Lunar ratios diagram

*The idea of this post is to try and show that the lunar apsidal and nodal cycles contain similar frequencies, one with the full moon cycle and the other with the quasi-biennial oscillation. *

There are four periods in the diagram, one in each corner of the rectangle. For this model their values will be:

FMC = 411.78443 days

LAC = 3231.5 days

LNC = 6798.38 days

QBO = 866 days (derived from 2 Chandler wobbles @ 433 days each)

The QBO period is an assumption (see Footnote below) but the others can be calculated.

The letters in the rectangle represent ratios of the periods on that side i.e.:

A = FMC:LAC = 1 : 7.8475526

B = QBO:LNC = 1 : 7.8503233

C = FMC:QBO = 1 : 2.1030421

D = LAC:LNC = 1 : 2.1037846

Obviously A and B are nearly the same, as are C and D (99.964% match, both pairs).

Hence the rectangle in the diagram.

The A (and B) numbers are also tropical years (TY):

7.8475526 FMC = 8.8475526 TY = 1 LAC (TY minus FMC = 1)

8.8475526 x 365.24219 days (1 TY) = 3231.5 days = 1 LAC

Phi aspect:

7.8475526 / 3 = 2.6158508 x 3

7.8503233 / 3 = 2.6167744 x 3

Phi² = 2.618034

So the two ratios are very close to 1 : 3 x Phi² (better than 99.9%).

– – –

**Footnote** – re the QBO period:

In Talkshop Suggestions 18 there are various references in the comments to a period of 2.369718 TY or 865.521 days.

If we use that instead of 866 days the ratio to LNC is 1:7.854668

7.854668 / 3 = 2.6182226 = > 99.992% match to Phi².

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Note that 3 x Phi² = (Phi²)² + 1 i.e. ‘4th power of Phi’ + 1

That wouldn’t work with any other number than Phi.

Also: Phi² = Phi + 1

People do things — and I’m interested in lunar + solar phenomena observable with naked eyes (preferably eclipses). For this I use DE 421 ephemeris which gives data points for each syzygy of observation intervals.

Taking 13 times the LNC, this gives an eclipse interval; it has 3248 nodical revolutions for 3235 sidereal moons (3248 – 3235 = 13 LNC), practically exact 242 years for 2993 synodic moons; quite observable that is with 4 x Inex + 7 x Saros (11 members eclipse series). It follows that 1 x LNC = between 6798.

85 and 6798.87 days. FWIW 242 + one more tropical year is the Venus transit pattern and practically 20½ Jupiter revolutions.And taking 213 times the FMC, this gives also an eclipse interval; it has 3283 apsidal revolutions for 2970 synodic moons (3283 – 2970 = 213 FMC) in practically exact 240 years. The corresponding eclipse interval is 22 x Inex – 22 x Saros (4 members in eclipse series). I’m working on a better setting where also the sidereal period is integer (and so the LAC length can be obtained by one division from eclipse interval).

Wikipedia says the LNC is 6798.3835 days.

http://en.wikipedia.org/wiki/Lunar_node

The derivation is from the tropical year (TY) and draconic year (DY) i.e. LNC = (TY x DY) / (TY – DY)

7 x the period of this model is ~766 Saros cycles.

Why stop there OB?

Get Ian over here and ask him and his friend Paul P. why LNC / (φ^4) looks so familiar.

hehe…

[ :

good fun

You see to be a quadruple agent you need to be 4-faced. To be a quintuple agent you need to be 5-faced …and people thought being 2-faced was

bad. Well, it’s actually good if you can face everyone and helpeveryone.Can’t wait to see the 2.71 look on Pukite’s face when he realizes what OB’s pointing out!

One day I met a religious man on the trail and convinced him the face of God can come to him through whatever channel. There are no limits. It’s multifaceted.

(28.095553)*(27.212221) / (28.095553 – 27.212221) = 865.5210016 days

By analogy:

(28.095553)*(27.32158236) / (28.095553 – 27.32158236) = 991.7882264 days

Make sense Ian?

Do you see what OB’s saying?

@oldbrew Re:

“Wikipedia says”Wikipedia says that “most Saros series have 71 or 72 eclipses”; thereafter apsidal + nodical + tropical are no longer aligned, i.e. ~766 Saros cycles does not exist 😦

P.S. for LAC I have a setting with 1 full-moon plus 6010 moons eclipse distance (28 x Inex – 18 Saros, 4 members); practically 486 years and less than 0.2° apsidal displacement per eclipse interval. Synodic 6011 moons are in alignment with 304 Venus conjunctions 😎

From PV’s comment – January 9, 2017 at 3:18 am

991.7882264 days / 865.5210016 days = 1.1458857

(Phi^4 + 1) / Phi^4 = 7.854102 / 6.854102 = 1.145898

That’s a 99.9989% match.

LNC / Phi^4 = 2.7156508 tropical years

2.7156508 TY = 991.87~ days

Comparing to PV’s number: 991.7882264 gives > 99.99% match.

Referring to the Sidorenkov post: 353 tropical years = 363 ‘lunar tidal’ years (1 LTY =

13trop. months)So 35.3 TY = 36.3 LTY

36.3 TM = 991.77342 days

13 x 27.321582 days (TM) = 355.18056 days

355.18056 x 36.3 = 12893.054 d

12893.054 d / 991.78822* = 12.999805 (

13)[* PV’s number]

So 13 occurrences per ‘Sidorenkov period’ of 35.3 tropical years.

tropical:

harmonic of 365.242189 nearest 27.3215823630557 is 28.095553

(28.095553)*(27.32158236) / (28.095553 – 27.32158236) = 991.7882264

991.7882264 / 365.242189 = 2.715426247 tropical years

nodal:

harmonic of 365.242189 nearest 27.212221 is 28.095553

(28.095553)*(27.212221) / (28.095553 – 27.212221) = 865.5210016

865.5210016 / 365.242189 = 2.369718033

anti-resonance:

3 * (2.369718033) / (2.715426247) =

2.618061937φ/Φ= 1.618033989 / 0.618033989 =2.618033989QBO stability just got a whole lot

lessmysterious.(φ/Φ)*(2.715426247) / 3 = 2.369692736 ~= QBO = 2.369718033

recognizing the

generalstructure ofstabletropical-nodalanti-resonantgeometry…(27.212221)*(27.32158236) / (27.212221 – 27.32158236) = 6798.387626 days

6798.387626 / 365.242189 = 18.61336897 tropical years = LNC

3 * ((φ/Φ)^1) * (2.369718033) = 18.61200706

1 * ((φ/Φ)^2) * (2.715426247) = 18.61180838

(18.61336897 / (3*(2.369718033)))^(2/2) = 2.61822556

(18.61336897 / (1*(2.715426247)))^(1/2) = 2.618143747

φ/Φ = 1.618033989 / 0.618033989 = 2.618033989

Lindzen must have known way back whenever…

It has been months since I looked at or even thought about any of this stuff. OB has managed to trigger a review of notes. One of the last results I had before I put all of this down to focus on other things was this:

(((φ/Φ)^2)+0) * (2.715426247) = 18.61180838

(((φ/Φ)^2)

+1) * (2.369718033) = 18.61200706+1 orbital

In case it isn’t obvious to everyone I’ll state this for context:

(2.715426247)*(2.369718033) / (2.715426247 – 2.369718033) = 18.61336897

It’s amazing what you can learn from kiwi leeks…

Via the Fibonacci numbers 13 and 34 (as a model):

340/130 = 34/13 = 2.6153846

13/130 = 1.3/13 = 0.1

sums to

340 + 13 = 353

34 + 1.3 = 35.3

2.6153846 + 0.1 = 2.7153846

35.3 / 2.7153846 = 13

2.7153846 tropical years = 991.773 days = 36.3 lunar tropical months

supplementary:

harmonic of 365.242189 nearest 29.530589 is 30.4368490833333

(30.43684908)*(29.530589) / (30.43684908 – 29.530589) = 991.7882264

991.7882264 / 365.242189 = 2.715426247

OB, I noticed your comment here:

https://tallbloke.wordpress.com/suggestions-24/comment-page-1/#comment-122298

35.3 is a subharmonic

by definition. It’s implicit in the derivation. It doesn’t tell us anything new.Remember that Pukite has 2.715 in his models too.

What’s new here is the way it all generalizes. Again we’re in that kind of territory where it’s so simple that for most people it won’t sink in …at all. It’s simply profound and profoundly simple at the same time.

Again the nodal stuff fits like a glove. We’ve seen that before. The nodal signal is loud, clear, and simple.

Above is the key. Now someone can write a landmark paper on why the QBO is stable. Same reason as why the asteroid belt is where it is: anti-resonance.

PV – re supplementary [January 9, 2017 at 4:37 pm]

Of course, thanks.

—

12 synodic months (SM) = 354.36706 days

13 tropical months (TM) = 355.18056 days

(13 TM * 12 SM) / (13 TM – 12 SM) = 423.60788 tropical years (TY)

423.60788 / 100 = ~Phi³ (4.2360678)

Also: 353 TY = 363 x 13 TM (363 lunar tidal years)

353 * 12 = 4236 TY

Why does the QBO have the period it has?

Because it can.

It’s the period with natural immunity from destructive resonance.

Lindzen suggests climate “science” defunding to purge rotten foundations:

http://joannenova.com.au/2017/01/richard-lindzen-axe-climate-science-funding-groupthink-has-destroyed-intellectual-foundations/

SYNCHRONIZATION OF TERRESTRIAL PROCESSES WITH

FREQUENCIES OF THE EARTH-MOON-SUN SYSTEM

N.S.Sidorenkov 2015

ABSTRACT. It is established that the frequencies of the

quasi-biennial oscillation (QBO) of atmospheric winds

and the Chandler wobble (CW) of the Earth’s poles are

synchronized with each other and with the fundamental

frequencies of the Earth–Moon–Sun system. The QBO

and CW frequencies are resonance combinations of the

frequencies of the Earth–Moon system’s yearly rotation

around the Sun, precessions of the lunar orbit, and the

motion of its perigee. The QBO and CW frequencies are

in a ratio of 1:2.

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

φnal exam solutions

x^2 + 1 = 3x

x^2 – 3x + 1 = 0(3±√5)/2

2 real roots

1 is the

reciprocalof the otherφ/Φ&

Φ/φDe-φ or not de-Φ?… Let’s watch for φ

not onlyin pentagon circles but also in the context of earthly QBO models. It’s a quick test of agency.…and it’s a quick test

of competence and integrity.Process of Elimination 101…

“Tell me I’m the only one— Bullet with Butterfly Wings — Smashing PumpkinsTell me there’s no other one …”

https://wikimedia.org/api/rest_v1/media/math/render/svg/d614aa87c81a24d192391c132504098b987e9ebd

Here α+β = φ/Φ + Φ/φ

= 3“[…]elementary symmetric polynomials– any symmetric polynomial in α and β can be expressed in terms of α + β and αβ The Galois theory approach to analyzing and solving polynomials is: given the coefficients of a polynomial, which are symmetric functions in the roots, can one “break the symmetry” and recover the roots? Thus solving a polynomial of degree n is related to the ways of rearranging (“permuting”) n terms, which is called the symmetric group on n letters, and denoted Sn. For the quadratic polynomial, theonlyway to rearrange two terms is to swap them (“transpose” them), and thus solving a quadratic polynomialis simple.To find the roots α and β, consider their sum and difference: […]”https://en.wikipedia.org/wiki/Quadratic_formula#By_Lagrange_resolvents

By the process of elimination it’s a test in earthly context of competence, agency, and integrity all rolled in

onesimple pack. As OB says: There’sno otherway.Real simpletest.(3±√5)/22 real roots1 is the reciprocal of the other

– – –

Good summary PV.

α+β = φ/Φ + Φ/φ = 3

αβ = (φ/Φ)(Φ/φ) = 1

Will this help people understand “

Why”3?Only a few lucid minds need to understand to make the educational effort worthwhile.

supplementary:

(29.530589)*(27.212221) / (29.530589 – 27.212221) = 346.6200854

(346.6200854)*(365.242189) / (346.6200854 – 365.242189) = 6798.387626

6798.387626 / 365.242189 = 18.61336897

Another one:

√5±2 = Phi³ or 1/Phi³

– – –

Re the ratios:

C = FMC:QBO = 1 : 2.1030421D = LAC:LNC = 1 : 2.1037846

The square root of half a lunar apsidal cycle (180 deg.) = 2.103278~ tropical years

That number falls between the two ratio figures.

LAC returns to ‘start’ position every 8.8475~ trop. years (ellipse exaggerated here)…

360 / QBO years + 360 / FMC years = 471.2275~

471.2275 / 360 = 2.6179304 / 2 (half of φ²)

471.2275 – 360 = 111.2275

360 /111.2275 = 3.23661 = 1.6183048 * 2 (2 * φ)

360 / LNC + 360 / LAC = 60.030172

360 / 60.030172 = 5.9969843

[180 deg. every ~3 years – see fig. 3 here:

http://onlinelibrary.wiley.com/doi/10.1111/j.2153-3490.1971.tb00568.x/pdf ]

5.9969843 / 3.23661 = 1.8528597 = 0.61762 * 3 (3 * Φ)

[Fibonacci numbers 21/34 = 0.617647 = 99.995% match]

GWPF:

In 1997-98, very warm El Niño conditions persisted for roughly 13 months, but were then followed by roughly 32 months of well-below-normal (La Niña) water temperatures in the eastern Pacific.http://www.thegwpf.com/weak-la-nina-follows-super-el-nino/

32 * 30.5 days = 976 days

So ‘roughly 32 months’ could be close to:

991.788 days = 2.7154~ years [LNC / Phi^4]

As an experiment I substituted the fields in the four corners with new ones, like this:

FMC becomes the Venus-Earth axial period [VE] = 0.3808806 years

QBO becomes the Earth’s orbit period [EO] = 1 year

LAC becomes the Jupiter-Saturn axial period [JS] = 8.456146 years

LNC becomes the solar Hale cycle [HC] = 22.14 years (variable)

So the graphic box now looks like this:

VE–A–JS

-C– | –D-

EO–B–HC

The new results for A,B,C, and D are:

A = VE:JS = 22.2:1

B = EO:HC = 22.14:1

C = VE:EO = 2.625:1

D = JS:HC = 2.618:1

Again A and B are nearly the same, as are C and D (99.73% match, both pairs).

VE axial period is very close to 8/21 years (99.98% match), and 2.625 is 21/8 (Fibonacci numbers).

JS axial period is very close to 2 * Phi³ in units of years i.e Earth orbits (99.81% match).

2.618 = Phi²

In tropical years (TY):

LAC * FMC = LAC + FMC

(TY / FMC) + (TY / LAC) = 1

The axial period of the LAC and the FMC is 1 tropical year.

SM = synodic month

TM = tropical month

AM = anomalistic month

(SM * TM) / (SM – TM) = TY

(SM * AM) / (SM – AM) = FMC [full moon cycle]

(AM * TM) / (AM – TM) = LAC [lunar apsidal cycle]

@oldbrew not bad 🙂 B.t.w. 391 calendar years (G. decimals .000~ J. decimals .99~) = 403 lunar years (each 12 moons); also 403 – 391 = 12 😎

Derived from (4 x Inex – 1 Saros) x 4 = eclipse cycle, has life expectancy 5083.00~ years (by v. Gent).

@ C: that would also be 5227 tropical months i.e. 403 * 12, plus 391.

@oldbrew good one. Let’s find a Golden Ratio: 5227 has 824 full draconic semesters [eclipse seasons] and 21 full revolutions of the Line-of-Nodes [LNC]; 5227 + 21 = 5248 draconic [nodical months].

Obtains 824 / 21 = 37 + √5 (very near) component of Golden Ratio.

Would like to see similar corollary for Line-of-Apse.

Line-of-Apse is related to the full moon cycle.

What happens in the chart below is:

728 = 104 * 7

5713 = 5720 – 7 (5720 = 104 * 55)

6441 = 6448 – 7 (6448 = 104 * 62)

6441 = 19 * 339 which gives the alignment with the Metonic cycle.

766 synodic months = 7 LAC i.e. a repeating cycle.

(79664 SM = 766 * 104)

6441 TY is the quarter precession period:

4 * 6441 = 25764 tropical years = 25763 sidereal years (difference = 1).

@oldbrew in my approach I have to match same apsis at both interval ends, e.g. 1285 apsidal – 1199 synodic = 86 FMC. For LAC add ½ moon (now 96.995~ years), then 1296½ sidereal – 1285½ apsidal = 11 full revolutions of Line-of.Apse.

“small” supplement: the difference between 6010 synodic (in the above: 486 yrs) and 4835 synodic (in the above: 391 yrs) is 5 x Metonic, and: this 117

5moons interval = 21 LNC, has virtually same declination at both interval ends, same coordinate as well 😎B.t.w. 6011 / 5⅟9 = 117

6= 4835 / 4⅟9 (beat period).365.24219 days (tropical year) / 411.78443 days (full moon cycle) = 0.8869742

0.8869742 * 6441 = 5713.0008

i.e. 5713 FMC in 6441 TY

5713 = 34 * 21 * 8, plus 1

1 – 0.8869742 = 0.1130258

0.1130258 * 6441 = 727.99917

i.e. 728 lunar apsidal cycles in 6441 TY (quarter precession)

728 = 13 * 8 * 7

5713 + 728 = 6441

5712:728 would be a 34*3:13 ratio so one ‘extra’ full moon cycle in the quarter precession period.

3,13, and 34 are Fibonacci numbers so the ratio equates to 3*Phi²:1

PV wrote:

In case it isn’t obvious to everyone I’ll state this for context:(2.715426247)*(2.369718033) / (2.715426247 – 2.369718033) = 18.61336897

And 18.61336897 + 1 = number of draconic years (in one nodal cycle)

18.61336897 / 2.715426247 = 6.8557524 (Phi^4)

19.61336897 /

2.5= 7.8453472 (~3 * Phi², or Phi^4 +1)2.5 draconic years = 866.5502 days = 2.3725358 tropical years

[…] we’ve explained elsewhere that the number of full moon cycles in one lunar apsidal cycle is very close to 3*Phi². We can see […]