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.
RichardLH
For the purposes of the piece would Willis’ simple model not suffice – just for fun. I know what you’re saying, and one could go further, for example the Earth’s density is not homogeneous creating slight variations in Earths own gravitational field across its surface. But at what point do you stop. The reductionist approach is to start as simple as possible – he’s done that not as complicated as possible otherwise you’d never start.
I don’t think Willis has shown what he set out to (personal opinion) but one has to start somewhere. Again, I think he need only add in the caveats you have raised.
I think the biggest problem facing Willis’ approach is that the signals are AM signals – as Greg pointed out earlier. The fundamental frequencies will not be immediately obvious from a simple Fourier approach.
A possible solution, although laborious:
for each window size
{
for location step window size
{
autocorrelation;
FFT;
output location and spectra;
}
}
Should be able to work out the fundamental frequencies from this approach.
Willis:
If you don’t get it yet – let me out it into language that an (ex?) rocket scientist might understand better.
On one hand you have a model/parameters for a generic elliptical orbit. On the other you have the need for knowing the possibilities of radio transmissions with a fixed bean, vertically angled radio from Cape Canaveral with a craft in that same orbit.
You rather do need a few more calculations that you have to hand so far.
CB:” Jupiter has a gravitational effect on the moon and on the earth but not a tidal effect on either.”
Indeed , way too far for an inv cube effect. However, it is one of factors causing all the complex variations in the lunar , so if there’s a lunar effect , it’s a player in that. It also a major player in the way the sun revolves amount the SS barycentre . We orbit the sun so get the same treatment.
It is quite feasible for there to be a Jupiter signal in climate.
“Only including the centrifugal force can you properly explain the second tide.”
I believed that to be the case a couple of years ago. But I’m pretty sure it’s incorrect now.
I made quite a detailed comment on this above. Perhaps you could comment on that if you see a flaw in the logic.
I would invite you to think this trough in the proper inertial frame without fictitious forces and see whether you come to the same conclusion. If you want to stick to a rotational frame of reference you need to apply Coriolis forces too. Not sure how that shapes up but it’s not simple nor whether its worth the head-scratching.
Unless I missed something of significance above, it actually seems simpler to treat it w.r.t. an inertial frame.
Let me know if you differ.
“I would argue that this isn’t the same effect as the lunar and solar tides on earth where the bodies are in orbit. ”
In a mathematical sense you could say falling into the sun is an orbit with infinite period.
What about a highly eccentric orbit like Haley’s comet?
Does it have to be a closed orbit ?
As soon as two bodies are close enough to interact the same laws should apply.
It may seem a bit contrived to call falling into a star an “orbit” but I’m not sure it changes the grad of the field. If you agree with bidirectional stretch in one case it should also apply to a stable orbit.
Now we cannot verify ocean tides which are incalculable. But the theory should be verifiable on solid bodies like the moon and probably the effect on solid earth.
The back-tide due to sol would be very different in the two cases due to the large separation, so presumably this aspect is readily verifiable.
Clive Best says:
February 12, 2014 at 7:05 am
“Jupiter has a gravitational effect on the moon and on the earth but not a tidal effect on either.”
If Jupiter alters the Moons orbit then that is a multiplier of the Jupiter effect here on Earth. If, say, the Moon’s orbit was to change 1 or 5 degrees N-S over time. Then that would affect how the vector field that the Moon exerts or Erath by a similar change also.
I don’t think anyone is expecting direct action. It is the indirect that matters. Sort of like servo action rather than direct.
The question is if this delta is so small that it is in the dust or not.
Willis Eschenbach says:
February 12, 2014 at 1:44 am
One other interesting thought came across my mind in contemplating all of this … suppose the sun was less bright, but this was balanced by the earth being only half the distance from the sun. In that case, we’d be just as warm, but the tides would be eight times as high as they are now … yikes!
=====================================================================
Back when the moon was near the Roche limit tides were worse than that. But that is a rather unstable situation: George Darwin calculated that the tidal braking effect is proportional to the sixth power of the distance, so the moon quickly quit spinning, and quickly moved away from the earth, while the earth quickly slowed down. The hypothesis being that formerly a day and month were of equal period. (Of course the tides would not begin till these periods became unequal.) Tides were stronger too back when lungfish evolved, and may have provided critical stimulus for the evolution of lungs. Before fish made their way to fresh water they were first stranded in hot, murky, oxygen depleted tidal pools, then rivers, then lakes. Then some fresh water lineage took its lungs back to the sea, and within a few million years most ocean fish had lungs, until reptiles went to sea and replaced the surface breathers (and displaced the coelecanths). –AGF
Greg
This is all getting pedantic.
In a mathematical sense you could say falling into the sun is an orbit with infinite period.
This sounds very technical but could be just waffle. In all practical sense, if you’re being pulled toward something then by definition you’re not in orbit – escape velocity etc. And if it is falling then one thing is for sure it isn’t infinite.
Does it have to be a closed orbit ?
It has a closed orbit.
As soon as two bodies are close enough to interact the same laws should apply.
This is nonsense. The ultimate effect of gravity on two bodies whose gravitational fields interact can be quite different: moving toward one another or one is relatively fixed and other not.
It may seem a bit contrived to call falling into a star an “orbit” but I’m not sure it changes the grad of the field
It sounds wrong. All this is getting quite pedantic but the point being made originally was correct but as I say pedantic. I think we need a physicist that deals with this to clear it up as it all sounds like semantics and arm waving.
For me the term tidal suggests a periodicity – as in tidal harmonic rather than a change in gravitational pull across something (which is basically everything). This periodicity, as mentioned by others, can be the result of elliptical orbits or the spinning of a body in orbit or even a falling object (as long as it spins while falling).
cd says:
February 12, 2014 at 6:05 am
“For the purposes of the piece would Willis’ simple model not suffice – just for fun. I know what you’re saying, and one could go further, for example the Earth’s density is not homogeneous creating slight variations in Earths own gravitational field across its surface. But at what point do you stop.”
For a non rotating or water only Earth – sure it is fine.
That is just such a LONG way from reality that it is just a child’s toy.
Just try to do a plot from the North Pole and see just how non-trivial this all really is.
Clive Best says:
February 12, 2014 at 12:43 am
Sorry, Clive, but that’s not true at all. For example, I personally have a tidal effect on Pluto. Strange but true. My body has a tidal effect on every planet. In fact, every body in or out of the solar system exerts a tidal effect on every other body. That’s not the question.
The question is … how much tidal effect? In the case of Jupiter acting on the moon, it is really, really small …
On earth, a kilogram of mass exerts a force of just under ten newtons.
I’ve taken up the habit of thinking about tidal forces (per unit mass) in micronewtons, a millionth of a newton. It’s a convenient unit, because the tidal force of the sun on the earth is about half a micronewton, and that of the moon on the earth is about 1.1 micronewtons.
On the other hand, the tidal force of Jupiter on the earth is 6.5e-6 microNewtons … and the tidal force of Jupiter on the sun is 3.7e-4 micronewtons.
And since tides on the earth are on the order of 600 mm or so, that means that tides on the sun from Jupiter are on the order of a millimetre …
Now, upthread tallbloke claims that although the Jupiter tide on the sun is very small, the horizonal component of the tide is much larger … but like most good cyclomaniacs, he offers neither calculations nor references to back it up …
w.
Willis,
You and I have a gravitational influence on Pluto but not a “tidal” influence on Pluto. Just because there is a formula for tides on earth: 2 * G * m1 * m2 * r / d^3 doesn’t mean it that it can be applied universally. Tides are not a fundamental law of nature.
When Greg says that a star or person in free fall into a black hole temporarily experiences tides he is correct. However, permanent tides can only apply to objects in orbit around each other. Orbits conserve angular momentum by balancing gravity against a “centrifugal” force.
A hammer thrower is also in balance until he lets go of the hammer !
Wilis:
Still no plot from the North Pole? Such a simple question really.
No plot from the Shetlands? Should be a matter of seconds surely.
Still no understanding that, for this to have any bearing on Climate reality, it is tidal FLOW that is the real question – not tidal height. And the flows N-S through the various Straits not in the deep ocean. Driven by the basin tide heights either side on a titled, rotating Earth as well as the tangential to the surface gravitational vectors at the same Latitude.
Greg Goodman says:
February 12, 2014 at 2:06 am
“I think the problem here is that since real tides are so far removed from all this talk of ‘bulges’ that none of it can be verified. People (especially academics teaching the stuff) are free to spout any hotch-potch “theory” of tides because it’s largely non verifiable.”
=======================================================================
This is more true of semidiurnal tides than of zonal tides, else how do you explain the c1ms fortnightly, reversible deviation in LOD?
tallbloke says:
February 11, 2014 at 2:26 am
I’ve shown above that Greg’s claim, which tallbloke merrily endorses without doing the math, is simply not true.
Nor is the horizontal component of the tidal force “much more extensive” than the vertical component. In fact, the tidal force is at its strongest at the points nearest and furthest from the body causing the tides. At the points the same distance from the body causing the tides as the sun’s center of mass, there are no tidal forces at all.
Not only that, but the horizontal component of the tidal force is never that large. Remember that at the points where the horizontal force would be largest, at the sides of the sun where a tidal pull would be horizontal to the surface, is where the tidal force is zero. And at the points where the tidal force is largest, nearest and furthest from the body raising the tides, the horizontal component of the tidal force is zero. In other words, the horizontal component of the tidal force varies as
sin(theta) * cos(theta)
where theta is the angle between a line from the point to the sun’s center of mass, and the line from the sun to the tide-raising body (e.g. Jupiter).
As a result, the horizontal component of the tidal force is largest at a 45° angle to the line between Jupiter and the sun, and at that point the strength of the horizontal component of the tidal force is sin(45°)*cos(45°) = 0.5 …
So rather than the horizontal tidal component being “much more extensive” than the vertical tidal component as tallbloke claims, the horizontal component is nowhere more than half as strong as the maximum vertical tide.
Run the numbers first, folks … it prevents you from being embarrassed by your claims.
w.
Richard, while poking about with numbers I’ve just found this. It look like you were correct about some resonant linkage between moon and Jupiter.
http://eclipse.gsfc.nasa.gov/LEcat5/figure.html
saros cycle=18.0308
2/( 1/8.85259 + 1/18.0308 ) = 11.8749
pJ = 11.8624
I looks like the moon has locked into an orbital resonance (not uncommon in solar system).
I’d suspected something there’d be some link to J in all this but never had any reason to spend time looking. I just stumbled across it while doing some other calculations.
The saros represents Earth-moon-sun alignments that Willis started off on this thread. 8.85 is the precession of the plane of the lunar orbit. It is usually stated that this precession is due to torque exerted by the sun on the E-M “gyroscope” angular momentum. But this must at least mean that Jupiter has enough influence to lock it in.
So that’s the proof of what I suggested to Clive earlier. That J may have an indirect effect via the moon, even though it’s tidal effect is negligible.
The other part of that combo , the modulation freq, has a period that I’ve often seen pop up in spectra but I had not idea of the cause.
2/( 1/8.85259 + 1/18.0308 ) = 34.782
Looks like that’s a lunar signal too.
This is rather cool since just a couple of days ago I was wondering why the precession didn’t just keep speeding up if it constantly had a torque applied by the sun. I meant to look into it but then forgot.
🙂
“I’ve shown above that Greg’s claim, which tallbloke merrily endorses without doing the math, is simply not true.”
“Run the numbers first, folks … it prevents you from being embarrassed by your claims.”
w.
What the hell do you think your water planet calculations have to do with reality of how earth tides work. Since no one in the world can model tides your suggestion “run the numbers” is pretty dumb. You’d do well to take you own advice on that one.
Since your last comment to me on the other thread was “Piss off you idiot” I think you’d better wind you neck in a bit.
You’ve indicated in personal communication that you don’t want to talk to me, which is fine, so don’t come in here and start making jibes.
Greg Goodman says:
February 12, 2014 at 9:54 am
“Richard, while poking about with numbers I’ve just found this. It look like you were correct about some resonant linkage between moon and Jupiter.
http://eclipse.gsfc.nasa.gov/LEcat5/figure.html
saros cycle=18.0308
2/( 1/8.85259 + 1/18.0308 ) = 11.8749
pJ = 11.8624
I looks like the moon has locked into an orbital resonance (not uncommon in solar system).”
Yea. Someone who is better than me with Google 🙂
All I ever claimed was that other planets alter the Lunar orbit. Not that any have any direct effect on tides here on Earth.
If the orbit changes – the tides change. Like a servo rather than a direct pull. Direct tidal influence. That’s a straw man if I ever saw one.
RichardLH says:
February 12, 2014 at 9:38 am
Richard, you seem to think that a) I read your posts with any regularity, and b) I have the slightest interest in acceding to your bizarre requests if I happen to see them (which in this case I didn’t). Neither one is true in the slightest. In my world, you’re a star, right up there with the rest of the folks that are too lazy to do your own math and want me to bail you out of your terminal innumeracy … not gonna happen.
If you want a plot of tidal effects from the north pole or the Shetlands, you know what to do, Richard … pull your thumb out of your fundamental orifice, stop whining about how I’m not doing what you want me to do, make your own fricken’ plot, and explain what you think it means.
Because I’m assuredly not fool enough to do your work for you …
w.
PS—I wrote this post to say folks, before you make your claims, do the math … and in response, you ask me to do your math for you.
Miss the point much, Richard?
“I’ve shown above that Greg’s claim, which tallbloke merrily endorses without doing the math, is simply not true.”
“Run the numbers first, folks … it prevents you from being embarrassed by your claims.”
w.
What the hell do you think your water planet calculations have to do with reality of how earth tides work. Since no one in the world can model tides your suggestion “run the numbers” is pretty dumb. You’d do well to take you own advice on that one.
Since your last comment to me on the other thread was “*iss off you idiot” I think you’d better wind you neck in a bit.
You’ve indicated in personal communication that you don’t want to talk to me, which is fine, so don’t come in here and start making jibes.
Willis Eschenbach says:
February 12, 2014 at 9:53 am
“Not only that, but the horizontal component of the tidal force is never that large”
Remind me again where on the [Earth’s] surface the greatest tidal FLOWS occur? What Latitudes?
“Yea. Someone who is better than me with Google :-)”
I didn’t Google it, I did it all by myself 😉
Willis Eschenbach says:
February 12, 2014 at 10:06 am
“If you want a plot of tidal effects from the north pole or the Shetlands, you know what to do, Richard … pull your thumb out of your fundamental orifice, stop whining about how I’m not doing what you want me to do, make your own fricken’ plot, and explain what you think it means.
Because I’m assuredly not fool enough to do your work for you …
w.
PS—I wrote this post to say folks, before you make your claims, do the math … and in response, you ask me to do your math for you.
Miss the point much, Richard?”
No I don’t miss your name calling either.
You started this thread partially as I can best tell in response to my posting an observation that the DATA shows some form of periodicity.
http://i29.photobucket.com/albums/c274/richardlinsleyhood/HadCrutMonthly11560Lowpass_zpsf542092e.png
Then you put together a toy example that had no real merit and claimed it proved your point.
I am just observing that until you bring your toy example down to a real, on Earth, example it is just that – a toy.
Climate is not related to some abstract 1D vector that actually follows part of the Saros cycle it is trying to (not) display. It is rotating, 3d, messy and a lot more complicated.
Greg Goodman says:
February 12, 2014 at 10:09 am
“I didn’t Google it, I did it all by myself ;)”
Bow, Bow.
Willis:
If you still don’t get it shall I repeat it in capitals?
TIDAL FLOW – NOT – TIDAL HEIGHT.
Willis Eschenbach says:
February 12, 2014 at 10:06 am
“In my world, you’re a star, right up there with the rest of the folks that are too lazy to do your own math and want me to bail you out of your terminal innumeracy … ”
Ironic isn’t it. On one hand you complain you don’t like the maths I do (because the averages are just TOO long don’t u know) then you think that toy orbital maths makes it all better.
Duh!
Greg Goodman says:
February 12, 2014 at 9:54 am
First off, the 8.85 is NOT the precession of the plane of the lunar orbit. It is the precession of the “line of apsides”, the imaginary line connecting the apogee and the perigee of the moon … so Greg doesn’t even know what his figures mean. However, they give the “right” answer, so he likes them.
In any case, let me get this straight. The frequency of the Saros cycle (1/18.0308), plus the frequency of precession of the lunar apsides (1/8.85259), is 0.16842 … so what? What is the physical meaning of this? Under what physical framework does adding the frequencies of the two cycles make any sense?
Let me give you an example. The frequency of the earth’s rotation around the sun is one rotation per year. The frequency of the moon’s rotation around the earth is 12.37 cycles per year.
But what is the physical meaning of the sum of those two frequencies, 13.27 cycles per year? That’s just numerology, it has no physical meaning. It’s like looking at three temperatures, say 30°C, 35°C, and 39°C, and saying “the total temperature is 104°C, wow, that’s above boiling” … the mere fact that you can add frequencies or temperatures doesn’t mean that the sum of two frequencies or two temperatures means something. At that point, you’re just playing with numbers.
To return to our story, Greg then takes this “pseudo-frequency”, which is the (meaningless) sum of two other frequencies, and converts it back into a pseudo-period. This gives him 5.937 years … then he multiplies the pseudo-period by two. Why multiply it by two?
Because cyclomania roolz, I guess, I see no reason to multiply by two … but then the whole procedure is totally illogical, so I suppose multiplying by two is just fine. Of course, if the number Greg was trying to match up with was three times his other number, he’d just multiply it by three instead of two. That’s the beauty of cyclomaniacal numerology, you don’t have to be bothered by ugly physics.
At the end of that, he gets 11.8749 years … which he compares to the orbital period of Jupiter, which is 11.8624 years, and declares (without further evidence, logic, or physical theory) that they are “phase locked” …
Phase locked? Say what? Not even close. In less than 500 years, the two cycles will be 180° out of phase … the moon is phase-locked to the earth, and we don’t see the back side of the moon every 500 years. The idea that those two numbers mean that the precession of the lunar apsides is phase-locked to Jupiter doesn’t even pass the laugh test.
Despite that, however, I’m sure RichardLH and tallbloke will be hailing Greg’s brilliant discovery as the next step in understanding the climate …
Folks, there are literally thousands of different astronomical “constants” up there. Finding two random numbers among them such that if you torture them enough you end up with two results that are kind of close means nothing except that the person doing the comparing has lost the plot …
w.
Greg Goodman says:
February 12, 2014 at 10:04 am
Take a deep breath there, son, you’re hyperventilating.
What I calculated in the head post is what the size of the tidal force is, and how it varies.
What I showed above is how it affects a water planet.
What does this have to do with earth tides? Well, I’m investigating your stupid claim that there is a big rush of water to the location of the tides, because the water has to come from somewhere … a claim you made without bothering to do the calculations. So I did them for you.
I’ve shown that no, it doesn’t take a big horizontal rush to move that amount of water, in fact it is very, very slow.
If you wish to dispute that, you’ll have to do more than wave your hands, make accusations, and spray spittle on your screen …
w.