[Yusup Hizirov (OP)]

The gravity of the moon does not reach the Earth:
Tidal forces are now successfully used, as a plug from many secrets of nature.
In fact, they exist only theoretically, but practically, they have not been fixed by any gravimetric device.
Acceleration of free fall at the equator is less than at the poles: at the equator is 978 gal, and at the poles - 983 gal.
There is no data on the fluctuation of tidal forces, if there were fluctuations, then there would be data.

The geostationary satellite of the Earth, acting as part of the Earth, is in zero gravity, at a distance of 35,000 km. from the Earth, and for some reason, is not exposed to the tidal influence of the Moon, despite the fact that the "tidal force" in the Earth-satellite system is 5 times greater than on Earth.
https://en.m.wikipedia.org/wiki/Geostationary_orbit
If the oceans and continents that are under the satellite, the moon raises, "by 0.5 m, then the satellite which is in zero gravity, the Moon should easily attract to itself, especially during the new moon, and the parade of the planets.
And it must not be forgotten that the oceans and continents are attracted by the Earth, and the satellite is in zero gravity ..

If the tidal forces existed, then the pressure of water and air would certainly react, especially during the parade of the planets.

[Colin2B]

The gravity of the moon does not reach the Earth:

It does. NIST and others measure the uplift of earth surface and measure gravimetric effect.

There is no data on the fluctuation of tidal forces, if there were fluctuations, then there would be data.

it is there if you search for it, but I suspect you won’t want to try and expect to be spoon fed.
There is a long history of measuring solid earth tides eg http://www.bidstonobservatory.org.uk/earth-tides/
Or
https://watermark.silverchair.com/125-1-106.pdf?token=AQECAHi208BE49Ooan9kkhW_Ercy7Dm3ZL_9Cf3qfKAc485ysgAAAbIwggGuBgkqhkiG9w0BBwagggGfMIIBmwIBADCCAZQGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMGT08BWeSTnQp5XrpAgEQgIIBZde8q6N4D3SLR6rpZesye1qyqgozQsoUwU5jFGJX63b283Jpn7glcxhnlUxEe811HvGOWMV3r6fFXM7CHU0BdSDpUvKC8JfebPc63xlRkPqPdtnLFnFLYjI_HFVkKCoX8o3IOec-48mdVr1XP_m104RSVDATU9I_1TpwrAXDnArwfGKsfx6qJhmmBtrwEKjgoRNgjIYuk87CJmuOPimbZ5qR6axPw_Q2ZsPed6lnf4zAZWpcZKs5qvKuGlJxD_O8aKDmyrXXClUKa3N-FEH1L92n9Fkw97ONkIFN3ezNKAuXdTVfDA9DFnXhiOnhb9ocHncJmnlzrohjoYAPjByyUIJ4kArAzqo21LLC5tsCG5mSRj8BI_IWtoWX37FIlV-kI18alfy2Ag6t8lNYGifeP3ZphPPfWkiyl87aquCmA7O6aKuCyKSVfrq6zKUXcpbsfh6TT_kWMHrf-X9J_d8wxDXhWl331A
You could also search for papers from NIST at their Boulder site.

The geostationary satellite of the Earth, acting as part of the Earth, is in zero gravity,

at a distance of 35,000 km. from the Earth, and for some reason, is not exposed to the tidal influence of the Moon,

It is a common misconception that satellites are in zero gravity, this is not true, if they were they could not maintain an orbit.
GPS satellites have to have timing corrected for influence of moon which deflects their orbit.

These posts are not only going off topic but are making ridiculous, unscientific suggestion and so will be moved.

You might also answer this question

To make calculations you need time.
First you need to look at the existing calculations.

We can't look at your calculations unless you post them.
If you are saying you have not done the maths, what are you using as the basis of your post?

If you state that the radius is too small, you must have made a calculation in order to reach that conclusion, please show it.

[evan_au]

The gravity of the moon does not reach the Earth

Incorrect.
Gravitational force follows an inverse-square law, so it reaches "to infinity", although it gets weaker with distance.
Because tides are created as the difference between the gravitational force at two points, tidal forces follows an inverse-cube law.

Acceleration of free fall at the equator is less than at the poles: at the equator is 978 gal, and at the poles - 983 gal

Earth is rotating, and so centrifugal force (and its effect on the equatorial bulge) means that surface acceleration is about 0.5% lower at the equator than at the poles.

The variation in Earth's surface acceleration due to gravity is about +/-5% at different points on the surface - this is much larger than the variation between equator and poles.
Columbia is about +5%, while the Yucatan area is about -5%. They are both fairly close to the equator, and each other.
To understand local variations, you must understand the local geology and history.
See: https://en.wikipedia.org/wiki/Gravity_of_Earth#Latitude

The geostationary satellite of the Earth... for some reason, is not exposed to the tidal influence of the Moon

Not correct.
The orbit of geostationary satellites must be continually adjusted to compensate for the gravity of Moon and Sun. When their stationkeeping fuel is exhausted, the satellite must be retired.
https://en.wikipedia.org/wiki/Orbital_station-keeping

If the tidal forces existed, then the pressure of water and air would certainly react,

Air and water do interact - if you mix air and water (eg during a rainstorm), the water moves down and the air moves up.
- The pressure in water (eg a lake) increases much more quickly as you go down than it decreases as you move up in the air.
- This is due to the fact that water has a density of 1 kg/liter, while air has a density of around 1g/liter.

Tides also occur in the atmosphere - but the pressure changes are dominated by the heating effects of the Sun.
See: https://en.wikipedia.org/wiki/Atmospheric_tide

the parade of the planets.

Due to their much greater distances from Earth, and the inverse-cube effect of tidal forces, the planets have a minimal effect compared to the Sun and the Moon.
- The next most likely is Jupiter, with 1% the mass of the Sun, but even at its closest, it is 4 times farther than the Sun.
- So the tidal influence of Jupiter = 0.01 / 4³ = 0.00015 of the tidal effect of the Sun
- And the Sun's tidal effect is less than the Moon

[Yusup Hizirov (OP)]

As the tidal forces of the earth stopped the rotation of the moon, if the side of the moon turned to the earth is concave, and the underside of the moon is convex.
Laser altimeters of the ships "Apollo-15, -17" showed that the visible side of the Moon lies below the average level, and the invisible side is higher than it.
In cosmonautics it is already a proven fact that the area of the attraction of the moon is limited to 10 thousand kilometers from the surface of the moon,
artificial satellites of the moon with an orbit radius of more than 10 thousand km. break from the orbit.

[Bored chemist]

In cosmonautics it is already a proven fact that the area of the attraction of the moon is limited to 10 thousand kilometers from the surface of the moon,

Are you  going to back up that assertion with evidence, or should we all just ignore it and laugh at you?
Here's a link to a page where they discuss the effect of the moon on satellites at much greater distances.
https://www.reddit.com/r/askscience/comments/52vinv/how_much_if_at_all_does_the_moons_gravity_effect/

[Yusup Hizirov (OP)]

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 formula A = V1 • V2 / t does not fully reflect the essence of the phenomenon of tides.
The amplitude of the tides also depends on the size of the whirlpools, the amount of water under the whirlpool, the distance from the coast to the whirlpool, and the inclination of the Earth's equator relative to the Earth's orbit.
You can predict the height of the tides and experimentally, catching the connection between the height of the tidal wave with the rotation speed of the whirlpool, in this area, using a remote tachometer

[Bored chemist]

You said this bit before

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).

And someone pointed out that the earth has more than one orbital velocity.

You seem not to understand that this makes your equation meaningless.

Why did you post it again?
Were you hoping we would forget?

[Kryptid]

In cosmonautics it is already a proven fact that the area of the attraction of the moon is limited to 10 thousand kilometers from the surface of the moon,
artificial satellites of the moon with an orbit radius of more than 10 thousand km. break from the orbit.

Wrong. Gravimeters on Earth can detect the passage of the Moon overhead by the way it changes the strength of the local gravitational field.

[Yusup Hizirov (OP)]

Also, there is no convincing evidence and facts, deformation of the Earth's continents under the influence of tidal forces.
Although deformation of the continents can be easily verified by aiming the telescope to the top of a mountain or to a geostationary satellite.

[opportunity]

This was moved to new theories, so I thought I would say something.

High tide close to moon has an effect on the opposite side of the planet.

Should have.

The question is how that oceanic fluidity works.

Apart from saying it works, we're asking "how" it works.

Is the Moon "directly" responsible for tides? Of course. Is there an opposite effect on the planet? Of course.
 Why?

[opportunity]

This process would have gone on for so many years before (or after?) it became harmonic, right?

[Colin2B]

This was moved to new theories, so I thought I would say something.

The thread you posted in is in Physics, Astronomy & Cosmology not in the lighter section! It hasn’t been moved.
This is the one that was moved https://www.thenakedscientists.com/forum/index.php?topic=73127.0

You are restricted to the lighter sections

[Yusup Hizirov (OP)]

The existence of tidal humpbacks from textbooks would mean a constant movement
scary trillion tons of sea water on the world's oceans.
This phenomenon in nature as something not observed.

[Colin2B]

As the tidal forces of the earth stopped the rotation of the moon, if the side of the moon turned to the earth is concave, and the underside of the moon is convex.
Laser altimeters of the ships "Apollo-15, -17" showed that the visible side of the Moon lies below the average level, and the invisible side is higher than it.

This makes no sense. No part of the moon is concave other than the craters.

In cosmonautics it is already a proven fact that the area of the attraction of the moon is limited to 10 thousand kilometers from the surface of the moon,
artificial satellites of the moon with an orbit radius of more than 10 thousand km. break from the orbit.

This is incorrect.
There is nothing to stop an object going into lunar orbit other than it’s speed, however, many lunar orbits are unstable. In low lunar orbits mass anomalies (mascons) make the orbits unstable, whereas orbits far out get into the influence of the earth’s gravitational field eventually pulling them off course.

Also, there is no convincing evidence and facts, deformation of the Earth's continents under the influence of tidal forces.

The evidence is only unconvincing if you choose to ignore it.
Plenty of studies on the net, search and ye shall find.
Here is one on the measurement of moon gravity on earth. https://academic.oup.com/gji/article-pdf/125/1/106/5889847/125-1-106.pdf
There are also regular measurements by a team working on Mt Etna who monitor deformations of the ground by the moon.
NIST at Boulder also measure these.

Although deformation of the continents can be easily verified by aiming the telescope to the top of a mountain or to a geostationary satellite.

Neither of these would work.

The existence of tidal humpbacks from textbooks would mean a constant movement
scary trillion tons of sea water on the world's oceans.
This phenomenon in nature as something not observed.

Have you ever worked out how much water flows with the tides? The numbers are scary because there is a lot of water in the oceans.
And yes it is observed in nature.

To make calculations you need time.
First you need to look at the existing calculations.

We can't look at your calculations unless you post them.
If you are saying you have not done the maths, what are you using as the basis of your post?

You still haven’t answered this!
As you don’t seem willing to back up your claims and insist on posting incorrect information, I see no point in continuing this discussion.

[Yusup Hizirov (OP)]

Mathematically, it is very difficult to express a theory.
I think the main physical expression of the theory.
That's what people write who understand the meaning of the theory:

All very accurately and competently, well done, it's amazing that this version was not announced and realized earlier, because circular movements of tidal waves along the coasts were known for a long time. You need to specify your full name at the beginning of the article, the country should know its heroes, discoverers, and also publish in the VIKI article in the TINS ​​section, I think your
The determination of the cause of tides should take place in all encyclopedias. And more boldly, formulate not as a possible version, but as the truth in the last resort ..
----------------------------------
I read honestly the whole article - not 15 minutes, of course, a little more to penetrate. Even in some links came with maps. For the "ordinary person" it will be uninteresting and pointless - unnecessary "garbage" in the head, but for those who are previously fond of or continue to be fond of astrophysics, astronomy or planetary climate, from the scientific point of view - very much.
That for me - this article and the hypothesis set forth in it seemed to me close, tk. I was once a lover of astronomy and re-read many scientific papers, "Astronomical annual calendars" and similar accompanying literature.
Virtually all the calculations in the article are quite logical. The article liked and appeared a vision of the causes of tides and tsunamis on the other hand. Moon tides - for a long time it is clear that this is a "hypothesized ears" hypothesis, turned into a physical law for schoolchildren. The author - offset for the depth of penetration into the topic! Very interesting! Thank you!

[Bored chemist]

Mathematically, it is very difficult to express the theory

Then perhaps we should stick with ordinary Newtonian physics which is expressible as maths and which has explained tides for a few centuries.

[Yusup Hizirov (OP)]

Galileo called the theory of tides frivolous, a sad return to the realm of mystical ravings and preferred to explain the tides, the rotation of the Earth.
Charles Darwin wrote in 1911: "There is no need to seek antique literature for the grotesque theories of tides." However, sailors manage to measure their height and use the opportunities of tides, without having an idea of the real causes of their occurrence.

[Bored chemist]

Galileo called the theory of tides frivolous,

Unless he could time travel, he didn't call the Newtonian theory frivolous.

Charles Darwin wrote in 1911:

Are you talking about the famous Charles Darwin who had died about 30 years earlier, or some other guy with the same name?

[Yusup Hizirov (OP)]

I apologize! Darwin George Howard.
https:/.../en.m.wikipedia.org/wiki/George_Darwin

[Bored chemist]

The wiki page cites this as one of his works, and it agrees pretty much with the Newtonian version.

http://www.1902encyclopedia.com/T/TID/tides-03.html

What point did you think you were making?