apse and the Earth-Sun line over DT days (= 1.3409308624 degrees) and divide it by the 360 degrees of movement of the angle between the lunar line-of-apse and the Earth-Sun line that has occurred over the previous FMC, it effectively has the same value as the incremental number of days between 15 anomalistic months and 1.0 FMC (=1.5338172 days) divided by 1.0 FMC i.e.

(S — L) / 360 degrees = (15 anomalistic months – FMC) / FMC = 0.0037248080            (1)

While this is not remarkable, what is remarkable, however, is that both of these fractions (whether they be measured in degrees or days) are precisely equal to the cumulative annual precession of the Perihelion of the Earth’s orbit (measured in days) over a period of 1.0 FMC!

= (11.723″/3600) deg. per yr x (365.256363004 days / 360 deg.) x (411.78443025 / 365.256363004)

= 0.0037248062 days per 1.127384686 sidereal years.

N.B. The current value for the precession of the Perihelion of the Earth’s orbit is 11.615 arc seconds per year. However it is increasing and will achieve a value of 11.723 arc seconds in roughly 2490 A.D.

Here’s the Rub [updated 25/04/2017]

1. The cumulative precession of the Earth’s orbit over a period of 1.0 FMC has the dimensions of days per year!

I must be missing something. Why the strange units of days per 1.127384686 sidereal years? Can anyone help me understand why I get these weird dimensionless units? [updates 22/04/2017]

1.  The increase in angle between the lunar line-of-apse and the Earth-Sun line as you move from 1.0 FMC to 15 anomalistic months (= 1.34093086degrees) seems to be almost precisely equal to the FULL annual precession of the Perihelion of the Earth’s orbit (= 11.723 arc seconds per year) PER DAY accumulate over 1.0 FMC i.e.

[(11.723 / 3600) deg per YEAR] x 411.78443025 days = 1.34093024 degrees

The question is, what angular motion associated with the movements in Sun-Earth-Moon system can cause the ANNUAL precession of the Perihelion of the Earth’s orbit to accumulate DAILY?

I am not aware of any mechanism that would produce a motion like this and I would appreciate if anyone could solve this interesting lunar puzzle for me!
Is it something to to do with the interaction between the mean and true anomalies of the Earth’s orbit and the Moon’s orbit? Could motion of the Earth and moon about the common centre-of-gravity have and effect? What about the effects of 18.6 year nutation of the Earth’s rotation axis?

[N.B. A Full Moon Cycle (FMC = 411.78443025 days epoch J2000.0) is the time required for the Sun (as seen from the Earth) to complete one revolution with respect to the Perigee of the Lunar orbit. In other words, it is the time between repeat occurrences of the Perigee end of the lunar line-of-apse pointing at the Sun. The value for the length of the FMC is equal to the synodic product of the Synodic (lunar phase) month (= 29.530588853 days) with the anomalistic month (= 27.554549878 days) for Epoch J2000.0. The anomalist month is the time required for the Moon to return to the Perigee of the lunar orbit.]

[N.B. The Lunar Anomalistic Cycle (LAC = 3232.60544062 days = 8.85023717 sidereal years) is the time required for the lunar line-of-apse to precess once around the sky with respect to the stars. This corresponds to a 0.11136528O per day movement of the lunar line-of-apse in a pro-grade (clockwise) direction. The value for the length of the LAC is equal to the synodic product of the anomalistic month (= 27.554549878 days) with the sidereal lunar month (= 27.231661547 days) for Epoch J2000.0. The sidereal month is the time required for the Moon to rotate once around its orbit with respect to the stars.]

1. oldbrew says:

I already commented on Suggestions 26 🙂
https://tallbloke.wordpress.com/suggestions-26/comment-page-1/#comment-125429

But TBH I’m still not sure exactly what the issue is – maybe I need another cup of tea.
Update: my initial thought was that there is just more than one way of calculating the same result.

I can see that:
1.5117449198° / (1.5117449198° – 0.17081406°) = 1.1273847 = number of sidereal years in one full moon cycle
– – –
In the blog post: ‘During these 1.5117449198 days‘ should read ‘1.53381792 days’ — as per preceding sentence.
– – –
Off topic (?):
1625 full moon cycles = 1832 sidereal years
‘Super Perigean Eclipse Tide(SPET) occurs every 1832 years (Perigean Eclipse Tide AND Earth/Sun at perigee)’

Quote is from Clive Best’s reply to IW re the Keeling & Whorf paper:
http://clivebest.com/?p=5464#comment-5356

1832y * 7 = 689 lunar nodal cycles = 1449 lunar apsidal cycles (207 * 7)
207 = 1832 – 1625

2. tallbloke says:

Stuart, I have some vague recollection that you found some other figures which seemed to tie in Earth’s perihelion precession period with some lunar period?

3. oldbrew,

Thanks for your initial comments and I agree with the correction that you made to the blog.

The critical equation is:

1.3409308624 degrees / 360 degrees = 0.003724807 +/- 0.000000001 =

(11.723″/3600) deg. per yr x (365.256363004 days / 360 deg.) x (411.78443025 /
365.256363004) (1)

If you cancel 360 degrees from both sides and, in addition, cancel the two incidences of 365.256363004 days from the right-hand-side you get:

1.3409308624 degrees = (11.723″/3600) deg. per yr x 411.78443025 days

This is saying that the drift between 15 lunar anomalistic month and 1.0 Full Moon Cycle (FMC) PER DAY seems to be occurring at exactly the total ANNUAL precession of the Perihelion of the Earth’s orbit about the Sun.

I do not have any obvious explanation for this phenomenon.

One wild suggestion is that the precession of the lunar line-of-apse is associated with the precession of the Perihelion of the Earth’s Orbit and that this drifts away from the time of the anomalistic month because it is associated with the 238 year cycle seen in the nodes of Venus’s orbit at the time of Venus transits, since:

3186 3168 lunar anomalistic months = 87,292.81401350 days = 238.9905361 sidereal years
149.5 Earth/Venus Synodic months = 87,296.1338258 days = 238.999625107 sidereal years.

Grasping at straws here.

3168.0 lunar anomalistic months = 87,292.81401350 days = 238.9905361 sidereal years
149.5 Earth/Venus Synodic months = 87,296.1338258 days = 238.999625107 sidereal years.

5. olbrew said:

Then 1.5117449198° / (1.5117449198° – 0.17081406°) = 1.1273847 = number of sidereal years in one full moon cycle [see diagram in above post]

This is true but it does not explain why:

1.3409308624 degrees = (11.723″/3600) deg. per yr x 411.78443025 days

which is saying that the drift between 15 lunar anomalistic month and 1.0 Full Moon Cycle (FMC) PER DAY seems to be occurring at exactly the total ANNUAL rate of precession of the Perihelion of the Earth’s orbit about the Sun.

6. Skeptikal says:

I’m thinking coincidence. It is what it is for no reason at all, and it will change in the future as the speed of precession changes…. just like the sun is about 400 times the size of the moon but is also about 400 times as far away from us, so the sun and the moon appear the same size in the sky. This will change as the moon slowly drifts away from us, but for now it gives us spectacular solar eclipses.

7. oldmanK says:

Please pardon my ignorance of the many numbers here. I have a question, based on the diagram above.

Where is the 0.0 days start point in the orbit of the earth? The diagram appears to be based on a circular orbit, which it isn’t. (other things that have ‘obsessed’ me have shown that the seasons, which presently vary by some days from one to another, require adjusting to.)

8. oldman K,

The zero point in the diagram is marked at the top of the diagram. You can arbitrarily take it to be the point where the Earth is a Perihelion, the Perigee end of the lunar line-of-apse is pointing at the Sun, and the Moon is located at New Moon, directly between the Earth and the Sun.

You are absolutely right that both the Lunar orbit and the Earth’s orbits are ellipses and this will have an effect on the angles and times involved. However, we are doing a first order approximation here that is valid in the sense that, in the long run, the periods and angles we are after with just oscillate around the values that are obtained with the numbers we are using.

When we take a second pass at this problem we will need to address the no spherical nature of the orbits. Life is cruel, however, since it may turn out that the no spherical nature of the two orbits plus the osculatory behavior of the Earth-Moon system (i.e. both bodies actually orbit the common centre-of-mass) might be the source of the problem.

9. oldmanK says:

astroclimateconnection: thank you. That will give me a glimpse of the ‘rules of the game’.

10. Rog and oldbrew,

If you go to this blog post of mine at:

you will see that you can rearrange equation (3) and (4) to get that:

[1 / (2 DY)] + [1 / (9 FMC)] = [1 / (SEV)] ___________________________________(1)

where DY = Draconic year (in days) ___________________________= 346.62007588 days
____FMC = Full Moon Cycle (in days) _________________________= 411.78443025 days
and _SEV = The synodic period of Of Venus and the Earth (in days) __= 583.92062760 days

The left hand side of the formula gives [1 / 583.99961227] days which only 1.89 hours of the correct synodic period.

Given that 5 x SEV = 2919.603138 days = 7.9932985 sidereal years which falls short of 8 sidereal years by 2.45 days, equation (1) could be rewritten as:

[1 / (10 DY)]+[(1 / (45FMC)] = [1 / (5SEV)] = [1/ 2919.99806135] = [1 / 7.99437972 sid yrs in days]

which just happens to be:

[2 / (MET)] + [1 / (45FMC)] = [1 / 2922.58327458 days] = [ 1 / 8.00145753 sidereal yrs in days]

where

MET = 235 Synodic months = Metonic cycle = 6939.68838046 days = 18.9995003054 sid yrs

[which is just 4.38 hours short of 19 sidereal years]

11. Rog and oldbrew,

Some people prefer the numbers to be on top:

[9 / (DY)] + [2 / (FMC)] = [18 / (SEV)] ___________________________________(1)

12. oldbrew says:

250 Full Moon Cycles = 297 Draconic Years
times 7 = 1973 tropical years

13. In addition, it is important to note that the relative orbital periods of Jupiter, Saturn and Venus, with respect to the Earth, most likely set the length of the Lunar Anomalistic Cycle (LAC), such that:

[(1 / SJS) + (1 / 10SVE)] = [1 / 8.85662] sidereal yrs = 1 / LAC

or

[(10 / SJS) + (1 / SVE)] = [10 / 8.85662] sidereal yrs = 1 / LAC

where SJS = the synodic period of the Jupiter and Saturn (in years) = 19.585 sidereal yrs
_____SVE = the synodic period of Vanus and the Earth (in years) = 1.598660 sidereal yrs
and__LAC = the lunar anomalistic cycle = 8.8502 sidereal yrs.

[with an error of 0.0064 sidereal yrs = 2.34 days]

14. oldbrew says:

Re – The critical equation is:

1.3409308624 degrees / 360 degrees = 0.003724807 +/- 0.000000001 =

(11.723″/3600) deg. per yr x (365.256363004 days / 360 deg.) x (411.78443025 /
365.256363004)
– – –
Can coincidence be ruled out [see note]? Alternatively, would a binary Earth-Moon system explain it?

‘some scientists view the Earth-Moon system as a double planet as well, though this is a minority view.’
http://en.wikipedia.org/wiki/Double_planet

‘Earth’s centre veers inside and outside the solar orbital path during each synodic month as the Moon moves in its orbit around the common centre of gravity’
http://en.wikipedia.org/wiki/Orbit_of_the_Moon#Path_of_Earth_and_Moon_around_Sun

Note: ‘The current value for the precession of the Perihelion of the Earth’s orbit is 11.615 arc seconds per year. However it is increasing and will achieve a value of 11.723 arc seconds in roughly 2490 A.D.’ – from this blog post

15. AlecM says:

Whatever it is, the answer is astronomical!

16. AlecM says:

Earth-Moon is a double planet with a centre of mass displaced from that of the Earth. The proof is that we have two tides a day instead of one.

17. oldbrew said: Can coincidence be ruled out [see note]?

The agreement is to nine decimal places using a Perihelion precession rate of 11.723 arc seconds per year. While this is a value that will apply in the not too distant future (2490 A.D.) it was chosen to produce almost perfect alignment. If, however, we use the current precession rate of 11.615 arc seconds we get a value of 0.00369049 compared to 0.00372481, this still produces an agreement to five decimal places (i.e. 0.000034 ~ 0.9 %) – I think we can rule out the possibility of coincidence with some confidence.

18. Yes, this is a well-known phenomenon since the Bronze Age. Precession of the plane of the moon’s orbit follows an 18.6-year cycle.

Some apparent changes in sea level linked to the major lunar standstill are mistakenly attributed to climate change. Which is how I twigged that your query is about the lunar standstill.

Wikipedia has an explanation, including the megaliths of the Bronze Age. If you don’t trust Wikipedia science, try the following URL.

https://en.wikipedia.org/wiki/Lunar_standstill

http://www.umass.edu/sunwheel/pages/moonteaching.html

“The 18.6-year cycle is caused by the precession of the plane of the lunar orbit, while this orbit maintains a 5° tilt relative to the ecliptic.”

19. My apologyies for not addressing the question of torques. The following paper cites Bills, B. G., and R. D. Ray (1999), See below.

Siegler, M. A., B. G. Bills, and D. A. Paige (2011), Effects of orbital evolution on lunar ice stability, J. Geophys. Res., 116, E03010, doi:10.1029/2010JE003652.

Bills and Ray state in their introduction that, “It has long been recognized that the present rate of tidal dissipation of energy, and associated transfer of angular momentum from the spin of the Earth to the orbit of the Moon is anomalously high [Gerstenkorn, 1955; Goldreich, 1966; Lainbeck, 1980; Touma and Wisdom, 1994; Munk, 1997].”

Bills, B. G., and R. D. Ray (1999), Lunar orbital evolution: A synthesis of recent results, Geophys. Res. Lett., 26(19), 3045–3048, doi:10.1029/1999GL008348.

http://onlinelibrary.wiley.com/doi/10.1029/1999GL008348/pdf

At least one article in Wikipedia does not distinguish between lunar axial precession and lunar orbital precession. There is also precession of axes of the orbital plane that cause variations in the position of the apogee and perigee.

20. Sera says:

Earth-Luna is a binary system, but does the barycenter move enough to make a difference?

21. AlecM says:

@Sera: two tides a day is the proof.

22. Sera says:

@ AlecM:

I’m speaking about the barycenter moving (position) based on lunar position.

23. Each of the following affects the rate of precession of the lunar line-of-apse (this is the long-axis of the ellipse of the lunar orbit). Care need to be taken, however, to distinguish whether the rate of precession is measured with respect to the stars (i.e. as seen from the centre of the Earth) or with respect to the Earth-Sun line (i.e the Sun as seen from the centre of the Earth).

**************************************************************************************************************
A. involve parameter that are measured with respect to the stars.

[(10 / SJS) + (5 / 5 SVE)] = 10 / LAC_______________________________________(1)

where

SJS = the synodic period of Jupiter and Saturn = 19.85810 sid yrs.
5 SVE = five times the synodic period of the Earth and Venus = 7.9932985 sid yrs.
LAC = the Lunar Anomalistic Cycle = 8.8502372 sid yrs.

The LAC is the time required for the perigee of the lunar orbit to move once around the centre of the Earth with respect to the stars.

N.B. Venus and the Earth realign with respect to the stars once every 5 SEV. only falling short by with an error of 0.0067015 sidereal years = 2.4125 degrees in the movement of teh Earth around the Sun. Interestingly, the left-hand-side of equation (1) is equal to (1 / 1.85663622 sid yrs.) which off the nominal value of the (1 / LAC) = (1 / 8.8502372 sid yrs) by 0.0063990 sidereal years. = 2.3036 degrees. This is very similar to the discrepancy between 5 SEV and 8.0 sidereal Earth years.

**************************************************************************************************************
B. Involve parameters that are measured with respect to the Earth-Sun line.

[9 / (DY)] + [2 / (FMC)] = [18 / (SEV)]_______________________________________(2)

where

DY = the Draconic year = 0.94897752 sid yrs.
FMC = Full Moon Cycle = 1.12738469 sid yrs.
SEV = the synodic period of the Earth and Venus = 1.5986597 sid yrs.

The Draconic year is the time required for a lunar node to move once around the centre of the Earth with respect to the Sun.

The Full Moon Cycle is the time required for the Perigee end of the lunar line-of-apse to move once around the centre of the Earth with respect to the Sun.

**************************************************************************************************************

If we rewrite equation (1) we can get:

[(1 / SJS) + (1 / (2 x 5 SVE))] = 1 / LAC_______________________________________(1)

\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$

This means that the length of the LAC is just half the harmonic mean of the Synodic period of Jupiter and Saturn and two pentagonal cycles of Venus and the Earth.

\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$

24. oldbrew says:

14 synodic months is very similar to 15 anom. months as this shows:

537 x 15 anom. months = 539 x 14 synodic months
537 x 15 anom. months = 539 full moon cycles = 7516 synodic months (537 * 14, -2)
[Validation: 537 * 15 = 8055, and 8055 – 7516 = 539]
539 / 537 = 1.0037243
15 AM / FMC = 1.0037248

As Ian W says: ‘(S — L) / 360 degrees = (15 anomalistic months – FMC) / FMC = 0.0037248080’
(effectively 1.0037248 – 1)

Whether that leads anywhere, I don’t know – maybe not.

Plus – multiplying the period of 539 FMC by 13 * 14 (= 182):
182 * 607.6604 sidereal years = 110594.2 years

360 * 60 * 60 = 1296000 arcseconds
1296000 / 110594.2 = 11.718516 arcseconds

This is very close to the year 2490 value (11.723) referred to by IW.
Which still leaves us to find a theory that explains the numbers, but there we are.

25. oldbrew says:

Re the last comment, I’ll refer to this from the blog post:
‘[(11.723 / 3600) deg per YEAR] x 411.78443025 days = 1.34093024 degrees’

Over 98060 full moon cycles:
98060 * 1.34093024 = 131491.61°
131491.61 / 360 = 365.25447 (1 sidereal year = 365.25636~ days)

From the last comment: 539 FMC * 182 = 98098 FMC
So not too far apart, only 38 FMC difference in nearly 100,000.
– – –
We did say it was a brainstorming session 🙂

26. J Martin says:

Sadly all this synodic, metonic stuff goes straight over my head. Can anyone point to an idiot’s guide to understanding these things ?

Whenever anyone mentions precession my first thought is always binary star. No doubt of no relevance to this discussion.

27. oldbrew says:

JM – nobody said it was straightforward 🙂

The Moon is especially tricky. You could perhaps start here:
http://en.wikipedia.org/wiki/Full_moon_cycle

Perigee,perihelion = closest point to the ‘parent body’ during an elliptical orbit

Re precession – there are several types (perihelion, axial, etc.) and the terminology can get a bit mixed up, depending where you look for advice.

This might give you a clue…
Precession of the perihelion of Mercury

…let me describe the way Mercury’s orbit looks. As it orbits the Sun, this planet follows an ellipse…but only approximately: it is found that the point of closest approach of Mercury to the sun does not always occur at the same place but that it slowly moves around the sun (see Fig. 7.20). This rotation of the orbit is called a precession.

The precession of the orbit is not peculiar to Mercury, all the planetary orbits precess.

http://physics.ucr.edu/~wudka/Physics7/Notes_www/node98.html

Fig. 7.20 [note – the graphic exaggerates the rate of precession]

28. Paul Vaughan says:

recap from https://tallbloke.wordpress.com/suggestions-25/ (Suggestions 25):

(4.42377107)*(1.184859016) / (4.42377107 – 1.184859016) = 1.618304218
antiresonant CW apse extremes

…said a different way:
QBO & LAC (lunar apse cycle)
(8.847542139)*(2.369718033) / (8.847542139 – 2.369718033)
= 3.236608436
~= 3.236067977 = 2φ = 1+√5

LAC refresher (for those who haven’t thought about it in years):
(27.55455)*(27.32158236) / (27.55455 – 27.32158236) = 3231.495658 days
(3231.495658) / 365.242189 = 8.847542139 tropical years

QBO (aliasing) reminder:
harmonic of 365.242189 nearest 27.212221 is 28.095553
(28.095553)*(27.212221) / (28.095553 – 27.212221) = 865.5210016 days
(865.5210016) / 365.242189 = 2.369718033 tropical years = QBO (quasi-biennial oscillation)
2.369718033 / 2 = 1.184859016 tropical years = 432.7605008 days = CW (chandler wobble)

where lunar months are
27.32158236 days = tropical
27.55455 days = anomalistic
27.212221 days = draconic

3rd eye on LAC extreme antiresonance:
(4.42377107)*(1.618033989) / (4.42377107 + 1.618033989) = 1.184714151 tropical years
= 432.7075898 days
~= CW (chandler wobble)

harmonic mean:
(4.42377107)*(1.618033989) / ( (4.42377107 + 1.618033989) / 2 ) = 2.369428302 tropical years
~= QBO (quasi-biennial oscillation)

where 1.618033989 = φ
& LAC extremes occur with period LAC / 2 = 8.847542139 / 2 = 4.42377107 tropical years

Generalized extension beyond the mean doesn’t end with volatility.
A comprehensive framework extends to aggregation criteria with infinite generality.

Stability -1Φ1: Antiresonance is a natural basin of attraction. (The ball rolls downhill.)
https://tallbloke.wordpress.com/suggestions-26/ (continued on Suggestions 26)

29. P.A.Semi says:

Note to original picture: pro-grade direction is counter-clockwise when lookup from above north pole, and also your diagram is drawn as if looking from below the south pole… (well, what is above and what is below somehow depends on which hemisphere one lives on, but usually the charts and directions are depicted as if looking from above north pole…)

If you find any connection between planets and Moon’s anomalistic precession (precession of Moon’s perigee/apogee), it is purely random and please ask your conscience, how did you get it there and what other numbers could play in your “rhymes” and “rhythms”…

By simple 3-body gravity attraction in Sun-Earth-Moon system it reproduces that 8-year cycle of perigee precesion quite well… (As compared to JPL ephemerides, my lay-man integration by 1 minute steps leads to 7.5 year precesion with similar inner structure, just departing from the JPL ephemerides and also my Moon tends to travel slowly (relatively fast!) away from Earth…)

My test integration:

(purple line is my interpolation of perigee azimuth, green line is perigee azimuth from JPL ephemerides)

My lay-man explanation is, that while Moon orbits arround Earth, the Earth swerves to the side by 4600km arround E-M barycenter (and/or due to orbit arround the Sun?), in prograde direction, which makes the Moon’s orbit ellipse axis swerve in that same direction…
(From the distance of Moon-Earth, that 2x 4600 km is 1.41 degree, by which Earth moves to side during Moon’s orbit… It looks somehow close to that 3 degree precesion per month, while whole Earth-Moon system turns cca 30degree on path arround Sun during a month, which does not look much similar…)
Actually it is 2x 4600km from one side, and 2x 4600 km from other side, which gives those almost 3 degrees…

Finding greatest common divisors between this moon cycle and planets is just playing with greatest common divisors of arbitrary numbers – or is the system better “tuned” artificially…?

Well – finding beat of any yearly phenomenon to Earth-Venus period, which is tuned 1.6 to Earth (13/8) should also be quite easy but misleading, same as finding beat with J-S period of 20 years to any yearly phenomenon…

On the other hand – finding tuning of nature to Moon’s cycles should be interesting and is reasonable…

30. P.A.Semi says:

My complaint above about playing with large numbers about FMC and/or anomalistic precession of Moon was probably rather for “astroclimateconnection”‘s articles (which relate Moon’s cycles to Earth-Venus and/or Jupiter-Saturn synodic cycles) than to original post here…

To the original post – it’s hard to tell, whereto rotates perigee axis during 1.5117449198 days, because the movement is not uniform at all, it is periodic. So statistically, one can say it moves some amount, but practically it has little sense…
1/0.0037248080*1.53381792 = 411.78442486163
1.127384686*365.25 = 411.7772565615

Earth’s precession can also be explained by 3-body system of Sun-Earth-Moon. Contribution of planets like Venus or Jupiter is quite minor, most important are Moon and Sun pulling Earth’s equatorial bulge to ecliptic, creating torque which makes the gyroscope precession…
So that would probably somehow explain, why is there that common number 1/0.0037248080 = 268.470 with other Sun-Earth-Moon interactions ?

31. P.A.Semi says:

“Dimensions of days per year” is a consequence, that you measure angles in days by implementing the conversion factor (365.256363004 days / 360 deg.) ,
which you then further numerically cancel by (411.78443025 / 365.256363004) …

(which just explains those “strange” units “days per year”… these “days” are unusual units of angle…)

32. Further P.A. Semi comment,

“To the original post – it’s hard to tell, whereto rotates perigee axis during 1.5117449198 days, because the movement is not uniform at all, it is periodic. So statistically, one can say it moves some amount, but practically it has little sense…
1/0.0037248080*1.53381792 = 411.78442486163
1.127384686*365.25 = 411.7772565615”

Not if the periodic alignment mark the passage of an underlying physical process e.g. We can observe eclipses of the Sun (an event) but still speculate about the underlying orbital movements that produce these events.

33. SLIGHTLY MODIFIED COMMENT SINCE MY ORIGINAL POST HAS NOT POSTED !
[Sorry, spam bin swallowed it – mod]

P.A. Semi,

Thank for your considered comments – they touch on many good points that need to be addressed.

************************
Comment 1: “If you find any connection between planets and Moon’s anomalistic precession (precession of Moon’s perigee/apogee), it is purely random and please ask your conscience, how did you get it there and what other numbers could play in your “rhymes” and “rhythms”…”

Comment 2: Finding greatest common divisors between this moon cycle and planets is just playing with greatest common divisors of arbitrary numbers – or is the system better “tuned” artificially…?

You are completely right in pointing out that when harmonic mean-like formulas [e.g. equations (1) and (2) of my May 1, 2017 at 12:05 pm post above] are presented, all they are really telling us is that it is always possible to approximate the relative orbital periods of two bodies by a whole numbered fraction. One good example is the 5:2 quasi-resonance of the orbits of Jupiter and Saturn.

2 x orbital period of Saturn = Ts = 2 x 29.4571 = 58.9142 sidereal years
5 x orbital period of Jupiter = Tj = 5 x 11.8617 = 59.3085 sidereal years
3 x synodic period of Jupiter and Saturn = SJS = 3 x 19.85810 = 59.5743 sidereal years

so someone could make the “discovery” that orbital period of Saturn is 1.5 the Synodic period of Jupiter and Saturn:

(3/2) x SJS = Ts
(3/2) x 19.85810 = 29.78777 sidereal years which is only 0.33067 yrs off 29.4571 yrs!

but all they would be doing is confirming that Tj/Ts = 0.402677 ~ 0.4 = 2/5

However, a slightly better approximation is that every 42 orbits of Jupiter there are 17 orbits of Saturn. So this could support the idea that:

(25/17) x SJS = Ts
(25/17) x 19.85810 = 29.203088 sidereal years which is only 0.25400 yrs off 29.4571 yrs!!

So someone could claim that their (25/17) fraction gives us a more fundamental understanding than the fraction (3/2).

And this process could go on and on without adding to anything to a better understanding of the underlying physics.

*****************************
comment 3: “My complaint above about playing with large numbers about FMC and/or anomalistic precession of Moon was probably rather for “astroclimateconnection”‘s articles (which relate Moon’s cycles to Earth-Venus and/or Jupiter-Saturn synodic cycles) than to original post here…”

No, I think that when the connections are very basic they have something fundamental to say about the underlying physics involved.

e.g. The length of the LAC is just half the harmonic mean of the Synodic period of Jupiter and Saturn and two pentagonal cycles of Venus and the Earth.

actually says something about the long-term connection between these five celestial bodies.

It is not far fetched to claim that the slow precession of the shape and tilt of the lunar orbit (over billions of years) may have been influenced by the Jupiter and Venus. It is misleading to point to the fact that the Sun is (by far) the dominant factor that sets the rate of precession of the line-of-apse and line-of-nodes of the lunar orbit. Why? Because the moon has been slowly receding from the Earth and so in the past its orbital period has occasionally come into close resonance with the orbital period of Jupiter and Venus. This means that at those time in history, Jupiter and Venus could have played a significant role in influencing the final precession rates of the lunar orbit. In addition, Jupiter (and to a lesser extent Saturn) have governed the slow changes in eccentricity, precession and tilt of the Earth’s orbit. Hence, modulating the way that the Sun’s gravitational influence sets the long term precession rates of the lunar orbit.

*****************************
Comment 4: “On the other hand – finding tuning of nature to Moon’s cycles should be interesting and is reasonable…”

I think that this is the point of the whole exercise. It doesn’t matter what has set up the current [temporary] “resonances” between the Moon, Sun and the planets but how they might be affecting
the Earth’s long term climate.

34. oldbrew says:

From Wikipedia: Orbit of the Moon

Perigee
(i.e. min. distance from Earth) 362,600 km (225,300 mi) (avg.)
(356400–370400 km)
Apogee
(i.e. max. distance from Earth) 405,400 km (251,900 mi) (avg.)
(404000–406700 km)

405400 / 362600 = 1.1180364
√5 / 2 = 1.118034
1.118034 + 0.5 = 1.618034 = Phi = (√5 + 1) / 2 by definition

35. P.A.Semi says:

I have updated my calculation, and now it almost perfectly matches:

(It showed me, that ignoring even millimeters / second ^2 accelerations beside kilometers / second speeds does not lead to good results)

It shows, that to explain almost-9-year prograde precesion of moon perigee/apogee and also the 18.6 year retrograde precesion of it’s orbital plane, ONLY Sun,Earth and Moon need to be calculated and all other bodies make some perturbations but do not contribute to main frequency… (as compared with JPL ephemerides). Moon distance to Earth is there just to show, that it does not recede or approach and so the equations seem legit…

For anyone interested:
Calculation iterates per minute steps (60 seconds), for each body P1, contributions from all peers P2, all calculated as 3D vectors in meters, seconds etc:
OldAcceleration = Acceleration
Acceleratoin = Sum ( GM2 / (r2-r1)^2 )
NewPosition = OldPosition + Time * OldVelocity
NewVelocity = OldVelocity + Time * Acceleration
NewPosition = NewPosition + Acceleration * (Time^2 / 2); // acceleration modifies velocity meanwhile
DiffAccel = Acceleration – OldAcceleration; // but acceleration direction also changes meanwhile – here simplified that it will change same as changed in last minute
NewVelocity = NewVelocity + (DiffAccel * (Time^2 / 2);
NewPosition = NewPosition + (DiffAccel * (Time^4 / 4);

Here “meanwhile” means “during that minute”, Time=60.
Omiting any step leads to wrong results (Moon recedes or approaches to Earth)…
(In my implementation, it runs million of minute steps per minute, the chart above was calculated 30 minutes. While I could add more planets to the algorithm, the goal of prooving that just Sun+Earth make the Moon orbit needs avoiding all other planets in the calculation instead…)

As “Azimuth” I mean – rotate vector to ecliptic, convert to polar coordinates and take X. As “orbital plane normal” I mean angular momentum vector L = r x p relative to some center (here Earth)…

On bottom chart you may estimate, how much is ever contributed by planets (difference between the green and purple curves) of the perigee precession…

36. P.A.Semi says:

About resonance – there is a quite good resonance of 5:2 in Jupiter/Saturn, and to this 20-year cycle is tuned Earth’s year (which is of the “fundamental” connections of J-S cycles with anything Earth-related). Also the resonance prevents Jupiter and Saturn – and other planets by 7:1 and 14:1 resonance with Jupiter from “travelling” arround, as they were travelling in early times of Solar system…

There is a PERFECT resonance of Venus/Earth of 13/8 with higher-order stabilization term (2 days per 8 years). Without this stabilization term, precise ratio would be destructive to orbits, while this way it helps to preserve them – the Milankovitch cycles are very probably (my guess) oscilation about optimum of Earth/Venus resonance.
There is a wrong lie in wikipedia, that this resonance has no dynamic meaning.
But showing frequency of meet-points between Earth and Venus (from JPL ephemerides) shows, that there are perturbations to Earth’s orbit by Jupiter/Saturn, but they are soon stabilized by meeting with Venus – as one planet is late to the meet-point, the other attracts it, and if it is too soon, the other brakes it a little. It may be shown, that the stabilization effect to Earth’s orbit by Venus resonance is significant… By a stable period, orbital distance is also stabilized…

There also is double-resonance from Venus, which is tidally locked to Earth (double resonance 13:12:8), similar as Moon, and every time when Earth and Venus meet, Venus shows the same face to Earth, and also when they are opposite arround Sun, that same face Venus shows to Earth and Sun…

There is another fine-tuning in Solar system – Earth’s Moon is tuned to Sun’s spin. From the region on Sun, that rotates in 27 – 29 days, every Sunspot cycle starts and travels toward equator. This synchronization of Earth’s Moon-pulse and Sun spin very probably re-charges our magnetosphere, that is said to already should have been faded away, but it’s still here…

http://astroclimateconnection.blogspot.com/2016/12/a-direct-connection-between-venus-earth.html
There is no sharp pentagram. The actual shape is very smooth curved…

(this is a trace of barycenter between Earth and Venus)

——–

Indeed.

But the influence is very probably NOT TIDAL.
Because Mercury at perihelion has bigger tidal influence than Earth. Adding Mercury into tidal calculations, the obvious 22-year cycle (2×11) of E-V-J disappears…
How would one explain omitting Mercury from tidal calculations on Solar surface ? (Surely not by claiming that it moves too fast. Mercury moves _slowest_ from point of view of Solar surface…)

I’m considering (still without any proof), that the influence is instead due to Earth magnetism, and perturbations to Earth orbit by Jupiter and Venus, because they show also the same E-V-J pattern… (And also because Sunspot cycles start from 27-29-day region, and they often start soon after perfect “touching” Earth-Jupiter heliocentric oppositions)
But yes, the heliocentric oppositions of Venus and Jupiter have almost immediate influence (possibly delayed until it rotates to our view), and also oppositions of Venus/Earth…

Meanwhile, I’m working on analysing Solar surface flows, and it seems very evident, that Jupiter brakes pole-ward meridional flow, with Jupiter above Solar equator, northward flow is stronger and south-ward flow is braked, and with Jupiter below Solar equator, northward flow is weaker and south-ward flow is boosted, it may or may not be Earth-modulated, still unsure.

There is still some problem, that Sun’s poles seem by 0.178° away from what is in SDO headers, creating Earth-yearly meridional “flows” on Solar surface, that are there only seemingly, since from Stereo spacecrafts they are also but on different times (sides)… So I’ll publish any more about the results when this will be corrected… (single calculation of flows on 60k HMI images for 7 years of data takes some 24 hours on 8 processors, and last correction did not seem right (while corrected Y flows it created more yearly X flows), it will need more such steps… Taking less input images does not work to statistically wipe away noise)

37. IT HAS SENT MY COMMENTS TO THE SPAM FOLDER AGAIN
[has your URL changed? – mod]

P.A. Semi,

You are missing the whole point because you are thinking too short term. Of course the Sun is the dominant torque acting upon the spin axis of the lunar orbit, and hence, almost the sole source determining its rate of precession. However, gravitational force of Jupiter is primarily responsible for making the Earth’s orbit precess once around the Sun every 112,000 years, it also causes a wobble of the Earth’s obit with respect to the invariant plane of the solar system roughly once every 71,000 years. The slow changes to the ellipticity of the Earth’s orbit are also primarily been driven by the gravitational perturbations of Jupiter, and all of these variations interact with the slow precession of the Earth’s rotation axis once every 25,700 years [which is driven by the solar and to a lesser extent lunar forces acting upon the Earth’s equatorial bulge that is tilted by 23 1/2 degrees to the Ecliptic] .

Over billions of years, all of these planetary driven perturbations to the Earth’s orbit modulate the gravitational force that is applied by the Sun to the Lunar orbit, simply by causing cyclical changes in the Earth-Sun’s distance and none of these would be apparent over the 50 year period you have shown in your graphs (2020 – 2070).

Indeed the first parameter that you plot – the Earth-Moon distance is constantly increasing, leading to epoch in which the period of the Moon’s orbit are perfect sub-multiples of the orbital periods of Venus, Jupiter and Mars. Models show that during these times the planets
can produce significant changes in the ellipticity of the lunar orbit and some scientists claim that
the planetary forces my have actually played a significant role in setting the current ellipticity.

38. P.A. Semi,

Thank you for updating us about the status of your current research activities, Your knowledge in this area is very impressive. However, I am afraid that most of your other comments are not helping solve the problem at hand. Here are a few things which are not adding to the discussion.

1. You have shown a plot of the geocentric distance between the Earth and Venus and said that it invalidates the pentagram that I used in my blog post. You are comparing apples and oranges.
The pentagram that I use in my post is not a plot of the Earth-Venus distance and the lines of the pentagram in the diagram on my blog post are just symbolic and they have nothing to do with the actual paths of these two planets.

2. It is not helpful to launch into a criticism of my ideas about what is driving the sun-spot cycle, This has nothing to do with the puzzle that I am trying to solve. I realize that you have some useful ideas to contribute on this important issue but I would appreciate it if our discussions could remain on topic.

39. P.A.Semi says:

I cannot integrate that long periods…

While Moon recedes due to tidal friction on Earth ocean bottoms, it is millimeters per year. (in early steps of my integration, it receded by 90,000 km in 20 years just by omitting modifying acceleration vector inside those minute steps…). Due to pulling by planets it would not recede anywhere, because what slows it on one side boosts it on other side an almost-same way, no friction, no energy lost… Almost. The difference of distance to Jupiter over 14 days is small and it also balances half a year later…

But look the page up and look to your pages to constants LAC and FMC and other such. You are not talking about 100-thousand year cycles, you are talking about daily cycles somewhere arround a yearly period.
8/Sjs + 8/LNC + 8/(2*LAC) = 3/DY – 3/Sev
This claims there is any relation between DY draconic year, LAC lunar anomalistic cycle and planet synodic periods… Other claim relations with planets and FMC full-moon cycle… Well, there is none, or it is random, or it happened long time in past.

It is a question, how Moon’s orbit became tuned this way it is now, but to maintain it there are just forces of Earth, Sun and Moon’s motion Inertia (i.e. “initial” state at present moment), with planets making only minor perturbations. It is a question and possibility, that these perturbations over long past periods made or modified Moon’s orbit to the way it is now…?

(While tuning of Moon to Sun spin and tuning Earth with both Venus resonances, and possibly also the Oldbrew’s phi-0.5 ratio of Moon’s orbit axes are too much obvious “intelligent design” to be random…)

About cycles in 50-100k year range (Milankovitch-related cycles, precession eccentricity etc), while I’ve not calculated it, I guess that it is an oscilation arround optimal resonance with Venus, and not caused just by Jupiter. (Or rather perturbed by Jupiter and stabilized by Venus). A guess…

I’ve extensively examined Chandler’s wobble, where projection of Earth’s rotation axis onto Earth surface makes a 400 – 439 day irregular cycle, while slowly shifting to one side.
http://semi.gurroa.cz/Chandler/Chandler.html
It’s somehow interesting: the Earth shape is assymetric, with Andes balanced by Himalayas, but Africa not balanced by Pacific, which makes the shape assymetric, a large “pot” with 1 big “handle” of Africa on the side. Simplifying geoid to 640 mass-points on the surface (the symmetric sphere beneath makes no contribution to any wobble) and integrating gravity influence (and tides), gives exactly OPPOSITE influence from Chandler’s wobble… (It does not play as “what slows it on one side boosts it on other side”, because meanwhile 12 hours later the Moon is more far/near and influence is smaller/larger, and this difference accumulates…)

What does it mean ? Chandler wobble is projection of rotation axis onto surface, and this axis seems assymetric from our knowledge of surface mass irregularities. But actually, the rotation axis is symmetric, and the outer assymetry of continents is balanced by opposite inner assymetry in core, so that the rotation axis and moment of inertia actually is symmetric. And by dragging the Earth’s core assymetry (opposite and counterbalance from surface assymetry), Sun and Moon and planets cause the Chandler’s wobble. By surface assymetry they instead pull it to the axial precession and nutation, as projection of axis of rotation onto outer stars… (mainly by Moon, Sun, little by Jupiter,Venus etc trying to align Earth’s equatorial bulge and continents to ecliptic plane, creating torque causing perpendicular movement instead)

—————
There is a Kepler’s law (3rd), that can be reformulated this way:
Omega = Sqrt( G * M / a^3 ) , T = 2*pi / Omega
Square of orbital angular frequency is G * central_mass divided by SEMI-MAJOR AXIS ^ 3.
Somewhere stated as T^2 / A^3 = constant = 4*pi^2 / (GM)

With Sun mass (GM=1.3271244e20) and AU distance (1.4959787e11) it gives Omega = 1.99098e-7 rad/s, which gives obvious T = 365.256898 days of 86400 seconds,
which btw is a method of weighing the Sun here or elsewhere a center of galaxy black hole etc. since it does not depend on weight of the orbiter (Earth arround Sun or stars arround center of galaxy etc)

But with Earth mass (GM=3.986e14) it does not play with Moon that well… With semi-major axis 387748000 gives Omega = 2.61483648E-6 rad/s and T = 27.811 days… Probably with semi-major axis shorter by 4600km (not to Earth but to EMB?) of 383148 km it gives T = 27.31788 days, somehow more familiar number… (I just have received yet another result yesterday which made me wonder and try the gravity integration written about above).

40. P.A.Semi says:

re “12:58 pm”
> “lines of the pentagram in the diagram on my blog post are just symbolic”
There is no ugly devil’s pentagram.
There is a nice gothic rosette with 5 round petals.

It depends how you look at it, but some “point of view” is vindicated by real physics…

I have not plot geocentric distance of Venus, but _Heliocentric_ distance of Earth-Venus barycenter, the barycenter of a resonant group, which I claim (or claimed earlier) has influence on the Sun. At the times of heliocentric opposition of Earth and Venus, the group’s barycenter is most near to Sun and almost stops moving for some 2 weeks and hovers on one place (note dot density on my picture – 1 dot per day). These times of heliocentric E-V oppositions are almost always visible in rise of Solar activity, and half-angle of Jupiter to this axis at these times (tidally adds or subtracts) gives outline of Sunspot cycle. SC24 started too early in 2008 and had to wait for this J-E-V stimullus until 2010. So I consider this “nice gothic rosette” a very important thing.

And meanwhile I really despise sharp pentagram, that possibly turns sometimes head down…

ad 2. – your article which I found just yesterday (due to this discussion here) was interesting to me, but the platform there prevents me from posting a comment there (without javascript), while WordPress is pretty script-free-enabled platform…
I do not absolutelly rule out tidal influence of J-E-V onto Sun, have spent a lot of time investigating that and now I just doubt the Mercury omision and look to other possibilities…
OK, let’s keep Sunspots aside for now…

41. oldbrew says:

19 Venus rotations = 169 (13²) lunar rotations

19 * 243.0185 d = 4617.3515 days
169 * 27.321582 d = 4617.3473 days

According to Sidorenkov: 13 lunar rotations/orbits = 1 lunar tidal year

‘Taking into account all these findings, we believe
that Rossby, Kelvin, and Yanai waves are visual manifestations
of tidal waves in the atmosphere. From year to year,
they repeat not with a tropical-year period of 365.24 days,
but with a period of 13 tropical months, which is equal to
355.16 days ≈ 0.97. It is called the tidal or lunar year.

http://oap.onu.edu.ua/article/viewFile/71141/66554

42. Oldbrew said:

537 x 15 anom. months = 539 full moon cycles = 7516 synodic months (537 * 14, -2)
[Validation: 537 * 15 = 8055, and 8055 – 7516 = 539]
539 / 537 = 1.0037243
15 AM / FMC = 1.0037248

****************
Thank you bringing this to my attention oldbrew – It is an interesting lead!
You are brilliant!
I will get back to you.

43. Sorry P.A. Semi, your a correct when you said:

“I have not plot geocentric distance of Venus, but _Heliocentric_ distance of Earth-Venus barycenter, the barycenter ” – my mistake.

44. You words are very true P.A.: “So I consider this “nice gothic rosette” a very important thing.”

It looks like it is a very useful way to following what is happening with regards to the influence of Venus and the Earth on the Sun.

45. P.A Semi,

You mention that:

“These times of heliocentric E-V oppositions are almost always visible in rise of Solar activity, and half-angle of Jupiter to this axis at these times (tidally adds or subtracts) gives outline of Sunspot cycle.”

You do understand that the tidal torquing model that I have proposed has the gravitational force of Jupiter at a maximum when Jupiter is 45 degrees to the tidally induced bulges that temporarily form on either side of the Sun along the Earth-Sun-Venus or Earth-Venus-Sun line.

In my model, the tidal forces (which fall off as one over distance cubed) of the Earth and Venus form the bulges on the surface of the Sun while the gravitational force of Jupiter (which falls off as one over distance squared) pulls on the Venus-Earth tidal bulges.

46. oldbrew says:

IW says: ‘Thank you bringing this to my attention oldbrew – It is an interesting lead!’

So brainstorming + pocket calculator can work 🙂

47. Paul Vaughan says:

Semi,

QB GPS vertical bounce ID by EEMD
http://www.mdpi.com/1424-8220/15/10/26096

“Apart from all other reasons, the parameters of the geoid depend on the distribution of water over the planetary surface.” — Nikolay Sidorenkov

48. Paul Vaughan says:

Free to go wildly off-topic, brainstorming can trigger lightning rods of insight anywhere — off-topic or on. (Example forthcoming on Suggestions 26…)

49. oldbrew says:

Re – astroclimateconnection says: May 2, 2017 at 4:19 pm
– – –
1.3409308624 degrees * 537 = 2 * 360 degrees (720.08~)

The ‘2’ corresponds to the ‘-2’ in the FMC number (14 * 537, -2), i.e. ‘losing’ 1 FMC per 360 degrees.
Some sort of precession perhaps?

50. oldbrew says:

In this comment : May 1, 2017 at 3:47 pm – the multiplier in the ‘update’ was 182.

But if the period is really a double period (see previous comment re 360 degrees) then the multiplier for the single period would be 2 * 182 = 364.

51. Paul Vaughan says:

People talk about a “2500 year” cycle or a “2400 year cycle”, but the data show a 2300 year cycle.

Given that the data show a 2300 year cycle, here’s something to consider:

(0.61519726)*(1.0000174) / (0.61519726 – 1.0000174) = 1.598689623

harmonic of 1.59868962272173 nearest 1.0000174 is 0.799344811360866
(0.799344811)*(1.0000174) / (0.799344811 – 1.0000174) = 3.98339766

harmonic of 3.98339765974742 nearest 1.0000174 is 0.995849414936855
(0.995849415)*(1.0000174) / (0.995849415 – 1.0000174) = 238.9324164

harmonic of 238.93241641449 nearest 22.1392998455063 is 21.7211287649536
(21.72112876)*(22.13929985) / (21.72112876 – 22.13929985) = 1149.985269

52. Paul Vaughan says:

Consider annual and semi-annual aliasing of LNC & LAC.

harmonic of 18.6133689658417 nearest 1 is 0.979650998202193
(0.979650998)*(1) / (0.979650998 – 1) = 48.14245966

harmonic of 18.6133689658417 nearest 0.5 is 0.503064026103829
(0.503064026)*(0.5) / (0.503064026 – 0.5) = 82.09199417

(48.14245966)*(82.09199417) / (48.14245966 – 82.09199417) = 116.4113316

Puzzle: Why is apse aliasing a harmonic of the node aliasing beat?

harmonic of 8.84754213940593 nearest 1 is 0.98306023771177
(0.983060238)*(1) / (0.983060238 – 1) = 58.03270559

harmonic of 8.84754213940593 nearest 0.5 is 0.491530118855885
(0.491530119)*(0.5) / (0.491530119 – 0.5) = 29.0163528

Suggested frame root:

(11.06964992)*(9.932517933) / (11.06964992 – 9.932517933) = 96.69017963

(29.447498)*(19.86503587) / (29.447498 – 19.86503587) = 61.04648218

(96.69017963)*(61.04648218) / (96.69017963 – 61.04648218) = 165.5999728

(11.862615)*(11.06964992) / (11.862615 – 11.06964992) = 165.5999728

(61.04648218)*(165.5999728) / (61.04648218 – 165.5999728) = 96.69017963

(96.69017963)*(165.5999728) / (96.69017963 – 165.5999728) = 232.360168

232, 116, 58, 29

I think people forget the ~Neptune-period cycle in JEV amplitude:

The JEV alignments alternate between stronger and weaker on that cycle. That’s from NASA JPL Horizons output. Also recall that Saturn swings the JEV frequency on 20 and 60 year cycles. (I’ve also illustrated that from Horizons output in the past.)

53. Paul Vaughan says:

Here’s another reminder from Horizons output:

Given the existence of such clear pattern, the deafening silence from the conventional mainstream on Chandler & QBO arouses razor sharp suspicion. Their prospects for securing credibility (future tense) hinge directly on taking prompt remedial steps (now). The blade was so sharp they didn’t even know (past tense) what happened to their credibility.

54. oldbrew says:

‘the data show a 2300 year cycle’

Timo Niroma has a 2289 year ‘supercycle’. Maybe I could do a post on that.

55. Paul Vaughan says:

Dimensional organization logically assures φ, but there’s a deeply entrenched culture of running integrations blindly without conceptual attention to multivariate spatiotemporal limits and balances.

Precise technicians are running perfect physical models but without philosophical awareness of simple geometric origins of aggregate stats. Regardless of whether the models are perfect there’s a failure to recognize geometric roots in output pattern.

It’s a fascinating insight into natural barriers blocking human perception.

Scores of individuals widely regarded as super-intelligent can be completely oblivious to something dead simple and universal. It’s a satisfying insight provoking conscious focus on natural limits.

Here’s something very simple to ponder as an exercise:

(6.4)*(6) / (6.4 – 6) = 96
(6)*(96) / (6 – 96) = 6.4

(6.4)*(96) / (6.4 – 96) = 6.857142857
(6.4)*(96) / (6.4 + 96) = 6

φ^4 = 6.854101966

2φ^4 Table 1
Detection of different-time-scale signals in the length of day variation based on EEMD analysis technique.

56. oldbrew says:

Re: (6.4)*(96) / (6.4 – 96) = 6.857142857

6.857142857 = 144 / 21 (Fibonacci numbers)

57. Sorry for not commenting for a while oldbrew,

I have been very busy with the clutter of life: The following comment that you have made is very pertinent:

*****************

Re – astroclimateconnection says: May 2, 2017 at 4:19 pm
– – –
1.3409308624 degrees * 537 = 2 * 360 degrees (720.08~)

The ‘2’ corresponds to the ‘-2’ in the FMC number (14 * 537, -2), i.e. ‘losing’ 1 FMC per 360 degrees.

Some sort of precession perhaps?

*****************

In its most fundamental form, I am asking why does the rate at which the lunar anomalistic month (i.e. when the Moon reaches the perigee of its orbit) drifts apart from the time of a Full Moon Cycle (i.e. the time for the Perigee of the lunar orbit to re-align with the Sun) as fraction of 360 degrees i.e.:

slippage / duration = 1.3409308624 degrees / 360 degrees = 0.0037248080?

so closely match:

{[(11.723 / 3600) deg per YEAR] x 411.78443025 days} / 360 degrees

= 1.34093024 degrees / 360 degrees?

*****************

The slippage factor (measured in degrees) over one FMC of 411.78443025 days is:

{[(11.723 / 3600) deg per YEAR] x 411.78443025 days}

i..e the precession of the Perihelion of the Earth’s orbit (measured in degrees) per year PER DAY TIMES the number of days of slippage.

******************

Hence, It is as though the ruler that is being used to measure events is stretching so that:

360 degrees always equals the DURATION of time considered [in this case 411.78443025 days)

and the SLIPPAGE / DURATION always equals 0.0037248080.

******************

After 537 x 15 anomalistic months = (537 + 2) FMC = ((537 x 14) – 2) Synodic months

or roughly 607.66 sidereal years, it corresponds to two complete cycles:

1.34093024 degrees x 537 = 720.080 degrees

Similarly,

after 268.5 x 15 anomalistic months = (268.5 + 1) FMC = ((268.5 x 14) – 1) Synodoc months

or roughly 303.83 sidereal years, it corresponds to one complete cycles:

1.34093024 degrees x 268.5 = 360.040 degrees

*******************

58. oldbrew says:

IW: re ‘{[(11.723 / 3600) deg per YEAR] x 411.78443025 days} / 360 degrees

= 1.34093024 degrees / 360 degrees?’

‘360 degrees’ is redundant in that equation.
– – –
Re: ‘the precession of the Perihelion of the Earth’s orbit (measured in degrees) per year PER DAY TIMES the number of days of slippage’

What does ‘per year PER DAY’ mean?

59. Fast says:

Astro and OB

Introducing 303.83 years and 607.66 years brings up the HIPPARCHIC cycle of 304 years.

Hipparchic cycle
The Greek astronomer Hipparchus introduced two cycles that have been named after him in later literature.

History
The first is described in Ptolemy’s Almagest IV.2. Hipparchus constructed a cycle by multiplying by 17 a cycle due to the Chaldean astronomer Kidinnu, so as to closely match an integer number of synodic months (4267), anomalistic months (4573), years (345), and days (126007 + about 1 hour); it is also close to a half-integer number of draconic months (4630.53…), so it is an eclipse cycle. By comparing his own eclipse observations with Babylonian records from 345 years earlier, he could verify the accuracy of the various periods that the Chaldean astronomers used.[citation needed]

The second is a calendar cycle: Hipparchus proposed a correction to the Calippic cycle (of 76 years), which itself was proposed as a correction to the Metonic cycle (of 19 years). He may have published it in the book “On the Length of the Year” (Περὶ ἐνιαυσίου μεγέθους), which is lost. From solstice observations, Hipparchus found that the tropical year is about  1⁄300 of a day shorter than the  365 1⁄4 days that Calippus used (see Almagest III.1). So he proposed to make a 1-day correction after 4 Calippic cycles, such that 304 years = 3760 lunations = 111035 days. This is a very decent approximation for an integer number of lunations in an integer number of days (error only 0.014 days). But it is in fact 1.37 days longer than 304 tropical years: the mean tropical year is actually about  1⁄128 day (11 minutes 15 seconds) shorter than the Julian calendar year of  365 1⁄4 days. These differences cannot be corrected with any cycle that is a multiple of the 19-year cycle of 235 lunations: it is an accumulation of the mismatch between years and months in the basic Metonic cycle, and the lunar months need to be shifted systematically by a day with respect to the solar year (i.e. the Metonic cycle itself needs to be corrected) after every 228 years.[citation needed] Indeed, from the values of the tropical year (365.2421896698 days) and the synodic month (29.530588853) cited in the respective articles of Wikipedia, it follows that the length of 228=12*19 tropical years is about 83275.22 days, shorter than the length of 12*235 synodic months, namely about 83276.26 days, by one day plus about one hour. In fact, an even better correction would be to correcting by two days every 437 years, rather than one day every 228 years. The length of 437=23*19 tropical years, about 159610.837 days, is shorter than that of 23*235 synodic months, about 159612.833 days, by almost exactly two days, up to only six minutes.

60. Paul Vaughan says:

Anomalistically reframing above sharpens antiresonance recognition:

(29.530589)*(27.55455) / (29.530589 – 27.55455) = 411.7844289 days

(411.7844289)*(365.259636) / (411.7844289 – 365.259636) = 3232.861904 days
= 8.851282797 tropical years

frequency of extremes:
8.851282797 / 2 = 4.425641398 tropical years

(4.425641398)*(1.618033989) / ( (4.425641398 + 1.618033989) / 2 ) = 2.3696965 tropical years

2.3696965 / 2 = 1.18484825 tropical years = 432.7565684 days

Checking these estimates on the nodal frame:
(1.184859016)*(4.425641398) / (1.184859016 – 4.425641398) = 1.618054067
(8.851282797)*(2.369718033) / (8.851282797 – 2.369718033) = 3.236108134

Anomalistic CW approximates nodal CW with 0.000908679% error. 32598.0994 tropical years of observations would be needed to determine whether the 2 estimates of CW period differ significantly.

61. oldbrew says:

110 anomalistic months = 7 * 433 days.

The Chandler wobble … has a period of 433 days.
http://en.wikipedia.org/wiki/Chandler_wobble

62. oldbrew says:

Fast: re. ‘the lunar months need to be shifted systematically by a day with respect to the solar year (i.e. the Metonic cycle itself needs to be corrected) after every 228 years.’
– – –
See the lunar chart below: 6441 * 4 = 228 * 113 = 76 * 339
It’s a model of the quarter cycle of Earth’s axial precession.
http://en.wikipedia.org/wiki/Axial_precession

In that period (6441 trop. yrs.) one synodic month is ‘lost’ compared to the ~235 per Metonic cycle.

25763 sidereal years = 25764 tropical years (6441 * 4)
Difference of 1 = precession

63. Paul Vaughan says:

Conventional mainstream EOP experts (e.g. at NASA JPL) can (it’s simpler than you can imagine) interpret and formalize what I’ve just outlined. CW is bivariate; the anomalistic basin of antiresonant attraction was missing from univariate nodal wisdom (e.g. Pierce Corbyn, Paul Pukite).

64. Fast says:

Astro

I suspect the answer to your questions maybe found in the mathematics of the Perihelion Precession of the Planets by Richard Fitzpatrick of the U of Texas and the associated chapters on Planetary motion. His simplification to consider the mass of all planets as a ring is understandable but never the less still results in arcane mathematics.
http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node115.html

65. oldbrew says:

Is arcane maths necessary? These ratios work quite well:

Perihelion precession : axial (or equinoctial) precession = 3:13
PP = The time taken for the Earth’s ellipse to revolve once, relative to the fixed stars.

Example: using IW’s figure of 11.615 arcseconds (present day) we get 111,579.85 years (perihelion).
111579.85 years / (3/13) = 25749.2 years, only about 20 years less than the standard estimate (~25770).

http://scienceworld.wolfram.com/physics/PrecessionoftheEquinoxes.html

In fact using the arcseconds figure from this link we can compare directly:

50.290966 / 11.615 = 4.329829
13 / 3 = 4.333 recurring
Match accuracy is 99.92%
– – –
https://tallbloke.wordpress.com/2016/02/01/why-phi-a-unified-precession-model/

66. Fast says:

OB
Yes you are right. The synchronization of all the bodies in our solar system is truly a marvel and I think the source of all climatic change.

67. Paul Vaughan says:

recapping CW bivariate framing sharpened here:
• nodal month on tropical year
• anomalistic month on anomalistic year

note well:
anomalistic month on tropical year does not align sharply,
but anomalistic month on anomalistic year does

68. oldbrew says:

PV – re: ‘anomalistic month on tropical year does not align sharply’

It does eventually, in theory at least.
85377 AM = 2352524.81~ days
86105 TM = 2352524.81~ days
6441 TY = 2352524.94~ days

NB ‘SM – AM’ should read: AM – SM

69. Paul Vaughan says:

OB: misinterpretation — review the outlined calculations — not referring to subharmonics — be very careful to not misunderstand as this is fundamental ….on the other hand misunderstandings are often or even usually a lot more fun than the truth …so why not contemplate misunderstandings just to make conversation more fun. So ok: good either way we can suppose — guaranteed either true or fun.

Next Up:
Simpler Than You Can Imagine…

70. Paul Vaughan says:

Φ = 1/φ

φ = (2A-2Y)-(N-(2D-26T))
Φ = (2A-2Y)-(2N-(2D-26T))

where frequency algebra terms are:
N = anomalistic year
T = tropical year
D = lunar draconic (nodal) month
A = lunar anomalistic month
Y = lunar synodic month
and compositely:
A-Y = full moon cycle
2A-2Y = half a full moon cycle
2D-26T = Chandler Wobble (CW)
2N = anomalistic semi-annual

Take the next step. Use the bold formula on line 1. Simplify …and solve.

71. Paul Vaughan says:

same thing said another way:

Chandler, anomalistic semi-annual, & 1/2 FMC:
(1.184859016)*(0.500023884) / (1.184859016 – 0.500023884) = 0.865110126
(0.563714217)*(0.865110126) / (0.563714217 – 0.865110126) = 1.618054067 ~= φ

Chandler, anomalistic year, & 1/2 FMC:
(1.184859016)*(1.000047768) / (1.184859016 – 1.000047768) = 6.411490792
(0.563714217)*(6.411490792) / (0.563714217 – 6.411490792) = 0.618055164 ~= Φ

same calculation for QBO (quasi-biennial oscillation) gives √5±1

Simpler Than You Think

72. P.A.Semi says:

It all just shows, that any two non-harmonic frequencies can be brought to a common number divisor, if you multiply them by sufficiently large numbers…

Paul Vaughan “May 4, 2017 at 2:43 pm” – if you consider 0.978 be a good harmony, then almost any can be a good harmony…
Once long ago I found, that all neighbour planet’s periods are some multiples of 1/27 from each other, just between Mars and Jupiter there is 1/27 ^ 2… 69/27, 44/27, 51/27, (68/27)^2, 67/27, 77/27, 53/27 … IF you ignore some small remainder… Problem here is, that “small” remainder when dividing by 27 is NOT 0.0030 … Ignoring remainders, that seem small but are not that much small, one can find anything what is not there actually… On the other hand, remainder (13/8=1.625)+0.000524 = 1.625524 _IS_ really small enough to be considered almost accurate, and the multipliers are small enough to be physically significant, more so if this is how it really “looks” repeating, and more so if you can show some observable significance of it. (here E/V resonance, observably damping noise in these two planet orbits)

The Nature does not listen to “”harmonies””, that make some beat after 437 years or after thousands of cycles… Because any two frequencies would beat after some thousands of cycles, if you ignore the small remainder… But after all it is just playing with numbers…

Oldbrew, 85377 AM vs 6441 TY with still some remainder is example of just that. Any two frequencies multiplied by sufficiently large numbers will make a single “beat” with some small remainder…

——-
To Paul Vaughan – Chandler wobble… It’s very futile to play with frequencies, because there is NO constant “”frequency”” with the Chandler wobble … The frequency itself is floating anywhere between one year, 400 days and 439 days… The times, when and how much does it float away are more interesting than the long-time average of the totter.
See this:
http://semi.gurroa.cz/Chandler/Chandler.html#Chandler
On Fig10a, top red line is Chandler wobble instantaneous period (inverse of frequency)…
On Fig10c it is for longer period from more chaotic data source…
(and there are links to source data , namely just below chapter header “Chandler wobble” files eopc04 and eopc01 )

They (Wikipedia) can tell you about some average period of X (439?) days, but there is no such thing really…

73. oldbrew says:

PA Semi – ‘any two non-harmonic frequencies can be brought to a common number divisor, if you multiply them by sufficiently large numbers’

No doubt, and that’s been said before. But in the chart I showed there are four matches (SM, AM, TM, and TY) and the total period is also an exact number of Metonic cycles (339).

If that can be replicated with smaller numbers that’s fine, but I have doubts about that.

The period is also very close to a quarter of the commonly quoted axial precession period (c. 25,770 years).
6,441 * 4 = 25,764 TY = 25,763 sidereal years
– – –
Note also that the Metonic cycle is 2 hours or so less than 235 synodic months.
http://calendars.wikia.com/wiki/Metonic_cycle

Therefore it’s plausible that it could take 339 Metonic cycles for the discrepancy to amount to one synodic month, as per the chart.
29.530589 days * 24 = 708.73413 hours
708.73413 / 339 = 2.0906611 hours (per Metonic cycle)

74. Paul Vaughan says:

Semi, you’re missing the obvious:

75. Paul Vaughan says:

Solar cycle length & CW period can rock away from perfect antiresonance, but the resultant rise in resonance means it’s naturally downhill back to antiresonance.

Antiresonance is the attractor.

(The source of perturbations away from antiresonance could be another line of exploration for anyone who choses to pursue it.)

The answer to the question’s simpler than confused conventional thinking can admit.

Why is the average solar cycle length what it is? Because that’s what it most easily can be. (Same for JEV.) For values increasingly further away there’s increasingly destructive resonance. Same for CW. It’s simple geometry.

Semi: You’re out of line.

76. Paul Vaughan says:

The JEV period observed matches a period affording minimal destructive resonance.

The Earth-Venus-couple’s relationship with Jupiter is antiresonant with Jupiter+Saturn sweeping (how long it takes their combined efforts to sweep half a circle).

Simple geometry sets the conditions for stability. Earth & Venus have settled stably into an accessible valley.

77. Paul Vaughan says:

Destructive Resonance Minimization:

QBO
√5±1 = (2Φ,2φ)

CW
(√5±1)/2 = (Φ,φ)

78. Paul Vaughan says:

Axial scaling’s defined by the unit-scaled period-frequency log-symmetry equivalence criterion:
difference = φ-Φ = 1 = beat = φΦ / (φ-Φ) = 1 / 1 = 1 = geometric mean = √(φΦ) = √(1) = 1

Fatal Communication Barrier: misunderstanding &/or misinterpretation (…&/or misrepresentation?)
Action Plan: Test integrity (…including honesty) with focused probe.

Pin-pointed diagnostics:

Semi:

1. Do you understand the aliasing derivations of 26D (CW) and 13D (QBO)? (repeated countless times over the years — outlined by Pierce Corbyn Sunday November 29, 2009 = earliest I ever saw the derivation)

2. Are you able to derive the JEV period in frequency algebra? Or are you only aware of the period via measurement of model output??

We can explore the root of misunderstandings &/or misinterpretations without ruling out the possibility of misrepresentation. Needle-sharp questions explore the basis for trust.

– –

History Review:
Tim Channon and I were discussing polar motion at another venue.
PC replied:

Sun, Nov 29, 2009 08:13 PM

Tim, Guys,
Yes
Thanks for this stuff. I had almost forgotten it.

It is an amplitude (envelope) beat and I would say it is a lunar tidal effect (according to some work we did some years back)
Note amplitude beats ALWAYS involve a phase change of the underlying (‘mean’) wave

Better do it the normal way (I know it comes to the same but working in frequency space is miles easier)

b = f1-f2
= 1/1 – 1/1.18 = 1-0.84746 = 0.1525 so 1/b = 6.55
or did you mean
= 1/1 – 1/1.19 = 6.26

I cannot get that pdf as shown as attached by the way

So where does it come from?

If D = 13.42199 is draconic month and 2D is either nodal crossing frequency (stronger tidal distortion)

The earth is at an equinox with period 2.0000000 (= better symetrical crustal tidal effect)

Here units are tropical year = 365.2422 days (where 1 day = 24hrs)

To find first beats of nodal forcing AND eqinoctal (or sollisticial if you prefer) crustal distortion note

(2D)/2 = 13.42…
FIRST beats B1 = 2D – 2×13 = 2D – 26 = 2(D-13) = 0.84398.. So 1/B1 = 1.185 yr

There will be a next beat from 2D/ 2 ~ 27/2 so Bnext = 2×27/2 – 2D = 27-2D = 6.41

OR that other beat can be something to do with annual asymmetry on beat B1 is 1 – 2x(D-13) = 0.1560
1/B* = 6.41yrs

All best
Piers

– –

Since then I’ve been reminding as follows:

QBO aliasing:
harmonic of 365.242189 (tropical year) nearest 27.212221 (draconic = nodal lunar month) is harmonic 13: 28.095553
(28.095553)*(27.212221) / (28.095553 – 27.212221) = 865.5210016 days
(865.5210016) / 365.242189 = 2.369718033 tropical years = QBO (quasi-biennial oscillation)

CW aliasing:
harmonic of 182.6210945 (tropical semi-annual = period between solstices = period between equinoxes) nearest 13.6061105 (period between opposite lunar node crossings) is harmonic 13: 14.0477765 (note that 13th harmonic of half-year is 26th harmonic of year)
(182.6210945)*(13.6061105) / (182.6210945 – 13.6061105) = 14.70143494
(14.0477765)*(13.6061105) / (14.0477765 – 13.6061105) = 432.7605008 days = CW (chandler wobble)

Are you familiar with Paul Pukite’s related empirical exploration?

The nodal aliasing derivation is years-old review.
What has been introduced in the current thread: aligned anomalistic antiresonance well.

Geometry (like mass) is a factor in stable limit’s.
Failure to acknowledge simple geometry triggers razor sharp suspicion.

Are we going to establish that we can trust Semi on antiresonant basins of attraction?
We’ll (perhaps silently) see what’s obvious.

79. Paul Vaughan says:

Is there anything symmetrical 2 note about √5±1 ?…

biennial
(√5+1) – (√5-1) = 2

difference
(√5+1)(√5-1) = 5-1= 4 = 2^2
of squares

(√5+1)(√5-1) / ( (√5+1) – (√5-1) ) = (2^2) / 2 = 2
beat

80. Paul Vaughan says:

Did anyone bother to check the calculation I suggested above? Here it is:
(1.236110328)*(3.236108134) / (1.236110328 – 3.236108134) = 2.000095537
Length of anomalistic year measured in tropical years = 365.259636 / 365.242189 = 1.000047768
2 anomalistic years = 2.000095537 tropical years

sorting the aliasing:
tilt: lunar nodes & tropical year
distance: full moon cycle & anomalistic (not tropical) year

81. tallbloke says:

Well this thread got interesting while I was dragged off to deal with General Election duties.

🙂

82. tallbloke says:

Paul V said:

Φ = 1/φ

φ = (2A-2Y)-(N-(2D-26T))
Φ = (2A-2Y)-(2N-(2D-26T))

Simply beautiful! Not just Phi in the sky. Chandler wobble and QBO explained in two linked lines 🙂

Take the next step. Use the bold formula Φ = 1/φ. Simplify …and solve.

Φφ = 1

and for binary completeness

(1/φ)-Φ = 0

and for complimentarity

Φ-φ = -1

Wikipedia:
“Mathematician Mark Barr proposed using the first letter in the name of Greek sculptor Phidias, phi, to symbolize the golden ratio. Usually, the lowercase form (φ) is used. Sometimes, the uppercase form (Φ) is used for the reciprocal of the golden ratio, 1/φ.”

83. Thank you Fast and oldbrew for your comments on May 6th – I will try to followup on these in the coming days.

84. oldbrew says:

PV: re Chandler wobble – rough guide…

(CW * 1 year) / (CW – 1 year) = 6.4 years
27 CW (5 * 5.4) = 32 years (5 * 6.4)
1 CW = 6.4 / 5.4 years

See – Paul Vaughan says: May 5, 2017 at 5:53 pm
(32 * 3 = 96)

85. Paul Vaughan says:

2300:
(18.61336897)*(8.851282797) / (18.61336897 + 8.851282797) = 5.99869949
harmonic of 5.99869948987323 nearest 0.5 is 0.499891624156103
(0.499891624)*(0.5) / (0.499891624 – 0.5) = 2306.287112

1800:
(18.61336897)*(8.847542139) / (18.61336897 + 8.847542139) = 5.996981151
harmonic of 5.99698115078654 nearest 1.00004776830423 is 0.999496858464424
(0.999496858)*(1.000047768) / (0.999496858 – 1.000047768) = 1814.352423

86. Paul Vaughan says:

1500:
harmonic of 5.99698115078654 nearest 0.500023884152113 is 0.499748429232212
(0.499748429)*(0.500023884) / (0.499748429 – 0.500023884) = 907.1762115
(2306.287112)*(907.1762115) / (2306.287112 – 907.1762115) = 1495.384536

87. oldbrew says:

Ian Wilson wrote: ‘it is important to note that the relative orbital periods of Jupiter, Saturn and Venus, with respect to the Earth, most likely set the length of the Lunar Anomalistic Cycle (LAC)’

Isn’t it just the beat period of the full moon cycle and the Earth’s orbit?
FMC * 1 / FMC – 1 = LAC (unit = sidereal year)
1.1273847 / 0.1273847 = 8.8502363

88. Paul Vaughan says:

Every good botanist sorts by simple similarity & difference.

1 / (T-(A-Y)) = 8.847542139 tropical years
1 / (N-(A-Y)) = 8.851282797 tropical years

89. Paul Vaughan says:

100:1 informal intro

Compare
contrast

907
9.07

20935
209.35

1986.5
19.865

This path leads to that view.

First, some review (covered at talkshop last year):
majority of terms in algebra flood cancelled
algebraic expression was simpler than you thought
harmony previously derived from the JSUN–EV pairwise beat matrix:

frequency:
Φ(J+S)-(U+N)

period:
(55.6462723)*(13.68233104) / (55.6462723 – 13.68233104) = 18.14345116 side

2:1 conway triangle ratio
2(Φ(J+S)-(U+N))
(27.82313615)*(6.841165521) / (27.82313615 – 6.841165521) = 9.071725582 side

my conventions reminder:
frequency algebra (convenient for terse, generalized sorting & classification)
with corresponding
period calculations (more convenient for human perception & memory)

Note the (increasingly familiar as our awareness grows) 13.68 year term.
period = 1 / frequency = 1 / (Φ(J+S)) = φ / (J+S)

It’s not the mystery it used to be.

D-A
(27.212221)*(27.55455) / (27.212221 – 27.55455) = 2190.350523 days = 5.996981151 trop

This quantity does not depend on Y (lunar synodic month frequency).

If you study the aliasing of the “6 year” (not exactly 6 but close to it) lunar cycle, you’ll discover long aliasing loops. Aliasing is sensitive to model frequency accuracy and precision …but long aliasing loops? not so much…

A little careful attention to aggregation criteria can simplify.
Long-run aliasing loops on familiar long cycles.

Would you believe there’s a 100:1 resonance between the solar system and the earth-moon system?

Φ(J+S)-(U+N)-(J-S)
(18.14345116)*(19.86503587) / (18.14345116 – 19.86503587) = 209.3538051 side

L

harmonic of 5.99698115078654 nearest 1.00004776830423 is
5.99698115078654 / 6 = 0.999496858464424
(0.999496858)*(1.000047768) / (0.999496858 – 1.000047768) = 1814.352423 trop

2L

harmonic of 5.99698115078654 nearest 0.500023884152113 is
5.99698115078654 / 12 = 0.499748429232212
(0.499748429)*(0.500023884) / (0.499748429 – 0.500023884) = 907.1762115 trop

T

harmonic of 5.99698115078654 nearest 1 is
5.99698115078654 / 6 = 0.999496858464424
(0.999496858)*(1) / (0.999496858 – 1) = 1986.512319 trop

2T

harmonic of 5.99698115078654 nearest 0.5 is
5.99698115078654 / 12 = 0.499748429232212
(0.499748429)*(0.5) / (0.499748429 – 0.5) = 993.2561593 trop

J-S
(11.862615)*(29.447498) / (11.862615 – 29.447498) = 19.86503587 side

2(J-S)
(5.9313075)*(14.723749) / (5.9313075 – 14.723749) = 9.932517933 side

J+S
(11.862615)*(29.447498) / (11.862615 + 29.447498) = 8.456145629 side

Φ(J+S) = ΦJ + ΦS
(19.19411427)*(47.64705265) / (19.19411427 + 47.64705265) = 13.68233104 side

harmonic mean of LAC & LNC:
(8.847542139)*(9.306684483) / ( (8.847542139 + 9.306684483) / 2 ) = 9.071307179 trop

This is the ONLY quantity in this post that is NOT independent of the length of the lunar synodic month. This unique constraint ties Y into a Y-free framework.

Convert everything to tropical (from sidereal), do the checks, and see 0.003% error from 100:1 resonance across the board. Tight hinging to J, S, & Φ is the solar system norm.

Where is this coming from?

Answer: This is another 1 of those simpler than you can imagines…

T-L
(365.242189)*(365.259636) / (365.242189 – 365.259636) = 7646485.299 days = 20935.3835 trop

Remember Conway: The aliasing loops close and algebra storms cancel.

1 simple long aliasing loop:
(6(D-A)-L) – (6(D-A)-T) = T-L
(1986.512319)*(1814.352423) / (1986.512319 – 1814.352423) = 20935.3835 trop

Notation note:
‘A’ already taken for anomalistic lunar month.
I’ve changed the notation for the anomalistic year.
Above I was calling it N, but N’s Neptune, so henceforth L = anomalistic year frequency to help (opportunistically) underscore anomaListic Links to eLLipse & eLLiptic.

One of the results presented above is extensible to generalized form. Maybe there will be time for that another day.

That concludes this informal 100:1 intro.

appendix #s:

tropical 365.242189
anomalistic 365.259636
sidereal 365.256363

27.212221
27.55455
29.530589

11.06964992 JEV +6V-10E+4J
9.007246722 SEV -6V+10E-4S
5.018891421 UEV -6V+10E-4U
4.492694707 NEV -6V+10E-4N
(generalized JSUN–EV derivations & pairwise coupling matrices covered in earlier talkshop discussions)

0.2408467 Me
0.61519726 V
1.0000174 E
1.8808476 Ma
11.862615 J
29.447498 S
84.016846 U
164.79132 N
247.92065 P
Seidelmann
1992

1/(J+S) (φ^0)/(J+S) 8.45614563
φ/(J+S) (φ^1)/(J+S) 13.68233104
(φ/Φ)/(J+S) (φ^2)/(J+S) 22.13847667
1/JEV 1/(3V-5E+2J) 22.13929985
4[(V-J)-(E-J)]-[(V-J)+(E-J)] = (4V-4E)-(V+E-2J) = 3V-5E+2J

90. Sparks says:

Paul Vaughan

Hi Paul, I just had a browse over this tonight, I noticed that you are pointing out the fact that the lunar orbit works out as phi. There is nothing incorrect about this, you don’t need help and there is no puzzle lol The lunar precession does actually equate to phi, Easter is actually based on this fact.

91. oldbrew says:

Re: ‘(27.212221)*(27.55455) / (27.212221 – 27.55455) = 2190.350523 days = 5.996981151 trop’
– – –
(Full moon cycle * Draconic year) / (Full moon cycle – Draconic year) gives the same result.

And (5.996981151 trop / lunar nodal cycle) + (5.996981151 trop / lunar apsidal cycle) = 1 (axial period)

Paper discussed here:
De Rop’s long-term lunar cycle
https://tallbloke.wordpress.com/2016/01/05/de-rops-long-term-lunar-cycle/

92. Paul Vaughan says:

OB, algebraically what you’re saying is (D-Y)-(A-Y) = D-A — which is independent of Y.

_
Sparks wrote: “the lunar orbit works out as phi. […] The lunar precession does actually equate to phi, Easter is actually based on this fact.”

Easter is based on the fact that “the lunar orbit works out as phi”? What does it even mean to say that “the lunar orbit works out as phi”? A prize for being vague. I respect and accept vagueness, but sharp clarification is welcome. Can you point to a link that sharply outlines what you’re trying to say?

Also, are you suggesting that 100:1 is common knowledge? If so, please point us everywhere you’ve seen it outlined as 100:1.

_
Sparks wrote: “you don’t need help and there is no puzzle lol”

Ian is asking for help with a puzzle (not Paul).

_
There’s plenty more to say about 100:1 after talkshoppers have a chance to digest the intro.

Your initial thoughts Rog? Have you finished preliminary digestion? Are you ready for next level?

93. Paul Vaughan says:

JSUN–EV pairwise beat martix review…
harmonic mean of all possible pairwise beats of {JEV,SEV} with {UEV,NEV}
= harmonic mean of 9.181885382, 7.561638477, 11.33459531, & 8.96363793
= 9.071449157

If you do the expansion you’ll independently rediscover an easily-tamed algebra storm (that reduces numerically to the preceding).

When people are ready and time’s available, I’ll derive the attractor algebra:
• short-term ( (J+S)-JEV ) – ( U+N )
• long-term ( Φ(J+S)-(J-S) ) – ( U+N )

94. Paul Vaughan says:

Simple symmetry exists with utility — imagine that….
https://tallbloke.wordpress.com/suggestions-27/comment-page-1/#comment-126070

95. tallbloke says:

Paul V: Your initial thoughts Rog? Have you finished preliminary digestion? Are you ready for next level?

As ready as I’m going to be in the next few weeks as we get to the general election over here. Apologies for the slow response, lots going on.

96. oldbrew says:

The full moon cycle in tropical years is very close to the fourth root of 21/13.

4th root of 21/13 = 1.1273765
FMC in trop. yrs. = 1.1273847

% match > 99.999
13 and 21 are consecutive Fibonacci numbers

Update:
Convert FMC to anom. years = 1.1273745 AY
1.1273745 ^4 = 1.6153728
1.6153728 * 13 = 20.999846
% match with 21 > 99.9992

97. Sparks says:

oldbrew

I was redirected here via a link Paul Vaughan posted on another page, my reply was in context… Also, Easter is a pagan festival, every Christian knows this. It celebrated the spiral of life, it comes from what historical observations they made. It’s understood and acknowledged, Phi and the precession of the moon however is also incorporated (because our ancestors were not brain dead knuckle dragging cave men) Observations of the moon from humans are like saying “how long is a piece of string” and this is very accurate, when measuring the moon, the piece of string overlaps, it always does ‘observed’ over thousands of years. Good Friday to Easter Monday represents PI, 3.141 The Christian calender is based on the moon and Easter Sunday is worked out according to the variable midway point of the vernal equinox and the closest Sunday becomes Easter Sunday.

98. P.A.Semi says:

Re: Paul Vaughan
> May 9, 2017 at 10:03 pm
> Semi, do you understand …
No, I do not understand a single formula from your post …
What is physical meaning of your equations?
What is “26D (CW) and 13D (QBO)” — “D” is not defined…

> frequency algebra
I do not understand any frequency algebra, it is physically a nonsense in this case, just playing with meaningless numbers…

As I noted with Chandler wobble, there is NO STABLE FREQUENCY !
Any calculation that relates Chandler wobble frequency or period with anything else is FALSE and void.

Most of all, the main component of CW frequencies is at 400 days period, not 432.

Why are you playing with an obvious nonsense: (φ/Φ) / (J+S) = 1/(J+S) … No need to put it there (there are more obvious ways to write 1, evenly meaningful)…
(Little reminds of PPN gravity using 0.5*(1+γ) and elsewhere using *β where both these numbers γ and β are =1, they spent 30 years of proving, how close is γ to 1 — very close now, so they are all the time pretending there is an obvious “complicator” item 0.5*(1+1) == 1 just so that the equations look more complicated with all those greek terms…)

> Needle-sharp questions explore the basis for trust.
If you were little less poetic, I could probably understand…

Yes, I know there was a Chandler Wobble discussion here, it is long since, and it was wrong. Just that, wrong.

And I strongly doubt there is any fixed frequency with QBO either.

————
> Tallbloke: May 11, 2017 at 8:29 am
Having defined this: Φ = 1/φ
Then all these: Φφ = 1 , (1/φ)-Φ = 0
are very basic thing for 3rd class basic school…
X * 1/X = 1, (1/x)-(1/x) = 0
holds for any X different from 0

Just this one is not trivial there :
Φ-φ = -1

> Oldbrew, May 20, 2017 at 5:44 pm
This looks somehow interesting, that
(FMC / year) ^ 4 == φ ((almost))
Does have the fourth power any “real” (physical) meaning ?

99. oldbrew says:

PA Semi asks ‘Does have the fourth power any “real” (physical) meaning?’

In fact the fourth root of 21/13 is almost 100% accurate. It’s an observation – the search for ‘physical meanings’ goes on.

What’s the physical meaning of 8 Earth orbits = 13 Venus orbits? Resonance is mentioned, or harmonics.

100. P.A.Semi says:

> Paul Vaughan May 9, 2017 at 10:03 pm
Technical note:
period between solstices =/= period between equinoxes
(the “year” is assymetric and the ratio is somewhere roughly near 179:186)

> can trust Semi on antiresonant basins of attraction
Is it about me? I haven’t told anything about “anti-resonant basins of attraction” ? Is it some poetry again or some oratorical turn ?

Meaning of those two yellow circles with blue lines escapes me… There is no equation of y=something * x to vindicate the slope line, why a circle at phi/2 ?
It also looks like a (dadaistic) mathematical poetry…

> May 10, 2017 at 12:02 am
> (√5+1)(√5-1) / ( (√5+1) – (√5-1) ) == 2 * (AnomalisticYear / TropicalYear)

That looks interesting and nice (and it would fit nice into “why phi” serie), but is there any “real” meaning to it?

————
> tallbloke May 19, 2017 at 2:55 pm
> mext few weeks as we get to the general election over here
A lot of work to do… How Europeans are planning to ROB Britain for the democratic decision of “Leave” is disgraceful and – can UKIP defend you somehow from it …?
The fight did not finish with last years “go ahead leave” Brexit vote, it was just a “Start!” command to begin the path from Euro-Hell toward freedom, but the path will be long and could be painful…

101. P.A.Semi says:

> What’s the physical meaning of 8 Earth orbits = 13 Venus orbits?
Beside it is small-integer resonant, it can be observed (on frequency of 8-year-spaced meet-points), that the irregularities in Earth or Venus orbit are damped by this, while Mars is not this well stabilised.

At the time of meet-point, mutual gravity is strongest.
If the Earth is too early to the meet-point, Venus brakes it a little, when the Earth is little late to the meet-point, Venus attracts it to it. The stable orbital period means stable orbital distance. Then there is second 8:12 synchronization of Venus day to this cycle – Venus always shows us the same face when meeting us.
This way, the possible irregularity caused by Jupiter,Saturn etc is damped and leveled out. Our orbital period and thereby distance is very well stabilised, and Mars or Mercury are much more irregular on centuries scale…
(as I observed in DE405 or DE422 JPL ephemerides)

I even speculate (did not verify it yet), that Milankovitch cycles in Earth perihelium distance and elipticity (the 40k and 100k year cycles) are oscilation about optimal harmonic beat with Venus planet…

Then there is non-constancy of Earth’s orbital angular momentum, which is most influenced by Jupiter (from helio-centric opposition to conjunction with J we are gaining energy, from conjunction to opposition we are loosing it), and second influence is Venus, to the meet-point we are gaining energy and from the meet-point we are loosing it. The changing modes of Earth’s orbital angular momentum seem to match the Sunspot cycle, which starts on Sun from the belt, that spins 27-29 days period at 40deg latitude, synchronized with Earth-Moon pulse (Earth orbits on tilde trajectory, pulsing in and out synchronized with Sun’s spin).
Earth is magnetic, unlike Venus, and although it is 100x weaker magnet than Jupiter (at Sun’s distance), the IMF (inter-planetary magnetic field – the solar wind plasma) is conductive and Sun’s magnetic strength here is also 100x stronger than it would be in vacuum, so Earth’s magnetic influence onto Sun may be comparable with Jupiter’s magnetic influence onto Sun…?
Or it is just another expression of that JVE cycle and the co-incidence with Earth’s angular momentum cycles and Sunspot cycles is mere coincidence without a direct causal link…

(People often claim tidal influence on Sunspots, but they always omit Mercury, which is tidally more important than Earth, and adding it to the equations looses the Sunspot-cycle link with tides…)

So this is, what is interesting on Earth-Venus 8:13 and 8:12 resonances physically and philosophically…
(philosophically – because it is too perfect to be random, and not seeing intelligent design in it is hard)

> fourth root of 21/13 is almost 100% accurate
There may be some physical explanation about fourth root (fourth power) ? It may appear in jerk&jounce equations… Anything else?
Or it is just another instance of “intelligent design too nice to be random” …

102. P.A.Semi says:

Correction:
> to the meet-point (with Venus) we are loosing energy and from the meet-point we are gaining it
because Venus runs faster… it reaches us to the meet-point from behind and then is escaping ahead of us from it …

103. Paul Vaughan says:

All the best Semi.

104. oldbrew says:

P.A. – we can note that fourth powers do come into play already in solar-planetary systems.

105. P.A.Semi says:

> as linked from May 13, 2017 at 4:41 pm

At the start it nicely explains, WHY Moon equatorial Z is important for LOD – because it drags the tidal bulges over the equator…
(beside an observation of coincidence of Moon Z with LoD – which I found too from frequency analysis, this one goes further and explains WHY, an intelligent and probably correct physically-based theory, nice and simple… then further linking longer terms in the cycles with Moon distance at Syzygy – again an observation connected with a physical explanation)

Match on the Figure 1 is little off the sinusoid (on right side) very probably because year is elliptic and not circular…??

106. Paul Vaughan says:

Semi: Several of the comments you have made are extremely rude and mathematically untrue. You have severely misrepresented and I suspect in bad faith.

107. Paul Vaughan says:

FOUL!
We can’t afford drivers breaking keys to stability in the ignition.

Axial scaling’s defined by the unit-scaled period-frequency log-symmetry equivalence criterion:
difference = φ-Φ = 1 = beat = φΦ / (φ-Φ) = 1 / 1 = 1 = geometric mean = √(φΦ) = √(1) = 1

Semi: Did you even check when you claimed it holds for all x not equal to zero?????

Trust completely obliterated.

I’ve been participating in climate discussion for nearly a decade and there have been plenty of times when people made me angry, but no one has ever made me more angry than Semi just did.

Semi: This is a serious failure. Your prejudice is unworkably intense. I’m filing a serious grievance.

You falsely claimed I said this:
“(φ/Φ) / (J+S) = 1/(J+S)”
I never said that. It’s mathematically untrue. φ/Φ 1.

What are you doing man????

You claim D is not defined. I defined it as follows: “D = lunar draconic (nodal) month”. You misrepresented and distorted on CW …but whatever, because in grander perspective these 2 don’t matter ….so staying resolutely focused on the highest-level grievance:

Take 1/2 and 2/1 and check Semi’s claim:
beat = (2)*(0.5) / (2 – 0.5) = 0.666666666666667
geomean = √( (2)*(0.5) ) = 1
difference = 2 – 0.5 = 1.5
ARE THEY EQUAL?
0.666666666666667 ≠ 1 ≠ 1.5

Try another one — say 1/3 & 3/1:
beat = (3)*(0.333333333333333) / (3 – 0.333333333333333) = 0.375
geomean = √( (3)*(0.333333333333333) ) = 1
difference = 3 – 0.333333333333333 = 2.66666666666667
ARE THEY EQUAL SEMI??
0.375 ≠ 1 ≠ 2.66666666666667

So WTF Semi is up to with such inflammatory LOGICALLY FALSE attempts to stir doubt and prejudice???

I think you can guess.

Review:

Axial scaling’s defined by the unit-scaled period-frequency log-symmetry equivalence criterion:
difference = φ-Φ = 1 = beat = φΦ / (φ-Φ) = 1 / 1 = 1 = geometric mean = √(φΦ) = √(1) = 1

beat = (1.61803398874989)*(0.618033988749895) / (1.61803398874989 – 0.618033988749895) = 1
geomean = √( (1.61803398874989)*(0.618033988749895) ) = 1
difference = 1.61803398874989 – 0.618033988749895 = 1

1 = 1 = 1

We’ve just discovered an unwelcome extension of the major western fault.
It’s infuriating to have to deal with such an intolerable intrusion.

TB: This is a serious matter of trust.

108. P.A.Semi says:

PV, beside you use english words that I do not understand in this context on first lines, the claim:

Having defined this: X = 1/y which also means y=1/X
Then all these: Xy = 1 , (1/y)-X = 0
are very basic things for 3rd class basic school…
X * 1/X = 1, (1/y)-(1/y) = 0

This is what holds for any X different from 0, despite you use φ instead of X symbol. That’s what I’ve written.
Only the other one subtracting them to get 1 is not trivial…

> Take 1/2 and 2/1 and check Semi’s claim:
I never claimed that and I do not understand what you mean by these equations…

—–
At this post May 9, 2017 at 5:16 pm is a chart, that has equation: (φ/Φ) / (J+S)

My error, the term (φ/Φ) =/= 1 …
But it is just a constant and any other constant would work there, since Y axis is scaled to some constant (also Halle cycle – in what units is it in the range -1..+1 ?). Is it just falsely complicating with greek letters … ?! Addingle apples and pears ?
After all, WHAT is (J+S) — you cannot add Jupiter + Saturn … Are you adding their WEIGHT or mass or attraction or what ?!
A frequency…? In which UNITS, so that you need to multiply them by those greek letters to get it in range -1 .. +1 ?! It surely would not be in SI units Hz…? After all, how do you display a constant frequency (J+S) in time ?

I’m not sure if your Jabberwocky english poetic language is more incomprehensible than your mathematic poems ?

—–
Your definition “D = lunar draconic (nodal) month” is many posts earlier before “26D (CW) and 13D (QBO)” and how could I understand, that it is still a same “D” ?

You probably mean that, which has average (mean) value 27.2418655057722 (during 20th century), but actually it oscilates with 6-year period in range 27.0063685456698 – 27.4773624658746 in an assymetric cycle ?
You probably did not mean ACTUAL instantaneous draconic month, but an average value, probably taken as it’s arithmetic mean ? Or which one ?

You made a foul of taking Chandler wobble be anything like 432 days regularly. It’s 399 days and 439 days mainly and many frequencies in between… Yes, it has some 6+ year envelope, that was a good claim.
And what relation is between 26D and CW ?!

QBO has hardly any stable frequency either.

The World is not an Algebra.
and the _differences_ from regularity are usually more interesting than theoretical beats of theoretical average values…

109. P.A.Semi says:

“It seems very pretty,” she said when she had finished it, “but it’s rather hard to understand!” (You see she didn’t like to confess, even to herself, that she couldn’t make it out at all.) “Somehow it seems to fill my head with ideas—only I don’t exactly know what they are! However, somebody killed something: that’s clear, at any rate.”

https://en.wikipedia.org/wiki/Jabberwocky#Lexicon
This is how I feel looking at some of those charts…

I even seem to understand every of these english words or can look it up in a dictionary:
“We can’t afford drivers breaking keys to stability in the ignition”, “Axial scaling’s defined by the unit-scaled period-frequency log-symmetry equivalence criterion” , “unwelcome extension of the major western fault”, “serious matter of trust”, “Needle-sharp questions explore the basis for trust.”

but what does it mean somehow escapes me… Sounds like a poetry… A Jabberwocky…
(It’s a pitty that I’m so ground-based without a sense for poetry…)

110. Paul Vaughan says:

“(φ/Φ) […] But it is just a constant and any other constant would work there”
Untrue.

Deliberately so? I suspect so.

111. P.A.Semi says:

And more particular
Draconic month – a distance between ascending nodes of Moon crossing ecliptic?

Average value (arithmetic mean) over 20th century seems 27.242 days, range 27.0 .. 27.477, with cycles of period 187.8 days and 5.79 years. (as calculated from DE422 ephemerides, which is very _probably_ same as used by “NASA JPL Horizons”…)

I guess (but did not verify yet), that the anomalistic month or syzygy month (Full Moon Cycle) will not be any constant frequency either…

112. P.A.Semi says:

> Untrue. Deliberately so? I suspect so.
It’s probably, because I do not understand, what you plot on X axis and what on Y axis, and what is a meaning of those cryptic equations. (Isn’t Y axis arbitrarily scaled to +-1 range by some constants?)

I doubt, that few other people here actually understand that, I suspect that many other have that jabberwocky feeling from that: “It seems very pretty, but it’s rather hard to understand

So that chart, comparing Hale cycle with (φ/Φ) / (J+S) , WHAT DOES IT MEAN ?

It looks pretty, but somehow I do not understand that equation. Not even a little…
(How can you add frequencies, or divide some non-dimensional constant by a frequency in unspecified units ? Beats are calculated by _subtracting_ frequencies instead of adding?? Is the red line somehow arbitrarily horizontally shifted sinusoid with that frequency mix? The Hale cycle – it’s probably very smoothed…)

113. Paul Vaughan says:

Remember:
arithmetic mean of frequency
harmonic mean of period (not arithmetic — you can’t pin the long-run central limit of a period with an arithmetic mean)

You’re missing foundations (including generalized means) and you’re failing to appreciate something fundamental and unique.

I only have time to help you realize what you don’t understand. Then it’s up to you if you’re going to take independent responsibility to understand.

Demand: Stop misrepresenting.

114. Paul Vaughan says:

100 back-and-forth exchanges will only waste our time. It’s extremely rude to bait conflict like this. Energy for moving forward is consumed moving backwards. This must stop now. Go to wuwt for that sort of political exercise.

115. oldbrew says:

Time to get back on topic (Ian Wilson’s puzzle) here, please.

116. P.A.Semi says:

There is nothing political about that. And neither google knows about “osculating cycle” (grey bottom of that chart May 9, 2017 at 5:16 pm )
Does it mean you will make me realize THAT I do not understand it, or does it mean, that you will help me to understand that what I do not understand ?

So once again, YOU addressed ME with a chart that seems somehow interesting, I appreciate it but do not understand it:
(φ/Φ) / (J+S)

(on-topic post will follow…)

117. P.A.Semi says:

FullMoon distance to Earth:

Note:
– 411.364 day cycle quite regular, but sparse. Unlike on the picture, probably the bottoms of the sinusoid (perigees) are more relevant than tops (apogees)…
– there is repeating pattern of 283.2 year sinusoids spaced cca 20.23 years apart — which beat may it be ?
The longer cycle’s sinusoids do not seem that much regular…

Moon eccentricity vector – here as instantaneous angle of perigee-apogee line to vernal equinox

Thanks to equation used there, the perigee-apogee line may be evaluated at any time during orbit, not only at distinct perigee-apogee time, it’s called the Eccentricity vector, whose direction is that line perigee-apogee and magnitude is eccentricity of the orbit…

As the 8.8 – year cycle is made from irregular steps, the anomalistic month would probably similarly quiver, and the original calculation is only an average beat, but not the instant beat, which can differ…?

118. P.A.Semi says:

I thought, that the anomalistic month would be irregular, but I’m quite surprised, that it is SO much irregular:

Time between following perigees range between 24.65 to 28.558, and average value I calculated as 27.57777 (taking first 100 perigees and divide by 100, I hope that isn’t some error like taking 101 or 99?)
It can be seen there is a 206 day cycle.
(planet&Moon position data from JPL ephemerides DE422)

So again – you measure beats with numbers on 10 decimal places, which one catches which cycle, but these are only theoretical averages. The actual beats vary over time, since the INSTANTANEOUS frequency also varies over time…

I mean, that the nature “feels” the instantaneous values and not the long-term averages:
https://en.wikipedia.org/wiki/Instantaneous_frequency

119. P.A.Semi says:

While calculating instant orbit period, from Kepler laws and JPL orbit vectors (position,velocity), I found orbital period in range 27.37 – 28.28 which is longer than expected 27.321. Also the “orbital energy” is quite inconstant.
Astromechanics – http://www.dept.aoe.vt.edu/~lutze/AOE4134/9OrbitInSpace.pdf
The problem is, that in case of Moon the “”simple”” Kepler laws are injust simplification, since the μ parameter, G * ME is only part of the influence, and the Moon’s orbit arround Sun is superimposed on the Moon’s orbit arround Earth.

So the “Moon eccentricity vector” chart above uses incorrect Kepler simplification, but on bottom part of my chart above May 2, 2017 at 11:54 am can be seen, that there is some irregularity in perigee azimuth, independent on “”simple”” Kepler laws.

The charts “FullMoon distance to Earth” and “Anomalistic month” above are measured on JPL ephemerides not using Kepler laws and simple μ = G * ME …

This also explains, why is the Anomalistic month so much anomalous, because the two orbits of Moon are superimposed (one orbit arround Earth and another independent orbit arround Sun)…

And back to original question – influences behind Moon orbit and behind Earth’s precession are the all same gravity attractions of Sun, Moon, Earth, little Jupiter and Venus, and other’s almost irrelevant, so finding some coincidences in precessions may be expected…

120. oldbrew says:

If lunar numbers were wrong wouldn’t eclipse forecasts be inaccurate?

121. P.A.Semi says:

I don’t mean they are “wrong”, I mean they are not SIMPLE, not simple Keplerian… One check for all:
a = 384748000, gm = 3.986e14 , Omega = sqrt( gm / a^3 ) = 2.6455e-6, T = 2*pi/Omega / 86400s
T = 27.48918 days
(all in SI units unless noted, “gm” is G * Earth.Mass, “a” is Moon’s semi-major axis from Wikipedia. Actual and instantaneous semi-major axis as measured with Earth as “”simple”” center of orbit varies significantly, same as orbital energy)

That slow would Moon orbit (Sidereal), if it orbited only arround Earth, by simple Kepler laws. But it orbits instead 27.321 days (sidereal orbit), because of Solar gravity and very little due to others… the “gm” in those equations is not only Earth’s gravity, the Solar “gm” must be considered too with “a” semi-major axis arround Sun and these two orbits superimposed…

So putting orbit vectors (position,velocity) into “”simple”” Kepler equations with one central body and one orbiting body, it does not work correctly…

And for calculating eclipse forecasts, the “”complicated”” equations for Moon must be performed, three-body or N-body problems…

Then the “anomalistic month” – distance between following perigee, really is nothing constant, as a “”simple”” wikipedia page would make you believe, it really oscilates +- 7 % which is a lot…

(I’ve naturally verified perigee from JPL ephemerides often, because it is usually noticeable by an “ugly” weather… The New Moon can be felt too, I feel more “”energy”” on those days, often I notice it and only later verifying it in a calendar. The Full Moon is noticed by clear-sky weather too often to be random – at least here far from ocean…)

Here another chart related to original post:

FullMoon time distance, oscilates arround 29.53 harmonic mean on this time span (thanks PV for teaching me which average type should be used, because I have been for ten years contented with Avg = (Min+Max)/2 without even noticing this old error in my program… It’s very often useful to dispute and own results are seldom verified really sufficiently…)

ranging from 29.276 to 29.833, with outer envelope arround 8.89 years (measured 10 cycles divided by 10), with high-frequency oscilating arround 393 to 443 days (as measuring times it crosses the Mean line)

(My time-span 1900-01-01 .. today leads to 29.5300149 days harmonic mean, but clipping to little shorter time-span to get complete sinusoid from mean to mean leads to 29.529856 harmonic mean, so take those values with a reserve)

So again – if you take FMC or AM with silly precision to 10 decimal places, it is just some average, but since both frequencies oscilate, the actual “beats” may be varying a lot…

122. Paul Vaughan says:

Graph instantaneous beat knowing harmonic mean’s central limit.

123. P.A.Semi says:

Not sure how frequency modulation works, but FFT sees just these frequencies in “FullMoon time distance” chart above (listed as day periods and amplitude): 411.652 (0.1365), 365.256 (0.06646), 205.889 (0.01219), 173.29 (0.01026), 193.503 (0.00391), … There is no frequency of 442 days… ? I’m confused.

I’m thinking how to get instantaneous frequency of that…
How to get instantaneous beat I do not know, what is “central limit of harmonic mean” I do not know at all…
((Sidenote: I have more thought on average – since I usually used it on sinusoid-like curves, the (Min+Max)/2 is usually quite near the expected value and is not dependent on horizontal boundaries so in these special circumstances – for sinusoids – it may even be better…))

But still it means, that thinking of “how far the E-M system rotates during 1 FMC cycle” is futile, because the cycle’s length is not constant, or it all works only with averages… And how far it rotates depends mainly on _where_ on that perigee-apogee ellipse it is, and little also on _where_ on the perihelium-apohelium ellipse it is…

124. oldbrew says:

P.A. – ‘if you take FMC or AM with silly precision to 10 decimal places, it is just some average’

I don’t think anyone is disputing that an average is an average.

The time interval between similar lunar phases—the synodic month—averages about 29.53 days.
http://en.wikipedia.org/wiki/Full_moon

125. P.A.Semi says:

They just somehow forget to mention, how large is the actual range… Reading these wikipedia values, I would somehow imagine, that it ranges just arround 29.525 and 29.535 or 29.52 and 29.54, and the actual range 29.276 to 29.833 somehow surprised me…
Even more so the anomalistic month (perigee) range 24.65 to 28.558 .

And the original post – how much it walks during difference (15 anomalistic months — FMC) – well, that is just on average, actual value depends on the phase of both cycles (i.e. where on the “yearly” ellipse it is).
It also considered the _maximum_ precesion value and not the current one or the average one, probably…

1 / 0.0037248080 = 268.47
How is it probable, that you find another value as reciprocal of 268.5, if you do not consider whole round cycles and can use partial cycles too ?

126. oldbrew says:

Which ‘whole round cycles’ are we looking at?

NASA: The mean length of the synodic month is 29.53059 days (29d 12h 44m 03s)
. . .
The duration of the lunation actually varies from its mean value by up to seven hours.
. . .
With an eccentricity of 0.0167, Earth’s orbit is about one third as elliptical as the Moon’s orbit. Nevertheless, it affects the length of the lunation by producing shorter lunations near aphelion and longer lunations near perihelion.

During the 5000-year period covered in this catalog, there are 61841 complete lunations. The shortest lunation began on -1602 Jun 03 and lasted 29.26574 days (29d 06h 22m 40s; 6h 21m 23s shorter than the mean). The longest lunation began on -1868 Nov 27 and lasted 29.84089 days (29d 20h 10m 53s; 7h 26m 50s longer than the mean). Thus, the duration of the lunation varies over a range of 13h 48m 13s during this time interval.

http://eclipse.gsfc.nasa.gov/SEhelp/moonorbit.html

This is public information.

127. Paul Vaughan says:

There’s only 1 central limit that balances the integral. This is mathematical proof territory. It’s either you understand this or you don’t. By definition (by equation), decomposition into central limit plus oscillation does not change integral balance.

If you can’t manage to prove this to yourself mathematically, just graph the unwrapped phase of your candidate central limits (if you’re exploring via Polya’s valid guess-and-test problem-solving method) against the unwrapped ephemeris phase. If the 2 don’t match, you know your conceptual understanding of central limit in the context at hand needs adjustment.

We each must take personal responsibility for developing sufficiently deep conceptual understanding of aggregation criteria to have a sensible discussion. It’s ridiculous that people are surprised when a universal geometric constant shows up in balanced integrals of circular (to first order) orbits.

Social intimidation doesn’t nullify geometry’s role in physics.

Everything that goes on at wuwt (and here when we get the occasional intrusion) is based on systematically false assumptions about spatial order and aggregation criteria. It’s a strategically chosen assumption designed to deliberately constrain discussion to an artificial (and thus controllable) framework. The proponents of the strategy do not care about what is realistic. They care about social and political control. They know there aren’t very many people smart enough to identify the false (disguised as implicit) spatial assumptions and identify the paralyzing implications for discourse.

If correction doesn’t come from inside the west, it will come from the east. Western bluffing on this file is a fatal error. The sensible and responsible thing to do is self-correct without further rude delay.

128. Paul Vaughan says:

Only people who think everything’s complicated are confused. For everyone else it’s simple:

Axial scaling’s defined by the unit-scaled period-frequency log-symmetry equivalence criterion:
difference = φ-Φ = 1 = beat = φΦ / (φ-Φ) = 1 / 1 = 1 = geometric mean = √(φΦ) = √(1) = 1

Let’s derive the roots of the axial scaling.

If you’re an adult along the major western fault:
Brace yourself, this is going to be the most difficult thing you’ve done this year.
If you’re a fit tiger mom’s brilliant 5-year-old anywhere in the far east: piece of cake!

The problem:

Solve
x-1/x=1

The solution:

x^2-1=x
x^2-x-1=0

(-b±√((b^2)-4ac))/(2a)

a=1
b=-1
c=-1

(-(-1)±√(((-1)^2)-4(1)(-1))/(2(1))
(1±√(1+4))/2
(1±√5)/2

φ

Check solutions
x-1/x=1

φ-(1/φ)=1
φ-Φ=1
1=1

(-Φ)-(1/(-Φ))=1
-Φ+φ=1
1=1

Antiresonance is the key to stability.

129. Paul Vaughan says:

“head in the clouds there’s no weight on my shoulders
i’ve got 1 less problem without you”
— Ariana Grande

130. Paul Vaughan says:

Polya problem-solving strategy:
Use symmetry.

if
F – (1-G) = φ
then
(2-G) – F =

131. tallbloke says:

Fascinating. I keep re-reading the three comments and something is happening. Not sure what yet. 🙂

132. Paul Vaughan says:

symmetric negative point in complex plane…
Golden quantum oscillator and Binet-Fibonacci calculus
…and generalizes further

thanks to OB for tip

133. tallbloke says:

Mostly a whoosh for me I’m afraid. How I wish I could share this exciting journey more fully.

134. Paul Vaughan says:

=
The concrete details of a given construction may be messy, but if the construction satisfies a universal property, one can forget all those details: all there is to know about the construct is already contained in the universal property. Proofs often become short and elegant if the universal property is used rather than the concrete details.
[…]
Universal properties occur everywhere in mathematics. By understanding their abstract properties, one obtains information about all these constructions and can avoid repeating the same analysis for each individual instance.
=
https://en.wikipedia.org/wiki/Universal_property