Time and the Tides Wait for Godot

Guest Post by Willis Eschenbach

I’ve been listening to lots of stuff lately about tidal cycles. These exist, to be sure. However, they are fairly complex, and they only repeat (and even then only approximately) every 54 years 34 days. They also repeat (even more approximately) every 1/3 of that 54+ year cycle, which is 18 years 11 days 8 hours. This is called a “Saros cycle”. So folks talk about those cycles, and the 9 year half-Saros-cycle, and the like. The 54+ year cycle gets a lot of airtime, because people claim it is reflected in a sinusoidal approximately 54-year cycle in the for example the HadCRUT temperature records.

Now, I originally approached this tidal question from the other end. I used to run a shipyard in the Solomon Islands. The Government there was the only source of tide tables at the time, and they didn’t get around to printing them until late in the year, September or so. As a result, I had to make my own. The only thing I had for data was a printed version of the tide tables for the previous year.

What I found out then was that for any location, the tides can be calculated as a combination of “tidal constituents” of varying periods. As you might imagine, the strongest tidal constituents are half-daily, daily, monthly, and yearly. These represent the rotations of the earth, sun, and moon. There’s a list of the various tidal constituents here, none of which are longer than a year.

day by day tidal force earthFigure 1. Total tidal force exerted on the Earth by the combination of the sun and the moon. 

So what puzzled me even back then was, why are there no longer-period cycles used to predict the tides? Why don’t we use cycles of 18+ and 54.1 years to predict the tides?

Being a back to basics, start-from-the-start kind of guy, I reckoned that I’d just get the astronomical data, figure out the tidal force myself, and see what cycles it contains. It’s not all that complex, and the good folks at the Jet Propulsion Lab have done all the hard work with calculating the positions of the sun and moon. So off I went to JPL to get a couple hundred years data, and I calculated the tidal forces day by day. Figure 1 above shows a look at a section of my results:

These results were quite interesting to me, because they clearly show the two main influences (solar and lunar). Figure 1 also shows that the variations do not have a cycle of exactly a year—the high and low spots shift over time with respect to the years. Also, the maximum amplitude varies year to year.

For ease of calculation, I used geocentric (Earth centered) coordinates. I got the positions of the sun and moon for the same time each day from 1 January 2000 for the next 200 years, out to 1 Jan 2200. Then I calculated the tidal force for each of those days (math in the appendix). That gave me the result you see in Figure 1.

However, what I was interested in was the decomposition of the tidal force into its component cycles. In particular, I was looking for any 9 year, 18+ year, or 54.1 year cycles. So I did what you might expect. I did a Fourier analysis of the tidal cycles. Figure 2 shows those results at increasingly longer scales from top to bottom.

Fourier analysis tidal forceFigure 2. Fourier analysis of the tidal forces acting on the earth. Each succeeding graph shows a longer time period. Note the increasing scale.

The top panel shows the short-term components. These are strongest at one day, and at 29.5 days, with side peaks near the 29.5 day lunar cycle, and with weaker half-month cycles as well.

The second panel shows cycles out to 18 months. Note that the new Y-axis scale is eight times the old scale, to show the much smaller annual cycles. There are 12 month and 13.5 month cycles visible in the data, along with much smaller half-cycles (6 months and 6.75 months). You can see the difference in the scales by comparing the half-month (15 day) cycles in the top two panels.

The third panel shows cycles out to 20 years, to investigate the question of the 9 and 18+ year cycles … no joy, although there is the tiniest of cycles at about 8.75 years. Again, I’ve increased the scale, this time by 5X. You can visualize the difference by comparing the half-year (6-7 month) cycles in the second and third panels. At this scale, any 9 or 18+ year cycles would be very visible … bad news. There are no such cycles in decomposition of the data.

Finally, the fourth panel is the longest, to look for the 54 year cycle. Again, there is no such underlying sine-wave cycle.

Now, those last two panels were a surprise to me. Why are we not finding any 9, 18+, or 54 year cycle in the Fourier transform? Well … what I realized after considering this for a while is that there is not a slow sine wave fifty-four years in length in the data. Instead, the 54 years is just the length of time that goes by before a long, complex superposition of sine waves approximately repeats itself.

And the same thing is true about the 18-year Saros cycle. It’s not a gradual nine-year increase and subsequent nine-year decrease in the tidal force, as I had imagined it. Instead, it’s just the (approximate) repeat period of a complex waveform.

As a result, I fear that the common idea that the apparent ~60 year cycle in the HadCRUT temperatures is related to the 54-year tidal cycles simply isn’t true … because that 54 year repeating cycle is not a sine wave. Instead, looks like this:

repeating 54 year tidal cycleFigure 3. The 54 year 34 day repetitive tidal cycle. This is the average of the 54-year 34-day cycles over the 200 years of data 2000-2200.

Now, as you can see, that is hardly the nice sine wave that folks would like to think modulates the HadCRUT4 temperatures …

This exemplifies a huge problem that I see happening. People say “OK, there’s an 18+ year Saros cycle, so I can divide that by 2. Then I’ll figure the beat frequency of that 9+ year cycle with the 8.55 year cycle of the precession of the lunar apsides, and then apply that to the temperature data …”

I’m sure that you can see the problems with that approach. You can’t take the Saros cycle, or the 54+ year cycle, and cut it in half and get a beat frequency against something else, because it’s not a sine wave, as people think.

Look, folks, with all the planets and moons up there, we can find literally hundreds and hundreds of varying length astronomical cycles. But the reality, as we see above, is not as simple as just grabbing frequencies that fit our theory, or making a beat frequency from two astronomical cycles.

So let me suggest that people who want to use astronomical cycles do what I did—plot out the real-life, actual cycle that you’re talking about. Don’t just grab the period of a couple of cycles, take the beat frequency, and call it good …

For example, if you want to claim that the combined tidal forces of Jupiter and Saturn on the sun have an effect on the climate, you can’t just grab the periods and fit the phase and amplitude to the HadCRUT data. Instead, you need to do the hard lifting, calculate the actual Jupiter-Saturn tidal forces on the sun, and see if it still makes sense.

Best regards to everyone, it’s still raining here. Last week, people were claiming that the existence of the California drought “proved” that global warming was real … this week, to hear them talk, the existence of the California floods proves the same thing.

In other words … buckle down, it’s gonna be a long fight for climate sanity, Godot’s not likely to show up for a while …

w.

THE USUAL: If you disagree with something that I or someone else said, please quote the exact words you disagree with, and tell us why. That way, we can all understand what you object to, and the exact nature of your objection.

CALCULATIONS: For ease of calculations, I downloaded the data for the sun and moon in the form of cartesian geocentric (Earth-centered) coordinates. This gave me the x, y, and z values for the moon and sun at each instant. I then calculated the distances as the square root of the sum of the squares of the xyz coordinates. The cosine of the angle between them at any instant is

(sun_x * moon_x + sun_y * moon_y + sun_z * moon_z) / (sun_distance * moon_distance)

and the combined tidal force is then

sqrt( sun_force^2 + moon_force^2 + 2* sun_force * moon_force * cos(angle))

DATA AND CODE: The original sun and moon data from JPL are here (moon) and here (sun), 20 Mb text files. The relevant data from those two files, in the form of a 13 Mb R “save()” file, is here and the R code is here.

EQUATIONS: The tidal force is equal to 2 * G * m1 * m2 * r / d^3, where G is the gravitational constant, m1 and m2 are the masses of the two objects, d is the distance between them, and r is the radius of the object where we’re calculating the tides (assuming that r is much, much smaller than d).

A good derivation of the equation for tidal force is given here.

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February 10, 2014 4:27 am

So it was this “spring” tide and not CO2 that gave Hurricane Sandy that extra push to help flood NYC and NJ?:
http://news.nationalgeographic.com/news/2012/10/121029-hurricane-sandy-path-storm-surge-full-moon-nation-weather-science/
I thought it was CO2 and so did the President, the Governors, and Al Gore…(sarc)

February 10, 2014 5:13 am

Incidentally, I still think it should be abs(cos(theta)) because there are two spring tides every month. The graph shows only one spring tide per month.
Sun – moon – earth. Theta =. O. Cos(theta) = 1.0. New moon
Sun- earth – moon. Theta = PI. Cos(theta) = -1.0. Full moon
Neap tides occur when theta = pi/2 and 3pi/2
The tidal bulge on the opposite side to the moon is mainly caused by the centripetal force of the earth rotating around the barycenter like a pair of scatters in a spin. When the sun is on that side the solar tide then increases this effect further. So we get a second spring tide when theta = pi.
IMHO the second bulge is NOT caused by an increased gravity on the near side surface to that on the centre of the earth. It is a pure rotational effect.

February 10, 2014 5:25 am

Of course that should be “a pair of skaters in a spin”
How I hate auto-correction on iPads !

Greg Goodman
February 10, 2014 5:38 am

As Willis says, this is just the theoretical tidal force or tidal potential. It is not in any way a measure of tides or the movement of water that actually happens.
The force calculated here exerts a force on the oceans , where and how the water actually moves is a whole other storey, that has as much to do with the 3D shape of the ocean basins and coastlines as it has to do this the primary driving force.
The hypothetical “bulges” get amplifies as the enter shallow waters , reflect of irregular coastlines and flow back out to sea. The passage of the moon is constantly moving on both a monthly and annual scale. The resulting tides are so complex that they still can not be modelled in anything but the vaguest terms and we still rely on empirical charts specific to each geographical locality as Willis discovered in the Solomans.
In fact tidal patterns progress in all directions and there are some points on the global called amphidromes that do not have ANY tides at all. Others have four tides a day, others just one.
http://en.wikipedia.org/wiki/Amphidromic_point
What Willis has plotted is the _magnitude_ of the tidal force. What is not shown is its direction.
The sun moves from one tropic to the other and back again in a year. The moon follows this but in addition moves +/- 5 degrees either side in cycle that takes 27.2 days. This “draconic” month is again different from the 27.55 day period I mentioned above.
The draconic cycle is the 2 ascending and descending moon people often confuse as being the same thing as the visible waxing and waning cycle. 29.53 days average.
Now all this really matters if you want to talk about real influence on the tides because force is a vector, with magnitude and direction, not a scalar as Willis has plotted.
When the sun is at it’s most southerly point 23.5 S the moon can go 5.1 degrees further south. This means tides will be at their most displaced from the equator and there will be a minimum of the semi-diurnal and a maximum of the diurnal components.
Willis linked to this paper yesterday which suggest 18.6 years is the relevant long period not the “saros” cycle of 18.01 that Willis is focusing on here.
http://www.jstor.org/discover/10.2307/621006?uid=3738032&uid=2&uid=4&sid=21103363205771
18.6 is the period of the precession of the lunar nodes. This is what affects the “declination angle” or the height of the moon in the sky , ie time between the 23.5+5.1 degree extremes I mentioned above.
Because of the ‘push-pull’ nature of tides it is the magnitude of the declination angle that determines whether sun and moon tides are focused on equator or pulling out to N,S extremes simultaneously.
Thus the period of water being draw towards or out of the tropics will be 18.6 / 2 years.
That will not be included in Willis plots or his spectra since he has explicitly ignored the directional component of the resultant force (though he did correctly us it to sum the forces).
So Willis found a little peak around “8.7” years. I would suggest more detailed examination would reveal it is 8.85 , the precession of lunar apsides. He did not find 9.3 and that is to be expected as I said from that analysis.
So far so good. We have picture of what long periods may be produced. Next step is to see whether there is any evidence of them in climate data.

RichardLH
February 10, 2014 5:39 am

Greg Goodman says:
February 10, 2014 at 3:58 am
“frequency modulation is another can of worms entirely , is that what you meant?”
It was an observation that frequencies always both add and multiply.
We just use that fact in different ways at different times. It can be very difficult to spot that they are there at times as it often just ends up as spreading the peak rather than them being visible as separate frequencies – but the maths says they are both there all the time.

RichardLH
February 10, 2014 5:46 am

Greg Goodman says:
February 10, 2014 at 5:38 am
“As Willis says, this is just the theoretical tidal force or tidal potential. It is not in any way a measure of tides or the movement of water that actually happens. ”
Indeed. The plot is for the vertical component only and for a water only Earth.
As I tried to point out above, it is how all this, the tidal bugle in a Sea/Ocean basin coupled with the tangential vector, into a tidal flow that affects the water/air movement through the Cills and Straights North/South is likely to be the main factor of interest. And that is a very much slower and complex cycle than the daily tides.
That tangential vector is almost never discussed or mentioned.
http://i29.photobucket.com/albums/c274/richardlinsleyhood/Tidalvectors_zps4fd5800f.png

Greg Goodman
February 10, 2014 5:47 am

CliveBest: “The tidal bulge on the opposite side to the moon is mainly caused by the centripetal force of the earth rotating around the barycenter like a pair of scatters in a spin. When the sun is on that side the solar tide then increases this effect further. So we get a second spring tide when theta = pi.
IMHO the second bulge is NOT caused by an increased gravity on the near side surface to that on the centre of the earth. It is a pure rotational effect.”
I was of that impression too at one stage but no. You need to understand that tides are caused by gradient of gravitational field ( grad operator in 3D) and not by gravitational attraction. That is confirmed by the 1/r^3 dependence not 1/r^2. They are essentially equal except for minute higher order corrections. I went into that in some more detail above.
Centrifugal force would be much stronger in the case of the sun and it would be very obvious in relation to the lunar tides. Solar is about 1/4 of lunar because of inv. cube. An inv. sqr effect would stick out a mile.

RichardLH
February 10, 2014 5:52 am

Greg Goodman says:
February 10, 2014 at 5:47 am
“They are essentially equal except for minute higher order corrections.”
I believe that the true filed is slightly ‘egg shaped’, pointy end towards the Moon. The Solar one is more equal. AFIK.

Henry Bowman
February 10, 2014 5:54 am

Willis writes

Perhaps … but Longman’s paper appears to be an ad-hoc, empirical method. Nothing against that, and for 1959, it was state of the art … It contains a number of formulas which give close values empirically, but are merely good approximations and not solutions of the underlying equations.

I don’t agree with your assessment, but if you are interested in more modern methodology, I suggest you check out the software package TSoft, developed and maintained by the Royal Observatory of Belgium, continuing Melchior’s decades-long efforts to study tides.

Greg Goodman
February 10, 2014 5:58 am

http://climategrog.wordpress.com/?attachment_id=757
Here we have the anaomalistic lunar month in arctic ice extent. Modulation is detected as 4.31 years
http://climategrog.wordpress.com/?attachment_id=774
frequency spectrum of Indian ocean SST reveals a strong peak at 9.32 years.
Looks a lot like 18.6 / 2 years.
The Indian ocean shows temperature records quite different from the other main oceans that are connected north and south. Here the declination angle seems to produce a prominent cycle.

Editor
February 10, 2014 6:04 am

Normally, I’m hesitant to jump onto the “planetary influence” bandwagon, but…
I notice that in your “monthly” tidal forces graph, there’s a peak at just over 13 months. My first thought was the Chandler wobble http://en.wikipedia.org/wiki/Chandler_wobble but that’s 433 days, over 14 months, so it’s not the answer. Now let’s look at at some planetary orbital data http://nineplanets.org/data.html
Earth orbits Sun in 365.26 days
Jupiter orbits Sun in 4332.71 days
The “beat frequency”, i.e. time between conjunctions, is…
1 / (1/365.26 – 1/4332.71) = 398.89 days
Can you “zoom in” on your analysis and see if that period matches the peak just past 13 months?

cd
February 10, 2014 6:07 am

Greg Goodman
Now fourier analysis, by definition only captures single frequencies of fixed amplitude. So if you have data with a modulation it can not detect it as such.
Fig. 1 looks like a modulated signal of:
high frequency carrier (pure sinusoid) * long range variation (structural signal)
Surprisingly, the spectra does not seem to show the “modulation” fingerprint.
But great points all the same.

Coldlynx
February 10, 2014 6:10 am

Atmospheric tide:
“Atmospheric tides are also produced through the gravitational effects of the Moon”
http://en.wikipedia.org/wiki/Atmospheric_tide
Do not forget the smaller solar tide.
Add to that the earth axial tilt and get a tide induced movement of air towards the poles in summer and from the poles in the winter.
The largest daily horisontal tidal force are during sunrise which accelerate air eastwards. Small force but rather long time of acceleration for some hours every day. In NH summer northeast due to earth tilt. During sunset is the largest tidal force deaccelerating the same airmass northwest in NH summer. Net force north but since day is longer than night and the net movement of the atmosphere be northeast.
In winter will the morning horisonal acceleration be southeast and evening retardation be southwest. The net force south and net movement southwest due to longer nights.
This will have an impact on winds patterns and climate. Not big but the effect will be there.
The Coriolis effect is also a small force with huge impact.
http://en.wikipedia.org/wiki/Coriolis_effect

Greg Goodman
February 10, 2014 6:23 am

North Atlantic SST shows 9.066 as main peak:
http://climategrog.wordpress.com/?attachment_id=217
Cross-correlation of N.Atlantic and ex-tropical N. Pacific shows a strong peak at 9.06 years
http://climategrog.wordpress.com/?attachment_id=755
As detailed in the text with the plot this could be a combination of 18.6 / 2 and 8.85 years.
In contrast to the Indian ocean there seems to be both declination and aspides cycles at play here.
A similar frequency was recently reported by BEST team by looking at cross correlation of AMO and PDO. I preferred actual SST to the processed PDO “index” but essentially the same period is found.

Greg Goodman
February 10, 2014 6:26 am

cd “Surprisingly, the spectra does not seem to show the “modulation” fingerprint.”
You won’t see it _directly_ in a spectral analysis.
That’s whole point of my comment which you read but apparently did not understand.

Greg Goodman
February 10, 2014 6:29 am

walterdnes says: Chandler etc.
Not planets. Probably 14 lunations. 29.53*14 = 413. = 1.13 years.
I haven’t got Willis’ code load properly but may be he can provide a central value for that peak.

Greg Goodman
February 10, 2014 6:40 am

Richar: That tangential vector is almost never discussed or mentioned.
http://i29.photobucket.com/albums/c274/richardlinsleyhood/Tidalvectors_zps4fd5800f.png
I did not get your point last time you posted that.
The point is , even without looking at the tangential vector, for a high tide to happen water has to come from somewhere else. When you look at water movement it does well-up from the deeps it is mainly a horizontal movement of water.
To create a high tide, surface water comes in from all around.
If there are temperature differences in SST, movement like that carries thermal energy.
As declination angle pulls tides towards or away from the equatorial zones in a 9.3 year cycle this will displace large amounts of heat energy. It is easy to see why this period is recurrent basin wide SST records.

February 10, 2014 6:40 am

Greg,

You need to understand that tides are caused by gradient of gravitational field ( grad operator in 3D) and not by gravitational attraction. That is confirmed by the 1/r^3 dependence not 1/r^2. They are essentially equal except for minute higher order corrections. I went into that in some more detail above.

Yes I am well aware that tides are caused by the gradient of the gravitational field. Hence the 1/r^3 dependence. Hence the the reason the lunar tide is about twice the solar tide despite the sun being 27 million times the mass of the moon. You haven’t understood what I am saying
It is the tractional component of the gravitational force of the moon acting on the oceans which gives rise to a tidal bulge. The tractional gravitational component is the projection onto the spherical surface of the earth which increases with angular distance from the vector joining the moon to the center of the earth. This works out as 1/r^3 effective tidal force and explains the cause of the bulge on the surface facing the moon. However the bulge on the opposite side has a different origin. The centripetal force is caused by the rotation of the earth about the earth-moon barycenter (located 4000km from the centre of the earth) during the lunar month. The change in centripetal force across the spherical surface of the earth also leads to a 1/r^3 dependence. A point on the earth perpendicular to the moon-earth vector feels no centrifugal force. A point opposite the moon feels maximum centripetal force. Somewhere in the middle the ocean feels a component of centrifugal force that is parallel to the earth’s surface. This parallel component then leads to the tidal bulge opposite the moon.
The centripetal force of the earth’s orbit around the sun is much smaller as the pivot point is 93 million miles away and the angular velocity is much smaller.
I just don’t buy the argument that the difference in gravity between the lunar facing surface, the centre of the earth and the opposite facing surface causes the second bulge.

RichardLH
February 10, 2014 6:41 am

Greg Goodman says:
February 10, 2014 at 5:58 am
“The Indian ocean shows temperature records quite different from the other main oceans that are connected north and south. Here the declination angle seems to produce a prominent cycle.”
I suspect that this is because the tidal flow is limited by the West – East land block to the North. This severely limits the effects in the Indian Ocean. You also need to consider how the orbital inclinations of both Sun and Moon interact with the normal ‘vertical’ globe we tend to think of. The maximum, central point is always some form of elliptical line running across the surface. In the case of the Indian Ocean this falls on land a lot of the time.

February 10, 2014 6:42 am

Important paper on tidal influence on climate:
Ray, RD, 2007: Decadal Climate Variability: Is There a Tidal Connection?. J. Climate, 20, 3542–3560. doi: http://dx.doi.org/10.1175/JCLI4193.1

RichardLH
February 10, 2014 6:46 am

Greg Goodman says:
February 10, 2014 at 6:40 am
Richard: That tangential vector is almost never discussed or mentioned.
http://i29.photobucket.com/albums/c274/richardlinsleyhood/Tidalvectors_zps4fd5800f.png
“I did not get your point last time you posted that.”
Look again at the diagram. The tangential to the surface force varies very slowly. It is at an angle to the orbit and only changes with that. It does not follow the normal daily pattern. It is much slower and more likely to be the ~60 year interaction.
This is a force that is horizontal to the surface. One that is likely to affect flows of all sorts North-South.
Sure flows caused by different vertical effects in basin North – South will also be in there, but they are on a much faster Daily timescale.

RichardLH
February 10, 2014 6:59 am

clivebest says:
February 10, 2014 at 6:40 am
This is always how I have believed the two tides explanation was supported in Physics.
http://i.stack.imgur.com/aE7Gd.jpg
from
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics
http://physics.stackexchange.com/questions/46792/tidal-force-on-far-side

February 10, 2014 7:00 am

Greg,

What Willis has plotted is the _magnitude_ of the tidal force. What is not shown is its direction.

No – he has plotted the magnitude of the “sub-lunar” tide and has ignored the “antipodal” tide. In other words when there is a full moon with the moon is on the opposite side to the sun the tidal force is again at a maximum. What he has plotted is just one tide per day and ignored the second tide. This error is compounded because at full moon he has the solar tide subtracting from the sub-lunar tide , whereas the opposite is the case. This subtle effect is now due to the alignment of the “antipodal” tide with the sun’s tidal force.

Greg Goodman
February 10, 2014 7:04 am

CliveBest: “The eccentricity becomes a maximum when the semi-major axis of the orbit lines up with the sun. This happens every 205.9 days – more than half a year due to the precession of the orbit every 18.6 years. ”
There’s the answer to the “13mo” peak.
Twice that value is almost exactly the 14 lunations that I suggested. It’s the alignment of max eccentricity with the visual lunar phase. ie full moon and max eccentricity (closest approach “perigee”) being at max lunar+solar tidal force alignment.
I’m a little curious why this is showing as a separate peak since the individual components should already be present in the rest of the spectrum. This implies a non linearity.
Since the ephemeris is essentially empirically based, perhaps it is picking up some slight variation in the E-M orbit due to ocean movement.

RichardLH
February 10, 2014 7:08 am

clivebest says:
February 10, 2014 at 7:00 am
“No – he has plotted the magnitude of the “sub-lunar” tide and has ignored the “antipodal” tide.”
And only for a water only Earth. Also one that doesn’t matter where the spin axis is. The real Earth has both of the complications to add as well.

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