[Yusup Hizirov (OP)]

What's wrong here, give more objective criticism.

I told you what's wrong.
You ignored it.

I answered all the questions (95 answers), but not all questions were correctly put, I could not answer them.

I have no problems answering your questions. I easily answer all your questions if they are correctly asked and translated.

[Kryptid]

Please specify the answer.
1. How many times a day are earthquakes?
2. What is the applied earthquake hour.
3. Are there earthquakes, on the other side of the Earth, to form a second tidal hump?

Earthquakes don't cause tidal humps, but that doesn't matter because your original question was about tidal waves. If you weren't talking about tidal waves, then you should not have used that term. The difference has already been explained to you.

[Yusup Hizirov (OP)]

In the Laptev Sea, the tide is well expressed, has the character of an irregular semidiurnal wave. A tidal wave enters, from the north and spreads to the shores, fading and deforming as they move towards them. The magnitude of the tide is usually small, mostly about 0.5 m. Only in the Khatanga Bay the sweep of tidal vibrations exceeds 2.0 m in syzygy. https://en.m.wikipedia.org/wiki/Laptev_Sea

How does the tidal wave move from the north pole to the south.
Who can constructively answer this question.

[Bored chemist]

How does the tidal wave move from the north pole to the south.
Who can constructively answer this question.

Anyone who knows what the words mean.
If there is an earthquake at sea somewhere north of you then there will be a tidal wave from the north.

[Yusup Hizirov (OP)]

1.Long ago when I was a boy, I already realized that, e.g., strongest high tides were always at same time, and when full or new Moon, in Atlantic coast of Spain where I was in summer holiday.
2. And nowadays, just seeing the Moon in Madrid sky at a certain moment, I can tell how high is the tide at mentioned coast, and also if the tide coefficient is high or low, without any complicated maths or any additional information.

This is too much.
During the tide on the Spanish coast of the Atlantic Ocean, the Moon must hang over Europe.
But the Moon can hang over America, or anywhere.
Focusing on the moon, no one can determine the time of the tide.

[Yusup Hizirov (OP)]

1.Long ago when I was a boy, I already realized that, e.g., strongest high tides were always at same time, and when full or new Moon, in Atlantic coast of Spain where I was in summer holiday.
2. And nowadays, just seeing the Moon in Madrid sky at a certain moment, I can tell how high is the tide at mentioned coast, and also if the tide coefficient is high or low, without any complicated maths or any additional information.

This is too much.
During the tide on the Spanish coast of the Atlantic Ocean, the Moon must hang over Europe.
But the Moon can hang over America, or anywhere.
Focusing on the moon, no one can determine the time of the tide.

Who can answer this question.

[The Spoon]

1.Long ago when I was a boy, I already realized that, e.g., strongest high tides were always at same time, and when full or new Moon, in Atlantic coast of Spain where I was in summer holiday.
2. And nowadays, just seeing the Moon in Madrid sky at a certain moment, I can tell how high is the tide at mentioned coast, and also if the tide coefficient is high or low, without any complicated maths or any additional information.

This is too much.
During the tide on the Spanish coast of the Atlantic Ocean, the Moon must hang over Europe.
But the Moon can hang over America, or anywhere.
Focusing on the moon, no one can determine the time of the tide.

Who can answer this question.

Probably somebody who believes the earth is flat...

[rmolnav]

This is too much. During the tide on the Spanish coast of the Atlantic Ocean, the Moon must hang over Europe.
But the Moon can hang over America, or anywhere.
Focusing on the moon, no one can determine the time of the tide.

Rubbish!!
From Madrid, just looking at the moon, I can say many things about the tides on the Spanish coast of the Atlantic Ocean, such as:
1) Rising full moon (always at sun set): about an hour latter, spring low tides ...
2) Full moon at midnight (always at local meridian zenith): about an hour latter spring high tides ...
3) First day visible new moon app. at sun set (some 360/14 degrees eastwards from the sun: in app. 3 hours low tides.
4) Quarter moon (when "growing": not sure now about its correct name) at local meridian zenith (always app. at 18:00): high neap tide in about an hour ...
Is that "too much" ?? Not at all whatsoever !!
Please keep in mind the given figures come directly from my head ... It could be some time earlier or later, especially because of the gap between moon meridian and high tide meridian, due to the fast spinning of earth, which doesn´t allow the high tide bulge to catch up with sublunar meridian ...
Apart from that gap, also maximum and minimum tide size are one or two days after full/new moon exact day, and quarter moon exact day ...

[Bored chemist]

This is too much.
During the tide on the Spanish coast of the Atlantic Ocean, the Moon must hang over Europe.
But the Moon can hang over America, or anywhere.

The Moon can be over Europe or over America- but not at the same time.
High tide today is due at about 7pm in Washington DC local time. Local time is 5 hrs behind UK time so that's about 12 midnight UK time
And high tide is due at about 02:45 on the West coast of Spain. They are 1 hr ahead of us so that means it will be high tide 03:45 UK time
So, very roughly, and ignoring things like daylight saving time, the high tides on the opposite sides of the Atlantic are something like 4 hours apart.

So, (broadly) when the Moon is over New York, it's high tide there, and when the Moon is over Spain, it's high tide there.
The Earth rotates under the Moon
In the 4 hours or so time difference between high tide on the 2 sides of the ocean the world will rotate about 4/24 of a circle that's about 60 degrees.
The longitude of Washington is about 77 degrees West and that of the  Spanish coast is about 8 degrees West
So the difference is about 70 degrees.
Given the rough nature of the calculation, that's close enough to the expected value.

Why do you think this is a problem?

[Yusup Hizirov (OP)]

В ago when I was a boy, I already realized that, e.g., strongest high tides were always at same time, and when full or new Moon, in Atlantic coast of Spain where I was in summer holiday.
And nowadays, just seeing the Moon in Madrid sky at a certain moment, I can tell how high is the tide at mentioned coast, and also if the tide coefficient is high or low, without any complicated maths or any additional information.

Questions.
You tried to determine the height of the tides by conducting the moon and the sun, so the forecast will be more accurate.
2. On which coast of Spain you determined the height of the tide, you can find out what the speed of the current is.

[rmolnav]

Questions.
You tried to determine the height of the tides by conducting the moon and the sun, so the forecast will be more accurate.
2. On which coast of Spain you determined the height of the tide, you can find out what the speed of the current is.

1) Only moon location counts for whether there is a high or a low tide, or the intermediate hights ... Sun works independently, ALWAYS with its own high tidal effect app. an hour after noon and midnight ... Those sun-related effects, to be added to moon´s, are the main cause of actual oscillation of tide size.
 2) Yesterday we were talking about tides at Spain Atlantic cost, and particularly at Gulf of Cádiz. From Madrid there is a small additional delay, which I didn´t consider ... At Galitzian atlantic coast everything happens a little later ...
ALL THAT demonstrates moon (and sun) pull, and inertial centrifugal forces, are the real cause of global oceanic tides ...
Another thing is local effects, whirlpools, currents generated, etc.   

[Yusup Hizirov (OP)]

The waters of the Bay of Cadiz rotate counter-clockwise, forming a huge whirlpool-gyroscope, which, precessing, reflects a tidal wave along the entire perimeter of the Bay of Cadiz.
A similar pattern of tides and tides is observed in all lakes, seas and oceans.

https://goo.gl/images/xQtQ7k
https://ru.m.wikipedia.org/wiki/Кадисский_залив
https://en.m.wikipedia.org/wiki/Ocean_current

[Yusup Hizirov (OP)]

The amplitude of the tides in the Bay of Cadiz depends on the strength of the eastern winds, in the north of the Gulf of Cadiz, the floods of the rivers, Guadalquivir, Guadiana, Odiel, Tinto and Guadal.

[opportunity]

The winds are funnelled by the equatorial play.


I agree.


Yet some would argue all this feeds back, namely a greater condensation on the equatorial region creating greater rains and thus more wind activity "on" the equatorial rim "outwards".

[Bored chemist]

Why do you think this is a problem?

[rmolnav]

The waters of the Bay of Cadiz rotate counter-clockwise, forming a huge whirlpool-gyroscope, which, precessing, reflects a tidal wave along the entire perimeter of the Bay of Cadiz.

You may be right, but still you can check that in a few days (4 or 5) both at Cadiz and Huelva will have spring tides (full moon on the 10th), app. 2.5 hours after noon and midnight (because of the previously mentioned gaps), and with only some 5 min. difference ...
Moon (especially) and sun locations in our skies clearly are the paramount cause of tides ... 

[Yusup Hizirov (OP)]

Moon (especially) and sun locations in our skies clearly are the paramount cause of tides ... 

The amplitude of the tides in the Bay of Cadiz depends on the strength of the eastern winds, in the north of the Gulf of Cadiz, the floods of the rivers, Guadalquivir, Guadiana, Odiel, Tinto and Guadal.
And the role of the Moon is indirect, creating an uneven orbital velocity of the Earth.

[Bored chemist]

creating an uneven orbital velocity of the Earth.

I think astronomers would have noticed.
Would you like to cite the evidence of this?
Alternatively, would you like to accept that you are simply wrong?

[Yusup Hizirov (OP)]

creating an uneven orbital velocity of the Earth.

I think astronomers would have noticed.
Would you like to cite the evidence of this?
Alternatively, would you like to accept that you are simply wrong?

This question has already been answered, and it was noted as the best answer.
https://www.thenakedscientists.com/forum/index.php?topic=73127.msg542160#msg542160
I would like to hear the opinion of rmolnav.

[Colin2B]

creating an uneven orbital velocity of the Earth.

I think astronomers would have noticed.
Would you like to cite the evidence of this?
Alternatively, would you like to accept that you are simply wrong?

This question has already been answered, and it was noted as the best answer.
https://www.thenakedscientists.com/forum/index.php?topic=73127.msg542160#msg542160
I would like to hear the opinion of rmolnav.

It was only marked as best answer by yourself, no one else believes it.

Your whirlpool theory is seriously flawed and you have yet to provide an answer.

The length of the tidal wave depends on the diameter of the whirlpool. And the height of the tidal wave depends on the rotation speed of the whirlpool of the orbital velocity of the Earth, and the time of the tilting of the whirlpool (12 hours).
A = V1 • V2 / t
where: A is the amplitude of the tidal wave (precession angle).
V1 - rotation speed of the whirlpool.
V2 is the orbital velocity of the Earth.
t - the time of tilting of the whirlpool (12 hours).

The first problem with this formula is that it does not return an amplitude, it produces a number of square metres per second. That is clearly wrong.

Next problem involves timing:

As is known, everything that rotates, including whirlpools, possess the property of a gyro (yule) to maintain the vertical position of the axis in space, regardless of the rotation of the Earth.
If you look at the Earth from the Sun, the whirlpools, rotating together with the Earth, turn over twice a day, due to which the whirlpools precess (swing by 1-2 degrees) and reflect the tidal wave around the entire perimeter of the whirlpool.

A rigid gyroscope on earth will indeed align itself to the sidereal day which is 4mins shorter than the solar day. However, that gives a number of problems for your theory.
Firstly, we know from observation that the main tide period is period is about 12hrs 25mins - half a lunar day - so two tides take 24hrs 50mins which is the average time for the Earth to rotate once relative to the Moon. This is a hugh difference from your difference of 4mins in 24hrs making your double tide period 23hrs 56mins.  Clearly your predictions can never work with this theory.