What is a Saros? Quoting Wikipedia:

‘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’

‘It takes between 1226 and 1550 years for the members of a saros series to traverse the Earth’s surface from north to south (or vice-versa)’

Only a few lines to go …

1226 / 2 = 613 years = 34 Saros

1 Saros = 223 synodic months by definition

34 Saros = 34 x 223 synodic months = 7582 SM

7582 + 613 = 8195 lunar orbits (sidereal months)

Since 8195 = 55 x 149

and 613 = 34 x 18, + 1

we can say:

34 Saros = 55 x 149 sidereal months = 34 x 18, +1 years

and

34 and 55 are consecutive Fibonacci numbers, which is the Phi connection (55 / 34 = 1.617647)

So 1 Saros = 55/34 x 149 sidereal months (= 223 synodic months)

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Nice work. 🙂

‘Related’ links should include:

https://tallbloke.wordpress.com/2014/04/23/why-phi-is-the-moon-a-phi-balloon/

@oldbrew: a very good one for starting the new year 😎

Related: http://www.alange-soehne.com/-/1RpXSO (zoom browser window)

“The motion of the moon [on a two-sided mechanical wrist watch]

In slightly more than 29.5 days, the moon orbits the earth once – in an anti-clockwise direction. A planetary wheel train assures that the moon phase is always correctly displayed. The mechanism is engineered to such a high degree of accuracy that it takes

1058years before a correction by one day is required.In the middle of the celestial disc, the earth disc rotates through 360 degrees once a day. It is daytime on the half that faces the sun (the balance) and night-time on the other. The peripheral 24-hour scale provides a rough time-of-day reference for regions in the northern hemisphere.”

If your Ferrari becomes boring on the UK roads, just sell it now and buy 2 Terraluna (send the 2nd to my address 😉

Related Inex period: 18 Inex = 521 years = 377 + 144 (Fibonacci numbers)

521 years was described by Professor George van den Bergh as ‘the basic period’:

https://tallbloke.wordpress.com/2014/12/12/eclipses-moon-cycles-and-inner-solar-system-observations-open-thread/comment-page-1/#comment-95247

It’s also 6965 sidereal months or 35 x 199.

35 sidereal months = 2.618092 years (phi² = 2.618034)

The 377 years and 144 years are 144 x 35 and 55 x 35 sidereal months respectively (144 and 55 are Fibonacci numbers, sum = 199).

This is getting really interesting now. Simple, remarkable observations. Well done OB!

So the number of Earth orbits x phi

^{2}= the number of lunar synodic months needed to multiply two fibonacci numbers separated by another phi^{2}relationship to obtain the ‘basic period’ which relates not only to the longest running eclipse cycle (inex), but also to several near-whole multiples of planetary alignments at 521 years.Can you list those conjunctions for us as a reminder?

Thanks TB.

Note that 1 Inex = 358 synodic months by definition.

358 x 18 = 6444, plus 521 = 6965 sidereal months (see previous comment).

PS to TB: see draft post.

OK, I’ll take a look now. See my update to previous comment.

‘Can you list those conjunctions for us as a reminder?’

The list already exists in the linked comment above (oldbrew says: January 6, 2015 at 10:40 am)

This one. Worth repeating here I think:

More Mars: there’s a 521-year lunar period that lines up with it, plus a lot of other stuff ‘nearby’.

If we put 18 Inex into the lunar Eclipse Cycle Calculator:

(http://www.staff.science.uu.nl/~gent0113/eclipse/eclipsecycles.htm#calculator)

521 years = 537 lunar years = 549 draconic years [DY] (= 1098/2 ‘eclipse seasons’)

= 18 Inex (18 x 358 synodic months = 537 SM x 12 = 716 SM x 9)

and

521 years = 277 Mars orbits = 244 Mars-Earth conjunctions

If 549 draconic years = 244 Mars-Earth conjunctions

then (divide by 61)

9 draconic years = 4 Mars-Earth conjunctions (ratio = 3² : 2²)

Note also:

2 Inex = 61 DY = 716 SM = 777 DM (draconic months)

times 9 gives

18 Inex = 521 years = [etc. – see above]

233 Jupiter-Mars = 520.814y

477 Jupiter-Earth = 520.912y

(477 – 233 = 244 Mars-Earth: see above)

803 Jupiter-Venus = 521.025y

326 Venus-Earth = 521.176y

(803 – 326 = 477 Jupiter-Earth)

847 Venus = 521.072y

2163 Mercury = 520.95y

44 Jupiter = 521.955y

1316 V-Me = 520.873y

2119 J-Me = 520.9286y

1642 E-Me = 520.934y

Interesting list OB. A few derivatives

233 Jupiter-Mars -Fibonacci number 233

326 Venus-Earth -Fibonacci numbers 233+89 = 322 (-4)

847 Venus -Fibonacci numbers 610+233 = 843 (-4)

2119 J-Me -Fibonacci numbers 1597+377+144 (-1)

Re: ‘9 draconic years = 4 Mars-Earth conjunctions (ratio = 3² : 2²)’

9 DY x 33 (297) = 250 full moon cycles (= 132 Mars-Earth)

More details of the lunar links to follow in new post.

TB: careful with those additions – almost any number can be the sum of three Fibonacci numbers 😉

Best to focus on neighbour planets for conjunctions, because others will be additions e.g. 244 Mars-Earth and 326 Venus-Earth means 570 Venus-Mars (847 Venus – 277 Mars).

OB: almost any number can be the sum of three Fibonacci numbersSure. I was looking for phi

^{2}linked pairs again.It might be an interesting exercise to see what pattern emerges out of the numbers which can’t be expressed as an addition of three Fibonacci numbers.

New post looks good. Go for it.

TB: Re post – may need to add a couple of lines to it first.

‘233 Jupiter-Mars – Fibonacci number 233’

Yes that’s an interesting one. These planets are neighbours obviously.

521 / square root of 5 = 232.99829, i.e. almost 233.

phi = ((square root of 5) + 1) / 2

Also there’s very little difference between the synodic periods of Earth-Mars and Jupiter-Mars (almost exactly 1/100th of a year). We don’t see that anywhere else in the solar system re. neighbouring planets.