Talk:Tide

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)

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)

A Question

I'm trying to make sense of this article but it's 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?

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

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)

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)

Tides and latitudes

Why is the tidal amplitude greater in the higher latitudes than at the equator? 65.166.78.66 14:38, 6 March 2005‎ (UTC)

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 fulfill 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)

Amplitudes do not follow a definite pattern with latitude. For example the semi-diurnal amp in the Gulf of Carpentaria is some 8m. Long period tides Mf and Mm do show a lat dependency, though. The analytic solution by Kelvin is larger at greater lats. The cutoff lat for M2 is 71N/S, as the inertial period (half pendulum day) is shorter than the tidal frequency. Mf and Mm can exist at such low frequencies since they are mixture of long and short Rossby waves (Wunsch et al, Progr Oceanogr, 1997). — Preceding unsigned comment added by 27.33.243.64 (talk) 10:30, 14 December 2015 (UTC)

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)

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 206.62.11.5 03:19, 13 April 2005 (UTC)

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)

A standing wave has zero group velocity and infinite phase speed; the usual G*p=c^2. If you are referring to amphidromic points, these are not quite standing waves as the phase speed is not infinite. The amp must be very small (but not necessarily zero when measured). The nearby phase lines all converge at a point. (phase lines must terminate on a boundary, a point, or have cyclic continuity on the sphere.) It is a wave going around a fixed point in the ocean, with a period that exactly matches the astronomic nodal period. In the wider scale ocean I suspect true standing waves are not possible.27.33.243.64 (talk) 10:57, 14 December 2015 (UTC)

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 shows 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)

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? 68.9.139.215 14:33, 5 June 2005 (UTC)

« 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, 5 Jun 2005 (UTC)

I can tell you what is unclear about this sentence:

″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.″

It is written for science majors, not average, reasonably educated folks. While it's obvious that tidal forces of the sun and moon reinforce each other when they're on the same side of the earth during a new moon, it's a confusing explanation when they're on the opposite sides during a full moon. How about explaining that

"When on opposite sides of the earth during a full moon, the sun and moon pull on it in opposite directions, deforming it and producing a bulge in the oceans on both sides."?

Genepoz (talk) 00:55, 29 October 2012 (UTC)

Bad idea. Both the Moon and the Sun "produce a bulge on both sides of the Earth" all the time, so not just during full moon. Wikiklaas (talk) 02:30, 9 April 2013 (UTC)

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)

In my opinion this article was very well written. The physics section was particularly well explained. If someone is in free fall above the earth, neglecting air resistance, they would feel no forces on their body (they are an inertial frame of reference). Since the earth is not an infinite plane this is a first approximation. In actuality there is a force difference between the head and the feet, pulling the person apart and another force compressing their body from the sides. An extreme example is known as spaghettification. As to the force pulling the body apart, the Taylor expansion of

$1/(R+\xi )^{2}\!\,$ is $R^{-2}-2R^{-3}\xi -3R^{-4}\xi ^{2}+\ldots \!\,$ so for relatively small $\xi \!\,$

${\frac {1}{R^{2}}}-{\frac {1}{(R+\xi )^{2}}}\doteq {\frac {2\xi }{R^{3}}}\!\,$

So the inverse-cube force law assertion is obviously correct. Scot.parker 18:23, 2 April 2007 (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:

$h_{2}={\frac {5}{2\left(1+{\frac {19\mu }{2\rho gr}}\right)}}\!\,$

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 05:13, 25 May 2006 (UTC)

Mistakes in tidal physics

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 Fundy 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. 155.246.33.18 (Hydrocomputer) 19:57, 08 Dec 2005 (UTC)

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). 80.229.15.164 23:57, 12 December 2005 (UTC)

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)

I see somebody has added that the 'Young Flood Stand' is due to the Isle of Wight. I would like to see proof of this statement. It sounds suspiciously like another effect of the four cycle a day tide at a time of the month when it is weak in comparison with the main semi-diurnal tide. David Webb (talk) 19:40, 6 December 2007 (UTC)

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)

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. 198.103.96.11 17:52, 23 June 2006 (UTC)

Also there are uses of the moon as a light source. In polar regions where the winter is without a visible sun, the moon provide light. Since everything is snow covered, the light can be quite bright, and when skiing (cross country, of course) it is often easier to just turn off your headlamp. --Tunheim 12:15, 7 February 2007 (UTC)

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)

Makes sense to me; and, I am a mariner (though not overly ancient.) Peter Ellis 03:46, 14 February 2007 (UTC)

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)

Problems with the dual tidal bulge model

Moved the following comment here from the main page:

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) Woodstone 07:34, 12 September 2006 (UTC)

While you are considering 'Bouncing off continents', should there also be some discussion of the Seiche effect -- standing waves in an enclosed body of water? Any body of water, even an ocean, has these. Peter Ellis 03:52, 14 February 2007 (UTC)

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. Lti 07:19, 24 October 2006 (UTC)

Doesn't the drawing contradict the description? The description says the spring and neap tides occur on the day of new, full, and quarter moons, the drawing says 2 days later. Djfeldman (talk) 17:32, 23 October 2009 (UTC)

It's not precisely "contradiction" but you have a point here. Wikiklaas (talk) 02:09, 10 April 2013 (UTC)

Etymology of the term "Neap"

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

My OR: I imagine it might come from the German word for it, "Nipptide" - "de:nipp" is pronounced very similar to "en:neap". The corresponding German verb "nippen" means "to sip", which would be appropriate as the neap tide only changes water level a little. -- Roentgenium111 (talk) 19:35, 7 November 2011 (UTC)

"neap²". Oxford English Dictionary (2nd ed.). Oxford University Press. 1989. Old English (example given from AD 469: forđganges nip - without the power of advancing). The Danish niptid is probably from the English. The English term neap-flood (from which neap tide comes) seems to have been in common use by AD 725. Martin of Sheffield (talk) 13:36, 6 July 2015 (UTC)

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)

But as you may have noticed in articles upon the topic, the Earth-Moon distance indeed increaeses over time, due to the dissipation of tidal energy. Wikiklaas (talk) 02:36, 2 April 2013 (UTC)

Don't confuse energy and angular momentum. Momentum (angular and linear) is always conserved. The angular momentum of the earth-moon has to be the same before and after. The angular momentum of the earth due to rotation about its axis is reduced. But the angular momentum of the orbits of the earth-moon system is increased.

Energy is another matter. Energy is also conserved. The energy due to the rotation of the earth about its axis goes into two places. It goes to increase the energy of the orbits of the earth-moon system and it goes into heating the earth via friction.(Anorderofmagnitude (talk) 15:23, 13 August 2013 (UTC))

Earth tide merge-in

Both this article and Earth tide carry a suggestion that it be merged into this.

• Strong Support. I support this. The part about tides within the crust of the earth, and presumably the underlying structures, deserves a mention. The paragraphs before that are mainly covered in the current document. Peter Ellis 03:56, 14 February 2007 (UTC)
• Merge The two articles are related. Lunokhod 09:22, 14 February 2007 (UTC)
• Should Do. I support the merging of Earth tide & others. Semiroyal 00:37, 17 February 2007 (UTC)
• Bad Idea The articles are related, yes, and some of the underlying physical causes are similar. However Earth Tide is clearly a distinct phenomenon from marine tides. Furthermore, the general usage of the term "tide" covers marine tides only; having a separate article for Earth Tides is not misleading.65.96.65.205 20:47, 4 March 2007 (UTC)
• Support - per above. Addhoc 14:02, 10 March 2007 (UTC)
• Strongly oppose The terrestial tidal mechanics are solid deformations (stress, strain, shear), which are completely different from the fluid mechanics of tidal flow. These tides cover the entire surface of the Earth. Earth tides corrections have been evaluated at GPS ground stations, many far inland. These corrections are essential, and independent of marine tide.rmo13 21:07, 18 March 2007 (UTC)

Please check revisions; no longer rated as stub.rmo13 04:32, 7 April 2007 (UTC)

• Oppose Rmo is correct in my opinion. You must also remember that these articles are read by many people of various education levels. If someone has a curiousity about ocean tides, an inclusion of earth's tides would cause confusion. If you wish to merge the concepts of ocean tides with earth tides, you might as well consider the effects of air tides :) . By the many comments in this discussion very many people seem to be confused about ocean tides. And you want to add to that earth tides??? Scot.parker 20:49, 2 April 2007 (UTC)

• Strong Oppose -- Short well targeted articles are best, not everything in the garbage can and kitchen sink can be buried here styles. There is a technical difference, and so should we recognize that. // FrankB 21:10, 10 May 2007 (UTC)
• Oppose. Cuvette 19:36, 16 June 2007 (UTC)

Proposed tables and pictures, moved

rmo13 01:39, 12 April 2007 (UTC)

Tidal Forces

The following section was recently added to the article, but is mostly a duplication of information already present and it does not fit in the flow and style of the article. I moved it here for the time being. The part about numeric values doesn't seem right, but after correction might be an addition.

Newtonian mechanics is adequate to explain the tide-generating forces of the Moon and Sun.For the Moon it leads to three basic concepts:

1) The only effective forces are the components tangential to the Earth’s surface (2,4). An upward radial (vertical) component merely decreases the effective density of the water column so that it rises higher than that which is not affected by tide forces. On average this rise adds less than one millimeter to a tidal bulge and may be ignored.

2) At Earth centre and at points several tens of kilometers to the Moon side of quadrature, the gravitational and opposing centrifugal forces, each amounting to 33.1 micronewtons, cancel one another precisely and at those points there is no tide-generating force (2,3,4). The centrifugal force due to the monthly rotation of the Moon-Earth couple about its barycentre is not to be confused with the centrifugal force due to the daily rotation of the Earth on its axis, which is one thousand times greater but has no influence on the tides.

3) At each of the two nodal points on the Moon-Earth axis (cis-lunar and trans-lunar), both gravitatonal and centrifugal forces are parallel to the Moon-Earth axis and can produce no tangential component (3,4).

Between quadrature and the nodal points there remain small opposing tangential forces: on the Moon side the gravitational excedes the centrifugal, giving rise to the cis-lunar bulge; on the opposing side the centrifugal is the stronger, producing the much-discussed trans-lunar bulge (3). The forces reach their maxima at 45° to the axis (2,3), with values of 0.56 and 0.54 micronewton respectively. These forces are all directed toward the axis, tending to produce axial bulges.

The tide-raising forces of the Sun are weaker than those of the Moon because of its greater distance.

Since the barycentre of the Sun-Earth couple lies very close to the Sun’s centre, the forces on both sides of the Earth are expressed by the combined effects of the gravitational and centrifugal gradients. At 45° from the axis, both differential forces have a value of 0.356 micronewton.

The table shows some significant force values for the Moon:

1	2	3	4	5	6
0°	32.07	0	0	0	78.29
45°	32.38	22.90	23.44	-O.54	68.59
90°	33.147	33.147	33.147	0	45.191
135°	33.94	24.00	23.44	0.56	21.79
180°	34.28	0	0	0	12.10

Column 1 : Longitude along any Earth equator passing through the axis, counted from the trans-lunar nodal point.

Column 2: The axial gravitational force of the Moon at longitudes from 0° to 180°.

Column 3: The tangential component of the gravitational force (value of Col. 2 multiplied by sin n).

Column 4: The tangential component of the centrifugal force (33.1 micronewtons multiplied by sin n).

Column 5: The tide-raising force.

Column 6: The radial centrifugal force, having no effect on the tides, is shown for comparison.

References :

1) Butikov, Eugene “A Dynamical Picture of the Oceanic Tides” Amer. Journ. of Physics 70, no.10, 2002: Ch. 2

2) Butikov ibid Ch. 4

3) Feynman, Richard “Lectures on Physics” vol. 1 (Addison-Wesley, Reading, MA,1966)

4) NOAA, National Oceanic and Atmospheric Admnistrtion “Our Restless Tides” Chap. 3.

−Woodstone 18:50, 5 April 2007 (UTC)

Major Reorganization

largely completed

• Introduction and basic terminology
  • Tidal variation
  • Phases and amplitudes

done, first pass fact review and polishing.

• Theory and physics
  • Tidal forcing
  • Laplace theory
  • Tidal dissipation
• Tidal observation and prediction
  • Constituents and Phases
  • Tidal analysis
• Navigation

seems to only really need wikification

• Intertidal ecology

New insertion

• Other tides and missapplications

rmo13 16:38, 14 April 2007 (UTC)

Tidal Physics (2)

It is going to be very difficult to come to a consensus on an article on Tide Forces. There are two conflicting schools of thought: one goes back to Newton’s explanation in which he applied gravitational forces alone. It was written when he was only 23 and had not yet worked out his laws of mechanics. The other, more recent, introduces centrifugal forces and relies on the basic laws of mechanics that Newton announced twenty years later.

The “classical” viewpoint has produced great confusion. Serious writers have used all possible combinations of concepts. Many take account of radial (vertical) forces, which were already shown to be without interest by Sir George Darwin at the end of the 19-th century; it is a question of the change in weight of ocean water (not its mass!) “under” the Moon. The situation is like liquids of different densities in the two arms of a U-tube. Getting rid of radial forces is essential to our subject, and it is so simple that all may be made clear on the back of a small envelope.

Other writers introduce axial forces at the nodal points. Several base their discussions on various reference frames; Butikov produces the delightful analogy of a frying-pan revolving in the hand of the cook, but not rotating. Butikov’s approach is very sophisticated, but he comes up with the results that the simple application of elementary mechanics does, namely no appreciable forces at the nodes nor at quadrature, but with tangential maxima at 45°, all directed toward the nodal points. This would suggest two tides per day at Temperate latitudes, and four smaller ones per day in the Tropics, at longitudes 45° and 135°.

The use of the word “bulge” seems rather unfortunate. It would be legitimate for the axial bulges that would occur, with big monthly tsunamis, if the Earth were to stop rotating. But what actually happens in the oceans is far more complicated and is more Oceanography than Physics. It is very well treated in other Sections of Wiki Tides.

I think part of the present article on Tidal Physics should be deleted. The Figure is misleading; it shows inward radial forces at quadrature, which in reality are negligible, and outward radial forces at the nodes which have practically no effect. Then the text that follows the Figures presents only the classical version.

It is a great pity that so few authors present concrete values of forces. They would be so much more eloquent than words! Johnhen 14:34, 6 April 2007 (UTC) Jhenniker

This is very nearly historical review of tidal force theory, which could well be broken out into its own section. You are right about the usefulness of numbers, but I think the ratio of vertical forces to horizontal may be easier to grasp. Gill writes the gravitational potential in a form equivalent to fictious height variations, which quickly summarizes fictious slopes, another way of writing the ratio of horizontal to vertical forcing. (these can be compared to river slopes). Many people seem to have some intuition about water being moved by horizontal energy changes. My prefered approach has been to look at Laplace tidal equations, and back out the physical meaning: sheet flow only sees horizontal forces, height change comes from horizontal velocity divergence, Coriolis terms turn flows, so waves "heap up" at boundaries and rotate around basins or around islands; and some dissipation (completely undescribed here). An equally bad fault, the physics section completely misses boundary conditions. Of less importance, a quick review of the literature (Cartwright) seems to indicate Euler (1740, Paris) first published the importance of horizontal forces. (just started Butikov) rmo13 05:41, 7 April 2007 (UTC)

References/footnotes

It seems that the references section has turned into a footnotes section. Perhaps it should be relabeled. Rracecarr 20:23, 7 April 2007 (UTC)

Misleading material removed

The Moon exerts its gravitational pull differently on different parts of the earth. The farther away the Moon, the weaker its pull. Imagine a shell of the outer Earth, this diagram shows the Moon's gravity differential over the thickness of the shell.

Caption and image seem misleading. None of my texts mentions shells, actually to the contrary.

Also: from Timing

This is because the tide originates in the southern oceans, the only place on the globe where a circumventing wave (as caused by the tidal force of the Moon) can travel unimpeded by land.

The resulting effect on the amplitude, or height, of the tide travels across the oceans. It travels as a single broad wave pulse northwards over the Atlantic. This causes relatively low tidal ranges in some locations (nodes) and high ones in others. This is not to be confused with tidal ranges caused by local geography, as can be found in Nova Scotia, the Bristol Channel, the Channel Islands and the English Channel. In these places tidal ranges can be over 10 metres. The Atlantic tidal wave arrives after approximately a day in the English Channel area of the European coast and needs another day to go around the British Isles in order to have an effect in the North Sea. Highs and lows of the Channel wave and North Sea wave meet in the Strait of Dover at about the same time but generally favour a current in the direction of the North Sea.

If these can be justified by some citations, we can re-insert. rmo13 00:59, 12 April 2007 (UTC)

Theory of tides (new article merger)

This page shows a pretty keen interest in the actual mechanics of tides along with the forcing terms. I have begun to rough out an article to address the technical matters discussed here for general reference. Eventual merging may be an option, but the mathematics and physics of tides should be made explicit for reference.

Suggestion to merge theory of tides into this article:

• oppose the technical discussion of physics and forcing seems to be interupting the flow of the article and seems to be a magnet for misleading and erroneous interpretations. If this can be taken else where, this article may be finalized into a good general interest article pretty quickly. rmo13 12:39, 12 April 2007 (UTC)

• Oppose -- What I said above: Short well targeted articles are best, not everything in the garbage can and kitchen sink can be buried here article styles. // FrankB 21:11, 10 May 2007 (UTC)

• Oppose Cuvette 19:37, 16 June 2007 (UTC)

Vorticity

The new paragraphs on vorticity are very unclear and contain some errors. Unless improved, they will need to be eliminated. −Woodstone 14:41, 19 April 2007 (UTC)

will do, I'll post here first:

When using a refence frame that rotates along with the Earth's surface, a mass will have rotation due to the Earth; so tracking this rotation (especially the rotation around a vertical axis) proves useful as the vorticity of the fluid. Thomson rewrote the Laplace tidal equations for vorticity to show vorticity stretching. This means that as a column of fluid lengthens, such as when the water height rises in a volume under a patch of water (or in a tornado), water moves in from the sides, and vorticity "converges" in the column, and the water rotates faster. Similarly if it shortens, the rotation diverges and slows, as if higher pressure was pushing down on the surface. Waves are alternating rising and falling of the surface, and so higher and lower vorticity. Putting these together shows that the waves will tend turn in a direction opposite to the Earth's rotation, (rightward in the Northern hemispere) and be turn into the coast on that direction. This explains the higher tidal amplitude along coasts and the direction of tidal propagation.

rmo13 00:25, 20 April 2007 (UTC)

Tidal Physics (3)

There is an often-repeated mistake that must be corrected. It is the statement that lunar centrifugal forces are equal all over the Earth’s surface. There are only two centrifugal forces in the Moon-Earth system. One, which is due to the daily rotation of the Earth on its axis, is indeed invariable at any given latitude, but has no influence whatever on the tides. The other centrifugal force is due to the monthly rotation of the Moon and Earth together around their barycenter. The radius that determines this force varies from (6378 – 4671) = 1707 km. on the Moon side, to (6378 + 4671) = 11049 km. on the opposite side, giving rise to small cis-lunar and large trans-lunar centrifugal forces respectively. It is not a uniform force. Combined with the (variable) gravitational forces (paying due attention to signs ! ), these lead to the very nearly equal tide forces tending toward the nodal points. 86.216.43.50 15:15, 23 April 2007 (UTC) Johnhen

It is actually not a mistake. I agree that it's a counterintuitive way to describe the motion, but it is a correct way, and seemingly a commonly used one. Instead of thinking of the earth revolving around the Earth-Moon center of gravity, you can think of the Earth undergoing circular motion with no rotation (like someone said, analogous to swirling a frying pan), with the radius of the circular motion equal to the distance from the center of the Earth to the Earth/Moon center of gravity. The centrifugal force associated with this circular motion is indeed equal everywhere on/in the Earth. It's all a matter of how you choose to break down the total motion into parts. Rracecarr 15:40, 23 April 2007 (UTC)

Content number 26

The explanation of Rraecarr very happily leaves the way open for two valid approaches to tide forces, one “expert” and the other “simple”. But the “expert” reasoning is not very clear: the centrifugal forces being equal, their radii must be equal and they must radiate from a single center; these must be the Earth’s radius and center? But without rotation? Are we not taking a frying pan to squash a fly! Would it not be kinder to many Wiki users to provide the “simple” reasoning, that is within reach of any first year high school physics student?

The same call for simplicity applies to the debate between those who are “for” and “against” centrifugal forces, as outlined in my post of 06 April 2007 (sec. 21 of Talk Tide). The case for centrifugal forces was the object of my first post of 26 Mar. 2007, that was meant to be a Wiki article, but got moved to Talk Tide (sec. 19). There is also a didactic version (48 kb) available, more suitable for a discussion group. Perhaps it should be shown in Talk Tide?

By the way, rmo13 was right about Euler being first to see that radial forces don’t count. My belated excuses to the memory of Euler. And my note about four tides in the Tropics may well be new: one has to imagine a pair of circular zones of force situated at 45° to the axis; each intersecting the equator at two points, making a total of four impulsions per day, alternately toward East and West. But at geographical latitude 45°, the circles would pass a given point only once each day. Do real tides show any such tendency? 83.113.71.200 14:06, 26 April 2007 (UTC) Johnhen

Units

In "tidal dissipation" it says "there is about 1 less day per 100 million years. See tidal acceleration for further details" - is that 1 less day in a year per 100 million years or is that 1 less day in a month per 100 million years? 81.108.225.22 21:13, 7 August 2007 (UTC)

Tides in rivers and lakes

I've noticed this article makes no mention of the effect of tides on the water levels of rivers and lakes, especially since large rivers can experience high and low tides several miles from their estuaries; the Willapa river, near me, experiences daily high and low tides as far as five miles inland from Willapa bay, and possibly farther (I've never traveled upriver to see how far these high and low tides go). Presumably, a lake very near an ocean, if connected to said ocean by a river, would also have a high and low tide. This isn't a matter of ocean water backing up into the rivers (and possibly lakes), since the water upriver never gets noticeably brackish, but a matter of the freshwater backing up because the high ocean water is impeding its flow (water currents are visibly slower in the Willapa river at high tide).

Also, I've noticed that large lakes (widths on the order of 2 or 3 miles) have what seem to be an intertidal zone and even tide pools on their beaches. Since these are presumably too small to experience significant tidal forces, what causes the high and low "tides" in these lakes?

Such information on freshwater tides or tidal-like behavior would make a valuable addition to the page, methinks.

Peace out, --GrammarDude 02:33, 21 October 2007 (UTC)

P.S. Also, as I understand it, the Nile river experienced a singnificant high and low tide in Ancient Egypt very, very far south (possibly as far south as extreme upper Egypt or lower Ethiopia?), which should have affected any lakes feeding the Nile's tributaries. —Preceding unsigned comment added by GrammarDude (talk • contribs) 02:45, 21 October 2007 (UTC)

I wonder; who tagged this "P.S."? and to whom on the English language Wikipedia would it not be obvious that a "P.S." is written by the person who signed his name just above it? especially when a "P.S." always comes aneath the signature. But, I see a bot tagged it, and bots are only as intelligent as their programmers think to make them. Though I can't help but think, 'surely there must be one who would have the common sense to program his bots so as not to tag something as unsigned when it has, in fact, been signed in the proper manner, rather than expecting a singature to appear in an incorrect place such as at the end of a "P.S." and tagging that "P.S." for not having found it there.' Queer, queer little stupid, queer bots. --GrammarDude (talk) 23:40, 31 January 2008 (UTC)

Yo, GrammarDude! Are you forgetting Wikipedia policy? Don't bite newcoming bots. --Þórrstejn [ˡθoɝ.staɪʲn]: Hammer of Thor ^talk 23:43, 31 January 2008 (UTC)

Could you cite a source for this Nile-tide? Are you perhaps confusing it with seasonal changes in Discharge? Or could it have been the Red Sea of which you "understood" it had a high and low tide? Wikiklaas (talk) 02:09, 10 April 2013 (UTC)

• Am I the only one who thinks the section on Tidal Power Generation, especially parts like "The most convenient situation is presented with Auckland city, which is between Manukau harbour and Waitemata harbour so that both power stations would be close to the load: nothing could be better!" are waaaaay biased. Get that stuff out. Needs a lot more info. —Preceding unsigned comment added by 76.114.19.58 (talk) 18:16, 28 November 2007 (UTC)

What does this have to do with tides in lakes and rivers? --Þórrstejn [ˡθoɝ.staɪʲn]: Hammer of Thor ^talk 23:43, 31 January 2008 (UTC)

Good article

I am surprised that this excellent article is only B class! It should be nominated for GA. Could someone fix the Example? The table, (if that is what it is meant to be) is not working, (at least in Explorer 7. I enjoyed reading this, it is well written and very informative, could be FA class with a little more work. --GrahamColm^Talk 11:19, 23 December 2007 (UTC)

• On closer examination I see it is a script. This section needs some attention. --GrahamColm^Talk 11:34, 23 December 2007 (UTC)

A question about tide charts

Is there a name for the tide chart used to show the tide height change over time? I know a flow over time chart is called a hydrograph. Is there something similar for the tide chart?? Theorangekombi (talk) 00:45, 8 February 2008 (UTC)

M2 tidal constituent period or tidal lunar day length

Tidal lunar day length is about 24 h 50 min, not 24 hour 48 min.

Half-day, or M2 constituent period, is therefore about 12 hour 25 min, not 12 h 24 min, as shown in the article.

More exact value would be 12 h 25.2 min; lunar day is 24 h 50.4 min.

The simplest way to verify that is to look at harmonic constituents for any of 100s of tide gages listed by National Ocean Service. Let's look, for example, at Nawiliwili gage The first line in the list is M2 constituent. Its period is 360 degree divided by Speed=28.9841042 degree/hour (see definition of Speed on the same page), which gives 12 hour 25.2 min.

98.203.251.116 (talk) 05:52, 13 February 2008 (UTC)

That's correct. M2 is 12 hours 25.23607 min Rracecarr (talk) 15:18, 13 February 2008 (UTC)

Laplace's Equations...

Umm... How come they arent in the article? Hamsterlopithecus (talk) 10:49, 28 April 2008 (UTC)

Thanks for pointing this out. I have added a wikilink to them. Crowsnest (talk) 12:16, 28 April 2008 (UTC)

Oh, a whole article! Thanks! The article is looking good btw... Hamsterlopithecus (talk) 05:46, 30 April 2008 (UTC)

Clarification

Like some other articles on physics, this would be off-putting to people with less than a degree in physics. It is not suitable for schools. I would prefer to see a basic explanation of the main concepts first before embarking on the complex material. In particular it should be possible to describe more clearly why there are two tides a day. Furthermore, no explanation is given why some places only have one tide a day. Just a freak caused by an island? No mention is made why tides alternate between slightly lower and slightly higher on a 12 hour cycle. JMcC (talk) 15:39, 12 June 2008 (UTC)

Giving a basic explanation is a most difficult thing to do, if one also wants it to be correct. That's why so many people embark on the complex material first. But you're welcome of course to add the basic explanation to the article yourself. More than welcome actually. Wikiklaas (talk) 02:09, 10 April 2013 (UTC)

Documentation of Centrifugal force

It appears that the article as written makes no attempt to incorporate the role of centrifugal force in the tides, an idea that goes back to Newton and is very well represented in modern publications. References have been added by me that introduce this idea. If there is some subtlety involved here, or some VERY recent developments that discredit these publications, a clear and well documented rebuttal of this school of thought should be presented to the reader. Brews ohare (talk) 03:32, 16 July 2008 (UTC)

It is with reason that not much attempt (there is some in the second paragraph of Tide#Forces) is made to use a non-inertial frame of reference, because that makes it more difficult to explain why there is a semi-diurnal tidal forcing of about equal strength, see e.g. here. Using an inertial frame, and tidal forces due to Moon and Sun in combination with the Earth's rotation gives an easier explanation of the phenomenon, see also elsewhere on this talk page.

In the description of the Earth's tides using the rotating frame of reference, there seem to be two forces responsible for the semi-diurnal tide: the centrifugal force due to the rotation of the Earth-Moon system about a common center of gravity and the pull by the Moon's gravity which is stronger at the side closer to the Moon. In the inertial frame of reference there is only the tidal forcing due to the gradients in the Moon's gravity field. Earth's rotation and the effects of the shape and depths of the oceans and seas is doing the rest in making it a time- and (more) space-varying phenomenon.

Further, a description in either frame of reference can be made valid equally well. You are of course right about the missing references. -- Crowsnest (talk) 07:39, 16 July 2008 (UTC)

Aside from whether we use centrifugal force to explain tides, the current wording in the intro bothers me:

...caused by the tidal forces of the Moon and the Sun acting on the oceans, and the effects of centrifugal forces caused in part by the Earth-Moon rotation...

That wording excludes the centrifugal force from the Earth-Sun rotation. That's inconsistent. Spiel496 (talk) 17:37, 17 July 2008 (UTC)

The entire discussion of the tides is unsatisfactory: First, it is largely undocumented, Second, it never really says what causes the tides: is it the bulging sea level, is it the tangential forces due to tidal forces, or what? It is my impression that the entire subject is way more complicated than these ideas and should be approached in layers. Either historically, or in terms of simplest model first, followed by each successive logical or pragmatic elaboration. Brews ohare (talk) 18:21, 17 July 2008 (UTC)

Yes, it's a big job. I have been thinking about having a go at it for over a year, but it's kind of daunting. The "bulge" idea is way oversimplified, but is a common image used to get the idea of tidal forces across. As you can see in some of the figures in the article, the M2 tide is not just a bulge, but varies at quite small scales all over the planet. Tidal forcing is very periodic and predictable, but the response of the ocean is much less so. I removed centrifugal force from the intro, because it is not separate from tidal forcing. In a rotating frame, centrifugal force is part of the tidal force. In an inertial frame, the tidal force is still there, but it is the result of the gradient in the gravitational field. Rracecarr (talk) 18:33, 17 July 2008 (UTC)

Don't forget that the forcing is only half the problem. The other half is the resultant motion as an effect of the complicated shorelines and depth variations of the oceans and seas, as described through e.g. Laplace's tidal equations. This gives rise to the strong differences in tides at different places (diurnal or semi-diurnal tides, tidal range, asymmetries in the tides, etc.) Crowsnest (talk) 16:39, 18 July 2008 (UTC)

I have never been happy with the use of "centrifugal force" in explanations, because it is not a real force (yes, I know about frames of reference in which it becomes real) and so it too has to be explained and that explanation is complicated. I'd favour an inertial frame throughout for explanation and then the introduction of rotating and other accelerated frames of reference should they help once equations are being flung about. Similarly, the inertial frame forces are one thing, with strictly astronomical periods and phases at assorted locations, but the observed tidal behaviour at those various places is dominated by the resonance aspect of the surrounding ocean basins. Thus, the M2 (or whatever) pattern as observed at some place, is not directly the M2 pattern of the forces at that position. NickyMcLean (talk) 21:08, 17 July 2008 (UTC)

My personal opinion is that an ideal encyclopedia article about tides for a general audience would never mention centrifugal forces once. After reading here, it should be obvious that it is 1) not needed, 2) prone to lead to misconceptions. After reading the "alternative development" in chapter 3 here, it should be obvious that a full treatment would need to involve two separate inertial systems for the Earth-moon and for the Earth-sun systems. All this for a fictitious force? Leave it out. The gradient concept in an inertial reference frame is all that should be presented. Flying Jazz (talk) 22:34, 17 July 2008 (UTC)

Leaving out centrifugal force might avoid some issues, but as most textbooks and web pages bring it up, the Wiki article would be incomplete and simply ducking the issue to omit it. Readers (like me) would wonder if the Wiki editors were simply out to lunch. Brews ohare (talk) 15:59, 2 August 2008 (UTC)

It is not necessary to give explanations in different frames of reference, in principle one will suffice. The description using the tidal force is, as far as I know, due to Lord Kelvin, see e.g. here where he explains tides including semi-diurnal tides to a lay audience (see also H. Lamb, Hydrodynamics, Cambridge Univ. Press, 1895, Appendix to Chapter VIII, p. 364). Just take the Earth and Moon. In an inertial frame of reference the gravitational force of the Earth on the Moon and vice-versa are equal but of opposite sign. So Earth is accelerated (falling) towards the Moon and v.v. As a result only the tidal forces -- i.e. corresponding with the difference between the acceleration at a certain point and the acceleration of the Earth as a whole, both due to the Moon's gravity -- will result in the tidal motion either in the landmasses or the seas and oceans (relative to the Earth's center).

In a rotating coordinate system with the origin at the center of gravity of the Earth-Moon system (inside the Earth, at a depth of about 1/4 Earth's radius), one gets the centrifugal forces. The centrifugal accelerations can be thought of as having two components:

• a "mean" component (at Earth's center) which one has to show to correspond with the mean acceleration by the Moon, and contributing to the tidal forces found in the inertial frame and
• an additional one, due to the rotating frame of reference, not due to the Moon's forcing. So that requires additional explanation.

Note that in the inertial frame the tidal forces of Moon and Sun can be added, which is not the case for the rotating frames having different centers of gravity for the Earth-Moon and Earth-Sun system.

In yet another rotating frame of reference with the Earth's center as the origin, it is quite difficult to explain the cause of the additional forces leading to the semi-diurnal tide, which are in this frame no longer simple centrifugal.

Crowsnest (talk) 21:16, 2 August 2008 (UTC)

After giving this some thought, I am also in favor of avoiding the centrifugal force explanation. It's not incorrect, of course, but I feel leads the novice astray, by implying that the orbital rotation is a key element. It's not. If one were to use the Moon's gravity to pull the Earth in a straight line, with constant acceleration, tides would still occur, right? If one force explains it, then we shouldn't invoke two forces. Another reason: No one even resides in this rotating frame. Well, Moon inhabitants would, but only because the Moon is in tidal lock. While I'm at it, I'm also going to weigh in on that figure (6) showing the Earth and Moon co-orbiting the center of mass. It doesn't work for me. If we must keep it, I would suggest adding the tidal bulge to the diagram. Even better, put some structure on the Earth (continents?) to show the spin relative to the tidal bulge. Otherwise, I doubt anybody is getting the point unless they already understand tides in the first place. Spiel496 (talk) 03:05, 3 August 2008 (UTC)

Free-fall

Encyclopedia Britannica gives a nice view on the cause of tides, in relation to free fall and tidal forces: http://www.britannica.com/EBchecked/topic/218241/free-fall -- Crowsnest (talk) 08:46, 2 November 2008 (UTC)

I believe that the unequivocal expert of freefalling is Tom Petty. — Preceding unsigned comment added by 70.166.197.2 (talk) 00:12, 22 September 2015 (UTC)

Tidal effects on distant inland caves and blowholes

Is there anyone here who is an expert on this subject? I have an interesting observation from the middle of Mindanao and would like to ask some questions and get some clarity on this, possibly by email. Please respond on my talk page. -- Fyslee (talk) 06:19, 22 March 2009 (UTC)

A.T. Doodson also analysed the tidal effect in lakes, so that might be a starting point. Presumably, the effect you are interested in is separate from a seiche in a lake (unconnected to the oceans) caused by wind or weather generally? NickyMcLean (talk) 20:38, 22 March 2009 (UTC)

It's fortunate I noticed this, since I have over 3,000 articles on my watchlist, plus their talk pages. Here's the story. As a child of missionary parents (about 9 years old at the time), I lived in the middle of Mindanao at Mountain View College. The college has (or had?) a logging industry and some of the older students were employed part-time there to help them pay for some of their tuition. One time I was out with them in the forest. They showed me a hole in the ground about a foot in diameter, with no apparent bottom. They dropped rocks into it and never heard them hit the bottom. (Nearby they caught a very small baby cobra which we played with!) Well, they had been working in that area for some time and had noticed that there was literally a wind that blew out of the hole during certain periods of the day, and it sucked into the hole at other time periods, and they surmised that there must be some connection to underground rivers that went into the ocean, and that tidal currents might be pushing and pulling, causing the air movements. We had fun throwing leaves at the hole and watching them get blown 6-10 feet into the air.

I have always wondered about that ever since (1960). What could be the explanation? Were they right, or was something else going on? -- Fyslee (talk) 02:57, 23 March 2009 (UTC)

Some possibilities are apparent.

The hole leads to an airtight cavern which has a lake in communication to an outside lake/ocean that is sloshing tidally (or to a river that rises and falls) and the cavern's lake rises and falls, thus huffing air in and out of the blowhole, as you suggest.

The cavern is airtight (with no connection to any lake) and the outside air pressure rises/falls due to local weather and wind so that air huffs in or out of the blowhole to communicate the same air pressure to the interior of the cavern.

The cavern is not airtight but has a communication to some opening on the far side of a hill or similar, and differing wind directions at each end mean that air wafts through.

The cavern is again a tunnel, but with ends at different altitudes; air in contact with the walls of the tunnel is warmed or chilled relative to the outside air and a chimney flow results.

Deciding what it might be would require an exploration of the cavern. Speleologists delight in such excursions, but I'd rather have the sky above me. NickyMcLean (talk) 02:48, 8 January 2011 (UTC)

I experienced the same rapacious dialog with a blowhard during a lunar climax. The peaks and valleys of the tides that washed over us were delightful. — Preceding unsigned comment added by 70.166.197.2 (talk) 00:10, 22 September 2015 (UTC)

Red tides and other distinctive patches of water

The original Saxon "Tide" meant "a distinctive patch of water" and was a very important word for a seafaring people which is still in evidence today, as in "Tidal Wave" or "Red Tide". These terms are not evidence that our linguistic ancestors didn't know the difference between the daily tides and an algae bloom, or that "Tidal Wave" is some kind of misnomer needing correction from a foreign borrowing, as if the Japanese understood the truth about them and the Saxons not. Chrisrus (talk) 05:37, 11 July 2009 (UTC)

Dubious measure

A 'dubious' tag has been added: the same point is discussed under Talk:Tidal_force#Dubious measure. Terry0051 (talk) 15:25, 6 October 2009 (UTC)

Reply has been given there. DVdm (talk) 16:12, 6 October 2009 (UTC)

Diurnal-semidiurnal tides

The image of physicalgeography is much clearer, perhaps the image now present can be swapped with this one ? 91.182.165.207 (talk) 11:35, 22 October 2009 (UTC)

More blatant racist exclusion on Wikipedia

So, apparently the history of tidal physics begins and ends with the west. So the Chinese, who had one of the largest navies in the world, knew nothing about tides? Africans, too? Middle Easterners? Could we get some inclusive histories here?

RR 00:42, 8 February 2010 —Preceding unsigned comment added by 75.70.143.123 (talk)

The protocols are in place for you to start adding materials regarding tidal internationalism. Thanks. — Preceding unsigned comment added by Slowclap (talk • contribs) 03:41, 7 January 2011 (UTC)

Introduction Section

I think the introductory paragraphs are too misleading to stay as they are. First they state that tides are caused by the rotation of the Earth and the gravitational forces of the Moon and Sun. It would be more correct to say that the tides are caused by the gravitational forces of the Moon and Sun, their periods then being determined by the rotation period of the Earth and the motion of the Moon and Sun relative to the Earth.

Without rotation of the earth around its axis, there would only be a very slowly shifting deformation of the earth. The solar tide would have a period of half a year if the earth would not rotate with respect to distant stars, the lunar tide would in that case have a period of half a sidereal month. It is the rotation of the earth which causes the tides as we experience them, with a main period of half a lunar day. Without rotation, the wavespeed of the very slowly propagating deformations would be much slower, causing a very different pattern of wave propagation. As this would be well within the limit of maximum wave speed foor deep water waves, it is hard to predict whether we would see a rise and fall of the water relative to the landmass at all. In this view, rotation of the earth is a vital cause for the tides as we know them. Wikiklaas (talk) 17:12, 16 April 2013 (UTC)

Secondly the paragraph states that the tides are influenced by the shape of the near-shore bottom. This is too restrictive. In fact they are influenced by the shape and bathymetry of the ocean everywhere.

Thirdly is the statement that "Most coastal areas experience two daily high and low tides". In fact this is true for most of the ocean - not just the coastal ocean - see any set of co-tidal charts. A few regions have just one high and low water each day and there are others where they may be one or two tides each day depending on the time of the (lunar) month.

Fourthly is the statement that "the point right "under" the Moon (the sub-lunar point), the water is at its closest to the Moon, so it experiences stronger gravity and rises". This is misleading on two counts. First although the force vectors will then be upwards, the actual response of the ocean depends on the bathymetry etc. As the earth rotates at some points below the Moon sea level will rise but at other points it will fall - again see the complexity of any set of cotidal charts. The explanation of low tide is equally wrong.

The second problem is a bit more sophisticated and goes back to - I forget, Laplace probably (1790??) - who showed that it was the horizontal component of the tidal forces that was important in driving the ocean tides and that in practice the vertical component could be ignored. A simple way of explaining this is to point out that when a tidal wave goes past the water particles only move up and down a few tens of centimeters, whereas their horizontal displacement will be a few kilometers. Thus the horizontal component of the tidal forces will do a lot of work pushing the tide during this period. The vertical forces may be of similar strength but the water moves such a small vertical distance that in comparison they do hardly any work.

It was Euler who, in 1740, as one of the winning contestants to the Paris prize, showed mathematically how the horizontal component of the force was of most importance in driving the tides. Laplace's dynamic theory was published in 1776. Wikiklaas (talk) 17:12, 16 April 2013 (UTC)

For the moment I am wary about changing the paragraphs myself - and would prefer from Woodstone and others on what is needed in this introductory section. I guess accuracy is one thing but what is the section trying to do for the readers?

David Webb (talk) 23:17, 5 March 2010 (UTC)

Biological Rhythms section

Does a "biological rhythm" necessarily correspond to the tide?

• The human cycle is nominally anywhere between 28 and 45 days
• The horse cycle is (according to Wikipedia) between 19 and 23 days
• Cats (according to Wikipedia) generally only ovulate after mating, and during the spring and summer months
• Birds, many herding animals, and even marine fish tend to breed seasonally, not with reference to the tides.

So it seems unreasonable to say that "vertebrates" have all genetically inherited tide-based rhythms when they're all completely different. The citation for this section is from 1871, which is not really up to date. —Preceding unsigned comment added by 144.36.24.43 (talk) 18:08, 22 September 2010 (UTC)

Again, a reference to a work from 1871 is not enough evidence to claim that the human menstrual cycle relates to the lunar tide when most other animals, including many that live in the ocean, do not have breeding cycles that relate to the tides. This is silly, stop putting it in. — Preceding unsigned comment added by 207.219.251.100 (talk) 17:52, 29 October 2012 (UTC)

Hmm... a more up to date ref (either supporting or otherwise is needed. You have one? If you had used an edit summary explaining the removal ... well I'd a looked here first maybe... Vsmith (talk) 18:35, 29 October 2012 (UTC)

I'd have thought that a multiple of 28 days (or ~56 tides) was a bit too much to count biochemically, especially when there is a much simpler relationship, the directly-observable visual effect of the moon at night in the absence of streetlights and televisions, etc. This cycle has one per month, an easily countable number. This is all speculation. NickyMcLean (talk) 20:25, 29 October 2012 (UTC)

Some misunderstandings about Lunar tidal waves?

It is said that these tidal waves are caused by the gravitational pull of moon. As we all know that gravity is dependent on the mass of a body. The mass of moon always remains same in every phase. Then how it happens that only in full moon and new moon the tidal waves are caused, and in other phases these waves are neap.

One possible solution to this query is that the moon's orbit is eliptical. So when the moon is on that part of the orbit that is nearer to the earth's surface the SPRING tides are caused. Similarly when the moon is that part of the orbit that is away from the earth's surface then the NEAP tides are caused. —Preceding unsigned comment added by 116.71.55.68 (talk) 15:27, 5 October 2010 (UTC)

NOTE, added much later. Interesting idea. Now you know (or you can find very easily) the period and the phase of the anomalistic month, so you know when the moon is in perigee and when in apogee. The first obvious point that would strike you is the moon being in perigee only once each (anomalistic) month, while we experience spring times twice a month, but for now, let's continue. The next thing you would have to do is find a tide table for the year for which you have data on the moon's orbit. Compare the actual occurence of spring and neap tides with the moments you predicted using the anomalistic cycle of the moon. You'll find they do not match (at least not over a period longer than let's say two months, and even if you find a match, you would miss the preceding and the following spring or neap tide - even to the effect that you'd find one when you predicted the other). So here's another failure to explain the spring-neap cycle in the tides. You could have known before because there's not a single work on tides in which you would have found this explanation, which also should have been a clue. Wikiklaas (talk) 17:12, 16 April 2013 (UTC)

Many things are being confused here. The tides are not caused by the gravitational pull per se, but by the difference of the pull locally and on average over Earth. Although the mass of the Moon is constant, the distance to a specific point on Earth varies. At a point on the surface facing the Moon, the gravity exerted by it is more than average, the resulting tidal force being towards the Moon (locally upward). On a point just opposite, away from the Moon, the force is lower than average, resulting in a tidal force away from the Moon (again locally upward).

The spring and neap tides are caused by the combined effect of Sun and Moon. When they reinforce each other (at full and new Moon), it's spring tide. When they partially cancel each other, at half Moon, it's neap tide.

−Woodstone (talk) 14:11, 18 October 2010 (UTC)

There is another factor which is ignored or misunderstood by most textbooks. Indeed, the only description I can find of it at the moment is in my book on local estuary fishing. This is earth-moon common center of gravity behavior. I hesitate to bring it up, as I have seen high school science teachers who teach just what this article does, get flushed and indignant when the subject is raised. But it is fairly easily understood. Does anyone want to discuss this? Rumiton (talk) 12:47, 7 January 2011 (UTC)

No! This issue has been brought up on this talk page several times already. This is when you try to explain the tides, using a rotating frame of reference. If your science teacher flushed when you brought it up, it probably was not out of shame but out of annoyance. It is not the way to make the tides "easily understood". Most often it is a simple but false way trying to "explain" the tides in as little words as possible. For instance, these texts hardly ever mention the difference between rotating and inertial frames of reference, which is vital to the theory. Read for example Tidal Misconceptions by Donald Simanek, a web page not overly beautifully formatted but what it lacks in aesthetics, it offers in clarity. Wikiklaas (talk) 17:12, 16 April 2013 (UTC)

Needs overhaul

This section puzzles me:

Because the oceans are fluid, but the Earth is mostly rigid, when the moon is directly overhead, it exerts a stronger gravitational pull on the fluid ocean than on the rigid Earth (since that pull is governed by the center of mass of the Earth, which is about 6300 kilometers further away from the moon than the surface ocean). Also, the Moon exerts a weaker gravitational pull on the ocean at the other end of the Earth than it does on the rigid Earth, thus pulling the Earth harder than it pulls the ocean.

First, the author notes that the moon exerts a stronger force on the oceans because the sea is fluid, and the Earth is solid. But then in the parenthesis, he seems to change his mind about this, and seems to state (very awkwardly) that the sea is CLOSER to the moon than the Earth which underlies it, and THAT is why we have tides. The whole thing is badly written and needs reworking. And I mean the whole thing.

To all the "experts" out there, especially those with long-winded analyses in the discussion pages, might I ask you to remember that this is a Peoples' Encyclopedia, not a formal scientific text on the subject. After nearly a decade, we are coming to see what the limits of community engagement in Wikipedia is. There are just far too many unnecessary points to be made, of unnecessary complexity and length. Cut it down, and remember the three golden standards: clarity, brevity, and accuracy. 125.7.71.6 (talk) 02:48, 13 January 2011 (UTC)

Actually, the parenthesis text does not contradict the preceding text. Perhaps the version at the Simple English Wikipedia would be more to your liking? See: simple:Tide. --- Barek (talk • contribs) - 02:58, 13 January 2011 (UTC)

I never even knew there WAS a simple Wikipedia version. See what you learn when wandering around the TALK section. I feel enlightened. And my reading will now be much simpler. — Preceding unsigned comment added by 70.166.197.2 (talk) 00:15, 22 September 2015 (UTC)

Location of High Tide/Low Tide Image

I removed the Hall's Harbour, Nova Scotia location from the caption to the title image which shows high and low tide views in a Bay of Fundy port. The boulder breakwater shown in the background does not exist at Hall's Harbour. See Hall's Harbour air views: http://www.hallsharbour.org/photoGallery/photoGallery-aerialViews.html The title image is a useful view and is certainly somewhere in the Bay of Fundy though. Maybe someone can come up with a more accurate location. Letterofmarque (talk) 17:47, 23 January 2011 (UTC)

Reverts

Three times the incorrect additions ([1], [2], [3]) by anon 72.177.55.143 (talk · contribs) aka new user JCM83 (talk · contribs) have been reverted. JCM83, next time you add this kind of content, you will need to provide a reliable source for your edit. Please have a careful read of wp:Reliable sources and wp:Verifiabilty DVdm (talk) 21:12, 14 February 2011 (UTC)

True, but doesn't he have a point though? I mean, it's simply not true that the surface of the earth is closer to the moon than it is to the center of the earth, right? You are right, though, he's not behaving properly, but the problem must lie with the summary of the citation, because the source couldn't possibly be saying that the surface of the earth is closer to the moon than it is to the center of the earth. It must be a miswording or something on our part, don't you think? Chrisrus (talk) 21:37, 14 February 2011 (UTC)

Quoting: "... the water furthest from the earth moves toward the moon more slowly than the earth itself ...", "... the water furthest from the earth is pulled toward the moon more slowly than the earth itself ...", "This is a fictitious force, similar to the Coriolis Effect". I'd really like to see a source that says that "earth moves toward the moon" or that "water ... is pulled toward the moon more slowly than..." or that "tidal force is a fictitious force, similar to the Coriolis Effect". DVdm (talk) 22:05, 14 February 2011 (UTC)

Agreed the referent of "than" is potentially ambiguous, and only one interpretation is right, the other silly, so it might be argued that readers should already have discarded the silly interpretation. But it is better to avoid ambiguity in the hope of forestalling misinterpretations. How about "The earth's surface on the side facing the moon is closer to the moon than the centre of the earth, which in turn is closer to the moon than the surface of the earth on the side facing away from the moon. Because the gravitational attraction of the moon reduces with distance, compared to the centre of the earth the side facing the moon has a slightly stronger attraction towards the moon, upwards, while the far side has a slightly weaker attraction, downwards. The result is a nett outwards force on both sides of the earth." Rather a lot of words, though. And I don't have a reference. NickyMcLean (talk) 20:24, 15 February 2011 (UTC)

Negative low, positive low, neutral good?

What is a "negative low tide" (as in the picture from Ocean Beach)? —Tamfang (talk) 23:58, 16 March 2011 (UTC)

Spring tides happen twice a month- at new moon and at full moon. Spring tides produce high tides that are higher than usual. Spring tides also produce low tides that are lower than usual.

Neap tides also happen twice a month but at quarter moons. Neap tides produce moderate tides. This is, tides that are not as high and not as low as regular tides. —Preceding unsigned comment added by 68.177.137.2 (talk) 19:13, 20 March 2011 (UTC)

So a neap-low would be "positive low"? Or what? —Tamfang (talk) 22:21, 20 March 2011 (UTC)

Not quite. Tide tables include predictions from models of not only "when", but also "by how much". The units of "when" are fairly simple. The date and time of day according to some standardized clock. But stating the least depth of water that my little harbour will experience during tomorrow morning's low tide in a universal way is not quite so obvious. Each predicted high and low tide has a height of water associated with it, that is a direct water height comparison with a historically calculated base-line for the region. In the USA I believe, and assuming diurnal tides (two full low high cycles every 25 hours or so), the base-line is the mean (summed average) of the lower of the two low tides each "day" over an extended period in the past. Implied by the use of an average for this, there will be some low tides whose depth falls below this historical arbitrary base-line. These are "negative low tide depths" (always around a spring tide), and are only an artifact of the chosen base-line measuring point. One could add "seven meters" to every predicted tidal water level and subtract "seven meters" from every depth indication on navigation charts, and everything would still work fine, BUT there would no longer be any special significance to those occasions formerly described as "negative low tide". Note: That "seven meter" adjustment was actually performed by 140 or so countries to produce the "Revised Local Reference" data used by PSMSL in Liverpool, UK (the tide data keepers of the world).ChrisJBenson (talk) 01:48, 29 June 2013 (UTC) [...]

Perhaps confusingly, there is a DIFFERENT phenomenon occasionally experienced when due to other longer term tide factors, the depth at low tide might actually exceed (be at a higher water level) than the next high tide. This is always around extreme neap tides, when the first order approximation of tidal range is overwhelmed by normally less significant second order terms in the tidal model for a specific location. This is rarer than the arbitrary base-line "negative low tide" but not impossible, and cannot be avoided by using a different measuring point. I have heard this called "negative tidal range", which seems reasonable to me. But obviously there is potential for confusion between the first described "negative low tide" (the water level compared with an arbitrary base-line), and the second described "negative tidal range" (comparison of the highest and lowest water levels within one tide cycle). ChrisJBenson (talk) 01:48, 29 June 2013 (UTC)

The "historical arbitrary base-line" has some significance in itself. On marine charts one can find soundings (local depth), giving the amount of water above the sea floor. When the mean low water at spring tide is used as the base-line (which at least in Europe is more common than the average over all low tides), this gives a fair indication of the minimum water depth at that spot, which of course is highly significant for sailors. Only during the rare occasions of negative low waters, they have to be carefull with respect to reading soundings from their charts, otherwise they can very quickly and easily find out whether it's safe to go there. Wikiklaas (talk) 12:35, 5 July 2013 (UTC)

I still don't understand what exactly that image is supposed to show. Could someone come up with a comprehensive caption?--FoxyOrange (talk) 10:01, 21 July 2013 (UTC)

I agree, if the term "negative low tide" appears in the caption then the article needs to explain what that means; otherwise use a different caption. -- Dr Greg  talk  10:43, 21 July 2013 (UTC)

Mixed tides

I am having difficulties with the concept of semidiurnal tides and mixed tides as described in the article. I see no difference in the figure between these two classifications. The text has "Semidiurnal range differences When there are two high tides each day with different heights (and two low tides also of different heights), the pattern is called a mixed semidiurnal tide.[7]" My understanding was that all semidiurnal systems had this feature. The higher high tide for the day gets lower over time and the lower high tide gets higher until one day they are about equal and then the reverse happens with the differnce between the high tide heights increasing.

I am trying to come to grips with tides at Portland on the south coast of Australia which appear to be semidiurnal for about 12 days and diurnal for two days (about half moon). Would this be what is meant by a "mixed tide"? 119.18.14.6 (talk) 01:25, 28 March 2011 (UTC)

Your interpretation of "mixed tides", as you stated it here, is correct. Wikiklaas (talk) 17:12, 16 April 2013 (UTC)

This entry is not written for a general audience

Iʻm surprised this concern hasnʻt been raised previously: this entry is an academic treatment of tides to its core and is among the most technical and scientific Iʻve ever stumbled across on Wikipedia. — Preceding unsigned comment added by 75.95.206.82 (talk) 07:40, 30 August 2011 (UTC)

History

This:

In the 2nd century BC, the Babylonian astronomer, Seleucus of Seleucia, correctly described the phenomenon of tides in order to support his heliocentric theory. He correctly theorized that tides were caused by the Moon, although he believed that the interaction was mediated by the pneuma.

must be wrong. You can't tell a heliocentric from a geocentric theory based on the tides. If his theory of tides supported his heliocentric theory, then his tidal theory was wrong William M. Connolley (talk) 21:17, 24 October 2011 (UTC)

There's a more detailed discussion at Seleucus of Seleucia#Tides. Saying twice that he was correct might be overkill. Maybe something like "largely correct" would be more accurate.   Will Beback  talk  21:26, 24 October 2011 (UTC)

I would go with 'mostly correct' in his assumptions of tital flow. Or something along those linesP0PP4B34R732 (talk) 21:28, 24 October 2011 (UTC)

What the above text says is that Seleucus correctly described "the phenomenon", which is basically about periods of high and low water and spring and neap tides. It doesn't say his theory on the phenomenon's causes was correct. Galileo Galilei did the same in 1616, using his theory of tides to support his heliocentric view. We now know that Galileo's theory of tides was utterly incorrect, but that doesn't mean he didn't use it to support his views. Wikiklaas (talk) 17:12, 16 April 2013 (UTC)

Tidal origin

Tides are actually not gravitational or centrifugal forces because gravity does not exist. Tides are a combination of light and magnetic forces at work. The moon's magnetic field is actually stronger than the earth! Even today, after the earth's magnetic field strengthened from its past because its spinning faster now. The magnetic field grazes the earth, but that is not the creation of tides, rather it is the creation light condensation. For instance, when the moon passes the equater at a close distance over the atlantic and pacific, the el nena and la nina effect come into being. El neno was 2 years ago, where in la nena and that is because of the light hitting the atlantic which evaporates the water and high cloud cover over earth's north pole. This only happens in the summer, so the weaker magnetic field is no longer the culprit for tidal creation. It is light itself and wind from the past. The tides have always existed because of the way the earth and moon's magnetic field touches. The moon is actually hit and damaged, going away from the earth because of the venusian collision, so is the sidereal earth. NO longer is the moon centrifugal like all other moons in our solar system. So the tides are not a part of the light evaporating effect, for magnetic fields cannot evaporate water as effectively as light. It is a combination of the magnetic field and moon's light because the earth does not change its path from the sun. The tides were here upon the creation of wind or upon the earth;s origin, when the moon was made. Air condenses, light evaporates, you get tides. Very complex, but true. Murriemir — Preceding unsigned comment added by 69.255.42.105 (talk) 19:01, 17 January 2012 (UTC)

Gravity Gradient

IMHO, "gravity gradient" is a likely search term, and so should be included in the article where convenient. 94.30.84.71 (talk) 12:26, 7 March 2012 (UTC)

Tides explained as simple orbital mechanics

Tides can be easily explained with no mathematics, based on what most people know of how satellites orbit the Earth. They are not caused by the moon's gravity "stretching" the Earth, or by centrifugal force. They are caused by the same laws that determine how quickly a satellite orbits the Earth.

Consider the Earth-Moon system. The Moon is approximately 2000 miles in diameter, while the Earth is approximately 8000 miles in diameter. The center of the mutual orbit lies inside the Earth's diameter, but the two planetoids are definitely orbiting one another.

Now consider two rocks on the opposite side of the Earth, one directly under the Moon and the other 8000 miles away. Left to themselves in space, the closer rock would move ahead of the farther rock. However, the Earth's gravity keeps them together at the same orbital speed. Taking the center of the Earth's mass into consideration, the rock under the Moon would move ahead of the Earth's center, while the farther rock would fall behind the Earth's center.

This is what the Earth's oceans try to do: the water directly under the Moon (allowing for friction with the Earth's sea bottoms and continents) tries to move ahead in orbit; the water 8000 miles farther away tries to fall behind; and the water at the same distance as the Earth's center flows from high point to high point as the Earth rotates. EndlessBob (talk) 01:36, 1 April 2012 (UTC)

Interesting notion but I honestly don't believe this is at all what is going on. While I have some knowledge of orbital mechanics, I do not have enough on the physics of tides or physics in general to properly point out all the flaws I suspect are in your above post. Btw, I was the editor who reverted your edits on the subject some time ago, just so everyone knows. I will let someone else more knowledgeable than I chime in on this one. Thank you. --RacerX¹¹ Talk to me^{Stalk me} 01:56, 1 April 2012 (UTC)

Yes, the two planetoids are orbiting their mutual centre of mass and the "further" rock (on the far side of the earth from the moon) is further out from the centre of mass than the "nearer" rock is and a rock/satellite in orbit moves faster the lower it is but on the face of it, a rock on the surface of the earth is not in orbit excepting in that the earth as a whole is in orbit. And the faster/slower aspect doesn't strike me as something everyone knows, especially knows without mathematics though I do recall a few SF stories in which this appears (John Brunner's The Sheep Look Up and Stand on Zanzibar drift into mind). And anyway, how would this interpretation handle the case of a moon being in retrograde orbit? The forces would be exactly the same (same distance). And the tidal motions are a consequence of the horizontal component (there's a diagram in the article) - suppose for simplicity that the moon is above the equator and consider the horizontal force effecton the equator at the two limbs of the earth as seen from the moon: on one limb the horizontal force is in the direction of rotation while at the other it is opposite. But the "speeding ahead/behind" notion is the same at both limbs. NickyMcLean (talk) 20:14, 1 April 2012 (UTC)

You can't explain tides by simple orbital mechanics, because tidal forces exist even if the two bodies aren't orbiting (either because they are infalling or because they are held apart by, e.g., a big stick) William M. Connolley (talk) 20:22, 1 April 2012 (UTC)

Tides

What is pulling the world's oceans on the side of the Earth that is not facing the moon? Nicpaul (talk) 16:55, 1 August 2012 (UTC)

The Moon pulls the Earth away from them. —Tamfang (talk) 17:53, 1 August 2012 (UTC)

Neap tide

When the moon is in the first quarter or the third quarter the positions of the moon and the sun with respect to the earth are at right angles to each other. The height of the tide produced by the moon's attraction is reduced by the sun's attraction on the waters in the low-tide regions. On such days, high tides are'nt as high usual and low tides are not as low. Such tides are called neap tides — Preceding unsigned comment added by Simransomani (talk • contribs) 13:13, 6 November 2012 (UTC)

The above is a paraphrase of a paragraph from the article. Are you suggesting a replacement? —Tamfang (talk) 18:07, 6 November 2012 (UTC)

Let's create a Tidal Constituent List!

There are articles with discography, movies, "best songs" etc, I think a list of the tidal constituents reported in scientific literature will be a valuable addition to Wikipedia. I'm new here, still don't know how to create an article, but I'll help if someone starts it. Or, I will if I manage to learn quick. — Preceding unsigned comment added by YtzikMM (talk • contribs) 13:06, 6 August 2013 (UTC)

Since you're new here, WP:WELCOME gives information on the site and editing. If you want to create a tidal constituents article, you can edit the Tidal constituents page – just replace the redirect with content. I personally have absolutely no idea what tidal constituents are, but if you want some content to start off the Theory of tides#Tidal constituents and Earth tide#Tidal constituents pages seem related. Welcome to Wikipedia. — Reatlas (talk) 13:49, 6 August 2013 (UTC)

Misleading primary cause of ocean tides

"Because the gravitational field created by the Moon weakens with distance from the Moon, it exerts a slightly stronger than average force on the side of the Earth facing the Moon, and a slightly weaker force on the opposite side. The Moon thus tends to "stretch" the Earth slightly along the line connecting the two bodies."

That is a *small* component... but the main component is due to the component of the moon's gravity that pulls toward the plane perpendicular to the line from the center of the Earth's mass to the center of the Moon's mass. It is NOT mostly due to the pull on the near compared to far. If it were, low tide would be AWAY from the moon... but it is not. I saw popular post based on this article and I hope to dredge up a PD image that shows the forces. Failing that, I hope to find some better wording. But it is complex and weird, and I am not sure it can be covered easily. *shrug*Unfriend13 (talk) 20:35, 2 December 2013 (UTC)

tidal force

See the picture from the article, copied here. The force at sub- and antilunar points is both away from Earth and the same size. The force at the "sides" is towards Earth and half the size of the sublunar one. At two circles in between, where the direction of the force changes from outward to inward, the tidal force is parallel to the surface. Some sources claim that the driving force is mostly the horizontal component, but that is not substantiated by theory. The tidal forces define equipotential planes around Earth, and the water will flow by pressure differences to fill one such contour (of course a steady state is never reach because of the Earth's rotation, and the presence of continents). −Woodstone (talk) 06:43, 3 December 2013 (UTC)

Exactly... this conflicts with the article text, quoted above. Actually though, the bulges are NOT the same size... so the text of the article is partly right... but the major effect is as you say... near and far bulges essentially identical... not a big bulge facing the moon, no bulge on the far side as stated in the article text quoted above.Unfriend13 (talk) 16:52, 3 December 2013 (UTC)

How about: "The Moon tends to "stretch" the Earth slightly in both directions along the line connecting the two bodies. Although the gravitational field created by the Moon weakens with distance from the Moon, and therefor exerts a slightly stronger force on the side of the Earth facing the Moon, and a slightly weaker force on the opposite side, this is not a primary driver of tides."Unfriend13 (talk) 20:07, 3 December 2013 (UTC)

Or: "The Moon tends to "stretch" the Earth slightly in both directions along the line connecting the two bodies. Although it is not a primary driver of tides, the moon does exert a slightly stronger force on the side of the Earth facing the Moon than on the opposite side." Unfriend13 (talk) 20:11, 3 December 2013 (UTC)

Or: "The Moon tends to "stretch" the Earth slightly in both directions along the line connecting the two bodies. Also, while it is not a primary driver of tides, the moon does exert a slightly stronger force on the side of the Earth facing the Moon than on the opposite side. Unfriend13 (talk) 20:23, 3 December 2013 (UTC)

Unless there is some objection, I expect to use one of those or something similar.Unfriend13 (talk) 06:58, 7 December 2013 (UTC)

I hoped to have made clear that any of those are not an improvement (and some even plain wrong). −Woodstone (talk) 16:46, 7 December 2013 (UTC)

No, you did not. Please do, by focusing on the wording in the article. The text leads people to believe that the primary cause of tides is "Because the gravitational field created by the Moon weakens with distance from the Moon, it exerts a slightly stronger than average force on the side of the Earth facing the Moon..." which is indeed "plain wrong". Please focus on the actual wording of the article. Please be specific about the wording, rather than making general statements. I hope some other interested editor will comment. The text currently is misleading the semi-technical press which STILL keeps confusing people into believing that tides are caused by the fact that the near side of the earth is pulled more than the far side. Unfriend13 (talk) 17:01, 7 December 2013 (UTC)

Here, for example is a fairly unhelpful image we can't use, [4]. Here is an interesting article that may be what you are driving at... that the article is correct [5].Unfriend13 (talk) 17:20, 7 December 2013 (UTC)

TIDAL

WHT IS TIDAL NEED INFOMATION STRAIGH AWAY NOT TIDE TIDAL — Preceding unsigned comment added by 94.2.244.177 (talk) 16:38, 12 February 2014 (UTC)

Amphidromic points

I came to learn about amphidromic points where this article states: Tide waves move around these points, counterclockwise in the N. Hemisphere and clockwise in the S. Hemisphere [15][16] (under the figure with amphidromic points).

On the other hand, the Wikipedia article entitled Amphidromic point states: Note that the rotational direction of tides around an amphidromic point bears no relationship to its location relative to the equator.

Furthermore, examples are given for clockwise and anticlockwise amphidromic points.

This is an apparent inconsistency, where the second article with examples appears to be correct.--Stm110 (talk) 10:21, 26 March 2014 (UTC)

Relevence of Bede's observations.

In the history section it is noted that Bede had described tides in Britain. But I don't understand the relevance of this observation. What about his observations were notable, who leveraged his observations, and how to they relate to, say, a theory of tides at all? My point, being, that if any random yokel just wrote "I saw the tides at my local beach and this is what they looked like", would this warrant inclusion in the discussion of the history of tides? I'm recommending that this sentence be removed as being simply irrelevant and unrelated to the history of tides. It would certainly not tell a less complete story about the history of the tides without Bede's "contribution". Qed (talk) 07:50, 30 March 2014 (UTC)

First principles derivation

I did a first principles derivation of the heights of tides on an earth with no land features. This is my own original work so I can not site it. I want to share this but I do not know where to put it. Right now it is at a website of mine. http://www.zen-ee.com/index.php/tide-height-derivation. Any ideas? — Preceding unsigned comment added by Anorderofmagnitude (talk • contribs) 17:32, 30 April 2014 (UTC)

Moon orbit and East-West vs North-South tides ?

This is a question. But the moon orbits the Earth , not above a certain latitude - but largely the orbiting still in East-West (or reversed ?) direction. And never from North to South (or reversed). Then I thougt - a well known location for huge amplitudes, is at the English channel. While, in a global perspective, not that far away , the Cattegat almost lackes tides. This is also a sea which is "in a North to South position". As the moon passes only very little water can be moved from east to west and back again. I presume that would have been different, if the moon passed along Cattegat in a north to south direction. Or i.o.w. are narrow straits in East-West directions more affected by the moon's gravity (although the moon's gravity has about the same strength everywhere), compared to a narrow strait in North-South direction ? I hope I could make myself understandable. Answeres appriciated. Boeing720 (talk) 09:47, 26 July 2014 (UTC)

Cause of indirect or opposite high tide: redux

I'm not an astronomer, but I think the explanation of tides in the subsection "Principal lunar semi-diurnal constituent" is incorrect:

Quote from the current article: "Because the gravitational field created by the Moon weakens with distance from the Moon, it exerts a slightly stronger than average force on the side of the Earth facing the Moon, and a slightly weaker force on the opposite side. The Moon thus tends to "stretch" the Earth slightly along the line connecting the two bodies. The solid Earth deforms a bit, but ocean water, being fluid, is free to move much more in response to the tidal force, particularly horizontally. As the Earth rotates, the magnitude and direction of the tidal force at any particular point on the Earth's surface change constantly; although the ocean never reaches equilibrium—there is never time for the fluid to "catch up" to the state it would eventually reach if the tidal force were constant—the changing tidal force nonetheless causes rhythmic changes in sea surface height."

This paragraph is also uncited. Since it implies that the source of tides is the "stretching" of the Earth along the Earth-Moon axis, this would imply that low tide occurs at the point when the Moon is at its zenith/nadir. But that's wrong, yeah? the high tide happens when the Moon is at its zenith/nadir, right? Plus, what force would "stretch" the Earth in the opposite direction from the Moon? wouldn't this force only operate on one side of the Earth, resulting in one principal lunar high tide per day?

The way that a friend explained it to me — which also seems to be the explanation at Myths about Gravity and Tides — is that the Moon (and, likewise, to a lesser extent, the Sun) exerts gravity on different bodies differentially based on the distance of the center of gravity of that body from the Moon. Thus, since the ocean on the near side of the Earth is closer, gravity acts more strongly on it than on the center of gravity of the Earth (i.e., the center of the Earth) — and conversely, the center of gravity of the water on the far side of the Earth is further from the Moon than the center of gravity of the Earth.

A neato explanation in lay terms is that on the near side of the Earth, the Moon pulls the water away from the Earth (high tide); on the far side of the Earth, the Moon pulls the Earth away from the water (also high tide). (You can copy that explanation if you want.)

Anyone with a stronger science background who can definitively speak to this? -CircleAdrian (talk) 01:16, 22 December 2014 (UTC)

There is no difference between your explanation and the one in the article. Under the differential gravitational force (= tidal force) the solid Earth strectches only a little, but the body of water around it stretches much more, because it is free to flow.−Woodstone (talk) 08:33, 22 December 2014 (UTC)