The idea here is to link up different lunar periods in a schematic intended to show how they integrate into a ‘bigger picture’, for lack of a better phrase.

This chart of selected lunar data is based on the numbers from this blog post comment:

https://tallbloke.wordpress.com/2014/12/12/eclipses-moon-cycles-and-inner-solar-system-observations-open-thread/comment-page-1/#comment-95315

The top line of the chart contains original data. The rest shows the differences between the original figures (e.g. 2079 – 1973 = 106) in a structured format. By using periods where there’s a good match with whole numbers, the patterns of the relationships can perhaps more readily be identified in diagrammatic form.

One point to note is that one more tropical year would give 1974 = 329 x 6 (also 987 x 2).

The figures at the bottom of each chart are validation only.

Now a chart using the same method, based on exactly seven times the period of the first chart:

This chart is a period of exactly 766 Saros cycles:

‘The saros is a period of 223 synodic months (approximately 6585.3211 days, or 18 years and 11+1/3rd days), that can be used to predict eclipses of the Sun and Moon. One saros period after an eclipse, the Sun, Earth, and Moon return to approximately the same relative geometry, a near straight line, and a nearly identical eclipse will occur’ – Wikipedia

http://en.wikipedia.org/wiki/Saros_(astronomy)

Note here that one more draconic year would be 14554 = 766 x 19.

Also that six more full moon cycles would be 12256 = 766 x 16.

These differences (as -6 and -1) are reflected in the top line of chart 2.

Compare with the Wikipedia notes:

‘The saros is an eclipse cycle of 223 synodic months = 239 anomalistic months = 242 draconic months. This is also equal to 16 full moon cycles.’

(239 – 223 = 16, 242 – 223 = 19)

Talkshop contributor Paul Vaughan quoted these figures from JPL data in another thread:

apse cycle – 6798.38 days

nodal cycle – 3231.5 days

Results from these charts (against tropical years):

apse cycle – 6798.329 days [using ‘APC’ in chart 1]

nodal cycle – 3231.4925 days [using ‘LP’ in chart 1]

The conclusion is that the charts closely follow lunar periods, so the relationships shown should be reasonably accurate over the long term.

Fibonacci footnote:

The basic period appears to be 1973 tropical years, which is 987 x 2, – 1 (see chart 1). 987 is a Fibonacci number.

The difference between the number of draconic years and full moon cycles in chart 1 is 329 (as shown), which is 987/3.

(The second chart is just a factor of 7 greater than the first one so the same comments apply).

Reference: http://www.cyclesresearchinstitute.org/cycles-astronomy/lunar.shtml

Something disabled comments, corrected now. Don’t all rush at once 🙂

Footnote to the Fibonacci footnote re. chart 2:

14553 DY x 5 = 72765 = 55 x 21² x 3 (all Fibonacci numbers)

As another possible link between full moon cycles and apsidal precession periods:

416 FMC = 469 years = ~53 (469-416) apse periods

@oldbrew (January 7, 2015 at 6:21 pm): double your interval to 342599 days and you find opposite perigee/apogee (commensurate offset) at opposite syzygy 😉

I will go this far and say the moon could moderate solar activity impacts on the climate to some degree but no further.

Let’s go 6 times the numbers:

37619 sidereal = 37301 anomalistic = 34805 synodic [all months]

37619 – 34805 = 2814 = 469 years x 6

37301 – 34805 = 2496 = 416 FMC x 6

2814 – 2496 = 318 = 53 apse x 6

@oldbrew wrote: Let’s go 6 times the numbers.

And, this is indeed one of the

best matching perigee intervals for same-phase syzygy, about one month slack (delta longitude is 20.~°) in the equinoctial year, 34804.98~ synodic (5 x 6961) and 1027812.06~ days (2 x 2 x 3 x 97 x 883), in my usual 7ka time-frame.No wonder (me thinks), since draconic has min (abs) inclination° to max (abs) inclination° at the interval’s ends.

Cross-reference:

https://tallbloke.wordpress.com/2014/12/06/ian-wilson-are-the-strongest-lunar-perigean-spring-tides-commensurate-with-the-transit-cycle-of-venus/comment-page-1/#comment-95666

The 4519 synodic months = 385 Draconic years equation relates to the above charts like this:

Chart 1: 2079 DY x 5 = 10395 = 385 DY x 27 (= 55 x 21 x 3² in Fibonacci terms)

Chart 2: 170818 SM x 5 = 854090 = 4519 x 27 x 7, -1 (854091 = 21 x 3² x 4519)

See also: http://en.wikipedia.org/wiki/Eclipse_cycle#Numerical_values

Under ‘The ratio of synodic months per half eclipse year and per eclipse year yields the same series:’ we find that the last entry in column ‘SM/full EY’ is ‘4519/385’.

The closing comment is: ‘Each of these is an eclipse cycle’.

One we may not have commented on before:

1925 (55×35) tropical months = 1781 (89×20,+1) synodic months = 144 tropical years

55, 89 and 144 are Fibonacci numbers. Maximum discrepancy less than 1 day.

@oldbrew who wrote: 55, 89 and 144 are Fibonacci numbers. Maximum discrepancy less than 1 day.

1] It’s 24.0~ times the 6 years wobbling plane, at both ends of the interval are draconic and apsidal in same phase.

2] checked in my 7ka time-frame: at 2 x the interval are eclipses at both ends (66%+ of the pairs).

@Chaeremon: thanks – more examples of the Moon agreeing with itself 🙂

Notes re the fact 1 Saros is almost 19 draconic years (DY) and almost 16 full moon cycles (FMC).

Lunar chart 2 says 766 Saros = 14553 DY, which is almost 100% true but the nearest match is:

754 Saros (377 x 2) = 14325 DY (19 DY x 377 x 2, -1)

One draconic year is ‘lost’ every 754 Saros.

For full moon cycles, where 1 Saros is almost 16 FMC the ‘adjustment’ is:

2040 Saros (34 x 60) = 2039 x 16 FMC

One group of 16 full moon cycles is ‘lost’ every 2040 Saros (= 2 FMC every 255 Saros).

Finally, comparing groups of 19 DY with groups of 16 FMC:

2380 (34 x 70) x 19 DY = 2379 (13 x 183) x 16 FMC

13, 34 and 377 are Fibonacci numbers.

@oldbrew, your 754 Saros and 2040 Saros: as I already tried to address earlier, Saros eclipses are physical phenomena which contradict mathematics by authenticatable observation. Citing NASA:

The exact duration and number of [members] in a complete Saros [series] is not constant. A series may last 1226 to 1550 years and is comprised of 69 to 87 [members].So, outside of this time-frame the Saros relation does not hold between draconic and anomalistic, and I’m afraid your calculation maintains Ex Falso Quodlibet 😦

Can your thoughtful approach be reduced to ‘loss’ of draconic months and ‘loss’ of anomalistic months, within the physical limit of 69 to 87 members, that would be great 😎

@ Chaeremon: ‘A series may last 1226 to 1550 years and is comprised of 69 to 87 ‘

1226 years is the period of 34 x 2 Saros. It starts and finishes with a Saros: 34 x 2, +1 = 69.

The previous comment was to explain the difference between ONE Saros and 16 FMC or 19 DY.

‘The problem with the 3-body problem is that it can’t be done’

http://www.askamathematician.com/2011/10/q-what-is-the-three-body-problem/

@oldbrew: Moon-Earth-Sun: The oldest three-body problem 😉 eg. in section F. The motion of the perigee and the node:

– http://scholar.google.com/scholar?q=Gutzwiller+The+perturbation+of+the+Sun+produces+a+closed+oval+orbit+that+is+centered+on+the+Earth+with+the+long+axis+in+the+direction+of+the+quadratures

Your 34 x 2, +1 = 69 Saros, thanks will pursue, my idea is continuation of two series separated by a dual eclipse (don’t hold breath).

Also re 87 Saros: that’s 3 x 29, and 13 x 29 = 377 – see earlier comment @ 9:38 AM.

A link to a physicist joke from ‘Mr Facetious’

’19 Draconic Years minus 223 Synodic Months is about 11 hours. This 11 hour gap is the key to explaining why Saros Families evolve over time.’

http://members.bitstream.net/bunlion/bpi/EclSaros.html#evolve

(14325 x 11.033 h) / 24 = 1 Saros in days

Re 14325: https://tallbloke.wordpress.com/2015/01/06/two-long-term-models-of-lunar-cycles/comment-page-1/#comment-96436

@oldbrew quoted:

’… is about 11 hoursBy cherry picking end dates and data points the mathematical theologist can obtain any value for the temporal distance between eclipse seasons, it just depends on for what bias the confirmation is wanted:

Plotted are 7ka which include 5ka NASA eclipse canon, can you tell which gaps have been averaged and which not?

@ Chaeremon: No, all I can say is that 1 draconic year is ‘lost’ every 14325 DY (or 754 Saros) compared to the idea that 1 Saros = 19 DY.

That averages to just over 11 hours per Saros, but it’s a 3-body question and obviously those figures say nothing about orientation.

Since ‘the initial and final eclipses in a Saros series fall near the north or south pole’ (from earlier link) the 754 Saros could be 377 each for N and S?

@oldbrew, you asked: Since ‘the initial and final eclipses in a Saros series fall near the north or south pole’ (from earlier link) the 754 Saros could be 377 each for N and S?

However one’s mathematical view on this, it’s simultaneously parallel Saros series going north and south at any time. I’ve connected all series (via their dual months which start/stop their same node series, therefore 1 x node = 1 x direction across series); this gives an impressive graph over the millennia, can send you .pdf (may want to print it as wall paper) 😉

This one is a bit odd, not sure if it means anything.

383 Saros = 371 lunar nodal cycles

383 + 371 = 754 = 377 x 2

It feeds into chart 2: 383 x 2 = 766 Saros

And into chart 1: 106 x 3.5 LNC = 371