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.
Figure 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.
Figure 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:
Figure 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.
clivebest says:
“That’s also about the average latitude where largest tidal currents are generated.”
I’m in no way disagreeing but I don’t immediately see where you get that.
In terms of temperature it’s where the sun spends most of it’s time. If we consider the d/dt(lattitude) is a sine with zeroes on the solstices and maxima on the equinoxes, it spends far longer within 20% of the extrema than it does within a similar margin of the equator.
On that evidence alone we cannot distinguish max overhead solar from any tidal drawing of water out of the equator towards tropics.
The spreading of the peaks to higher latitudes in lines below 200m looks it could be accounted for by diffusion.
Of course the moon tags along for the ride and also adds its own +/-5.1 degrees. This is the major tidal player but to a fair degree is correlated with the above.
Lunar declination seems to have max extent around 2006.76 AD so 2004 used for the M2,K1 maps is near max lunar declination. Smallest declination amplitude around 1997.25, prev max 1987.75 (the smaller roughly semi-annual cycle is latching the non-integer part).
It may be interesting to compare the 1997-2007 dates to the free-fall in Arctic ice coverage:
http://climategrog.wordpress.com/2013/09/16/on-identifying-inter-decadal-variation-in-nh-sea-ice/
here’s a quick plot of lunar range cubed * declination angle. Not rigorous , no solar. Just a quick guide to lone term cycles in lunar tide raising forces.
http://tinypic.com/view.php?pic=29w8xmx&s=8
Looks like they chose 2004 (perhaps indirectly because it gave nice standing waves) because it was a max in K1 component. For example 2002.7 and 2006.7 would be the most pure M2 with little K1.
I really would like to have similar plots for those years.
Hmm, I think I need to check the workings on that graph and label it up properly.
There is a 17.7 ( 8.85 * 2 ) year cycle on the pairs of peaks (2004-1880)/7
Also if we look at the alternating high and low peaks on the top, there is a phase change in 1969 (two equal peaks either side reverses the low,hi.low pattern).
Also larger magnitude swings in late 19th c. which is also found in the SST and land records.
so much angst and noise. I hesitate to even say this…
The basic problem with this analysis is that it looks only at the scalar quantity of tidal force, while tides are a result of a vector force (size AND direction). That vector part matters. It has an 18.6 year period due to the precession of the lunar orbit. more here:
http://chiefio.wordpress.com/2014/02/16/tides-vectors-scalars-arctic-flushing-and-resonance/
Just looking at size of the scalar is not enough. The position over the geography of the Earth matters, as does the change of the vector direction over time. Even how this interacts with which season gets more tidal flushing of the North Pole matters. It is just not a simple scalar problem. So a scalar analysis is insufficient.
E.M.Smith says:
February 16, 2014 at 8:23 pm
Thanks, Chiefio, always good to hear from you.
Actually, that limited subject matter is the basic advantage of this analysis. It starts at the start, with no attempt to delve in the the full complexities. People have taken things apart and studied the individual parts for centuries … why is it suddenly unacceptable when I do it?
In addition, when you try to discuss every aspect of a complex issue in one post, the conversation tends to scatter and shatter.
I note that over at your blog, you have more than one post about the tides … given the point of view you urge on me above, why didn’t you look at the entire field in one single post as you say I should have done?
w.
E.M.Smith says:
February 16, 2014 at 8:23 pm
“The basic problem with this analysis is that it looks only at the scalar quantity of tidal force, while tides are a result of a vector force (size AND direction). That vector part matters.”
It is strange that when you arrive and point out what I have been pointing out, you get civil treatment as opposed to the flack I have been getting.
Willis Eschenbach says:
February 16, 2014 at 11:42 pm
“People have taken things apart and studied the individual parts for centuries … why is it suddenly unacceptable when I do it?”
Because a 3-4 year old child with a 10 inch circular saw may well be able to create the 3 inch wood building blocks as the toy he wishes to use, but there is a large likelihood that he will chop off his own or other peoples limbs in the process of doing so?
EMSmith: “The basic problem with this analysis is that it looks only at the scalar quantity of tidal force,”
And even manages to get wrong.
CliveBest pointed out the error early on and I provided a fix for the code.
Rather than correct it. Willis prefers to start another thread the try bluster through with his mistakes. OH well.
Cross posted from the other thread because it is relevant here as well.
—-
Willis:
“On my planet, people commonly study simplified models of complex situations, precisely to draw conclusions from them Happens all the time, I’m in mystery why you think we can’t learn valuable lessons from simple models.”
RLH.
OK. Then at least do it correctly. So as to allow understanding instead of confusion. I obviously need to put my teaching hat on and explain how I think that this would best have been done so that there would have been a lot less squabbling all round.
Let us start from a very simple concept. A nearly featureless Globe, oriented vertically in the orbit, with a constant depth ocean, and a big ‘pin’ stuck down into the surface at 0 Lat, 0 Long.
Derive the gravitation forces that apply from the field diagram as below
http://upload.wikimedia.org/wikipedia/commons/d/d8/Field_tidal.png
Fig 1.
Let us then take some pictures of that from 3 directions starting from the longest possible intervals and working down towards human time frames.
The pictures are
1. looking down on the Earth from above at right angles to the orbit around the Sun,
2. looking along the orbit with the Poles North/South
3. looking out along the line drawn fro the Sun though the Earth.
This gives us x, y, z
Let us just deal with the Sun first, no Moon at all.
And we start taking pictures. At 4 year intervals to start with and at the Periapsis in the below.
http://en.wikipedia.org/wiki/File:Seasons1.svg
Fig 2.
Now we can see the ‘pin’ in the centre for 3, and pointing at the Sun for 2 and 1.
We can now deal with that oblate spheroid that the Earth’s rotation gives. We can also point out that most of the flattening is in the rock because the water skin is so thin. So all the centripetal forces, spin and orbit can be dealt with before we get into other stuff.
Now we speed up to once a year. And the first complication shows up. The ‘pin’ does not stay steady. It moves around those field diagrams in Fig 1.
So we draw out the first small cycle. One at 4 years. Then we move to 4 times a Year. Describe what happens there.
Now we have done the Solar components. Time to add in the Moon.
I am sure you can get the rest. Or do I need to do the full slide and description set?
Now is that a better way to do it or not? You tell me.
RichardLH says:
February 17, 2014 at 4:53 am
You can’t find a single thing I’ve said in the head post that is wrong, not one solitary error … so instead, you call me a child?
That’s your idea of dialogue?
You are a nasty, underhanded, passive-aggressive piece of work, Richard—not a good man in any sense, and no fun to play with at all.
I’m done with it, I’ll leave you to play with yourself, you’re probably good at that at least …
w.
RichardLH says:
February 17, 2014 at 6:50 am
No thanks. The day I take a man who calls me a child and a dog as my teacher hasn’t come yet. Take your ugliness elsewhere. Not interested, sorry.
w.
RichardLH says:
February 17, 2014 at 6:50 am
I’ll tell you, but I’m sure you won’t pay any more attention than you have in the past. Here it is, anyhow, are you ready … I haven’t a clue if that’s a better to explain the tides or not.
But there’s a way to find out. As I’ve told you before, if you’re so damn brilliant, then how about you write up your crystal-clear and oh-so-insightful explanation of the tides in your “better way”, and publish it on this or some other blog. Then we can see if people can follow your vague ramblings or not. I can’t, but maybe that’s just me, and perhaps you are right and you know the better way.
But until you actually get up off the couch and produce something, until you actually do the hard yards and deliver your finished explanation, as we used to say on the ranch … podner, you’re all hat and no cattle …
w.
Wliis: Thank you for your careful and reasoned analysis of the point of view I offered.
As usual your megaphone amplified mutterings mean that your hearing aid is in need of some adjustments you have failed to apply.
You cannot, apparently, see anything from any other place that your specific, terrifically narrow, point of view.
Your curiosity level is pretty close to zero.
Others have pointed it out too.
Have you ever wondered why?
Willis:
“But until you actually get up off the couch and produce something, until you actually do the hard yards and deliver your finished explanation, as we used to say on the ranch … podner, you’re all hat and no cattle …”
Well it would appear that the chances are that the above approach which deals with the various problems in a logical and careful way so as to enlighten rather than confuse is too far beyond your comprehension.
I came to help and got insulted, continuously, by someone who’s power of language, curiosity and understanding is more limited by his own personality than his potential capability.
“We can now deal with that oblate spheroid that the Earth’s rotation gives. We can also point out that most of the flattening is in the rock because the water skin is so thin. So all the centripetal forces, spin and orbit can be dealt with before we get into other stuff.”
I think that no-where have you ever offered this rather important point in your ramblings to date.
That oblate spheroid is mostly rock. The thin skin of water hardly notices the centripetal as a force at all. The very slight difference that an even layer of water would be distorted to can be calculated. If you wish. Or are you going to suggest that the poles would be dry?
Willis Eschenbach says:
February 17, 2014 at 10:40 am
“I’m done with it, I’ll leave you to play with yourself, you’re probably good at that at least …”
Again with the invective.
Greg says:
February 17, 2014 at 5:24 am
Your “fix for the code” involved using abs( 2* sun_force * moon_force * cos(angle)). This gives horribly unphysical results. As I pointed out above, this leads to a ludicrous situation where, when equal unit forces are opposed to each other, rather than summing to zero they sum to 2 … but the vector part of the sum disappears.
You have never responded to my clear scientific objections to your loony “fix”, nor has anyone else come along to support that goofy claim. In fact, I doubt very greatly whether Chiefio agrees with that kind of nonsense, he’s a smart guy … Chiefio, do you agree with Greg’s “fix”?
Hogwash. I raised my clear scientific objections to the “fix” when you proposed it. it was nonsense then and it’s nonsense now. At the time I said:
Looks like my guess was right … again, I encourage you to read any text on vector addition if you doubt my formula. The sum of two vectors a and b with an included angle theta is
sqrt( a2 + b2 + 2 a b cos(theta)
and there are no absolute values involved.
Or, you could bring in a citation that shows your formula is correct. But my guess is that you’ll do neither …
w.
RichardLH says:
February 17, 2014 at 11:08 am
You call a man a dog, you call him a child … then you bitch, whimper, and whine when he slaps your face for it.
You’re a piece of work, all right.
w.
Willis Eschenbach says:
February 17, 2014 at 11:39 am
Again with the invective.
“You call a man a dog, you call him a child … then you bitch, whimper, and whine when he slaps your face for it. You’re a piece of work, all right.”
Oh, I stood for your petty childlessness for a very LONG time before I started to responded in kind.
I do so dislike all this bitchy, slapy, talk. But I can dealt it out with the best if called for.
I got to hone my skills in writing up some ones skillset in as few as words as possible a long time ago.
As I said before –
I came to help and got insulted, continuously, by someone who’s power of language, curiosity and understanding is more limited by his own personality than his potential capability.
“Looks like my guess was right … again, I encourage you to read any text on vector addition if you doubt my formula. The sum of two vectors a and b with an included angle theta is
sqrt( a2 + b2 + 2 a b cos(theta)
and there are no absolute values involved.”
Looks like you do not understand that summing two vectors from a 2D space creates a 1D vector in the same space in the process because you managed to turn that into a positive only scalar, as demonstrated above, even if you didn’t get it then – or possibly – now.
RichardLH says:
February 17, 2014 at 11:44 am
Whine on, Richard, whine on …
w.
RichardLH says:
February 17, 2014 at 11:49 am
Say what? That has nothing to do with the subject under discussion, which (as evidenced by your quote of my words) is whether we should use the absolute value or not … it was a nice try at misdirection, though, the judges awarded it a score of 9.7 for the number of flips and flops …
w.
“Say what? That has nothing to do with the subject under discussion, which (as evidenced by your quote of my words) is whether we should use the absolute value or not ”
Hmmm. Well as the vector sum in question is using gravity to calculate magnitude and the concept of the resultant 1D vector going negative has rather large implications…….
Willis Eschenbach says:
February 17, 2014 at 11:57 am
“Whine on, Richard, whine on … ”
Dog’s whine so…..
Some interesting facts about tides (the short version)
The Earth is an oblate spheriod.
The Earth’s equatorial radius is the distance from its center to the Equator and equals 6,378.14 kilometers
The Earth’s polar radius is the distance from its center to the North and South Poles and equals 6,356.75 kilometers
A 21.39 kilometers difference.
Angular velocity of Earth’s axial rotation in radians per second = 7.29 x 10^-5 rad/s
Angular velocity of Earth around the Sun in radians per second = 1.99 × 10^-7 rad/s
Centripetal (outwards) acceleration at Earth’s surface due to Earth’s axial rotation = 0.034 m/s^2
Centripetal (outwards) acceleration at Earth’s surface due to Earth’s orbit around the Sun =~ 0.0000952 m/s^2
Gravity at Earth’s surface = 9.8 m/s^2
The tiny differential between 9.8 m/s^2 (Polar) and 9.834 m/s^2 (Equatorial) creates the 21.38 kilometres difference in radii above.
The Oceans average depth is 3.79 kilometres so 3.79 / (6,378.14 – 3.79) =~ 1 / 1682 so the difference is mostly in the rock.
The Solar tidal force is 46% as large as the Lunar. More precisely, the Lunar tidal acceleration (along the Moon–Earth axis, at the Earth’s surface) is about 1.1 * 10^-7 * 9.8 m/s^2,
while the Solar tidal acceleration (along the Sun–Earth axis, at the Earth’s surface) is about 0.52 * 10^-7 * 9.8 m/s^2.
The theoretical amplitude of Oceanic tides caused by the moon is about 54 centimetres at the highest point, which corresponds to the amplitude that would be reached if the Ocean possessed a uniform depth, there were no landmasses, and the Earth were rotating in step with the Moon’s orbit. The Sun similarly causes tides, of which the theoretical amplitude is about 25 centimetres (46% of that of the Moon) with a cycle time of 12 hours. At Spring tide the two effects add to each other to a theoretical level of 79 centimetres, while at Neap tide the theoretical level is reduced to 29 centimetres. Since the orbits of the Earth about the Sun, and the Moon about the Earth, are elliptical, tidal amplitudes change somewhat as a result of the varying Earth–Sun and Earth–Moon distances. This causes a variation in the tidal force and theoretical amplitude of about ±18% for the moon and ±5% for the sun. If both the Sun and Moon were at their closest positions and aligned at New Moon, the theoretical amplitude would reach 93 centimetres.
Real amplitudes differ considerably, not only because of depth variations and continental obstacles, but also because wave propagation across the ocean has a natural period of the same order of magnitude as the rotation period: if there were no land masses, it would take about 30 hours for a long wavelength surface wave to propagate along the equator halfway around the Earth (by comparison, the Earth’s lithosphere has a natural period of about 57 minutes). Earth tides, which raise and lower the bottom of the ocean by less than 1 metre, and the tide’s own gravitational self attraction are both significant and further complicate the ocean’s response to tidal forces.
In most locations, the four largest amplitude tidal components turn out to be:
M2 Principal lunar 12.42 hr
K1 Luni-solar diurnal 23.93 hr
S2 Principal solar 12.00 hr
O1 Principal lunar diurnal 25.82 hr
S2 is largest at mid-latitudes and vanishes at the Equator and the Poles.
M2 is largest at the Equator and vanishes at the Poles.
The Long-period tide (not listed above) is largest at the pole and (with reversed sign) at the equator.
A list of other components can be found in Knauss (1978) table 10.1. Particularly important is the fortnightly (2 week) tide, often written Mf.
See Figure 10.15 in Knauss (1978) for plots of partial tides.
Tides in different locations are classified based on the predominant frequency of the tide using a function called the form ratio which measures the relative strength of the diurnal and semi-diurnal tides.
F = (K1 + O1) / (M2 + S2)
F > 3 Diurnal 1 High, 1 Low per day
0.25 < F < 3 Mixed 2 Highs, 2 Lows per day, but of different strength
F < 0.25 Semidiurnal 2 Highs, 2 Lows per day, similar strength.