Earth’s orbit [credit: NASA]

We’ll assume the diagram is self-explanatory but if not,

this should help (see opening paragraphs).

We’re looking at Aphelion minus Perihelion (A – P) distances of the giant planets.

Figures are given in units of a million kms. (lowest value first), using Jupiter as a baseline.

J: 76.1 = Jupiter

N: 101.22 = Neptune

S: 161.95 = Saturn

U: 262.32 = Uranus

Data from NASA Orbital parameters here.

Comparing each planet to the next in ‘A-P’ size:

Ratios (accuracy in brackets)

J:N = 1:1.3301 = 3:4 (99.76%)

N:S = 1:1.59998 = 5:8 (~100%)

S:U = 1:1.61976 = 21:34 (99.956%) – near 1:Phi (99.893%)

(or 8:13 (99.677%))

and

if N:S = 5:8 and S:U = 8:13 then…

N:U = 1:2.5916 = 5:13 (99.677%)

We find ratios closely following consecutive Fibonacci numbers 3,5,8,13, 21 and 34 (and 4 = 2 x 2).

——-

Notes:

Mars semi-major axis (average distance from the Sun) is 227.92

Mars SMA 227.92 / J(A-P) 76.1 = 2.995 (99.83% of 3:1)

Earth Aphelion (furthest distance from the Sun) is 152.1

Earth Aph. 152.1 / J(A-P) = 1.9987 (99.93% of 2:1)

——-

The figures can be interpreted – or not – but the pattern in the ratios is at least interesting.

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Caution TB:

The units of √5 / (2(J-S)) (in frequency algebra) are years.

The √5 answer I gave above has no units. (same as for (E+V)/(J+S) ~= 22)

one

step

at a time

You can get a number very close to √5 when dividing a Lucas number by a Fibonacci number e.g.:

521 / 233 = 2.2360515

322 /144 = 2.2361111

76 / 34 = 2.2352941 (very close to J-Mars period)

Then the ratio of J-S to Hale would be half that, so divide the Lucas by 2 or multiply the Fibonacci by 2 e.g.:

38 J-S = ~34 Hale

1/(cos(2Pi/5)) = 1 + √5 = 2φ

PV: (φ/Φ) / (J+S) = 22.13847667y

—

Cross-check of orbits per expected Hale cycle

22.13847667 / J = 1.8662389

22.13847667 / S = 0.7517948

Sum of J + S = 2.6180337

phi² = 2.6180339~

Gentlemen, I’ve now nailed this to an accuracy of

22 minutesabsolute deviation (0.000383075%) every Schwabe cycle:11.09012746 (based on Seidelmann (1992))

11.09016994 = φ^5

Please be patient as I struggle to find time to relay insights in comprehensible format. Thank you.

Ian Wilson, some very friendly advice:

Please be prepared to stay ahead of the pack understanding all of this. You could end up playing a very important role communicating this to a wider audience. I may be gifted with tenacious exploratory instinct, but I cannot foresee a day when I’ll have the financial and time freedom to translate insight into adminese.

Curiosity about systematic patterns in ERSSTv3b2 is the reason I’ve pursued this despite unfairly prejudiced (and at times downright malicious) obstruction. It remains unacceptable that NOAA has vandalized the recordings of nature that motivated this exploration of natural beauty.

For me exploration of natural pattern has turned into a philosophical lesson on the difference between law and justice. Defiant administrators’ adamant refusal to be sensible is a serious problem.

tallbloke (January 19, 2016 at 2:09 pm) wrote:

“1/(cos(2Pi/5)) = 1 + √5 = 2φ = √5(φ)”1/cos(2π/5) = √5+1 = 2φ

≠φ√5typo?

Thanks Paul, φ√5 is φ^2 + 1 , not √5 + 1.

Corrected.

TB: There’s no “0.07” deviation in aggregate.

Matrix deviations balance out to 2/(J+3S-U+N) ~= φ^5.

It’s a

jointconstraint. You can walk straight to this intuitively (it’s an easy algebraic derivation once you see) by looking at the patterns in the matrix and noticing how they balance. By reviewing the notes in this thread, most (possibly all?) people probably won’t see this. Organizing + (axial), – (synodic), – – (origin alignment), & + + (harmonic mean) in matrix format makes it intuitive to notice thecollectivestructure.obstacle: so much to communicate, so little time each day

as/when time permits: ongoing elaboration

Note that 2/(J+3S-U+N) can be rewritten as 2/((J+3S)-(U-N))

Take care to notice that 3S is the S harmonic nearest J.

Back during my discussion with wayne when I first noticed φ^5 I didn’t see this.

same as first noticing 2 and later learning from the remaining 0.009176129 in 2.009176129

hierarchically progressing insight

Wonderful. We always knew the whole system was involved, and that the big four gas giants act in concert causing slowdowns in solar activity on longer timescales. To get the confirmation of tight coherence at the decadal level is a breakthrough. Congratulations.

I get a figure for U+N of 55.65136y (50 U-N / 154 (52 N, 102 U))

No. of U+N in U = 1.5096997

No. of U+N in N = 2.9611373

Sum of these two = 4.470837

Arithmetic mean = 4.470837 / 2 = 2.2354185

Result is 99.971% of √5

(NB U+N based on single orbits returns 99.961% match)

Paul: Note that 2/(J+3S-U+N) can be rewritten as 2/((J+3S)-(U-N))Thanks for the helpful hint.🙂

I get that to be:

11.09013

Which compares very well to

11.09017 =~ phi^5

periods based on Seidelmann (1992)

S & U

(84.016846)*(29.447498) / ( 2 * (84.016846 – 29.447498) ) = 22.66919063 = 1/(2(S-U))

(84.016846)*(29.447498) / (84.016846 + 29.447498) = 21.80496372 = 1/(S+U)

(22.66919063)*(21.80496372) / ( (22.66919063 + 21.80496372) / 2 ) = 22.2286803 = 2/(3S-U)

J & N

(164.79132)*(11.862615) / ( (164.79132 + 11.862615) / 2 ) = 22.13204008 = 2/(J+N)

J, S, U, & N

(22.2286803)*(22.13204008) / ( (22.2286803 + 22.13204008) / 2 ) =

22.18025492 = 2/(J+3S-U+N)2φ^5 = 22.18033989

~3/4 hour difference(0.000383075%)every ~Hale cycleYep, I got an identical result when I cross checked the previous formulation.

22.18025492

Outstanding work Paul.

“[…] 3S is the S harmonic nearest J.”(January 19, 2016 at 6:17 pm)That statement was wrong.

2S is the harmonic nearest J

I show a more intuitive approach immediately above (January 20, 2016 at 8:38 am).

I agree it scans much more intuitively.

You’ve nailed the big four – what a tremendous achievement!

Hale-SUN = 2φ^5 = 2/(J+3S-U+N)The Hale cycle runs on double-phive-power🙂

Only a small nit, maybe I don’t understand how you came to this conclusion:

2S is the harmonic nearest JOn simple differences:

J

_{orbital}– S_{orbital }/3 =~ 2.04S

_{orbital }/2 – J_{orbital}=~ 2.86What am I missing?

Anyway, both calcs came to an identical value, so it’s not a biggy, and doesn’t detract from your breakthrough.

170 x 22.18025492y fits the J-U-N long period model very well.

https://tallbloke.wordpress.com/2015/08/20/why-phi-a-jupiter-uranus-neptune-model/

170 (34 x 5) Hale = 273 (21 x 13) J-U = 22 U-N = 295 (273+22) J-N

Of course 5,13,21 and 34 are Fibonacci numbers.

Re: ‘2S is the harmonic nearest J’

2S = 4.9647565 J = ~99.3% of 5J

Ah of course. 2S not 2/S, thanks OB.

educational aside

TB asked:

“What am I missing?”The definition of nearest-harmonic.

This is a big deal so it’s necessary to invest whatever time & effort is needed get it. It’s impossible to appreciate and understand volatility and variance envelopes without

perfectlyandexactlyunderstanding this. For example, people could not possibly understand the JEV derivation, nor could they understand the various lunisolar envelopes we’ve discussed over the years. Another example is the “60 year” cycle.Let’s go over some basics.

Remember:

• frequency = 1/period

• y = x = x^1 is a straight line (frequency), whereas y = 1/x = x^(-1) is nonlinear (period)

• the algebra is

frequencyalgebra (not period algebra)Adding and subtracting period algebra (fractions) gets messy because before you can add or subtract for example 1/a & 1/b you first need to re-express with a common denominator b/ab ± a/ab = (a±b)/ab wheres adding frequencies is just a±b. Frequency algebra is as Piers Corbyn puts it “miles easier”.

The S harmonic nearest J is 2S. (Remember that these are frequencies, not periods. The periods would be written algebraically as 1/S, 1/J, & 1/2S.)

To find the nearest harmonic, multiply by 1, 2, 3, etc. and see which is closer.

J = 1/11.862615 = 0.084298445

S = 1/29.447498 = 0.033958742

|J-S| = 0.050339703

|J-2S| = 0.01638096

|J-3S| = 0.017577782

|J-2S| = 0.01638096 is the minimum — i.e. it’s closest.

If you want to think in periods (where things are nonlinear), the way to identify the nearest harmonic is to find the one (dividing by 1, 2, 3, etc.) that gives the longest beat period. Mathematically this is equivalent because the closer frequency (on the linear frequency scale) will

of coursegive a longer beat period. The longer beat periodmeansthe frequencies are closer.(29.447498)*(11.862615) / (29.447498 – 11.862615) = 19.86503587 = 1/(J-S)

(14.723749)*(11.862615) / (14.723749 – 11.862615) = 61.04648218 = 1/(J-2S)

(11.862615)*(9.815832667) / (11.862615 – 9.815832667) = 56.88999848 = 1/(3S-J)

61.04648218 = 1/(J-2S) is the maximum (the familiar “60 year” cycle)

This 101 lesson is aside from what we’re discussing, interjected for educational value.

How did this even come up?

I was exploring equivalent reordering of the algebra to see if that would assist with interpretation, but it’s actually less intuitive to think about it that way, so it’s the common theme of knowing where you’re from better after looking around and I’ve summarized the most intuitive perspective at January 20, 2016 at 8:38 am.

…but it’s a good thing it came up, because I’ve been assuming for years that all readers have the background to fathom derivations like that for JEV. If they didn’t know how to identify firsthand with absolute confidence a nearest-harmonic and appreciate & fully understand that it

definesthe variance envelope, then I knowfor surethat they never understood. So it’s useful for me to know that for 7 years people may have been just skimming envelope isolations without fathoming.OB quoted and added:

“Re: ‘2S is the harmonic nearest J’2S = 4.9647565 J “

No. Careful here everyone.

I’m writing frequency algebra, not period algebra.

The 2 are not interchangeable.

When I write frequency algebra it can

notbe interpreted as if I’m writing about periods.Important note to readers:

Different writers are using different notation conventions. It won’t be practical to force everyone to use the same convention. There are good reasons for using different conventions. You will have to be sufficiently aware to confidently infer from context what’s really meant. That’s easily manageable.

Many thanks for the frequency algebra lesson Paul. Along with the earlier JEV derivation, I’m better able to see the scheme now. It has been confusing in the past, where periods and frequencies have been interchanged invisibly behind planet initial letters.

How did I notice?

You may remember the φ-power tables I shared on comment page-1 of this discussion.

1/(S-N) & 1/(J-U) fell close to the pattern, so I intuitively explored as follows:

(35.85455172)*(13.81290444) / (35.85455172 – 13.81290444) = 22.46907821

(35.85455172)*(13.81290444) / (35.85455172 + 13.81290444) = 9.971428676

(35.85455172)*(13.81290444) / ( (35.85455172 + 13.81290444) / 2 ) = 19.94285735

I compared with nearest 2*φ^n beats:

(35.88854382)*(13.70820393) / (35.88854382 – 13.70820393) = 22.18033989

(35.88854382)*(13.70820393) / (35.88854382 + 13.70820393) = 9.91934955

(35.88854382)*(13.70820393) / ( (35.88854382 + 13.70820393) / 2 ) = 19.8386991

Then I explored with the question:

What would match (φ/Φ)/(J+S), 1/(2(J-S), & 1/(J-S)?

(36.03007699)*(13.71275318) / (36.03007699 – 13.71275318) = 22.13847667

(36.03007699)*(13.71275318) / (36.03007699 + 13.71275318) = 9.932517933

(36.03007699)*(13.71275318) / ( (36.03007699 + 13.71275318) / 2 ) = 19.86503587

The basic frame was obvious right away without any delays and I could effortlessly see the 22ish symmetry in the matrices. There are a cluster of periods that hover around 22 & harmonics & subharmonics when flipping through the + +, +, -, – – matrices. It’s a breeze to recognize that it’s an attractor when looking at it from that perspective.

All I had to do was isolate the attractor core. It was easy.

The symmetry is centered here:

J S • U N

We’ve long been well-aware of the harmonic mean of J & N.

There was a moment when my awareness of a

trivialsymmetry between + + & – – for S & U became lucid. Seconds later I had the core defined.This was no triumphant spark.

The sparks were scolding & painful.

Imagine for comparison you were responsible for leaving a baby in a stroller on a slope and you turned around to see a bus colliding with the stroller.

It felt like that. It was that terrifying feeling of becoming aware of one’s inexcusably ignorant responsibility for an important failure to notice something of consequential simplicity.

You’re too hard on yourself. “It’s easy when you know how” in hindsight.

I think the presentation of the calcs is where some confusion has arisen for others. All the numbers presented are periods, because you’ve implicitly solved the inverse, but the equations are frequency algebra.

Paul, have you got the email I sent you this afternoon?

For derivations, period algebra blows up into a nightmare because of combinatoric cross-terms. (It’s nonlinear.)

Frequency algebra derivations are easy. (It’s just straight lines.)

However,

afterthe derivation, taking the reciprocal to get back to period is easy._ _

Here’s something to think about:

√√(1/(U-N)) = √√171.4062162 = 3.618317245 ~= φ+2 = 3.618033989

log_(φ+2) (1/(U-N)) = 4.000243518 ~= 4

(φ+2)^4 = 171.3525492 ~= 171.4062162 = 1/(U-N)

I have to stop there for now. (Other responsibilities call…)

PV: ‘√√(1/(U-N)) = √√171.4062162 = 3.618317245 ~= φ+2 = 3.618033989’

Another thing with 3.618033989: subtract it from 5 and the ratio of the remainder to 3.618033989 is 1:phi².

And whether you add or multiply the two numbers, the answer is 5 both times.

The difference between the two is √5.

That would be because φ√5 = 3.618 in the first place.🙂

As I mentioned here:

https://tallbloke.wordpress.com/2015/12/21/why-phi-an-orbital-parameters-test/comment-page-2/#comment-112628

” φ√5 is φ^2 + 1″ (and φ+2).

Good to see Paul found some use for it.

I do like φ√5 much better. I was planning on figuring out a bunch of the other φish things φ+2 equals when time would permit (knowing full well that surely there would be

manyoptions), but thanks for saving me the trouble by pointing directly to the 1 I know I like best.I’m doing a combination of sorting & further exploring. I would say the sorting is important before we run a thread devoted specifically to this topic. Were the thread to start say now, I would not be prepared and for sure I would be very dissatisfied with my preparedness to contribute at the level the topic deserves. Somewhere in between will be the optimal balance, as I think it’s guaranteed that I’ll never in a whole lifetime be able to finish more than a tiny fraction of what would be possible with liberation. Financial slavery sucks.

OK, we’ll carry on sharing ideas and results here until we agree we’re at a stage where we can present a reasonably tidy new post. I’m still buying a lottery ticket every month. When I win, I’ll set up a research foundation and you can give up the day job.🙂

I’m working on a graphic for the headline of the new post. Suggestions for improvement welcome.

TB: could it be simplified e.g. like this Moon one?

OB; nice Kepler triangle. Relative densities of Earth and Moon are very nearly 1:phi too. My image could be as simple if there were only two planets…

“the Earth has an average density of 5.5 gm/cm3 and the Moon has a much lower overall density, 3.34 gm/cm3, indicating that it lacks the iron core. The densities of Mercury, Venus, and Mars are 5.4 gm/cm3, 5.24 gm/cm3, and 3.94 gm/cm3 respectively. Even the majority of meteorites have densities greater than the Moon and many are mostly iron”

http://webcache.googleusercontent.com/search?q=cache:GwG8CLYfIUQJ:hyperphysics.phy-astr.gsu.edu/hbase/astro/moonform.html+&cd=4&hl=en&ct=clnk&gl=uk

Meanwhile, Mike ‘Pluto-killer’ Brown thinks he’s discovered planet 9

http://www.motherjones.com/kevin-drum/2016/01/scientists-discover-most-planet-y-planet-solar-system

The ‘sciency bit’ is interesting

I’m a bit suspicious of autonomous Hamiltonians😉

The things I’m finding are too easy.

It makes me strongly suspect we may simply be rediscovering the wheel …maybe not though — at least sort of — because way back whenever, they didn’t have precision like Seidelmann (1992) as an aid.

Can anyone point to a list of where Kepler saw φ?

I’m not suggesting he had the info back then to be aware of the Hale-scale stuff, but surely at least some of the things we’re discussing were known or perhaps even extremely well known to him?

Well, Kepler’s famous illustration in Harmonice Mundi showing the orbits as being bounded by the 5 regular solid polyhedra had phi all over it. I’m certain he was aware of how phi could be found in the geometry of those polyhedra. He never explicitly linked phi with the Sun and planetary orbits so far as I know though.

We are breaking new ground.

Pentagonal phi

Kepler: see no. 4 The Planets

http://www.keplersdiscovery.com/Harmonies.html

The inner pentagon has side Φ/φ so the ratio of the outer pentagon circumference to the inner pentagon circumference is φ/Φ.

I wonder what that does for the ratio of the square root of the cube of the radii of the circumscribing circles. I’ll get to that later.

Once again OB has influenced timing.

I had a list of things I was going to contemplate. Pentagon geometry was on the list. I would have gotten to it whenever, but the pentagon image OB posted naturally triggered a reordering of priorities.

Thank you OB. Just like with the Mae-Wan Ho link, a leap followed quickly. I can now illustrate all of the geometry we’ve been discussing with simple pentagons. It’s so f**king simple. I’m finding it very hard to believe all of this isn’t already known. In fact, I don’t believe it isn’t known. It seems ridiculous to think something this simple isn’t known and regarded as trivial and obvious. Kepler must have known most of this stuff and surely were he living in our times (with our more precisely determined orbital periods and solar cycle records) he would have nailed the rest of it long ago.

TB: This is what I meant by illustration. The illustration you’ve posted has value in a different quite helpful way, but what I meant was I wanted to illustrate the geometry of system orbital periods. I’ve verified that as I suspected, it can be done with pentagons. For example 1/(J+S) on the inner pentagon and (φ/Φ)/(J+S) on the outer pentagon. Other examples of pairings are E+V with E and V with V-E. It’s a simple fractal. For the period doubling it’s just a harmonic mean of inner & outer pentagons in period or an arithmetic mean in frequency. The φ/Φ pentagon perimeter ratios are the same for pentagon circumcircles and their radii.

…So illustration of the φ power ladder is resolved. When opportunity & inclination coincide, I’ll contemplate √5 power ladder geometry.

Paul: Kepler must have known most of this stuffKepler got sidetracked by the ratios of the adjacent planetary distances at aphelion/perihelion, because they resolve to the notes on a major scale (quite closely). And he worked with his elliptical orbits rather than the circles we’re currently contemplating.

TB, the relationship between radius & circumference is linear but certainly there are associated areas and volumes to consider. For example the area of 5 squares attached to OB’s outer pentagon is 5 and reorganizing that total area into a single square would give a side of √5. By analogy the cubed root of 5 shows up if we do the same thing for volume. The radius in the geometry I’m describing is measured in sidereal years, not AU. Conversion between years & AU is determined by pentagon perimeter area:volume ratios. Introducing that at this stage would probably just confuse most people. That’s part of why I didn’t want to rush into a new thread just yet. The basics need to be clearly illustrated first. Due to time constraints it could be months or years before the picture is extended. I admit I find it very annoying the way things that should take a few weeks get dragged on for years by time & resource constraints …but that doesn’t mean I’m going to stop. It’s natural to continue. We just won’t get very far in a lifetime this way, but still it’s natural to explore as far as we can.

Per Ardua, ad Astra, as the Royal Air Force used for its motto. Don’t worry, we’ll find some research money eventually.

Check the geometry of the wing, circle crossing points.🙂

The Pentagram & The Golden Ratio

http://www.contracosta.edu/legacycontent/math/Pentagrm.htm#StarArea

Some interesting problems near the end of the page.

—

From the LOD thread – RJS says:

‘5.018891421 = UEV

4.492694707 = NEV’

UEV/NEV ≈ √5/2 (99.84% match)

Let’s get these 2 side-by-side…

[(φ/Φ)/(J+S)] / [1/(J+S)] =

φ/ΦOB,

log_√5 (1/(1JEV)) ~= 3

log_√5 (1/(4SEV)) ~= 1

log_√5 (1/(1UEV)) ~= 2

log_√5 (1/(2NEV)) ~= 1

1/JEV ~= 1(√5^3)

1/SEV ~= 4(√5^1)

1/UEV ~= 1(√5^2)

1/NEV ~= 2(√5^1)

It looks like the main asteroid belt is centered at

(1/(2(

J+S)))^(2/3) = (8.456145629/ 2)^(2/3) = (4.228072815)^(2/3) = 2.614738763 AU~= ((φ^3)/2)^(2/3) = (8.472135955 / 2)^(2/3) = (4.236067977)^(2/3) = 2.618033989 = φ/Φ AU

…or

( ((Φ/φ)/2) / (J+3S-U+N)) )^(2/3) = ((Φ/φ)(22.18025492) / 2)^(2/3)

= (8.4721035 / 2)^(2/3) = (4.23605175)^(2/3) = 2.618027303 AU

background:

1/(J+3S-U+N) was derived above (January 20, 2016 at 8:38 am).

The man’s on fire! We can’t keep up.🙂

PV: ‘8.456145629 / 2’ is close to phi³ in years, or 55/13 in Fibonacci numbers.

How much of what we’re looking at depends on regarding the Sun-Earth distance of 1AU as some kind of specially privileged number/unit of measure?

There is a phi relationship between the positions of Earth-asteroid belt and Jupiter.

1:phi:2phi in AU

The Asteroid belt is halfway between Jupiter and the Sun. ~2.618 AU vs ~5.2 AU

If we defined the asteroid belt as distance 1 from the sun, Jupiter would be 2 and Earth would be 1/phi^2

It’s pretty easy to make steady progress on this. The only hard part was opening the mind before the beginning. Nasty gatekeepers had done some partially successful brainwashing. Lesson learned.

Cracked √5 power ladder.

Key: 1/(radius^2).

Reveals φ/2Φ scale.

more details forthcoming, possibly tomorrow

Cheers

Another one to ponder:

Draw a tangent from the small circle through the other two, crossing points A and B and extending to G.

The ratio of the length of segment AG to segment AB is Phi:1

http://www.goldennumber.net/circles/

This is worth reporting right away:

The inverse square (1/radius^2) key opens the √5 door with regression (not to be confused with radius) r^2 = 0.9999982167 = 99.99982167%, leaving only 0.00000178328 = 0.000178328% to support the dissatisfaction of extreme perfectionists.

I haven’t

yetfinished sorting out the implications of this for other parts of the whole-system puzzle.Somewhere on a wiki page I saw an approximation of π based on φ. I vaguely remember a √ sign being in the equation with neither sines nor cosines. I couldn’t relocate the approximation formula with a reasonable amount of search effort yesterday. If anyone spots that equation, please report it. It may help explain something I’ve noticed.

Try the Kepler triangle page. Pi ≈ 4/√φ

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

I now have r^2 = 1 …so the perfectionists are going to have to look for something else to overemphasize.

The neverending harassment and torment at wuwt chased countless intelligent people away from this relatively easy puzzle. There are bright well-resourced people with lots of free time who could almost certainly crack this whole problem in about 2 hours. Those free hares fell for the brainwashing that dissuaded them from looking deeper, so it’s up to obstructed tortoises. Better late than never.

Miles Mathis showed this triangle in his first ‘golden ratio’ paper.

https://tallbloke.wordpress.com/2013/08/26/miles-mathis-the-physics-behind-the-golden-ratio/

Now match the top left of that triangle to the angle at the tip of the blue arrow in the pentagon posted earlier.

https://tallbloke.wordpress.com/2015/12/21/why-phi-an-orbital-parameters-test/comment-page-2/#comment-112761

The Mathis triangle fits exactly into the pentagon – in fact it fits at any of five points.

Line BC would be any side of a pentagon.

Move BC round 108 degrees to create the next side, repeat until the pentagon is complete.

Thanks TB. That’s exactly where I remember seeing it:

https://en.wikipedia.org/wiki/Kepler_triangle#A_mathematical_coincidence

φ = 1.61803398874989

√φ = 1.27201964951407

4 = 4

4/√φ = 3.14460551102969

π = 3.14159265358979

π-(4/√φ) = -0.00301285743989999

(π-(4/√φ))/π = -0.000959022308782551 = -0.0959022308782551%

My attention is elsewhere right now but I’m noting this for later reference:

(1/JEV)^(2/3) ~= 5

(1/SEV)^(1/2) ~= 3

(1/UEV)^(1/1) ~= 5

(2/NEV)^(1/2) ~= 3

1/JEV ~= (5^(3/2))/1

1/SEV ~= (3^(2/1))/1

1/UEV ~= (5^(1/1))/1

1/NEV ~= (3^(2/1))/2

__ __ __

period / radius

Schwabe / Schwabe^(2/3) ~= √5

circumference / period

2π*Schwabe^(2/3) / Schwabe ~= 2√2

Schwabe^(1/3) ~= √5

Schwabe^(-1/3) ~= 1/√5

This is about

• how long the thing takes to circle

• how far out it is

• how far it travels while cycling

• etc. (i.e. scaling of other physical quantities more generally)

Relative scaling of different physical quantities share a sweet spot at Schwabe-timescale (5th φ power) because:

φ^(5*(2/3)) ~= 5

2π*5/(φ^5) ~= 2√2 = 2^(3/2)

Another option on a Kepler triangle is instead of sides 1, √φ, φ you could have sides of √φ, √φ^2, √φ^3

Paul: φ^(5*(2/3)) ~= 52S(orbital)/J(orbital) is nearer at 4.9647565. It looks like J and S are trying to square the circle.🙂

TB suggested:

“Another option on a Kepler triangle is instead of sides 1, √φ, φ you could have sides of √φ, √φ^2, √φ^3”More generally it’s a fractal (e.g. hierarchically nested pentagrams).

TB suggested:

“2S(orbital)/J(orbital) is nearer at 4.9647565. It looks like J and S are trying to square the circle.”Well put. Everyone may be well aware of the former, but perhaps not so much the latter.

_

Rather than ruin my health further, time for a break. Today’s revelations can wait in support of the health needed for future revelations.

Good idea. Stuart and I had a stroll out to a remote loch along a nice forest trail today. My back was killing me after we got back, but it was great to get out for a few hours.

kayaked out to some islands, had a hike around, caught some spectacular orange lighting on the evergreen mountains through a sunset cloud gap. Refreshed and noting…

π = 3.14159265358979

4 = 4

4/π = 1.27323954473516

(4/π)^2 = 1.6211389382774

φ = 1.61803398874989

((4/π)^2)-φ = 0.00310494952750973

(((4/π)^2)-φ)/φ = 0.00191896434135394 = 0.191896434135394%

doesn’t work as well that way around, but fuller awareness is good.

Now I’ve got a piece of work on my hands sorting out concise summary of the √5 & φ power ladders and their Schwabe intersection (given paid work). As you can guess, it will end up taking weeks to reach the standard for which I aim, but I’ll keep up the brief updates along the way.

Eventually it will all be done to minimally tolerable standard. Only then will open minds be able to make

overallsense of it.I think at this stage anyone following along should at least be aware of the inner system φ fractal and it’s connection to Schwabe, but I would be surprised if anyone has more than a vague sense of how it hooks to the √5 frame. Even though I have r^2 =1 for the latest √5 scaling (inverse square), I haven’t yet even given any thought to how to illustrate it.

…and to me there’s (effectively) nothing if there’s no illustration. Words are just to convey thoughts on the trail to a picture. Before there’s a clear concise picture, there’s only unfinished thinking.

…but in this

particularlysimple case, once there’s a picture it’s going to looksotrivial as toagainland back at nothing …and it isfrom the nothing withinthat the wheel derives itsinfiniteutility.What’s the picture going to look like? See if you can guess before seeing.

Inex cycle 521 years / π = 165.83945 y

https://tallbloke.wordpress.com/2015/03/20/why-phi-the-inex-eclipse-cycle-part-2/

44 Jupiiter = 521.955~y

14 Jupiter = 166.0766y

44/14 = 22/7 = ~π

2 (J+S) = ~15 full moon cycles (15.0013~)

34 x 2 (J+S) = ~13 x 5 lunar apsidal cycles (5 x 115~ years)

5 lunar apsidal cycles = ~2 Hale cycles

PV says: ‘It looks like the main asteroid belt is centered at’ [2.614 – 2.618 AU]

5.2026 AU = semi-major axis of Jupiter

5.2026 / 2 = 2.6013

So the central zone of the asteroid belt (AB) is about halfway between the Sun and Jupiter’s SMA.

With Earth at 1 AU, the approx. relative distances of Earth and Jupiter to the AB centre would be:

(5.2026 – 2.616) / (2.616 – 1) : 1 = 1.60062~ : 1 = ~8:5

trouble choosing between models — r^2 set:

0.999991869, 0.999987017, 0.999987844, 0.999989971, 0.999992955

so I’ve been looking at physical parameters to help discern

something I noticed:

mass

J / S = 189.9 / 56.846 = 3.34060444

1/(radius^2)

J / S = 0.036962337 / 0.010996863 = 3.361171098

and of course there’s the obvious one we’ve seen pointed out countless times forever:

period:

S / J = 29.447498 / 11.862615 = 2.482378295 ~= 5/2

Given the √5 coming up, notice that

(2 * 2.482378295)^(3/2) = 11.06233845

Of the models based on √5, φ, & 11.06233845 none are uniformly better. Each performs better in a certain segment of the solar system. With r^2 so high, the general pattern is nailed, but even though the residuals are small, they’re sufficiently systematic to demand more scrutinizing attention.

I’m not one to get carried away with details. I like exploring first-order aggregate structure. So I may just report all models side-by-side and leave it there for others to go explore the roots of the subtle but systematic structural nuances. I want to work out and illustrate the geometry before reporting just tables, which are by orders of magnitude inferior to illustrations on the information assimilability scale.

I’m going to rewrite this from above since it’s probably more intuitive reorganized this way:

(φ^5)^(2/3)) ~= 5

2π*5/(2√2) = 2π*5/(2^(3/2)) = 11.10720735

~= φ^5 = 11.09016994

(11.10720735)*(11.09016994) / ( (11.10720735 + 11.09016994) / 2 ) = 11.09868211

2*(11.09868211) = 22.19736421

(1/(J+S)) / (1/(V+E)) = 8.456145629 / 0.380883098 = 22.20141997

5 = 5

2 = 2

φ = 1.61803398874989

1/φ = Φ = 0.618033988749895

2φ = 3.23606797749979

1/2φ = 0.309016994374947

arccos(1/2φ) = 1.25663706143592

(5/2)*arccos(1/2φ) = 3.14159265358979

π = 3.14159265358979

2π = 6.28318530717959

2π/5 = 1.25663706143592

cos(2π/5) = 0.309016994374947

1/cos(2π/5) = 3.23606797749979

2cos(2π/5) = 0.618033988749895

1/(2cos(2π/5)) = 1.61803398874989

1/(2cos(2π/5)) = φ(5/2)*arccos(1/2φ) = πPaul: something I noticed:mass

J / S = 189.9 / 56.846 = 3.34060444

1/(radius^2)J / S = 0.036962337 / 0.010996863 = 3.361171098

This also relates to the density-distance relation for E, J and S and noticed a long time ago by Le Verrier.

For all the planets, density distance looks like this:

Paul: 1/(2cos(2π/5)) = φ(5/2)*arccos(1/2φ) = π

I knew a smart man like yourself would be able to re-arrange the equation I posted here to isolate π and φ:

January 19, 2016 at 2:09 pm

1/(cos(2Pi/5)) = 1 + √5 = 2φ

🙂

I found the le Verrier quote in one of my comments on the Jackpot thread:

“In comparing the masses of the Earth, Jupiter and Saturn to their volumes, one remarks that the

densities of these planets are, to some degree, inverse to their mean distances from the Sun.”

– le Verrier – 1859

For Uranus and Neptune:

Mars 0.17

Venus 0.26

Mercury 0.485

So it seems J and S have a special balance in this system, with E and Ma close by.

So:

Paul has found the 1/r^2 identity for J & S masses

I have got le Verriers 1/r identity for J & S densities

There’s something important going on with that. But what?

Then there’s something OB spotted a couple of years ago that I mentioned in my ‘Hum’ paper in PRP:

The summed diameters of J&S are φ^2 the summed diameters of U&N

Just now he tells me he’s summed the masses of Me&V and E&Ma and found a √φ relationship between them!

More mysteries to solve.

Here’s yet another mystery:

The theoretical rates of Perihelion Precession given at http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node115.html have Jupiter and Saturn in a 4/φ of 1:2.4722 ratio. 4/√φ is a close approximation to π

The observed rates for J & S have a ratio of 1:2.97

p=2

^{0}, 2^{1}, 2^{2},… 2^{n}for a sequence of planets and dwarf planets starting at Mercury.Dist(AU) = p

^{1/φ}(1/φ^{2})(φ+(1/φ))/2)`Not a bad heuristic. I get the feeling the equation might simplify a bit, but I can't do algebra the way I used to.`

My awareness of the √5 geometry has improved to the level of an r^2 = 0.999993167 model (right column) of JSUNEV timing.

more detail

r^2 = 0.999993166910473

Seidelmann (1992)

X = parameter from √5 geometric relations

√9 = 3

5 = 5

√5 = 2.23606797749979

√√5 = 1.49534878122122 (geometric mean of √5 & 1)

2√√5 = 2.99069756244244

√√5+√5 = 3.73141675872101

(√√5+√5)/2 = 1.86570837936051

√((√√5+√5)/2) = 1.36590935986269

2√((√√5+√5)/2) = 2.73181871972538

φ = 1.61803398874989 (arithmetic mean of √5 & 1)

√5/φ = 1.38196601125011 (harmonic mean of √5 & 1)

√9/√√5 = 2.00622091492927

See if you can draw the geometry.

J+N, S+N, & U+N fit into the √5 picture with r^2 even closer to 1. Details forthcoming.

T^2=r^3period-frequency spectrum

More details forthcoming.

On a log scale that’s a straight line with r^2 = 0.999998059. (regression r^2 here, not radius — confusing maybe, because above it is radius squared in the table …but should be easy enough to carefully infer from context)

r^2 without logging:

0.999994974 for φ^(5n/3)

0.999995616 for (√5)^n

0.999995762 for their harmonic mean

There are implications.

(to be continued….)

typo

-n values in the mean column should be:

1.000000000

0.447817981

0.200541309

0.089806331

0.040217110

0.018010076

0.008065309

frequency arithmetic mean = 1 / period harmonic mean

so the column would be better labeled “mean” rather than “harmean”

r^2 values change negligibly

(√5)^(-4) = 1/25

J-S = 1/19.86503587

Deduction:

(J-S)-(1/25) = 1/

96.71457943(J-S)+(1/25) = 1/

11.0693302((J-S)+(1/25)) / 2 = 1/

22.13866041(25)*(19.86503587) / (25 – 19.86503587) =

96.71457943(25)*(19.86503587) / (25 + 19.86503587) =

11.0693302(25)*(19.86503587) / ( (25 + 19.86503587) / 2 ) =

22.13866041How did I get there?

(29.447498)*(25) / (29.447498 – 25) = 165.5284499

In frequency: 1/25-S

(165.5284499)*(11.862615) / (165.5284499 + 11.862615) = 11.0693302

In frequency: 1/25-S+J = J-S+1/25

(165.5284499)*(11.862615) / ( (165.5284499 + 11.862615) / 2 ) = 22.13866041

In frequency: (J-S+1/25)/2

Compare that back to

January 22, 2016 at 11:57 pm

(

22.13847667 & 22.13929985)…and now you know the

rootof the deviation fromJanuary 20, 2016 at 8:38 am

(22.18025492)

Got it?

There’s a

perfectly goodreason why they differ.The important thing to recognize is that there’s symmetry

across dimensions.TB is on the road and I’ve just logged in. Thanks to P V for all the new info.

More to think about:

(22.18025492)*(22.13847667) / (22.18025492 – 22.13847667) = 11753.41394

(22.18025492)*(22.13866041) / (22.18025492 – 22.13866041) = 11805.43105

(22.18025492)*(22.13929985) / (22.18025492 – 22.13929985) = 11990.09688

Tabulating dyadic harmonics:

22.18025492 = 2/(J+3S-U+N)

further derivations forthcoming

Just got home. Amazing stuff Paul, keep it coming at your own pace. I have to take the missus to hospital for a minor op early in the morning. I’ll check in as I can. I’ve updated my earlier Bode killer with the error, which averages out at 5% across 13 bodies now I’ve substituted Ceres with Vesta at the densest region of the asteroid belt (2.37AU). Pearson r^2 is 0.9995. Not bad. I’m thinking about ways to improve the fit with a ‘wobbler’ consisting of a set of fib ratios. Using p

^{1/φ}*(1/√5) is similar to p^{1/φ}*(1/φ2)*((√5)/2) but not quite as accurate and may lose information. I’ll handle the rough’n’ready engineering approximation across the full system while Paul finesses the tight fits and identities in the outers and inners. Together we’ll crack this big walnut.🙂Tallboke and Paul Vaughan. I know this isn’t the area for this, however you guys can do the math. On What Up with that web site, the reference is about incoming and out going w/m^2. More than 10 years ago the IPCC stated that incoming was 343 w/m^2 out going was 103 w/m^2 and retained was 240 w/m^2. That number is now changed from the satellite data down to a difference of 0.6 w/m^2 retained. That 240 w/m^2 is extremely important as it is worked into the formula to give the rise in temperature from co2 alone. I think there is something wrong with the formula. When I plug in a higher retained w/m^2 I get a lower Kelvin and a lower retained w/m^2 a higher K.

delta Ts = (-Ts/4) (delta E/E)

E(oTs ^4) =240 w/m^2

(delta E) (oTs^4 = -4 w/m^2 delta E /E = (-4/240)

Ts = 288 K

delta Ts = (-288/4) (-4/240) = 1.2 K

Which came from oE(T)^4 + 4 + L E + G which became E (oTs^4) 4 (pi) a ^2

I don’t dare say anything, however after the 1.2 K was = 0.5 C . I need a second opinion. or third..

the 0.5 C is suppose to be the observed increase in temperature.

Hi Rishrac. The experts in this field are currently commenting on the black-body thread, I suggest you enquire there. It’s not my area of knowledge, and we’re a bit busy at the moment as you can see!

Try this too

http://www.john-daly.com/forcing/hug-barrett.htm

http://eae.sagepub.com/content/17/4/603.full.pdf+html

‘Lord Carrington determined the solar rotation rate by watching low-latitude sunspots in the 1850s. He defined a fixed solar coordinate system that rotates in a sidereal frame exactly once every

25.38days’http://umtof.umd.edu/pm/crn/CARRTIME.HTML

The lunar axial period = 19 tropical years / 273 (or: 254 TM + 19 TY) =

25.42daysOr: (TY x TM) / (TY + TM) = 25.420059

(25.420059 / 27.321582 (TM)) + (25.420059 / 365.24219 (TY)) = 1 (axial)

1 J+S axial period = 55/34 x 70 solar sidereal rotations (~99.996% match)

Ok.. thank you

In the complex plane the geometry reduces to z^5=1 (the fifth roots of unity):

For example the orbital period circumcircles of Me+J, V+E, V, E, & V-E fit the fractal pentagon vertices:

The slight difference in period of JEV & J+N is no longer a mystery.

(φ/Φ)/(J+S) & (J-S+1/25)/2 are so close to each other that astronomers would need

more than half a billion yearsof observations to confidently test the hypothesis that they are not equal.(φ/Φ)/(J+S) & (J-S+1/25)/2 are sufficiently close to 1/(3V-5E+2J) that astronomers would need 150000 to 200000 years of observations to confidently test the hypothesis that 1/(3V-5E+2J) differs from (φ/Φ)/(J+S) & (J-S+1/25)/2.

“(φ/Φ)/(J+S) & (J-S+1/25)/2 are so close to each other that astronomers would need more than half a billion years of observations to confidently test the hypothesis that they are not equal.”More precisely:

They would need 2/3 of a billion years.

Wow! That’s beautiful!

We have a new side-shoot Why Phi? thread where the law of fives is in operation too.

https://tallbloke.wordpress.com/2016/01/26/why-phi-sylvia-sons-swing-serenely-by-shining-stars/

It looks like an example where we might take a look at relative masses and a possible different approach to the three-body problem.

More strong pentadal and bi-pentadal structures abound throughout the solar system too. I’ll post some illustrations later.

Examples from ‘Signature of the Celestial Spheres: Discovering Order in the Solar System’

By Hartmut Warm – my copy arrived last Thursday.

Available (very) inexpensively here:

http://www.abebooks.co.uk/servlet/SearchResults?an=hartmut+warm&sts=t

Preview on google books

Recall:

(√5)^4 = 5^2 = 25

φ+1 = φ^2 = φ/Φ

φ+2 = φ√5

Refining:

√√(1/(U-N))= √√171.4062162 = 3.618317245 ~=φ√5= 3.618033989log_(φ√5) (1/(U-N)) = 4.000243518 ~= 4

(φ√5)^4= 171.3525492 ~= 171.4062162 = 1/(U-N)(5(φ/Φ))^2= 171.3525492 ~= 171.4062162 = 1/(U-N)25(φ/Φ)^2= 171.3525492 ~= 171.4062162 =1/(U-N)Apologies if this is a repeat:

U+N / J+S = ~phi^4 ((13/5)² is nearer)

45 U+N = ~296 J+S = ~41 x 61.05y = ~126 J-S

Derived from above (in frequency, so invert to get period):

S = ((((2/φ)+(1/25))/(φ^5))+(φ+1)/25)/((φ+3)+3/φ)

When time permits I’ll derive expressions for all other planets. J, N, U, Me, & Ma will be relatively easy. For E & V I’ll have to solve a system of equations. At this point I don’t expect people to understand how I’m doing this, but some of the insights I’ve posted above are orders of magnitude more solid than others. I know the difference and although it’s algebraically tedious, it’s possible to stitch together the more solid results. It should be possible to approximate the entire system with minimal information.

Thanks Paul. It’ll be fascinating to see what patterns emerge from the completed set of derived expressions. I’ll be watching the ongoing revelations with interest.

In the meantime I’m revisiting Kotov’s 160min wave and thinking about a possible refinement of my heuristic full system equation: for p=2

^{n}AU=p^{1/φ}*(1/φ^{2})*((√5)/2)I had to go and look up dyadics, as it rang a vague bell from my engineering maths classes. I’ll try to remember the specific application we were looking at.

J = (-(φ+1)/25+(2/3)(φ+3)/(φ^5)+(φ+3)/(75(φ^4)))/(Φ+φ/3+1)

I don’t like messy algebra, but the equations I’ve just given for J & S in terms of φ

exactlypreserve the Hale geometry I’ve outline above.Given the countless options φ gives for re-expression, something more terse &/or more aesthetic can be derived as time permits.

What’s different about the approach I’m using is that it focuses on beats rather than orbital periods of single bodies. The geometry suggested this approach.

There were simple patterns in the beats … but the single bodies?… not so much. I wanted to capitalize on the simple part. Can the above expressions also be reduced to more terse, more aesthetic expressions via identities? It’s a worthwhile pursuit, but it’s not what really matters.

What really matters is the simple patterns in the beats. I chose this (

mindful of coupling) approach because it (bystitching) gives a better hope of reproducing the entire set of system beats (+ +, +, -, & – -). I want r^2 highnot onlyfor single bodies, but for all possible combinations of beats. I’m sure I’ve lost some people at this stage, but that’s where this is going. People may think I was on the right track and then lost it. But no. It’s the same track and I knew at the outset that it was a set of coupled equations with a fewsimpleconstraints. The only thing I wasn’t sure of at the outset was whether there were enough sufficiently rigid constraints to solve the system. Recently I became convinced that there are.Recall that I’m looking for a solution

that preserves JEV-Hale geometry. That couplingmustbe respected in order to have a realistic model. Weeks from now the overall philosophy may be more understandable after there has been time to write down a lot more notes and then iteratively refine for concise presentation. I’m showing bits of the work at the snail’s pace paid work and other obligations allow.The messy algebra has been the least enjoyable part of the exploratory journey, but it was a relatively easy proof of concept (to verify the path chosen). Outcome: Proof of concept tests passed

exactly(preserving all decimal places). The next step is further messy algebra to extend the proof of concept beyond the Hale core. Next up: U & N, then Ma & Me, and finally the trickiest: E & V). Then there will be tests on the 4 matrices. It will probably be a month before I will even be ready to consider writing up insights and iteratively refining the write-up, but I knew weeks ago from the steady progress that this was a viable track. Things were continually falling together with high precision without too much effort. It doesn’t go like that if one is on the wrong track.Seeing how messy the algebra is for single bodies is actually somewhat satisfying, as it underscores the relative simplicity accessible to a mind

open enoughto look atpairsand combinations of bodies.I’ll never forget the deserved scolding words of a mature, appreciated friend who once strategically accosted with

“I thought you people had an open mind! — I was wrong”before storming out provocatively. Point taken. Lesson (later) learned.Paul, this is a well considered roadmap. I agree that the beats are the primary key and have been saying for ages that it’s the energy passed between pairs via orbital resonance that determines the conjunction period and hence the orbital distances and periods, not the orbits that determine the conjunction periods. That’s why we’re getting simpler expressions for the conjunction periods and axial periods.

The other main aspect is the Sun’s characteristic periods and the link between those and the planetary beats. My whole-system-heuristic isn’t so useful as a precision map as your expressions are, but the aberrations between it and reality may be revealing something important. The model deviates from the actuals at ~0.4AU (Mercury), ~0.7AU (Venus), ~5AU (Jupiter) and ~10AU (Saturn). These are the very same distances as the well known solar 5 minute and 160 minute light speed waves Kotov is interested in. Ray Tomes’ take is here

The following diagram shows the tendency for planets to lie at multiples or fractions of various distances from the Sun. In the range shown the 5.01 and 10.06 AU distances are the strongest periodicities and equal 41.7 light minutes and 83.7 light minutes for internodal distances of waves, or 83.3 and 167.3 minutes for the period of the waves. These longer waves correspond to the outer planets.I’ll work on that problem and see if Stuart and I can derive a wave function expressed in terms of the ratios and resonants we’re using to reconcile the residuals and see if that gives any further hints for integrating the results of my outside-inwards and your inside-outwards approaches. Ut supra, ita inferius.

The ratio of half the precession of the equinoxes to precession of the perihelion is ~8:13.

20938 / (25771/2) = 1.6249272

1.6249272 x 8 = 12.999417 (~13)

Based on this from Dr Bob:

‘Here’s the basic data for these effects. We define years of different lengths, and here are three key ones:T1 = 365.242190 days = tropical year (equinox to equinox)

T2 = 365.256363 days = sidereal year (fixed star to fixed star)

T3 = 365.259635 days = anomalistic year (perihelion to perihelion)

The length of the cycle for precession of the equinoxes is

T2/(T2-T1) tropical years = 25771 tropical years

(Over this length of time, the number of tropical years will exceed the number of sidereal years by one.)

Similarly, the length of the cycle for perihelion is

T3/(T3-T1) years = 20938 years

So in half this time (10469 years), the times of perihelion and aphelion in the calendar will be the reverse of what they are today. That will make northern-hemisphere seasons a bit more extreme, and southern-hemisphere seasons a bit less extreme.

note: These numbers (25771 and 20938 years) are given to a little too much precision. I don’t think they’re correct to the nearest year, but they’re very close. Wikipedia says the precession cycle is 25765 years long.’http://answers.yahoo.com/question/index?qid=20080122112444AAz6n6y

Re last comment: it follows that the Moon-Earth quarter precession cycle of 6441 tropical years (link below) is in a 13:8 ratio with half the precession of the perihelion i.e. the period Dr Bob refers to:

‘Similarly, the length of the cycle for perihelion is

T3/(T3-T1) years = 20938 years

So in half this time (10469 years), the times of perihelion and aphelion in the calendar will be the reverse of what they are today.’

6441 TY = 6440.75 sidereal years

https://tallbloke.wordpress.com/2015/11/09/why-phi-some-moon-earth-interactions/

Update: so if the quarter is 13:8 with the half, then the half is 13:8 with the whole (perihelion precession cycle).

Something immediately noticeable from Ray Tomes plot above is that Jupiter and Saturn are pulled together (by gravity and resonance), between the peaks in the Kotov wave nodes at 5 and 10AU. Jupiter is 0.2AU

outsidethe 5.01AU peak, and Saturn is 0.5AUinsidethe peak at 10.06AU. This is a 2:5 ratio of differences, in the expected direction (Saturn pulled further off the node point by more massive Jupiter). What does that remind you of? 2 S orbits ~= 5 J orbits.Likewise, Mercury and Venus lie respectively

outsideandinsidethe Kotov wave nodal points at 0.376AU and 0.734AU, both by around 0.14AU.By inspection of the √5 & φ tables above for S+N & S-N respectively:

N = 1/50 – 1/(4(φ^6))

This is such a tight fit that astronomers would need close to a billion years of observations to confidently test the hypothesis that N does not equal this.

Recall that

1/U-N = ((5+√5)/2)^4 = (φ√5)^4 = (5(φ/Φ))^2

U = 1/50-1/(4φ^6)+1/((φ√5)^4)

Astronomers would need over 150000 years of observations to confidently test the hypothesis that U does not equal this.

The expressions I’ve given for J, S, U, & N together reproduce all of the major results I’ve given above. (These aren’t just guesses. They’re solutions to a system of equations that preserves the geometry.)

I’ve actually already given the Me & Ma derivations way above.

Remaining derivations: E & V.

Double check:

JPL Siedelman value for N = 164.79132

http://ssd.jpl.nasa.gov/?planet_phys_par

1/50 – 1/(4(φ^6)) = 164.799556

dividing: = 0.99995

Excellent fit!

[…] tallbloke on Why Phi? – an orbital pa… […]

Wikipedia values for Saturn:

Aphelion 10.086 AU (1,509 Gm)

Perihelion 9.024 AU (1,350 Gm)

Semi-major axis 9.554909 AU (1,429.39 Gm)

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

1509 / 1350 = 1.1177777

Saturn’s Perihelion:Aphelion ratio is 1:√5/2 (99.977% accurate)

Update:

Jupiter’s Perihelion:Aphelion ratio is 1:√(34/7)/2 (99.9994% accurate)

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

It’s not remarkable that N (alone) fits, but rather that whole beat matrices fit while maintaining ALL Hale identities.

It’s proving challenging getting acceptable constraints on E & V. (It’s because they move so fast.)

Everything else is nailed and E & V are coupled by an equation, but I need one more well-constrained (by sufficiently precise observations) identity involving E & V.

How about E = 1 ? 🙂

The other neat thing about the E+V period is it splits into the why phi pi slice I outlined.

finished

r^2 for matrix of ALL pairwise + & – beats in:

period

0.999999973524346

0.999999997406611

frequency

Wow. Quick work.

Pairwise solar system beats model:

(J+3S-U+N)/4 = 1/(2(φ^5))

(J+S)/(φ/Φ) = (J-S+1/25)/2 = 3V-5E+2J

U-N = ((5+√5)/2)^(-4) = (φ√5)^(-4) = (5(φ/Φ))^(-2)

N = 1/50-1/(4φ^6)

E = 1/1.0000174

Ma = (φ^(-3))-J

Me = √5+J

r^2 = 0.999999997408816

In frequency:

Observed:

Solution of model system of equations (stated at January 29, 2016 at 11:50 pm):

matrix structure:

upper-right triangle excluding diagonal: – beats

lower-left triangle including diagonal: + beats

Astronomers would need observations spanning 3/4 of a billion years to test the hypothesis that (J+3S-U+N)/2 ≠ φ^(-5).

What I did:

I solved a system of algebraic summaries of √5 & φ geometry (joint constraints) evident from matrix structure.

Recognizing the boundary conditions was the trickiest part.

Matrix structure clarified further:

Residuals Commentary

Ma residuals stand out as systematic (average absolute value 0.02358673%).

Although an order of magnitude smaller, Me residuals also stand out as systematic (average absolute value 0.006184066%).

So there’s room for targeted improvement …but does it really matter?

That of course depends on context. For some purposes yes …but in the context

currentlyof highest interest theHale Coreof the constraining system of equations doesNOTdepend on Me & Ma. This work is first and foremost about defining the Hale Core, so it’s encouraging that the worst first-round model offenders play no role in core definition …and anyone broadening the context knows exactly where to start.U’s residual (0.012802007%) is the one that attracts my attention.

How to constrain U & N more generally and how to rationalize it deserves more attention.

For round 1 modeling it was clear that there’s a

boundary conditionwithuniqueproperties …but do we really have observations precise enough to make it worthwhile to try to do any better with this part of the model at this point in time?Maybe. Maybe not. It isn’t a big worry, but it’s (along with many other things) worthy of consideration if/when people have time.

Right now I see more immediate priorities.

Meanwhile it has been established that

even crudely identified boundary constraintsfacilitate solution of the system of equations withhighly notableprecision.I don’t doubt the Hale core …at all.

(11.09016994)*(11.09012746) / (11.09016994 – 11.09012746) = 2895031.475

Astronomers need to observe a quarter cycle:

(2895031.475) / 4 = 723757.8688

That’s how long it would take to test the hypothesis that they differ. It’s not practical, so effectively they’re equal.

(J+3S-U+N)/2 = 1/(φ^5)

That’s the core model assumption (model error for this expression = 0% by model definition) that makes it possible to push r^2 so high with help from the other Hale-timescale model equation.

(J+S)/(φ/Φ) = (J-S+1/25)/2 = 3V-5E+2J

Then all that’s needed is a boundary condition at U-N and somewhere to start (I chose E). Tying Me & Ma in as afterthoughts did little to erode the overall model. It’s a 101 level Hale Core model. Bright students will take note of the core and with awareness of it go well beyond. Aside from tidying up notes to assist with efficient understanding, my contribution may end here.

Breaking down model system-of-equations

by categoryto ease conceptualization:Hale Core:

(J+3S-U+N)/2 = φ^(-5)(J+S)/(φ/Φ) = (J-S+1/25)/2 = 3V-5E+2JBoundary conditions:

U-N = ((5+√5)/2)^(-4) = (φ√5)^(-4) = (5(φ/Φ))^(-2)

N = 1/50-1/(4φ^6)

(needed to solve system of Hale Core equations for J & S; note that with U-N & N, can solve for U; then with 2 Hale equations and 2 remaining unknowns, can solve system for J & S; the constraint on U-N is not random, but rather an intersection of √5 & φ geometry; N was derived from the √5 & φ tabulations for S+N & S-N (see way above); it’s a starting point to at least recognize that there are boundary conditions with an aim to motivate a more generalized approach to deriving a system’s boundary conditions)

home base:

E = 1/1.0000174

(without another equation constraining E & V, need somewhere to start; other E & V constraints (e.g. based on V+E, V, E, V-E φ-fractal) although clearly nonrandom are too loose for an equation; perhaps they hold only in some form of long-run aggregate related to the Milankovitch cycles; for the purposes at hand the 1 available equation represents coupling and home base pins the model to present observation to facilitate exhibition of Hale Core insights)

afterthoughts:

Ma = (φ^(-3))-J

Me = √5+J

(the simplest obvious thing; just quick parsimony to facilitate the core Hale lesson; better can be done as others’ awareness of aggregate geometry goes deeper)

Above I tabulated observations & model in frequency matrices. Even though technically a log-transform standardizes perception of a period-frequency spectrum, it’s the periods that most readers will most easily recognize & digest:

Period = 1/frequency

Seidelmann (1992) (I’ve been calling these “observations”):

Round 1 Hale Core model:

r^2 = 0.999999995004976 with log-transform

Everything’s jointly coupled and we can work our way up & down the chain and from one cluster to another.

Do we really need more terse algebraic expressions for J & S in terms of φ? It’s effortless to directly calculate numeric values

withoutgoing to all that fuss, so I wouldn’t describe it as “necessary”, but if someone discovers more terse expressions, I’m curious.Hale Core

model + residuals

at a glance:

As you can see from the

vanishingV+- residuals, (J+S)/(φ/Φ) = (J-S+1/25)/2 = 3V-5E+2J works wonders. A φ-alone model assuming only E = 1/1.0000174 (to absolutely ground V+E, V, E, V-E relative scaling) comesnowhere nearthe model performance derived from theHaleequations.The Hale Core’s the big news here.

E = (3/5)V+(2/5-Φ/5φ)J-(Φ/5φ)S

V = (5/3)E-(2/3-Φ/3φ)J+(Φ/3φ)S

E = (3/5)V+(3/10)J+S/10-1/250

V = (5/3)E-(1/2)J-S/6+1/150

What an achievement! There’s much to digest here.

On the boundaries:

https://tallbloke.wordpress.com/2016/01/18/length-of-day-modelling-the-lunar-and-annual-effect/comment-page-1/#comment-113095

With iterative refinement it should be possible to keep substantially simplifying & clarifying how specific exploratory insights led

directlyto derivations.Hopefully others will try to race me in the direction of

more generalizeddiscovery that applies toanyplanetary system. I’ve no doubt we’re just rediscovering a simple wheel that’s trivially well known to more advanced civilizations.TB suggested:

“There’s much to digest here.”It’s actually pretty simple. Just isolating variables algebraically and then plugging them into other equations and solving. It reminds me of level 1 Fluid Mechanics.

The only thing novel was to look at all bodies pairwise on dyadic log scales in matrix format to isolate to a bare minimum the few key backbone structures needed for first order solution. Sure I’ve posted tables, but does that make it clear what tabulated insights led directly to derivations? I think it might help to review in the weeks ahead a few examples with the benefit of hindsight.

Meanwhile trying to rearrange the equations to solve for different variables may help people understand the model. Just go about the process of solving for each

single-body from the set of derivedjointconstraints. It may at this stage seem alien to have a model stated in terms of a system of constraints. It may not be what people are used to.1/XEV for X = J, S, U, N

r^2 = 0.999999998979721

N derivation elaboration …because I suspect it may look unfathomably mysterious (

“how did he get that equation for N??”) before people realize it’s just trivial subtraction of one straight line from another:)From the φ power tabulation far above, I noticed that:

2φ^6 = 35.88854382

is close to Seidelmann (1992) 1/(S-N)

(164.79132)*(29.447498) / (164.79132 – 29.447498) = 35.85455172

From the √5 power tabulation not quite as far above, I noticed that:

(√5)^4 = 25

is close to Seidelmann (1992) 1/(S+N)

(164.79132)*(29.447498) / (164.79132 + 29.447498) = 24.98312189

You don’t need to go guessing & testing. The easy way to notice the systematic structure is to tabulate dyadic beat logs in sorted matrices. For example in Excel you can use conditional formatting to light up the nearest integers (ranked by remainders) as you roll though the base scale. Systematic patterns (scale discontinuities) JUMP out at φ &√5.

Generally I don’t hesitate to do manual method prototyping when a pattern suggests it. (Cookbooks listed in textbook tables-of-contents limit exploration to a tiny fraction of what’s possible.) A more generalized algorithm that goes straight to the underlying geometry looks feasible and such an approach could be a staple. Methods get rediscovered independently countless times and ability to independently rediscover circumvents reliance on catalogs.

From a thorough comparative inspection of the tables, a sharp observer will note (I noted this far above) that J-U & J+U also fall near rungs on the φ & √5 power tabulations respectively …but not as close, which means a tighter constraint can be derived via S-N & S+N.

(35.88854382)*(25) / (35.88854382 – 25) = 82.39977818 = 2/N

2*(82.39977818) = 164.7995564 = 1/N

In frequency that’s just a halving of

(√5)^(-4) – (1/2)φ^(-6) = 5^(-2) – (1/2)φ^(-6) = 1/25 – (1/2)φ^(-6)

to

1/50 – (1/4)φ^(-6) = 0.006067977

1/frequency = period

1/0.006067977 = 164.7995564

Similarly from beat matrix logs (logarithms) it was easy to see that U-N is just an intersection of φ^4 and (√5)^4 ladder rungs.

With these system boundary conditions clarified it was a trivial exercise to substitute for U-N and rearrange core equations to solve for S & J. (A system of 2 equations with 2 unknowns can be solved.)

That in turn defined the coupling of E & V to the Jovians.

It’s actually pretty simple. Stuff’s coupled and the observations point straight to the coupling.

Paul: “It’s actually pretty simple.”When you know how.🙂

I’ll be poring over your tables and comments for weeks.

Probably nothing will help more than organizing your beat calculations in matrices.

Time to dig out my old engineering maths books and see if I can learn simultaneous equations and matrix dyadics again.

No need to tangle with the messy algebra. Just walk through it with numbers. It leads to the same conceptual understanding — actually probably

betterintuition.I’ll walk you through it.

Let’s focus only on the

corecomponents.Equation 1: (J+3S-U+N)/2 = (J+3S-(

U-N)) = φ^(-5)Equation 2: (J+S)/(φ/Φ) = (J-S+1/25)/2 = 3V-5E+2J

Equation 3:

U-N= ((5+√5)/2)^(-4) = (φ√5)^(-4) = (5(φ/Φ))^(-2)Equation 4: E = 1/1.0000174

Just note that without even knowing N, I can plug U-N into equation 1 from equation

~~4~~3. Then I’m just solving for the 2 unknowns J & S in the system of equations 1 & 2. Then with J & E from equation 4 it’s trivial to solve the tail end of equation 2 for V.With numbers it’s just a few quick beat calculations. The algebra above is

frequencyalgebra. Keep sharply in mind at all times that in the following beat calculations I’m working withperiodsto make it allmore intuitive…since people tend to think in periods (just how they’re wired).Start with The Jovian Balance:

φ^5 = 2/(J+3S-U+N) = 11.09016994

(φ^5)/2 = 1/(J+3S-U+N) = 11.09016994/2 = 5.545084972

(Review January 20, 2016 at 8:38 am to see where this comes from, noting that the “2/” there should be “4/” — that was a typo — going too fast because of neverending time stress — apologies for the impact.)

(φ√5)^4 = 1/(U-N) = 171.3525492

(171.3525492)*(5.545084972) / (171.3525492 + 5.545084972) = 5.371267117

(√5)^4 = 25

(Recognize from earlier √5 tables. Also see January 25, 2016 at 8:53 pm for

essential background.)(25)*(5.371267117) / (25 – 5.371267117) = 6.841077247

2*(6.841077247) = 13.68215449

13.68215449

φ= 1.618033989*(13.68215449) = 22.138191016.841077247

φ= 1.618033989*(6.841077247) = 11.0690955122.13819101 / (

φ/Φ) = (22.13819101) / 2.618033989 = 8.456036517 = 1/(J+S)(Bells should be going off now… Recognize from earlier (φ/Φ)/(J+S) tables & illustrations.)

(25)*(11.06909551) / (25 – 11.06909551) = 19.86428001 = 1/(

J-S)(19.86428001)*(8.456036517) / (19.86428001 – 8.456036517) = 14.72383345 = 2/S

(19.86428001)*(8.456036517) / ( (19.86428001 + 8.456036517) / 2 ) = 11.86237286 = 1/

J2*(14.72383345) = 29.4476669 = 1/

S(Remember those algebraically horrendous equations I gave for J & S in messy terms of φ? This is much easier to follow with numbers, isn’t it?!?! Same numbers, just expressed in a more familiar form. People (I’m thinking mostly of academics) like algebra because it’s generalized, but at times it becomes a cumbersome intuition-breaker! …but past that now I hope?…)

We’re not quite done. We have to deal with V, so review equation 4. We have what it equals:

22.13819101 = 1/(3V-5E+2J)

We also have E & J. We just need to chop Earth’s period by 5 and Jupiter’s period by 2 to beat a path to Venus’ period chopped by 3.

(1.0000174) / 5 = 0.20000348

(11.86237286) / 2 = 5.931186432

(22.13819101)*(0.20000348) / (22.13819101 + 0.20000348) = 0.198212763

(5.931186432)*(0.198212763) / (5.931186432 – 0.198212763) = 0.205065803

3*(0.205065803) = 0.615197409

If you double-check the validity of equations 1 & 2 with the quantities derived, you’ll find they pass.

Remember:

log_(

φ√5) (1/(U-N) = log_(3.618033989) (171.3525492) = 4.000243518Overlooking that essential clue would have dead-ended derivation.

I know

exactlyhow to design an algorithm that will pick out such keys. (Crudely put: just measure deviations from integer for logs of all dyadic harmonics of matrix elements across a log spectrum and look for systematic structures. Here I prototyped manually, but automation’s feasible.)the few loose ends outside the core…

3 more simple beat calculations finish deriving the beat matrix backbone.

The frequency algebra looks like this:

N = 1/50-1/(4φ^6)

Ma = (φ^3)-J (

Note that there’s a typo wherever I’ve written the frequency as φ^(-3). φ^(period–3) is the. (φ^3) is the frequency.)Me = √5+J

I gave the N derivation at January 30, 2016 at 6:34 pm.

(11.86237286)*(0.236067977) / (11.86237286 – 0.236067977) = 0.240861254

(11.86237286)*(2.236067977) / (11.86237286 + 2.236067977) = 1.881418832

That’s everything needed to flesh out the whole model beat matrix for comparison with the Seidelmann (1992) beat matrix.

The model isn’t stated most tersely as a formula for each system body, but rather as

a system of equationsthat precisely constrains all individual bodies and their beats. With nothing more than a terse statement of the model system of equations, all periods can be derived (and without need to resort to messy algebra, as I’ve clarified).One thing I’ll need to do is feature (J-S+1/25)/2 more prominently in graphics because it’s important. Same for (J+3S-U+N)/2 = (J+3S-(U-N))/2 = φ^(-5). (Note the typo (second “/2” missing) where I wrote “(J+3S-U+N)/2 = (J+3S-(U-N)) = φ^(-5)”. Hopefully people notice from context when time-stress is inevitably a never ending source of typos.)

Can you follow the numeric beat-period streams better than the cryptic frequency algebra Rog?

Despite what I’ve shown about 25 (e.g. January 25, 2016 at 8:53 pm) including it’s appearance as a factor in U-N, it’s appearance in the √5 power ladder, and it’s nearness to S-N, here’s yet more help with intuition about the 4th power of the square root of 5…

(φ/Φ)/(J+S) = 22.13847667

1/JEV = 1/(3V-5E+2J) = 22.13929985

(22.13929985)*(22.13847667) / (22.13929985 + 22.13847667) = 11.06944413

(22.13929985)*(22.13847667) / ( (22.13929985 + 22.13847667) / 2 ) = 22.13888825

(29.447498)*(11.862615) / (29.447498 – 11.862615) = 19.86503587

(19.86503587)*(11.06944413) / (19.86503587 – 11.06944413) =

25.00058109(25.00058109)*(19.86503587) / (25.00058109 – 19.86503587) = 96.70588381

(25.00058109)*(19.86503587) / (25.00058109 + 19.86503587) = 11.06944413

(25.00058109)*(19.86503587) / ( (25.00058109 + 19.86503587) / 2 ) = 22.13888825

(√5)^4 =

25 is the intermediary in the relationship between J-S & the solar cycle.Let 25 sink into permanent awareness. 25’s a core member of the constraining system of equations.

Maybe if you consider the estimate of 1/JEV = 1/(3V-5E+2J) = 22.13929985 to be sensitive to fast motion, consider 25 relative to just (φ/Φ)/(J+S) = 22.13847667 to further awareness.

(22.13847667) / 2 = 11.06923834

(19.86503587)*(11.06923834) / (19.86503587 – 11.06923834) =

24.9995314(24.9995314)*(19.86503587) / (24.9995314 – 19.86503587) = 96.72159311

(24.9995314)*(19.86503587) / (24.9995314 + 19.86503587) = 11.06923834

(24.9995314)*(19.86503587) / ( (24.9995314 + 19.86503587) / 2 ) = 22.13847667

25’s what matches J-S with (φ/Φ)/(J+S).To solve the system core you need the jovian balance (equation 1) and both sides (J+S & J-S) of the heavyweight splitter (the first 2 parts of equation 2 — note the plus sign between J & S in one part & the minus sign in the other part (the lightweight 3rd part (with E & V) just couples because it’s forced to)).

___

For comparison:

Something to stop and consider at least briefly…

Does anyone think the Titius-Bode “law” represents coupling? It pays

noattention to it whatsoever…and it gets called a “law” even though it’s errors are

orders of magnitudelarger: https://en.wikipedia.org/wiki/Titius%E2%80%93Bode_law#DataNever mind it’s beat matrix (

BIGerrors).triple-typo alert:

Under “Parsimony” on the hale core model & residuals summary should read:

=

Me = (φ^3)-J

Ma = √5+J

=

While reviewing the summary, please carefully consider the role of

(a) the core constraining equations,

(b) the boundary conditions, and

(c) the matrix approach

in the (orders of magnitude) performance differential with Titius-Bode “law”.

TB, let me know if you can walk through the simplified calculations I outlined at January 31, 2016 at 12:19 am. If you can follow that, you understand (at the conceptual level, even if perhaps not algebraically) how the matrix backbone is derived from the model equations.

The model system of equations

exactly and uniquely specifiesthe matrix backbone and consequently every one of its elements. By walking through the calculations you’re seeing that for those equations the matrix can’t turn out any other way. This is about realizing implications. Once the equations are specified, the matrix is specified.This is just an exercise in seeing how the equations

translateinto matrix elements. You’re doing thetranslationby walking through the calculations I outlined. Once done, you’ll be able to recognize theequivalenceof the statement in 2 languages:A. system of equations language

B. matrix full of numbers language

It’s the same thing, relayed in familiar versus alien mother tongues.

Have you succeeded following the step-by-step translation I outlined numerically?

typo at January 31, 2016 at 12:19 am

written:

“Just note that without even knowing N, I can plug U-N into equation 1 from equation

4.”intended:

“Just note that without even knowing N, I can plug U-N into equation 1 from equation

3.”Paul: Can you follow the numeric beat-period streams better than the cryptic frequency algebra Rog?

Yes! But I appreciate it’s time consuming for you. As an exercise I’m going to calculate out where you have pointed out typos as I correct them, so I understand what I’m correcting.

Should I edit the big graphic summary with image editing software and re-upload and link it here?

Some provocative precession questions for you guys to parse and dice with…

“We know the length of the cycle for precession of the equinoxes is

T2/(T2-T1) tropical years = 25771 tropical years

(Over this length of time, the number of tropical years will exceed the number of sidereal years by one.)

However a web site makes some interesting observations which appear to be true. Below is just a sample, but the overall impact is a stronger then expected case for a very radical idea. Below are just a few of the many interesting concepts presented. The actual web site http://www.binaryresearchinstitute.org/bri/research/introduction/theory.shtml has far greater detail.

==========================================================================

“Long term predictions of changes in the earth’s orientation to VLBI sources have been historically unreliable. The IAU has found that current methods are “not consistent with dynamical theory”.

Part of the problem appears to be that measurements of the precession observable are made to points outside the moving frame of the solar system yet do not account for motion of the solar system relative to those reference points. Another problem with current theory is the moon is thought to be the principal force acting upon the oblate earth. However, the moon is slowly receding from the earth (thereby theoretically producing less torque) whereas the precession rate is slowly speeding up (an indication of a greater force at work.

And of course the biggest failure of the current lunisolar theory is it makes no allowance for the different reference frames (a moving solar system versus fixed stars) and therefore requires that the earth change orientation relative to all objects, near and far, at the same rate. Such is not the case.”

(My comment here. This is fascinating. Apparently we use two very different methods of tracking objects outside our solar system, vs. inside, as precession does not appear to apply to objects within our solar system)

“The simplest way to produce the precession observable is the binary model. In this common stellar system, the observable of a moving equinox (or stars appearing to move relative to the equinox when viewed from earth on the date of the equinox) is simply the geometric effect of a solar system that curves through space as part of a binary system. As the solar system curves through space it gently changes the orientation of the Earth relative to the fixed stars (but not relative to objects within the moving solar system such as the sun or moon). This model requires little or no local wobble and is fully consistent with observations that show little or no precession relative to nearby objects and a full 50 arc seconds of precession relative to distant objects.”

(It also accounts for acceleration of precession. My comment.)

“It is the missing motion of the solar system curving through space that modern scientists have failed to calculate in their lunisolar precession theory. But the Moon does not lie. Its movement is exact and acts like a witness to the Earth’s motion. The only way the Sun can appear to move around the Earth, and be confirmed by lunar data, is because the Earth is spinning on its axis. Likewise, the only way the Earth’s axis can appear to precess or wobble relative to inertial space, and not wobble relative to the Sun as confirmed by lunar data, is if the solar system (the reference frame that contains the Sun and Earth) is curving through space. Furthermore, the only way the solar system can be curving through space at a rate of 50 arc seconds per year, is if it were gravitationally affected by another very large mass: a companion star.

I really think you guys will, at the very least, find this to be an interesting site.

One particular section of interest to you guys maybe the study of the orbital resonance of the outer dwarf planets, and their harmony with a 24,000 year precession cycle, and the observable steady increase in the rate of precession looks more like an orbit pattern following Kepler’s Laws than any local wobble phenomenon,

I think you guys will also find the work done on Sirius to be seriously interesting as well.

http://www.binaryresearchinstitute.org/srg/SiriusResearch.shtml

Hi David. Thanks for the comments, which might be better placed on the gravity/LOD thread. With a period of 24kyr, wouldn’t the hypothetical binary pair be a mere 3.5 light days apart at closest approach? Surely such a closeby object sufficiently massive to be capable of producing the Sun-barycentre radius required would be spotted by its occultation of other stars, even if it were a brown dwarf?

Tallbloke, thanks for the comment and please take the time to read the site. Yes, the not finding it yet is really the biggest criticism. However we have been finding more solar dwarf stars, possibly cooler then brown dwarfs, and not easily located depending on their vicinity.

The Sirius link is interesting, what with your sites study of orbital resonances and potential phi relationships to these observations, and this portion http://annesastronomynews.com/double-stars-mysterious-connection/ about forces linking binaries is likewise interesting.

Occam’s razor does apply in that many mysterious factors, including missing solar angular momentum, are solved by one simple explanation.

The part I find hard to dismiss concerns the acceleration of precession, and the non observation of precession to objects within the solar system. If it was a 24,000 year earth wobble, this would not be possible.

Sirius is over 8Lyr from the Solar system, or at least, that’s what current theory says. If that was our binary, the orbital period would be a lot more than 24kyr wouldn’t it?

David A: ‘the non observation of precession to objects within the solar system.’

How would such observation be done, i.e. is it a lack of observation or just a lack of attempts?

‘do not account for motion of the solar system relative to those reference points.’

Isn’t that begging the question?

Oldbrew,

PRECESSION MEASUREMENT PARADOX

Studies show that changes in earth’s orientation relative to objects “inside” the SS (i.e. Sun, Moon, Venus, etc.) are negligible (less than an arc second or two p/y), whereas changes in earth

orientation relative to objects “outside” the moving frame of the SS (fixed stars, quasars, etc.) are over 50˝p/y.

In fact, rotation time equivalence studies and lunar studies show the earth hardly

“precesses” at all relative to objects within the SS.Most astronomers acknowledge this

in practice by using a non-precessing tropical frame to locate objects “inside”

the SS, whereas they require a precessing sidereal frame (or T[J2000]+PxY) to find

objects “outside” the moving SS.

365.24219878 x 86400s = 366.24219878 x 86164.0905382s = 31,556,925.97s

This equation describes Earth’s complete 360° period of revolution of 31,556,925.97s relative

to a fixed frame of reference, implying that the position of the vernal equinox remains fixed

with respect to the orientation of Earth’s axis in space. The total number of rotations of Earth

in such a complete orbit is expressed by the equations:

1 ÷ (1- (86164.0905382 s ÷ 86400 s)) = 366.24219878

86400 s ÷ 235.9094618 s = 366.24219878

No precession with respect to the SS frame.

You may find the lunar witness portion of the site interesting as well.

Tallbloke says, “Sirius is over 8Lyr from the Solar system, or at least, that’s what current theory says. If that was our binary, the orbital period would be a lot more than 24kyr wouldn’t it?

=====================================================

My thinking initially ruled out such a possibility as well, nor is it central to BRI studies, yet my mind is not completely closed…

However, there are other possibilities, including the possibility that the solar system itself is moving much faster than any of the planets and therefore our companion star may be a nearby visible star.

While the solar system speed is difficult to measure (the question is always “compared to what?”), astrophysicist Reg Cahill of Australia has suggested that the solar system is moving in excess of 430km/s, relative to the cosmic microwave background (CMB), and this opens the possibility that our solar system might be orbiting anyone of a number of local stars, but more than likely one of the larger masses that lies not too far inclined to the plane of the solar system. Such a scenario might seem improbable given our current understanding of gravity and visible star distances, however there are compelling theories, such as MOND theory, and there is unusual evidentiary information, such as the data from Voyager 1 and 2 or the anomalous acceleration of the Pioneer 10 and 11 spacecraft that make this scenario attractive to investigation.

In addition to the odd behavior of distant satellites the most distant dwarf planets also display the signature of a potential companion star in their strange orbits. Mike Brown, Professor of Planetary Astronomy at Caltech, and the discoverer of Sedna, said simply:

“Sedna shouldn’t be there. …There’s no way to put Sedna where it is. It never comes close enough to be affected by the Sun, but it never goes far enough away from the Sun to be affected by other stars.” (Space.com, “Sun’s Nemesis…” 3-11-10)

Other astronomers, including John Matese, Emeritus Professor of Physics at the University of Louisiana at Lafayette and physicist Daniel P. Whitmire, of the same institution, and Richard Muller of the University of California Berkeley, have come to similar conclusions (based mostly on comet data), albeit without correlating orbit periodicity to the precession observable. Given such research we urge the scientific community to keep an open mind on possible forces that might be affecting the sun’s motion, and in turn the earth’s orientation.

Thanks David. Please move continuation of the discussion to the gravity/LOD thread so we can remain focussed on our phi work here. Repost my comment and your response there if you like and I’ll remove them here after you do that. Cheers.

Certainly.

The range of years on Timo Niroma’s ‘probability distribution of the sunspot lengths’ chart is a factor of 8/5:

8.7y x 8/5 = 13.92y

The difference is 13.92 – 8.7 = 5.22y

Taking the J+S value as 8.4561445y we get:

8.4561445y x phi = 13.682328y

The difference is 5.226184

So the same time gap, but with a start point at a slightly lower value (a time difference very close to one Mercury orbit or one Jupiter-Mercury conjunction).

Update 2/2/16:the ratio of 13.92:8.7:5.22 = 8:5:3I’ve been working on the Mars anomaly.

Again I’m reminded:

Math is a big world.

Have you ever started linking through wikipedia math pages exhaustively following all connections attempting thorough awareness extension? It’s a turbulently looping radial explosion.

I’ve just suffered another lightning rod.

φΦ/(φ+Φ) = 1/√5The Mars anomaly is a big clue.

Remember the equivalence criterion?

If anyone thought we’re done: we are not done.

Paul: φΦ/(φ+Φ) = 1/√5and way above:

Ma = (φ^(-3))-JMe = √5+J

Bearing in mind the Me eq.

(φΦ/φ)+Φ = 2Φ = √5-1

Not sure where it’s going, but the symmetry looks nice.🙂

You’re going to like this TB …but time-stress is murdering me. I appreciate your patience.

Remember: That’s a typo Rog:

“Ma = (φ^(-3))-JMe = √5+J”

Quoting the correction:

January 31, 2016 at 6:45 amtriple-typo alert:

Under “Parsimony” on the hale core model & residuals summary should read:

=

Me = (φ^3)-J

Ma = √5+J

=

There are 3 corrected typos (φ^3 not φ^(-3); Me not Ma; & Ma not Me (Me & Ma swapped position)) in that one short quote.

If you accidentally err in trying to edit in typo corrections, I may never notice. I did notice one error you made attempting a correction. It may be wise to leave the record unedited. I know I’m under too much time-stress. That’s why I wanted to delay running a new feature article. It will be a long time before there’s time to clean everything up, but the typo correction notes are being recorded.

I’ve been refining the matrix format I introduced above and with the refinement the intersection of φ & √5 at Mars crystallized further.

To first order √5-J ~= φ+J ~= Ma,

which means that 2(√5)φ/(√5-φ) ~= 1/J

and 2(√5)φ/(√5+φ) ~= 1/Ma.

Since frequency algebra’s not edible for everyone, number translations to ease digestion:

(2.236067977)*(1.618033989) / (2.236067977 – 1.618033989) = 5.854101966

(2.236067977)*(1.618033989) / (2.236067977 + 1.618033989) = 0.938748902

(2.236067977)*(1.618033989) / ( (2.236067977 + 1.618033989) / 2 ) = 1.877497804 = harmean

2*(5.854101966) = 11.70820393

Compare with Seidelmann (1992):

(2.235252542)*(1.623446114) / (2.235252542 – 1.623446114) = 5.9313075

(2.235252542)*(1.623446114) / (2.235252542 + 1.623446114) = 0.9404238

(2.235252542)*(1.623446114) / ( (2.235252542 + 1.623446114) / 2 ) = 1.8808476

2*(5.9313075) = 11.862615

This symmetry is across the matrix diagonal at the ± intersection of Ma & J. In matrix format the structure is absolutely systematic and it extends outwards in the matrix from that focal point.

The next task is to identify the balance that defines the residual (derivations analogous to those for JEV).

When all of this is done I’ll aim to illustrate the matrix with color coding that might help people see at least on some vague level where intuition originated.

I’m also pursuing verification of system boundary conditions via balancing symmetry of S+N & J-S with S-N & J+S and that exploration is converging nicely on tight coherence from an alternate perspective.

Landscheidt wrote:

‘The Golden section has left its mark, too, upon the 11-year sunspot cycle. Reliable data are available since 1750. They show that the ascending part of the cycle has a mean length of 4.3 years [73]. The mean cycle length amounts to 11.05 years. The minor of the mean length falls at 4.2 years (11.05 years × 0.382 = 4.22 years). This is close to 4.3 years. Thus, the maximum of the 11-year cycle falls at the minor of the Golden section. The descending wing of the cycle has the length of the major. This contributes to the stabilization of solar activity which is characterized by phenomena generated by instability.’

SOLAR ACTIVITY: A DOMINANT FACTOR IN CLIMATE DYNAMICS

by Dr Theodor Landscheidt

http://www.john-daly.com/solar/solar.htm

NB 4.2 or 4.3 years looks a lot like J+S/2 (8.456/2 = 4.228)

It has been years since I looked at Landscheidt. I recall that he considered J-S, but I don’t recall him looking at J+S. Certainly the sun’s behavior suggests it’s just sitting in a dynamical valley downslope from resonance, exactly where you’d expect it to be timing-wise regardless of physics. It’s such a no-brainer with the benefit of hindsight.

Landscheidt also wrote: ‘35.76 years × 0.618 = 22.1 years’

‘Saturn-Neptune conjunction / Hale cycle’ returns a figure around ~phi e.g.:

35.85455 / 22.1385 = 1.6195564

Fibonacci equivalent: 34/21 = 1.6190476

Note also the ratio of Saturn-Neptune to Saturn-Uranus conjunctions is very close to 1:√1.6 (8/5).

The S-N thing was clear early on. As mentioned above, I’m using it to verify the boundary conditions and there’s only 0.000256741% error.

It crossed my mind a week or 2 ago to go back and review Landscheidt’s mentions of golden section in light of growing recent insights. It’s on the long list of things to do that will take months to years.

PV: something else along the same lines…

(U-N / J-S) / (E-Ma / V-Me) = 1.59941 = ~8/5 (1.6)

Pluto-Eris / U-N = ~2.603 = ~13/5 (2.6)

Just like S+N doesn’t exactly align with 25 = (√5)^4, S-N doesn’t exactly align with 2φ^6. They’re not sufficiently free of overriding constraints to align exactly, so they’re free only to

symmetrically balanceas closely as possible to 25 & 2φ^6 …and this verifies the boundary conditions.Demonstrating this by extension of the simplified numerical derivations I outlined above for TB (January 31, 2016 at 12:19 am)…

Going from the Hale estimate:

22.13819101

φ= 35.820345512φ^6 = 35.88854382

(35.88854382)*(35.82034551) / ( (35.88854382 + 35.82034551) / 2 ) = 35.85441223 = harmonic mean

= estimate of S-N …which you can see has residual -0.000389041% on the summary

which ties back to the boundary condition exactly:

(35.85441223)*(29.4476669) / (35.85441223 – 29.4476669) = 164.7995564

So as you can see, Landscheidt was close (good instinct), but with his speculated relationship between S-N & Hale the model would drift out of alignment too fast to be tenable. This is analogous to noticing the near φ scaling of V+E, V, E, & V-E but needing constraints tying to J, S, U, & N to model it with sufficient precision.

For someone not seeing the constraints, the nearness to φ scaling must look tantalizing …but a φ scaling only model drifts out of phase

waytoo fast. It must drive people nuts looking at this and not being able to pin it down because they don’t see the constraints.I looked over a bunch of Landscheidt’s golden section notes and I do wish he had used phase instead of all of those little triangle symbols. What a mess.

…but Landscheidt’s right about φ being dynamically the most stable, antipolar to instability.

The sun’s pacing

naturally(it’s dynamicallydownhill) fallsAWAYfrom the high ridges of destructive resonance into the low valleys of stability.I find it astonishing that so many seem to think there’s resonance when there’s what you would expect naturally:

antiresonance.Resonance is the

unstable condition and whenever the system drifts towards it a bit, even the tiniest arising bit of resonance pushes the timing back AWAY FROM resonance.Resonance evasion is natural. It’s literally as natural as a ball rolling DOWNhill …and it’s ridiculous to think the ball would balance on sharp ridges of resonance; rather it would fall right off at the slightest provocation.

It’s not clear to me how all of these misconceptions about resonance entered the discussions years ago and persisted, but I suspect something different in our comparatively cross-disciplinary backgrounds. Back then I hadn’t yet extended my background far enough to know what was wrong with some of the speculation, but now it’s as clear as a bell.

φ indicates not resonance but rather antiresonance and destruction evasion. φ indicates stability. Resonance causes the opposite: instability. In hindsight it’s simple.

A few sentences of Mae-Wan Ho’s writing

instantlycorrected my φ vision and I once again thank OB for the link. Without that link the insights shared above mayneverhave arisen. I was ignorant of themost crucialpiece of information before Mae-Wan Ho’s sparkling communication talenteffortlesslytriggered conceptual correction.I’m leaning towards leaving the Mars anomaly in the model. At this stage it has educational value that should be exploited.

There is at least one more graphic I need to prepare (color-coded beat matrix). Then there’s the arduous process of iteratively refining presentation. I may never have time to do it right. I may not even do it. Lol. I’ll probably at least do the graphic. But with

otherthings to do, it seems like such a stupid perfectionist’s waste-of-time. Maybe it will happen whenever… and maybe not!!planet orbital period beat matrix:

suggestion: Independently build firsthand understanding of color-coding structure.

for comparison, here’s the hale core model beat matrix:

From the orbital matrix: (J+N / J+S) x 2 = 2.6172666 = ~89/34 (Fibonacci)

These numbers show what portion of the semi-major axis (SMA) of Saturn, Uranus and Neptune would fit into the SMA of Jupiter.

S = 0.5452

U = 0.2711

N = 0.17303

Sum = 0.98933 i.e. close to 1.

1/(J+S) = 8.456145629

(φφ/Φ)/(J+S) = 35.82080771

(35.82080771)*(8.456145629) / (35.82080771 – 8.456145629) = 11.06923834

(35.82080771)*(8.456145629) / (35.82080771 + 8.456145629) = 6.841165521

(35.82080771)*(8.456145629) / ( (35.82080771 + 8.456145629) / 2 ) = 13.68233104 = φ/(J+S)

2*(35.82080771)*(8.456145629) / (35.82080771 – 8.456145629) = 22.13847667 = (φ/Φ)/(J+S)

(22.13847667)*(13.68233104) / (22.13847667 – 13.68233104) = 35.82080771

(22.13847667)*(13.68233104) / (22.13847667 + 13.68233104) = 8.456145629

(22.13847667)*(13.68233104) / ( (22.13847667 + 13.68233104) / 2 ) = 16.91229126

It’s not 1/(S-N) like Landscheidt suggested.

1/(S-N) = 35.85455172

…but harmonizing with 2φ^6 approximates 1/(S-N):

(35.88854382)*(35.82080771) / ( (35.88854382 + 35.82080771) / 2 ) = 35.85464377 (0.000256741% off)

So this is just second order refinement of what Landscheidt noticed to first order.

I took a quick look at Sevin & Kotov and the first thing I notice is that 24*60 = 1440 = 2*2*2*2*2*3*3*5 and that 1440/9 = 160 where 9 = 3*3 and 160 = 2*2*2*2*2*5. Note that with daily sampling 1/2 = 0.5 and 1/5 = 0.2 but 1/9 = 0.111111111 so I’m naturally inclined to at least initially suspect a sampling &/or aggregation root of the 160min universality.

Something else entirely that I noted from the paper is something they call a “commensurability spectrum”. That concept is related to the nearness-of-logs-to-integers method I prototyped to solve the system to first order with the hale core model, which transforms (irrational) φ to (integer) 1. Whether or not something’s irrational depends on the scale you put it on.

A paper refuting Kotov claimed a harmonic atmospheric oscillation was responsible for his 160min solar observation. Kotov claims the period is slightly different to 1/9 of an Earth day. I don’t know the truth of either claim.

Sights along the √5 visualization trail:

These are Conway Triangles in a Pinwheel Tiling.

https://en.wikipedia.org/wiki/Pinwheel_tiling#The_Conway_tessellation

Tiling on the φ trail:

https://en.wikipedia.org/wiki/Penrose_tiling#Features_and_constructions

That sheds some light on

2cos(2π/5)=Φ=1/φ

clarifying that the 2 & 5 in 2π/5 come from reflection & rotation symmetry respectively.

Some of what’s outlined there will look familiar to OB.

Those are Robinson triangles.

Also familiar to OB:

https://en.wikipedia.org/wiki/Golden_triangle_%28mathematics%29#Golden_gnomon

The Conway Triangle can be put in a Conway Circle:

For breadth:

Aperiodic Tiling:

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

It gets interesting. There are connections with:

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

https://en.wikipedia.org/wiki/Markov's_inequality

Where this trail started:

https://en.wikipedia.org/wiki/Square_root_of_5#Geometry

The trail went to a bewildering array of other fascinating places, but focus is narrowing in on the linked subset. My instinct is that this material may bring a merger of solar system φ & √5 visualization within striking distance.

643/430 is very near the 4th root of 5. Whether that’s useful is another question.

The side-shoots and branches of the φ number-space are always fascinating. The fractal nature of these simple polygon tilings are a real delight. How do we connect the self-similarity scaling to our solar system quest? There’s a bewildering array of choices to investigate.

There are 8 significant planets, four small rocky and four large gassy.

How hard can it be? 😉

Focusing in…

The Pinwheel Tiling of Conway Triangles represents the scaling of the axial period when the equivalence criterion (difference = beat = geometric mean) is met:

φΦ/(φ+Φ) = 1/√5

“…infinitely many orientations…”“They are the first known non-periodic tilings to each have the property that their tiles appear in infinitely many orientations.”— https://en.wikipedia.org/wiki/Pinwheel_tiling#The_Conway_tessellationSo the √5 pinwheel is an

even more special caseof the following special case:“No tiling admitted by such a set of tiles can be periodic, simply because no single translation can leave the entire hierarchical structure invariant.”— https://en.wikipedia.org/wiki/Aperiodic_tiling#Aperiodic_hierarchical_tilings— —

OB asked:

“There are 8 significant planets, four small rocky and four large gassy.

How hard can it be?😉 “

Exactly. Understanding of the

simplepinwheel tiling matured only22 yearsago. The formal connection with the “undecidable problem” is enlightening.Perhaps this image symbolizes best why awareness of Hale geometry evaded human minds for so long:

Can you imagine that humans would design a city with such a layout?

Consider the surveying & building challenges for a city with:

“…infinitely many orientations…”And can you imagine how well (or perhaps not so well) some would fair navigating it?

“They are the first known non-periodic tilings to each have the property that their tiles appear in

infinitely manyorientations.” — https://en.wikipedia.org/wiki/Pinwheel_tiling#The_Conway_tessellationRealize that this is

an implication ofthe equivalence criteriondifference = φ-Φ = 1 = beat = φΦ / (φ-Φ) = 1 / 1 = 1 = geometric mean = √(φΦ) = √(1) = 1

because when it’s met the axial period

φΦ/(φ+Φ) = 1/√5

gives rise to

“…infinitely many orientations…”.

Are you starting to see?

Something to consider

in parallel:Human instinct is better with counting numbers.

Log base φ transform tamed irrational φ powers to tangible integers {-1,0,1}.

Were humans confronted with immersion in Conway Pinwheel-layout cities with

“…infinitely many orientations…”, what analogous simplifications would theylearn?Exploring such questions may enlighten us on the nature of Hale core recognition delays. As always this is about exploring

nature, including human nature.Paul:”Exploring such questions may enlighten us on the nature of Hale core recognition delays. As always this is about exploring nature, including human nature.”Well to be fair, you decided we should hold back on a new post while we completed the matrices and tied some more things together. What OB and I have discovered over the last few years, is that there’s always more in the phi number-space to be included, but you have to keep each why-phi post to a manageable subset, or it overwhelms the reader.

I’m loving this foray into tiling patterns. Geometry is something I can visualize and imaginatively transform more easily than algebra.

I’m not sure if this is relevant or of interest but the Conway triangle has sides of 1,2 and √5. If you take a triangle with sides at right angles of the inverse of those lengths i.e. 1/1, 1/2, then the hypotenuse is √((√5)/2). This is because √(5/4)=(√5)/2. This is also equal to φ-(1/2) because φ=(√5+1)/2

Conway developed the Leech lattice. I read a book a while ago that covered that and the further development of the ‘monster group’ and Monstrous Moonshine.

Can’t say I understood it all.🙂

Taster of the book here:

https://books.google.co.uk/books?id=wDjD0PowhIwC&redir_esc=y

‘φ=(√5+1)/2’ = arithmetic mean of √5 and 1

φ=(√5+1)/2 contains the values of the sides of the Conway triangle 1,2,√5

The Conway Triangle reshaped my perception of axial periods when the equivalence criterion is met.

1/(U-N) = φ^8+4φ^6+6φ^4+4φ^2+1

Note that the coefficients are from Pascal’s Triangle.

I get 171.352549 from Paul’s equation.

Note that the coefficients are from Pascal’s Triangle.Please identify the coefficients for us dummies. Are you referring to the powers of φ ?

φ^8 = 46.98

φ^6 = 17.94

φ^4 = 6.85

φ^2 = 2.62

I don’t see how these relate to Pascal’s triangle. They are each φ^2 bigger than the one below of course. And if you divide the powers by 2 you get 4,3,2,1 which does appear in Pascal’s triangle. Please walk us through so we can follow your train of thought. – Thanks

U-N = 5 x phi² measured in units of one Earth year.

Fourth root of U-N = ~phi² + 1 (3.6183~)

4N = 3U x phi² relative to each other.

The coefficients are 1, 4, 6, 4, and 1. Could it help to have the last term in the equation written out as 1*(phi^0)?

Thanks Sphene, 1,4,6,4,1 ; the 4th row of Pascal’s triangle. The left side of the equation was sitting in my algebra blindspot🙂

This is a constraint on what’s physically possible imposed by geometry:

It’s an

embeddedtiling.Suggestion for Open-Minded Luminaries:

Think through the perceptual implications in terms of group theory.

1/(U-N) = 1(φ/Φ)^4 + 4(φ/Φ)^3 + 6(φ/Φ)^2 + 4(φ/Φ)^1 + 1(φ/Φ)^0

Something to consider in conjunction with the beat matrix & Pascal’s Triangle:

(φ+Φ)^n

(That should be subtle enough.)

In 2503 years I’m seeing:

113 Hale

296 axial periods (J+S)

183 harmonic mean (296 – 113)

126 conjunctions (J-S)

211 Jupiter

85 Saturn

41 x 360 degrees of retrograde movement (126 – 85)

The difference defines unity.

The sum defines scaling.

Conway Triangle shows up in pentagon construction too:

https://upload.wikimedia.org/wikipedia/commons/2/29/Pentagon_construction.svg

And here’s a useful summary depicting φ,1,Φ,1 vs. 1,φ,1,Φ in Penrose tiling kites vs. darts:

__ __ __

Could there be something as revolutionary as Euler’s equation on the φ front? It’s a thought that has increasingly crossed my mind in recent weeks. As of yesterday I strongly suspect yes.

I’ve never before encountered a need or even an inclination to develop new algebras &/or types of numbers. This is interesting territory. Arriving at this juncture made me pause to realize how dependent our civilization has become on spoonfeeding.

It’s so bad! We’re conditioned to just assume that whatever’s needed (like types of numbers) is already known. Who ever stops to think to develop a new type of number? It’s a very specialized subset of the population.

For sure I have no experience with this, so I’ve found myself philosophically contemplating what kinds of questions do I even need to be asking myself? How would I know if 100 people have already independently proposed or even fully developed this?

For sure there won’t be time to go everywhere, so the best possible is to hope instinct goes the right way. I don’t think I’ll be defining the new type of numbers, but I won’t be surprised if someone else doing it either has had or is having some from the same series of revelations.

For example, look at that last graphic I put up the next day. You can even see the potential for a merger with Euler on square roots of negative differences …and then there’s a whole other axis (the scale axis) that has reciprocal symmetry on a log scale.

So is this something “beyond Euler?” Even if not, I can honestly say that I’m

pleasedthat thisopenedmy mind in this andseveralother ways (that I wouldn’t be able to state concisely if I tried).Certainly I no longer see it as the pentagon, but now rather the (φ+Φ)^2agon.

I just searched “(φ+Φ)^2agon” to see if anyone’s going by that handle.

Google responded:

“Including results for (φ+Φ)^wagon”Melbourne: Federation Square

http://www.panoramio.com/photo/18289243

The Conway triangle can contain four self-similar Conways instead of five.

The red lines represent the set of four, each has sides half that of the major triangle.

In this version the red-blue line is in phi proportion i.e. blue x phi = red.

The blue triangle is a Kepler triangle, ratio of sides = 1:√phi:phi

Where are you getting that last claim OB?

–

With the division into 4, keep in mind that you lose the extra-special property:

“…infinitely many orientations…”(Instead you’d have only 4.)PV: the hypotenuse is half of √5, the short blue side is (√5/2)/phi, so the ratio of the short side to to the hypotenuse is 1:phi, i.e. the definition of a Kepler triangle. That was my logic anyway.

Also: ‘the red-blue line is in phi proportion i.e. blue x phi = red’

The red part is the same length as the hypotenuse of the blue triangle.

UPDATE: the angle wouldn’t be 90 degrees, will re-work the diagram.

This might be it. The ratio of mid-blue to dark blue is phi:1

The vertical blue line is length 1, the same as the black line it meets.

So the mid-blue lines create a mirror-image of the original Conway triangle.

“This might be it.”If you work it through (more carefully) you’ll see that it’s not a Kepler Triangle. (We’re in the realm of geometric proofs so we can follow through to check initial hunches.).

Exploring though is good and in this case it causes me to even more deeply appreciate the extra-special property of the Pinwheel Tiling:

“…infinitely many orientations…”

–

There was an unfortunate misinterpretation above.

I wrote:

”Exploring such questions may enlighten us on the nature of Hale core recognition delays. As always this is about exploring nature, including human nature.”TB, it’s not a put-down of hosts &/or readers. This has been a respectful, clean thread (free of nasty climate discussion invective). It has been rather refreshing.

The comment’s not about whether people are getting it here & now. It’s musing about why something so simple wasn’t noticed decades ago.

The perspective:

The pinwheel tiling has only been known for 22 years. Looking at how simple it is, isn’t it a bit mind-boggling that it wasn’t known since ancient times? Our ignorance is beyond profound. Relatively we know nothing.

Hale & J+S define a tiling with

infinitely manyorientations.What currently commonly-used investigative methods (e.g. unwindowed Fourier) would detect them all?

None.

For example unwindowed Fourier would see noise.

It’s

perfectcamouflage.Is there some brilliant (simple) transform that brings (simple) crystal focus? I think it’s a question worth exploring. If there is such a transform, how many generations will pass before someone discovers it? Maybe it is known and used for stealth operations (e.g. to scatter radar) and therefore not publicized (since it gives a security edge) …and maybe not.

This Pinwheel Tiling is a mind-opener. It clarifies to me that relatively we’re seeing nothing of what there is.

“Penrose originally discovered the P1 tiling in this way, by decomposing a pentagon into six smaller pentagons (one half of a net of a dodecahedron) and five half-diamonds”https://upload.wikimedia.org/wikipedia/en/7/7f/Pentagon_with_half_dodecahedral_net.svg

For about a week I waited for a sufficiently well-rested opportunity of adequate duration to burst start-to-finish through an adequately focused check of that geometry.

It works out to:

(4/(φ(φ+Φ)))π^(φ/Φ) = 22.13821562

For me at this stage that raises more questions than it answers.

The question I was exploring:

What’s the outer:inner circumradius ratio?

With the 6 smaller pentagons having sides of length 1, the big pentagon has sides of length φ/Φ and the base of the filler white isosceles triangles has length Φ. (Keep in mind that φ/Φ = 1+1+Φ.)

Was Penrose exploring solar system geometry? I have no idea….

The easily-accessible Penrose awareness step here perhaps for anyone & everyone is recognizing:

P1 big pentagon perimeter

:P1 small pentagon perimeter=Hale:Jupiter + SaturnBeyond that fascinating questions arise about the

embeddedPinwheel Tiling andthat(not this) is what has mereallyfascinated because the implications ofinfinitely many orientationsareso far beyond delectablethat what can I say?……Just that it’s that type of

extremelyspecial case where reality and sweet dreamsare one.I still can’t believe that no one knew about the Pinwheel Tiling until 22 years ago. That’s just

crazy…How could somethingsosimple not even be known? You see?…PV: I remembered it now – it’s not a Kepler triangle, it’s a 3:4:5 Pythagorean triangle that can be drawn.

That’s because twice the smallest angle of the Conway triangle = the middle angle of the 3:4:5 triangle.

It can be confirmed by The Right-angled Triangles Calculator here:

http://www.cleavebooks.co.uk/scol/calrtri.htm

edge a = 3

edge b = 4

returns: 53.130 102 4 (angle B)

edge a = 1

edge b = 2

returns: 26.565 051 2 (angle A)

2 x 26.565 051 2 = 53.130 102 4

Back-to-back Conways gives us twice the small angle.

Pinwheel tiling reminds me of the math used to discover the shortest distance between 4 or more points. I believe bell labs under the old at&t did a lot of math on this. There was a very practical reason. 1st it saved a lot of money on installing the cable, and 2nd, they tax the circular mill on the copper. Applications can in one area be used in another. This is an excellent blog, though I haven’t contributed anything. The math on frequency and harmonics, and pinwheels especially under the old analog systems, probably wasn’t copyrighted, just remained proprietary.

Which also got me to start thinking about interference patterns, standing waves, feedback, and reflected impedance mismatches.

Implications of Pinwheel Tiling for Signal Processing

are profound. Fledgling human awareness at first Haleversary (like anniversary but 22): At age 1 do we collectively fathom the implied perceptual ignorance? Clearly not. This is quite a spectacle. The arrogance and the hubris of the ignorantly blind. What a joke it reveals us to have been thusfar. We’re just learning that we can’t see yet. The birth of our vision has not yet occurred. It certainly puts things into apriority-alteringperspective…Paul if I have offended you it was not meant to be. Truly excellent work. If I am ignorant and blind, I’ll go away. I was merely trying to add, not diminish.

No offense rishrac. I submitted the last comment before yours even appeared! Your appreciation of the Pinwheel Tiling is clear and I certainly welcome and appreciate such respectful commentary.

OB: Yes: 3:4:5. (You don’t need angles to figure that out — just “similar triangle theorem” or some-such-named thing I instinctively recall from grade 10 math, luckily with a really good teacher.)

Here are some helpful illustrations for deriving rhomb, kite, & dart proportions:

Those are from here:

http://jwilson.coe.uga.edu/emat6680fa05/schultz/penrose/penrose_main.html

Once 5 is perceived as (φ+Φ)^2 it can be seen to underpin

everythingwe’re discussing.Writing all this up is going to be a

monstrousprocess …especially given that because we’re on the right track the revelations keep pouring in faster than there’s time to go back and summarize. (a good “problem” to have! this certainly isn’t a complaint — it’s rather adescriptionof whatis.)Conway vs. 3:4:5

http://www.quadibloc.com/math/til04.htm

Let’s get these 3 up side-by-side to expedite efficient recognition of pentagon Φ,1,φ proportions:

When time permits I want to redraw the latter to more clearly emphasize the 1 by 2 rectangle containing 2 Conway Triangles, the diagonal of which together with the unit circle defines Φ+φ and thus the regular polygon of its square.

As reliably happens frequently now, something from an OB link clarifies something else (something

keyin this case) yet further.“Incidentally, it was Lorenzo A. Sadun who proposed that the pinwheel tiling could be generalized to other right triangles if one allows a tiling with different sizes of triangles in it […] In the diagram, the original 1:2:sqrt(5) triangle is shown, with a different triangle, the 3:4:5 triangle, below it. The two acute angles are still not integer parts of a circle; triangles whose angles are integer parts of a circle could also be chosen, to create triangle tilings that could be the basis for tilings with any desired order of symmetry.”Pinwheel Tiling was already a special case (infinitely many orientations) of a special case (hierarchical aperiodic).

Sadun’s generalization puts

Conway’sPinwheel Tiling (not to be confused with Sadun’s generalization of pinwheel tiling, which I might argueeliminates the pinwheel!!) in the perspective needed to make it clear that Conway’s Pinwheel Tiling is a special case of a special case of a special case (triply-special let’s say).Conway’sPinwheel Tiling defines the axial scaling when the equivalence criterion is met. This really is something on par with Euler and someone who’s able to communicate it and it’s broad-ranging implications formally is going to secure the very highest level of enduring distinction.The Conway Triangle both temporally scales and spatially axially references unit geometric beat differentials.

Sadun’s generalization takes the pin out of Conway’s pinwheel.

Just doodling…

See Conway triangle OMP above [PV: February 13, 2016 at 12:02 pm]

A line from point R parallel to AB (or OM) will bisect the vertical OP.

PR:RS = 1:phi

OP = OQ = RS

Update: it’s 2 parallelograms forming 8 Conway triangles.

Some random notes on this version of PV’s pentagon…

AEDZ is a parallelogram

OY:PY = 1:phi

OY:RS = 1:phi² (RS = OQ)

Paul: (4/(φ(φ+Φ)))π^(φ/Φ) = 22.13821562For me at this stage that raises more questions than it answers.

Doodling with numbers:

22.13821562/π = 7.046813

7.046813 * 3 +1 = 22.14043869

22.14043869/22.13821562 = 1.0001

simple illustration to help fathom 1/(U-N) algebra:

Now can you see the root of the letter φ?

__ __ __

Those are back-to-back Conway Triangles and this is the root of the pentagon construction.

1/(U-N) has been written many different ways during this thread. This unifies them.

That’s brilliantly simple!

The square roots of all of the first 5 fibonacci numbers are trivially easy to construct on Paul’s simply brilliant diagram.

Diameter 1 circle cuts Conway’s Hypotenuse into φ & Φ.

Like I said above:

I no longer think of it as √5, but rather φ+Φ.

pentagon construction:

PR = PA = PB = Φ

PS = PE = PC (the C with a dot, not the other c) = φ

RS = OQ = 1

The line PRMS can be continued outward from S by Φ to complete a pair of back-to-back Conway Triangles.

The image that got me thinking more clearly about all of this:

…but it bugged me that the arc of length Φ (from the bottom left corner, intersecting the diameter 1 circle at it’s bottom left) was not marked on the image.

That’s from here:

http://blog.world-mysteries.com/science/numbers-dont-lie-ancient-metrology/

Another graphic from there that will attract the curiosity of talkshoppers:

The bottom line here:

The φ√5 root of 1/(U-N) is no longer a mystery.

It’s just the scaling I’ve illustrated.

Some recap:

φΦ = 1

φ+1 = φ^2 = φ/Φ = 2.618033989

√5 = φ+Φ

φ√5 = φ(φ+Φ) = φ^2+1 = φ/Φ+1 = φ+1+1 = φ+2 = 3.618033989

(√5)^4 = 5^2 = 25

√√(1/(U-N)) = √√171.4062162 = 3.618317245 ~= φ√5 = 3.618033989

log_(φ√5) (1/(U-N) = log_(3.618033989) (171.4062162) = 4.000243518 ~= 4

(φ√5)^4 = (5(φ/Φ))^2 = 25(φ/Φ)^2

=

1φ^8 +4φ^6 +6φ^4 +4φ^2 +1φ^0=

1(φ/Φ)^4 +4(φ/Φ)^3 +6(φ/Φ)^2 +4(φ/Φ)^1 +1(φ/Φ)^0= 1/(U-N)

When I first saw φ√5 in the root of 1/(U-N) I didn’t recognize it as a

trivial scaling of the Conway Hypotenuse from Φ+1+Φ to 1+φ+1because I was not thinking of √5 as φ+Φ = Φ+1+Φ and so I was not seeing φ√5 as φ(φ+Φ) = 1+φ+1.Once I saw that I could reexpress the Pascal’s Triangle representation in terms of φ/Φ to knock the exponents (in terms of φ) down by a factor of 2, that’s when the lights came on at a higher level.

None of the insights that have arisen in this thread would have occurred (including the Hale Core Model) had I not noticed this early on:

√√(1/(U-N)) = √√171.4062162 = 3.618317245 ~= φ√5 = 3.618033989

log_(φ√5) (1/(U-N) = log_(3.618033989) (171.4062162) = 4.000243518 ~= 4

All of the other insights hinge on that pin (as in pinwheel).

In the second diagram at February 14, 2016 at 9:25 am:

If you make it a cube, there’s φ³ in the middle and two cubes of 1³ each.

φ³ – 2 = √5 = φ+Φ

The letter looks like it’s visual definition and fi phi φ five phive φve isn’t much of a stretch to suspect shared roots.

Note that we can classify on the criterion of whether

1fallsinside the circle oroutside.3.618033989² = 5 (φ/Φ)

With newfound clarity based on whether 1 falls inside or outside the circle I’ve identified another symmetry in the beat matrix. It’s not just x+N for x=J,S,U &

1+Φ+1, but also symmetrically |x-U| for x=J,S,N & φ+1+φ. That solves a riddle that lasted for too long.Was something simple overlooked? Recorded but lost? Destined for repeated rediscovery?

That’s a fundamental definition.

Sound rooms built 10′ * 16′ * 26′ suggest awareness in practice at least, but can anyone point to the theoretical definition anywhere? If this definition is not a staple of mainstream awareness, that could explain some of the controversy surrounding φ.

PV: SOUND 101: The “Golden Acoustic Ratio”

‘1 x 1.6 x 2.56 (famous golden ratio) – invented by the Greeks &

.62 x 1 x 1.62 – sound engineers use this.’

http://radiobombfm.wordpress.com/2012/09/15/sound-101-the-golden-acoustic-ratio/

Also: http://www.sweetwater.com/insync/golden-ratio/

“Here is perhaps the most interesting manifestation of phi in nature. It has to do with efficient packing. When you look straight down at a tree from above, the tree is most efficient if as many leaves as possible are visible and not shaded by other leaves. As a stem grows, it follows a genetic formula to know how often to produce a leaf and at what angle from the preceding leaf. If it produced each leaf at intervals such as 1/4 turn, every leaf would end up shaded by the fourth leaf above it. In fact,no matter what integer fraction of a full turn we use, we end up eventually repeating a pattern and shading leaves. So evolution eventually resulted in a more efficient genetic instruction: φ. Produce φ leaves per turn — just over 137.5° between successive leaves — and no two leaves will ever shade each other. This angle is called the golden angle.[…]

This tendency for ratios based on φ to eliminate repetition has engineering applications. One of the most familiar is in the design of sound rooms for listening to music or watching movies, rooms in which we want to

cancel out standing audio waves and resonances. Audio engineers refer to the golden room ratio, which establishes the basic ideal dimensions of a sound room to be 10 × 16 × 26. The height of the room 10 × φ ≈ 16, giving the length of the room, and 16 × φ ≈ 26 which gives the width of the room.Any diagonal straight-line path traveled inside a golden rectangle will reflect infinitely without ever repeating its course, and so sound waves inside a room of such dimensions are always dispersed as efficiently as possible.Despite many books and articles claiming otherwise, the exact history of man’s understanding of the golden ratio is not known.”— https://skeptoid.com/episodes/4325The author goes on to suggest:

“Today the concepts and relationships are well understood, and they are now common mathematical devices.”On the contrary, the Pinwheel Tiling was only discovered 22 years ago. This is rather unbelievable because of how incredibly simple it is, but simple things sometimes get overlooked.

Nowhereto date have I seen the simple axial scaling definition I’ve given above (February 20, 2016 at 10:14 am) for the conditions when the equivalence criterion (difference = φ-Φ = 1 = beat = φΦ / (φ-Φ) = 1 / 1 = 1 = geometric mean = √(φΦ) = √(1) = 1) is met.I challenge every human being on the planet to point to that definition somewhere.

It appears to be the overlooked key to more sound theoretical understanding & appreciation of φ. Much (or perhaps even most) φ controversy appears at least partially (& in some cases fully) rooted in ignorance of fundamentals.

Upon deep and careful study of the definition outlined, bright students & sound judges should fathom that this simple geometry has profound (perceptual) implications.

As provocation I’ve alluded to The Doors “Five to One” and “Break on Through”, inviting sufficiently adept minds to open the perceptual door circumcircled by φ. This is virgin territory, the flooding prime radiance of which may be too much for many (and perhaps most) mainstream palettes.

key thing to note:

in frequency:trivialproportioning of astraight line segment(axial frequency φ+Φ=√5).however:

in period: reciprocal cross-terms blow out another axis (1,1/(φ+Φ)=1/√5,2/(φ+Φ)=2/√5 triangle).

Re: φ+2 = 3.618033989 = φ√5

In the diagram PV posted (shown below):

PM/PR = √5 / (√5 – 1) = 3.618033989 / 2

PS/2 = (√5 + 1) / 2 = φ

(where OMP is a Conway triangle and OM = 1)

OB: Extend the line out from from S a distance PR. Then you can compare directly with the new illustrations I’ve posted (and you won’t need to divide by 2).

PV: I’ve been looking at that idea, with extended PS meeting a horizontal from D to make another Conway triangle.

I was also looking at this, where the 4 small Conway triangles (with 2 red sides and a black hypotenuse) can be used to fill the central red square. Therefore the red square is one fifth of the black square by area.

Of course there are also four large overlapping Conway triangles (each with 1 red hypotenuse and 2 black sides).

[…] post stems from comments on an earlier Talkshop thread e.g.: https://tallbloke.wordpress.com/2015/12/21/why-phi-an-orbital-parameters-test/comment-page-2/#commen… […]