Archive for the ‘Maths’ Category

Image credit: naturalnavigator.com


The contention here is that in the time taken for 14 lunar nodal cycles, the difference between the number of Saros eclipse cycles and lunar apsidal cycles (i.e the number of ‘beats’ of those two periods) is exactly 15.

Since 15-14 = 1, this period of 260.585 tropical years might itself be considered a cycle. It is just over 9 Inex eclipse cycles (260.5 years) of 358 synodic months each, by definition.

Although it’s hard to find references to ~260 years as a possible climate and/or planetary period, there are a few for the half period i.e. 130 years, for example here.

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The Kepler-42 system as compared to the Jovian system [credit: NASA/JPL-Caltech]

The headline was NASA’s joke about both the size and the short orbit periods (all less than two days) of the three planets in the Kepler-42 system.

The discovery of this system dates back to 2012, but there don’t seem to be any numbers on resonant periods, so we’ll supply some now.

Wikipedia says:
‘Kepler-42, formerly known as KOI-961, is a red dwarf located in the constellation Cygnus and approximately 131 light years from the Sun. It has three known extrasolar planets, all of which are smaller than Earth in radius, and likely also in mass.’

‘On 10 January 2012, using the Kepler Space Telescope three transiting planets were discovered in orbit around Kepler-42. These planets’ radii range from approximately those of Mars to Venus. The Kepler-42 system is only the second known system containing planets of Earth’s radius or smaller (the first was the Kepler-20 system). These planets’ orbits are also compact, making the system (whose host star itself has a radius comparable to those of some hot Jupiters) resemble the moon systems of giant planets such as Jupiter or Saturn more than it does the Solar System.’

The three planets in order of distance from their star (nearest first) are c,b and d. They all have very short orbit periods ranging from under half a day to less than two days, and the star has only 13% of the power of our Sun.

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Image credit: interactivestars.com


In 2015 this post discussed long-term lunar precession from an apsidal, or anomalistic, standpoint.

We saw that all the numbers related to an exact number (339) of Metonic cycles (19 tropical years each, as discussed below).

Here we show the equivalent from a nodal, or draconic, standpoint.

Again, all the numbers relate to an exact number (337 this time) of Metonic cycles.

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Credit: NASA’s Goddard Space Flight Center


First the report, then a brief Talkshop analysis.

NASA’s Transiting Exoplanet Survey Satellite (TESS) has discovered a world between the sizes of Mars and Earth orbiting a bright, cool, nearby star, reports MessageToEagle.com.

The planet, called L 98-59b, marks the tiniest discovered by TESS to date.

Two other worlds orbit the same star.

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Kepler Space Telescope [credit: NASA]


Star Kepler-102 has five known planets, lettered b,c,d,e,f. These all have short-period orbits between 5 and 28 days. Going directly to the orbit period numbers we find:
345 b = 1824.0012 d
258 c = 1824.4263 d
177 d = 1825.1709 d
113 e = 1824.4629 d
(for comparison: about 1-2 days short of 5 Earth years)

For the purposes of this post planet f (the furthest of the five from its star) is excluded, except to say that in terms of conjunctions 8 e-f = 11 d-e. Now let’s look for some resonances of the inner four planets.

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Jupiter – the dominant planet in the solar system

The aim here is to show a Lucas number based pattern in five rows of synodic data, then add in a note on Mercury as well.

There’s also a strong Fibonacci number element to this, as shown below.

The results can be linked back to earlier posts on planetary harmonics involving the Lucas and Fibonacci series (use ‘search this site’ box on our home page).

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Continuing our recent series of posts, with Uranus-Neptune conjunction data an obvious starting point for the table is where the difference between the number of Neptune orbits and U-N synods is 1.

647 U-N takes a long time (~110,900 years) but the accuracy of the whole number matches is very high.

Lucas no. (7 here) is fixed, and Fibonacci nos. follow the correct sequence (given their start no.).
Full Fib. series starts: 0,1,1,2,3,5,8,13,21…etc.
Multiplier: 0,1,1,2,3
Addition: 1,1,2,3,5

The Neptune orbits are multiples of 26 with the same Fibonacci adjustment:
Add 0,1,1,2,3 to the Neptune column numbers to get an exact multiple of 26 (which will be the pattern number in the last column).

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Lunar evections and the Saros cycle

Posted: May 7, 2019 by oldbrew in Maths, moon, solar system dynamics
Tags:

Credit: Matthew Zimmerman @ English Wikipedia


The Saros cycle can be used to predict eclipses of the Sun and Moon, and is usually defined as 223 lunar synodic months, or about 11 days over 18 years.

But there are a few other lunar-related periods which can used to arrive at 223.

One Saros cycle can be said to be the difference between the number of:
— anomalistic months and full moon cycles (239 – 16)
— draconic months and draconic years (242 – 19)
— tropical months and tropical years (241 – 18)

That may be fairly well known, but then there are the lunar evections.

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Distances not to scale.


This is an easy data table to interpret.

The Uranus orbits are all Fibonacci numbers, and the synodic conjunctions are all a 3* multiple of Fibonacci numbers.
[Fibonacci series starts: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …etc.]

In addition, the difference between the two is always a Lucas number. And that’s it for Saturn-Uranus, which would make for a very short blog post.

But it’s possible to go further.

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We’re now looking for a pattern arising from the Jupiter-Saturn synodic conjunctions and the orbit periods.

Focussing on the numbers of Jupiter orbits that are equal, or nearly equal, to an exact number of Saturn orbits (years), a pattern can be found by first subtracting the number of conjunctions from the number of Saturn orbits.

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A simple pattern emerges when looking at the Earth-Mars synodic conjunctions.

Focussing on the numbers of Mars orbits that are equal, or almost equal, to an exact number of Earth orbits (years), the pattern can be found by subtracting the number of conjunctions from the number of Mars orbits.

The difference between the two sets of numbers follows the Fibonacci series, which is strongly related to the golden ratio.

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Jupiter – the dominant planet in the solar system

The aim here is to show a Lucas number based pattern in seven rows of synodic data.
There’s also a Fibonacci number element to this, as shown below.
The results can be linked back to an earlier post on planetary harmonics (see below).

The nearest Lucas number equation leading to the Jupiter orbit period in years is:
76/7 + 1 = 11.857142 (1, 7 and 76 are Lucas numbers).
The actual orbit period is 11.862615 years (> 99.95% match).
[Planetary data source]

It turns out that 7 Jupiter orbits take slightly over 83 years, while 76 Jupiter-Earth (J-E) synodic conjunctions take almost exactly 83 years. One J-E synod occurs every 1.09206 years. (83/76 = 1.0921052).

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Image credit: NASA


Researchers have an ambition to use ‘new mathematics’ to try and predict where and when these extreme events will occur.

Florida State University researchers have found that abrupt variations in the seafloor can cause dangerous ocean waves known as rogue or freak waves—waves so catastrophic that they were once thought to be the figments of seafarers’ imaginations, Phys.org reports.

“These are huge waves that can cause massive destruction to ships or infrastructure, but they are not precisely understood,” said Nick Moore, assistant professor of mathematics at Florida State and author of a new study on rogue waves.

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Vertical line shows planetary conjunction with the Sun [credit: Wikipedia]


Numerous studies have found evidence of an apparently regular and significant climate event every 1,470 years (on average), which seems to show up most clearly in glacial periods. They speak of a ‘robust 1,470-year response time’, ‘a precise clock’, ‘abrupt climate change’ and so forth.

However they also say things like: ‘The origin of this regular pacing…remains a mystery.’

A couple of example studies here:
Possible solar origin of the 1,470-year glacial climate cycle demonstrated in a coupled model (2005)

Timing of abrupt climate change: A precise clock (2003)
– – –
Now we can relate this to the half period of the Jupiter-Saturn (J-S) conjunction cycle, i.e. one inferior or superior conjunction, as explained at Wikipedia.

The average J-S half-period is 9.932518 years.
The nearest harmonic to that period in Earth years is 10.
1470 = 148 * J-S/2
1470 = 147 * 10y
148 – 147 = 1 Dansgaard-Oeschger cycle

We find also that Jupiter, Saturn and Neptune conjunctions are such that:
148 * J-S/2 = 74 J-S = 41 S-N = 115 J-N = 1,470 years. [74 + 41 = 115]

Therefore 3 of the 4 major planets have a 1,470 year conjunction cycle.
(Planetary data from JPL @ NASA here)

So that’s the concept.
– – –
The graphics below are from Carsten Arnholm’s Solar Simulator software tool.
The interval between left and right sides is 1,470 years (May 501 – May 1971).

Each one shows a Jupiter, Neptune and Earth syzygy with Saturn opposite.
Note the similarity of the positions (red lines cross at the solar system barycentre).


This started as a search for a period when the Sun and the Moon would both complete a whole number of rotations.
The result was:
Solar: 25.38 days * 197 = 4999.860 d
Lunar: 27.321662 * 183 = 4999.864 d
(data sources: see reference notes at end)

Taking these as equivalent, we have 197-183 = 14 ‘beats’.
197 = 14*14, +1
183 = 13*14, +1
4999.864 / 14 = 357.13314 days
357.13314 days * 45/44 = 365.2498 days
45 * 14 (630) beats = 44 * 14 (616) calendar years, difference = 0.022 day

So the beat period of the two rotations is 44/45ths of a year, i.e. the difference in number of rotations is exactly 1 in that length of time.
630 beats = 616 years (630 – 616 = 14)
616/45 = 13.68888 calendar years = 4999.8663 days
184 lunar sidereal months (rotations) = 4999.864 days

Then something else popped up…

The Phi factor:
‘We recover a 22.14-year cycle of the solar dynamo.’ (2016 paper)
See: Why Phi? – modelling the solar cycle

Solar Hale cycle = ~22.14 years (est. mean)
13.68888 * Phi = 22.149~ years
22.14 / 13.68888 = 1.61737 (99.96% of Phi)
(55/34 = 1.617647)

From the same post:
Jupiter-Saturn axial period (J+S) is 8.456146 years.
That’s when the sum of J and S orbital movement in the conjunction period = 1

13.68888 / 8.456146 = 1.618808
Phi = 1.618034

Conclusion:
This cycle of solar and lunar sidereal rotation (SRC) sits at the mid-point of the Phi²:1 ratio between the J+S axial period and the mean solar Hale cycle, i.e. with a Phi ratio to one and inverse Phi to the other.
SRC = (J+S) * Phi
SRC = Hale / Phi
SRC = Hale – (J+S)
(Mean Hale value is assumed)

In a period of 616 years there are 45 SRC.
The period is 44 * 14 years = 45 SRC = 45 * 14 beats.
SRC * (45/44) = 14 years.

Cross-checks:
Carrington rotations per 616 y = 8249
8249 CR / 45 = 4999.865 days

Synodic months per 616 y = 7619
7619 SM / 45 = 4999.856 days
8249 – 7619 = 630 = 45 * 14

45*183 sidereal months = 8235
8235 – 7619 = 616
8249 CR – 8235 Sid.M = 14
Beat period of CR and Sid.M = 616/14 = 44 years = 45 * (13.6888 / 14)
Every 44 years there will be exactly one less lunar rotation (sidereal month) than the number of Carrington rotations.

8249 CR – 7619 synodic months = 630 = 45 * 14
630 – 616 = 14
– – –
The anomalistic year

The beat period of the tropical month and solar sidereal rotation * 45/44 = the anomalistic year.
(27.321582 * 25.38) / (27.321582 – 25.38) = 357.14265 days
45 * 357.14265 = 16071.419 days
44 * 365.259636 = 16071.423 days

The anomalistic year is the time taken for the Earth to complete one revolution with respect to its apsides. The orbit of the Earth is elliptical; the extreme points, called apsides, are the perihelion, where the Earth is closest to the Sun (January 3 in 2011), and the aphelion, where the Earth is farthest from the Sun (July 4 in 2011). The anomalistic year is usually defined as the time between perihelion passages. Its average duration is 365.259636 days (365 d 6 h 13 min 52.6 s) (at the epoch J2011.0).
http://en.wikipedia.org/wiki/Year#Sidereal.2C_tropical.2C_and_anomalistic_years
– – –
Data sources

— Carrington Solar Coordinates:
Richard C. Carrington determined the solar rotation rate by watching low-latitude sunspots in the 1850s. He defined a fixed solar coordinate system that rotates in a sidereal frame exactly once every 25.38 days (Carrington, Observations of the Spots on the Sun, 1863, p 221, 244). The synodic rotation rate varies a little during the year because of the eccentricity of the Earth’s orbit; the mean synodic value is about 27.2753 days.
http://wso.stanford.edu/words/Coordinates.html

— The standard meridian on the sun is defined to be the meridian that passed through the ascending node of the sun’s equator on 1 January 1854 at 1200 UTC and is calculated for the present day by assuming a uniform sidereal period of rotation of 25.38 days (synodic rotation period of 27.2753 days, Carrington rotation).
http://jgiesen.de/sunrot/index.html

The sidereal month is the time between maximum elevations of a fixed star as seen from the Moon. In 1994-1998, it was 27.321662 days.
http://scienceworld.wolfram.com/astronomy/SiderealMonth.html

Lunar precession update

Posted: October 15, 2017 by oldbrew in Fibonacci, Maths, moon, Phi, solar system dynamics
Tags: ,

Credit: NASA


I found out there’s an easy way to simplify one of the lunar charts published on the Talkshop in 2015 on this post:
Why Phi? – some Moon-Earth interactions


In the chart, synodic months (SM) and apsidal cycles (LAC) are multiples of 104:
79664 / 104 = 766
728/104 = 7

The other numbers are not multiples of 104, but if 7 is added to each we get this:
86105 + 7 = 86112 = 828 * 104 (TM)
85377 + 7 = 85384 = 821 * 104 (AM)
5713 + 7 = 5720 = 55 * 104 (FMC)
6441 + 7 = 6448 = 62 * 104 (TY)

TM = tropical months
AM = anomalistic months
SM = synodic months
LAC = lunar apsidal cycles
FMC = full moon cycles
TY = tropical years


Here’s an imaginary alternative chart based on these multiples of
104. [Cross-check: 828 – 766 = 62]

In reality, 55 FMC = just over 62 TY and 7 LAC = just short of 62 TY.
For every 7 apsidal cycles (LAC), there are 766 synodic months (both chart versions).

In the real chart:
For every 104 apsidal cycles, all numbers except SM slip by -1 from being multiples of 104. So after 7*104 LAC all the other totals except SM are ‘reduced’ by 7 each.

In the case of tropical years, 6448 – 7 = 6441 = 19 * 339
19 tropical years = 1 Metonic cycle

If the period had been 6448 TY it would not have been a whole number of Metonic cycles.
Also 6441 * 4 TY (25764) is exactly one year more than 25763 synodic years i.e. the precession cycle, by definition.

Fibonacci: 104 is 13*8, and the modified FMC number is 55 (all Fibonacci numbers).

Phi: we’ve explained elsewhere that the number of full moon cycles in one lunar apsidal cycle is very close to 3*Phi².
We can see from the modified chart that the FMC:LAC ratio of 55:7 is 3 times greater than 55:21 (55/21 = ~Phi²)
– – –
Note – for more discussion of the ~62 year period, try this search:
site:tallbloke.wordpress.com 62 year
[see Google site search box in grey zone on left of this web page]

Credit: NASA


This is from a Q&A on a website linked with Sydney Observatory. We add brief notes at the end.

Lionel asks: Congratulations on your Venus book.

Excellent. I notice that there is a 243 year cycle for Transits of Venus
243 x 365.242 = 224.7 x 395
So far so good. The axial rotation period for Venus is 243.1 days.
Is this a coincidence or is there some underlying geometrical fact that I cannot see?
well-done,

Answer: An interesting and complex question that I address below.

Patterns in the transits of Venus
Let us first look at the patterns in the transits of Venus. We need to note that Venus and the Earth line up with the Sun every 583.92 days or 1.59872 years. This is called the synodic period.

If there was a transit, say the one in June 2004, for another transit to occur, the two planets must not only line up with each other and the Sun, but do so after an integer number of years so that they are back in the right places on each of their orbits.

Venus and Earth fulfil these requirements after five synodic periods = 7.9936 years as this is almost, though not quite, equal to the integer eight. Thus transits of Venus generally occur in pairs eight years apart. However, because of the slight inequality there is no third transit after another eight years.

A more accurate relationship occurs after 152 synodic periods = 243.00544 years or ~395 Venus years. The pattern of Venus transits thus repeats at 243 year intervals (This is the cycle quoted by Lionel in his question above). For example, the first pair of June transits after 8 June 2004 begins on 11 June 2247. Of course, in the meantime there is also a pair of December transits beginning in 2117.

The rotation of Venus
Scientists using radar observations from the 1960s onwards discovered that Venus spins backwards, that is in the opposite direction to its motion around the Sun, at the slow rate of 243.02 days.

They soon realised that means that Venus, almost but not quite, shows the same face towards the Earth each time the planets are lined up with each other and the Sun. Somehow there is a resonance between the motion of the Earth around the Sun and Venus’ spin around its axis. Scientists are unsure why this is the case, but one suggestion is that Venus is more massive on the face turned towards the Earth at those times and consequently it was gravitationally captured by the Earth.

How is it worked out that Venus shows the same face towards the Earth each time they line up? The quoted value of 243.02 days is with respect to distant stars. With a little arithmetic (taking inverses) we can easily convert that value to the rotation period with respect to the Sun or, in other words, to the day on Venus. It is 116.75 (Earth) days. Five of those periods equal 583.75 days, which is almost the same as the 583.92 day synodic period. So each time the planets line up Venus shows almost the same face to the Sun and hence the same face to the Earth, which is always on those occasions on the opposite side of Venus.

Coincidence or not
As Lionel points out it is interesting that transits of Venus repeat in a cycle of 243 years while the rotation period of Venus with respect to the stars is 243 days, The above detailed discussion indicates that there is no obvious connection that gives rise to the same number in each case. However, the calculations all depend on many of the same factors such as the orbital periods of Venus and the Earth so maybe there was a chance that the same number should recur.

Note the values quoted above are from the NASA Venus Fact Sheet.

Source: Are transits and the rotation of Venus linked? – Observations
– – –
Talkshop notes

Re: ‘Five of those periods equal 583.75 days, which is almost the same as the 583.92 day synodic period.’ [‘Venus and the Earth line up with the Sun every 583.92 days or 1.59872 years’]

Note 1: 23 solar rotations @ 25.38 days = 583.74 days
This also looks like a resonance, this time between the Sun and the Venus day.
. . .
Re: Venus and Earth fulfil these requirements after five synodic periods = 7.9936 years
A more accurate relationship occurs after 152 synodic periods = 243.00544 years or ~395 Venus years.

Note 2: using their own data, 157 synodic periods is more accurate, i.e. closer to a whole number of Earth orbits.
1.59872 * 152 = 243.00534 years (as stated in their notes)
1.59872 * 157 = 250.99904 years (~408 Venus years)
Of course that would be an ‘extra’ five synodic periods = 7.9936 years.

That may contradict the official ‘wisdom’ but there it is. It was discussed in some detail in this 2015 Talkshop post (some readers may find the comments to be of interest):
Why Phi? – a Venus transit cycle model

Why Phi? – the rainbow angle

Posted: September 3, 2017 by oldbrew in Maths, Measurement, Phi, weather
Tags:

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.

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Some Pythagorean triples [credit: Cmglee / Wikipedia]


Could Babylonian base-60 maths be about to make a comeback? The tablet has been dated to between 1822 and 1762 BC and is based on Pythagorean triples, as Phys.org reports. It uses ‘a novel kind of trigonometry based on ratios, not angles and circles’.

UNSW Sydney scientists have discovered the purpose of a famous 3700-year old Babylonian clay tablet, revealing it is the world’s oldest and most accurate trigonometric table, possibly used by ancient mathematical scribes to calculate how to construct palaces and temples and build canals.

The new research shows the Babylonians beat the Greeks to the invention of trigonometry – the study of triangles – by more than 1000 years, and reveals an ancient mathematical sophistication that had been hidden until now.

Known as Plimpton 322, the small tablet was discovered in the early 1900s in what is now southern Iraq by archaeologist, academic, diplomat and antiquities dealer Edgar Banks, the person on whom the fictional character Indiana Jones was based.

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The Catch-22 of Energy Storage

Posted: July 27, 2017 by tallbloke in Analysis, Energy, Maths, wind
Tags:

.

H/T @hockeyschtick1 for this great article on the non-viability of wind/solar as large-scale replacement for fossil/nuclear. Now can we scrap the CCA please?

 

Brave New Climate

Pick up a research paper on battery technology, fuel cells, energy storage technologies or any of the advanced materials science used in these fields, and you will likely find somewhere in the introductory paragraphs a throwaway line about its application to the storage of renewable energy.  Energy storage makes sense for enabling a transition away from fossil fuels to more intermittent sources like wind and solar, and the storage problem presents a meaningful challenge for chemists and materials scientists… Or does it?


Guest Post by John Morgan. John is Chief Scientist at a Sydney startup developing smart grid and grid scale energy storage technologies.  He is Adjunct Professor in the School of Electrical and Computer Engineering at RMIT, holds a PhD in Physical Chemistry, and is an experienced industrial R&D leader.  You can follow John on twitter at @JohnDPMorgan First published in Chemistry in Australia .

Several recent analyses of the…

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