## Christmas Puzzle: Fibonacci spirals and the third dimension

Posted: December 26, 2012 by tallbloke in Measurement

Here’s an interesting brainteaser I’d like some help in confirming the answer to. The Fibonacci series is of interest because it crops up in all sorts of disparate natural phenomena such as snail shells, flower seedpods, leaf growth around stems and the arrangement of the inner planets. It is a series of numbers which is easily generated by adding the previous number to the current number to obtain the next number. So, if we take the first pair to be 0 and 1 we get:

0,1,1,2,3,5,8,13,21,34,55,89,144

The ratio between adjacent numbers settles down to be around 1.6180399:1 if you divide the larger number by the smaller, or around 0.6180399:1 if you divide the smaller number by the larger. The fraction after the decimal point is the same in both cases. These two numbers are designated as the greek letters Phi and phi, the upper case being used for the bigger number. The ratio is known as the Golden Section. Kepler believed this ratio and pythagoras’ theorum held the great secret of the universe. Together they form the Kepler triangle, which has interesting properties.

We can make another picture showing the Fibonacci numbers 1,1,2,3,5,8,13,21,.. if we start with two small squares of size 1 next to each other. On top of both of these draw a square of size 2 (=1+1).

We can now draw a new square – touching both a unit square and the latest square of side 2 – so having sides 3 units long; and then another touching both the 2-square and the 3-square (which has sides of 5 units). We can continue adding squares around the picture, each new square having a side which is as long as the sum of the latest two square’s sides. This set of rectangles whose sides are two successive Fibonacci numbers in length and which are composed of squares with sides which are Fibonacci numbers, we will call them Fibonacci Rectangles.

Here is a spiral drawn in the squares, a quarter of a circle in each square. The spiral is not a true mathematical spiral (since it is made up of fragments which are parts of circles and does not go on getting smaller and smaller) but it is a good approximation to a kind of spiral that does appear often in nature. Such spirals are seen in the shape of shells of snails and sea shells and, as we see later, in the arrangement of seeds on flowering plants too. The spiral-in-the-squares makes a line from the centre of the spiral increase by a factor of the golden number in each square. So points on the spiral are 1.618 times as far from the centre after a quarter-turn. In a whole turn the points on a radius out from the centre are 1.6184 = 6.854 times further out than when the curve last crossed the same radial line.

So, onto the puzzle:

Suppose we are standing in a large tube, say it’s 4 metres in diameter, and there is a spiral inscribed inside it, such that as perspective makes it get smaller as we look further down the tube, it appears to form the proportion of the fibonacci spiral shown above.

Q . What is the ratio of the pitch of the spiral (the distance down the tube between two points where the spiral crosses a line drawn straight along the tube), and the diameter of the tube?

Hint: Objects twice as far away appear to be half the size.

1. graphicconception says:

I am going to stick my neck out here and say that it cannot be done without making an assumption.

My reasoning is that we know nothing about the horizontal distance into the tube.

A similar but more simple problem would be: Imagine you are standing next to a post in a row of evenly spaced 4 metre fence posts, in terms of post height, how far away is the next one?

2. tallbloke says:

Hi GC and welcome. I’m not sure what you mean by ” horizontal distance into the tube”.

If it helps, Imagine you are in the tube, and the spiral is continuing down it from behind you and off into the distance ahead.

3. OzWizard says:

There is no real solution to your posed question, assuming a ‘cylindrical’ pipe.

Consider the top points of the first three revs of your hypothetical spiral sequentially, moving away from your vantage point (eye at pipe centerline?). If the spiral has uniform pitch (p), those points must be equidistant from each other.

The first radius (r1) will be 2 m (directly above your head), but whatever the ‘apparent height’ of the second radius (r2’) may be, the third radius (r3’) – being twice as far away as the second – must have an ‘apparent height’ half as large (following your ‘Hint’). The ratio between apparent heights of those particular adjacent radii (r2’:r3’) must be 2:1 (or 0.5:1 in the reverse direction).

The assumption built into your question is that each ‘revolution’ of your spiral corresponds to each ‘fourth point’ in the Fibonacci series (e.g. 2, 13, 89, 610, … OR 3, 21, 144, 987, … etc.), but the apparent heights of all sequential pairs of ‘vertical radii’ as you move down the Fibonacci spiral are in a ratio which ultimately converges to 6.854:1 (or 0.1459:1 in reverse direction), so your r2 and r3 hypothetical spiral radii need to be much more than twice as far apart as the second (r2) is from the first (r1).

In the absence of equal pitch, your ‘spiral’ cannot be a spiral.

4. Sparks says:

Is this a trick question?

If calculated in 2d assuming a particular vanishing point the pitch should also have an equal ratio of Phi.

It is not possible for a physical 3d model of a fibonacci spiral of only 4 meters in diameter to be grater in pitch than the golden number in length so it can be worked out (I think) by multiplying 4 meters by phi to get the length and then dividing the length by Phi to get the pitch. I forget what the actual formula is, but funny enough, when I was studying electrical engineering I read a book on the golden ratio and thought the formulas between Ohm’s law and the golden ratio were very similar.

(Diameter) 4 * phi = (Length) 6.472135955 meters.

6.472135955 / 4 =1.61803398875

Pitch = 1.61803398875

And to get the actual diameter

4/ golden ratio = 2.472135955

1.61803398875 + 2.472135955

diameter = 4.09016994375 meters

The values should now be interchangeable.

5. Sparks says:

Just for clarity!!

If you use the actual diameter that fits the fibonacci spiral of 4.09016994375 meters

(Diameter) 4.09016994375 meters * phi = (Length) 6.61803398875 meters.

(Length) 6.61803398875 / (Diameter) 4.09016994375 = (Pitch) 1.61803398875

6. wayne says:

Seems to be 3.
Being log( fib(n) / fib(n+3) ) / log(Phi), n–>inf.

7. wayne says:

How many times do I have to tell myself… don’t just pull it from the mind !!!
Test, test, test.
Try instead: log( fib(n+3) / fib(n) ) / log(Phi), n–>inf

8. tallbloke says:

OK, thanks for the answers so far. They all disagree, so I shall sit back and await the development of a consensus. 🙂

After all, this is on the face of it a numerical problem with only one correct answer. So the challenge now becomes one of convincing the other contributors of the correctness of your own calculation through brilliantly clear exposition of what your working out means.

Good luck!

9. tallbloke says:

Sparks says:
December 27, 2012 at 6:24 am
Is this a trick question?

It’s not meant to be. I just want the right answer.

Having asked everyone else to ‘explain their workings out’ it’s only fair I should have a go myself.

I was thinking along these lines:
If the apparent diameter of the tube 4 metres ahead is, say, 2 metres, then by the linear rule I gave, 8m ahead it will have an apparent diameter of 1m, and 16m ahead it will have an apparent diameter of 0.5m. So concentric circles drawn on the inside of the tube at these distances will have apparent radii of 1m, 0.5m nd 0.25m respectively. So the third circle, which is 4 times further away than the first circle, will have an apparent radius 4 times smaller than the apparent radius of the first circle. If we add a fourth circle twice as far away as the third, (16m away), circle at 32m, it will be 8 times as far way as the first circle and will also have an apparent radius 8 times smaller at 0.125m.

Now, we know from the headline post that rather than getting two times smaller in a full turn, a 2D fib spiral ‘radius’ gets 6.854 times smaller. So rather than being twice as far away for a full turn, the 3D ‘in tube’ spiral ‘next turn’ would need to be 6.854 times further away in order for perspective to ‘shrink’ the 3D ‘in tube’ spiral enough to fit the 2D fibonacci spiral that is drawn on the clear plastic sheet we are holding up in front of us to check we got it right. This is because something twice as far away appears to be twice as small, something 4 times further away appears to be 4 times smaller, so for something to appear to be 6.854 times smaller, it needs to be 6.854 times further away.

Does everyone agree so far? If not, why not?

10. wayne says:

TB, I misread what you were looking for, the ratio of the pitch; I thought it being the distance dropped down the tube to the ‘apparent’ diameter as seen from far above.

Seems you want the drop of one revolution to the actual tube’s diameter which is always 4 feet. The number of one rev drop is slightly less than Phi^4 for we are not looking straight down the tube but our eye-point are off center slightly. Instead of a vertical line down one side of the tube being perfectly straight, it is at a slight angle looking at the spiral at the top of the post. So, you can’t really use exact increments of fib(x) to form that ratio. Can you see the same angle? It’s about 20 degrees clockwise and we’ve got a bit of geometry of the squares involved it seems. Seems that is atan(5/13) degrees off clockwise and using the limit as ninfinity would be atan(sqrt(phi)) or atan(.381966) = 20.905 degrees.

For instance, look at the top square, it’s 13×13 and one full rev touches both sides of that square so you might want to say the visual diameter difference is 13. But it is not, it’s slightly more than 13 when you dray a line where each touches that 13×13 square.

Next, is that even the answer you are seeking, the exact value?

That might help getting it closer.

11. tallbloke says:

Wayne, you’re running on ahead of me. Do you agree with my analysis so far?

And, if your eye is slightly off centre, stand on a box. I don’t need to 😉

And, we can live with an approximation for now, so long as we know where the fudge is.

12. wayne says:

Ok, agree so far. 😉

13. tallbloke says:

Cool, I’ll give others a chance to catch up.

In the time it takes Jupiter to complete 2/3 of an orbit, Earth and Venus will meet 5 times, as Earth orbits 8 times and Venus orbits 13 times. Meanwhile, Mercury will pass Venus 21 times as it orbits 34 times.

And also have a think about the relative distances of these planets from the Sun and each other.

14. OzWizard says:

To calculate the ‘foreshortening’ effect of the perspective view on receding parts of a spiral you first need to specify an eye position (EP) and the position of the picture plane (PP) relative to the end of the 4m diameter pipe in order to define the vanishing point (VP).

My “Reductio ad absurdum” proof that your Fibonacci construction is not a spiral avoids the need for actual perspective calculations; it assumes only what is known for certain about a true spiral and perspective construction. The assumption of a pure spiral allows a deduction which shows that the Fibonacci construction CANNOT be such a spiral, no matter what ‘pitch’ you try to apply to the Fibonacci ‘spiral’.

Step 1: A true spiral drawn on a (transparent) cylinrical pipe, viewed side-on in 2D, will be a pure sine wave, so successive crests MUST be the same distance apart, that distance being the pitch (X) of the spiral. No argument there? The starting points of the first three loops of the true spiral are at distances 0, X, 2X from the end of the pipe. This is exactly correct and easily visualised. Let me set up vertical radii at each of these points, R0, R1, and R2. We know R0 = R1 = R2 = 2m in real life, but don’t know the perspective effect on R1 or R2.

Step 2: So, let the apparent (perspective) heights of R0, R1 and R2 be r0, r1 and r2, respectively. Obviously, r1<R1, r2<<R2 due to perspective shortening of more distant objects. For convenience, if we fix the PP at the end of the pipe, then r0 (the apparent height of R0) will actually be 2m.

Step 3: The apparent height of R1 (or r1) compared to r0 is arbitrary (because it depends on the selection of EP as well as position of the PP) but we can say with certainty that r2 (the apparent height of the starting point of loop 3) will be EXACTLY half of r1, because it is twice as far away from the pipe end as R1., i.e r1:r2 = 2:1. This is only true for the first three radii, but it is true irrespective of the size of X (pitch). Changing X has no effect on the ratio r2:r1. It will always be EXACTLY 2:1 for a true spiral.

Step 4: It is sufficient to prove that the Fibonacci construction fails at one point only and the whole theory fails. So, since the 'apparent radii' at the top points in successive loops of the hypothesised Fibonacci 'spiral' have a known and precise ratio to each other of 6.854102:1 and this ratio is the same for ALL such successive pairs of top point radii in a Fibanacci 'spiral', this 'spiral' cannot be a true spiral drawn on a cylindrical pipe of any diameter, irrespective of its pitch. If it were such a true spiral, at least two successive radii would have a ratio of 2:1. Since no successive radii have that ratio, the Fibonacci sequence cannot be represented as a perspective distortion of a true spiral on a cylindrical surface.

15. tallbloke says:

Ozwizard: I accept we’re not going to get a perfect solution, nature doesn’t do perfect anyway. But I think we can get sufficiently close to find some interesting ‘spin-off’ results as it applies to what we observe in nature. My particular interest is in the orbits of the planets, as outlined in my reply to Wayne above.

Engineers don’t get hung-up on imperfection, if it’s close enough, it’ll do.

16. graphicconception says:

“If the apparent diameter of the tube 4 metres ahead is, say, 2 metres …”

There is the assumption I mentioned!

If you know that information it becomes possible.

17. Sparks says:

RE:
“In the time it takes Jupiter to complete 2/3 of an orbit, Earth and Venus will meet 5 times, as Earth orbits 8 times and Venus orbits 13 times. Meanwhile, Mercury will pass Venus 21 times as it orbits 34 times.”

2/3 of Jupiter’s sidereal period is approximately 8 (7.913) years.

If Saturn, Uranus and Neptune appear not to conform into The Fibonacci series that is likely because they would be part of their own Fibonacci series, with larger variables, but as part of the same sequence. Like I’ve said to here before, the variable distance from Neptune to Jupiter matches the the variable length of the 11 year solar cycle, It should be interesting to note; as both the outer planets Uranus and Neptune recently had their closest approach which happens approximately once every hundred and eighty or so years (I don’t have the figures with me) The sun experienced a series of more active cycles, up-to the time their orbits passed each-other then solar activity gradually declined (although the peek of activity remained high), I think there is a geometric planetary effect on the sun, dominated by the outer planets, they may have a stretching and and contracting (which can be tied to Large scale geometric moments ) influence on the sun’s natural physical process. It is an extremely complex issue, discussing it and airing ideas is step 1 step 2 will be when everyone in the discussion is familiar with the basics and comfortable with the math.

The third dimension RE of the Fibonacci sequence is like an early form of relativity, where ones perspective is relative, where we can look at tiny detail on small a body or look at vast mountain ranges on a large body, the Fibonacci series repeats it’s self on a similar principle this is called a sequence, In programing the 3d math of a Fibonacci sequence inverts it’s self as it’s a 2d construct, but in our physical reality, when the Fibonacci series repeats it’s self in nature as a sequence, it has a physical limit which begins and ends according to the scale of it’s possible physical dimensions.

18. graphicconception says:

Phi = Golden Mean (approx 1.618).

1 turn is 4 Fibbonacis so Phi^4 (approx 6.854) is the size ratio of the arcs for a turn.

Assuming that:

1. The tube looks half as tall in 2m distance and
2. We are standing at the start of a turn then

We need to pass approx 6.854 times that distance to find the start of the next turn.

ie (approx) (6.854 – 1) * 2m
or (approx) 13.708 metres

More exactly: ((Phi^4) – 1) * D where = D is the distance to halve apparent height

19. graphicconception says:

Oh dear, subtracting 1 seems to be a bit hit or miss at this time of day. The 13.708 above should be 11.708.

20. tallbloke says:

Excellent, we all seem to be in agreement so far.

The fibonacci series starts

1,1,2,3,5,8,13,21,34,55,89

The whole number approximate ratios of the distances of the planets from the Sun in terms of Mercury orbital distances is:

1 Mercury
2 Venus
3 Earth
4 Mars
7 Ceres
13 Jupiter
24 Saturn
49.5 Uranus
77.5 Neptune

Mars and Ceres don’t quite fit the sequence, being out by 1/5 and 1/8 respectively, Saturn is out by 1/8, Uranus by 1/9 and Neptune by 1/7. but the general idea is just about OK.
So why might the planets be arranged this way?

The sprouting angles of leaves around the stems of some plants follow a similar pattern in order to maximise the amount of sunlight falling on them by minimising shading from the leaves above. Are the planets similarly receiving the maximum energy from the Sun and minimising the amount of (Electro-dynamic) ‘interference’ between themselves and other planets by being in this configuration?

As well as an approximate Fibonacci relationship in the orbital distances, there is a Fibonacci relationship in the orbital periods, as noted in the comment I made above:

In the time it takes Jupiter to complete 2/3 of an orbit, Earth and Venus will meet 5 times, as Earth orbits 8 times and Venus orbits 13 times. Meanwhile, Mercury will pass Venus 21 times as it orbits the Sun 34 times.

And we also know there is a relationship between orbital distance and orbital period from Kepler’s third law:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

So how would the ‘idealised’ solar system’s orbital periods look if we arranged the planets orbital distances to be in line with the Fibonacci series? Would we still get the Fibonacci numbers popping out in the relative orbital periods and conjunctions? Or is the current arrangement the best compromise which most nearly fits the Fibonacci sequence for both distances and periods? I’m hoping we might be able to add something to Kepler’s squares and cubes with the fourth power factor in the proportion of the Fibonnaci spiral, and the distance squared relationship in the Parker spiral. That would tie the Sun’s revolution rate in to the scheme.

I’ll get the calculator warm as I find time today.

21. wayne says:

“minimising the amount of (Electro-dynamic) ‘interference’ between themselves and other planets by being in this configuration?”

I’ve never been so hot on “(Electro-dynamic) ‘interference’” and such. Those terms are not present in long-term ephemeris code that back fit just fine with just Newton, Kepler and Einstein’s relations.

But you might look into a minimization of gravitational interference, there you’ve probably have got something. Put any of the planets too far off their current orbits and you usually have an unstable solar system in the long run. Might also need to play in their masses or neighboring masses and that might be why the asteroids and Mars don’t quite fit in and the combo might fall in line to that Fibonacci sequence better.

Keep in mind there are also factors of 16, eight, four and two when you go to scale to fit such a numeric sequence in period, orbit, and masses (densities). Those might cancel some of the apparent variances at a different scale.

22. tallbloke says:

Thanks Wayne,
The orbital eccentricities and perturbations certainly speak of gravitational effects. I think there are electrodynamic effects too. I’m mindful of Miles Mathis’ interesting work in teasing apart Newton’s gravitational equation, and finding buried within it terms which are to do with charge, as well as mass.

http://milesmathis.com/ug.html

In fact, in looking for the link to that paper, I’ve just come across this one, which may be highly relevant to my speculative investigation here.

http://milesmathis.com/phi.html

Taster:
“Mathematicians have long known that it requires constraints to create the golden ratio. The golden ratio is a sort of feedback mechanism between two variables, forcing them into the ratio. They can show these constraints mathematically, but they cannot show them physically. Put simply, the golden ratio is the only solution to x – 1 = 1/x. Mathematicians can show you tricks with that all day long, but they can’t tell you why phi shows up in nature, although we know that it does. That is what I am doing here. The short answer is that pure math doesn’t have the complexity necessary to solve this. That last equation has only one variable in it. The more popular form of the golden ration has two. But that still isn’t enough. We have to give the constraints variables, also, in order to show the mechanical cause of the ratio. And once we have given the constraits variables, we have to assign those variables to something. That is the other historical problem here: without the charge field, physicists had nothing to assign the constraints to. Gravity is completely incapable of causing the golden ratio, and most physicists in history came to see that pretty quickly. So they gave up. That is why the problem still exists today. “

23. Baa Humbug says:

Oh brother, Theodore Landscheidt and the Golden Section…my head hurts.

Alignment of the planets has nothing to do with receiving energy from the sun. It is all about stability.

Assume for a moment you are an Olympic hammer thrower. Spinning that heavy weight at the end of a chain puts enormous strains on the athlete. Unless the athlete has stability, he’d spin out of control and end up on his ass.
The Sun is like that hammer thrower, except that the Sun has many heavy weights at the end of many chains spinning around it. Imagine the forces created by the giant planets.

As the solar system is stable, it is safe to assume a harmonic relationship has been established between the Sun and the planets revolving around it (it has been billions of years after all). This harmonic is explained by the Golden Section.

In fact, 11 year solar cycles exhibit this harmonic whereby the first part of the cycle (the rising) is 0.3819 x 11yrs = 4.2 years long, and the second part (the falling) is 0.618 x 11yrs = 6.8 years long.

The golden section is exhibited all over nature as well as stable man made structures e.g.

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

As illustrated in Figure 11, the Golden section divides a frame structure like a line segment, a surface, a cycle, or any other delimited feature so that the ratio of the whole to the larger part (major) equals the ratio of the larger part to the smaller one (minor). Point G represents the irrational Golden Number
G = 0.618… It divides the unit height of the temple into major (0.618…) and minor (0.3819…). To find the major of a line segment, a cycle etc., it has to be multiplied by 0.618. Multiplication by 0.382 yields the minor. As the fundamental oscillation of the sun about CM depends on the masses and the positions of the giant planets, the relationship with the Golden section extends to the whole solar system. A. N. Kolmogorov [47], V. I. Arnol’d [1], and J. Moser [79] have proven theoretically, that the stability of the solar system hinges on the Golden section. This is crucial, as we know from publications by G. J. Sussman and J. Wisdom [110] as well as J. Laskar [67] that the orbits of all planets are chaotic. In my paper “The Cosmic Function of the Golden Section” [64] I have shown in practice how the Golden section, which stands for stability in polar opposition to instability, keeps the chaotic planetary orbits stable. The mean of the ratios of the perihelion distances of neighbouring planets from Mercury to Pluto, including the mean radius vector of the planetoids, turns out to be very close to the Golden number G. The difference between this mean and G is as small as 0.002. Fivefold quantities have deep roots in Nature. There are not four, but five physical forces. We merely have forgotten that electromagnetism is composed of different forces. First Maxwell unified electricity and magnetism and later on electromagnetism and the weak force was unified to constitute the electro-weak force [44].

24. tallbloke says:

Baa: It’s an interesting topic, and a fine one for a Christmas time speculative brainstorming session.

Somehow, the golden section, expressed in various aspects of natural phenomena, is an underlying force which shapes the proportion of things. Mathis’ paper on Phi linked above makes a good stab at what is going on I think.

http://milesmathis.com/phi.html

25. oldbrew says:

If you can apply the Fibonacci numbers you’ll have no trouble converting kilometers to miles (the previous number) or vice versa (the following number), even if it’s not quite a ‘natural phenomenon’ 🙂

E.g…13 miles = 21 km.

26. oldbrew says:

TB: FYI re the Golden Ratio, Miles Mathis has gone further courtesy of an ‘assistant’.

‘The golden ratio is so prevalent in nature—not because the number has any real magic—but because the ambient charge field causes feedback mechanisms that make this number a stable ratio.’

http://milesmathis.com/mhphoton.pdf

His new concept is the “silver ratio.”

27. oldbrew says:

Correction: silver ratio is not a new concept, but this application of it is.

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

28. tallbloke says:

Thanks Oldbrew, I’d not come across that before.

29. J Martin says:

Sparks said ” the variable distance from Neptune to Jupiter matches the the variable length of the 11 year solar cycle

So that would imply that one (you or someone) can predict not just the length of this solar cycle but also the next one. Because presumably the orbital motions of these two planets for years ahead can be obtained from JPL and the consequent relative distance between Neptune and Jupiter could be worked out by someone or some organisation with the right knowledge and tools (JPL presumably).

Or better still Sparks already has this information and will post it.

30. Sparks says:

@J Martin

I actually have been researching that very idea, I originally wanted to find away to hind-cast the sunspot cycle back to observations made by early astronomers as far back as 28 BC, the Neptune and Jupiter orbital variation give a promising match to recent observations to work from, I have also been working on a technique to add or remove the influence of all the other planets at inferior conjunction and opposition etc… for better resolution. The interesting thing about studying the planetary orbits and the possible influence on the sun, is that the planetary orbits are very well known and can be easily modeled to high accuracy with a high degree of confidence years into the past and years into the future.

It’s a lot of hard work and can be monotonous at times building the orbital data, extracting the sunspot data and programing a flat-file database, then working out the beginning and end of each cycle and comparing that data to the orbital data. Looking at certain planetary data runs, the orbits appear to be happening slightly out of sync, as if the beginning and end of an entire orbital cycle had taken place just before the beginning and end of each sunspot cycle I think that this timing could be important, there is also a slight shift in this from other planetary orbits which I hope to nail-down as I mentioned above. So, at the moment it’s a learning and discovery process for me, there maybe no value whatsoever to any of this work or maybe there will be.

It’s important to acknowledge that looking into and studying a subject in some depth is a positive thing regardless of the result or where it leads.

31. oldbrew says:

@ Sparks

FYI…Interview: Is climate change caused by solar inertial motion?

32. Sparks says:

@ oldbrew

Thanks, great interview, How bizarre is it when the same science that is dismissed by climate alarmists about our own Sun, is being used to discover new worlds and has actually proved itself useful.

@ Roger, I found three of what looks like Fibonacci spirals in the four outer giants, you probably know about them, I’ll have more time to take a snapshot for you tomorrow, the timing is of these are very interesting, one happened recently and they seem to be related to the orbital sidereal distance between Neptune and Uranus when they are passing each other which only happens once every 170+ years. The configuration of the planets at these points in time, does indeed seem to have a greater effect on the Suns activity. I’m wondering if there is a change in Neptune’s north pole recently as it usually faces towards the inner solar system, Could Neptune right it’s self at times become unstable and/or have a greater angular momentum or have another direct influence on the sun? A planetary mechanism like this may explain how Ice ages begin without having external solar system factors or catastrophic terrestrial events involved. It’ may simply comedown to Neptune becoming temperately/slightly unstable in regular intervals. I know there are no actual observational records for Neptune’s orbit going back to the last Ice age and that it’s assumed to have a stable orbit, but it’s obvious that if any of these planets were to become temperately/slightly unstable the most likely candidate is the outermost planet and in this case Neptune. Do you think my pet theory is plausible? I call it the Neptune Theory!! It’s not a complete theory, but I’ve been working on it for awhile, as you are interested in this area, I thought I’d bounce the basic idea off you.

33. oldbrew says:

To stir the [Fibonacci] pot a bit more perhaps, some more Neptune info.

“Pluto lies in the 2:3 mean motion resonance with Neptune: for every two orbits that Pluto makes around the Sun, Neptune makes three. The two objects then return to their initial positions and the cycle repeats, each cycle lasting about 500 years.”

http://en.wikipedia.org/wiki/Pluto#Relationship_with_Neptune

Wikipedia also notes that “a small region of Pluto’s orbit lies nearer the Sun than Neptune’s” (due to its eccentricity).

34. Roger Andrews says:

Don’t know whether it means anything, but I find it intriguing that the Fibonacci series defines a lognormal distribution almost exactly even though it predates logarithms by a thousand years.

35. oldbrew says:

Blogger AGW-not proposes a 13,20,13,20 [etc.] pattern of the number of solar cycles between deep and shallow solar minima.

He then notes: ‘Also 13 cycles at average length of 10.68 years and 20 cycles at average length of 11.23 years is the golden proportion (38.2:62.8).

http://www.agwnot.blogspot.co.uk/2010/11/climate-and-solar-regularities-and_16.html