A remarkable discovery: All Solar system periods fit the Fibonacci series and the Golden Ratio. Why Phi?

Posted: February 20, 2013 by tallbloke in Analysis, Astronomy, Astrophysics, climate, Cycles, data, Gravity, Natural Variation, Ocean dynamics, Solar physics, solar system dynamics, Tides
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Many other people have noticed Phi relationships in the solar system in the past, from Kepler onwards, and there are several websites which cover this interesting topic. But up until now, so far as I know,  no-one has been able to find a single simple scheme linking all the planets and the Sun into a harmonious whole system described by the basic Fibonacci series. A couple of weeks ago while I was on holiday, I had a few long ‘brainstorming sessions’ with Tim Cullen, and decided to roll my sleeves up and get the calculator hot to test my ideas. What I discovered is laid out below in the style of a simple ‘paper’. Encouraged by an opinion from a PhD astrophysicist that this is “a remarkable discovery”, I will be rewriting this for submission to a journal with the more speculative elements removed and some extra number theory added to give it a sporting chance of acceptance. For now, this post establishes the basics, but there is much more I have discovered, and I will be using some of that extra material in more posts soon.


Relations between the Fibonacci Series and Solar System Orbits

Roger Tattersall – February 13 2013


The linear recurrence equation: an = an-1 + an-2 with the starting conditions: a1 = a2 = 1 generates the familiar Fibonacci series: 1,1,2,3,5,8,13… This paper will use the first twenty terms of the sequence to demonstrate a close match between the Fibonacci series and the dynamic relationships between all the planets, and two dwarf planets in the Solar System. The average error across the twenty eight data points is demonstrated to be under 2.75%. The scientific implication of the result is discussed.


Since it was noticed that five synodic conjunctions occur as Earth orbits the Sun eight times while Venus orbits thirteen times, many attempts have been made to connect the Fibonacci series and it’s convergent ‘golden ratio’ of 1.618:1 to the structure of the solar system. Most of these attempts have concentrated on the radial distances or semi-major axes of the planet’s orbits, in the style of Bode’s Law, and have foundered in the inner solar system.

The present paper adopts a different approach, in order to simultaneously study the dynamic relations between planet pairs as signified by the frequencies of their synodic conjunctions in addition to their individual orbits. A static analysis of semi-major axes is inadequate to an understanding of a dynamic solar system in the same way that statically balancing a flywheel cannot reveal the out of balance forces which will cause vibration when it is rotated at high speed.


The highest number in the series used (6765) is allowed to stand for the number of orbits of the Sun made by Mercury, the innermost planet. The number of orbits made by the other planets and dwarf planets during the same time period of ~1630 years taken by Mercury to complete 6765 orbits are calculated. Additionally, the number of synodic conjunctions between adjacent planet pairs made in the same period is calculated using the method derived by Nicolaus Copernicus:

asteriod-corePeriod = 1/((1/faster orbit)-(1/slower orbit))

Additionally, the harmonic periods associated with the Power Spectral Density (PSD) study made of the sunspot number by talkshop contributor ‘Bart’ and used in the subsequent posting on Jupiter and Saturn’s relationship with the solar cycle and independently confirmed by  Scafetta 2012a[7] are included. The results are then compared to the descending values of the Fibonacci series and the deviations from the series calculated.

Juno is selected as representative of the Asteroid Belt as it lies near the middle of the main core at a distance of 2.67 AU. By Kepler’s third law this object has an orbital period of: P=(SQR)2.673=4.36yr.


Results are tabulated in table 1. The hypothetically vanished planet ‘Vulcan’ is shown in order to demonstrate the interesting phi relationships  which would have existed given its 2.67 year orbit.



This is a startling result. There is no currently accepted physical mechanism which can explain the clear and strong link between the Fibonacci sequence, the dynamic motion of the solar system, terrestrial cyclic phenomena at around 60 years and 205 years and solar activity levels. The underlying ratio is Phi, known as the golden section or ratio. This ratio does manifest itself elsewhere in nature. In plant biology, Phi is well known to appear in the spacing of leaf stems and the packing of seed heads. The leaf stem spacing maximises sunlight exposure and the seed packing maximises abundance[1]. In Geology, Phi relationships are evident in atomic, quasi-crystalline and other chemical structures[2].

Space has no crystalline structure. However it does have gravitational fields and electromagnetic fields permeating it. What kind of interaction of these fields with matter could bring about a situation whereby, approximately 4.5 billion years after the formation of the solar system, such close relationships to Phi are found to link every planet and two dwarf planets in the solar system? Evidently, harmonic and other periodic perturbations between planets and planet pairs have helped shape the system, and continue to maintain its internal relationships.

The average deviation from the Fibonacci series for the eight planets plus two dwarf planets orbits is 2.75%. This compares well with Bode’s Law which exhibits a 15% average deviation.  Solar activity cycles are represented by the inclusion of results from a Power Spectral Density (PSD) analysis which finds sunspot (SSN) activity peaks at 19.86 and 23.72 years, generating harmonics at 10.8 and 122 years[3]. This suggests that there is a link between planetary motion and solar activity levels.

Because the Sun’s gravity diminishes on an inverse square law, perturbation between Jovian planets will affect their orbits more strongly than the inner planets. Consequently, the Jovian planets excepting Saturn show a bigger deviation from the Fibonacci series than the three innermost planets.

It is suggested by Miles Mathis that inside Newton’s gravitational equation: F=GM1M2/r2 and Coulomb’s similar charge equation: F = kq1q2/r2  there is a unified field rather than two separate forces described by the two equations[4]. Mathis demonstrates that with a minimum of postulates, a fully mechanical ‘pool ball physics’ can be developed.  As well as providing a gravitational acceleration bringing extended bodies together, it also contains a repulsive electromagnetic force which although weak in our everyday experience, can become significant at the scale of astronomical bodies when they are in proximity. Importantly, the gravitational acceleration and the repulsive force scale differently as distance changes due to the different properties of the bodies they relate to.

This may explain why empty viable orbits are free of formation debris; the changing of the planetary orbits to create the most efficient order has over time traversed and swept the solar system clear of debris. The exception is the Asteroid Belt between Mars and Jupiter. Some evidence suggests its formation may be recent (3.2Ma).

Additionally, Mathis’ ‘foundational E/M field’ pervades space at varying densities (dependent on the proximity of emitting bodies), providing a ‘background’ against which the ratios of forces exerted by bodies will operate.  Mathis suggests that rather than trying to understand Phi in isolation, we can only appreciate the way that the two quantities which form the ratio can operate mechanically, by understanding the way in which they are relative to the ambient field in which they operate[5]. This is not a proposal for a ‘Universal Aether’, but for an interplanetary space which contains a density varying field of charge and spin bearing photons being constantly emitted and absorbed by matter. In a solar system with over 98% of the mass at the centre in a strongly magnetic star, and planets exhibiting Phi relationships in their orbital and synodic periods, this implies a relationship between Phi and the inverse square laws governing gravity and electro-magnetism.

phi-tendrilUPDATE 27-8-2013 Since writing this paper, Miles Mathis has demonstrated that Phi contains the inverse square within itself .[8] Consider boxes 2 and 3 in the Fibonacci spiral construction (Right). “The radius of the second box is r = 1/φ. The radius of the third box is r = 1/φ2. I would call that an inverse square relationship. If we then compare box four directly to box three, we get the same relationship, and so on. If we look at each box as a field component (or fractal) instead of as part of a series, we do have an inverse square fall-off.  We know that box 4 is to 3 as 3 is to 2, so if there is an square relationship between 3 and 2, there is a square relationship between them all. The reason we don’t find that inverse square relationship between box 1 and 2 is that box 1 is arbitrarily assigned the number 1. But our series is not based on the number 1, it is based on the number 0.618. That is why box 2 is our foundation, not box 1. This is also why we don’t find a square relationship between box 4 and 3, with the given numbers. The given numbers are written as functions of 1, not of .618. In other words, if we divide 1/φ2 by 1/φ3 , we don’t find a square. But again, that is because the series is not afunction of 1/φ2 . It is a function of 1/φ. So the only relationship that will directly tell us that the series IS based on the square law is the relationship between 3 and 2, as I showed.”


The logical conclusion is that feedback is present via perturbations between the planets and Sun which arranges the planets into an order which minimises work done, enhances stability and maximises entropy. This calls to mind the constructal law, stated by Bejan in 1996 as follows: “For a finite-size system to persist in time (to live), it must evolve in such a way that it provides easier access to the imposed currents that flow through it.”[6]

A true system contains cybernetic feedback. The Phi relationships demonstrated here are evidence that the solar system truly is a system in the full sense of the word.


[6] Bejan, Adrian (1997). “Advanced Engineering Thermodynamics,” (2nd ed.). New York: Wiley.


[7] Scafetta, Nicola (2012a) “Multi-scale harmonic model for solar and climate cyclical variation
throughout the Holocene based on Jupiter–Saturn tidal frequencies plus the
11-year solar dynamo cycle”  JASTP

[8] http://milesmathis.com/phi2.pdf

  1. oldbrew says:

    Mars and asteroids are in the next bit of the plan. Not sure about Mercury / Venus idea as we need a group of five in my theory, but it could just be about how we look at it. I’m seeing more and more that there can be different levels, some are becoming clear, others are possibly not even in view yet – unknown unknowns and all that.

    Must go now, the Mars update will be a bit later than advertised but it’s ready for print.

  2. oldbrew says:

    Had a slight re-think, so taking a time-out on the Mars question, if there is one.

    Let’s say Venus was once flying around the Sun with no Mercury in sight, then due to unknown forces it broke in two, as suggested. I can’t get at that Mathis paper due to the malware message unless there’s another way, but here’s the question. If the theory has it that a break-up led to one of the two new bodies – let’s say the ‘main’ one – slowing down and/or rotating the ‘wrong way’, does that imply Neptune/Uranus was a similar scenario? If so, did one lead to the other so the system would still be in balance? Or could it even have been that a 4-planet system was inherently unstable due to not being ‘Fibonacci compliant’.

    The other thought I had (well, one of them…) was: could the break-up of Venus/Mercury, if it did happen, in some way led to Mars becoming unstable and having its own break-up, shedding Ceres and the asteroid belt? OK, enough speculation for now, I could go on…and on…but no.

    Btw – from yesterday 3:55PM this needs a fix:
    ‘E:Mars = 19.11:26.11 = 1.372 (2-Phi) [actual is 1.3715]’

    It should have read ‘2-phi’, which is 1.382 – close enough? If not I’ll have to have another go at it.

  3. tallbloke says:

    OB: I don’t think we should speculate about Uranus on the basis of whatever Venus got up to. Different end of the solar system, very different energy/gravity regime.

  4. oldbrew says:

    Yes, but there is only one set of principles applying to the whole thing. I didn’t mean that N:U reacted to Venus having a bad day :-)

  5. tallbloke says:

    OB: Thresholds. These matter to matter.

  6. oldbrew says:

    You got me there. I’m not a physics expert, more computing with a bit of linguistics thrown in.

    Just looking at this, one line jumped out at me.

    Quote: ‘Jupiter’s magnetosphere extends seven million km towards the Sun’s and to the edge of Saturn’s orbit.’

    Funny how it just goes to the edge. Being a bit naive in such matters, I ask : how does it know? ;-)

    Also: ‘Jupiter’s magnetic field is thought to interact with nearly every body in the Solar System in some way.’

    Is that it? In some way ;-)

  7. oldbrew says:

    Turning to Saturn, Wikipedia says:

    ‘Saturn has a very hot interior, reaching 11,700 °C at the core, and the planet radiates 2.5 times more energy into space than it receives from the Sun. Most of this extra energy is generated by the Kelvin–Helmholtz mechanism of slow gravitational compression, but this alone may not be sufficient to explain Saturn’s heat production.’

    Interesting. Also:

    ‘Together, Jupiter and Saturn hold 92% of the total planetary mass in the Solar System’

    I’m no expert but that could be said to point in the general direction of our (or my?) ideas, or at least not eliminate them.

  8. Gray says:

    A couple of notes on the spiral equation. The equation gives a false though related sequence.

    Because Fibonacci starts 0,1,1, The first turn calculates Phi which essentially seems wrong.

    Numbers in the sequence are created by adding the last two numbers in the sequence 1+1=2, 1+2=3, 2+3=5, 3+5=8 etc.

    Multiplying the last number in the sequence by 1.618 creates the next number in the sequence
    5* 1.618033 = 8.090169

    However 2 * 1.618033 = 3.2360679775 If you use that product as the next term then the error grows. This tells us the Power needs to be the number of rotations.

    r = ae ^b

    1 * 1.618033988749895 = 1.61803398875
    2 * 1.618033988749895 = 3.2360679775
    3 * 1.618033988749895 = 4.85410196625
    5 * 1.618033988749895 = 8.09016994375
    8 * 1.618033988749895 = 12.94427191
    13 * 1.618033988749895 = 21.03444185375

    Thus the equaton is r = a, a single rotation, e, 1.61803398875 and b, the power, a count of the rotations

    This will produce a Phi spiral except for the problem at the start, 0, I can understand as 0, The first 1 produces 1.618033, the second 1 produces 1x Phi ^2 = 2.618033. 2 becomes 1 x Phi ^3 = 4.2360679775, which isn’t 3.

    I’m not sure what to make of this. Perhaps if 1 indicates either an x-axis or y- axis then area doesn’t exist until after the second 1 in the series.

    Still pondering this somewhat… The square root of Phi is 1.27201964951 this could also explain the 1s at the beginning ot the sequence and may throw up some suggestions.

  9. oldbrew says:

    Gray: don’t know if this helps but one theory is that it’s a logarithmic spiral advancing one quarter turn at a time, then Phi is invoked again. So it’s just a particular type of that process, if it’s true.


  10. J Martin says:

    The asteroid belt is surely part of this work you gents are valiantly undertaking.

    How did the solar system form anyway ?

    It must have coalesced out of a giant version of Saturn’s rings, but one that ringed the sun, with natural forces and patterns shaping, dictating the size composition and position of each planet. The hand of Phi, so to speak. If that were so then the layout of our solar system could prove to be a fairly common pattern in the Galaxy and Universe.

  11. Gray says:

    Hi Oldbrew

    Yep, that’s how it’s mapped. The other issue was the question of two 1s at the start. That’s solved if the square root of Phi is the first 1,

    If the first rotation of the Fibonacci Series is the square root of Phi the two axis are 1.272 and 1.618. What did you expect, Phi to use a gridded reference :)


    Phi-based -2.0022334732293 5.5848781529840

    So here’s a question, how are they Phi based?

    Second bit of info, tallbloke, did you see the two decay diagrams halfway down this page. Especially the first. Could those be your polarity channels as per the magnetic Sun.


    I also had a few issues with this page crashing my browser, I used the stop loading command this time and it worked.

  12. tchannon says:

    Moderator message.
    Complaints there is a problem with the website, PHI thread especially.

    I’ve removed three videos which were going into Abobe Flash loops, recursive calls to javascript, crashing.

    Anyone see an improvement?

    Comments about this will be DELETED LATER. –Tim

  13. oldbrew says:

    J Martin says:
    ‘The hand of Phi, so to speak. If that were so then the layout of our solar system could prove to be a fairly common pattern in the Galaxy and Universe’

    I had a similar thought. What I’m trying to imagine is a basic formula for the structure. It would give a number of variations, mutations even, but within the system parameters.

  14. Gray says:

    Tim, page seems to be behaving better. Thanks

  15. tallbloke says:

    Thanks Tim, I’ve restored the comments with the youtube url’s doctored and added instruction on how to fix them.

    Gray: Apologies, too much on tonight. Will study your spiral comments tomorrow.

  16. oldbrew says:

    Ninderthana does some interesting calcs and discusses synodic periods, harmonics and the gas giants’ role in solar cycles here.


  17. oldbrew says:

    Gray says: ‘Phi-based -2.0022334732293 5.5848781529840

    So here’s a question, how are they Phi based?’

    Like this?

    We could ask: is it valid to say 2xPhi or Phi^2? We wouldn’t say 2 x ‘1001’(binary digits) for example.

  18. oldbrew says:

    tallbloke says:
    March 3, 2013 at 2:14 pm

    ‘Thought: could it be that the planets with strong magnetospheres are being driven round by the right hand rule and the planets that don’t are being tugged along by the magnetospheric planet’s gravitational perturbations acting on them?’

    Check this out – the Phi symbol is near the bottom to the right. What’s that about?



    Going back a bit…

    ‘OB: Jupiter is Phi, Saturn is phi, the other side of the triangle is 1 = the Phi:1:phi triangle.

    Phi is the inverse of phi, they are not in a Phi ratio to each other’

    Correct, should say: ‘Jupiter is Phi, Saturn is 1, the other side of the triangle is sqrt(Phi)’

  19. oldbrew says:

    I said (Feb 28): ‘(Earth:Mars orbital ratio 1:1.881) ^4 = 13 (1.889^4)

    Where does the fourth power fit in?’

    It doesn’t. A better derivation is:
    Phi + (sqrt(Phi) -1) = 1.618 + (0.2721-1) = 1.89 = the non-squared orbital period ratio

    The squared orbital period ratio of Earth:Mars orbits is 1:1.37
    The corresponding figure for Uranus:Neptune is 1:1.40

    I’m seeing the same Phi ratio there. TB noted earlier:

    ‘To visualise this we could try thinking about the numerical relationships between two sphere’s surrounding the gravitational mass with radii of 1 and 1.37, because at those distance the period of the orbits will be in the relation 1:Phi’

    And there it is.

  20. tallbloke says:

    The Phi symbol is used in physics for the angle between the force and the position vector in angular momentum problems. e.g. http://faculty.ycp.edu/~jforesma/educ/pchem/lec09.pdf

  21. oldbrew says:

    It’s the symbol for magnetic flux.
    [scroll down to ‘Magnetic Flux’]

    There’s the answer to the ‘why Phi’ question IMO. The rotating planets are magnetically driven, so the Phi relationships are showing up everywhere. Magnetic charge is all around in the solar system.

    Even at the outer edges there’s a ‘magnetic highway’.

  22. tallbloke says:

    Oldbrew: Venus – ‘sideways axis’

    Can you explain?

    The Phi symbol used in AM calcs and E/M calcs is interesting. I’m going to try to find out more about why Phi was used for this. I think it unwise to jump to conclusions about physical properties on the basis of choice of symbols, some are very arbitrary. In this case, some bright spark may simply have decided that the Phi symbol looks like a bar magnet with flux lines running between the poles.

  23. oldbrew says:

    Saturn/Venus, Earth/Mars, Uranus/Neptune: all are pairs.
    There are 2 planets with ‘sideways’ axes, Pluto and Venus, one on each side of the S/J axis.

    Mercury is iron rich and Venus is iron poor. That could be due to its proximity to its giant neighbour, the solar magnet. The axial tilt of Mercury is virtually zero, unlike the other (less iron rich) planets. There’s a strong chance IMO that axial tilt is a function of iron content and its distribution within the planet i.e magnetism-related.

    That lists two other planets with low axial tilt values: the Sun(7) and Jupiter(3), the rest being 23 or more, plus dwarf Ceres(4) and the Moon(1.5). The Moon is known to have abundant iron.

  24. oldbrew says:

    Sorry (again!), should be ‘upside down’ for Venus in that style of description. But that’s a matter of opinion: what it is really is reversed magnetic polarity. That’s why it spins the ‘wrong’ way.

    If anyone is in doubt about the mechanism, see ‘Reversing the field’ in these GCSE notes.

    TB: I understand your caution so I won’t push it too much, but I think we’re going in the right direction. I firmly believe certain things but everyone else probably does too, so we have to accumulate evidence of course.

  25. oldbrew says:

    This might be of interest. I think with an electro-magnetism approach it all falls into place pretty well independent of the Phi/Fibonacci aspect.

    I can see how it resolves (for me at least) the gaps either side of Jupiter i.e. magnetic repulsion. It may be the real reason Saturn outputs more energy than it receives from the Sun. And so on.

  26. Gray says:

    Hi OldBrew

    Mercury is 1.272 Proton
    Venus is 1.618 = Electron
    The Neutron is 1

    If you have a logarithmic/Phi relation with the quantum number it’s probably correct.

    It most definitely does fit with Fibonacci…


    If you build a a triangle with sides 1.272, 1.618 and 2, next
    bisect the triangle and you get.a length 1.

    Thus the three strands 1, 1.272, and 1.618 create 2

    Thereafter the three strands create all of the Solar System harmonics through Fibonacci.

    The next 1,2,3 are a straight line creating 3.

    The Square Roots are 1 =1, 2=1.414, 3 =1.732
    The Squares are 1 =1, 2= 4, 3 =9
    The product of the Squares should also be added 1+4 =5, 4+9 =13, 1,9 =10

    The next are 2,3,5 a straight line creating 5.

    The Square roots are 2=1.414, 3 =1.73, 5 = 2.236
    The Squares are 2 =4, 3 =9, 5 =25
    Added 4+9 =13, 9+25 =34, 4-25 = 29

    The next are 3,4,5 a triangle
    The Square Roots are 3= 1.732, 4=2, 5= 2.236
    The Squares are 3=9, 4=16, 5=25
    The products of the squares become 9+16=25, 16=25=41, 9+25=34

    And so on

    I can’t work on this tomorrow, long work day. Err, welcome to the atom.

  27. Gray says:

    The next in the series is 3,5,8

    The Square Roots are 3=1.732, 5=2.236, 8=2.828
    The Squares are 3=9, 5=25, 8=64
    The products of the squares become 9+25= 34, 25+64=89, 9+64=73

    The next 5,8,13
    The Square Roots are 5=2.236, 8=2.828, 13=3.605
    The Squares are 5=25, 8=64, 13=169
    Products of the squares 25+64=89, 64+169=233 169+25=194

    The next 8,13, 21

    The Square Roots are 8=2.828, 13=3.605 21=4.582
    The Squares are 8=64, 13=169, 21=441
    Products of the squares 64+169=233, 169+441=610, 441+64=505

    5,12,13 a 2nd Pythagorean Triple Branch

    The Square Roots are 5=2.236, 12=3.464, 13=3.605
    The Squares 5=25, 12=144, 13=169
    Products of the squares 144+25=169, 169+144=313, 169+25=194

  28. oldbrew says:

    Gray: that’s the kind of thinking we need, the question of course being – how does all that relate to the solar system data? We’ve at least started trying to do that and got some promising results. I think there could be more Kepler triangles ahead…

    Recap here – as a reference point (first attempt):

    Phi – phi = 1
    Phi + phi = Phi^2 – phi^2 = sqrt(5) = 2.236
    Phi = phi+1 = 1/phi
    phi = Phi – 1 = 1/Phi

    Phi^2 = Phi/phi = Phi + 1 = 2.618
    Sqrt(Phi) = 1.272

    phi^2 = phi/Phi = 1 – phi = 0.382
    sqrt(phi) = 0.786

    Phi^2 + phi^2 = 3 = [A]
    2.236^2 = 5 = [B]
    [B] – [A] = 2

    NB: A and B have no meaning, just notation.
    Any Fibonacci number above 5 can be derived from the above, e.g. 5+3 = [A]+[B] = 8
    The ratios of the Kepler triangle are 1: sqrt(Phi): Phi

    So, re-ordering your K triangle:

    The Neutron is 1
    Mercury is 1.272 = Proton
    Venus is 1.618 = Electron

    The above should be re-checked for bugs, typos, missing info. etc.

  29. oldbrew says:

    TB: ‘I think it unwise to jump to conclusions about physical properties on the basis of choice of symbols’

    True. Getting some useful data from the spin rates of the planets though (rotation period in Earth days), fastest first:

    J:S = 0.41:0.44, U:N = 0.72 (both), E:Ma = 1:1.03, P = 6.38, Me = 58.65, V = 243

    The two most massive planets are spinning nearly twice as fast as the next two, and a lot faster than all the others. Interesting ;-)

    Also, the planets I’ve paired in the list have moons as follows:
    J:S = 63:62 (1:1), U:N = 27:13 (2:1), E:Ma (1:2).

    Then there’s magnetic field strength. If Earth = 1, Jupiter = 19000, Saturn = 1000.
    Putting that together with the fast spin rates of J and S, what have we got?

    Footnote: as well as Saturn’s rings (275,000 km. rim to rim including the gaps), Uranus has 11 and Neptune at least 2. One of Saturn’s is described as ‘twisted’ and braided’. What is keeping them in place? This perhaps…

  30. oldbrew says:

    The Saturn paper referred to at 12:43pm says:

    ‘During both encounters of the ‘Voyagers’ with Saturn planetary radio-astronomy, the experiment (PRA) has shown fixed, non-polarized, very broadband radio radiation, through all observable ranges of the experiment (20,4 kHz-40,2 MHz). These incidental radio discharges are called Saturn’s electrostatic charges (SED). The average period of SED was determined by Voyager I and II to be 10 hour 10+-5 min’

    The mean rotation period of Jupiter/Saturn is ((0.41+0.44)/2) days x 24 = 10.2 hours = 10h. 12min.
    Again, interesting.

  31. oldbrew says:

    Gray says:

    ‘A constant: Kepler’s Third Law: The square of the period of a planet’s orbit is proportional to the
    cube of its semimajor axis. Can this be re-jigged in some way to describe your constant, possibly with the inclusion of Phi.’

    Is this any good…

    The ratios of the squares of the Kepler triangle are 1: Phi : Phi^2

    So: 1 + Phi = Phi^2, meaning 1 + 1.618 = 2.618
    The sum of the 2 largest sides = 4.236 = Phi^3

    Ulric’s constant (= 1) could be the third side.

  32. Ulric Lyons says:

    Fee Phi fo

    (Phi^5+Phi^8)/Phi^7 = 2
    (Phi^5+Phi^13)/Phi^9 = 7
    (Phi^5+Phi^21)/Phi^13 = 47
    (Phi^5+Phi^13+Phi^21)/Phi^13 = 48
    (Phi^13+Phi^21)/Phi^17 = 7

  33. Gray says:


    The Parker Spiral contains twisted +/- flux tubes, I think I’m right in saying. The Parker Spiral is regular and a straight line to the Sun. Embedded in the Parker Spiral, I assume then, are harmonics matching the Fibonacci sequence.If you like, the tonal note of the Sun and all its harmonics.

    It would be interesting to know what permutations of Phi exist including the Mercury orbit and Venus orbits. Especially at the 1, 1.272, 1.618 ratios. I am working again tomorrow, another long day, so I expect to pop by in the evening.

  34. tallbloke says:

    Gray: The curvature of the Parker Spiral varies with the solar wind speed/density. I think this fact is key to the whole solar system feedback issue. I have a link to a paper on the equations which I’ll scoop off the other PC tomorrow for you.

  35. Gray says:

    Ulric and Oldbrew.

    On the formula question I can only answer you may be right at the moment. I have generated triangles at various dimensions to aid in the process. I need to compare what you have written with an actual triangle and work out where in the sequence it fits. From this we should be able to predict which planetary dimensions are referred to by which Fibonacci plot.

  36. Gray says:

    Hi tallbloke,

    Think of the Sun as 1 the Neutron pulsing as a ‘constant’, It creates An Archimedes Spiral
    1.272 is another Fibonacci Spiral
    1.618 is the other Fibonacci Spiral

    The three create the interference pattern known as Phi, and extended Phi. So varying windspeed may be created this way by jumping from one state to another. It’s a possibility.

    I’m sorry, I’m extremely tired at the moment as I’m working and trying to keep tabs on this simultaneously.

  37. Ulric Lyons says:

    tallbloke, study the magnetic connections on the “Top View Ecliptic-Plane Field Plot” model here:

  38. […] l’ho trovata su Tallbloke, blog climatico spesso ricco di spunti interessanti. So che piacerà a molti lettori di CM che […]

  39. oldbrew says:

    Using basic calcs based on the distances between planets, from the Sun to Eris inclusive, and counting Jupiter as equivalent to 2 planets (due to the ‘missing’ planet between J and Mars – see below), I’m getting a figure of around 1.68 for the average ratio of planet:planet distances.

    The lowest is Mercury:Venus at 1.39, the highest excluding J is Saturn:Uranus at 2.01. Note here the mass of Venus is 14.7 times that of Mars, and Saturn is 6.6 times that of Uranus, which could be a factor in the ratios, i.e. forces deriving from relative mass.

    The J:Mars figure is an anomaly at 3.412, roughly double the average. Given the facts that Jupiter has nearly 3000 times the mass of Mars, and there is an asteroid belt including dwarf planet Ceres between J and Mars, it seems a plausible inference to say 3.412 represents 2 planet’s worth of data (mean = 1.706), for the purpose of obtaining the final figure.

    Comparing 1.68 to Phi (1.618), the discrepancy is under 4%. A possible explanation for the variation might be the force of the solar wind, and/or minor errors in the calcs.

    Gray says: ‘It would be interesting to know what permutations of Phi exist including the Mercury orbit and Venus orbits. Especially at the 1, 1.272, 1.618 ratios.’

    See here and the post below it.

  40. […] oldbrew on A remarkable discovery: All So… […]

  41. oldbrew says:

    oldbrew says: March 7, 2013 at 10:05 am
    ‘Comparing 1.68 to Phi (1.618), the discrepancy is under 4%. A possible explanation for the variation might be the force of the solar wind, and/or minor errors in the calcs’

    Further analysis points to Jupiter as the main suspect for the small discrepancy. The inner 4 planets average to 1.625 on my basic figures, with the outer 4 (Uranus, Neptune, Pluto, Eris) coming in at 1.63. The ‘Jupiter group’ I’ve estimated at 1.75 allowing, as stated earlier, for the missing planet where Ceres (‘dwarf planet) is.

    NB Eris is bigger than Pluto, but is in a clear harmonic relationship with it according to this…

  42. If…
    The “time” is astronomy…
    the astronomy is fibonacci…
    fractals, golden sections, spirals

    What is the “time” ?

  43. oldbrew says:

    What is the “time” ?

    Phi o’clock :-)

  44. oldbrew says:

    @ Michele Casati

    We all know the time/distance formula that gives the year i.e. one Earth orbit of the sun.

    For a possible Fibonacci / golden section link, see here:

  45. oldbrew says:

    Photo here of Saturn’s north polar hexagon.

    Hexagons have a known Fibonacci link. The six angle points (ABCDEF) can be connected to form 3 identical rectangles: ABDE, BCEF, CDFA. The long and short sides of the rectangles are always in the golden ratio of 1.618:1, so this Saturn phenomenon could well be a manifestation of Phi i.e. 1.618.

    Perspective on the size of the hexagon: NASA says ‘The view was acquired at a distance of approximately 403,000 miles (649,000 kilometers) from Saturn.’

  46. oldbrew says:

    Gray says:

    ‘A constant: Kepler’s Third Law: The square of the period of a planet’s orbit is proportional to the
    cube of its semimajor axis. Can this be re-jigged in some way to describe your constant, possibly with the inclusion of Phi.’

    See here.

    Pi = 1.2xPhi

  47. tallbloke says:

    OB: Pi = 1.2xPhi

    Errr, no.

    Pi=4/SQRT phi to within 0.1% though.

  48. oldbrew says:

    Thanks TB, well spotted. The power of 2 got left behind: should read 1.2xPhi^2 (as per link).

    The discrepancy with Pi is about 0.000015.

  49. tallbloke says:

    OB: OK, so is there anything which makes 1.2 of especial interest?

  50. oldbrew says:

    Good question, but sadly no answer at present. Do you or anyone you have contacts with, have a knowledge of sine waves at all? I’m fishing for info there.

    I’ll record what I found about a feature of Saturn’s rings. Various sites note what is described here:

    ‘Saturn’s outermost ring, the F-ring, is a complex structure made up of several smaller rings along which “knots” are visible. Scientists speculate that the knots may be clumps of ring material, or mini moons. The strange braided appearance visible in the Voyager 1 images (right) is not seen in the Voyager 2 images perhaps because Voyager 2 imaged regions where the component rings are roughly parallel. They are prominent in the Cassini images which also show some as yet unexplained wispy spiral structures.’

    Knots…strange braided…unexplained wispy spiral structures – mystery or not? Look at this.

    Under ‘Characteristics’ we find:
    ‘Birkeland currents are also one of a class of plasma phenomena called a z-pinch, so named because the azimuthal magnetic fields produced by the current pinches the current into a filamentary cable. This can also twist, producing a helical pinch that spirals like a twisted or braided rope, and this most closely corresponds to a Birkeland current.’

    Proof enough that magnetism is alive and well in Saturn’s rings?

  51. oldbrew says:

    TB: ‘is there anything which makes 1.2 of especial interest’

    Take a Pythagoras triangle whose sides are 2:sqrt(5):3 i.e. 2:2.23606~:3.
    Squared nos. for that are 4:5:9
    (2×3) / 5 = 1.2 (short side x long side / medium side^2)


  52. tallbloke says:

    Hi OB:
    Nice. Now consider that Phi itself is equal to (1+SQRT5)/2 and Pi=4/SQRT Phi (<0.1%)

    Can you rework those equations to find out why it all fits together?
    I think it'll have something to so with Phi^2 being the same as Phi+1

  53. oldbrew says:

    I’m looking at it but my brain hurts, thought I’d nailed it but it didn’t quite come out right. Try again.
    There may be a way to derive a specific Pythagoras triangle based on Phi.

    NB the ‘^2’ should be inside the bracket i.e. ‘medium side^2)’

    [Reply] Fixed

  54. oldbrew says:

    OK, two equations to test. (Note: 2.618~ = Phi + 1 = Phi^2).

    #1 – the circle: diameter is d, radius is r (2r=d).
    Let d = r1 + r2 where r1 : r2 = 2.618:1 – i.e. split the diameter into two related values.

    k = r / sqrt(5)
    r1 = r + k
    r2 = r – k

    Golden ratio formula for d = 2 x r1 : 2 x r2 which is as shown…[(2 x (r + k)) : (2 x (r – k))]

    #2 – Pythagoras triangle (Short, Medium, Long sides): notation as above i.e. r1 : r2 = 2.618:1.

    S = 2 x (r1 – r2) = sqrt(L^2 – M^2)
    M = r1 + r2 = sqrt(L^2 – S^2)
    L = 3 x (r1 – r2) = sqrt(M^2 – S^2)

    Test with two Fibonacci numbers. To get 2.618:1 ratio, select alternate F numbers e.g. 55 and 21 (larger number = r1 in the equations, smaller = r2).

    Hopefully no typos, omissions or other gremlins.

    [Reply] “L = 3 x (r1 – r2) = sqrt(M^2 – S^2).” shouldn’t this be sqrt(M^2+S^2)?

  55. oldbrew says:

    Correct, just checking who was paying attention ;-)

    In the triangle, imagine a line drawn from the angle of the 2 shorter sides (S,M) that intersects side L at right angles at point X. Call the two parts of L thus created: L1 (longer part) and L2.

    Then: sqrt(L1) / sqrt(L2) = 1.2

    That’s the result I got anyway.

  56. tallbloke says:

    OB: In the triangle, imagine a line drawn from the angle of the 2 shorter sides (S,M) that intersects side L at right angles at point X. Call the two parts of L thus created: L1 (longer part) and L2.

    Then: sqrt(L1) / sqrt(L2) = 1.2

    That’s the result I got anyway.

    Cool. So L1/L2 will be 1.2^2 = 1.44

    Where does that get us?

  57. Gray says:

    hi tallbloke, I have some illustrations that I’ve done, how can I upload them. I’ll try to sort a web link for the time being.

    [reply] paste the image URLS and I will turn them into inline images in your comment. Better still how about a guest post?

  58. Gray says:

    Hi Moderator, I cannot get the images online as per normal, so no URL. Can I email the page and then maybe consider whether it makes a guest post.

    [Reply] Gray, my incoming email server is down today. I’ll see if Tim Channon can receive and upload them here. If so, he’ll mail you at the address you use here. Cheers, Rog.

  59. oldbrew says:

    TB: it means you can build a Kepler triangle if you have the info for one side only, as long as you know which side it is. Once the other two sides are known, matches for them can be investigated. Plus that triangle will have the characteristics already described, which could help in spotting potential new ones (e.g ‘short side:long side’ is always 2:3).

    Anything else? Well, if it’s the diameter or radius of a known circle (e.g. of a planet), that would (perhaps) point to it being the ‘medium’ side. See these comments and the Wikipedia link on the Bulliades thread: a ‘golden ratio’ ellipse has a diameter that splits 1.618:1.
    [oldbrew says: March 8, 2013 at 1:13 pm and 5:37pm]

    In the Wikipedia animation, one might ask what the relevance of the outer circle is to all this. If it’s a sort of parameter within which a planetary orbit has to operate, how can that value be obtained? It would presumably be something to do with the distance between the orbits of neighbouring planets and/or distance from the Sun? Maybe there are some stats around that could be used.

    Just for amusement, Wikpedia says the SI units formula for Newton’s gravitational constant has a ‘relative standard uncertainty 1.2×10−4’. There’s that 1.2 again ;-)

  60. Gray says:

    Hi tallbloke the illustrations are at this link:


    I’ll let Tim know so that he can embed the images as required.

  61. tallbloke says:

    Thanks Gray, some more text needed for a post I think. I can probably put it together from discussion here if I can get you to say some more about the plots.

  62. Gray says:

    Hi tallbloke,

    I agree, this thread contains the setting for the piece, and there’s also some further explanation due.

  63. oldbrew says:

    @ 7:53 pm: ‘In the Wikipedia animation, one might ask what the relevance of the outer circle is to all this.’

    This might help? FIgure 5: Geometric construction for Kepler’s calculation of θ.

  64. Gray says:

    Hi tallbloke,

    I think for the three dimensional explanation we should start with your right hand rule. Oldbrew’s formula should then snap into place in a logical sequence.

  65. tallbloke says:

    Gray: I suspect you are three steps ahead of me, so please slow down and take me through it slowly. :)

  66. oldbrew says:

    Ulric Lyons says:
    March 2, 2013 at 9:06 pm

    ‘I found a good 89 on Saturn-Jupiter synods that fits the orbit periods:
    60*10759.22 = 149*4332.589 = 89*7253.455 (60+89=149)
    error ~2.5 days.’

    Using the square roots of Ulric’s orbit periods we get:

    S=103.726, J=65.8224

    5*S = 518.633
    8*J = 526.58
    error < 0.002.

    8:5 is the best J:S fit because the actual ratio of their orbit periods is 1.576:1, and the two Fibonacci numbers that deliver the lowest ratio are 8:5 = 1.6:1

    Another stat. – Wikipedia gives Saturn's rotation rate as 0.44403 days (deep interior), and Uranus r.r. as 0.71833 days, giving U:S = 1.61775:1 = Phi:1.

    The Neptune equivalent is N:S = 1.512:1 = 3:2
    TB says: 'Perhaps the best I can do right now is use the image on the wiki page on the Kepler Triangle which shows the squaring of the circle. But that result is inaccurate, though only by <0.1%'

    Would the result be better or worse if Phi replaced pi in the calculation?
    pi=3.14159265359 approx.
    1.2*(Phi^2)= 3.1416407865 approx.
    Difference is 0.00001532 approx.

  67. oldbrew says:

    Basic rotation period round-up of planets with comments.

    Near 1:1 matches — E/Mars 1:1.03, J:S 1:1.07, S:J 1:0.93, U:N 1:0.93, (and Ceres:J 1:1.09)

    Near 1: Phi — S:U 1:1.618, N:Eris 1:1.612 (if Eris rotn. = 26hrs – may vary a bit), S:N 1:1.512
    (Also Ceres:Mars 1:2.714 = 1:Phi^2)

    Pluto (6.38d), Mercury (58.6d) and Venus (243d) rotate much slower than the rest – no obvious connections with nearby planets. Pluto also has an anomalous orbit – big variations in distance from Sun.

    Other — S:Earth 1:2.245 (Phi+phi=2.236), S:Mars 1:2.311

    Links to Saturn all give good results apart from the ‘slow rotators’.

  68. Gray says:

    Hi tallbloke,

    OK… I’ll slow down and review the posts so far…

  69. tallbloke says:

    Gray, thanks mate. I’ll find some time off from policing greenhouse debate to join you presently.

  70. Ulric Lyons says:

    oldbrew says:
    “8:5 is the best J:S fit because the actual ratio of their orbit periods is 1.576:1”

    10759.22 / 4332.589 = 2.4833:1

    You are going off on the wrong tangent. Try changing the number of synods to 144 or 233 instead of 89 and then see how many orbits of Sa and Ju there are.

  71. tallbloke says:

    popup factoid:

    111,111,111 x 111,111,111 = 12,345,678,987,654,321

  72. Rog,
    This comment produce :
    I-ching/hexagram meditation….

  73. oldbrew says:

    @ Ulric L

    That wasn’t about the synodics, it was comparing the lengths of the rotation periods, not the orbit periods.

    I might do a separate list of those, but most of the ground has probably already been covered. There might be some merit in then comparing the different sets of ratios to see if any pattern emerges?

  74. Ulric Lyons says:

    oldbrew says:
    March 12, 2013 at 5:24 pm
    “That wasn’t about the synodics, it was comparing the lengths of the rotation periods, not the orbit periods.”

    No, you did the square root of the orbit periods for some reason. The first meaningful orbital ratio for Saturn and Jupiter is 2:5.

  75. oldbrew says:

    ‘you did the square root of the orbit periods for some reason’

    That was earlier – let’s say it’s experimental. But some interesting results have emerged.

    Of course that was for moons not planets, but it’s still minor bodies orbiting a major body.

    March 11, 2013 at 2:11 pm:
    — Here I referred to a ‘Basic rotation period round-up of planets.’ That was nothing to do with orbit data, sorry if I caused confusion.

  76. oldbrew says:

    Ulric – re orbital period:

    Quote: ‘So what this is really telling us is that the relationship of Venus to the Earth and the Earth to Mars is really based on the orbital period, or rather the square root of the orbital period.’

  77. oldbrew says:

    TB: on reading one of Kotov’s papers I thought of another Fib. view of the planet structure.

    Me 3, V 2, E 1, M 1, J 0, S 0, U 1, N 1, P 2, Er 3

    We already know E&M , J&S, U&N have matching rotation periods.
    Also , P&Er are in a 2:3 ratio based on ‘sqrt(orb.prd)’.

    Not sure about Me&V yet – anomaly?

  78. tallbloke says:

    OB: Yep and that echoes the ‘pairing from opposite ends’ that Sevin tried, which is what kicked Kotov off in the first place.

  79. oldbrew says:

    Re 9:18pm:

    I’m getting Me:V ‘square root of orbit period’ ratio of 9.32:15.03 = 1:1.612 (Phi = 1.618).
    1:1.612 is close to 2:3 in Fib terms :-)

  80. oldbrew says:

    Orbital periods:
    Sqrt-o.p. for Venus:(Earth+Mars) = 1:3.01
    Sqrt-o.p. for Uranus:Neptune = 1:1.4 exact
    Sqrt-o.p. for Pluto:(U+N) = 1:1.398 (= 1:1.4)
    Sqrt-o.p. for Saturn:(U+N) = 1:4.054

    Rotation periods (addition to earlier list):
    Mercury:Venus 1:4.15 (= 1:4)

  81. oldbrew says:

    ‘Sqrt-o.p. for Uranus:Neptune = 1:1.4 exact
    Sqrt-o.p. for Pluto:(U+N) = 1:1.398 (= 1:1.4)’

    3 (planets) x 1.4 = 4.2 = Phi^3 (4.236)

  82. Ulric Lyons says:


    There is no reason for them to slip the Phi in there like that, and the Venus diameter root 2 is miles off. Just sloppy yoghurt weaving methinks.

  83. oldbrew says:

    @ Ulric L

    You say ‘The first meaningful orbital ratio for Saturn and Jupiter is 2:5.’

    But we already know the synodic ratio of J:S is 1:3?

    Quote: ‘Jupiter and Saturn have three synodic cycles to a basic unit of three conjunctions taking 59.5742 years (see diagram)’.

    I’ll post some more data a bit later about all this, following on from the comment on March 12, 2013 at 5:24 pm. However I agree you do have a point about the use of ‘phi’. I’ll try and clear that up later too with a more comprehensive explanation, or an attempt at one at least.

  84. Ulric Lyons says:

    oldbrew says:

    “But we already know the synodic ratio of J:S is 1:3?”

    Synodic ratio with what ?

    “Quote: ‘Jupiter and Saturn have three synodic cycles to a basic unit of three conjunctions taking 59.5742 years”

    Yes that’s 2 Saturn orbits and 5 Jupiter orbits.

  85. oldbrew says:

    Since J&S are over 90% of the mass of all solar system orbiting planets, their conjunctions must be a key factor re the planetary system resonances. More so than the orbit count in terms of harmonics surely? Sorry if the wording was misleading.
    ‘There is no reason for them to slip the Phi in there like that, and the Venus diameter root 2 is miles off’

    The proposition is that Phi is integral to the system as manifested in Fibonacci ratios. I said Venus ratio was 15.03 – from another data source I get 14.99, hardly ‘miles off’?

    So the Earth:Venus ratio is 1:15 and Mercury:Venus is 1:1.6.
    Mercury 9.379 (sqrt. 87.969) x Phi = 15.175

    14.99 and 15.175 are both saying 15, then it’s personal opinion whether that is of interest or not I suppose.

    Re Fibonacci and Phi: a general note, looking at the ‘0,1,1,2,3,5’ end of the scale (the rest being increments of Phi = 1.618)

    If we say that 1 breaks down into 0.382 and 0.618, (0.618/0.382=1.618 =golden ratio), these become the basic units of the system:
    0 = 0.382 (rounded)
    1 = 0.618 ( ” )
    1= 0+1
    2= 1+1 (or 1.618 rounded)
    3= 1+2 (or 2.618 rounded)
    5= 2+3

    Units like 1.618 and 2.618 get rounded to 2 and 3 in Fibonacci. An additional 0, null or similar is implied at the start, no F number for that. From 5 onwards it’s ‘F number’ x Phi (rounded). The ‘Phi-nary’ system uses a ratio instead of powers of 2 as in binary.

  86. Ulric Lyons says:

    oldbrew says:
    “Since J&S are over 90% of the mass of all solar system orbiting planets, their conjunctions must be a key factor re the planetary system resonances. More so than the orbit count in terms of harmonics surely”

    The synod’s of three or four bodies will usually meet before their orbits do.

    “The proposition is that Phi is integral to the system as manifested in Fibonacci ratios. I said Venus ratio was 15.03 – from another data source I get 14.99, hardly ‘miles off’?”

    I was talking about their Venus diameter figure, not the square root of the orbit period. What they do to the Kepler equation is nonsense.

    I’m outahere…

  87. oldbrew says:

    UL: agree there’s some data-mangling there – too many brackets, fractions, sq. root of 6(?) etc. But the basic interpretation of the Kepler law has logic I believe.

    Further data analysis: signs of resonances in the relative planetary distances from the Sun, as measured in Astronomical Units (AU). First the raw data, including Ceres, from Sun to Jupiter:

    Sun-Me-V-E-Ma-C-J = 0.24/0.62/1/1.5/2.77/5.2

    Second, relative distances between planets:
    Sun-Me-V-E-Ma-C-J = 0.24/0.38/0.38/0.5/1.25/2.53/(5.2)

    Third, convert the above to units of 0.125 AU (rounded):
    Sun-Me-V-E-Ma-C-J = 2/3/3/4/10/20 = total: 42
    Sun-J = 5.2 = 42

    Seen that number before somewhere… ;-)

    1.625 / 0.125 = 13
    1.625 / 1.6180339 (Phi) = 1.0043

    Therefore this conversion unit (1 AU = 8 x 0.125) can be used to linked to Phi values.

  88. oldbrew says:

    ‘Sun-Me-V-E-Ma-C-J = 0.24/0.38/0.38/0.5/1.25/2.53/(5.2)’

    Subtracting the Sun-Mercury unit value (0.24 = 2 units – see above) and re-converting to the units gives:

    Sun-Me-V-E-Ma-C-J = 0/1/1/3/4/9/25

    The square roots of that are 0/1/1/1.732/2/3/5

    Earth is 1.732, which can be resolved as:
    Sqrt(5) – 1/2 = 2.236 – 0.5 = 1.736 (where 5=Jupiter, 2=Mars,1=Venus)

    If the expected Fibonacci sequence is 0-1-1-2-3-5, Earth appears to be a sort of balance in the middle (sqrt(3)).

  89. oldbrew says:

    Re March 14, 2013 at 11:56 am

    Note Sun-Mercury=2, Sun-Jupiter=42 (using the 1/8th ‘AU’units), so M:J = 1:21
    Fibonacci stats. for outer planet group.

    — Planetary distances from the Sun:
    Sun-J-S-U-N-P-Er = 5.2/9.54/19.18/30.06/39.44/67.67

    — Distances as 8*AU (1/8 AU = 0.125 AU)
    Sun-J-S-U-N-P-Er = 42/34/77/87/75/226

    — Relative distances between the planets:
    Sun-J-S-U-N-P-Er = 42/34/77/87/75/226

    Looking for Fibonacci patterns here is trickier than the Sun-Jupiter group but there are ways.

    J = 42 (2×21 – J links the two planet groups)
    S = 34 (F no.)
    J+S = 76 = U+P/2 (or: both U and P)
    U/N = 76/34 = 2.236 = sqrt(5)
    Er/P = 226/75 = 3 (226 close to F no: 233)
    U:P = 1:1

    Rt. angle triangle: J/N/U = 42/76/87 (or 42/78/89 – F no. 89)
    (N,U have same rotation period)

    Rt. angle triangle: J/S/[(N+U)/2] = 42/34/54*
    (*actual is 54.66 = F no. 55)

    Rt. angle triangle: P/Er/U+N+P = 75/226/238*
    (*actual is 238.12 – U+N+P = 239 taken from rounded values)

    Rt. angle triangle: N/P/(Er/2) = 87/75/115*
    (*Eris orbit overlaps 2 dwarf planets, see below: Er / 2 = 113)

    Er 226 / 2.618 (Phi^2) = 86.3 (N=87)

    Er 226 + 239 = 465 / 2 = 232.5 (F no. 233)

    Neighbours of Eris (with Eris they have overlapping variable distances from the Sun, so direct AU comparison is not possible).

    — Orbital periods

    Pluto=248y / Haumea=285y / Makemake=309y /

    Pluto + Haumea + Makemake = 842
    (842 / 3) x 2 = 561.3

    561.3 / 557 (error margin) = 1.0078

    — Square root of o.p.- ratios

    P:E 2:3
    P:H+M 2:3
    H+M:E 1:1

    — Rotation period – ratios

    M:E 1:1
    H:M,H:E 1:2
    P:H+M 1:13 (H+M orbit in the same zone)
    P:E 1:19 (19+10% = 21 – Eris orbit overlaps H + M about 10%)

    That’s enough stats for one day.

  90. oldbrew says:

    Amendment to 6.54pm:
    – ‘Distances as 8*AU (1/8 AU = 0.125 AU)
    Sun-J-S-U-N-P-Er = 42/34/77/87/75/226’

    Line 2 should read:
    Sun-J-S-U-N-P-Er = 42/76/153/240/315/541

  91. oldbrew says:

    A bit more on the outer ‘dwarf’ planets, whose orbits are described as ‘eccentric’.
    Take the square root of the orbit periods:

    P – Pluto 15.74
    E – Eris 23.6
    H – Haumea 17.6
    [M – Makemake 16.89]

    Input 15.74(P) and 23.6(E) into Pythagoras calculator.
    Value returned is 17.584 (H).
    So right angle triangle P/H/E is validated.

    (15.74 + 23.6) / 17.584 = 2.237 [sqrt(5) = 2.236]
    This is the ‘2:3:sqrt(5)’ ratio.

  92. tallbloke says:

    OB: Excellent work!
    We need to bring your stuff together with Gray’s stuff in a new post soon. Sending you mail.

  93. oldbrew says:

    Got that thanks, will have a look soon. Just had a surprise result but it may lead somewhere.

    Attempting another triangle with Neptune and Pluto orbit periods (NOT the square root this time).
    N = 164.8, P = 248 (note: U=84, so U+N=P – clear link but no triangle)
    P calculator says 3rd side is 297.75

    The actual orbit periods of Haumea and Makemake are 285.4 and 309.88
    (285.4 + 309.88) / 2 = 297.64

    The same linkages don’t work with the square root values (as expected).

    Another quirk:
    N + (H+M / 2) = 164.8+297.64 = 462.53
    462.53 / 248 = 1.8645
    sqrt(5) – (sqrt(5) / (2×3)) = 1.8634

    Might try and see if that appears anywhere else – maybe too obscure?

    For the 2/sqrt(5)/3 triangle the following applies:
    Let A=2, B= sqrt(5)= 2.236, C=3

    (A+C) / B = sqrt(5) = B / (C-A)
    (B+C) / A = Phi² = A / (C-B)
    (A+B) / C = sqrt(2) = 1.414 (actual is 1.412)

    C/(B-A) = sqrt(Phi) x (5×2) = 12.72 (actual is 12.71)

  94. oldbrew says:

    Add to that: 12.72 / 1.414 = 3²

  95. oldbrew says:

    Distance from Sun(‘dS’) – formula where (planet2 = previous planet) is: divide the distance of planet1 from the Sun by the difference between that and the distance to the previous planet, e.g.
    Earth dist. / Earth dist. – Venus dist

    result = ds(planet1) / (ds(planet1) – (ds(planet2))

    Me-V-E-Ma-C-J-S-U-N-P-Er =
    (Me=1 because no previous planet)

    Plausible best matches for the sequence:
    1.63 ~ Phi (1.618)
    2.91 ~ 3
    2.225 ~ sqrt(5)=2.236
    1.413 ~ sqrt(2)=1.414
    2.19 ~ sqrt(5) ?
    1.99 ~ 2
    2.63 ~ Phi²
    4.2 ~ Phi³
    2.414 ~ sqrt(2)+1

    Re Eris it was said earlier: 61.3 / 557 (error margin) = 1.0078

    Another website quoted 560 not 557, even closer.

  96. oldbrew says:

    An alternative Fibonacci sequence up to 5:

    0 = 0.382 (rounded down)
    1 = 0.618 (rounded up)
    1 = 0.382 + 0.618 (= sum of 0 plus previous 1)
    [Phi = sum of both values of 1 = 1.618]
    2 = 1.618 + 0.382
    3 = 2.618 + 0.382
    5 = 2+3

    Phi² x ((3×2) / 5) = 3.14164 [Pi is 3.14159 rounded]
    5 / 0.618 = 8 (8.09)
    5 / 0.382 = 13 (13.09)
    [0.382 x 0.618 (0x1) = 0.236]
    5 / 0.236 = 21 (21.18)
    sqrt(5) = 2.236 = Phi² – 0.382


  97. oldbrew says:

    Gray says: ‘This results in a 3, 4, 5, triangle in which the 4 is not a part of the central Fibonacci Series’

    4 could be computed as (Phi+0.382)² – see previous post. I try to steer away from using 1 as it’s a bit ambiguous in my version of Fibonacci, where the first 1 (‘1A’) is phi and the second (‘1B’) is decimal 1. In the Kepler triangle 1 is 1B.

    Another triangle is 2, 3, sqrt(5) [3 is the longest side]. Squares of the sides are then 4, 9, 5.
    2 means 0+1A+1B, 3 means 2+1B, sqrt(5) = 2+1A


    Note the 3 is common to both triangles.

  98. oldbrew says:

    Consider Saturn’s north polar hexagon.

    There’s a spiral shape in the middle of it, so what’s going on? Pondering that I found something that might be interesting – try this (it’s easy):

    Draw a circle with two intersecting diameter lines in the shape of a cross (North/South, West/East).
    The centre point is C, the four points on the edge are N,S,W and E. Diameter NS = WE.
    Draw a line from N towards W, length NC (= radius), that ends at the edge of the circle (point H).

    You have now drawn the first side of a hexagon whose corner points all touch the circle edge.
    Now draw a line from H to S, which will go through line WC, and call that position point G.
    You should now find the ratio of length(WG): length(GC) is 1:1.618 (GC larger than WG).

    Join H to C and N to G, intersecting at X. The angles at X are four right angles
    Right angled triangles are now NSH, NCG, NCW (same as SCW), NHX, NCX, GHX, GCX.
    NCH is equilateral. Six of those triangles make up the hexagon.

    In NSH the ratio of NH:SH is 1:1.618 and NS = 2x NH (diameter = 2x radius).
    In NCG the ratio of CG:NC is 1:3236 (1:2Phi).

    The ratio of WG:CG also defines the golden ratio ellipse (see hypotrochoid animation).
    The distance between the two focii of the ellipse should be 2x CG.

  99. kab says:

    quote: “But up until now, so far as I know, no-one has been able to find a single simple scheme linking all the planets and the Sun into a harmonious whole system described by the basic Fibonacci series”

    I am not really knowledgeable enough to participate in this advanced discussion. However I am sure I have seen this described in a few places on websites discussing financial investing using heliocentric astrology & possibly also the mayan calendar system. If you can’t find similar references, I can look it up in my notes.

  100. tallbloke says:

    Hi Kab and welcome. he references I’ve found on financial sites to fibonacci involving astrology mostly relate to lunar periods and solar cycle periods. It seems a bit piecemeal and ad hoc to me, but maybe you can find us a link to something more extensive?

  101. oldbrew says:

    More re Jupiter’s three linked moons, in order: Io, Europa, Ganymede.

    Semi major axis figures:- 421,800km.(I) — 671,100 km.(E) — 1,070,400 km.(G)

    Ratios of S.M.A are:- E:G 1:1.595 — I:E 1:1.591 — E:G 1:2.5377
    (cf. Phi = 1.618, Phi² = 2.618, deviation for single moons < 1.7%)

  102. tallbloke says:

    Gray: And they used to be covered in marble laid to geometric perfection.

  103. oldbrew says:

    More re Jupiter’s three linked moons, in order: Io, Europa, Ganymede – part 2.

    Ratios of rotation periods for these moons:-
    I:E 1:2 (2.007), E:G 1:2 (2.015)

    And for Ganymede to Callisto, further away from J than the others:-
    G:C 1:2.33 (sqrt(5) = 2.236)

    The square roots of the orbit period of all four moons – Io, Europa, Ganymede, Callisto are:-
    I=1.33, E=1.884, G=2.674, C=4.085

    So I:G:C = 1:2:3 and I:E = E:G = sqrt(2) [1.416, 1.419]

    NB they all have synchronous rotation/orbit periods. Kepler triangles anyone?

  104. oldbrew says:

    Further to 7:16pm, it probably boils down to this.

    Rotation: I-E-G-C = 1-sqrt(2)-2-3

    Orbital (sqrt): I-E-G-C = 1-2-2²-3²

    Now I think of it, U:N was 1:1.4 orbital (sqrt) so that must be sqrt(2) as well.
    Likewise S:U+N was 1:4 so that’s 1:2².
    And U:N+P = 1:3 (3.117)
    And J:S+U = 1:4.238 (2.618+1.618=4.236)
    And J:S+U+N = 1:8 (7.965)

    Maybe there’s more mileage yet in these stats ;-)

  105. oldbrew says:

    Last word: divide Jupiter’s S+U+N ratio by its S+U ratio: 8/4.238 = 1.888 ….
    Sqrt(Phi) 1.272 + phi 0.618 = 1.89

    Mercury:Venus = Phi (1.607)
    Mercury:Earth = 1:2
    Mars:Me+V+E = 1.666
    E: Me+V = 1.277 (sqrt(Phi) = 1.272)
    J:Ma+E = 1 (1.02)

  106. tallbloke says:

    Excellent work again OB. I’m looking forward to having some time to review and digest yours and Gray’s findings so I can put the next post together. Gray has given us a good start already

  107. Gray says:

    Hi tallbloke,

    Looks like the Ancient Egyptians got there first…

    Still gently editing the article I think there will be a few more revises as we learn more. Does anyone have a spiralmaker on their computer, mine only has Kidpix.

  108. tallbloke says:

    Gray: You could try google sketchup and these instructions, and screenshot the result

  109. oldbrew says:

    Animation of a spider making its own spiral a.k.a. web…

  110. oldbrew says:

    A run through the differences (below) in the outer planet orbital periods turned up a few numbers.

    S-J = 17.8y, U-S = 54.55y, N-U = 80.79y, P-N = 82.9y, Er-P = 312y (true no. not exactly known)

    JS:SU = 1:3 (3.06), SU:UN = 2:3, UN:NP = 1:1, NP-PE = 1:2² approx.

    Also, SU:(UN+NP) = 3 (3.0007), UN:(NP+PE) = 1:5 (4.889)

  111. oldbrew says:

    A similar effort on the inner planets works better with the distances from the Sun.
    (Some of this has been mentioned before).

    (Me+V):E = 1:1 (1.11), (Me+V+E):J = 2:5 (2.466), (Me+V):J 1:5 (4.687)

    E:Ma = 2:3 (1.52), Ceres:J = 1:2 (1.88), (E+Ma+C):J = 1:1 (1.017)

    (Me+V):(E+Ma) = 2²:3² = 4:9 (2.27), (Me+V):(E+Ma+C) = 1:5 (4.82)

    NASA says dwarf planet Ceres and the asteroid Vesta ‘form a bridge from the rocky bodies of the inner solar system to the icy bodies, all of which lay beyond in the outer solar system’.

  112. oldbrew says:

    Wikipedia: ‘asteroid belt has 1:4 [or, 1:2² – my insert] resonance with J’.

    ‘Computer simulations suggest that the original asteroid belt may have contained mass equivalent to the Earth. Primarily because of gravitational perturbations, most of the material was ejected from the belt within about a million years of formation, leaving behind less than 0.1% of the original mass.’

    Radius of Jupiter’s moons:

    Io = 1821.6, Europa = 1560.8, Ganymede = 2631.2, Callisto = 2431.2
    (cf. Mercury = 2439)

    Io²+Eu² = Ca² (diff.= 1.011%)

    Ga is the ‘balancer’ as shown below. That gives us a right-angled triangle ‘ICE’:

    hypotenuse = Ca
    medium side = Io
    short side = Eu

    The biggest square that can be fitted into triangle ICE has sides equal to half the radius of Ga (according to my scale drawing).

    DGL = length of corner-corner diagonal of the ‘Ga square’ = 1.414 x side of square
    [i.e. 1.414 x (Ga radius / 2)]

    Ca radius / 2 = DGL [i.e. half the hypotenuse = DGL (1.013% diff. = very close to diff. above)]

    Therefore it’s possible that where a figure of 1.414 [= sqrt(2)] shows up in calcs, it represents the diagonal of a square inside a (Kepler?) triangle whose hypotenuse equates to twice the value of whatever the 1.414 represents. Not the full answer to everything but getting a bit closer.

    A diagram to scale – using those radius numbers – makes it easier to grasp. I’ll attempt to build on that tomorrow, got some irons in the fire.

  113. oldbrew says:

    (Radius of Jupiter)² / (Radius of Saturn)² = 1.416 = sqrt(2) [= 1.414]

    But J and S seem too big for Kepler triangles, new ideas needed.

    Outer planets – radius values:

    J =71.4, S=60, U=25.56, N=24.76, P=1.3, Er=1.8

    The inner planets are small compared to the rest. Total of all their radii is about 18.3.
    Total of all radii / total J + S = 1.54

    The hunt for a ‘solar system triangle’ goes on.

  114. tallbloke says:

    OB: Keep at it please, I’m being distracted elsewhere but will be back…

  115. oldbrew says:

    This might mean something useful, or not.

    Triangle using radius values:

    Hypotenuse = J = 71.4
    Medium = S = 60
    Short = (calc, below) = 38.7

    J² – S² = ((U+N+P+Er+’dwarfs’) – (Me+V+E+Ma+Ce))² [approx*]
    71.4² – 60² = 38.7²

    (U+N+P+Er+’dwarfs’) = 55.5
    (Me+V+E+Ma+Ce) = 18.75
    55.5 – 18.75 = 36.75 [*Discrepancy 5%].

    NB Wikipedia says the mean radius of Jupiter is 69.9 and
    equatorial is 71.5, so the discrepancy may be in the input data.
    That doesn’t alter the basic result IMO.

    Note: 60 (S) / Phi = 37.08 (i.e. between 36.75 and 38.7)

  116. oldbrew says:

    Re. above: if a square is drawn inside the triangle so two sides meet at the right angle and the other two sides at the hypotenuse, the length of the sides should be (J-S) x 2, and the diagonal within the square should be (J-S) x 1.414. That ties in with J² / S² = 1.414 as per the 11:42 am post.

    The divisions of the sides as created by the square also have interesting (Phi-related) properties.

  117. oldbrew says:

    Ian Wilson’s post on Feb. 20 this year says:

    ‘The spin-orbit coupling mechanism is based upon the idea that
    periodic alignments of Venus and the Earth (once every 1.5993
    years) produce temporary tidal bulges along the Earth-Venus
    -Sun line, on opposite sides of the Sun.’

    We know there are 365.25 days in a year, give or take a fraction, but there are also 360 degrees in one orbit.

    1.5993 x 365.25 = 584.144325

    584.144325 / 1.6180339 = 361.021 (cf. 360)
    584.144325 / 360 = 1.6226~ (cf. Phi)

    The introduction to this blog also post covered it:
    ‘Since it was noticed that five synodic conjunctions occur as Earth orbits the Sun eight times while Venus orbits thirteen times, many attempts have been made to connect the Fibonacci series and it’s convergent ‘golden ratio’ of 1.618:1 to the structure of the solar system.’

    Note also a close link to the ‘Jose cycle 178.72 years’ (Wilson).

    360 x (360/2) = 64800
    64800 / 365.25 = 177.413

    The 4628 year cycle breaks down as:
    4628 / 178.72 = 25.895 (years)
    4628 / 177.413 = 26.086 (degrees)
    Average = 25.991 (26 = 2 x 13)

    360 = 2³ x 3² x 5
    180 = 2² x 3² x 5

    To get an exact number of days with an exact number of degrees, multiply by 4 (years).
    365.25 x 4 = 1461 (years)
    360 x 4 = 1440 (degrees)

    177.413 x 1461 = 259200 years (actual is 259200.39)
    259200.39 / 1440 = 180 degrees (actual is 180.00027)

    ‘A grand alignment of the Jovian planets (Jupiter, Saturn, Uranus, Neptune), with all of the planets arranged in a line on the same side of the Sun, occurs roughly every 4628 years’ (Wilson)

    259200 / 4628 = 56 years (actual is 56.006914)
    56 x 360 = 20160 degrees
    20160 / 365.25 = 55.195~ (55 is a Fibonacci number)

    Not a bad match for a period of 259,200 years?

  118. Gray says:

    If you take 89 Earth orbits at 89 years and divide by the orbital time of Venus 0.61519 = 144.67.

    Then divide 89 years by the synodic period of Venus and Earth 1.598 = 55.69

    VE synodic 55
    Earth Orbit 89
    Venus Orbit 144

    Sun Synodic Earth 1202 (or 89 / 0.074 = 1202.70)

    Further, if you divide into 89 the orbital values for the following planets
    you get:

    Mercury 369.53
    Venus 144.67
    Earth 89
    Mars 47.31
    Jupiter 7.5
    Saturn 3.02
    Uranus 1 approx
    Neptune 0.5 approx

  119. tallbloke says:

    Gray: Yes. :)
    And as I showed in my original article, the apparent ‘odd one out’ – Mars at 47, lies at a golden section within the golden section. i.e. diff between 55 and 34 is 21. 21 is made up of 13 and 8. 47 is 34+13 or 55-8

  120. Gray says:

    Hi tallbloke,

    Excellent, I noticed 47.31 is close to 4 orbits of Jupiter 4 x 11.861 = 47.4472 also

    Mars and Jupiter

    The Sqrt 5 = 2.2360
    Synodic cycle MJ = 2.2354

    MJ Synodic 178 / 2.2354 = 79.627 (80)
    Jupiter 178/11.86 = 15.007 (15)
    Mars Orbit = 178 / 1.8808 = 94.64 (95)

    This is the keystone relationship in my view. It should be a 3, 4, 5, triangle with Jupiter 3, MJ Synodic 4, and Mars Orbit 5. But then again I could have it the wrong way up…

    Be back later….

  121. Gray says:

    Following on from the identification of the Mars/Jupiter triangle, three other triangles were calculated with Fibonacci relationships proving, I think, the format of much of the inner Solar System. The figures were derived by dividing a whole period of 178 orbits of Earth, 178 years, by the various orbital and synodic values of Venus, Earth, Mars and Jupiter.

    The three triangles are:

    Venus 288
    Earth 178
    VE Synodic 110

    Earth 178
    Mars 95
    EM Synodic 83

    Mars 95
    Jupiter 15
    MJ Synodic 80

    Finally the relationship between the planets becomes clear. By adding the values derived for the outer planet orbital time and the synodic value, a pattern emerges. The two lower values in each case equal the value of the higher figure.

    Venus 288 = Earth 178 + VE Synodic 110 = 288
    Earth 178 = Mars 95 + EM Synodic 83 = 178
    Mars 95 = Jupiter 15 + MJ Synodic 80 = 95

    The following scaled schematic shows the total relationship between all four planets in the series.

    Note that the original triangle is an Isosceles triangle.

  122. oldbrew says:

    Guys: think of a starting point of 89/8

    Half the assumed 178 year cycle is 89 years
    89 year period = 8 x 11.125 years = 8 solar magnetic reversals aka sunspot cycles
    11.125 = 11 and one eighths = 89 / 8 (again)

    11.125 years = 133 and a half months = 267 / 2 = 89 x 3 / 2 = 89 x 1.5 months
    8 x 1 .5 months (89 /8 formula) = 1 year

    Not sure yet if that can be taken further in the ‘micro’ direction, but note that:
    1.5 months = 45.65625 days (365.25 / 8) and also 45 degrees of orbital movement (360 / 8),
    and that 89 / 2 = 44.5, so any ideas welcome.

    On the ‘macro’ side, 89 x 52 = 4628. The significance of 52 (2x2x13) years is not clear, but consider 8 x 52 (89/8 formula). That is also 2 x 4 x 52 = 2 x 208 = 2 de Vries cycles?

    ‘a ~ 208 year-cyclicity, named de Vries or Suess cycle (Damon and Sonett, 1991, Stuiver and Braziunas, 1993 and Wagner et al., 2001), is documented in various Holocene records (e.g. Schimmelmann et al., 2003, Raspopov et al., 2008, Taricco et al., 2009, Incarbona et al., 2010 and Di Rita, 2011). It might as well be present from historical sunspot observations (Ma and Vaquero, 2009). Its influence on several climatic parameters has been discussed by Raspopov et al. (2007), who document a non-linear response of the climate system in various geographic regions.’


    Btw re the 4628 cycle: 2 x 2 x 34 x 34 = 4624 (the next 4-year [2×2] period up to 4628 is not completed)

  123. oldbrew says:

    TB says: ‘Mars at 47, lies at a golden section within the golden section’

    The square root of 47 is 6.85565~
    Phi^4 = 6.8541~

  124. Gray says:

    The rest of the series…

    Mercury orbits = 739.061
    Venus orbits = 288
    Synodic Venus = 449

    incidentally tallbloke 6785 / 739.061 = 9.180

    Jupiter orbits 15
    Saturn orbits = 6.042
    JS Synodic – 9

    Saturn orbits = 6.042
    Synodic Uranus = 3.92414235995
    Uranus Orbit = 2.11878024916

    Uranus orbit: 2.11878024916
    Synodic Neptune : 1.03488372093
    Neptune orbit = 1.08019540614

  125. Gray says:

    Correction : 6765 / 739.061 = 9.15350695004

  126. oldbrew says:

    Re. ‘starting point of 89/8’:
    360 degrees / 89 = 4.044944 x 11.125 = 44.9 recurring = 45 degrees

    Which proves that 89 / 8, i.e. Fibonacci, directly links to degrees of orbit and rotation (45 x 8 = 360 of course).

    Re. ‘The significance of 52 (2x2x13) years is not clear’:
    52 = 5² + 3³

    89 x 5² = 2225
    89 x 3³ = 2403
    2403 – 2225 = 178 = 89 x 2

    If anyone wants to talk about a 1000 year solar cycle, we can offer:
    89 years x 11.125 = 990.125 years

    That would be a one-eighth part of a 7921 (89²) cycle.

  127. Gray says:

    …and one for the road.

    Mercury figure 6765 divided by 449 Mercury/Venus Synodic periods

    6765 / 449 = 15.066

    So 15 and 9 scales to:

    Jupiter orbits = 15
    Saturn orbits = 6.042
    JS Synodic = 9

  128. oldbrew says:

    Using 89 as a base gave:
    4628 / 11.125 (i.e. 89 /8) = 416 (solar cycles in 4628 years)

    4628 / Phi = 2860.26
    2860.26 / 16 = 178.75
    178.75 / 16 = 11.171875
    (16 = number of solar cycles in the 178~ year period)

    These results are closer to the numbers we actually know of, e.g. Jose 178.7 year cycle.
    Gray’s website refers to ‘the recognised sunspot frequency of 11.171 years’.

    Calculating in the other direction:
    11.171 x 16 x 16 x Phi = 4627.215~
    4627.215 / 11.171 = 414.22~ (solar cycles in 4627~ years)

    Previous discussions on the Talkshop refer to a 4627 year cycle (not 4628), e.g.

  129. oldbrew says:

    More triangles for the schematic list?

    Mercury 369, Venus 144, 369-144 = 225 = 15×15 (Jupiter 178 /11.86 = 15)

    Earth 89, Mars 47.3, Jupiter 11.86² (=141 approx = 3.45% above exact match) OR
    Earth 89, Mars 47.3, Venus 144 (5.6% above exact match)

    Also: Me – (E + V) = 369 – 233 = 136 = E + Ma (136.3)

  130. oldbrew says:

    ‘Mercury 369, Venus 144, 369-144 = 225 = 15×15 (Jupiter 178 /11.86 = 15)’

    The Venus sidereal year is also 225 (224.7 days).

    The creationists have some interesting stats here (just sticking to the figures):


  131. Max™ says:

    Something which may help make sense of the Fibonacci resemblance:

    For Earth, every 97,000 miles we travel around the Sun, we travel 447,000 miles roughly towards Vega.

    Fibonacci sequences naturally describe spirals, the planets follow spirals as they orbit the Sun.

  132. oldbrew says:

    This paper: ‘The ∼ 2400-year cycle in atmospheric radiocarbon concentration: bispectrum of 14C data over the last 8000 years’…

    …reports that:
    ‘A principle feature of the time series is the long period of ∼ 2400 years, which is well known. The
    lines with periods of 710, 420 and 210 years are found to be the primary secular components of power spectrum.’

    The 2400 year period is 89 x 27 = 2403 years (27 = 3³)

    89 x 8 = 712 (cf. 710) = 4 Jose cycles of 178 years (= 2 x 89 years)

    210 has already been identified in this thread as the de Vries cycle (208), and 420 = 2 x 210.

  133. oldbrew says:

    More on the 89-year period: re. the reported 160-minute solar oscillation.

    Take a notional solar cycle of 11.125 years ( = 89 years / 8) and convert it to hours.
    11.125 x 365.25 x 24 = 97521.748 hours

    NASA data says the sidereal rotation period of the Sun is 609.12 hours
    97521.748 / 609.12 = 160.1 hours (160.10268)

    160.10268 x 60 = 9606.1608 minutes

    9606.1608 / 60 = 160.10268 minutes



  134. tallbloke says:

    160.10268 x 60 = 9606.1608 minutes

    9606.1608 / 60 = 160.10268 minutes

    Hmmm, this doesn’t seem all that informative at first glance.

  135. oldbrew says:

    Sorry, it should have been the other way round because there’s no reason to divide by 60 on the last line. If we assume the 160.01 and use it as a divisor:

    9606.1608 / 160.01 = 60.03475~

    (The other calculation returns 160.1 not 160.01)

    There are some interesting comments here about the Sun, Jupiter and Kotov’s wave, as they call it. The link was in the last comment on the ‘160-minute’ thread mentioned above.

    ‘Strange Coincidences between the Sun, Jupiter and OJ 287’

    See: Relations with the 160 mn wave of Kotov.

    The author claims to have found proof that, just as the Moon is very gradually moving away from the Earth, so for the same reasons Jupiter is moving away from the Sun.

  136. tallbloke says:

    Earth-Moon masss ratio ~80:1 Lunar orbital distance ~60 Earth radii

    Sun-Jupiter Mass ratio ~1000:1 Jovian orbital distance ~1000 solar radii

  137. oldbrew says:

    The 609.12 hours figure is based on an arbitrary decision to measure at 26 degrees from the equator, ‘approximately the point where we see most of the sunspots’

    Read more: http://www.universetoday.com/60192/does-the-sun-rotate/#ixzz2OkmPNFwP

    So with a very slight tweak the 160.01 division could return 60 exactly. Obviously we don’t know what that indicates but it doesn’t do the 11.125 solar/Fibonacci hypothesis any harm.

  138. oldbrew says:

    Let’s try it another way.

    11.125 x 365.25 = 4063.4062 Earth days
    609.12 hours = 25.38 Earth days = 1 Sun rotation
    4063.4062 / 25.38 = 160.1 solar rotations

    So there are approximately 160.1 (could be 160.01) solar rotations in a solar cycle.

    If the incoming ‘wave’ acting on the Sun occurs every 160.01 Earth minutes, and Earth minutes derive from the time taken to orbit the Sun, the evidence for a ‘solar clock’ seems to be there.

  139. oldbrew says:

    Using Roy Martin’s harmonics list…

    …the total of all Jupiter synodics is:
    9+276+164+728+80+13+14 = 1284

    One solar rotation takes 25.38 days (NASA figure). Using Roy’s J/S base period of 178.732 years:
    (178.732 x 365.25) / 25.38 = 2572.94

    2572.94 / 1284 = 2.00385~

    That equates to an average of 1 Jupiter synodics per 2 solar rotationsi.e. a 1:2 ratio.

  140. oldbrew says:

    One solar rotation takes 25.38 days.

    25.38³ = 16348.384 days
    16348.384 days / 365.25 = 44.76 years
    44.76 x 4 = 179.04 years (Jose cycle)

    So the cube of the solar rotation period = a quarter of the Jose cycle.

  141. tallbloke says:

    Good one!
    I had some thought s on solar rotation related to planetary periodicity a while back. I got panned for it, but I still think the inverse relationships are real somehow.

  142. Gray says:

    The final triangle in the series is:

    Sun rotation 0.069 years 178 / 0.069 = 2,579.7
    Mercury orbit .240846 years 178 / .240846 = 739.06147496741
    Sun synodic Mercury .096739 years = 178/.096739 = 1,840.002

  143. oldbrew says:

    Found something in the link that might be of interest. You said:

    ‘So for the orbits of Jupiter(11.86 years) and Saturn(29.46 years) we find that the squares (multiplication by itself) of the orbital periods are 140.67 and 867.3. The cube roots of these values are 5.2 and 9.54’

    Compare those cube root values with the ‘mean distance from Sun (AU)’ figures for J and S here:

    Any comment?

  144. oldbrew says:

    Re. the cube root values (as per Newton / Kepler, ignore previous post) – there seems to be a correlation between planetary pairs.

    Earth:Mars = 1:1.52
    Uranus:Neptune = 1:1.56

    Mercury:Venus = 1:1.86
    Jupiter:Saturn = 1:1.83

    Looking at the links between these pairings, we have:

    Venus:Earth = 1:1.385 (1.382 = 2 – 0.618)
    Saturn:Uranus = 1:2.01 (2)

    For the outer planets, the cube root ratios and the ‘distance from Sun’ ratios are identical.

    For the inner planets, the exception is Earth:Mars, where the distance ratio is 1:1.88 (similar to M:V and J:S) but the cube root ratio of 1:1.52 is like U:N, as shown. That may well relate to the fact that Mars is the last inner planet before Jupiter, and/or the existence of the asteroid belt between the two.

  145. oldbrew says:

    Correction – the Earth:Mars distance ratio IS 1:1.52, so it’s not an exception.
    (The actual Mars orbital period is 1.88 years, copied in error)

    Extending the analysis to dwarf planet Ceres (between Mars and Jupiter), we get:

    Mars:Ceres 1:1.81 (very close to Jupiter:Saturn = 1:1.83)
    Ceres:Jupiter 1:1.88 (very close to Mercury:Venus = 1:1.86)

    Looking at it as a whole, we have:

    1.86 — 1.385 — 1.52 — 1.81 — 1.88 — 1.83 — 2 — 1.56

    The planetary pairs with the lowest ratios (Mars/Earth, Neptune/Uranus*) orbit either side of the Jupiter/Saturn ‘axis’ (ignoring the dwarf planet Ceres).

    *Quoting myself (March 4 @ 8:36am):
    ‘Consider the other ‘half’ of the system: imagine we fold it over so all the planets are back-to-back.
    Now, Mars/Earth = Neptune/Uranus, and so on.’

  146. oldbrew says:

    @ Gray: improvements to the ‘speculation’ here…

    Saturn’s orbit period cubed (sopc) = 29.457³ = 25560.276 years
    Jupiter/Saturn synodic 19.86 / 2 is 9.93 (conjunctions+oppositions)
    25560.276 / 9.93 = 2574.05 (J/S synodics in sopc)

    Other results from that:
    25560.276 / 178.74 = 143.00255 (no. of Jose cycles in sopc)
    25560.276 / (143 x 16) = 11.1715 (one solar cycle, 16 per Jose cycle)
    25560.276 / 19.86 = 1287.02 = 143 x 9

    9.93 x 365.25 = 3626.9325 days
    3626.9325 / 25.38 = 142.9 (no. of solar rotations in 9.93 years)

    Solar rotations = (2574 x 25.38 days) / 365.25 = 178.86
    2574 / 18 = 143 rotations per J-S half-synodic (confirms previous calc.)
    18 x 9.93 periods per Jose cycle = 178.74

    (21 x 9.93 = 208.53 = one de Vries cycle?)

    Another view:
    143 x 19.86 (9.93 x 2) = 2840
    25560.276 / 2840 = 9

    (1) no. of s.r. per Jose cycle = no. of J-S conjns. + opposns. in Saturn s.o.p.³

    (2) no. of s.r. in 9.93y (J/S conjns. + opposns.) and
    no. of J-S conjns. + opposns. in Jose cycle* =
    no. of Jose cycles in Saturn s.o.p.³

    [(1) = 2574, (2) = 143]

    NB 9.93 x 9 = 89 (89.37)
    (143 cf. Fibonacci 144 ?)

    (* 2:1 s.r. to all J synodic conjunctions already identified)

    Conclusions at this stage:

    The cube of the orbit period seems significant.
    Definite link between s.r. and orbit period.
    S.r. / Jupiter synodic conjunctions link confirmed.

    Number of solar rotations per (ideal) Jose should be 2574 (143×18)

    J-S half-synodic x 18 = 9.93 x 18 = 178.74
    (Jose cycle – J-S figure : Sun figure = 1:1.0006713)

    Now we turn to the cube of the orbit period for the other ‘gas giants’.

    Jupiter s.o.p.

    11.86³ = 1668.2228 years
    1668.2228 / 9.93 (J-S/2) = 168 = 21 x 8
    (21 x 9.93 = 208.53 = de Vries cycle?)

    No. of solar rotations in 1668.2228 years:
    (1668.2228 x 365.25) / 25.38 = 24007.816
    24007.816 / 168 = 142.904 (143)

    NB 1668.2228 / 3 = 556.0743 years
    One Eris orbit = 557 years approx.

    Uranus s.o.p.

    84.01³ = 592915.67 years
    592915.67 / 143² = 29

    Not sure if this means anything, but:
    9.93 x 29 = 287.97 (9.93 = J-S half-synodic period)
    287.97 / 2 = 143.985

    Neptune s.o.p.

    164.7885³ = 4474872.7 years (nearly 4.5 million years)
    4474872.7 / (143 x 175) = 178.8161 years = Jose cycle

    4474872.7 / 12.78279 (J-N synodic) = 350070.1 conjunctions
    350070.1 / 2574 (143 x 18) = 136.00236 (136 = 34 x 2 x 2)

    All the gas giants plus the Sun have clear links to multiples of 143.

    143 solar rotations take 9.936 years (based on 25.38 days each).
    One J-S half-synodic takes 9.93 years.
    These two figures may in reality be exactly the same period.
    (The difference between them is around 0.0006%)

  147. Gray says:

    Hi Oldbrew

    End ‘speculation’, this should answer your question…

  148. oldbrew says:

    Hi Gray

    Sorry, I’m having trouble joining the dots there. Where does the 55 come from, for example?

    More speculation probably…

    Jupiter-Saturn synodic period = Jupiter orbit period³ / Uranus orbit period
    19.859 = 1668.223 / 84.01 (19.857)

  149. Gray says:

    Hi OldBrew.

    I’m sorry, It’s an abstract concept which I’m trying to convey as logically as I can.

    I’ll work some more on it…

  150. oldbrew says:

    I’ve expanded the data for the constant that Ulric Lyons showed earlier.
    This is where the ratios of the synodic periods of 3 contiguous planets are compared.

    Mercury — Venus — Earth (Me-V 0.3958, V-E 1.5987, Me-E 0.31726)

    V-E / Me-E = 5.0391
    V-E / Me-V = 4.0391

    Venus — Earth — Mars (V-E 1.5987, E-Ma 2.1353, V-Ma 0.91423)

    E-Ma / Ma-V = 2.3356
    E-Ma / E-V = 1.3356

    Jupiter — Saturn — Uranus (J-S 19.859, J-U 13.812, S-U 45.363)

    S-U / J-U = 3.2843
    S-U / J-S = 2.2843

    Saturn — Uranus — Neptune (S-U 45.363, U-N 171.39, S-N 35.869)

    U-N / S-N = 4.7782
    U-N / S-U = 3.7782

    Orbit periods from NASA used to calculate synodic periods here:

    The only rule is that the largest figure is divided by the two smaller ones.
    The difference between the two results is always 1.

  151. oldbrew says:

    In order to make more sense of the available data, there may be a case for ‘superimposing’ a template on it. Take the Jose cycle:

    The synodic period figure is 178.72 years, so let’s suppose a template figure of 180.
    The difference is: 180 / 178.72 = 1.00716 (180 is a multiple of 2,3 and 5 in Fibonacci)

    Jupiter-Saturn half-synodic = 9.93 x 1.007 = 10 exactly (2 x 5 in Fibonacci)
    Full J-S synodic 19.86 x 1.007 = 20 exactly (9 x 20 = 180)

    9.93 = 143 x 25.38 day solar rotations
    143 x 1.007 = 144 exactly (Fibonacci)

    Jupiter orbital period = 11.862 years
    11.862 x 1.007 = 11.95 (12 = square root of 144 in Fibonacci)

    Saturn orbital period = 29.46 years
    29.46 x 1.007 = 29.666 = 89 / 3 (Fibonacci, 144 / 89 = Phi)

  152. tallbloke says:

    Oldbrew: Nice job, and starts to tie in with Gray’s Jose period approach too. This is looking really good. I had a chat with Gray on the phone last night and agreed a way forward. I will review his comments and tie them in with his diagrams as best I can, then email him my summary for him to annotate and return. We’ll include you in the email loop for comments too. Then I’ll write a summary post, run it by you both for final comments, amend and publish.

  153. oldbrew says:

    I’m sending a triangle diagram by e-mail, re this:

    ‘Since it was noticed that five synodic conjunctions occur as Earth orbits the Sun eight times while Venus orbits thirteen times, many attempts have been made to connect the Fibonacci series and it’s convergent ‘golden ratio’ of 1.618:1 to the structure of the solar system.’

    It’s an attempt to show a Phi link (the Phi link?).

  154. tallbloke says:

    Excellent, I’ll forward a copy to Gray when it arrives.

  155. James says:

    I find this discussion absolutely fascinating as it ties in with some research I have been doing with regard to the fibonacci sequence and it’s apparently universal relationship/connection to the laws of nature. With this in mind,( and with regard to the ‘big picture’), I would be interested in any comments which may shed any light on as to why the fibonacci sequence and the laws of nature seem to be so inextricably connected?

  156. tallbloke says:

    James: tell us more. My intuition tells me the relationship is to do with the principle of least action, entropy, and the constructal law.

  157. oldbrew says:

    Re Ulric Lyons:

    ‘Uranus-Saturn-Jupiter resonance:
    Ur-Sa synod divided by Ur-Ju and Sa-Ju synodic periods is 3.2843 and 2.2843. Multiplying by 34*3 gives the number of synods for the whole 4627yr cycle.
    3.2843 and 2.2843 times 7 gives 23 and 16, the 7:16:23 ratio is at 317.714yrs, one seventh of the 2224yr period. There is a little slip in this harmony, which gets mostly resolved at the 4627yr grand synod.’

    233 x 317.714yrs = 74027.36 years
    16 x 4627.15yrs = 74034.4 years

    16 Grand Synods = 233 JSU
    (1 G.S. = 102 SU + 233 JS + 335 JU)

  158. oldbrew says:

    One more point re the ‘Uranus-Saturn-Jupiter resonance’ (see above comment).

    3 x Uranus-Neptune synod (171.38y) / 317.714y (7:16:23 period) = 1.618 (golden ratio)

  159. tallbloke says:

    I’ll give you a ring tonight.

  160. Bart says:

    First I’ve seen this. Very interesting…

  161. tallbloke says:

    Thanks Bart. Much more to come, we’ve been making some really interesting discoveries following on from this one, not yet published.

  162. oldbrew says:

    TB I’ve got some more stuff too. For example the ‘1.007’ factor can be simply explained with exact numbers – two of them to be precise. It can also be used to shed light on a number of planetary synodic harmonics etc. that might otherwise be unclear or overlooked altogether (I suggest!).

    Plus Jupiter and Saturn can now very easily be linked to phi if we haven’t done that already: using the correct ‘1.007 factor’ only 2 other numbers are needed to produce the exact orbit periods for both planets (corresponding precisely with the NASA figures).

  163. […] Oldbrew and I have been working on the orbital configurations and have some news related to the Phi planetary discovery made earlier in the year here at the […]

  164. crikey says:

    Great number crunching here
    Thought l would leave a small contribution

    The Heliospheric current sheet
    ( 2 fibonnacci spirals ?) one for each magnetic pole perhaps

    The Venus earth dance

    A ‘new age course’ in Arizona teaches this stuff

    You have got me hooked playing with the calculator looking at ratios.. LOL

    Keep up the great work..

    Googling ” picture fibonnacci “yields some mind blowing stuff

  165. Ian Middleton says:

    Fascinating stuff. I’ll have to read about 4 times over to get it all to sink in. It makes you wonder what an ideal solar system would look like if it adhered exactly to the fibonnacci series. If it comes anywhere close to the system we have here, and on what I have read, it will, we could expect all those thousands of other solar systems out there to be pretty similar. Give or take the odd jovian or two.

    Thanks TB for the post

  166. […] the coming weeks and months. It consists of some of the observations gathered since February when I published my discovery that the fibonacci series and the Golden Ratio – Phi connect the planetary […]

  167. Asynsis says:

    Hi TB,
    I’ve passed your fascinating blog post on phi in the solar system directly to Professor Adrian Bejan for his further information as we are having an ongoing discussion about phi in nature and this is a pretty compelling example.
    I concur with you that the reasons for the phi geometries are derived from Constructal behaviours, where perturbations and gravitational/electro-magnetic resonances minimise the local entropy-production of this particular open system, it’s a principle of least-action, robustness-enhancing, optimisation signature.
    This video reveals an aesthetic feeling for why in intuitive, vivid relief:

    It was a relationship I speculated on in earlier work, including elementary particle rest mass relationships with each other. Worth a look too?
    On the sole equation in this new post of mine on why the Constructal law yields these emergent
    Asynsis geometries:
    it is trivial unless we can find behaviours where phi emerges as an asymptotic convergence, which I also conjectured was the case in the Feigenbaum logistic map diagram in 1995 – here’s the QED in the PDF:
    I suggest we have the beginnings of a new paradigm here, where universal Constructal thermodynamic behaviours explain emergent optimal, analogical Asynsis geometries.
    Or is it vice versa? That’s perhaps one for the philosophers.
    All the best with your own work!

  168. tallbloke says:

    Hi Nigel,
    Thanks for bringing our work to prof Bejan’s attention, I hope something fruitful comes from that.
    We have more recent posts investigating the ‘Why Phi?’ Question. Heres the list
    I’ve also submitted a paper, which has been accepted pending revision.

    “emergent optimal, analogical Asynsis geometries. Or is it vice versa?”

    Actually, it’s both, you’ll love this as an architect:

    The Fibonacci series converges to phi. Everybody knows this. As the numbers get bigger, the ratios get closer and closer to 1.618…
    But it works the other way too. As well as phi ’emerging’ from the Fibonacci series as it evolves, the Fibonacci series emerges from phi as a quantisation effect, forming a fractal.

    A simple (simplest possible?) example is shown in this comment:

    How that plugs into your e^-(i Pi)=1 I’m not sure. I’ll have a think later, but I have to finish a referees report.

    Regarding connecting the macro-cosmological to the micro-subatomic; we are aware of a few links and I’ve seen some interesting work on golden means and the silver mean family in relation to light and photosynthesis. No time yet to integrate that, but it’s on my to-do list. :)



  169. Hi Rog, No problem, thanks for the links and happy to help spread the word, speaking of which, yes – you should definitely write a formal paper and submit to a reputable, peer-reviewed science journal. Your work certainly appears to have sufficient merit.
    Nice iron horse btw, mind how you go!

  170. tallbloke says:

    Nigel: The paper is written and submitted, and accepted pending revision – a work in progress.

  171. oldbrew says:

    TB: I’ve also found a possible way to reconcile the ‘variants’ of the Fibonacci series, using other Fibonacci numbers. For example:

    89/55 = 1.61818 recurring (’18’ recurs)

    A higher Fibonacci number is 10946 (there’s a simple maths process to find the right number).
    Multiplying 1.61818~ by (10946 – 1) gives 17711 = the next Fibonacci number after 10946.
    Then dividing 17711 by 10946 gives 1.6180339~, a lot closer to Phi than 1.61818~.

    Added later (all whole numbers are Fibonacci):
    89/55 = 1.61818~ x (10946-1) = 17711
    55/34 = 1.617647 x (4181+1) = 6765
    34/21 = 1.6190476 x (1597-1) = 2584
    21/13 = 1.6153846 x (610+1) = 987
    13/8 = 1.625 x (233-1) = 377
    8/5 = 1.6 x (89+1) = 144
    5/3 = 1.666~ x (34-1) = 55

    From the top, each line reverses the +/- sign and reduces by two powers of Phi.
    The result is the ‘omitted’ Phi number.

  172. tallbloke says:

    OB: Very interesting. Can it be pushed further to see what happens if you apply the same technique to the remaining difference between phi and the resulting number? It might develop yet another Fibonacci series. If so, you can reduce it to a general rule and earn a place on Wolfram Alpha’s maths website.

  173. wayne says:

    I learned something new here on this thread. Thanks TB!

    One of the most amazing characteristics of Phi itself is that you can just take any two real numbers, not both zero, to form a sequence that n[+1] = n[0] + n[-1] like 1, .1, 1.1, 2.1, 3.2 … or 1, 999, 1000, 1999, 2999, … or 9, 4, 13, 17, 30, …, even negatives, and by the 27th iteration the ratio of the two consecutive numbers there are always Phi to at least 10 digits, 1.618033989…, quite surprising. Didn’t realize that before this unique type of thread.

    Phi, not necessarily just the familiar Fibonacci sequence after all, there are an infinite number of sequences all leading to Phi, just always sum the two preceding numbers enough times!

  174. tallbloke says:

    Wayne, yes, and plotting the deviation from phi as the series converges towards it produces interesting ‘decay curves’ too. I’m wondering if those could be used as diagnostics in some way.

  175. oldbrew says:

    ‘plotting the deviation from phi as the series converges’

    See expanded comment above at 9:40 a.m. today. If the original number is greater than phi, subtract one from the phi multiplier, otherwise add one.

  176. wayne says:

    Think I see what you mean TB. The plots of the differentials always creates the same strongly convergence to zero, dampening the differences, and the ratios of consecutive pairs of the first differentials always converges, though more slowly, to -φ², no matter what two numbers you begin with. It is always negative due to the vacillation of the dampening about zero.

    I still can’t quite see how orbits graviationally could lock or syncronize the rotations of other planets since if moons are involved (mostly) those rotation periods are always slowly transferring angular momentum. So for right now, I’m sticking to just the orbit to orbit aspect between the planets/moons systems, that I can nearly visualize a possible interaction. Like maybe starting with the ratios and work backwards instead of forwards to see if the small numbers at the beginning of such sequences may hold something there.

  177. tallbloke says:

    Hi Wayne. Good thinking, and interesting about the second differentials converging on -φ² too. I’m beginning to get the glimmer of an idea about mechanism for orbit-spin coupling. Magnetics. If it’s right, then the planets are being driven, which means they won’t gradually slow down as the tidal action of moons takes effect. The timings we are now discovering for Earth’s rotation, and that of the Earth-moon system mean its all coupled.

  178. wayne says:

    Nice! I’ll keep low-key then more on the programmable search side, still randomly waiting for that “well lookie there” moment. Sounds like you may have found it. Can’t wait to read it. Ok, I have all of the data on both+ so will keep looking at the rotations and the orbits at each step to maybe add support.

  179. tallbloke says:

    Brilliant, thanks. Good to have a your mind working independently on this. I thinks there’s a lot of benefit in having your approach as a check on mine and Oldbrew’s. Your method gives a sort of ‘impartial overview’ while we ferret out the ratios by hand on intuitive moves.

  180. wayne says:

    TB, one last thought, this may seem screw-ball to you and OB and mostly it seems that way to me too at times but I keep being drawn back to some thoughts that kinda automatically began to manifest in those few years when I was messing with solar systems and ephemerides since I then had it all programmed to allow such ad hoc n-body system creation. Though I might not just keep my mouth shut. And close encounters and how they were created especially drew my attention in n-body systems.

    You keep looking for synchronizations and repeated patterns but what I tended to see is harmonics in a n-body systems is exactly what you don’t want to be stable. If angular momentum due to harmonics is ever feed positive between two bodies that is exactly leads over time to the destabilization of the system and usually close encounters would eventually occur. It’s almost like Phi may be the answer as to how a n-body system stabilizes into a minimum a.m. transfer and the least synchronization between all and every pair of the bodies involved. !! And there may be another almost magical imprint of Phi being the relationships. Kind of weird.

    See the counter-intuitive aspect of that thought? Kind of a backwards view of what an unstable system should be to be stable. I was never able to get anything conclusive there, that was about five or six years ago, my math at the very abstract level fails me there and I became bored never to find a way to advance on what I “tended” to be seeing. Well, that’s just one far-fetched idea that keeps drawing me to what you may have also found in the relations with Phi. Curious that nothing immediately seems to synch and maybe that is exactly what every stable solar system needs the most, no simple harmonics and least a.m. transfer between pairs and Phi is the answer, I have to say hmm… maybe. Just a thought you might also stick in the back of your mind on this subject.

  181. tallbloke says:

    Wayne: That’s what I was getting at with the ‘Why Phi Pi Slice‘ post. Phi seems to be the angular proportion which doesn’t return to it’s start point, yet keep the inter-relations in the almost-but-not-quite resonant ratios which maintain order and stability. If you look at fig 1 in this paper, you’ll see that the planetesimals in the Kuiper belt definitely do cluster around resonant ratios which are also Fibonacci pairs; 2:1, 3:2, 5:3 and also Lucas numbers ratios 4:3, 9:4. The secondary mechanism which prevents collision is orbital eccentricity. Then there is the precession of synodic cycles, as Ian Wilson discussed in a recent post, and which I go into detail in the paper I’ve submitted.

    You’ve put your finger on a key issue, and I’m also thinking about it. Hopefully, the light will dawn before too long.

  182. Gents,
    I would say that what we are looking at here is no less than (in Constructal law terms): the solar system’s (and of course all other stellar menageries’), freedom to morph and flow more easily while homogeneously distributing multiscale imperfections (perturbations and gravitational resonances), more efficiently to optimise system robustness.
    Phi thereby (as RT correctly surmised), being the geometric signature of the new universal Constructal design law of nature, a relationship I’ve described as the Asynsis principle (asymmetric, asymptotic, analogical, synthesis – if you will): http://asynsis.wordpress.com/2013/10/15/is-there-a-new-geometric-law-of-thermodynamics/

  183. oldbrew says:

    @ wayne

    What we find is a lot of ‘almost’ harmonics where the relationships are in some way ‘just off’ an exact mean motion resonance number when examined closely enough.

    An obvious example is the Jupiter-Saturn synodic of 60 Saturn = 149 Jupiter orbits = 89 J-S synodic periods. 60:149 can appear to be 2:3 at a glance but the maths show the exact result.

    Likewise Neptune-Pluto is said to be 3:2 but in fact is nearer to 1503:1000. And so on – we have so many examples, it appears this is the rule not the exception.

  184. tallbloke says:

    I think what is happening is that energy is being transferred by resonance, which pushes the planets slightly out of exact resonsant ratio. That means the energy transfer drops off again, and the planets try to fall back into their preferred orbital ‘valleys’. A balance point is reached. Depending on the strength, number and timings of the perturbations, eccentricities are increased, or decreased.

  185. wayne says:

    TB&OB (hey, I like the sound of that):

    OB, that is basically what I was saying, they are not exact and that may be the way they *must* be to be stable. I sit and look at the sequences that program spit out and it rang a bell with me (but keep in mind I may be completely wrong). The more precise the ratios became the larger the number of orbits it would take to reach a perfect synch. As I loosen up then time period limit they get closer and closer but then it was taking many thousands or millions of orbits before they would synch at a ratio and I almost visualize that as a spirograph-like pattern of points that never comes quite back to the exact point and keeps that one pair always out of synch with a neighbor pair since it slowly rotates because it off a bit but always close to a Phi ratio. That same pattern would precess around and in some 1000+ repeats would then be back to where it was long ago but the neighbor pair would always be at a different place in its precession.

    It’s like they match (close) but don’t match all at once. And that was my question, is there any way except a guaranteed irrational relationship like Phi since any two numbers can start such sequence that is always irrational that would guarantee a close yet randomized state of n bodies. Possibly could be something there, I have never read anything even close along those lines.

    Simply, could it be Phi being irrational but such sequences can start at any pair of numbers that divide stable systems from unstable systems? Had to call that thought intriguing.

    Hmm, don’t know how you would ever even test such. need to know more if anyone has concrete info on when close encounters occur, or not, and if anyone has come up as to why if analysis of n-body solar systems has advanced that far. I don’t know, been years since I was interested in that area.

  186. tallbloke says:

    Wayne, I was looking at a paper the other day which modeled the evolution of a planetary system. Plenty of collisions. Of course, whether this model was any good or not is another question. Probably not because it didn’t mention lognormal distribution or treat resonance in anything but oversimplistic terms. :)

  187. oldbrew says:

    Phi is irrational but its cousin Fibonacci is not.

  188. wayne says:

    Of course. The 0,1,1,2,.. sequence and the furthrer you go… its consecutive ratios converge to Phi. But also any two in the same manner do the same, new to me. The title of this post says “Why Phi?” and I’ve just been exploring. Maybe I should have made all of this by email, I sure don’t want anything I say to affect what you are currently doing, just something to think on a year or so down the road, maybe you and TB’s paper two. The solar system and Phi, got to admit it is interesting.

  189. Roger Andrews says:

    OB says: “Phi is irrational but its cousin Fibonacci is not.”

    Any numerical sequence where the next number is the sum of the previous two will converge on Phi. It doesn’t matter what the two seed numbers are. Thus it works for the Fibonacci sequence:

    0, 1, 1, 2, 3, 5, 8, 13 etc.

    For the Lucas numbers:

    2, 1, 3, 4, 7, 11, 18, 29 etc.

    For sequences where the seed numbers are plucked out of the air:

    2, 4, 6, 10, 16, 26, 42, 68 etc.
    42, 0, 42, 42, 84, 126, 210, 336 etc
    1234, 5678, 6912, 12590, 19502 etc.

    And even when one of the seed numbers is negative:

    -22, 23, 1, 24, 25, 49, 74, 123 etc.

    With an effectively infinite number of numerical sequences that converge on Phi to choose from, the question isn’t “why Phi?”, it’s “why Fibonacci?”

  190. oldbrew says:

    RA has asked this question (“why Fibonacci?”) before. Earlier today I wrote:

    ‘An obvious example is the Jupiter-Saturn synodic of 60 Saturn = 149 Jupiter orbits = 89 J-S synodic periods. ‘ 89 is a Fibonacci number.

    8 Earth orbits = 13 Venus orbits = 5 synodic periods. 5,8, and 13 are Fibonacci numbers.

    The list goes on.

  191. tallbloke says:

    OB: I’m not sure that actually answers Roger A’s question. But what does answer it is the fact that Fibonnaci ratios are resonant. Especially 2:1, 2:3, 3:5. and 5:8. So the question becomes: what brings about the love/hate relationship planets have with resonant ratios. It’s like they need the power they transfer in order to maintain their places in the system, but have to shy away a little so it doesn’t burn them too much.

  192. Chaeremon says:

    @ob, @tb: if the only prime divisors in the ratio figures would be 1, 2, 3, 5, 7, 11 (sorry) and not 13 or higher, then it could be possible there is a physical cause which we eartlings observe in the maths. I’ve noted that for later.

  193. Roger Andrews says:

    OB: I went back and rechecked my analyses of the relationship between Fibonacci numbers and the synodic conjunctions of planetary pairs and confirmed to my satisfaction that it’s random. But I’ve already shot my mouth off enough about this and will say no more.

    TB: “So the question becomes: what brings about the love/hate relationship planets have with resonant ratios.” Saturn and Uranus both complete a full orbit once every 50th synodic conjunction and Uranus and Neptune both complete a full orbit once every 25th. Is this an in phase (i.e. love) or out of phase (i.e. hate) relationship?

  194. tallbloke says:

    Roger A: Wayne has run a test with random orbital periods and found the ratio numbers are around 50% lower with the real orbital periods for a given precision.

    There’s a bit of a mystery to be solved with the ‘full orbits’ obs. I also got 25 for U-N’s synodic conjunction precession cycle from calculation, but when I checked with an ephemeris, it was 21 and a bit. Similarly with J-S there’s a discrepancy. I think the reason is that the average figures given in NASA data are for ‘current epoch’ and actually change over timescales of hundreds of years due to interaction between the gas giants.

    The S-U synodic conjunction period is 45.36yr
    The U-N synodic conjunction period is 171.389yr
    The beat period of these two synod pairs is 61.68yr

    The secondary beat period of S-J is 61.02yr


  195. oldbrew says:

    Another aspect is that the longer the orbit period is, the more the barycentric movement of the Sun during that period must become a factor in the ‘shape’ of the orbit. Whether that is factored in to the given orbit period data, I don’t know.

  196. oldbrew says:

    @ Chaeremon

    We find prime divisors of 34 (Saturn-Uranus) and 233 (Jupiter-Saturn) in the Grand Synod cycle.

    Also as mentioned earlier Jupiter and Saturn converge in terms of whole numbers of orbits at 60(S) and 149(J) in 89 conjunctions. 34,89 and 233 are Fibonacci numbers.
    89 J-S is 1767.44 years. The number of Jupiter-Earth conjunctions in that period is 1767.44 – 149 (= no. of J orbits) = 1618 = Phi x 1000.

    Uranus and Neptune converge every 25 conjunctions. After every 50 conjunctions Uranus has completed 34 x 3 orbits and Neptune 13 x 4.

  197. Chaeremon says:

    @oldbrew: thanks, I’ll put that into my Nuremberg Funnel and wait ;) for what comes out.

  198. oldbrew says:

    Consider also that in the period of 863 x 233 Jupiter-Saturn conjunctions there should be 23299 Uranus-Neptune conjunctions, based on the average length of each one.

    That can be broken down further:
    863 = 144 x 6, minus 1.
    23299 = 233 x 100, minus 1.

    These numbers may look dull, but remember the context: each U-N conjunction takes 171.39 years and every set of 233 J-S conjunctions takes just over 4627 years. Note also that 233 is a prime number so cannot be a multiple of another figure.

    Footnote: 23299 U-N conjunctions = nearly 4 million years

  199. Roger Andrews says:


    Here’s the plot of the S-U and U-N pairs that got me intrigued. (Note that we get the match only when the synodic conjunction number is plotted on the x-axis. It falls apart when we plot elapsed time.)

    Now some thoughts based on the above. Here’s the same plot for Venus-Earth.

    The fifth conjunction is close enough to a whole number of rotations for both planets (12.993 Venus, 7.993 Earth) to appear on Oldbrew’s list of Fibonacci suspects, but by the 10th conjunction the Venus and Earth orbits have drifted apart to the point where OB’s rotational criteria aren’t met. The divergence then gets worse with each following 5-year conjunction to the point where at conjunction 200 the orbits are 264 degrees out of whack.

    Maybe this drift is related to barycentric motion or some other physical effect, but maybe it’s simply a result of errors in the V or E orbital period estimates and would go away if we had precise numbers. So I began to tweak the orbital periods to make it go away and got a match to within 0.001 degrees at the 200th conjunction by increasing Venus’s orbital period by 0.03% from 224.701 to 224.77329 days while keeping Earth’s constant at 365.256 days. With more tweaking and enough decimal points I could get an effectively exact match that went on for ever.

    This result highlights the sensitivity of this kind of analysis to small errors in orbital period, but we already knew about that. What I was wondering was whether I might not have stumbled across a way of making better orbital period estimates?

    Just asking. :-)

  200. Chaeremon says:

    Roger Andrews wrote: So I began to tweak the orbital periods to make it [this drift] go away and got a match …

    Oh, God [pun intended] but, you pointed to a possible method for which I thank you very much.

    Let’s find out how many orbits (revolutions), per planet, have been tallied and since when (work like the ancients). From this information it looks to be possible to grasp some of these mismatches.

    B.t.w. just a few years ago I wouldn’t have thought that such research seemed to be necessary in the 21th century. And now, I’m reluctant about my own expectations. Anybody have an account on ask.com?

  201. Roger Andrews says:


    You’re very welcome.

    While I’m here let me correct a minor mathematical error. The Venus orbital period should have been 224.772923 days, not 224.77329 days. The revised number gives a match within 0.0001 degrees at the 200th conjunction.

  202. oldbrew says:

    Astrophysicist Ian Wilson noted in one of his recent posts:

    ‘the E-V-S line slowly revolves about the Sun, taking 150 EV alignments (= 239.7990 years) to move backwards [clockwise in the above diagram] by one point in the five pointed star or pentagram pattern.”


    My calcs are based on the NASA figures.

    The overall patterns can be over long periods, often many hundreds or thousands of years, but they are there according to the numbers that emerge.

  203. oldbrew says:

    @ Roger A

    There should be a link between every 9 U-N and 34 S-U in about 1542.5 years (also 43 S-N).

  204. Roger Andrews says:


    Thanks for you comments. I’m having trouble getting a handle on the VE realignments, which are very complex, but I can replicate your 9 U-N and 34 S-U alignments every 1542.5 years.

  205. oldbrew says:

    @ RA

    Re U-N/S-U: that’s a good sign your method can work. Note that 3 x 1542.5 years = a shade over 233 Jupiter-Saturn conjunctions. I believe that extra fraction is resolved as described here:


    For Venus-Earth at a guess it might be worth trying 1199 years (5 x 239.8) = 750 V-E.

    Re this: ‘Saturn and Uranus both complete a full orbit once every 50th synodic conjunction and Uranus and Neptune both complete a full orbit once every 25th. Is this an in phase (i.e. love) or out of phase (i.e. hate) relationship?’

    The 50 S-U is 77(S):27(U) and the 25 U-N is 51(U):26(N), so the U part of that is 27:51 or 9:17, so it’s slightly out of phase if we take 9:18 (1:2) as being ‘in phase’.

  206. Roger Andrews says:


    Finally back, having reviewed the question of VE orbital alignments. First let me confirm the numbers I’m using:

    Earth orbital period: 365.256 Earth days, or 1 Earth year
    Venus orbital period: 224.701 Earth day,s or 0.6151877 Earth years.
    Synodic period: 583.9236 Earth days, or 1.5986696 Earth years

    According to the synodic period we have the first realignment after 1.5986696 years, at which time the Earth has completed 1.5986696 orbits and Venus 2.598670. But if you’re looking for Fibonacci matches you can’t use this realignment. You need whole numbers, which you get only when both planets realign after completing a whole number of orbits. And when you start looking for whole-number-of-orbit VE conjunctions things get kinda complicated.

    Take, for example, the first whole-number VE realignment, which you’ve identified as occurring after 13 Venus and 8 Earth orbits. It’s indeed very close to 13/8 but it’s not exact. The actual numbers are 13 and 7.997440, a difference of 0.0026 Earth orbits, representing 0.92 degrees of angular divergence. There’s another close match at the 26/16 realignment, but the difference is now up to 0.0051 Earth orbits, or 1.84 degrees, and at the 130/80 realignment the difference reaches 0.0256 Earth orbits, or 9.22 degrees. These results prompt the question; how exact does a realignment have to be to qualify as a realignment?

    Well, here’s what I get using different limits-of-error criteria for VE realignments taken out to 5,000 Earth years:

    Absolutely, positively, unequivocally exact realignment: Probably never happens

    Realignment within 0.1 degrees:

    0, 721, 1442, 2163, 2884, 3605, 4326 years. (Regular 721 year cycle).

    Realignment within 0.5 degrees:

    0, 243, 478, 721, 964, 1199, 1442, 1685, 1920, 2163, 2406, 2641, 2884, 3127, 3362, 3605, 3848, 4083, 4326, 4569, 4804 years. (Note the absence of a regular cycle. The periods between realignments run in the sequence 243, 235, 243, 243, 235, 243, 243, 235 years etc.)

    Realignment within 1 degree (going out only a limited number of years here to save space):

    0, 8, 243, 478, 486, 713, 721,729, 956, 964, 1199, 1207 years. Now we’ve added a few of the near-miss realignments that occur eight years either side of the 0.5 degree realignments listed above, such as those at 1, 486, 713 and 729 years.

    Realignment within 3.6 degrees, or 1% of a full orbit (going out even fewer years here):

    0, 8, 16, 24, 211, 219, 227, 235, 243, 251, 259, 267 years. Now we’ve added a whole bunch of flanking realignments.

    The figure below shows what all these cycles look like when plotted up. Note the phase shift that occurs every ~240 years in the 8-year cycle. This probably explains the 243/235 year variation:

    In summary, the time it takes Venus and Earth to complete a whole-number-of-orbits realignment can be 721 years, 243/235 years or eight years depending on how we choose define a realignment. Like I said, kinda complicated.

    But the source of all the complication is the fact that the VE synodic period is 1.5986696 Earth years, not 1.6000000. If it was 1.6000000 we would get an exact VE realignment once every eight years, representing exactly 8 Earth and 13 Venus orbits. So how reliable are our orbital period measurements? We humans have been watching the Earth go round the sun since before the Druids built Stonehenge so we can be fairly sure that the 365.256-day number for Earth is good, but what about Venus’ orbital period? Could we have it 0.03% off, which is all it would take?

  207. oldbrew says:

    Thanks for that RA. I’ve only just read it so for now I’ll be brief re your last paragraph.

    The easiest conversion to 1.6 for V-E is 1267/1266, confirmed by:
    1267 x 5 V-E is around 1266 x 8 years.

    It’s quite normal to find this type of ‘offset’ in the planetary data. I gave an example earlier:

    The answer is usually to scale up to bigger numbers to try and find the overall pattern, bearing in mind no two planets interact in isolation from the rest of the system, especially their own other neighbour and/or neighbour pair.

    Btw that period (1266 x 8y) also = 34 x 5 x 3 Jupiter-Saturn conjunctions (all Fibonacci numbers).

    Venus precessional pentagram:

  208. oldbrew says:

    @ RA

    Here’s the value of what I called the ‘offset’ of Venus:Earth in terms of orbits:
    1 Venus orbit x (3294/3293) / 365.25 = 1.625 = 8/5

    3293 Venus orbits = 1267 V-E and 102 Jupiter-Saturn conjunctions
    1267 – 102 = 1165 = 233 x 5

    3293 = 89 x 37
    3294 = 54 x 61

  209. Roger Andrews says:


    All due respect and all that, but I’m not sure I’m getting my point across. So let me try another tack:

    By using a 99% match (i.e. within 0.01 of a complete orbit/rotation or 3.6 degrees of angular difference) as the threshold for defining a realignment we get the following Fibonacci numbers:

    First orbital realignment for VE: 13 V orbits, 8 E orbits
    First orbital realignment for NP: 3 N orbits, 2 P orbits
    First rotational realignment for JU: 13 J rotations, 8 U rotations
    First rotational realignment for SN: 21 S rotations, 13 N rotations

    But when we tighten the threshold to 99.9% (i.e. within 0.001 of a complete orbit/rotation or 0.36 degrees of angular difference) we get no Fibonacci numbers at all:

    First orbital realignment for VE: 395 V orbits, 243 E orbits
    First orbital realignment for NP: 750 N orbits, 499 P orbits
    First rotational realignment for JU: 502 J rotations, 289 U rotations
    First rotational realignment for SN: 577 S rotations, 357 N rotations

    And choosing a threshold is, of course, a subjective exercise.

    I also can’t find anything about potential errors in the Venus orbital period estimate on the web, so maybe you or someone else could assist me by filling in the blank:

    Venus orbital period = 224.701 +/- _ _ _ _ _ _ days.

    Thanks :-)

  210. tallbloke says:

    Roger A: The difference between your S-N rotation rate ratios is 0.053%
    This is the sort of error climatologits can only dream of.

  211. oldbrew says:

    RA : Venus orbit is confirmed as 224.701 days on the NASA fact sheet.

    You’re saying it’s hard to find a 100% match and I’m not surprised, so we seem to agree?

    What we have is a lot of ‘almost matches’ to Fibonacci at low numbers, which on closer inspection require higher numbers to show the true pattern (within 0.001 of an orbit or very near to that).

    The higher numbers still won’t be 100% Fibonacci matches and if that’s how the system works I don’t have a problem with it. For example there are about 1503 Neptune orbits to 1000 Pluto, but it’s mostly described as a 3:2 ratio. The time period for that is nearly 250,000 years.

    Precession probably plays a part in the results too.

    PS ‘First orbital realignment for VE: 395 V orbits, 243 E orbits’

    Looks like confirmation of the Venus transit period :-)

    ‘Transits of Venus are among the rarest of predictable astronomical phenomena.They occur in a pattern that generally repeats every 243 years, with pairs of transits eight years apart separated by long gaps of 121.5 years and 105.5 years. The periodicity is a reflection of the fact that the orbital periods of Earth and Venus are close to 8:13 and 243:395 commensurabilities.’


  212. Roger Andrews says:

    Climatolo-gits? Sounds appropriate. :-)

    Back soon with a big long table of numbers.

  213. Roger Andrews says:


    “What we have is a lot of ‘almost matches’ to Fibonacci at low numbers, which on closer inspection require higher numbers to show the true pattern (within 0.001 of an orbit or very near to that).”

    I calculated the number of orbits or rotations needed for each of the 36 single-planet pairs to realign within 0.001 of a full orbit or rotation, and FYI here are the results:

  214. wayne says:

    Roger Andrews, the 365.256 is really something like 365.256363898 (from memory years ago so check current value). Does that do much the same (or part of) the correction you were making to Venus’s period?

    I agree, we need closer values. IIRC those values to 8-12 digits are buried somewhere on JPL’s site but I can’t seem to find them now. Might have been in the header of one of the DE-40x ephemeris files. Looking now but they seem to only list the GMs and state vectors for an epoch.

    Need the accurate orbit parameters and century deviations of each sure would help here.

  215. Roger Andrews says:


    Adding more digits to the Earth’s orbital period won’t make any significant difference to the periodicity of the Earth-Venus pair. But when I change it from 365.256000000 to 365.256363898 the first realignment at 395/243 orbits no longer meets my 99.9% orbital match threshold and gets replaced by 777/478 orbits.

    Yup, it’s that sensitive.

  216. wayne says:

    Right RogerA, I just don’t want to spend too much more time, the programs are all written and ready, just using such rough data that, as you said, might not be producing correct results once accurate data is used instead. My programs also hinge on error terms that determine the inclusion/exclusion from consideration.

  217. oldbrew says:

    That’s a useful list RA, thanks. Re your discussion with Wayne about the sidereal period, it appears you’ve used 24 hours for the rotation period as indeed we did on the rotations post. However NASA use 23.9345h as the sidereal rotation period and refer to 24 hours as LOD.

    Re Venus-Earth I find that 750 V-E = 1199y = 1949 Venus orbits (within 0.00002%)
    The true 5:8:13 ratio would be 750:1200:1950

    Once past six decimal places we’re talking billionths which won’t make much difference to the results unless you’re looking at really vast timescales.

  218. tallbloke says:

    As I’ve pointed out, planets get pushed a little way off the perfect ratios by resonance. This means the precision thing, while interesting, doesn’t inform us as to whether the Fibonacci idea is ‘right or wrong’. It just gives us clues about how much energy is being transferred in the resonant couplings. Even then the issue is complicated by the number of perturbing bodies and the strength of their actions on a given body.

    I’m satisfied that there are real relationships here. It’s not random, and Roger A’s precision argument can’t show that it is anyway, for the reasons given above.

    It’s better to consider this from the big scale downwards in order to see the non-random nature of the system.
    1) The Neptune-Uranus synodic precession cycle is in a 2:3 with the Halstatt cycle.
    2) The halstatt cycle is in a 1:2 with the V-E synodic precession cycle and a 1:1 with the J-S synodic precession cycle.
    3) The V-E synodic precession cycle is in a 2:3 with the ‘star point’ precession of the J-S synodic precession cycle
    4) The V-Me synodic precession cycle is in a 5:1 with the ‘star point’ precession of the V-E synodic precession cycle

    These long term resonant relationships which operate on very subtle and intermittent applications of force are proof in themselves that the more obvious shorter term close to resonant ratios are mechanically effective and non-random.

  219. wayne says:

    RogerA: Should you need some accurate SS figures see:

    TB: Jeesh, I had that Chap8 file on my own machine all along and had forgotten it also had th orbit elements all listed within. Bong. Think it was used in solving that SSB question month ago. Short term memory not on all eight cylinders or just plain overloaded. ;)
    As you say this added accuracy is not necessary for what you are doing but I’m a stickler for leaving such a set of programs in the best shape if I should come back years from now needing them for some odd reason. Just takes a bit more to leave them complete. (See, once again I’d probably not be able to locate that file again :) )

  220. Chaeremon says:

    @wayne: the Chap8 file mentions The observational basis for JPL’s DE 200, the planetary ephemerides of the Astronomical Almanac, wish we had at least this level of detail, reference and (mathematical) accuracy for geocentric climate data …

  221. Roger Andrews says:

    “I’m satisfied that there are real relationships here. It’s not random, and Roger A’s precision argument can’t show that it is anyway, for the reasons given above.”

    TB, you’re misreading me. I agree there are real relationships here and that they’re not random. In fact I’ve already presented a number of examples that strike me as very real and non-random, in particular the log plot of the orbital distances of the Galilean moons of Jupiter, which has a trend very close to Phi.

    I also agree that my “precision argument” doesn’t prove these relationships don’t exist. It was never my intention to prove they don’t exist. I’m simply pointing out the difficulty of expressing them in terms of whole numbers, which don’t seem to occur very often in the solar system.

  222. tallbloke says:

    Roger A: OK, that’s good to hear. I think the value of ‘doing the checks by hand’ is in discovering the interesting coincidences pop up. Like the near whole number of rotations involving V & E that was the same number as the orbits in a subdivision of the synodic precession cycle. It’s those sorts of observations which have enabled Stuart and I to make some sense out of all this. Still some way to go before we can generalise ‘rules’ though…

  223. oldbrew says:

    @ RA

    The synodic relationships offer another route to finding the Phi links, e.g.:

    89 Jupiter-Saturn = 1618 Jupiter-Earth
    9 J-S = 89 Saturn-Mars
    233 J-S = 34 x 3 Saturn-Uranus

    89 J-S may also be the nearest we get to a whole number match (60 S and 149 J) – 2.5 days difference in 1767+ years. Probably not a coincidence that these are the two biggest planets and of course neighbours.

  224. Chaeremon says:

    Rational Numbers Distribution is linked to resonance phenomena and also to stability of oscillating systems

    Note this author arrives at phi (his number M1) and (quote) takes Laplace’s classic works forward (unquote). Ah, he puts number distribution at work for orbital radii of planets and also orbital periods.

  225. Greysen says:

    oldbrew you made a remarkable work, especially your calculation using conjunction Jupiter-Saturn, I can assure you this is not only conjecture you touch it with a niddle especially this part :

    “25560.276 / 178.74 = 143.00255 (no. of Jose cycles in sopc)
    25560.276 / (143 x 16) = 11.1715 (one solar cycle, 16 per Jose cycle)
    25560.276 / 19.86 = 1287.02 = 143 x 9

    9.93 x 365.25 = 3626.9325 days
    3626.9325 / 25.38 = 142.9 (no. of solar rotations in 9.93 years)
    NB 9.93 x 9 = 89 (89.37)
    (143 cf. Fibonacci 144 ?)
    24007.816 / 168 = 142.904 (143) ”

    Thank you TB to open this new door here, with new opportunities. I look forward to read more about your discoveries of this subject oldbrew.

  226. oldbrew says:

    @ Chaeremon

    Not sure what to make of that paper, but as you say the author does include phi in his workings.

    @ Greysen

    Thanks for your interest. You can read some newer posts on this topic by putting ‘Phi’ in the ‘search this site’ box at the top left of the page.

  227. From the recent exchanges it looks increasingly like we are witnessing an edge of chaos/self-organised criticality phenomenon here where the solar system dynamical symmetries are poised in a phase-transition/cantor set zone that allows for universal computation as it fine tunes optimal period, spin/angular-momentum rates, gravitational/EM resonance-minimisation and the like in the face of the resistances it encounters as it coils around our undulating galactic disk (whose spiral arms are BZ-like reaction waves of auto-catalytic star-formation).
    As Adrian Bejan said to us in London this May 2013 when we visited ARUP’s Foresight & Innovation team, just look for what is flowing in the system as it seeks optimal freedom to flow and to morph and you will solve your problem.
    The optimal, analogical geometric signatures of these Constructal law processes over time will reveal nested φ relationships because this is an innate, fundamental systems evolution constant of nature, a geometric manifestation of the principle of least action. http://asynsis.wordpress.com/2011/05/26/dynamical-symmetries-asynsis/

  228. wayne says:

    @ Chaeremon, in Dombrowski’s article, thanks for the interesting read, that is a strange sequence he is using indeed! You can find in that integer set that, for one thing, each is the continued fraction of that single integer repeated. Neat. But the more surprising fact is that every odd integer as the power of the silver ratio can be found as the fraction portion plus the integer entry itself. For example:
    1 is 1 + Φ^1
    4 is 4 + Φ^3
    11 is 11 + Φ^5
    29 is 29 + Φ^7
    76 is 76 + Φ^9
    199 is 199 + Φ^11
    and so on. Quite close on the periods. At first I thought it was popping out the primes for I misread the 9th root entry at 76 but it ended up being just all odd powers. I’ll say again, that’s one curious sequence.

  229. wayne says:

    Hmm… and running backward into the negative integers it is just Φ^p, p being all odds. But the bases grow very rapidly, nearly exponential or maybe a form of a polynomial?
    -1 is Φ^1
    -4 is Φ^3
    -11 is Φ^5
    -29 is Φ^7
    -76 is Φ^9
    -199 is Φ^11

  230. oldbrew says:

    ‘Place your mouse over this image, and you will see the Fibonacci spiral at work!’ – it says here:


  231. oldbrew says:

    3 Uranus-Earth conjunctions = 2 x Phi Neptune-Earth conjunctions

    because (using NASA data):
    63 U-E @ 396.66d = 24989.58 days
    68 N-E @ 367.49d = 24989.32 days

    63 = 21 x 3 and 68 = 34 x 2
    21 x 3 U-E = 34 x 2 N-E
    divide both sides of the equation by 21
    3 U-E = 34/21 x 2 N-E
    34/21 is Phi
    3 x U-E = 2 x Phi x N-E