Talk:Tide

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[edit] Tides and latitudes

Why is the tidal amplitude greater in the higher latitudes than at the equator?

Is it? Probably because the Moon's orbit is inclined with respect to Earth's equator. You'll probably find the equinoctial tides (when Sun and possibly Moon are in Earth's equatorial plane) are highest at the equator.
Urhixidur 15:37, 2005 Mar 6 (UTC)

Clearly the forces causing tides are biggest at points on the line through the earth's and moon's centres. So the hight can also be expected to be more. Since the moon's orbit is practically in the ecliptic, the forces are highest between the tropics. −Woodstone 19:02, 2005 Mar 6 (UTC)

It may look like the tidal amplitude is greater in regions in the higher latitudes of the earth but there's no causal relation between the two. The inclination of the moon's orbit to the ecliptic is slightly over 5°. Depending on the position of the moon's nodes, the inclination of the moon to the earth's equator (and the maximum tidal force) is at some point between 28.5°N and 28.5°S when the ascending node of the moon's orbit coincides with the vernal equinox (like it will do at about the summer solstice of june 2006; you'll notice at that time that at the days around new moon, the moon wil be very high above the horizon round midday). The inclination of the moon to the earth's equator (and the maximum tidal force) will be at some point between 18.5°N and 18.5°S when the ascending node of the moon's orbit coincides with the autumnal equinox (like it will do in about 9.3 years after june 2006). This is what is often called the 18.6 year cycle of the moon's orbit. You'll have noticed that the maximum inclination of the moon to the earth's equator varies from 18.5° (N or S) to 28.5° (N or S) over time. The maximum tidal force due to the moon's gravitation will be between these boundaries. At present the maximum tidal force due to the moon will be between 28.5° N and 28.5° S. In about 9.3 years the maximum tidal force will be between the boundaries of 18.5° N and 18.5° S. But the actual tidal amplitude depends on many more factors than just the tidal force. Most important is the depth of the ocean basin. If a tidal wave travels from deep water into a shalow basin, the wave is slowed down but the amplitude increases. Also if the wave travels into a funnel shaped basin (like the English Channel or the Bay of Fundy) the amplitude increases much. It's only a coincidence that sea basins that fulfil such conditions are found in the higher latitudes. To think that the tidal amplitude is only a function of tidal force would be an enormous simplification of this quite complicated fenomenon. Wikiklaas 23:32, 13 March 2006 (UTC)

[edit] Cause of indirect or opposite high tide

This edit by Woodstone on 24 Dec 04 was titled "simplified explanation of two high tides". In fact by removing all mention of centrifugal force, this revision is fundamentally misleading.

A useful explanation must make clear that two directly opposing forces act on the Earth: the Moon's gravitational pull, and centrifugal force. Please see an excellent explanation at the Center for Operational Oceanographic Products and Services, especially point 3.

I'm editing appropriately and would welcome further comment. Cheers --Air 11:47, 23 Feb 2005 (UTC)

Quote from the current article: "On the lateral sides water and Earth have no net movement, as the gravitational and centrifugal forces are in balance." This is incorrect, where do you think the water to form the high tides comes from? On the lateral side there is no direct tidal force, but there is surely a movement. The waterlevel sinks.
You're right, this is of course low tide. I'll correct. --Air 12:30, 25 Feb 2005 (UTC)
Apart from this I think the centrifugal explanation may be correct, but it is unnecessarily complex. And how does is apply to the earth-sun system, where the centre of mass practically coincides with the sun's centre. The referenced article contains a several errors; I will comment more when I find the time. --Woodstone 13:26, 2005 Feb 24 (UTC)
I feel it is critical to explain exactly what force causes the 'opposite' tide - previously this was left undefined. I titled the section "Lunar Tides" to indicate only the Earth/Moon system was being discussed. Another section explaining the secondary effect of Earth/Sun would be useful. --Air 12:30, 25 Feb 2005 (UTC)

In your current explanation you say the tides are caused by the difference between the gravitational force and the centrifugal force at one place. That is not correct. The cause is a difference between forces at the centre of mass (and the attached solid part of earth) and at the surface (where the water can move about). This applies both to the gravitational component and the centrifugal component. Both factors have equal signs. −Woodstone 19:55, 2005 Mar 2 (UTC)

The current state of the article is quite wrong. I guess we've all been suckered by some publications from official looking sites that have it wrong. I found excellent sources that explain the mistakes. Essential fact is that the centrifugal component plays no role, since it is uniform across earth (different locations on earth rotate with the same radius, but displaced centre). See for example Bad Physics and especially Myths about Gravity and Tides and for a more informal discussion Our Restless Tides. I will update the article later, unless someone else takes it up. −Woodstone 13:24, 2005 Mar 28 (UTC)

In my renewed explanation, based on gravity alone, I have added a paragraph making clear how an explanation using the centrifugal force can yield the same result.−Woodstone 12:28, 2005 Apr 3 (UTC)

[edit] Use of terms "apogee" and "perigee"

Those terms are only appropriate with reference to Earth as the orbiting body. The correct terms in this case are "aposelene" and "periselene," respectively. Alternatively, you can use the generic terms: "apoapsis" and "periapsis." More info: apsis

Vessbot 06:22, 27 Jan 2005 (UTC)

This change is a mistake. Apogee refers to the place farthest from earth of an object (e.g. the moon) in an orbit around earth. Not the "orbiting" body is named, but the "centre" of orbit. Aposelene applies to an object orbiting around the moon. Aphelion applies to an object (e.g. earth) orbiting around the sun.
All of this is of course relative because bodies orbit around their common centre of mass, not one around the other. In the given cases however one of the masses is clearly dominant. --Woodstone 13:28, 2005 Jan 27 (UTC)
Consider my foot to be in my mouth, all the way down to the knee. I just looked it up again. Sorry for *beep*ing up the article. Next time I'll try to be more careful. Vessbot 21:42, 27 Jan 2005 (UTC)

User:Hydrocomputer 14:50, 08 Dec 2005 (UTC)

I believe there are some mistakes in the physics explanation. I disagree with Woodstone's characterization of the centripetal force as irrelevant, but I'm not prepared to argue that one. For now, the amplitude of the tide is largely dominated by resonance, not currents. The reason for the bay of Fundee being such a violent tidal regime is that the length of the path travelled by the water at a speed of sqrt(g h) for shallow-water waves is such that it is close to 12.42 hours, the lunar M2 period, and thus is nearly in resonance.

The solar tide is approximate 10 percent, not 50 percent of the lunar, but again, given a resonant period close to 12 hours, it might be more like 50 percent for a particular region, which is hardly the same thing.

[edit] A Question

I'm trying to make sense of this article but its fairly difficult and while a number of points are still somewhat fuzzy, the following statement has me baffled "This causes relatively low tidal ranges in some locations (knots)..."

Should this be clearer? Is it saying that a "knot" is a term denoting a location where relatively low tidal ranges exist? Or is this the familiar nautical rate knot, thrown in here for no apparent reason?

I agree that much of this article is very casually technical and could definitely be made clearer (I have tried to make the Lunar Tides section readable). Knot has no directly related tidal meaning (according to answers.com). I can find this ref which states:
"The velocity (rate) of a tidal stream is expressed in knots (i.e., nautical miles per hour)"
which equates to what you mentioned (nautical knots). The article still makes no sense in its use of (knots) though. --Air 17:22, 2 Mar 2005 (UTC)

TIA Mwanner 15:40, Mar 2, 2005 (UTC)

My guess is that "knot" is a mistaken spelling for "node" (the Dutch translation for both words is "knoop"). For standing waves in a string that is the place where the amplitude is zero. In a more complex system like the earth's oceans a similar, but less exact effect could occur for a forced frequency wave. −Woodstone 19:55, 2005 Mar 2 (UTC)

I'm quite sure that what is meant here are the amphidromic points: these are points in mid ocean where the cotidal lines come together and where the actual amplitude is zero. If it is high tide at one side of the amphidromic point, then it's low tide at an angle of 180°, which is in the opposite direction. This has nothing to do with the nodes in standing waves. It has to do with the coriolis effect and the fact that the tidal wave does not travel the ocean basins as a long front but in fact rotates round these amphidromic points. Wikiklaas 00:12, 13 March 2006 (UTC)

[edit] Swapping the pictures

Earlier I swapped the pictures because, from how my browser (Firefox) displays them, they were backwards. It turns out that IE and Firefox display the images in opposite orders. I'm reverting the change since I suspect that more people use IE than Firefox. Sorry about the trouble. One-dimensional Tangent 18:00, 7 Apr 2005 (UTC)

[edit] Standing wave

I certainly welcome the recent major revision of this page that brought the explanation of physics of tides into agreement with the original argument by Sir Isaac Newton.

Minor remark: The author wrote "It is known that it travels as a standing wave...".

From a physicist's point of view this is an oxymoron - a standing wave does not travel - it just "stands", i.e. it vibrates with fixed location of nodes and antinodes, like a wave on a string. The author probably wanted to say that the shape of the wave does not change as it travels across the ocean, but the proper term for such a fixed traveling shape is a wave pulse.

Care to revise the wording?

Regards

Student

Wave pulse is not completely correct either. It is a periodic phenomenon. It can be seen as a forced frequency vibration of a complex object. −Woodstone 12:01, 2005 Apr 13 (UTC)

[edit] Animation

The animation is nice, but a little broken (in my IE6 browser). The moving bulge is not centered correctly and the centre moves as well. Also it is misleading. The travel of the bulges is not caused by the Moon moving, but by the Earth rotating. Care to improve? −Woodstone 20:08, 2005 Apr 27 (UTC)

The animation is broken for me when I view the main page but seems OK when I open the full version.
I understand that it show how the declination of the moon affects the position of strongest force - imagine that the Earth graphic is spinning like crazy and we're just watching the Earth Moon line to see how the tide is affected in more than one dimension. Joffan 07:43, 4 May 2005 (UTC)

[edit] Tidal Ranges

I am having trouble understanding some of the content of this article. What I am asking is why do tidal ranges increase and decrease in the time span of a month?

« The height of the high and low tides (relative to mean sea level) also varies. Around new and full Moon, the tidal forces due to the Sun reinforce those of the Moon, due to the syzygy found at those times - both the Sun and the Moon are 'pulling the water in the same direction.'[...] When the Moon is at first quarter or third quarter, the forces due to the Sun partially cancel out those of the Moon. At these points in the Lunar cycle, the tide's range is at its minimum.»
What is unclear with this?
The simplest way to look at it is thus. There are two distinct tides, the solar and the lunar. In the course of a month, they drift with respect to each other by a full 720º in phase, changing from full addition (at new and full Moon) to full subtraction (at first and last quarters). Note that I said 720º (twice 360º) because each of the two tides has double natural frequency; in other words, there are two tidal bulges on opposite sides of Earth, not a bulge on one side and a dip on the other.
Urhixidur 15:10, 2005 Jun 5 (UTC)

[edit] Double High Tide

The article currently states that the double high tide in Southampton Water is caused by the flow of water around the Isle of Wight. I have seen many sources dispute this theory. For instance read this article (search for Southampton).

The article is wrong! The double high tide at Southampton is caused by a four-cycles a day wave generated by non-linear effects where the main semi-diurnal tide reaches the shallow waters near Dover (and probably the southern North Sea and the Thames Estuary). The four-cycle a day wave propagates westwards along the English Channel as a Kelvin wave with its maximum amplitude along the south coast of the UK. Its amplitude and phase can be determined from the analysis of tidal records. The phase relation with the incoming semi-diurnal tide is roughly fixed, so it regularly produces double high waters at Southampton and double low waters in the Weymouth/Portland region. David Webb 22:39, 18 August 2006 (UTC)

[edit] Tidal Physics

In showing that the Sun has 46% of the Moon's effect, the author makes use of an inverse cube law. There is no such law. The force from the Sun cannot be one thing for all other effects and another thing for tides. Besides, an inverse cube law would give the Sun .00000012% of the Moon's force. Do the math. 81.164.255.103 01:16, 2 October 2005 (UTC)

The above comment is mistaken. The tidal effect is related to the difference in gravity experienced across the radius of the Earth. The force itself follows an inverse square law (Newton's law). The difference follows an inverse cube law. Furthermore, not only the distance but also the mass counts. The Sun's mass is much larger than the Moon's, compensating most of the larger distance in the tidal effect. Now please redo your maths. −Woodstone 10:19, 2 October 2005 (UTC)

I finally see what is trying to be shown with the math. I had to go to the discussion page of Tidal Force, and from there to the French site. The proposed math is comparing rates of change of the fields, not the strength of the fields. In this way the number 46% is derived. But there is still a big problem. This math treats the tidal force of the Sun on the Earth as simply an expression of the static gravitational field. It is admitted that the Earth would be feeling this force, or change in force, even if it were in freefall--if it were not orbiting. But the Earth IS orbiting, therefore you have to add the force of circular motion to the force of a static gravitational field. When books and websites explain tidal forces of the Earth on the Moon, they seem to understand this. They offer us tidal forces that are a summation of the static gravitational field and the fact that the Moon is in circular motion. Both situations are tidally positive. But with the Earth and Sun, the math stops without including the circular motion, which would create a variable pseudo-field of centrifugal forces. The math stops because it needs to keep that number 46%, which keeps the effects of the Sun less than the Moon. But the Moon can hardly be feeling the effects of circular motion while the Earth is not. Tidal theory is inconsistent. It seems to be squeezing the math to fit the data. 81.164.129.137 05:22, 3 October 2005 (UTC)

The Earth and Moon are in freefall around each other. An orbit is nothing else than freefall. Some explanations of tides use the apparent centrifugal force, but this is an unnecessarily complex way of looking at it from a rotational frame of reference. The centrifugal force is uniform across all Earth and does not contribute to the tidal generating force. The alternative explanations subtract this homogeneous force from the varying gravitational force. Since at the centre of Earth both forces are equal in size but opposed in direction, this produces just the difference in the gravitational field. Tides would also exist (for a very short time) if the Moon would be falling straight to Earth. −Woodstone 11:26, 3 October 2005 (UTC)

The Earth is not in freefall relative to the Sun. If it were it would be falling directly toward the Sun. Are you seriously claiming that the Earth is not in circular motion around the Sun? My point in the previous post is that the Earth's circular motion around the Sun must add to the 46% figure. 81.164.229.69 18:53, 3 October 2005 (UTC)

All heavenly bodies are in freefall. A circular orbit is just a free fall with the right initial velocity. The Earth is constantly accelerated towards the Sun because of its gravity. If not, Earth would follow a straight line and disappear into space. The formula for calculating the Moon's and the Sun's effect on the tides is of course identical. The formula for the tidal acceleration is: 2·G·M·R/d³, where G is the gravitational constant (about 6.67E-11 m³/(kg·s²)), M is the mass of Sun or Moon, d the distance and R the Earth's radius. The rate of revolution does not play any role, but of course depends on the gravity formulas as well.

Freefall with a tangential velocity is not freefall, it is called circular motion. You are either in orbit or in freefall, you cannot srictly be both. An orbit requires a centipetal force, which you have admitted is gravity. All circular motion causes centrifugal forces, and the rate of revolution must play a role. If the orbit does not create centrifugal forces, then you must explain how the orbiter avoids Newton's Third Law and a=v^2/r. 81.164.229.69 02:15, 4 October 2005 (UTC)

If the physics is described in an inertial frame of reference there is only one force: gravity. It completely describes the movement of heavenly bodies (except for electromagnetic and relativistic effects). Centrifugal force is a virtual force, used when describing movements in a rotating frame of reference. This apparent force is an expression of inertia as it exists in the non-rotating frame. Please detach your mind from these unnecessarily complex explanations and look at it from a purely newtonian view. −Woodstone 09:45, 4 October 2005 (UTC)
Woodstone is right and 81.164.229.69 is wrong. Freefall does not mean that no forces act on you, it means you don't resist the forces at all. Enough said.
Urhixidur 13:32, 4 October 2005 (UTC) (M.Sc. Astronomy, 1984)

No, not enough said. I never claimed that freefall meant that no forces were acting on you. Freefall means that ONLY the centripetal force is acting on you, so that you are NOT in circular motion. The math on the Tidal forces page, which is what we are discussing here, admits that there is a centrifugal force in the tidal equations, or its mathematical equivalent, that is why the term with w^2 is added to the total equation. This adds another 50% to the total tidal force, as the page admits. The angular frequency is a measure of the speed in orbit. It takes into consideration the time. So you cannot claim that the orbit is not important, or that the force differentials from orbiting all cancelled. If they all cancelled, you wouldn't need that extra term containing the angular frequency. You (or Woodstone, at any rate) don't seem to understand what your own variables imply. Let's be clear, once I found the derivation of the math at the French link (this derivation needs to be included on the English site, under tides, why is it not?), I agreed with it, at least regarding the tidal force from the Sun. The equation yields the correct differentials and the correct accelerations. But regarding the tidal effect on the Earth from the Moon, it doesn't logically apply. The main term works, but the term containing the angular frequency does not. The Earth is orbiting the barycenter, not the Moon, so that the radius R in that term (contained in the w variable) must be 4671 not 384000. If you correctly calculate the differentials and accelerations, you find that the barycenter tides totally swamp the lunar and solar tides. The barycenter tide is 48 times as large as the lunar tide and 72 times as large as the solar tide.

Regarding the "inverse cube law", I still think it is misleading to call it a law, since it does not apply to the field itself. It applies to the differential field and is therefore not directly comparable to the inverse square law. It is just the derivative of the inverse square law. 81.164.229.69 15:14, 4 October 2005 (UTC)

I was referring to your statement that « You are either in orbit or in freefall, you cannot srictly be both », which is clearly incorrect. On a different note, what is the address of the "French site" you mention? It isn't fr:Marée nor fr:Force de marée...
Urhixidur 18:21, 4 October 2005 (UTC)

Before continuing this discussion you ought to read the link 'Misconceptions about tides' given in the article. The idea that the "barycenter tide" swamps the "lunar" tide is utterly wrong (but I confess we had that statement in this article for a while too.) −Woodstone 18:43, 4 October 2005 (UTC)

The French article is linked from the Tidal Forces discussion page and is Sphere de Hill. I think you, Urhixidur, placed the link yourself. I know that the barycenter tides can't swamp the lunar tides, since that is not what we see. More to the point, Solar tides would be too small to mix in the right way with Barycenter tides, using these equations. You would not find spring tides varying correctly. I am just following the math you have given and it doesn't work when you compare the Lunar differential field to the Solar differential field. The total Solar differential field, using the equation you have published, including the angular velocity term, yields the number da = 7.61 x 10^-7 m/s^2. I confirmed this number using my equation, in which I do the differentials by hand instead of finding the derivative. This equation is da = ω^2r + GM[1/R^2 – 1/(R-r)^2]. The first term yields half the second term, confirming your math in that way.

When we look at the lunar tidal field on the Earth, we can apply only the second term, since the Earth is not orbiting the Moon. The angular velocity must apply to the velocity of the Earth about the barycenter. The line of tides will still match up, so that we can add the effects if we like. But the numbers are a big problem. GM[1/R^2 – 1/(R-r)^2] applied to the lunar field gives us da = 1.14 x 10^-6 m/s^2 while ω^2r applied to the barycenter gives us two different differentials: 5.5 x 10^-5 m/s^2 on the inside and 4.5 x 10^-5 m/s^2 on the outside.

As for the "alternative explanations" paragraph, that looks like gobbledygook to me. I have no idea what the author means by saying that the Earth is not rotating, it is just "displacing." It is either in circular motion or it isn't. If it isn't, then the Earth certainly isn't in circular motion around the Moon and we can throw out the angular velocity term altogether, regarding the Earth-Moon system. If it is, then we have to use the barycenter radius to calculate ω, in which case we have a huge differential. In either case, the given math seems doomed. Without the ω term, the Sun's total tidal field becomes 67% of the Moon's. This doesn't fit the data. With the ω term, the Sun's total tidal field is 100 times too small. 81.164.229.69 19:52, 4 October 2005 (UTC)

Ah, yes, the Hill Sphere bit. I know, I wrote the derivation myself. The subject matter is somewhat different, mind you. In looking at tides, we're not trying to orbit a third body in co-rotation with the Earth-Sun or Earth-Moon systems. Still, the forces present in a co-rotating frame are useful to look at. In comparing the Earth-Sun and Earth-Moon systems, we see the Sun's pull on the Earth, in absolute terms, is about 179 times that of the Moon. However, the differential acceleration (nearside to farside) is much smaller, because the barycentre of the Earth-Moon system is inside the Earth's radius, flipping some signs around. The computation shows the lunar differential acceleration is about 61 times stronger than the solar. Now, consider the centrifugal effect of Earth's own rotation. It is about 5.7 times larger than the solar centrifugal term, and about 2800 times larger than the lunar one. The Earth's gravity is even stronger than that (obviously): 1650 vs. the Sun, some 286,000 vs. the Moon.
We're misleading ourselves if we think tides are only a second order effect. More to the point, they're not a static effect. Even if you had the Earth with no obliquity or eccentricity, as well as the Moon with no inclination or eccentricity, you'd still have to deal with the Moon and Solar tidal effect coming in as "pulses" with two slightly different periods; as the article mentions it, the oceans' natural frequency is such that we're no longer dealing with simple harmonic motion.
In 1911, mathematician Augustus Edward Hough Love (1863-1940) of Oxford University formulated partial differential equations for calculating how the gravity of one body affects the shape of a second. One set provides the amount of deformation (tidal amplitude) for a compressible, uniform sphere. Love's solution to this equation depends on only two factors. One is the ratio of a body's density times its radius times its surface gravity to its rigidity. The other is the ratio of a body's rigidity to its compressibility (as expressed by the so-called Lamé constant). Love looked at two cases, using values for the constants that would be reasonable for characterizing Earth. He got plausible results for tidal amplitudes. The well known solution ([1] Love, A. E. H., A Treatise on the Mathematical Theory of Elasticity, New York: Dover Publications, 1944) for tidal amplitude of an incompressible, homogeneous sphere of density rho, radius r and elastic shear modulus mu (Basalt: ~2,4x10^10 Pa; Standard Rigidity of the Earth’s crust = 3x10^10 Pa) is 2:
\delta r = \frac{h_2}{g} \frac{GM}{R^3} r^2 \left ( \frac{3}{2} cos^2 \theta - \frac{1}{2} \right ) \!\,
where g\!\, is the surface gravity, M\!\, is the mass of the tide raiser whose center is a distance R\!\, from the center of the homogeneous sphere of radius r\!\, and:
h_2 = \frac{5}{ 2 \left ( 1 + \frac{19 \mu}{2 \rho g r} \right ) } \!\,
is a tidal "Love number". Theta is the longitude. If you compute the Love-predicted tidal amplitudes, you get 0.22 m from the Sun and 0.49 m from the Moon. These figures are for the land tides, mind you. For water, whose shear modulus is zero, we get 0.4 and 0.9 m. Those values are completely wrong, not only because we're keeping rho, g and r the same, but mostly because water doesn't respond as a rigid body: it flows. So Love's assumptions don't apply. I'm pretty sure Love also assumes a static solution, which is valid only if the solid Earth's relaxation time is short enough compared to the twice-daily tidal pulse —and it turns out it is. I do not know whether or not Love factored in centrifugal effects; you'll have to look it up, I guess...
Urhixidur 18:58, 7 October 2005 (UTC)
The confusion is that the calculations above by 81.164.229.69 assume that Earth moves as if there is an invisible bar connecting Earth and Moon as a halter. The bar would rotate about the barycenter. Points in (or on) Earth close to barycentre rotate with a smaller radius than points far away from the barycenter. This is not how Earth moves. The rotation of Earth itself has a completely different omega. To carry the picture futher, it is more like Earth is pinned to this bar at its centre, but stays in the same orientation all the time. In this case every point in (and on) Earth moves with the same radius (the distance between centre and barycentre). That is what is meant in the article by "displacing, not rotating". So if you insist on using the centrifugal explanation, it is a homogeneous force across all Earth (which is equal with opposite sign to the gravity at the centre). −Woodstone 20:56, 7 October 2005 (UTC)
It is instructive to apply Love's formula to some other object pairs:
  • Mercury-Sun: the predicted tidal amplitude is 0.40 m, a little less than the Moon's land tides on Earth.
  • Io-Jupiter: 300 m (if using the Earth rigidity); using half Earth rigidity gives 580 m - zero rigidity limit is 7800 m
  • Phobos-Mars: 1 or 2 mm - negligible, zero rigidity limit is almost 3 km, however
  • Triton-Neptune: 6.8 m - no wonder it is synchronous
  • Nereid-Neptune: zero - zero rigidity limit is a mere 4 cm
  • Charon-Pluton: 0.41 m, zero rigidity limit 340 m (very cold ice may indeed be very rigid, however)
Urhixidur 21:16, 7 October 2005 (UTC)

I am getting two different answers here, neither of which seems very satisfactory. I am curious, first of all, if Urhixidur agrees with Woodstone in his analysis of the barycenter rotation of the Earth. To me it is clear that it doesn't matter what the Earth's "orientation" is. If the Earth is rotating in any way about the barycenter, then uncancellable forces must result. Besides, we already know exactly how the Earth is rotating about its own axis, and therefore the barycenter. There need be no speculation. The Earth's daily rotation does not cancel tide-causing differentials from the Sun or Moon, how could we think its daily rotation would exactly cancel a barycenter tide? In fact, we can calculate the real position of points on the Earth at each dt, relative to the barycenter. The forces at these points do not cancel, they TRAVEL, just like with other tides.

As for Love's equations, even Urhixidur here admits they have many problems. They appear to be the throwing of more math at tides, to little effect. The biggest problem, which is never addressed, is that Love's equations again ignore the barycenter. He is trying to fine-tune a theory that doesn't work in the first place. If Love's equations had been the least bit convincing, Feynman wouldn't have been playing with barycenter tides much later. I am not a big fan of Feynman, but he was savvy enough not to publicly test-drive equations in the face of superior math.

I am not promoting the barycenter tide. You guys can have whatever theory you want to have here. That is why I am not attempting to edit the main page. I just wish physics could be honest enough to admit that tidal theory is currently very incomplete. It is not just that we don't understand the complexity of it, as Urhixidur implies. It may well be that we don't understand the fundamental forces at work. All we know is that our numbers don't work. It is difficult to extrapolate from that fact-- to know whether it is fundamental postulates that are wrong or whether it is that we are not including all the minor variables. If we were within a small fraction of the right answer, with a consistent theory unlying it, then I would agree that a combination of minor variables might be the culprit. But nothing I have ever read, including Love, leads me in that direction. 81.164.229.69 01:25, 11 October 2005 (UTC)


LC writes: Let me describe what I gathered you are thinking, to point out what I think is wrong with the "centrifugal"-explanation: You think about Earth and Moon as mass-points. These mass-points circle around each other around a center which is somewhere between Earth and Moon. Now you imagine the Earth as a sphere of rock around the Earth-mass-point with some water at the surface. The Earth-gravitation pulls the water towards the center of Earth. Then you animate this model by

  • 1.) rotating Earth around its own mass-point
  • 2.) rotating Earth and Moon around the common center of mass

While you imagine the 2nd rotation (of the Earth-Moon-System) you imagine Earth-mass-point ATTACHED to Moon-mass-point by gravitation. Then you see the Earth as the rock sphere and the water on top. Might think: Hey the Earth is rotating around the common center of Mass and thats some thousands of kilometers away from earth center so the water on the surface must run away from the center and accumulate towards the direction thats faces away from the common center of rotation.

Here comes what is wrong: There is no Earth mass-point attached to Moon mass-point. Mass points are just a way to make physical equations easier. In fact each atom of Earth attracts each atom of Moon, the water included. The Earth is a sphere rotating in a gravitation field.

Furthermore the Earth-Moon System is in balance. That means the centrifugal force on Earth is exactly as big as the gravitational force of the moon. Thats why they circle around each other instead of flying away in a curve.

So the gravitational force of the moon is as big as the centrifugal force on earth EXCEPT the difference of the gravitation field on the side of Earth that is facing the moon and the one opposite to it. Furthermore the centrifugal force grows stronger the more you move away from the common center of mass.

This common center is INSIDE Earth. That means we have a centrifugal force facing away from earth-center on both sides of the earth. The one facing away from moon is bigger but there is a still second one, facing towards the moon. Keep in mind that Earth's total centrifugal force equals the gravitational force of the moon.

In Short:

  • 1.) We have a difference in the gravitational field on both sides of Earth.
  • 2.) We have centrifugal forces facing away from earth center with two maxima on the Earth-Moon line.
  • 3.) The system is balanced.

LC the cutter 01:34, 15 November 2005 (UTC)

This is a very complex topic and there must be a lot of forces involved. But about the other bulge on the opposite side of the Earth, I find the model using barycenter not convincing. Try to set up an experiment with rotation off the center like that and you find the water sphashes everywhere. It won't be gentle and uniform like the tides. Furthermore any model citing the system of Moon and Earth revolving, dancing or orbiting around each other would directly lead to the perception that the Moon must be doing a complete orbit around the Earth each day to get 2 tides per day. At this speed, you would have tsunamis every day of the year. This is clearly wrong. In fact the Moon is moving around the Earth so slowly each day, that I would prefer considering it stationary and find other more influential factors. In my opinion, the mathematics only complicates the picture. My own explanation is very simple. The Earth's own rotation about its own axis naturally creates uniform bulging of the water body toward the equator. This is due to the centrifugal force of the Earth's own rotation about its axis. The gravity forces of the Moon and the Sun only alter this uniform pattern. These forces deplete, accentuate and soften the existing uniform bulging of water body to create tides. The Moon's gravity is the most influential of all forces. On the Earth side closest to the Moon, the water is pulled strongly by the Moon's gravity to create extra bulging on top of the existing natural bulge. On the other side of the Earth, the Moon's gravity is much weaker, so it could not cancel out the centrifugal force of the Earth's rotation, the natural water bulge retains its shape. On the other two sides of the rectangle that envelops the Earth, the natural bulging is depleted as water is drawn out. So the water body around the Earth is a bit like the egg shape with the smaller end points to the Moon. As the Moon - Earth relative positions change over the year we have some variations. Also the gravity of the Sun at different times of the year add some influence along to create a complex pattern for tides. Other influential factors are land barriers of the continents. In short, my opinion is that the Moon's gravity is the main force creating tides by interfering with the natural bulging of the water body toward the Earth's equator. This natural bulging is originally uniform due to the centrifugal force of the the Earth's rotation around its own axis. And the perceived bulge on the other side of the Earth furthest from the Moon is simply that the Moon's gravity could not deplete the original bulging. dmaivn 05:15, 25 May 2006 (UTC)

[edit] Paragraph about Lucio Russo

I think the whole paragraph about Lucio Russo is out of subject. Moreover, his theories are controversial to say the least. For a commentary on The Forgotten Revolution, See this. I propose to eliminate the paragraph. 200.50.123.220 03:09, 23 May 2006 (UTC)

[edit] What do they do for us?

Do tides do anything useful for the population of Earth? If, say, the moon disappeared overnight, or was stolen by aliens to be mined for cheese: Would any species be destroyed or inconvenienced by the perpetual mid-tide? Would a perpetual mid-tide be the actual result? boffy_b 23:06, 14 June 2006 (UTC)

Yes, some species would be destroyed. Many marine species are intertidal meaning they live within the tidal range. If the tidal range disappears, where do they live? No, a perpetual mid-tide wouldn't be the actual result. The sun also causes a tidal force. Taking away the sun would destroy even more species.

[edit] Timing

Could this paragraph be used anywhere in the article?

The tides are an hour later each day because the moon orbits the planet in the same direction as the planet spins. Since the moon advances in its orbit during the course of a day, it takes about an hour extra for your location to catch up to the moon and the tide that follows it. Since the phases of the moon concern its position relative to the lit side of the earth, the timing of the tide reflects the lunar cycle. In the course of a week following a full or new moon, high tides occur during the middle of the day and the middle of the night. For instance, at the beginning of the week following a full moon, high tide is in the middle of the morning. Since the tide returns about an hour later than it did the day before, by the end of the week, high tide is late in the afternoon, as a low tide week begins. --Maradja 03:40, 26 July 2006 (UTC)

[edit] Dutch verb?

I removed the claim that the English verb 'to spring' is derived from Dutch. My dictionary states "Middle English, from Old English springan; akin to Old High German springan to jump and perhaps to Greek sperchesthai to hasten"


Paul Tracy|\talk 11:47, 23 August 2006 (UTC)

[edit] Problems with the dual tidal bulge model

Moved the following comment here from the main page: (Woodstone 07:34, 12 September 2006 (UTC))

This simple model has been used for centuries to calculate tide levels all over the world but it has a number of insurmountable oceanographical problems.

  • Wave speed: since the tide is just another gravity wave travelling along the ocean's surface, it must satisfy the laws for surface waves. For a wave to travel along the equator of 40,000 km in 25 hours, requires a speed of around 1600 km/hr, which is not sustainable. The maximum wave speed in a 'channel' of 5000m deep is about 800 km/hr. Average depth of the ocean is around 3800m, demanding a lower speed still.
  • Bouncing off continents: As the tide wave reaches a continent, most of it will be bounced back off the continental shelf, causing a tide wave of almost equal height to run in the opposite direction. This is not observed in real life.
  • Starting and stopping: as the tide wave apparently needs to start at one continent and stop at the other, it would be larger at the continent where it arrives and smaller where it came from. During the starting and stopping, far too much energy would be wasted. This is not in accordance with tidal movements world-wide.
  • Zero, one and two tides each day: there are places without tide, with one tide and most with two tides each day. This cannot be explained with the simple model.
  • Tide height: the height of the tide, the difference between high and low tide, does not follow the two-bulge idea which suggests that the tide should be maximal around the equator or on opposite sides of a large ocean. Near the equator one can find places without tides and places with near-maximal tides.
  • Tide timing: high tide occurs at different times of the lunar cycle, depending more on one's place on Earth than on the position of the moon.

There is obviously a better explanation of how tides move around the world. That explanation came in the early seventies, as computers were able to model wave behavior, they were able to show that tide waves would run in ways to prevent loss of energy. Instead of running east to west, tide waves run around in circles (clockwise and CCW on both hemispheres) around islands, and certain points in the sea, called nodes. This early model has been shown to agree with reality and, for its time, has been a remarkable achievement of computational mathematics. From the map below one can see that some places in the world (the nodes) have no tides, others two (12 lines) and a few places one (24 lines).

Since the TOPEX/Poseidon satellite has been measuring surface height data, the oscillating surface of the oceans due to moon tides could be measured and mapped. Note that the nodes correlate with areas of no tide change, except where these rotate around islands such as New Zealand and Madagascar. Highest tidal ranges are found where continental coasts distort the tide wave. http://svs.gsfc.nasa.gov/stories/topex/images/TidalPatterns_hires.tif (1MB)


[edit] Spring and Neap

Shouldnt a section be made up explaining the occurance of Spring and Neap tides. These may be mentioned but are lost in the horribly confusing text.

[edit] Etymology of the term "Neap"

Where the heck does it come from? Jachra 05:43, 7 November 2006 (UTC)

[edit] Changing Earth to Moon distance

If tides dissipate energy, then taking energy out of the Earth-Moon system should cause the Moon to drop to a lower orbit (decreasing the separation), should it not? The article (Other Tides section) currently states the opposite.

Silverchemist 12:54, 7 November 2006 (UTC)

No. While the sum of the earth's and moon's orbital kinetic and potential energies increases as the distance between them increases, this is more than compensated for by the loss in the earth's rotational kinetic energy as a result of the decrease in its rate of rotation.
David Wilson 16:43, 13 November 2006 (UTC)
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