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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Greg Goodman
February 10, 2014 9:29 am

That’s interesting Clive. Could tie in with Scaffeta’s paper I mentioned above. He finds 9.1 years in JPL data that is attributable to the moon’s presence. I had not thought that it may be related directly to change in insolation but to distance. I knew the perihelion was notable but not the Earth’s orbit around the moon 😉
One main reason is probably that TSI is usually shown “corrected” for 1AU, ie this cycle is actively removed.
I think what you are plotting is a manifestation of the fine scale of what Scaffeta investigated. I also see the variations in TSI sometimes match your red line in amplitude other times its less or broken up. Symptoms of another cycle.
Obvious choice is the perigee cycle. 8.85 . I’ve already suggested several times the Scaffeta’s 9.1+/-0.1 is in fact 9.08 which is produced by superposition of 9.3 (declination / 2) and 8.85.
18.631 / 2 + 8.852591 => 9.078 modulated by 356 years.
Cross-correlation of N.Atl and N.Pac SST shows same thing.
http://climategrog.wordpress.com/?attachment_id=755
18.631 /2 = 9.3155
Indian Ocean:
http://climategrog.wordpress.com/?attachment_id=774
Good idea of Willis’ to start this thread, it’s bringing a lot of things together.

RichardLH
February 10, 2014 9:31 am

Gail Combs says:
February 10, 2014 at 8:50 am
“I do not see anyone mentioning this paper:”
I did reference Fig 1. from there above as to the very complicated state of the Moon/Earth gravitational interaction and asked if Willis was attempting to refute it.

RichardLH
February 10, 2014 9:34 am

clivebest says:
February 10, 2014 at 9:22 am
“Wikipedia has also got it wrong http://en.wikipedia.org/wiki/Tide
So too has : http://physics.stackexchange.com/questions/46792/tidal-force-on-far-side
Here is someone who explains it properly. ( http://www.moonconnection.com/tides.phtml)”
I believe that the ‘egg shaped’ resultant field is correct though.

Greg Goodman
February 10, 2014 9:44 am

“Here is someone who explains it properly.”
Properly because…? who are they apart form the provider of some naff app ? They don’t even say who they are beyond “moonconnection.com”.
I don’t even bother reading WP for shit like this any more because it’s like global bar-talk. Everyone’s an expert, and those that argue the longest prevail on WP.
Where did your egg plot come from earlier? Was that derived from “centrifugal” ideas or gravity gradient?
I may buy the idea that there is some inertial component but I want something solid with numbers. I thought that was the case a while ago but the relatively small solar tide argues against it.

dan
February 10, 2014 9:54 am

Way to not even look at the link and see what the hell was even being referred to, Mosh

RichardLH
February 10, 2014 10:09 am

This is probably the best reference for explaining the slight egg shape to the tidal bulges (one tide is larger than the other on most days)
http://co-ops.nos.noaa.gov/restles3.html
And this for explaining why it is not due to centrifugal forces 🙂
http://www.lhup.edu/~dsimanek/scenario/tides.htm

February 10, 2014 10:23 am

Greg,
Years ago I went through the calculations to “derive” the 1/R3 dependence. I also convinced myself that the differential centrifugal force on the opposite hemisphere to the moon was indeed the cause of the second bulge. I don’t have my notes here but will try and reproduce them.
One other known climate effect : Moonshine !
Don’t laugh – but reflected sunlight from the moon at night is the only direct energy source on earth. Globally this is expected to add vary about 0.004C of warming between new and full moon. What is even more interesting is the effect that the moon has in polar regions during the long dark winters. The warming effect is then proportionately much more, and lunar atmospheric tides bring in circulation in from higher latitudes. Any effect must depend strongly on the 18.6 year cycle as the tidal bulge moves to higher latitudes.

RichardLH
February 10, 2014 10:26 am

clivebest says:
February 10, 2014 at 10:23 am
“I also convinced myself that the differential centrifugal force on the opposite hemisphere to the moon was indeed the cause of the second bulge.”
You might want to re-think that explanation.
http://www.lhup.edu/~dsimanek/scenario/tides.htm
(I am sorry I picked the site I did for the egg shape – I did not check the text – only saw the image – bad boy!)

Reply to  RichardLH
February 10, 2014 11:10 am

Richard,
What I wrote was indeed wrong – sorry!
The centrifugal force is the same for all points on the earth including that at the centre of the earth. The gravitational force of the moon on the centre of the earth is exactly balanced by the centrifugal force. Oceans on the surface of the earth experience different gravitational forces according to their distance from the centre of the moon. These are now not in balance with the centrifugal force. The differential of this net force causes the tides. For the surface facing the moon gravity “wins” resulting in a bulge. On the other side the centrifugal force “wins” causing the opposite bulge.
This gives the correct description : http://co-ops.nos.noaa.gov/restles3.html
Clive

RichardLH
February 10, 2014 11:49 am

clivebest says:
February 10, 2014 at 11:10 am
“What I wrote was indeed wrong – sorry! …This gives the correct description: http://co-ops.nos.noaa.gov/restles3.html
Yes I know – our posts must have crossed.

Greg Goodman
February 10, 2014 12:18 pm

Thanks for the two links Richard.
This whole thing is very inadequate. the NOAA description is quite good but when they show a “centrifugal” force pointing towards the centre of rotation, I see we’re not out of the woods. That may work for the solid earth on the basis that its all connected. It does not work for the fluid part. BTW when you bring in one ‘fictitious force’ you usually need the other : Coriolis.
I prefer the lhup.edu presentation working in an inertial frame of reference.
Luckily, for the purposed of this thread we can tell the eggs to beat it.

February 10, 2014 1:56 pm

I was slow to come around to this super tide theory, don’t have much invested in it, but don’t see much reason to back off of it yet, because: 1) it depends on zonal tides, capable of moving water north and south; 2) the fortnightly zonal tide is the strongest tide, at least as far as LOD is concerned. See: http://hpiers.obspm.fr/eop-pc/index.php?index=realtime&lang=en
Choose LOD; don’t remove “tidal variations” except for comparison.
One reason zonal tide is so strong is that its bulge can easily keep up with a fortnightly gravitational pull, unlike diurnal or semidiurnal tides, whose bulges can only move a few hundred mph in the shallow ocean. The latter are limited to idiosyncratic oscillations governed by basin bathymetry. The fortnightly zonal tide is conspicuous by its absence on Willis’s chart.
–AGF

Greg Goodman
February 10, 2014 2:07 pm

” The fortnightly zonal tide is conspicuous by its absence on Willis’s chart.”
Indeed, and that’s because he dropped the direction part of the resultant vector he calculated and just plotted the magnitude of the tidal force. The zonal (north/south) flow is primarily determined by the declination angle, which as what produces the 18.6 year cycle.
Because of the presence of the ‘opposite’ tidal force as well, its the deviation from the equator that matters, resulting in 9.3 year variations in the zonal tides.
AGF, was it you I discussed this with last year on another thread about the Stuecker paper and you did some back of envelop calculations on heat transport?

Greg Goodman
February 10, 2014 2:26 pm

Thanks for the EOP link, useful tool.
However it looks like that is the anomalistic cycle that is coming up clearly. Plot for last 136 days and you get pretty clearly ten bumps, or 5 full cycles.
This appears to be perigee cycle not tides. Would you agree?

February 10, 2014 2:58 pm

Yes, GG, you witnessed my reluctant conversion, and yes the zonal tide you see is not very fortnightly. –AGF

Greg Goodman
February 10, 2014 3:18 pm

Unfortunately I can’t get Willis’ code to load properly. Maybe he’ll pop in later to correct it.
It would be interesting to plot the deviation of the resultant force vector’s angle from the equator.
That will be similar to the lunar declination angle but taking into account sun’s contribution too.
Cross-correlate that with the Indian ocean SST and it may start to get interesting.

February 10, 2014 3:25 pm

Willis Eschenbach says:
“However, after three Saros cycles, the three bodies line up again (of course), but this time the points under the earth are (again approximately) the same. So regarding your question, viz:
How long does it take for the Moon to return to the same point in the sky at the same time of month, year, etc.
… the answer is, three Saros cycles.”
The exeligmos cannot occur at the same time of year as the cycle is not a whole number of years, but the eclipse will be at a similar Earth location. One Saros is 6585.32 days so one eclipse occurs ~120° ahead of the previous on the Earth’s surface, so it takes three Saros cycles for an eclipse to reoccur at the same surface location.
I would have looked at Lunar precession rather than eclipse cycles, not that I can see it forcing climate cycles.
I’m curious about the 13.44 month spike in your analysis (lunar phase perigee cycle), surely there is a king tide at half of that frequency as alternate full and new Moon coincide with lunar perigee?

Greg Goodman
February 10, 2014 3:32 pm

“…. surely there is a king tide at half of that frequency as alternate full and new Moon coincide with lunar perigee?”
cliveBest has pointed out that Willis should have done abs(cosines) in his code but Willis hasn’t been by since to comment on that.

charles the moderator
February 10, 2014 3:49 pm

David L. Hagen,

I understand Scafetta to say that he documents his use of publicly available data, and fully describes his method in his peer reviewed papers sufficient for others to replicate his results.
While I would encourage him to show his code as well, I thought data and a full published method to be sufficient for the scientific method.
Is the data or his method not sufficiently clear?

That is almost verbatim the same excuse used by CRU to avoid FOI requests before Climategate. And no, no one I know of is able to replicate his work. Until you apologists drop your double standards and apply ethics equally to all sides, your hypocrisy will continue to rule the day.
It is a scientist’s job to make it as easy as possible for critics to scrutinize their work. This is how science grows and self-corrects. Science is not a bunch of trade secrets.
As long as Scafetta refuses to produce data and code, his work is nothing more than clever, albeit kinda boring, anecdotes.

February 10, 2014 5:49 pm

Greg Goodman says:
February 10, 2014 at 2:26 pm
This appears to be perigee cycle not tides. Would you agree?
===============================================================
Back to the keyboard. Not sure what you’re getting at. Perigee is neither zonal nor fortnightly. Declension is both. That is, perigee could affect diurnal earth and sea tides over a 4 week period, while the zonal bulge both grows and moves poleward according to lunar declension. So now it’s a matter of lining up the LOD graph with declension data. Haven’t done that. –AGF

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