[rmolnav]

In the phase of the new moon or full moon, the pendulum clock will go faster.

I do agree with what said by Colin2B on #102. It´s a not working theory ...
And, apart from that, what quoted is erroneous. No honest experiment could show it is correct.
If seen otherwise somewhere, we all should be able to read the analysis of the experiment, and challenge it ...

[Bored chemist]

The most sensitive gravimeter, this is a conventional pendulum clock.

No
https://arxiv.org/pdf/1008.2884.pdf

In the phase of new moon and perihelion, the clock will go fast or slow.

We have known about the effect of tide on clocks for decades.
It agrees with real world physics rather than your fairy tale.
http://trin-hosts.trin.cam.ac.uk/clock/theory/rcl_report.pdf

[Bored chemist]

The waters of lakes, seas and oceans of the northern hemisphere rotate counterclockwise, while the waters of the southern hemisphere rotate clockwise, forming giant whirlpools.

Define "clockwise".
The real currents look something like this


Currents.png (39.87 kB . 343x376 - viewed 2785 times)

As you know, everything that rotates, including whirlpools, has the property of a gyro (yule) to maintain the vertical position of the axis in space, regardless of the rotation of the Earth.

It may not be vertical.


Temperature (and other effects) have a great influence in oxygen/ H2S levels.

[Yusup Hizirov (OP)]

The waters of lakes, seas and oceans of the northern hemisphere rotate counterclockwise, while the waters of the southern hemisphere rotate clockwise, forming giant whirlpools.

Define "clockwise".
The real currents look something like this


Currents.png (39.87 kB . 343x376 - viewed 2785 times)

As you know, everything that rotates, including whirlpools, has the property of a gyro (yule) to maintain the vertical position of the axis in space, regardless of the rotation of the Earth.

It may not be vertical.


Temperature (and other effects) have a great influence in oxygen/ H2S levels.

Steep turns of rivers and currents also have a radius, and as a consequence, possess the properties of a gyroscope.

[Colin2B]

As you know, everything that rotates, including whirlpools, has 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 side of the Sun, the whirlpools, rotating together with the Earth, overturn, due to which the whirlpools precess, and as a result there is a vertical movement of oceanic waters.

I’m afraid you have neglected to do your calculations and clearly do not understand gyroscopic precession.
Any force on the 2 coasts would be constant and would not result in a regular oscillation, and certainly not one that matched the observed relationship with moon and sun. Also, as @bored points out the actual currents don’t go anywhere near many of the coastal areas where tides occur.
I’m afraid your ‘best answer’ has already sunk your theory and adding to that error does nothing to help.

[rmolnav]

Quote from: Fermer05 on Yesterday at 08:27:39
In the phase of new moon and perihelion, the clock will go fast or slow.
We have known about the effect of tide on clocks for decades.

Again, you are saying not fully coherent things:
"We have known about the effect of tide on clocks for decades": correct ...
But that doesn´t necessarily mean that "In the phase of new moon and perihelion, the clock will go fast or slow", as I said on #103.

[Bored chemist]

Quote from: Fermer05 on Yesterday at 08:27:39
In the phase of new moon and perihelion, the clock will go fast or slow.
We have known about the effect of tide on clocks for decades.

Again, you are saying not fully coherent things:
"We have known about the effect of tide on clocks for decades": correct ...
But that doesn´t necessarily mean that "In the phase of new moon and perihelion, the clock will go fast or slow", as I said on #103.

Part of the problem is that "In the phase of new moon and perihelion, the clock will go fast or slow" doesn't actually tell you anything.
Do you mean it will be fast or do you mean it will be slow?
Not that it matters.
We know the effects of tides on clocks.
We know that answer to your question.
There's nothing new there.
It's all conventional science so it fails to back up the implausible claim made earlier.

the conventional explanation of tides, ..., does not correspond to reality.

In what way?

[Yusup Hizirov (OP)]

Are you able to show in detail how your whirlpool theory accounts for spring and neap tides and their periodicity?
I don’t think you can. Until you do there is very little point in posting more as you are just wasting our time.

It is believed that the maximum tides occur in the new moon for the reason that the Moon and the Sun affect gravity to Earth in one direction. And in reality the Moon revolving around the Earth, it accelerates that slows down the orbital velocity of the Earth, due to which in the new moon, the orbital velocity of the Earth is maximal, and in the full moon, the minimum.
The moon's gravity does not affect the ebb and flow.
The Moon's role is indirect, creating an uneven orbital velocity of the Earth.
The maximum orbital velocity of the earth happens three days after the new moon.
The minimum orbital velocity of the earth is three days after the full moon.

Once a year, the Earth as close as possible to the Sun (perihelion), while maximizing the Earth's orbital velocity and, as a consequence, the height of the tides increases.

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 amphidromic point is the center of the maelstrom
https://en.m.wikipedia.org/wiki/Amphidromic_point
this is the point in the ocean where the amplitude of the tidal wave is zero, the tidal wave "runs around" this point around the clockwise or counter-clockwise direction.

[Bored chemist]

The length of the tidal wave depends on the diameter of the whirlpool.

As far as I recall, you have no supplied any evidence of the existence of this whirlpool.

[rmolnav]

Part of the problem is that "In the phase of new moon and perihelion, the clock will go fast or slow" doesn't actually tell you anything.
Do you mean it will be fast or do you mean it will be slow?

Well, I supposed he meant respectively: faster when new Moon, and slower when perihelion ...
That is not exactly so, and as I had previously said on #103, is something different than "We have known about the effect of tide on clocks for decades", which is certainly correct ...

[Yusup Hizirov (OP)]

The length of the tidal wave depends on the diameter of the whirlpool.

As far as I recall, you have no supplied any evidence of the existence of this whirlpool.

The amphidromic point is the center of the maelstrom
https://en.m.wikipedia.org/wiki/Amphidromic_point
this is the point in the ocean where the amplitude of the tidal wave is zero, the tidal wave "runs around" this point around the clockwise or counter-clockwise direction. 
https://ru.pinterest.com/pin/279856564317644726/
https://ru.pinterest.com/pin/310537336784446596/
http://goo.gl/eYVTo6
The scheme of motion of a tidal wave, along the perimeter of North Atlantic planetary maelstrom.
https:/.../youtu.be/ZEhm_ONTQKc

[Colin2B]

The amphidromic point is the center of the maelstrom
https://en.m.wikipedia.org/wiki/Amphidromic_point
this is the point in the ocean where the amplitude of the tidal wave is zero, the tidal wave "runs around" this point around the clockwise or counter-clockwise direction.
The scheme of motion of a tidal wave, along the perimeter of North Atlantic planetary maelstrom.

You are confusing tidal currents with tidal waves. This is a tidal wave, which is a rotating standing wave not a current, the currents do not flow in the same direction as the wave. As there is no rotating current, whirlpool, maelstrom, there is no gyro effect.
There are currents in an amphidromic system but they flow across the amphidromic point and perpendicular to the rotating wave, these currents are turned to the right ie clockwise (N hemisphere) by Coriolis effect which is why the standing wave rotates counter-clockwise.
Even if there were rotating currents your gyro equation is incorrect and does not result in an oscillating tide.

[Yusup Hizirov (OP)]

Scheme of circular motion of a tidal wave along the perimeter of the North Sea
https://upload.wikimedia.org/wikipedia/commons/9/97/North_Sea_Currents.svg
https:/.../youtu.be/JNOpJxfI_Mg

[Bored chemist]

Very pretty.
You have shown that water, pushed past the uneven shoreline of the UK will swirl round.

So what?

[Colin2B]

Scheme of circular motion of a tidal wave along the perimeter of the North Sea

Yes, exactly what I was pointing out, wave not a current/whirlpool, hence no gyroscopic effect.

[Yusup Hizirov (OP)]

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.

To calculate the height of the tides, you need to know the rotation speed of the whirlpools and the depth of the sea.
To calculate the height of the tides of the Bay of Fundy, you need to know the speed of flow in the north of the Gulf of Maine, but this data is not.
In the Bay of Fundy, for the formation of tides 18 meters high, the flow in the north of the Gulf of Maine should be about 15 km / h.
http://www.gulfofmaine-census.org/wp-content/images/circulation/fig4.jpg

[Bored chemist]

At a height of the tides of 18 meters, the flow speed should be about 15 km / h.

Why?

[Yusup Hizirov (OP)]

To calculate the height of the tides in the Bay of Fundy, you need to know the speed of the whirlpool and the depth of the Gulf of Maine.
If anyone has this information, please write them.
https://goo.gl/images/cWTVG4
http://www.gulfofmaine-census.org/wp-content/images/circulation/fig4.jpg
https://www.tide-forecast.com/locations/Fundy-Offshore-1-Nova-Scotia/tides/latest

[Colin2B]

To calculate the height of the tides of the Bay of Fundy, you need to know the speed of flow in the north of the Gulf of Maine, .........To create tides 18 meters high, the speed of the current should be about 15 km / h.

But it isn’t.  Current in N of Gulf of Maine is around 2km/h. Another coffin nail in your theory.
I too would like to see your calculation, because the tide ht depends on more than the current and sea depth.

Again however, you are confusing 2 different things, these currents are not the same as the tidal wave circulating around the amphidromic points. The example you showed of North Sea is a different phenomena to the currents in the Gulf of Maine, try not to confuse the two.

[Yusup Hizirov (OP)]

To calculate the height of the tides of the Bay of Fundy, you need to know the speed of flow in the north of the Gulf of Maine, .........To create tides 18 meters high, the speed of the current should be about 15 km / h.

Current in N of Gulf of Maine is around 2km/h.

Correctly!
The height of the tides is now about 2 meters.
https://www.tide-forecast.com/locations/Fundy-Offshore-1-Nova-Scotia/tides/latest