Sun’s core rotates four times faster than its surface

Posted: August 1, 2017 by oldbrew in research, Solar physics
Tags:


Something new for solar theorists to ponder. Of course the surface itself doesn’t have a uniform rotation speed – at the equator it rotates faster than it does at the poles. ‘The idea that the solar core could be rotating more rapidly than the surface has been considered for more than 20 years, but has never before been measured’.

The sun’s core rotates nearly four times faster than the sun’s surface, according to new findings by an international team of astronomers. Scientists had assumed the core was rotating like a merry-go-round at about the same speed as the surface, says Phys.org.

“The most likely explanation is that this core rotation is left over from the period when the sun formed, some 4.6 billion years ago,” said Roger Ulrich, a UCLA professor emeritus of astronomy, who has studied the sun’s interior for more than 40 years and co-author of the study that was published today in the journal Astronomy and Astrophysics.

“It’s a surprise, and exciting to think we might have uncovered a relic of what the sun was like when it first formed.”


The rotation of the solar core may give a clue to how the sun formed. After the sun formed, the solar wind likely slowed the rotation of the outer part of the sun, he said. The rotation might also impact sunspots, which also rotate, Ulrich said. Sunspots can be enormous; a single sunspot can even be larger than the Earth.

Continued here.

Comments
  1. oldbrew says:

    ‘After the sun formed, the solar wind likely slowed the rotation of the outer part of the sun, he said.’

    So the sun slowed itself?
    – – –
    Abstract: ‘Here, P0 is measured to be 34 min 01 s, with a 1 s uncertainty. The previously unknown g-mode splittings have now been measured from a non-synodic reference with very high accuracy, and they imply a mean weighted rotation of 1277 ± 10 nHz (9-day period) of their kernels, resulting in a rapid rotation frequency of 1644 ± 23 nHz (period of one week) of the solar core itself, which is a factor 3.8 ± 0.1 faster than the rotation of the radiative envelope.’ [bold added]

    http://www.aanda.org/articles/aa/abs/2017/08/aa30460-17/aa30460-17.html

  2. oldbrew says:

    ‘the Earth’s inner core is rotating faster than its surface by about 0.3-0.5 degrees per year’

    Not like the solar result, but interesting.

  3. JB says:

    “The most likely explanation is that this core rotation is left over from the period when the sun formed, some 4.6 billion years ago,” said Roger Ulrich, a UCLA professor emeritus of astronomy.

    I don’t believe him. An inertia driven universe just doesn’t fit.

  4. Curious George says:

    Nothing to do with solar wind. Any movement with a vertical component on a rotating body tends to rotate slower when going up, faster when going down. Same as Coriolis force.

  5. Am I the only one that thinks the sun is driven by something? That the sun has an external source? Everything we observe has a source of energy, why is the conclusion that stars don´t have a source? Because we don´t see it? I don´t like to base my assumptions on what I don´t see.

    Everything else we observe is driven by a source in shape of a force, heat or mass flow, from something with higher energy, right? Is it rational to make an exception for stars? Don´t ask me about black holes and their relationships with stars, I only have fantasies about that;)

    Who is confident in math here? Just hijacking for a moment, considering sources and stuff. Some may have seen my hillbilly-calculations using only TSI as a single driver for earth energy flow. Now I think that my calculations causes infinite number of solutions for g, within the limits of the heat flow. For example: TSI/(4/3)=4/3*8g^2. For anyone interested, 4/3*g^2 is equal to the mean emissive power of the tropopause, and 4g^2 is equal to surface emission, calculated from 1/2*TSI(4/3). There seems to be no end to the solutions when using the volume function and the inverse square law for the flow. The inverse square law can be folded through the surface, a point of irradiation seems to lie next to a point at altitude of emission, if you understand how I mean. Vectors, scalar field, divergence and Maxwell equations seems to arise from geometry inside the sphere. I am learning on the fly.

    What do you think? Anyone? Infinite number of solutions, what should I make of that?

    Sorry, TB, for hijacking. But you run the best comment field on climate science and stuff. Rational and good people here. I guess I am the looney one.

  6. Gaah… 1/2*TSI/(4/3)^2, that is what I should have written. To much numbers in my overheated brain. But that´s natural, Life is thermal.

  7. oldbrew says:

    Even galaxies rotate.
    http://www.universetoday.com/30710/galaxy-rotation/

    Where’s the galactic dynamo – down a black hole perhaps 😉
    – – –
    Wikipedia’s ‘solar dynamo’ page has ‘multiple issues’ – they say.
    http://en.wikipedia.org/wiki/Solar_dynamo

    They say it’s ‘essentially a naturally occurring electric generator in the Sun’s interior’. How many of those did the Big Bang create?

  8. tallbloke says:

    OB: Indeed. Magnetic stuff rotating in a magnetic field generates both a magnetic and an electric field. Wheels within wheels, driven round by,,, the next bigger wheel outside. Galaxies rotate, inside rotating galaxy clusters. Where the mind starts to boggle is when you ask whether the universe as a whole is rotating, and if it is, what’s it rotating relative to? The biggest spinning turtle you could conceive of maybe…

    “It’s spinning turtles all the way down professor…”

  9. tallbloke says:

    But to get back to the topic, If the Sun’s core is spinning 4 times faster than the surface, isn’t that likely to be due to a coupling between the Sun and the more slowly spinning IMF mediated by the solar wind? That would slow the part of the Sun in direct contact with the solar wind, i.e. the surface. Viscous drag would then slow the subsurface layer, and so on down to the core. The planets orbit more slowly than the Sun spins. Their magnetospheres are coupled to the IMF too…

  10. oldbrew says:

    Then there’s the photon-braking effect…

    Escaping Photons Slow Down the Surface Rotation of the Sun
    February 7, 2017

    http://www.insidescience.org/news/escaping-photons-slow-down-surface-rotation-sun
    – – –
    ‘Our universe may exist inside a black hole. This may sound strange, but it could actually be the best explanation of how the universe began, and what we observe today.’

    http://www.insidescience.org/news/every-black-hole-contains-new-universe

    A spinning black hole presumably 😐

  11. oldbrew says:

    This looks interesting…

    New clue to solving the mystery of the Sun’s hot atmosphere
    August 3, 2017

    The elemental composition of the Sun’s hot atmosphere known as the ‘corona’ is strongly linked to the 11-year solar magnetic activity cycle, a team of scientists from UCL, George Mason University and Naval Research Laboratory has revealed for the first time.
    . . .
    Through its 11-year cycle, the Sun moves from relatively quiet periods at solar minimum, to intense magnetic activity at solar maximum, when large numbers of sunspots appear and there is an increase in radiation.

    “Previously, many astronomers thought that elemental composition in a star’s atmosphere depended on the properties of the star that don’t change, such as the rotation rate or surface gravity. Our results suggest that it may also be linked with the magnetic activity and heating processes in the atmosphere itself, and they change with time, at least in the Sun,” said the study’s lead author, Dr David H. Brooks (George Mason University).
    . . .
    “Why the Sun’s corona is so hot is a long-standing puzzle. It’s as if a flame were coming out of an ice cube. It doesn’t make any sense! Solar astronomers think that the key lies in the magnetic field, but there are still arguments about the details,” added Dr Brooks.

    Read more at: http://phys.org/news/2017-08-clue-mystery-sun-hot-atmosphere.html
    – – –
    Where does that leave ‘the solar constant’?

  12. p.g.sharrow says:

    Astro-physicists have a problem with understanding “temperature” of energies. Thermal in degrees are not the same as radiation energies in degrees. Solar mater is heated thru conduction while it also carries energy of radiation. The Sun has a solid core, a “liquid” body, a gaseous atmosphere with a tropospheric “surface” photosphere, and a plasma corona. The last, it’s material; too defuse for conduction, is heated and cooled only thru radiation. Most of the radiation just bangs about and only some of it escapes into space.The wispy plasma material appears to be super heated while the escaping “heat” energy is of much lower temperature. Plasma, a 4th state of mater, is super energized protons and their force field repels all other particles, even electrons.
    The spiraling out of AM ,Angular Momentum, from the sun’s massive core drags the planetary system thru it’s gravity/EMF coupling with them. This transfer of energies tends to keep them in their order. The Inertial drag of the Aether slows while the AM of the Sun speeds the planetary energy exchange…pg

  13. p.g.sharrow says:

    @Tallbloke;
    I look forward to your summation of solar system energy transfers that regulate the motions of the bodies involved.

    From the largest to the smallest or the smallest to the largest, the physics is all the same. GOD is not a mathematician but works in applied science. K.I.S.S. Complex answers mean you haven’t fully grasped the question. 😉 …pg

  14. p.g.sharrow says:

    In the beginning there was only CHAOS. Then there came the word “GOD” and there was order.
    Light and darkness, Mater and void. Complexities built from simplicity.

    Kind of old fashion I know but, it works for me. Energy in chaos, or dark mater/energy in chaos, without organization. Organize some of it and the first true singularity, a proton is born with it’s electron shell, Hydrogen! Mater, Creation, GOD! expanding fields of Electro-Motive-Force as well as Mass/Inertia and gravity. All creatures of organized energy. All follow the same simple principals of physics…pg

  15. tallbloke says:

    The rotation rate of the Sun’s surface has obviously reached a dynamic equilibrium between something trying to slow it down and something trying to speed it up. The core rotates faster, and viscous drag will resist shear between the differential rotation rates, so that’s what is trying to speed the surface up.

    So what’s slowing the surface down? It has to be a coupling between the surface and something in the surrounding space. What’s in the surrounding space? The Sun’s own magnetic field and the planetary magnetospheres it interacts with. The planets all orbit much slower than the Sun spins on its own axis, so that magnetic coupling will create drag. Jupiter has by far the biggest magnetic field of any planet.

    Something I noticed long ago which has no obvious explanation is that measured in Earth years, the inverse of Jupiter’s orbital period is equal to the Sun’s rotation rate, which seen from Earth, is similar to the orbital period of our moon.

    Wheels within wheels. If only we could see all the invisible gears that connect them…

  16. oldbrew says:

    8 Jupiter orbits = 99.9% of 5 Metonic cycles (about one week short per 19-year Metonic)

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

  17. p.g.sharrow says:

    We exist in an ocean of energy in chaos. We only “see” perceive, events or conditions of organization in that ocean. EMF, gravity, mass/inertia, light etc. are all manifestations of organization in that energetic fluid. Aether…pg.

  18. tallbloke says:

    That may well be so p.g.
    Through observation and the testing of hypotheses, we attempt to make that ‘chaos’ intelligible. Going back to what I said about the average rotation rate of the Sun being the inverse of the Jupiter orbital period (in Earth years); I just noticed that the inverse of the Jupiter-Saturn synodic period is the Sun’s polar rotation rate of around 36 days.

  19. oldbrew says:

    TB – re. ‘the inverse of the Jupiter-Saturn synodic period’

    1/19.865y = 18.386~days
    * 2 = 36.77~ days

  20. tallbloke says:

    Sorry, quite right. I meant the half (tidal) period: 9.93 years

  21. Paul Vaughan says:

    V = [5E-√5T(ΦΦ)^e+1/25]/3 = 1 / 0.615196915778351
    V = [5E-Φ√5(J+S)+1/25]/3 = 1 / 0.615197163689601
    V = [5E-φJ+ΦΦS]/3 = 1 / 0.615197069188456

    Tightening up lunisolar-Hale frequency algebra by an order of magnitude yesterday the preceding simplifications turned up.

  22. p.g.sharrow says:

    Examine Archimedes pry-bar. The Barycenter of the solar system is your fulcrum and the mass/inertia of the various bodies along that bar must be in balance because the solar system is a unit. Conservation of energy requires it. Your Fibonacci spiral is the direct result of this balancing of Angular Momentum of mass/inertia over distance…pg

  23. Paul Vaughan says:

    “balanced multi-axial differential” (repeated for many years) = same …(language barrier maybe)

  24. Paul Vaughan says:

    It’s a simple minimization problem. The context is spatiotemporal, not temporal. In the old paradigm, the role of geometry (a boundary condition) in the physical minimization was overlooked.

    I would have thought (based on standard childhood brainwashing that everything simple is already known) that all of this sort of stuff would have been well known, for example to engineers looking to exploit trivial geometry to minimize vibrations in machines or for geographers looking to simply engineer peace and stability …but a vacuous question mark arises trying to fathom why it’s up to us on volunteer hours to rediscover from scratch what’s obvious to sharper minds, as surely so many others have before and will later (wheel reinvention).

    This exploration and learning exercise is underscoring how dependent our sensory apparatus is on linearity. What can we do to increase by several orders of magnitude human ability to perceive nonlinear simplicity? With what general mathematical strategy can we linearize perception of ALL simple patterns, such that we don’t get lost in intersections of nonlinear rings sorting through disordered rafts of transformations and identities?

    Our machines may perfectly solve only within their algorithmic span; they have not the imagination needed to go beyond their logical limits. Teaching machines open-mindedness might be more tractable than teaching intuition. Guess-and-test is a valid learning method, so it can be an AI component even if the search time can’t be minimized by mechanized intuition. Another valid learning method is mutation and recombination — (misunderstandings are often more fun than “the truth”).

    Our exploratory race for stability hinges crucially on rediscovery of methods for exploring and discovering orders of magnitude faster. (Let’s have some really good fun not assuming recorded history is anything more than a laughable model of what really happened.)

    The primary factors limiting artificial stability engineering efforts are sluggish and failed perception and comprehension of simple nonlinear stability in nature. The major western fault is unnecessary dependence on linearity.

    Natural stability’s simpler than the complex imagination of the major western fault. A more naturally principled exploratory race is ripe for orders of magnitude faster fundamental discovery.

    In the immediate context we have a tractable step on a longer evolutionary path.

  25. tallbloke says:

    Paul V: V = (5E-φJ+ΦΦS)/3
    derive 5J ~= QBO.

    OK. we need to rearrange for J and multiply by 5 on both sides of the equation.

    5J = 5((5E+ΦΦS-3V)/φ) ~= 1/2.372 ~= QBO

    Is that right?

  26. Paul Vaughan says:

    I’ve learned to be more principled. Saving time for exploration is vital. Answering questions when the time’s unripe? …not so much!!

    V = [5E-(3J+S-1/25)/2]/3 = 1 / 0.615197116439025

    All of these expressions are derived systematically. Note that they are nearly equal. This is a minimization problem.

    The conventional mainstream hasn’t bothered with comparative exploration and classification of the (geometric) limits of (temporal) polynomial integration. The way they represent this stuff in the computer is a mess. They haven’t bothered with reduction. They just let the machine sort out the convergence numerically (brute force method — i.e. try to walk through a cement wall rather than looking for a door).

  27. tallbloke says:

    Paul V: Saving time for exploration is vital. Answering questions when the time’s unripe? …not so much!!

    Well you could think of it as marking the homework you set then. It’d save more time for exploration to give a simple “correct” or “incorrect because…” than type the above.

    It’d show more simple courtesy too.

    Paul V: It’s a simple minimization problem. The context is spatiotemporal, not temporal.

    Yes. That’s why the abstract of my PRP main paper states:
    “The forces of gravity and magnetism and the principles of energy conservation, entropy, power laws, and the log-normal distribution which are evident are discussed in relation to planetary distribution with respect to time in the solar system.”

  28. Paul Vaughan says:

    Nature sets the path and pace.

    Hale = Φ(T/E)(ΦΦ)^e = 1 / 22.1383572294165
    J+S = φ(T/E)(ΦΦ)^e = 1 / 8.45610000655013
    S = -(1/2)φ(T/E)(ΦΦ)^e+(Φ^5)+(1/2)[φ(φ+Φ)]^(-4) = 1 / 29.4472819261743
    J = +(3/2)φ(T/E)(ΦΦ)^e-(Φ^5)-(1/2)[φ(φ+Φ)]^(-4) = 1 / 11.8625602808965
    J-S = +2φ(T/E)(ΦΦ)^e-2(Φ^5)-[φ(φ+Φ)]^(-4) = 1 / 19.8649807488765
    V = [5E-(φ+Φ)(T/E)(ΦΦ)^e+(φ+Φ)^(-4)]/3 = 1 / 0.615197275210489

    Compare with observations:

    model period
    observed period
    %error
    %error log scale

    22.1383572294165
    22.1384769776823
    -0.000540906%
    -0.000174637%

    8.45610000655013
    8.4561457463176
    -0.000540906%
    -0.000253365%

    29.4472819261743
    29.4474984673838
    -0.000735347%
    -0.000217391%

    11.8625602808965
    11.8626151546089
    -0.000462577%
    -0.000187022%

    19.8649807488765
    19.8650360864628
    -0.000278568%
    -0.000093199%

    0.615197275210489
    0.615197263396975
    +0.000001920%
    -0.000003953%

    r^2
    0.999999999990233 period
    0.999999999999921 frequency
    0.999999999998969 log scale (same for both period and frequency)

  29. Paul Vaughan says:

    Polya problem solving strategy: Use symmetry.

    Ma = J+1/{[2(J+S)]^(1/3)+[2(J+S)]^(-1/3)}

    a simple balance of powers

  30. Paul Vaughan says:

    NOW you know why it’s not quite Ma = J+1/√5 = J+1/(φ+Φ).
    Thanks for your patience while questions were evaded for years.

  31. Paul Vaughan says:

    Recapping:
    Ma = J+1/{[2φ(ΦΦ)^e]^(1/3)+[2φ(ΦΦ)^e]^(-1/3)}

    Any formal summary omitting this would have been premature.

  32. Paul Vaughan says:

    Nothing’s OT.
    “Sun’s core rotates four times faster than its surface”

    Relative frequencies:
    (4/5) / (1/5) = 4

    Axial Frequency
    = Integral of Pareto Principled Division of Unity
    = 4/5 + 1/5 = 5/5 = 1

    5 = (Φ+φ)^2
    √5 = Φ+φ

    φ = 1.61803398874989
    Φ = 0.618033988749895

    [2(J+S)]^(-1/3) = 1.61701248631139
    [2(J+S)]^(1/3) = 0.618424414446623

    [2φ(ΦΦ)^e]^(-1/3) = 1.61703340324967
    [2φ(ΦΦ)^e]^(1/3) = 0.618416414893069

    φ-Φ = φ^(1)-φ^(-1) = Φ^(-1)-Φ^(1) = 1
    φ^3-Φ^3 = Φ^(-3)-Φ^(3) = 4

    φ^3-φ = φ^2
    Φ^3-Φ = -Φ^2

    φ^2-Φ^2 = (φ-Φ)(φ+Φ) = (1)(φ+Φ) = φ+Φ = √5

    Reminder: Me = φ^3-J = Φ^(-3)-J
    Suggestion: Review yesterday’s notes comparatively. Note symmetry of Ma with Me.

    √5 is the arithmetic mean of φ^3 and Φ^3.
    1/√5 is the harmonic mean of φ^3 and Φ^3.
    1 is the geometric mean of φ^3 and Φ^3.

    2(J+S)]^(-1/3) ~= [2φ(ΦΦ)^e]^(-1/3) ~= φ

    V/E is suggestively close to but not quite equal to φ.
    2(J+S) is suggestively close to but not quite equal to Φ^3.
    Exploration removed several orders of magnitude of fog, eliminating these mysteries.

    Me is suggestively close to but not equal to φ^3.
    Ma is suggestively close to but not quite equal to J+1/√5.
    Exploration has defogged these mysteries.
    More than an order of magnitude of fog has been removed thusfar.

  33. Paul Vaughan says:

    Review of JEV (Jupiter-Earth-Venus) derivation:

    (11.8626151546089)*(1.00001743371442) / (11.8626151546089 – 1.00001743371442) = 1.0920796543202
    (11.8626151546089)*(0.615197263396975) / (11.8626151546089 – 0.615197263396975) = 0.648846557532906

    (1.0920796543202)*(0.648846557532906) / (1.0920796543202 – 0.648846557532906) = 1.59868955949705
    (1.0920796543202)*(0.648846557532906) / (1.0920796543202 + 0.648846557532906) = 0.407020193867456

    harmonic nearest 0.407020193867456 is 1.59868955949705 / 4 = 0.399672389874261
    (0.399672389874261)*(0.407020193867456) / (0.399672389874261 – 0.407020193867456) = 22.1392314983839

    Closing in on Mars:

    √5 = 2.23606797749979
    φ = 1.61803398874989

    (11.8626151546089)*(1.88084761346252) / (11.8626151546089 – 1.88084761346252) = 2.23525255532127
    (11.8626151546089)*(1.88084761346252) / (11.8626151546089 + 1.88084761346252) = 1.62344612704193

    Note the similarity with √5 & φ and let your imagination follow parallel paths that focus like a lens.

    JEV harmonic nearest 2.23525255532127 is 22.1392314983839 / 10 = 2.21392314983839
    (2.21392314983839)*(2.23525255532127) / (2.21392314983839 – 2.23525255532127) = 232.011969669376

    JEV harmonic nearest 1.88084761346252 is 22.1392314983839 / 12 = 1.84493595819866
    (1.84493595819866)*(1.88084761346252) / (1.84493595819866 – 1.88084761346252) = 96.6272194493165

    JEV harmonic nearest 1.62344612704193 is 22.1392314983839 / 14 = 1.58137367845599
    (1.58137367845599)*(1.62344612704193) / (1.58137367845599 – 1.62344612704193) = 61.0203365856337

    Notice how they fit together:

    (61.0203365856337)*(232.011969669376) / ( (61.0203365856337 + 232.011969669376) / 2 ) = 96.627219449317 = harmonic mean

    Remember that we’ve seen this framework before:

    Ian Wilson: Help Needed to Solve an Interesting Lunar Puzzle

    This is about long JEV cycles and long JS cycles. Somewhere up this hierarchy lies a compact little set of simple expressions for Milankovitch Cycles.

    Some basic review plus a few more steps:

    (11.8626151546089)*(29.4474984673838) / (11.8626151546089 – 29.4474984673838) = 19.8650360864628

    (11.8626151546089)*(29.4474984673838) / (11.8626151546089 + 29.4474984673838) = 8.4561457463176

    φ/Φ = φφ = 1/(ΦΦ) = 2.6180339887499
    8.4561457463176 * 2.6180339887499 = 22.1384769776823
    22.1384769776823 / 2 = 11.0692384888412
    19.8650360864628 / 2 = 9.93251804323141
    29.4474984673838 / 2 = 14.7237492336919

    (11.8626151546089)*(14.7237492336919) / (11.8626151546089 – 14.7237492336919) = 61.0464822565173

    (11.0692384888412)*(9.93251804323141) / (11.0692384888412 – 9.93251804323141) = 96.7215918741308

    harmonic nearest 61.0464822565173 is 96.7215918741308 / 2 = 48.3607959370654
    (48.3607959370654)*(61.0464822565173) / (48.3607959370654 – 61.0464822565173) = 232.723433067725

    Focus derived from parallel insight:

    (1.58137367845599)*(61.0464822565173) / (1.58137367845599 – 61.0464822565173) = 1.62342762859098
    (11.8626151546089)*(1.62342762859098) / (11.8626151546089 – 1.62342762859098) = 1.88082278407335
    100*(1.88082278407335 – 1.88084761346252) / (1.88084761346252) = -0.00132011700412112%

    (1.84493595819866)*(96.7215918741308) / (1.84493595819866 – 96.7215918741308) = 1.88081189266319
    100*(1.88081189266319 – 1.88084761346252) / (1.88084761346252) = -0.00189918625339757%

    (2.21392314983839)*(232.723433067725) / (2.21392314983839 – 232.723433067725) = 2.23518672250026
    (11.8626151546089)*(2.23518672250026) / (11.8626151546089 + 2.23518672250026) = 1.88080100137918
    100*(1.88080100137918 – 1.88084761346252) / (1.88084761346252) = -0.00247824879619476%

    Compare the focused (second-order) estimates with first-order instinct:

    Target:
    (11.8626151546089)*(2.23525255532127) / (11.8626151546089 + 2.23525255532127) = 1.88084761346252

    First shot based on √5:
    (11.8626151546089)*(2.23606797749979) / (11.8626151546089 + 2.23606797749979) = 1.8814249265745
    100*(1.8814249265745 – 1.88084761346252) / (1.88084761346252) = 0.030694305474224%

    Target:
    (11.8626151546089)*(1.62344612704193) / (11.8626151546089 – 1.62344612704193) = 1.88084761346252

    First shot based on φ:
    (11.8626151546089)*(1.61803398874989) / (11.8626151546089 – 1.61803398874989) = 1.87358704127241
    100*(1.87358704127241 – 1.88084761346252) / (1.88084761346252) = -0.386026605140269%

    I introduced the sharper Mars derivation based on 96 before, but this comment underscores that derivation based on longer Jupiter–Saturn (JS) cycles is generalizable.

    Moving beyond first-order instinct we’re developing deeper instinct about the hierarchical geometry of offsets from base level central limits.

    I’ll leave it there for now and spell out some of the frequency algebra in a follow-up comment sometime. Then beyond that there are layers of work to sort concisely insights from trails explored more generally. The structure of my freedom from complementary pursuits guarantees this will take years.

    Summary

    First-order Mars estimates based on:
    φ:
    1.87358704127241 (-0.386026605140269%)
    √5:
    1.8814249265745 (0.030694305474224%)

    Note that they bracket (pointing to a simple inequality) the real thing:
    1.88084761346252 = 1 / Ma (Seidelmann 1992)

    Refined estimates (sharpened by 1-2 orders of magnitude) based on JEV and long JS cycles:

    1.88080100137918 (-0.00247824879619476%)
    1.88081189266319 (-0.00189918625339757%)
    1.88082278407335 (-0.00132011700412112%)

    The sharpest estimate is based on JEV and the “60 year” JS cycle.

  34. Paul Vaughan says:

    Link back to the core trail:

    Suggestions-28

  35. Paul Vaughan says:

    The most naive model for Mars’ sidereal orbit:
    • period = harmonic mean of √5 & φ.
    • frequency = arithmetic mean of 1/√5 & Φ.

    Inner Solar System First-Order Model Recap

    1 / 0.240846697327135 = 1 / Me
    1 / 0.23606797749979 = φφφ

    1 / 0.615197263396975 = 1 / V
    1 / 0.618033988749895 = φ

    1 / 1.00001743371442 = 1 / E
    1 / 1.00000000000000 = 1

    1 / 1.88084761346252 = 1 / Ma
    1 / 1.8774978038635 = (Φ+1/√5)/2

    r^2 = 0.999976601526245 (period)
    r^2 = 0.999912913357706 (frequency)

  36. Paul Vaughan says:

    OldBrew will probably be quick to note the Fibonacci pair most nearly approximating Mars:
    frequency = (8/13+1/(13/8+8/13))/2
    period = 2/(8/13+1/(13/8+8/13))

  37. oldbrew says:

    Mars orbit in years can also be expressed as (48/35)². This is accurate to less than an hour, maybe even less than half an hour, but may well just be numerology.

    – or (144/(21*5))² using Fibonacci numbers

  38. tallbloke says:

    Paul V: [2φ(ΦΦ)^e]^(-1/3) = 1.61703340324967

    More beautiful symmetry.

    Apropos nothing but if instead of (-1/3) we raise [2φ(ΦΦ)^e] to the power of -0.3337623775 we get 0.618033988724 vs φ = 1.618033988749…

    Just noting it here in case it turns out to relate to anything.

  39. Paul Vaughan says:

    It’s about second order balance — exploring the geometric balance of departure from first order. Not only does it not equal φ: The way in which it does not equal φ tells something very specific about EV. It’s not just J; it’s J and S and coupled E & V clued us in. Mars was (past tense) the biggest anomaly in v1. Now we have inner clarity as our exploration of outer limits and stability more generally endures.

  40. oldbrew says:

    Synodics: 3 Earth-Mars = 4 Venus-Earth = 7 Mars-Venus works pretty well (> 99.9%).

  41. tallbloke says:

    Sweet. Lucas numbers.

    Symmetrical relationship between Fibonacci and Lucas numbers

    F(n) =
    Phi^n – (–phi)^n
    ___________
    Phi– (–phi)

    L(n) =
    Phi^n + (–phi)^n
    _____________
    Phi + (–phi)

  42. oldbrew says:

    F(n) + F(n+2) = Lucas number
    e.g.
    1 + 3 = 4
    2 + 5 = 7
    3 + 8 = 11
    5 + 13 = 18
    8 + 21 = 29
    etc.

  43. oldbrew says:

    Wikipedia: Mars has an orbit with a semi-major axis of 1.524 astronomical units (228 million kilometers), and an eccentricity of 0.0934. The planet orbits the Sun in 687 days and travels 9.55 AU in doing so, making the average orbital speed 24 km/s.

    The eccentricity is greater than that of every other planet except Mercury, and this causes a large difference between the aphelion and perihelion distances—they are 1.6660 and 1.3814 AU. [bold added]
    http://en.wikipedia.org/wiki/Orbit_of_Mars

    1.666 = 5/3
    1.3814 = 3 – Phi, or (Phi² + 1)/Phi²
    [ratio to Earth at 1 AU]
    – – –
    Venus:Earth semi-major axis ratio = 1(V): ~1.383(E)

  44. Paul Vaughan says:

    It’s also a scaling of √5:
    Φ√5 = 1.38196601125011

    Helliocentric Mars angular momentum with 20/20 first-order frequency hindsight:

    2/√5
    3/√5
    4/√5
    φ/2
    φ/(2√5)

    Challenge: Point to every existing documentation
    that to first order Mars’ sidereal orbital period is 2 / ( Φ + 1/√5 ).

    Is there a search engine that can do that??

    The red planet exposes a red line crossed by conventional mainstream failure (the major western fault) to recognize, appreciate, and respect the roots of natural stability.

    Imagine the limits of human design ensure cycles of lost and found discovery. If civilization’s stability dependently hinges on balance, we may have to lose to gain.

  45. p.g.sharrow says:

    Paul, sometimes stability is in the eye of the beholder. All indications that I see point to very slow change. Perhaps orderly change. I see nothing that indicates the planetary orbital distances from the sun are fixed, but the ratio of distance may well be a fixture of the physics involved…pg

  46. oldbrew says:

    34 Mars orbits = 40 (8 * 5) Venus-Earth conjunctions
    (34 * 686.98 = 23357.32 d, 40 * 583.93 = 23357.2 d)

    5 V-E = one Venus pentagram = ~8 Earth orbits (years) = ~13 Venus orbits
    http://en.wikipedia.org/wiki/Venus#Pentagram_of_Venus

    34 Mars = 8 Venus pentagrams
    5,8,13 and 34 are Fibonacci numbers
    34/8 is a proxy for φ³

  47. Paul Vaughan says:

    pg, “fixed”? No one is saying fixed. This is about centrally limiting geometric boundary conditions, including oscillations and discrete steps. Misrepresentation, misinterpretation, and/or misunderstanding strikes again.

  48. Paul Vaughan says:

    2 / ( Φ + 1/√5 )

    Challenge: Point to everywhere that it has been documented.

  49. tallbloke says:

    Paul V: Helliocentric Mars angular momentum with 20/20 first-order frequency hindsight:

    2/√5
    3/√5
    4/√5
    φ/2
    φ/(2√5)

    I don’t understand. These are 5 different quantities. How are they all “Helliocentric Mars angular momentum”? What am I missing here?

  50. Paul Vaughan says:

    Easily-addressed question TB = a good question.
    You can quickly review Semi’s notes from 8 years ago:

    Click to access 0903.5009.pdf

    Look just under Figure 69 on pdf p.41.

  51. Paul Vaughan says:

    TB, a thorough search should be undertaken to catalog all recorded instances of the following quantities:

    2/(Φ+1/√5) = 1.8774978038635
    (Φ+1/√5)/2 = 0.532623792124926

    It may be peripherally informative to broaden the search to:

    1/(Φ+1/√5) = 0.938748901931751
    Φ+1/√5 = 1.06524758424985

    I’m not sure if the needed algorithms will be available to searchers.

  52. Paul Vaughan says:

    quickly found this even though I’m only devoting a few minutes to searching today:
    https://pds-imaging.jpl.nasa.gov/data/msl/MSLMST_0002/DATA/RDR/SURFACE/0138/0138ML0008190130104269E01_DRCL.LBL

    The search was for “0.9387489”. It’s a NASA page and “MARS” appears 4 times on the page.

  53. oldbrew says:

    If Mars = (48/35)² years:
    (35/48)² = 0.5316839 = ~ 1/(Mars in years)
    i.e. the amount of a Mars orbit completed during one Earth orbit.

  54. tallbloke says:

    APOD:

    I have a last minute flight out to Mallorca for a hike through the Tramuntana. Back in a fortnight.

  55. oldbrew says:

    Mars:Earth — the radius ratio is the inverse of the orbit ratio.

    Order of data: Mars, Earth, Mars/Earth ratio
    Volumetric mean radius (km) 3389.5 6371.0 0.532
    Sidereal orbit period (days) 686.980 365.256 1.881
    [1/1.881 = 0.532]
    Max. orbital velocity (km/s) 26.50 30.29 0.875 = 7:8
    Min. orbital velocity (km/s) 21.97 29.29 0.750 = 3:4

    http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

  56. tallbloke says:

    I may be wrong, but I have a vague recollection of doing a calc which resulted in 0.532. It may have been something to do with the falloff in gravity between two planets whose semi-major axes are in a phi ratio. or something.

  57. Paul Vaughan says:

    Semi wrote 8 years ago: “Main frequency of angular momentum of Mars relative to Sun is 1.11 year, 2.75 year, 271.8 days, 11.7 years, 2.12 years, 333 days, 1.23 years, 203.8 days and other.”

    Spelling it out a little more clearly (as always, doing this in incremental slow-walking steps to help any sensible people realize their own previous prejudice and/or inability to recognize simple pattern intuitively and independently):

    1.11
    √5/2

    2.75
    (2√5)/φ

    271.8 / 365.25
    √5/3

    1.23
    2/φ

    203.8 / 365.25
    √5/4

    I’m leaving a few of the quantities as a learning exercise: Explore how quick (or not) you are at deriving the √5 forms.

    The quantities listed above are not from guess-and-test (a logically valid method of exploration); they’re all derived. All the info needed to derive these first-order quantities was presented further above on this page.

    “the ratio of distance may well be a fixture of the physics involved…pg”

    The ratios aren’t fixed. Suggestion: Review Cuk’s video on JEV evection resonance, Milankovtich, and attractors more generally.

    _
    “sometimes stability is in the eye of the beholder.”

    The suggested subjectivity is imagined.
    Bristling at the simplicity is a political exercise.
    Geometric proofs are simple, unambiguous, and strictly objective.
    _
    Now, to have some good fun stirring stability with stability (I’ll accept your “eye of the beholder” comment if you try to put it here)…

    Another fact:
    The major western fault is disinclined towards natural stability …and it’s symbolized by a really clear red line.

    Red lines, red planets, red flags, and red armies — lots of symbolic learning opportunities if we’re ready to understand, appreciate, and respect the roots of stability …the *right* thing to do when we’re *left* with no option …and isn’t it neat that a pentagon and flags with 5-point stars are there to remind (whether polarized east, west, φ, Φ, left or right) of our dependence on √5 stability.

  58. tallbloke says:

    OB: ah yes, 0.526 not 0.532. What’s 0.006 between friends? 🙂
    Maybe the 0.532 would appear if a simple fraction was used instead of phi?

    Paul (quoting pg)
    “the ratio of distance may well be a fixture of the physics involved…pg”

    The ratios aren’t fixed. Suggestion: Review Cuk’s video on JEV evection resonance, Milankovtich, and attractors more generally.

    I think pg might have meant ‘feature’ rather than ‘fixture’. And he’d be right, since resonance (or avoidance of it) is involved.

    Paul: I’m leaving a few of the quantities as a learning exercise: Explore how quick (or not) you are at deriving the √5 forms.

    11.7
    √5*φ(2φ) = 11.7082

    2.12
    √5/2+(φ-(1/φ)) = 2.118
    (√5/2)+1 = 2.118
    (φ^3)/2 = 2.118
    1/(2*φ^-3) = 2.118

  59. tallbloke says:

    Going back to my 2013 comment OB linked above:

    Consider two planets A and B with orbital periods in the golden ratio Phi. Their synodic period as given by Copernicus’ theorum will be 1/( 1/((1/A)-(1/b))=1/((1/1)-(1/Phi))=1/(1-0.6180399)=2.618=Phi^2

    Recalling that Keplers third law states that the cube of the semi-major axis is proportional to the square of the period we find that the cube root of 1 squared is 1. The cube root of Phi squared is 1.378.

    Recalling that gravity falls in strength proportionally with the square of the distance, if the strength of the field at a distance of 1 is 1, then the strength of the field at a distance of 1.378 will be 1/(1.378*1.378) =1/1.899=0.526

    A third planet C with an orbital period which is equal to the synodic period of planets A and B will by Keplers third law be orbiting at a radius of CuberootPhi^2*Phi^2=1.899 and the strength of gravity at that distance will be 1/(1.899*1.899)=1/3.608=0.277

    The ratio of 0.526:0.277 is also 1.899:1

    I think this is how Kepler deduced an inverse square law for gravity before Newton.

    I’m beginning to get a glimmer that the quantisation of planetary orbits may be a function of gravity and the resonance involved in neighbour synodics.

    This page may come in handy at some point
    http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/KeplersLaws.htm

  60. oldbrew says:

    Re ‘A third planet C with an orbital period which is equal to the synodic period of planets A and B’

    Mars is close to that : 17 Mars = 20 Venus-Earth (= 4 ‘Venus pentagrams’)

    20/17 = 1.17647 (i.e. no. of V-E in 1 Mars orbit)
    Mean orbital velocity ratio V:E = 1.176:1
    (same as 4th root of solar irradiance ratio 1.911)

    http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html

  61. Paul Vaughan says:

    Perspective reload:
    Let’s be careful to distinguish between first-order and second-order models.

    The remarkable thing arising out the current thread is that internet search engines don’t find 2/(Φ+1/√5) = 1.8774978038635.

    Bigger Picture:
    Let’s be mindful of what we’re trying to do.

    Big Reminder:
    There’s no JEV cycle matching Hale with the first-order model.
    (Derive the JEV cycle from the first-order model if you haven’t already: 44.05508683 years.)

    Why are we bothering to study the properties of the first-order model?
    To help us appreciate and understand the baseline from which the second-order model deviates.

    Take care to ensure the models aren’t getting artificially mixed in half-baked conceptual awareness. They are distinct, complementary entities. They don’t mix. They sit alongside one another for comparative learning. This is sorting and classification the way a botanist does it with compare and contrast to catalog similar and different features.

    The really neat thing about the first order model is that it’s so simple …and it also makes us appreciate that an order of magnitude more care is needed (the second order model) to get a JEV cycle matching Hale.

    Of course the bigger goal is simply to trail-blaze towards compact algebra for Milankovitch cycles. Do we have all the info we need? Are the measurements sharp enough? There’s enough info to get things started. Then technicians can finish the work while explorers explore something new.

  62. Paul Vaughan says:

    To both provoke and clarify further:

    The EV couple’s relationship with Jupiter & Saturn is NOT stable in the FIRST-order model.
    The EV couple’s relationship with Jupiter & Saturn IS stable in the SECOND-order model.

    The simplest way to look at the stability is hierarchically.

    The first-order model of the inner system accounts for LOCAL stability while the second-order model ADDITIONALLY accounts for GLOBAL stability.

    Jupiter & Saturn nudge the inner system from its local minimum.
    The local hills and valleys are in the context of global hills and valleys.

    In local context inner system members have slid downhill into antiresonance with one another while simultaneously in broader context they have also slid downhill into antiresonance with Jupiter & Saturn.

    It’s both a hierarchical and joint minimization problem. The inner system has naturally slipped towards minimal instability both locally and globally.

  63. Paul Vaughan says:

    The Mars angular momentum component asides above were first-order, but also presented above (incoherently if/when/as it happened and if/when/as complementary pursuits afforded commentary) is all the info an industrious technician needs to pursue second-order parallels. The methods are extensible to other bodies.

  64. tallbloke says:

    Paul V: Jupiter & Saturn nudge the inner system from its local minimum.
    The local hills and valleys are in the context of global hills and valleys.

    Hi Paul. Yes, this is why the VE synodic cycle precession of 1199 years is in a 2:1 resonance with the JS synodic cycle precession of ~2400 years. They are coupled. The JS pair is nudging the inner system, as you say. The remaining imbalance shows up in the Halstatt periodic variations in Earth’s climate records. If OB has got the numbers right, the JS synodic cycle precession is also in ~2:1 with the UN synodic cycle precession ~4627 Years as well. These numerical relationships are a strong hint telling us that resonance is at work not just in planet pairs, but between pairs of pairs too. My impression is that Mars is further down the hierarchy, and gets shoved around more than most. Hence its high orbital eccentricity. Same goes for Mercury.

  65. oldbrew says:

    The 27 U-N (4628 years) period is not an exact number of Jose cycles. However the difference between the ~2400 year period and the rest of the 4628 years is very close to one U-N.

    4628 – 2400 = 2228 y
    2400 – 2228 = 172 y = ~1 U-N

    So it could be seen as a 14 to 13 split of U-N periods, which is also ~233 J-S in total.

  66. tallbloke says:

    OB: I think it’s good to consider your ‘whole numbers’ models alongside Paul’s phi models. At least while we’re still grappling with the forces underlying the ‘cogging’ evident in the cycle relations. I find them all to be good food for thought anyway.

  67. oldbrew says:

    Don’t forget the ‘1800 year’ lunar cycle, in fact 1799 anomalistic years according to de Rop [1971].
    Halfway between 1199 and ~2400 years.

    De Rop’s long-term lunar cycle

    Keeling and Whorf [2000] found the same tidal cycle:

    The 1,800-year oceanic tidal cycle: A possible cause of rapid climate change
    http://www.pnas.org/content/97/8/3814.long

    2 tidal cycles = 21 U-N

  68. Paul Vaughan says:

    TB the observations suggest stable antiresonance (and natural evasion of unstable resonance).
    Did you mean 1150 years where you wrote 1199?

    Ian Wilson: Help Needed to Solve an Interesting Lunar Puzzle

  69. Paul Vaughan says:

    OB, as with the EV aliasing, “1800 year” lunisolar is sensitive to measurement error:

    Ian Wilson: Help Needed to Solve an Interesting Lunar Puzzle

  70. Paul Vaughan says:

    Ready to leap to the end?
    It’s time to introduce the missing quantum link:

    2(J-S)(J+S) = 2(J^2-S^2)
    Note that this is a difference of squares.

    Suggestion: Explore the implications.

  71. Paul Vaughan says:

    Reminder: log_φ(φ+Φ) 1/(U-N) = 4
    Connect the dots.

  72. Paul Vaughan says:

    So that provides a stable footing for second-order refinement of an outer system first-order model:

    100*(83.9908202014941 – 84.016845922161) / 84.016845922161 = -0.030976788501514%
    100*(164.740800341812 – 164.791315640078) / 164.791315640078 = -0.0306541021717814%
    100*(171.352549156242 – 171.406220601552) / 171.406220601552 = -0.0313124256057426%

  73. oldbrew says:

    PV – de Rop argues:

    Thus it can be said that after exactly 17.99
    centuries the coincidence of the perigee and the
    ascending node occurs on the same day of the
    anomalistic year.

    So, if this coincidence also coincides with the
    moment of perihelion passage of the Earth in
    its orbit, this situation of double coincidence
    will repeat after exactly 1799 years.

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

    That’s the summary of his lunar tidal cycle maths.
    It’s also 300 lunar wobbles as per his figure 3 showing the half period of ~3 years.

  74. oldbrew says:

    Re: 2(J-S)(J+S) = 2(J^2-S^2)
    – – –
    Works for Fibonacci numbers e.g. 5 and 2, 13 and 8
    The leading 2s cancel out?

    Re ‘Did you mean 1150 years where you wrote 1199?’

    No, see PA Semi’s paper (750 V-E = 1949 V = 1199 E), page 2.
    Ref: Paul Vaughan says: August 20, 2017 at 6:19 pm

    One orbit of both V and E ‘lost’ every 1199 years, compared to 13:8 ratio.

  75. oldbrew says:

    Paul Vaughan says:
    August 20, 2017 at 6:43 pm
    TB, a thorough search should be undertaken to catalog all recorded instances of the following quantities:

    2/(Φ+1/√5) = 1.8774978038635
    (Φ+1/√5)/2 = 0.532623792124926

    – – –
    Mars data – Wikipedia:

    Equatorial radius
    3,396.2±0.1 km
    0.533 Earths
    Polar radius
    3,376.2±0.1 km
    0.531 Earths

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

  76. Paul Vaughan says:

    From comparison of Hale core model variants which radiated (in the sense of evolutionary adaptive radiation) from exploration of redundant insight, a simple Hale Core model sharing convergence properties with Ramanujan identities has emerged. The model has the lowest r^2 (0.99999999), but it’s simpler.

  77. oldmanK says:

    From PV’s link “Ramanujan credited his substantial mathematical capacities to divinity, and stated that the mathematical knowledge he displayed was revealed to him by his family goddess. ‘”An equation for me has no meaning,” he once said, “unless it expresses a thought of God.”

    Some are more blessed than others, perhaps? With the right connections – of neurons and gods. Thanks PV, a good read, even if its all beyond me. https://faculty.math.illinois.edu/~berndt/articles/aachen.pdf

  78. oldbrew says:

    There’s a golden ratio ((√5+1)/2) on page 19 of that pdf next to ‘G65’.

  79. For anyone interested in new climate physics, https://www.journals.elsevier.com/journal-of-atmospheric-and-solar-terrestrial-physics. This journal has some really interesting new articles on the theme “global electric circuit”. One article found significant correlations between SST and solar wind speed, http://www.sciencedirect.com/science/article/pii/S1364682616300360. The greenhouse seems to be broken.

    [reply] thanks

  80. Paul Vaughan says:

    “When asked about the methods Ramanujan employed to arrive at his solutions, Hardy said that they were “arrived at by a process of mingled argument, intuition, and induction, of which he was entirely unable to give any coherent account.”” https://en.wikipedia.org/wiki/Square_root_of_5#Identities_of_Ramanujan

  81. Paul Vaughan says:

    Speculation: Records provide a fragmented glimpse of no more than 1/5 of Ramanujan’s revelations.

    scale (fractal) symmetry review:
    recursive Pareto Principled division of unity

    Math’s a big world where everything’s connected simply.

    “For |q| greater than 1, in contemporary language, Ramanujan asserted that, amazingly, the odd approximants of R(q) approach q^(1/5)/R(-1/q), and the even approximants approach R(q^4)/q^(4/5), proved for the first time in [1992]. If q = exp(2πim/n), where (m,n) = 1, Ramanujan asserted that R(q) diverges if n is a multiple of 5 and converges otherwise.”

    “[…] it seems to us that no person in the history of mathematics possessed the skills that Ramanujan had in determining continued fractions for various functions or finding closed form representations for continuing fractions.”

    Matrix summary of all pairwise solar system beats and harmonies hinges compactly on 2 Ramanujan insights plus another from the “remarkable limit” in the golden quantum oscillator paper.

  82. Paul Vaughan says:

    2nd order Ramanujanian matrix model of current solar system configuration:

    Across rows sum coefficient * column header for all columns to get frequency.
    Then period = 1 / frequency.

    Learning Opportunity:
    Compare and contrast rows and columns of the matrix for intuition about solar system morphology.

    The matrix is the solution a system of equations that includes the Jupiter Earth Venus (JEV) cycle:
    (ΦΦ)^e = Φ(J+S) = φ(3V-5E+2J)
    U-N = (φ√5)^(-4) = J+3S-2Φ^5
    U = 2(J+S)(J-S)
    1 / E = 1.00001743371442
    Me = (φ^3)-J
    Ma addressed above (comparatively for 1st vs. 2nd order)

    First-order models fail to account for the JEV cycle.
    More precise second-order models are feasible.
    The Ramujanian stability model’s strength is simplicity.

  83. Paul Vaughan says:

    1 / Me
    0.240846697327135
    0.24086077691279
    +0.0058458703447504 %

    1 / V
    0.615197263396975
    0.615195620579235
    -0.000267039181943022 %

    1 / E
    1.00001743371442
    1.00001743371442
    0%

    1 / Ma
    1.88084761346252
    1.88075732040493
    -0.0048006577960707 %

    1 / J
    11.8626151546089
    11.8635293416618
    +0.00770645461352669 %

    1 / S
    29.4474984673838
    29.4452917180706
    -0.00749384303606059 %

    1 / U
    84.016845922161
    84.0087142122654
    -0.00967866599414566 %

    1 / N
    164.791315640078
    164.809655321902
    +0.011129034168239 %

    r^2 = 0.999999990130865
    r^2 = 0.999 999 990 130 865

    ___

    On natural log scale:

    1 / Me
    -1.42359465857151
    -1.4235362015767
    -0.00410629489596221 %

    1 / V
    -0.485812309115007
    -0.485814979510391
    +0.000549676353362477 %

    1 / E
    0.0000174335624575433
    0.0000174335624575433
    0%

    1 / Ma
    0.63172253346057
    0.631674525730257
    -0.00759949626149007 %

    1 / J
    2.47339187133752
    2.47346893291434
    +0.00311562343633316 %

    1 / S
    3.38260896480509
    3.3825340235667
    -0.00221548630553749 %

    1 / U
    4.43101732542891
    4.43092053408483
    -0.0021844045500945 %

    1 / N
    5.10467991972812
    5.10479120387749
    +0.00218004167003543 %

    r^2 = 0.999999999053696
    r^2 = 0.999 999 999 053 696

  84. Paul Vaughan says:

    format of preceding:

    sidereal orbital period in Julian years = 1 / planet frequency
    Seidelmann (1992)
    Ramanujanian model
    % error

    2nd list on natural log scale

  85. oldbrew says:

    Mars:
    Aphelion 1.6660 AU
    Perihelion 1.3814 AU
    http://en.wikipedia.org/wiki/Mars

    Ratio to 1 AU:
    Aphelion 3:5
    Perihelion 1:√5/φ

  86. Paul Vaughan says:

    Asteroid Belt
    2φ(ΦΦ)^e

    so
    radius = {[2φ(ΦΦ)^e]^[-2]}^{1/3} = 2.614797027 AU

  87. oldbrew says:

    5/2 * Pi = Phi^4 + 1 (> 99.999% true)
    Phi^4 + 1 = 3 * Phi²

    ‘the angle in a semicircle is always a right angle’
    http://www.bbc.co.uk/schools/gcsebitesize/maths/geometry/circles2hirev5.shtml

    So for a 3,4,5 Pythagorean triangle:
    The circle enclosing such a triangle has a diameter of 5, and a radius of 5/2.
    – – –
    The ratio of the lunar apsidal cycle to the full moon cycle is 1: 3 * Phi² (> 99.95% true).

    Mars:Earth orbit ratio is 1: 4th root of 5²/2 (> 99.97% true).

  88. oldbrew says:

    Re Kirkwood gaps:
    Semi major axis ratio of Mars (1.52 AU) to start of K gaps (2.1 AU) = 1:√5/φ

    Same as perihelion:aphelion ratio of Mars

    Sun’s core rotates four times faster than its surface

    PV wrote:
    Closing in on Mars:

    √5 = 2.23606797749979
    φ = 1.61803398874989

    (11.8626151546089)*(1.88084761346252) / (11.8626151546089 – 1.88084761346252) = 2.23525255532127
    (11.8626151546089)*(1.88084761346252) / (11.8626151546089 + 1.88084761346252) = 1.62344612704193

    Note the similarity with √5 & φ and let your imagination follow parallel paths that focus like a lens.

    Sun’s core rotates four times faster than its surface

  89. Paul Vaughan says:

    Scaling the Ramanujanian Edges of the Milankovitch Pyramid
    with recursive Pareto Principled Division of Unity

    Imagine scaling stability (celestial or otherwise) trivially hinged to iconic Pyramids.

    V5 is a third-order Hale Core model:

    Hurricanes, AMO , And Sahel Droughts

    a key clue from pyramid design and Ramanujan identities:
    √(φ√5) = 2sin(2π/5)

    The Ramanujanian outer limit without Milankovitch scaling:
    U-N = [√(φ√5)]^(-8)

    The Ramanujanian outer limit with Milankovitch scaling:
    U-N = [√(φ√5)(T/E)]^(-8)

    (T/E)^(-8) balances Uranusian variance from the difference of Jupiter and Saturn squares:

    U = 2(J-S)(J+S)[(T/E)^(-8)] = 2(J^2-S^2)[(T/E)^(-8)]

    V5 simplifies the Hale Core Model centered on (ΦΦ)^e, sharpening estimates by orders of magnitude.

    (84.01684525 – 84.01684592) / 84.01684592 = -0.000000794 %

    V5 is the first hale core model with sufficiently sharp estimates to bridge to lunisolar.

    Properties of V1 (initial second-order model):
    • Earth-Venus couple stability based on Jupiter-Earth-Venus antiresonance with Jupiter Saturn sweep rate
    • r^2 (for the matrix of all pairwise beats and harmonies) = 0.999 999 995

    Properties of V5:
    • Earth-Venus couple stability based on Jupiter-Earth-Venus antiresonance with Jupiter Saturn sweep rate
    • Ramanujanian unit scaling encoded in pyramid design identifies Milankovitch cycles
    • r^2 (for the matrix of all pairwise beats and harmonies) = 0.999 999 999 99

  90. Paul Vaughan says:

    Methodological note:

    Exploratory adaptive radiation (in the evolutionary sense) has been pursued with a botanist’s approach to sorting and classification.

    Ramanujanian relaxation of constraints (recall the simple exploratory model) radiated exploration over a saddle point enabling detection of a key lower order valley in the hierarchical landscape of stability. Relaxing lock on higher-order target facilitated more precise identification of a key lower-order feature.

    One step back, 2 steps forward.

  91. Paul Vaughan says:

    The potential for alternate expression is limitless.

    φ = (1/2)√[1+tan²(2π/5)]
    Φ = (1/2)√[1+tan²(4π/5)]

    Explore endless ways to rewrite φ, Φ, & √5 in terms of e & π.
    https://en.wikipedia.org/wiki/List_of_trigonometric_identities

    Φ = crd(π/5)
    where crd = chord function
    https://en.wikipedia.org/wiki/Trigonometric_constants_expressed_in_real_radicals

    V1’s core was initially derived from the fifth roots of unity in the complex plane, but let’s keep in mind Hardy’s comments on Ramanujan, who had profound capacity to reduce infinite series to closed form despite having never heard of complex numbers.

    φ√5 = 3.6180339887499
    Φ√5 = 1.38196601125011
    √5 = 2.23606797749979

    The beats & harmonies shape a Conway Triangle:
    (1.38196601125011)*(3.6180339887499) / (1.38196601125011 – 3.6180339887499) = √5
    (1.38196601125011)*(3.6180339887499) / (1.38196601125011 + 3.6180339887499) = 1
    (1.38196601125011)*(3.6180339887499) / ( (1.38196601125011 + 3.6180339887499) / 2 ) = 2

    Imagine extension of vision with identity omniscience.

    [2sin(2π/5)]^2 = φ√5
    [2cos(2π/5)]^2 = Φ√5

    Tip: of iceberg.

  92. oldbrew says:

    φ√5 + Φ√5 = 5
    φ√5 – Φ√5 = √5

  93. Paul Vaughan says:

    Awareness grows with view from many perspectives.

    OB’s last comment clarifies:
    sum of squares = √5 * difference of squares

    The statement’s equivalent is a familiar 1:

    The unit-scaled period-frequency log-symmetry equivalence criterion defines the axial scaling:
    difference = φ-Φ = 1 = beat = φΦ/(φ-Φ) = 1 / 1 = 1 = geometric mean = √(φΦ) = √(1) = 1
    = √5 * axial period = √5[φΦ/(φ+Φ)] = √5/2 * harmonic mean = 2/√5 * arithmetic mean

    _

    V5 Powers of √5 Prelude

    2nd-order Mars stable frequency estimate derived from JEV antiresonance with J+S, the “60 year” Jupiter Saturn cycle, & the Compound Pareto Principle — (details of derivation can be deduced from earlier comments and will be reviewed to aid communication at a later date):
    5(J-S)+7/25 = 1 / 1.88076509466002

    the 3rd-order variant (that accounts for Milankovitch):
    5(J-S)+7(√5/E)^(-4) = 1 / 1.8808341622151

    Observed:
    Ma = 1 / 1.88084761346252

    1st-order review (see somewhere far above in this thread):
    (Φ+1/√5)/2 = 1 / 1.8774978038635

  94. Paul Vaughan says:

    Compound Binomial Expansion Review

    5
    = (√5)^2
    = (Φ+φ)^2
    = (Φ+φ)(Φ+φ)
    = ΦΦ+Φφ+φΦ+φφ

    25
    = (√5)^4
    = (ΦΦ+Φφ+φΦ+φφ)(ΦΦ+Φφ+φΦ+φφ)
    = (ΦΦΦΦ+φφφφ) + 18
    = 7 + 18

    All of the terms (the full expansion was presented earlier) are whole numbers except ΦΦΦΦ & φφφφ, which add up to the whole number 7.

    7/25 = (ΦΦΦΦ+φφφφ)/25

    Bookmark this note for later review when factorial connections of puzzle pieces start crystallizing with broadening and deepening awareness of simple components of stability.

  95. oldbrew says:

    7 and 18 are Lucas numbers, and 18/7 is a Lucas near match to φ².

  96. Paul Vaughan says:

    5(J-S)+7/25 = (Φ+φ)(Φ+φ)(J-S)+(ΦΦΦΦ+φφφφ)/[(Φ+φ)(Φ+φ)(Φ+φ)(Φ+φ)]

    Border 101

    Compound Pareto Principled Division of Unity naturally stabilizes with recursive boundary conditions expressed in close form rather than as infinite series.

  97. Paul Vaughan says:

    Venus estimates

    1st order:
    V = φ
    0.618033988749895 = 1 / V
    absolutely fatal JEV failure

    2nd order:
    V = [5E-Φ√5(J+S)+1/25]/3
    0.615197163689601 = 1 / V

    3rd order (accounts for Milankovitch):
    V = (5/3)E-(1/3)[1+2(ΦE)(ΦE)-(ΦE)(ΦE)(ΦE)(φT)][(ΦΦ/T)(φ/E)(φ/E)(φ/E)(ΦΦ)^e]+(1/3)(√5/E)^(-4)
    0.615197242927538 = 1 / V

    observed:
    0.615197263396975 = 1 / V

    The 2nd-order models I’m sharing now are derived from the 3rd-order models. It’s a 20/20 hindsight thing. By writing the models in a standard format (just like statisticians do with extreme value theory), compare-and-contrast sorting and classification and equivalence recognition are expedited …and there in the rear-view mirror are simpler 2nd-order models, some of which were never noticed before they were surpassed.

  98. oldbrew says:

    φ√5 + Φ√5 = φ√5 * Φ√5 (= 5)

    Re: ‘All of the terms (the full expansion was presented earlier) are whole numbers except ΦΦΦΦ & φφφφ, which add up to the whole number 7.’

    The 7 result is part of the Lucas series.
    An equation of φ^n +/- 1/φ^n will return a Lucas number (the +/- will alternate):
    φ – 1/φ = 1
    φ² + 1/φ² = 3
    φ³ – 1/φ³ = 4
    φ^4 + 1/φ^4 = 7
    etc.

    http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/lucasNbs.html#section5

  99. tallbloke says:

    I’m in awe. Superb work gents.

  100. Paul Vaughan says:

    “All of the terms (the full expansion was presented earlier) are whole numbers except ΦΦΦΦ & φφφφ, which add up to the whole number 7.”

    The unique property is actually uniformity of components, so the pair of terms adding to 7 is unique but for a reason different than the incorrect one stated hastily in a moment lacking the time needed for full care.

    For the current moment rather than stop to present third-order V5, exploration of fourth-order is commencing beginning with generalization from Φ^e to Φ^e^x including sums and differences of Φ^(e^ix) & Φ^(-e^ix).

  101. Paul Vaughan says:

    Simpler Than You Think…

    The Golden Difference Squarely Made by Stabilizing Powers

    Recall that
    Φ+φ is a difference of squares:

    Why Phi? – the rainbow angle

    The square of that difference of squares neatly partitions a provocative (provocative only because of the current context of dark ignorance and/or deception) first-order model of the inner solar system.

    With a comparative eye (you needn’t get lost in the details if you can squarely spot the differences), recall:
    ___

    √5 = φ+Φ
    5 = (φ+Φ)(φ+Φ)
    5 = φφ+φΦ+Φφ+ΦΦ
    5 = φφ+ΦΦ+φΦ+Φφ
    5 = (φφ+ΦΦ)+φΦ+Φφ
    5 = 3+1+1

    √5 = φφ-ΦΦ
    5 = (φφ-ΦΦ)(φφ-ΦΦ)
    5 = φφφφ-φφΦΦ-ΦΦφφ+ΦΦΦΦ
    5 = φφφφ+ΦΦΦΦ-φφΦΦ-ΦΦφφ
    5 = (φφφφ+ΦΦΦΦ)-φφΦΦ-ΦΦφφ
    5 = 7-1-1
    ___

    5 = ΦΦ+Φφ+φΦ+φφ
    25 = (ΦΦ+Φφ+φΦ+φφ)(ΦΦ+Φφ+φΦ+φφ)
    25 = ΦΦΦΦ+φφφφ+4ΦΦ+4φφ+6
    25 = (ΦΦΦΦ+φφφφ)+4(ΦΦ+φφ)+6
    25 = (7)+4(3)+6
    1 = (7/25)+4(3/25)+6/25

    5 = φφφφ-φφΦΦ-ΦΦφφ+ΦΦΦΦ
    25 = (φφφφ-φφΦΦ-ΦΦφφ+ΦΦΦΦ)(φφφφ-φφΦΦ-ΦΦφφ+ΦΦΦΦ)
    25 = φφφφφφφφ+ΦΦΦΦΦΦΦΦ-4φφφφ-4ΦΦΦΦ+6
    25 = (φφφφφφφφ+ΦΦΦΦΦΦΦΦ)-4(φφφφ+ΦΦΦΦ)+6 (Pascal’s triangular tip: of Sierpinski’s iceberg)
    25 = (47)-4(7)+6
    1 = (47/25)-4(7/25)+6/25
    ___

    Take the stabilizing golden square power difference 3 simple steps further.

    1 = 1.88 – 1.12 + 0.24
    1.12 = 1.88 + 0.24 – 1
    0.62 + 0.5 = 1.88 + 0.24 – 1
    1/V+1/(2E) ~= 1/Ma+1/Me-1/E ~= √5/2

    Easy as 1, 2, 3.

    0.24 ~= 0.240846697327135 (observed) = 1/Me
    0.62 ~= 0.615197263396975 (observed) = 1/V
    1 ~= 1.00001743371442 (observed) = 1/E
    1.88 ~= 1.88084761346252 (observed) = 1/Ma
    1.12 ~= 1.11803398874989 = √5/2

    golden hindsight (left as an exercise for readers who can do the simple algebra) sharpens first-order inner solar system model estimate for Mercury:

    Me = 1/[√5/2-2/(Φ+1/√5)+1]
    1/Me = √5/2-2/(Φ+1/√5)+1 = 0.240536184886392

    eclipses original 1st order Mercury estimate by factor ~15.5
    sharpened 1st-order model stats:
    0.999993810334498 = r^2 period
    0.999991799260955 = r^2 frequency
    0.999988916677216 = r^2 log

    As higher-order sight sharpens, lower-order hindsight crystallizes.

    Conclusion:
    First order inner stability rests squarely on the power of a golden difference.

  102. Paul Vaughan says:

    mods: comment posted a few minutes ago vanished

    comment restored – wordpress objected for some reason [mod]

  103. Paul Vaughan says:

    partial solution key for anyone independently deriving the frequency algebra of the tuned inner system first-order model:

    1/V+1/(2E) ~= 1/Ma+1/Me-1/E ~= √5/2
    1/V+1/E+1/(2E) ~= 1/Ma+1/Me ~= √5/2+1/E

    1/Ma+1/Me ~= [1/V+1/E]+1/(2E) ~= √5/2+1/E
    1/Ma+1/Me ~= φ+1/2 = √5/2+1

    1/Ma+1/Me ~= 1/V+[1/E+1/(2E)] ~= √5/2+1/E
    1/Ma+1/Me ~= Φ+3/2 = √5/2+1

    1/Ma ~= 2/(Φ+1/√5)
    1/Me ~= √5/2+1-2/(Φ+1/√5)
    1/Ma+1/Me ~= [2/(Φ+1/√5)]+[√5/2+1-2/(Φ+1/√5)] = √5/2+1

    different ways to look at the same thing
    1/Ma+1/Me ~= φ+1/2 = 3/2+Φ = √5/2+1= (φ^3)/2= 2+(Φ^3)/2

    regular practice looking from lots of different ways help expedite identity recognition

  104. Paul Vaughan says:

    Reminder:
    Why study first-order inner system models (whether with or without golden hindsight)?
    To understand the exact nature (comparatively) of their failure to account for JEV & Milankovitch.

    It’s a taxonomy lesson. Evolutionary tree-branching isn’t random. Ignoring the geometric legacy (of recursive boundary conditions) inherent in systems — whether biological, physical, or whatever — isn’t sensible.

    The conventional uniformity assumption is devilishly ignorant and/or deceptive, casting a dark, evil shadow over God’s brilliant geometric legacy.

  105. Paul Vaughan says:

    “properties much less restricted” “Such a theory obviously only makes sense in two-dimensions, where clockwise and counterclockwise are clearly defined directions.” “In between we have something different.”

    Looking back (to lower order) and looking forward (to higher order)…

    Φ=φ^e^(πi)
    φ=φ^e^(2πi)
    φ^e=φ^e^(-ii)
    Φ^e=φ^e^(ii)
    √5=φ^e^(πi)+φ^e^(2πi)=[φ^e^(2πi)][φ^e^(2πi)]-[φ^e^(πi)][φ^e^(πi)]
    1/Hale=[φ^e^(2πi)][φ^e^(-ii)]/[φ^e^(ii)]

    https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Relation_to_the_complex_exponential_function

    …puts third-order comparatively in hierarchical context.

  106. Paul Vaughan says:

    1 more glance backward…

    2/V+2/E+1/E ~= 2/Ma+2/Me ~= √5+2/E ~= 2φ+1 = 3+2Φ = √5+2 = φ^3 = 4+(Φ^3)

    …and forward…

    φ^(-e^x)=Φ^e^x
    1/Hale=[φ^e^(2πi)][φ^e^(-ii)][φ^(-e^(ii))]

    …before refocusing on V5.

  107. oldbrew says:

    Venus day (116.75 d) * 5 = 23 solar rotations (25.38 d)

    Carrington Solar Coordinates: http://wso.stanford.edu/words/Coordinates.html

    Solar rotation * 23 = 583.74 days
    Venus day * 5 = 583.75 days
    Venus-Earth synod = 583.94 days

  108. Paul Vaughan says:

    φ^(iie^x)=Φ^e^x
    1/Hale=[φ^e^(2πi)][φ^e^(-ii)][φ^(iie^(ii))]
    1/Hale=[φ^e^(2πi)][φ^e^(iiii)][φ^(iie^(ii))]
    1/Hale=[φ^e^(2πi)][φ^iiiie^(iiii)][φ^(iie^(ii))]

  109. Paul Vaughan says:

    V5

    Me = [(φ/E)(φ/E)(φ/E)] – [(1/2)+(ΦE)(ΦE)](J+S) + (1/2)(√5/E)^(-4)
    V = (5/3)E – (1/3)[1+2(ΦE)(ΦE)-(ΦE)(ΦE)(ΦE)(φT)](J+S) + (1/3)(√5/E)^(-4)
    E = E
    Ma = [(10(ΦE)(ΦE)](J+S) + 2(√5/E)^(-4)
    J = [(1/2)+(ΦE)(ΦE)](J+S) – (1/2)(√5/E)^(-4)
    S = [(1/2)-(ΦE)(ΦE)](J+S) + (1/2)(√5/E)^(-4)
    U = 2[(T/E)^(-8)](J+S)(J-S)
    N = 2[(T/E)^(-8)](J+S)(J-S) – [√(φ√5)(T/E)]^(-8)

    J+S = (Φ/T)(φ/E)(φ/E)(φ/E)[Φ(ΦΦ)^e]
    J-S = 2(ΦE)(ΦE)(J+S) – (√5/E)^(-4)
    U-N = [√(φ√5)(T/E)]^(-8)

    __

    V5 stripped down (stripped of Milankovitch but with JEV intact)

    Me = φφφ – [(1/2)+ΦΦ][φ(ΦΦ)^e] + (1/2)(1/25)
    V = (5/3)E – (1/3)[1+ΦΦ][φ(ΦΦ)^e] + (1/3)(1/25)
    E = E
    Ma = 10ΦΦ[φ(ΦΦ)^e] + 2(1/25)
    J = [(1/2)+ΦΦ][φ(ΦΦ)^e] – (1/2)(1/25)
    S = [(1/2)-ΦΦ][φ(ΦΦ)^e] + (1/2)(1/25)
    U = 2[φ(ΦΦ)^e][2Φ(ΦΦ)^e-(1/25)]
    N = 2[φ(ΦΦ)^e][2Φ(ΦΦ)^e-(1/25)] – (ΦΦΦΦ/25)

    J+S = φ(ΦΦ)^e
    J-S = 2Φ(ΦΦ)^e – (1/25)
    U-N = ΦΦΦΦ/25

  110. p.g.sharrow says:

    Paul Vaughan says:
    September 26, 2017 at 1:34 am “God’s brilliant geometric legacy.”

    Nice turn of phrase…pg

  111. Paul Vaughan says:

    V5 introduces scaling of Milankovitch parameters E & T.
    Hale Core Model extends to lunisolar context with Milankovitch parameters included.
    E & T are nearly 1. Ripping them out (replace with 1) facilitates comparative insight.

    Alternate expression of stripped-down components:

    V
    = [5E-√5(ΦΦ)^e+1/25]/3
    = (5/3)E – (1/3)(Φ√5)[φ(ΦΦ)^e] + (1/3)(1/25)
    = (5/3)E – (1/3)(φ√5)[Φ(ΦΦ)^e] + (1/3)(1/25)
    = 5E/3 – Φ√5(J+S)/3 + 1/75
    = 5E/3 – φ√5Hale/3 + 1/75

    Ma
    = 10Φ(ΦΦ)^e + 2/25
    = 10ΦΦ(J+S) + 2/25
    = 10Hale + 2/25
    = 5(J-S) + 7/25

    Compound Pareto-principled division of unity simply stabilizes.

    Challenge: Extend recursive boundary (fractal edge) generalization from circles to ellipses and other shapes, from 2 dimensions to higher dimensions, and from our solar system to systems in general.

    Euler-league quest for bright open minds: compact expression of simple fundamentals.

  112. Paul Vaughan says:

    V6 conceptual provocation:
    Pareto compounding of nested exponentials (parallel context anyon math structure tip above):
    period & frequency of solar cycle are real (-1 & 1) so what’s the imaginary part?

  113. Paul Vaughan says:

    derived from pared down V5 identities

    J = (1/2){(1+φ√5)[Φ(ΦΦ)^e]-(1/25)}
    J = (1/2)(Hale+φ√5Hale-1/25)
    J = (1/2)(2Hale+φφHale-1/25)
    J = (1/2)(4Hale+ΦHale-1/25)
    J = (1/2)(φφφHale+ΦΦHale-1/25)
    J = (1/2)(φφφφHale-√5Hale-1/25)

    S = (1/2){Φ[Φ(ΦΦ)^e]+(1/25)}
    S = (1/2)(ΦHale+1/25)

  114. Paul Vaughan says:

    test comment

    J = (1/2)(φHale+φφHale+ΦΦHale-1/25)

  115. Paul Vaughan says:

    1Φ1

    Towards V6 above I started introducing functions of form:

    φ^(-e^x)=Φ^e^x

    In haste I posted a few mathematically suggestive but untrue statements. Corrections are left as an exercise for alert readers ready to explore mathematical architecture and design.

    Rear-view mirror:

    Suggestions-29

    My instinct included Euler and now I’ve designed what I call Flucasnacci polynomials (complex-valued), which equate to the following for integers:

    …,5,3,2,1,1,Φ,1,3,4,7,11,…

    At 0 the function (details forthcoming) switches abruptly from 0 via Φ to 2.

  116. Paul Vaughan says:

    The sequence
    …,8,5,3,2,1,1,Φ,1,3,4,7,11,18,…

    arises from integer
    …,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,…
    evaluations of the complex-valued function:

    Φ(x) = {[sgn(x)][φe^(2πix)]^[xe^(2πix)]+[φe^(πi(2x-1))]^[xe^(πi(2x-1))]}{H(x)-H(x)e^(πix)+e^(πix)} / {φ+Φ[sgn(x)][e^(πi(2x-1))]}

    where
    sgn(x) = sign function = 2H(x)-1
    H(x) = Heaviside step function = [sgn(x)+1]/2

    By convention sgn(0) = 0, H(0) = 1/2, & Φ(0)=Φ.
    By design tradeoff the evaluation at 0 can be flipped like a switch between F(0)=0 & L(0)=2.

    Refining Euler’s imagination:

    i^2 = -φΦ = Φ-φ
    √(Φ-φ) = √(-φΦ) = i

    Note well Pythagoras:
    i^2 + (√5)^2 = [L(0) – F(0)]^2

    …to be continued…

  117. Paul Vaughan says:

    polynomial notation follow-up
    L(x) = Lucas
    F(x) = Fibonacci
    Φ(x) = Flucasnacci

    ___

    Compare and contrast:
    √(Φ-φ) = √(-φΦ) = i
    √(φ-Φ) = √(φΦ) = 1

    It’s not your imagination.

    φ’s 1 of a kind.
    No other number can do that.

    ___

    next up:
    Imaginary Triangles …including the Flucasnacci.

  118. Paul Vaughan says:

    This comment is a test.

  119. Paul Vaughan says:

    WordPress is blocking comments with:
    square root of five
    lower case phi
    upper case phi

    [mod] sorry, not sure what the problem is – maybe repetition of symbols or not enough English?
    Approved a comment on the Trump admin thread

  120. Paul Vaughan says:

    Of course the problem’s a familiar 1 oldbrew: oldschool pattern recognition algorithms founded on false assumptions.

    I suggest we have good cause to believe Xi, Trump, and Putin have the enlightened capacity to triangulate a precisely principled division of excessive unity with nothing more than a stable balance of imaginary and irrational powers:

    Trump administration to terminate Obama’s climate plan 

    Misunderstandings can be a whole lot more fun than the truth:
    Maybe wordpress doesn’t accept mathematical proofs based on irrational and imaginary powers.

  121. oldbrew says:

    tallbloke says: August 3, 2017 at 7:59 am

    Where the mind starts to boggle is when you ask whether the universe as a whole is rotating, and if it is, what’s it rotating relative to?

    Relative to its own axis perhaps?
    – – –
    All Space is Not Equal: Physicists Find Axis that Gives the Universe Orientation
    April 17, 1997

    “This work defies the notion that there is no ‘up’ or ‘down’ in space,” says Nodland, research fellow at Rochester’s Theory Center for Optical Science and Engineering.

    http://www.rochester.edu/news/show.php?id=1294

  122. Paul Vaughan says:

    Does the presence of words before and after algebra relax auto-prejudice?
    We’ll see if automatic pressure against phi and square root of five lands this in the spam bin.

    Critics lacking imagination (and a sense of fun) mistake as closed nationalism minds opening worldwide to the simpler powers of irrationalism.

    √(Φ-φ) = √(-φΦ) = i
    √(φ-Φ) = √(φΦ) = 1

    Across all landscapes, imaginative exploration of stability irrationally drives towards simple generality.

    ΦΦ-1 = -Φ = (Φ-1)(Φ+1)
    ΦΦ+1 = Φ√5 = (Φ-i)(Φ+i)
    φφ+1 = φ√5 = (φ-i)(φ+i)

    φφ-1 = φ = (φ-1)(φ+1)

    Review: These identities are encoded in Pyramid design.
    Plugging clues together, irrationally imagine how Jupiter & Saturn really compare & contrast:

    J = [(1/2)+ΦΦ][φ(ΦΦ)^e] – (1/2)(1/25)
    S = [(1/2)-ΦΦ][φ(ΦΦ)^e] + (1/2)(1/25)

    J = [(φ-Φ)(Φ-i)(Φ+i)-(1/2)][φ(ΦΦ)^e] – (1/2)(1/[(√5)(√5)(√5)(√5)])
    S = [(Φ-φ)(Φ-1)(Φ+1)-(1/2)][φ(ΦΦ)^e] + (1/2)(1/[(√5)(√5)(√5)(√5)])

    J = [(φ-Φ)(Φ-√(Φ-φ))(Φ+√(Φ-φ))-(1/2)][φ(ΦΦ)^e] + (1/2)(Φ-φ)(1/[(φ+Φ)(Φ+φ)(φ+Φ)(Φ+φ)])
    S = [(Φ-φ)(Φ-√(φ-Φ))(Φ+√(φ-Φ))-(1/2)][φ(ΦΦ)^e] + (1/2)(φ-Φ)(1/[(φ+Φ)(Φ+φ)(φ+Φ)(Φ+φ)])

    A simple identity challenge to check:
    Are you left without a clue or right on the ball?

    A red line on maps indicates what’s written in stone.
    The concentration of mistaken identities is highest along the major western fault.

    Can we learn from nature (including our own)?

    Principled division of unity stabilizes.
    Unprincipled fracture’s a gamble.

    Bets on false uniformity assumptions are devilish gambles.
    Stability may be left to the imagination even if it’s irrationally staring us right in the face.

  123. Paul Vaughan says:

    Complex exponential growth on the unit circle is rotation.

    Refining Euler’s gem:

    [(Φ-φ)^(3/2)][(φ-Φ)^(0/2)] = (Φ-φ)√(Φ-φ) = e^[(3/2)πi] = -i
    [(Φ-φ)^(2/2)][(φ-Φ)^(1/2)] = (Φ-φ)√(φ-Φ) = e^[(2/2)πi] = -1
    [(Φ-φ)^(1/2)][(φ-Φ)^(2/2)] = (φ-Φ)√(Φ-φ) = e^[(1/2)πi] = i
    [(Φ-φ)^(0/2)][(φ-Φ)^(3/2)] = (φ-Φ)√(φ-Φ) = e^[(0/2)πi] = 1

    It’s a multiplicative spin on a ladder of complex powers either way:

    1. With clockwise rotation note
    decreasing powers of the negative unit multiplied by increasing powers of the positive unit.

    2. With counterclockwise rotation note
    increasing powers of the negative unit multiplied by decreasing powers of the positive unit.

    Equivalently:

    [(-φΦ)^(3/2)][(+φΦ)^(0/2)] = (-φΦ)√(-φΦ) = e^[(3/2)πi] = -i
    [(-φΦ)^(2/2)][(+φΦ)^(1/2)] = (-φΦ)√(+φΦ) = e^[(2/2)πi] = -1
    [(-φΦ)^(1/2)][(+φΦ)^(2/2)] = (+φΦ)√(-φΦ) = e^[(1/2)πi] = i
    [(-φΦ)^(0/2)][(+φΦ)^(3/2)] = (+φΦ)√(+φΦ) = e^[(0/2)πi] = 1

    φ’s the only number that can do this.

  124. Paul Vaughan says:

    Generalized extension (from discrete) to continuous:

    [(Φ-φ)^(2p)][(φ-Φ)^(2(1-p)-1/2)] = [(-φΦ)^(2p)][(+φΦ)^(2(1-p)-1/2)] = e^(2πpi)
    (Φ-φ)^(2p) = (-φΦ)^(2p) = (-1)^(2p) = e^(2πpi)
    0≤p≤1

    φ’s the only number that can do this with Euler’s gem.

    ___

    Review:

    √(φ+2) = √(φφ+1) = √[(φ-i)(φ+i)] = √[(φ-√(Φ-φ))(φ+√(Φ-φ))] = √(φ√5)
    √(φ+1) = √(φφ±0) = √[(φ-0)(φ+0)] = √[(φ-√(φ-φ))(φ+√(φ-φ))] = √(φφ)
    √(φ+0) = √(φφ-1) = √[(φ-1)(φ+1)] = √[(φ-√(φ-Φ))(φ+√(φ-Φ))] = √(φ)

    Hindsight’s 20/20.

    J = [(φ-Φ)(Φ-√(Φ-φ))(Φ+√(Φ-φ))-(1/2)][φ(ΦΦ)^e] + (1/2)(Φ-φ)(1/[(φ+Φ)(Φ+φ)(φ+Φ)(Φ+φ)])
    S = [(Φ-φ)(Φ-√(φ-Φ))(Φ+√(φ-Φ))-(1/2)][φ(ΦΦ)^e] + (1/2)(φ-Φ)(1/[(φ+Φ)(Φ+φ)(φ+Φ)(Φ+φ)])

    The expression for Saturn’s frequency is identical to the expression for Jupiter’s frequency with 4 exceptions specified by only (unit) sign changes — i.e. swaps of (φ-Φ) for (Φ-φ) or vice versa.

    Recall:

    1 = difference = φ-Φ = beat = φΦ/(φ-Φ) = geometric mean = √(φΦ)
    = √5*(axial period) = (φ+Φ)[φΦ/(φ+Φ)] = (√5/2)*(harmonic mean) = (2/√5)*(arithmetic mean)

    Now anyone with sufficiently irrational intuition simply imagines the extension to complex numbers.

    Psychologists not left in the dark can rightly imagine new tests measuring irrationality quotients.

    Entrepreneurs with optometric instinct can get busy inventing and marketing new lines of corrective lenses (may be a pharmaceutical product) for those lacking naturally irrational vision.

    Imagine the impact on elections and possibly a new era of democratic stability.

    With everyone seeing so clearly, imagine the possibilities to leverage necessity for superior exploration and invention.

    Step 1 may be a complete overhaul of our math education systems.

  125. Paul Vaughan says:

    One last exercise to cement irrational lessons on complex differences of squares:

    φ = Φ+1 = (√Φ-i)(√Φ+i) = (√Φ-√(Φ-φ))(√Φ+√(Φ-φ))
    φφ = φ+1 = (√φ-i)(√φ+i) = (√φ-√(Φ-φ))(√φ+√(Φ-φ))
    Φ = φ-1 = (√φ-1)(√φ+1) = (√φ-√(φ-Φ))(√φ+√(φ-Φ))
    ΦΦ = 1-Φ = (1-√Φ)(1+√Φ) = (√(φ-Φ)-√Φ)(√(φ-Φ)+√Φ)

    This is one last sentence to make sure the press doesn’t bin this comment for the deadly sin of being a minor algebra storm without words.

  126. Paul Vaughan says:

    The devil got his fingers on the switchboard of human learning and employed “experts” to compound learning delays by beating humans with intellectual sticks for daring to read God’s notes.

    There was red tape over some of the switches but we burned some of it off by firing the operator.

    The edges of the switchboard (remember the equator’s an edge) have some special statistical properties that may not be so well understood by just anyon (that’s not a typo).

    “In between we have something different. […]
    [anyons] can have any phase when particles are interchanged.”

    Imagine Earth’s equator as “a topological quantum computer”.

    Φ & φ together specify real or imaginary units, addition, subtraction, multiplication, and division.

    They’re special building blocks capable of representing not only all other numbers but also operations and complexity.

    Imagine a design (and/or divine) thought: “I’ll flip 4 switches.”

    J = [(φ-Φ)(Φ-√(Φ-φ))(Φ+√(Φ-φ))-(1/2)][φ(ΦΦ)^e] + (1/2)(Φ-φ)(1/[(φ+Φ)(Φ+φ)(φ+Φ)(Φ+φ)])
    S = [(Φ-φ)(Φ-√(φ-Φ))(Φ+√(φ-Φ))-(1/2)][φ(ΦΦ)^e] + (1/2)(φ-Φ)(1/[(φ+Φ)(Φ+φ)(φ+Φ)(Φ+φ)])

    Imagine also that a mature understanding and appreciation of God’s brilliant geometric legacy hinges on your irrationality quotient.

    Ignoring the infinitely deep roots of the Farey tree you’d be left with primitive notions of units and unity and right to open your mind to natural stability principles.

    Globalism isn’t the problem but rather a primitive view of a problem (or opportunity) deeply understood by very few. A more enlightened description of the problem is ignorance of the stabilizing power of the Pareto Principle and key identities derived from it. Suggestion: An enlightened 1/5 of globalists can learn to respect and apply the Pareto Principle.

  127. oldbrew says:

    The ratio of Jupiter-Saturn to Saturn-Uranus conjunctions is 2.2823206:1

    That’s 1/3rd of 6.8469618:1
    6.8469618 = 2.61667² = ~Phi^4
    Phi² = 2.618034

    89 J-S = 39 S-U = 128 J-U (89+39=128)
    39 = 13*3
    89/13 = ~Phi^4
    3,13 and 89 are Fibonacci numbers

    39, 80, 89 is a Pythagorean triple.
    In 89 J-S or 39 S-U there will be about 80 solar Hale cycles.

  128. oldbrew says:

    Tricky stuff…

    Missing link between new topological phases of matter discovered
    October 18, 2017
    http://m.phys.org/news/2017-10-link-topological-phases.html

  129. Paul Vaughan says:

    WordPress curiously resists simple symbolic expression unless it’s accompanied by logically unnecessary words. Conclusion: Just like the stupendously inefficient circus of western politics, WordPress artificially necessitates logically unnecessary words.

    Recap:

    J = [(1/2)+ΦΦ][φ(ΦΦ)^e] – (1/2)(1/25)
    S = [(1/2)-ΦΦ][φ(ΦΦ)^e] + (1/2)(1/25)

    J = [(1/2)+(φ-Φ)ΦΦ][φ(ΦΦ)^e] + (1/2)(Φ-φ)(1/√5)^(5-1)
    S = [(1/2)+(Φ-φ)ΦΦ][φ(ΦΦ)^e] + (1/2)(φ-Φ)(1/√5)^(5-1)

    Jupiter’s frequency is easily switched to Saturn’s.

    Can you imagine that this has been known for nearly a century. “It’s easy if you try.” — John Lennon

  130. oldbrew says:

    Still relevant today…

    SOLAR ROTATION, IMPULSES OF THE TORQUE IN THE SUN’S MOTION, AND CLIMATIC VARIATION

    T. LANDSCHEIDT (1987)
    Schroeter Institute for Research in Cycles of Solar Activity

    http://bourabai.kz/landscheidt/rotation.htm

  131. Paul Vaughan says:

    Stability derives from a principled division of unity, not reckless fracture.
    Here I’ve left the final step to axial generality as a trivial exercise:

    Juno finds that Jupiter’s gravitational field is “askew”