Why Phi? – the rainbow angle

The rainbow angle [credit: Hong Kong Observatory]

The minimum deviation angle for the primary bow [of a rainbow] is 137.5° according to Wikipedia. This is known as the rainbow angle. A circle is 360 degrees, so the ratio of the rainbow angle to the circle is therefore the square of the golden ratio i.e. 137.5:360 = 1:2.61818~.
– – –
Hong Kong Observatory has some useful explanatory text and graphics (rounding 137.5 to 138 degrees) titled:
Why is the region outside the primary rainbow much darker than that inside the primary rainbow?
Written by : SIU Kai-chee (summer intern) and HUNG Fan-yiu

Let’s first look at Figure 1, which shows sun rays entering a water drop and going through refraction and reflection.

The ray (ray no. 1) passing through the centre goes directly backward on reflection, i.e. a change in direction of 180 degrees.

For ray no. 2, this angle becomes smaller, following the rules of refraction and reflection.

For the next (ray no. 3) the angle continues to decrease, so on and so forth. This trend does not continue for long, however.

Why Phi? – a lunar ratios model

Lunar ratios diagram

The idea of this post is to try and show that the lunar apsidal and nodal cycles contain similar frequencies, one with the full moon cycle and the other with the quasi-biennial oscillation.

There are four periods in the diagram, one in each corner of the rectangle. For this model their values will be:

FMC = 411.78443 days
LAC = 3231.5 days
LNC = 6798.38 days
QBO = 866 days (derived from 2 Chandler wobbles @ 433 days each)
The QBO period is an assumption (see Footnote below) but the others can be calculated.

Why Phi? – Pluto’s eccentric orbit

Pluto’s non-standard orbit [credit: Wikipedia]

‘Pluto’s orbital period is 248 Earth years. Its orbital characteristics are substantially different from those of the planets, which follow nearly circular orbits around the Sun close to a flat reference plane called the ecliptic. In contrast, Pluto’s orbit is moderately inclined relative to the ecliptic (over 17°) and moderately eccentric (elliptical). This eccentricity means a small region of Pluto’s orbit lies nearer the Sun than Neptune’s.’ – Wikipedia

Why Phi? – lunar eclipses at Stonehenge

Bluestone Horseshoe at Stonehenge – 19 Stones

Stonehenge Visitors Guide – under ‘Eclipse Cycles’ – says:

‘Now, it’s widely accepted that Stonehenge was used to predict eclipses. The inner “horseshoe” of 19 stones at the very heart of Stonehenge actually acted as a long-term calculator that could predict lunar eclipses. By moving one of Stonehenge’s markers along the 30 markers of the outer circle, it’s discovered that the cycle of the moon can be predicted. Moving this marker one lunar month at a time – as opposed to one lunar day the others were moved – made it possible for them to mark when a lunar eclipse was going to occur in the typical 47-month lunar eclipse cycle. The marker would go around the circle 38 times [2 x 19] and halfway through its next circle, on the 47th full moon, a lunar eclipse would occur.’

Why Phi? – solar rotation notes

Carrington Rotations = CarRots [credit: dreamstime.com]

Tallbloke recently acquired a book by Hartmut Warm called ‘Signature of the Celestial Spheres: Discovering Order in the Solar System’ which offers this gem:
588 solar Carrington rotations (CarRots) = 587 lunar sidereal months
We’ll call this the HW cycle, about 43.91 years.

‘Richard Christopher Carrington determined the solar rotation rate from low latitude sunspots in the 1850s and arrived at 25.38 days for the sidereal rotation period. Sidereal rotation is measured relative to the stars, but because the Earth is orbiting the Sun, we see this period as 27.2753 days.’ – Wikipedia

Picking this ball up and running with it, we find there are 308 CarRots (27.2753 d) per 331 solar sidereal days (25.38 d) in 23 years (331 – 308). This period, or a multiple of it, can be found in certain identified solar-planetary cycles (as discussed below).

Why Phi? – a unified precession model

Combined precession cycle [credit: wikipedia]

‘Because of apsidal precession the Earth’s argument of periapsis slowly increases; it takes about 112000 years for the ellipse to revolve once relative to the fixed stars. The Earth’s polar axis, and hence the solstices and equinoxes, precess with a period of about 26000 years in relation to the fixed stars. These two forms of ‘precession’ combine so that it takes about 21000 years for the ellipse to revolve once relative to the vernal equinox, that is, for the perihelion to return to the same date (given a calendar that tracks the seasons perfectly).’ – Wikipedia

Here we’ll fit the three precession cycles into one model and briefly examine its workings.

Why Phi? Sylvia & Sons Swing Serenely by Shining Stars

Out at the unfashionable end of the Asteroid Belt, lies a seldom seen squashed spud of rock known as Sylvia. NASA has this:

Composite image showing the two moons at several locations along their orbits (shown by red dots). Image Credit: NASA

Discovered in 1866, main belt asteroid 87 Sylvia lies 3.5 AU from the Sun, between the orbits of Mars and Jupiter. Also shown in recent years to be one in a growing list of double asteroids, new observations during August and October 2004 made at the Paranal Observatory convincingly demonstrate that 87 Sylvia in fact has two moonlets – the first known triple asteroid system. At the center of this composite of the image data, potato-shaped 87 Sylvia itself is about 380 kilometers wide. The data show inner moon, Remus, orbiting Sylvia at a distance of about 710 kilometers once every 33 hours, while outer moon Romulus orbits at 1360 kilometers in 87.6 hours. Tiny Remus and Romulus are 7 and 18 kilometers across respectively. Because 87 Sylvia was named after Rhea Silvia, the mythical mother of the founders of Rome, the discoverers proposed Romulus and Remus as fitting names for the two moonlets. The triple system is thought to be the not uncommon result of collisions producing low density, rubble pile asteroids that are loose aggregations of debris.

Why Phi? – the resonance of Jupiter’s Galilean moons

Io, Europa and Ganymede – three of Jupiter’s four Galilean moons

The resonance of three of the four Galilean moons of Jupiter is well-known. Or is it?

We’re usually told there’s a 1:2:4 orbital ratio between Ganymede, Europa and Io, but while this is not far from the truth, a closer look shows something else.

Phi and the Great Pyramid of Khufu

Great Pyramid of Giza from a 19th-century stereopticon card photo [credit: Wikipedia]

Let’s have a look at some numbers for the Great Pyramid.

Source: Building the Great Pyramid (aka Cheops)
Copyright 2006 Franz Löhner and Teresa Zuberbühler

Dimensions as designed (in Egyptian royal cubits):
Length: 440
Height: 280
Slope: 356

Original dimensions as built (a,h and c in the pyramid diagram below):
Length: 230.36m (half = 115.18m)
Height: 146.59m
Slope: 186.42m

Why Phi? – some Moon-Earth interactions

Tallbloke writes: Stuart ‘Oldbrew’ has been getting his calculator warm to discover the congruences in various aspects of the Lunar orbit around Earth, and its relationship to Earth-Moon orbit around the Sun. Emerging from this study are some useful insights into longer periods, such as the ‘precession of the equinoxes‘.

Some matching periods of lunar numbers:
86105 tropical months (TM) @ 27.321582 days = 2352524.8 days
85377 anomalistic months (AM) @ 27.55455 days = 2352524.8 days
79664 synodic months (SM) @ 29.530589 days = 2352524.8 days

These identical values are used in the chart on the right (top row). The second row numbers are the difference between the numbers in the first row (TM – AM and AM – SM).
The derivation of the third row number (6441) is shown on the chart itself [click on the chart to enlarge it].

The period of 6441 tropical years (6440.75 sidereal years) is one quarter of the Earth’s ‘precession of the equinox’.
Multiplying by 4: 25764 tropical years = 25763 sidereal years.
The difference of 1 is due to precession.

Why Phi? – Neptune’s moon Triton and its neighbours

The orbit of Triton (red) is opposite in direction and tilted −23° compared to a typical moon’s orbit (green) in the plane of Neptune’s equator [image credit: Wikipedia]

Triton is the seventh largest moon in the solar system. Not only that, it has over 99% of the mass of all Neptune’s moons combined. Its retrograde orbit makes it unique among the large moons of the solar system, and it is also the coldest known planetary body at -235° C (-391° F).

Turning to the orbit numbers, and looking at Triton’s closest ‘inner’ (nearer to Uranus) neighbour Proteus and the next two ‘outer’ moons, we find these values (in days):
1.122d Proteus
5.877d Triton
360.13d Nereid
1879.08d Halimede

We’ll treat Proteus and Triton as a pair, and the same for Nereid and Halimede.
Nereid is over fifteen times further from Uranus than Triton is, so hardly a neighbour at all.

Looking at the orbit ratios (which are also the rotation ratios, as usual with moons):
T/P = 5.877 / 1.122 = 5.238
H/N = 1879.08 / 360.13 = 5.218

The first thing to say is that the two results are very similar. One is about 99.62% of the other.

Why Phi? – a Jupiter-Neptune model

Neptune (top), Uranus, Saturn, Jupiter (bottom)

Continuing our long-term series researching Fibonacci and/or Phi based ratios in planetary conjunction periods, it’s time for a look at the inner- and outer-most gas giants of our solar system: Jupiter and Neptune.

Initial analysis shows the period of 14 Jupiter orbits is close to that of one Neptune orbit of the Sun, and even closer to the period of 13 (14 less 1) Jupiter-Neptune (J-N) conjunctions.

It also turns out that there’s a multiple of 13 J-N that equates to a whole number of Earth orbits:
Jupiter-Neptune(J-N) average conjunction period = 12.782793 years
221 J-N = ~2825 years (2824.9972y)
(221 = 13 x 17)

But this period is not a whole number of either Jupiter or Neptune orbits.
This is resolved by multiplying by a factor of 7.

Why Phi? – a Jupiter-Uranus-Neptune model

From the top: Neptune, Uranus, Saturn, Jupiter
[image credit: NASA/JPL]

Continuing our quest to understand more about planetary frequencies, we turn to links between the largest planet Jupiter and the two ‘outer’ giant planets, Uranus and Neptune.

This model is based on a match of synodic periods, which is found to be:
22 Uranus-Neptune (U-N) = 273 Jupiter-Uranus (J-U) = 295 Jupiter-Neptune (J-N)

The period is just under 3771 years (~3770.93y).
To find a link to Fibonacci numbers we can look first at Jupiter-Uranus:
273 J-U = 13 x 21 (13 and 21 are Fibonacci numbers)

Why Phi? – Moons of Pluto

See main post for details [image credit: Wikipedia / WolfmanSF]

In this extract from Wikipedia we’ve highlighted the relevant part in bold, so without more ado:

Resonances
Styx, Nix, and Hydra are in a 3-body orbital resonance with orbital periods in a ratio of 18:22:33. The ratios are exact when orbital precession is taken into account. This means that in a recurring cycle there are 11 orbits of Styx for every 9 of Nix and 6 of Hydra. Nix and Hydra are in a simple 2:3 resonance. The ratios of synodic periods are such that there are 5 Styx–Hydra conjunctions and 3 Nix–Hydra conjunctions for every 2 conjunctions of Styx and Nix.

Why Phi? – a Venus transit cycle model

[credit: F.Espenak / NASA]

NASA: 2004 AND 2012 TRANSITS OF VENUS – Introduction says:
‘Transits of Venus across the disk of the Sun are among the rarest of planetary alignments. Indeed, only six such events have occurred since the invention of the telescope (1631, 1639, 1761, 1769, 1874 and 1882). The next two transits of Venus will occur on 2004 June 08 and 2012 June 06.’

Obviously there are three pairs of transits (eight years apart per pair) shown in the brackets, plus the fourth pair that occurred in 2004 and 2012. The model we use here is structured as per this graphic:

This is where it gets a little bit tricky perhaps.

Why Phi? – a triple conjunction comparison

[image credit: imagineeringezine.com]

Only two questions are needed here:

(1) What is the period of a Jupiter(J)-Saturn(S)-Earth(E) (JSE) triple conjunction?
JSE = 21 J-S or 382 J-E or 403 S-E conjunctions (21+382 = 403) in 417.166 years (as an average or mean value).

(2) What is the period of a Jupiter(J)-Saturn(S)-Venus(V) (JSV) triple conjunction?
JSV = 13 J-S or 398 J-V or 411 S-V conjunctions (13+398 = 411) in 258.245 years (as an average or mean value).

Since JSV = 13 J-S and JSE = 21 J-S, the ratio of JSV:JSE is 13:21 exactly (in theory).

As these are consecutive Fibonacci numbers, the ratio is almost 1:Phi or the golden ratio.
Golden ratio: relationship to Fibonacci sequence

Why Phi? – a Neptune-Pluto-Eris model

Comparison of the eight brightest TNOs [credit: Wikipedia]

As Pluto is getting some media attention due to the impending ‘fly-by’ of a NASA space probe, let’s take a look at its orbital relationship with its neighbours.

Why Phi? – a Jupiter-Venus-Mercury model

Planetary conjunction [image credit: EPA / Daily Mail]

For the Jupiter-Venus-Mercury (JVMe) model, we start with this basic synodic conjunction relationship:
61 Jupiter-Venus (J-V) = 100 Venus-Mercury (V-Me) = 161 Jupiter-Mercury (J-Me) conjunctions in 39.58 years.
Orbit numbers per 39.58y: 64.337~ Venus, 164.337~ Mercury, 3.3365~ Jupiter

[3 x 39.58 years = 118.74 years]

Since the ratio 61:100:161 is only one conjunction different from 60:100:160 (= 3:5:8), there is a very close match to a Fibonacci-based ratio as 3,5 and 8 are all Fibonacci numbers.

In the model we convert the orbits to whole numbers using a multiple of 3, to obtain a triple conjunction period where there are (very close to) a whole number of orbits of the relevant planets, as per the chart [right].

Why Phi? – a Mars-Earth model and more

Click on image to enlarge

The Mars-Earth model is based on 34 Mars orbits. This equates to 64 years, which is 8². Since Venus makes 13 orbits of Earth in 8 years, we can easily add it to the model.
2,3,5,8,13 and 34 are Fibonacci numbers.

The story doesn’t end there, because as the diagram shows this results in a 3:4:7 relationship between the 3 sets of synodic periods. This was analysed in detail in a paper by astrophysicist Ian Wilson, featured at the Talkshop in 2013:

Ian Wilson: Connecting the Planetary Periodicities to Changes in the Earth’s Length of Day

Why Phi? – a simple Venus-Mercury model

The model is ~99.78% accurate

The model is in the diagram, so here’s the explanation.
Divide the orbit period of Venus by that of Mercury:
0.61519726 years / 0.2408467 years = 2.554310522

To get to whole numbers, round the result up to 2.56 then:
2.56 x 5 = 12.8
12.8 x 5 = 64
64 / 25 = 2.56

64 = 8² and 25 = 5²
Therefore the approximate ratio of Mercury:Venus orbit periods is 8²:5².
The number of conjunctions in the period is the difference in orbit numbers:
8² – 5² = 64 – 25 = 39 = 13 x 3

Phi link: 2,3,5,8, and 13 are all Fibonacci numbers.

2.554310522 / 2.56 = 0.99777755~ so the accuracy of the model is around 99.78%.

An even more accurate model would be:
626 Venus = 1599 Mercury.
1599 / 626 = 2.554313 i.e. almost the same as 2.554310522 = the true ratio.

Note that 1600 / 625 = 2.56 which is the same as 8² / 5².
So there’s one more Venus (626) and one less Mercury orbit (1599) in reality, every 385.11 years, compared to our model.

Footnote:
1600 = 8² x 5²
625 = 5² x 5²
(The common 5² is redundant in the ratio, leaving 8²:5²)