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wiki has this picture of the tidal bulge:

enter image description here

but says that:

The average tidal bulge is synchronized with the Moon's orbit, and Earth rotates under this tidal bulge in just over a day. However, Earth's rotation drags the position of the tidal bulge ahead of the position directly under the Moon. As a consequence, there exists a substantial amount of mass in the bulge that is offset from the line

On the other hand, this academical article states the opposite:

as we move with respect to both tidal bulge and moon, the moon crosses our meridian before we experience the highest tide.] How early? Some books show misleading diagrams with the symmetry axis of the tidal bulges making an angle of 30° or more with the moon. In fact, the angle is only 3°, so the tides are late by about 24(3/360)60 = 12 minutes

through the centers of Earth and the Moon

Can you say what is the real position of the bulge? it is surely a hard fact not open to interpretations. Do you have access to sites that say where is now the moon and where is the tidal bulge?

According to the basic laws of conservation of momentum and energy, a mass moving to a greater radius should be slowed downm and if we add friction the lag of 12 minutes or more makes sense. If the highest point of the bulge is forward wrt to the vertical of the moon, can you explain what stronger force pushes the water in the direction of the spin?

I found this site that gives the position of sun and moon, but doesn't show tides enter image description here

This animation, on the other hand, shows that there is no bulge

Thered areas experience a rise over 1 m, the blue one a depression of over 1 m: enter image description here

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  • $\begingroup$ Just making sure, but you understand that they aren't saying the opposite? The bulge moves ahead of the Moon viewed from space, but the Moon apparently moves ahead of the bulge, viewed from the surface of the Earth. They're saying the same thing but from two different perspectives. $\endgroup$ – userLTK Aug 1 '16 at 11:38
  • $\begingroup$ @userLTK, can you explain this: the bulges move to the right, also the earth is moving to the right, are the bulges moving faster than the earth? $\endgroup$ – user13628 Aug 2 '16 at 4:40
  • $\begingroup$ Look at the top picture and imagine the Earth turns, following the arrow. For practical purposes the Moon is stationary though it does orbit, but much more slowly than the Earth rotates. The bulge rotates with the Earth. Now imagine you are standing on the Earth and as the Earth turns (West to East) the Moon in the sky moves East to West. The view is reversed simply by the position from which you look at it. It's like feeling the wind blow right to left, then you turn around and you feel it blow left to right. Both are correct but the change depends on where you view the movement. $\endgroup$ – userLTK Aug 2 '16 at 12:12
  • $\begingroup$ Tides move across the Earth, so I'm not sure what you mean when you ask, is the tide moving faster than the Earth. The Earth turns West to East, Tides move across the Earth East to West, same general direction as the Moon and stars though the shape of land and ocean floor does prevent Tides from being simply longitudinal. $\endgroup$ – userLTK Aug 2 '16 at 12:18
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    $\begingroup$ That specific question might be better suited for Earth Science. Tidal movement is complicated and driven by the shape of the Ocean floor and land as well as the apparent movement across the sky of the Moon and Sun. I'm very far from an expert, just a hobbyist astronomer. $\endgroup$ – userLTK Aug 2 '16 at 12:36
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There is no great contradiction. The offset of the tidal bulge is about 3 degrees. It is exaggerated in diagrams for clarity. The diagram is correct, but not to scale.

This causes the tides to be slightly late. Imagine a person standing on the Earth of the diagram, with the moon directly overhead. The tidal bulge is on their left. The rotation of the Earth will take them towards the left (the moon is also orbiting but its motion is much slower), so a little later (12 min later) they will reach the maximum of the tide. The maximum is delayed by about 12 min.

Actual flows of water around the coast are driven by this tidal bulge, but are complex effects of local topography. The actual flows of water are highly non-linear, including multiple locations at which there is no tide.

enter image description here

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As the same wikipedia article states

An equilibrium tidal bulge does not really exist on Earth because the continents do not allow this mathematical solution to take place. Oceanic tides actually rotate around the ocean basins as vast gyres around several amphidromic points where no tide exists. The Moon pulls on each individual undulation as Earth rotates—some undulations are ahead of the Moon, others are behind it, whereas still others are on either side. The "bulges" that actually do exist for the Moon to pull on (and which pull on the Moon) are the net result of integrating the actual undulations over all the world's oceans. Earth's net (or equivalent) equilibrium tide has an amplitude of only 3.23 cm, which is totally swamped by oceanic tides that can exceed one metre.

So the cartoon that is drawn is an idealised scenario that gets messed with due to the real physical geography of the Earth and its oceans. The tides are also affected by the Sun. However, any asymmetry in the same sense as is shown in the cartoon will result in tidal torques and tidal accelerations in the sense that the Earth's rotation is slowed whilst the Earth-Moon system is tidally accelerating to larger separations and longer orbital period.

The "bulge" leads the meridian position of the Moon in your diagram because there is a lag in the response to the tidal force. This is akin to a driven oscillator where the damping leads to a lag in the response with respect to the driving force. And the Earth simply rotates by a certain amount during the lag time. The periodicity of the driving force and response is 12h25m. The actual lag and phase shift will depend on local geography but should be longer than 12 minutes. This source (which I strongly recommend you read) says that in the Southern ocean - which has the "cleanest response" to the tidal driving - the bulge rises about 2-3 hours after the Moon crosses the meridian.

I don't understand the rest of your question; the tides are raised by tidal forces due to the Moon (and Sun).

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  • $\begingroup$ So, the article is right, and the lag is even greater. Why is it represented ahead of the moon? If the main response lags by 12-25 minutes, then the bulge(s) should slow down the moon and not accelerate it (increasing AM and the radius of the orbit). That is my question: if the bulge is ahead, what pushes it ahead?. $\endgroup$ – user13628 Aug 1 '16 at 9:49
  • $\begingroup$ @user104372 It is just the rotation of the Earth. The tides raise the bulge under the Moon but the spin of the Earth carries it away (to some extent). $\endgroup$ – Rob Jeffries Aug 1 '16 at 10:00
  • $\begingroup$ Can you say (roughly) how many Km is the bulge ahead of the vertical? $\endgroup$ – user13628 Aug 1 '16 at 10:42
  • $\begingroup$ We have to also consider : friction, inertia, the slowing down due to the greater radius, the depressions ahead of/after the peak, and the fact that for 25' the pull is in the opposite direction, the drag when the moon goes on a continent, etc; after all that, is there enough left to accelerate the moon? Is there any quantitative study to assess the real mass and height of the water ahead of the vertical, or we just guess it must be so, since the moon is receding? $\endgroup$ – user13628 Aug 1 '16 at 11:15
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    $\begingroup$ Unfortunately, the key part of that wikipedia article ("Earth's net (or equivalent) equilibrium tide has an amplitude of only 3.23 cm) should have been flagged with "citation needed." I would love to know the source of that extremely precise value. A tenth of a millimeter precision is ridiculous. Current tidal models are not anywhere close to that precise. They're not good to the centimeter level, AFIAK. Nonetheless, plus one. $\endgroup$ – David Hammen Aug 1 '16 at 13:38

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