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  4. Why do we have two high tides a day?
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Why do we have two high tides a day?

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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #160 on: 02/08/2018 20:18:48 »
Quote from: Le Repteux on 02/08/2018 17:42:01
but it is not clear how a small vertical motion in the middle of the oceans can transform into huge ones on the coasts.

The Earth's rotation moves the bulges towards the coasts which block their way, so they bunch up there. It's not nearly that simple though, as I'll explain further down.

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That's what tsunamis do, but they are considered as a fast wave that breaks on shallow waters, so they can break in any direction from the impact point, while the two opposing tidal bulges are only going west and could only break on eastern coasts if they were a wave, which is not what is observed.

These bulges are tiny and don't add up to the size of tsunamis. What you actually get is resonance in ocean basins and seas, and in different parts of the world you can get no tides at all, one tide per day, two tides per day, four tides per day, ... 128 tides per day. Think about a young child on a swing - you can push them each time they come back to you, or you can push them every second time, but they swing at the same rate regardless - the basin determines how many high tides a day it can handle, and the rotation of the earth puts in regular nudges to maintain the oscillation.

Now let's think about the size of the bulges. To make a big bulge, you have to move a lot of water, but we simply don't see any significant amount of water rushing around as a bulge crosses an ocean. The bulge hardly shows up at all. What must actually happen is that you get a reduction of pressure as the sea tries to lift, but water is so incompressible that it can change in pressure a lot without any measurable change in volume. These "bulges" thus manifest themselves as pressure changes rather than actual bumps. These pressure changes then serve to drive currents involved in the tides that we see at coasts where the water does move about and change in height a lot, but this water is only moving at a few miles per hour and not at the speed of the "bulges" which travel at the speed of an aeroplane.
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #161 on: 02/08/2018 20:50:55 »
Quote from: David Cooper on 02/08/2018 20:18:48
The Earth's rotation moves the bulges towards the coasts which block their way, so they bunch up there. It's not nearly that simple though, as I'll explain further down.
Wouldn't they only bunch on the east coasts of the emerged lands if it was the case?

Quote from: David Cooper on 02/08/2018 20:18:48
What you actually get is resonance in ocean basins and seas
If I understand well, the tidal wave would get amplified in a basin the same way a sound wave is amplified in a trumpet. If the trumpet or the basin is longer, the frequency is lower, and vice versa. It makes sense, but it's hard to imagine. It means that if we were to begin such a cycle, the first one would give a low tide, and the following ones would give higher and higher tides until friction losses equalize the gravity nudges. I think it also answers my first question.
« Last Edit: 02/08/2018 20:53:05 by Le Repteux »
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Online Colin2B

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Re: Why do we have two high tides a day?
« Reply #162 on: 03/08/2018 08:30:24 »
Quote from: Le Repteux on 02/08/2018 20:50:55
Wouldn't they only bunch on the east coasts of the emerged lands if it was the case?
No, various reasons. Firstly the pull of the moon/sun system passes over (or the earth rotates from under, depending on how you view it) and releases the bulge so it can fall back. Secondly, the sloshing backwards and forwards can due to the distance result in a wave superposition that results in the waves performing a seesaw movement, a form of resonance.  Also the movement of the bulge isn’t straight across a basin west-east but Coriolis effect turns the movement to the right (northern hemisphere). For example in the English Channel when the bulge moves west to east you get slightly higher tides on the south (french) coast and then on the north (english) when it goes back east to west. Motion in a basin is more difficult to explain without pictures, but it results in a bulge that rotates counter clockwise around a null point - look up amphidromic circulation and dynamic theory of tides for details.   

Quote from: Le Repteux on 02/08/2018 20:50:55
If I understand well, the tidal wave would get amplified in a basin the same way a sound wave is amplified in a trumpet. If the trumpet or the basin is longer, the frequency is lower, and vice versa. It makes sense, but it's hard to imagine.
This only happens in some very specific places where the geology works to produce a resonance - look up Bay of Fundy where length of bay causes a resonance with 12 hr tides hitting the entrance.
In many other places eg Atlantic the bulge is amplified by the wave hitting shallow water of continental shelf just like waves on a beach increasing in ht as the water shallows. It is also amplified due to amphidromic effects as the further it is from the null point, the higher the bulge will be.
As David says, the geography is responsible for the specific local tides and these will deviate from the equilibrium 2 bulges/day eg Solent has a double peak high tide due to water entering first the west then the east entrance.

However, as David pointed out in an earlier post, I think this is moving away from the objectives of the OP which is to discuss the specifics of the moon/earth effects of the equilibrium theory and the terminology used.
« Last Edit: 03/08/2018 09:27:02 by Colin2B »
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #163 on: 03/08/2018 17:37:02 »
More thoughts:-

Going back to the mention of tsunamis, they are started at a point location and then spread from there carrying all the energy in a compact wave, but tidal "bulges" are generated across thousands of miles rather than at a single point, so you don't have the same intensity of impact at the coast - it's a much gentler business even if it involves a lot more total energy.

On the (undiscussed) issue of how much of the energy is newly added by the moon and sun's pull and how much might be a remnant of resonance from previous cycles, the variation between spring and neap tides shows that a lot of energy must be lost from the system on each cycle because there is a big variation depending on how the bulges generated by the moon and sun combine.

Note to rmolnav:-

If your mechanism was correct about the centripetal/centrifugal effect flinging the water out in two bulges at opposite sides as the Earth rotates around the barycentre, wouldn't that generate a single bulge on the side furthest out from the barycentre with the sea lowest on the side nearest to it? That would lead to most places having one tide per day instead of two. (You may have discussed this earlier, but due to the quantity of your writing here, it's hard to find out in a reasonable length of time.)
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #164 on: 03/08/2018 21:47:58 »
Quote from: David Cooper on 02/08/2018 20:04:52
Quote
...even if moon´s pull were uniform across our planet,
1) earth, as it´s not “free to move” (moon and earth c.g.s have to keep their distance, and the barycenter has to follow its orbit around the sun), it would experience additional internal stresses ...
Of course it's free to move! The Earth's sitting in space without being moored to anything - it will simply accelerate towards anything that applies a gravitational force to it. The Earth, moon, sun, other planets, etc. are all falling through space without any strings tied to them - they are free to go wherever they're flung. They don't have to keep their distance from each other, but simply fall uncontrolled, but fortunately, if they miss each other at closest approach, they naturally produce orbits which keep repeating the same pattern.
You keep repeating erroneous things, I´m afraid that due to basic misconceptions …
If an artificial satellite is orbiting around the earth, it is due to suitable speed vector perpendicular to earth pull, what makes it follow an elliptical path. The whole earth pull is the centripetal force, and the satellite (considered as a whole) neither experiences nor exerts any centrifugal force. It´s reaction to earth pull (3rd Newton Law) is just to pull back on earth … what practically doesn´t affect our planet …
Earth-moon dynamics is something quite different, especially as far as our planet is concerned. Moon not only pulls earth, what makes its linear speed change direction (proportionally to the pull)… The moon also keeps moving in the opposite sense, what makes earth mentioned “answer” (the continuous  "bending” of its intended rectilinear path, due to inertia) “insufficient” … Earth path gets more curved than if it were going to orbit around a static moon: it is a much smaller circumference, and around the barycenter instead of around the moon ...
That´s why I consider that, rather than talking about “free falls” within a given gravitational field, it is more realistic to consider earth-moon interactions as if they were part of a single extended object, with no massive parts apart from where both celestial objects actually are, and rotating/revolving about the common center of mass, maintaining the distance between them due to their dynamic equilibrium.
Quote from: David Cooper on 02/08/2018 20:04:52
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2) if due to initial conditions earth parts were somehow following not rectilinear paths, centripetal and centrifugal forces would be occurring ...
Why? With an even force applying, there is no centrifugal force involved because every particle is accelerated the same amount in the same direction, thereby ensuring that no stress is generated.
Repeating erroneous things does´t make them right !!
I don´t know how old are you (I´m 74), but you seem to have forgotten basic Physics boys usually learn when teenagers ...
Without a centripetal force an object can´t get a curved trajectory whatsoever !!
If you “ … use such an even force field to make the object travel in a circle”, that would be insufficient: you also need a certain initial speed of the object, and the field force has to have a component perpendicular to initial speed… As soon as that field started acting on the object, that very field force (or part of it) would become the centripetal force, whatever caused that field !! 
And if, for any reason, at some parts of the object the required centripetal force doesn´t match with the field force there, interactions with the rest of the object appear, that is, internal stresses !!
The bottom line would be that those internal stresses, exerted on considered part by contiguous material, and/or by part of gravitational pull exerted by own earth, would have to compensate the mismatch between moon pull there and required centripetal force (in one or in the opposite sense).
Those internal stresses would include "moonward" forces (centripetal), and outward forces (centrifugal, in a non restricted sense of the term).
Quote from: David Cooper on 02/08/2018 20:04:52
Quote
3) if those paths were as they actually are (circumferences with ALL earth parts continuosly at maximum distance from the moon), inertia would tend to make water follow the tangent, earth own gravity would avoid it … and we would get the further hemisphere bulge ... as I said even if moon pull there were the “uniform” average !![/b]
Why should water follow the tangent when the Earth doesn't follow the tangent? Why are you applying one rule to the Earth and a different rule to the water sitting on it?
Please, don´t twist my words. I didn´t say “water (on earth) follow the tangent”, but "inertia would tend to make water follow the tangent”, and exactly the same happens to earth as a whole … But neither of them is fully free, and their dynamics features (movement, deformation, inertia related reaction forces …) depend on each “scenario” ...
And, as water is relatively more "free" to move than earth solid parts, as I´ve said many times, bulges happen, as a result of both actual moon pull at each location, and also previously mentioned inertial effects !!
As I´ve also said many times, the "information" of differential moon pulls, between not contiguous parts of the earth, can exist in our intelligent minds, but cannot reach and directly be exerted on those parts !!
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Online Colin2B

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Re: Why do we have two high tides a day?
« Reply #165 on: 03/08/2018 22:52:31 »
Quote from: David Cooper on 03/08/2018 17:37:02
Going back to the mention of tsunamis, they are started at a point location and then spread from there carrying all the energy in a compact wave, but tidal "bulges" are generated across thousands of miles rather than at a single point, so you don't have the same intensity of impact at the coast - it's a much gentler business even if it involves a lot more total energy.
Yes, the wavelengths are quite different. The way a tsunami is generated produces a wave pulse of a crest preceded by a trough which is why the sea recedes a considerable distance from the beach just before the crest hits. Again the shallowing causes the crest to increase in size.

Quote from: David Cooper on 03/08/2018 17:37:02
the variation between spring and neap tides shows that a lot of energy must be lost from the system on each cycle because there is a big variation depending on how the bulges generated by the moon and sun combine.
Water creates very high damping in this system.
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #166 on: 04/08/2018 00:25:45 »
Quote from: rmolnav on 03/08/2018 21:47:58
You keep repeating erroneous things, I´m afraid that due to basic misconceptions …
If an artificial satellite is orbiting around the earth, it is due to suitable speed vector perpendicular to earth pull, what makes it follow an elliptical path. The whole earth pull is the centripetal force, and the satellite (considered as a whole) neither experiences nor exerts any centrifugal force. It´s reaction to earth pull (3rd Newton Law) is just to pull back on earth … what practically doesn´t affect our planet …

You're failing to recognise the key difference. When you stop a ball on the end of a string from going round and round just above the ground and just let it sit on the ground at the end of the string, the deformation disappears. If you do the same with an orbiting object, the deformations remain because they are not driven by the orbit. That is why centrifugal/centripetal force cannot validly be regarded as the cause of tides - they remain in full without centrifugal/centripetal force acting at all.

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Earth path gets more curved than if it were going to orbit around a static moon: it is a much smaller circumference, and around the barycenter instead of around the moon ...

It hardly moves at all, but it doesn't matter how much it's moved by gravity, that doesn't make any force that can generate bulges. The bulges only occur because of the change in gravitational strength over distance.

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That´s why I consider that, rather than talking about “free falls” within a given gravitational field, it is more realistic to consider earth-moon interactions as if they were part of a single extended object, with no massive parts apart from where both celestial objects actually are, and rotating/revolving about the common center of mass, maintaining the distance between them due to their dynamic equilibrium.

If you're going to ignore existence of bulges during free falls, you're making it harder for yourself to see the key point that the are not caused by the rotation. All the orbit does is enable the Earth and moon to continue to operate for a long length of time without colliding. The bulges are not caused by the path that either body is moving in. If you want to create a bulge, you could actually do it by grabbing the Earth and forcing it to move in a straight line while the water (free to move) tries to continue to follow the orbit.

Quote
Quote from: David Cooper on 02/08/2018 20:04:52
Quote
2) if due to initial conditions earth parts were somehow following not rectilinear paths, centripetal and centrifugal forces would be occurring ...
Why? With an even force applying, there is no centrifugal force involved because every particle is accelerated the same amount in the same direction, thereby ensuring that no stress is generated.
Repeating erroneous things does´t make them right !!

Then you need to learn to stop doing it.

Quote
I don´t know how old are you (I´m 74), but you seem to have forgotten basic Physics boys usually learn when teenagers ...

It should not be classed as centrifugal/centripetal force in an orbiting system because the force continues to act in full if you stop the rotation and is therefore entirely independent of that rotation, rendering the rotation incidental. (Physics got rid of the term centrifugal force because it was an incorrect way of looking at things, and if it is still common to use centripetal force in the description of orbits, that is another error which should be corrected - it is only acceptable as an analogy.) Most importantly though, the rotation has no role whatsoever in forming a bulge.

Quote
Without a centripetal force an object can´t get a curved trajectory whatsoever !!

I don't accept that it's centripetal force other than as an analogy. That's entirely a side issue though, because the only point that matters here is that whatever we call it, it doesn't generate bulges It is not the mechanism for tides.

Quote
If you “ … use such an even force field to make the object travel in a circle”, that would be insufficient: you also need a certain initial speed of the object, and the field force has to have a component perpendicular to initial speed… As soon as that field started acting on the object, that very field force (or part of it) would become the centripetal force, whatever caused that field !!

Same irrelevant disagreement. The important point is that it doesn't generate bulges.

Quote
And if, for any reason, at some parts of the object the required centripetal force doesn´t match with the field force there, interactions with the rest of the object appear, that is, internal stresses !!

If you apply an even force, it will match everywhere and there will be no stresses. The mechanism for tides is that the force is not even, as is demonstrated by the straight-line case with no rotation.

Quote
The bottom line would be that those internal stresses, exerted on considered part by contiguous material, and/or by part of gravitational pull exerted by own earth, would have to compensate the mismatch between moon pull there and required centripetal force (in one or in the opposite sense).
Those internal stresses would include "moonward" forces (centripetal), and outward forces (centrifugal, in a non restricted sense of the term).

The stresses are caused by the force not being even, and as I keep pointing out, they continue to apply if the rotation is eliminated, thereby revealing that they are not centrifugal/centripetal by any viable definition at all. If you have a planet with an ocean travelling in a straight line through space with no bulges in the ocean as there is no stress applied to it, if a spaceship then uses a tractor beam to cause the path of the planet to curve by applying an even force to it, no bulges appear despite the change in direction. If no change occurs between the beam not being applied and being applied, no part of the direction change can have anything to do with generating bulges.

Yes, this is repetition, but it's necessary because it's correct and you continually refuse to accept reality.

Quote
Quote from: David Cooper on 02/08/2018 20:04:52
Quote
3) if those paths were as they actually are (circumferences with ALL earth parts continuously at maximum distance from the moon), inertia would tend to make water follow the tangent, earth own gravity would avoid it … and we would get the further hemisphere bulge ... as I said even if moon pull there were the “uniform” average !![/b]
Why should water follow the tangent when the Earth doesn't follow the tangent? Why are you applying one rule to the Earth and a different rule to the water sitting on it?
Please, don´t twist my words. I didn´t say “water (on earth) follow the tangent”, but "inertia would tend to make water follow the tangent”, and exactly the same happens to earth as a whole … But neither of them is fully free, and their dynamics features (movement, deformation, inertia related reaction forces …) depend on each “scenario” ...

I'm not twisting anything. If you imagine that you have centrifugal force involved in making the Earth and moon orbit each other (in their dance around the barycentre) and you also have it involved in making the water of the ocean go along with the Earth, you have no mechanism for inertia to move the water any differently than the Earth. Your idea that the water should follow the tangent should naturally lead to the Earth needing to follow the tangent too, and because it isn't tied to the moon, it would be free to do so. You have failed to provide a mechanism for the water to behave differently from the Earth other than by applying different rules to each. The Earth is governed by gravity such that it follows a curved path, while the water is magically not governed by the same gravity such that it tries to follow the tangent. NO - both will follow the same path (unless the gravitational force being applied is not an even one, which is of course the case, and it's that unevenness that causes the tides).

Quote
And, as water is relatively more "free" to move than earth solid parts, as I´ve said many times, bulges happen, as a result of both actual moon pull at each location, and also previously mentioned inertial effects !!

The water and Earth are exactly as free as each other to follow the same path as each other. You only get bulges from an uneven force, and then what you get is a change in pressure in the water (which we're calling a bulge) with stresses also generated right through the Earth.

Quote
As I´ve also said many times, the "information" of differential moon pulls, between not contiguous parts of the earth, can exist in our intelligent minds, but cannot reach and directly be exerted on those parts !!

There is a gravitational force being applied to each particle of Earth and sea from the moon, and it tries to move each of those particles. Because the force isn't evenly applied, it pulls at some with greater strength than others, and that's what generates the bulges. Simple as that - not driven by any rotation.
« Last Edit: 04/08/2018 00:32:44 by David Cooper »
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #167 on: 04/08/2018 13:52:19 »
@ Molnav,

In case it hasn't been presented, here is my two cents picture of the phenomenon.

Its uneven acceleration at different distances from the pulling body that creates the bulges. Put three balls side by side, accelerate them at three different rates in the same direction, and the distance between the balls is going to increase with time. That's what creates the bulges in the case of the earth, and what prevents them to increase forever is the earth's own gravity that also accelerates the bulges towards one another.
« Last Edit: 04/08/2018 13:57:52 by Le Repteux »
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #168 on: 04/08/2018 14:15:49 »
Quote from: Le Repteux on 04/08/2018 13:52:19
Put three balls side by side, accelerate them at three different rates in the same direction, and the distance between the balls is going to increase with time.
What quoted is right ... but not what you say afterwards.
If you have a look at my #149, you´ll see that, using same analogy, I said:
"Imaging three tennis ball size pieces of earth stuff, one were earth c.g. (E),  another just under the moon (M), and the other at the antipodes (A).
If we had them without the rest of our planet, they would be free to react to forces such as moon gravity … M would accelerate the most, and A the least. Distances between M and E, and between E and A, would increase, though M and A don´t even “know” moon pull on E. And certainly without needing any rotation or revolving.
But in the real case, M and A neither are free whatsoever, nor “know” how much E is pulled by the moon, let alone that material stuff “knows” how to subtract pull vectors, in order to directly react to gravitational differences !!
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #169 on: 04/08/2018 14:52:44 »
I regret to have to say: more, and more, and more rubbish ...
One of the many differences between our stands is that you say ONLY differential pull causes the bulges, but I don´t say ONLY inertial effects related to earth revolving can cause the bulges ...
And with your absurd logic, if you find another scenario where deformations similar to the bulges occur without any curved movement, you wrongly deduce your stand is the right one!!
E.g.:
Quote from: David Cooper on 04/08/2018 00:25:45
The bulges are not caused by the path that either body is moving in. If you want to create a bulge, you could actually do it by grabbing the Earth and forcing it to move in a straight line while the water (free to move) tries to continue to follow the orbit.
Whatever the imagined way you did that, it would be a different case, where same reason (inertia) would have similar effects because you change the speed vector of a massive body ...
But that doesn´t mean those inertial effects wouldn´t happen changing only speed vector direction, what is continuously happening to the revolving earth !!
As I thought yesterday, perhaps it would be good for you (and for many others, I´m afraid ...) to go back to basics.
If e.g you google "phisics.ohio-state.edu" and go to "Dynamics of Uniform Circular Motion" (Chapter 5, 5.3 Centripetal Force), you can read (apart from many other things and formulas):
"...Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration.
The centripetal force is the name given to the net force required to keep an object moving on a circular path
".

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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #170 on: 04/08/2018 15:30:00 »
Quote from: rmolnav on 04/08/2018 14:15:49
What quoted is right ... but not what you say afterwards.
Your example is more complicated than mine even if it is similar, so I will keep mine and add gravitation to it instead. Let my three balls be accelerated by the earth's gravitation while falling directly to it, and add mass to them so that they are also accelerated towards one another. We can give them a large mass so that they stay together while falling, then reduce it so that they get away from one another just a bit, or reduce it so much that they get away from one another a lot. The just a bit case represents the tides' case, with no input from forces due to rotation.
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #171 on: 04/08/2018 16:13:03 »
Quote from: rmolnav on 04/08/2018 14:52:44
"...Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration.
The centripetal force is the name given to the net force required to keep an object moving on a circular path".
There is certainly a motion produced by the earth in the direction of the moon and vice versa, but this motion is not driven by a rigid link like a rope for instance, so contrary to the tangential one, it is not purely inertial. I have a theory that says particles must execute small steps to justify their inertial motion in space, but those steps cannot justify their gravitational motion, so some other steps have to be executed by the gravitating particles towards one another, and those steps have to be driven by the information exchanged between the two gravitating bodies, contrary to my inertial steps that are driven by the information exchanged between bonded particles. The information exchanged between gravitating particles nevertheless transforms immediately into steps similar to my inertial motion's steps, so that there is no more force involved than for a body on inertial motion. Force develops only when the steps between bonded particles get out of sync, which is far from being the case for orbiting particles since they have all the time they need to adjust to their orbital changes without getting out of sync.
« Last Edit: 04/08/2018 20:43:38 by Le Repteux »
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #172 on: 04/08/2018 18:40:02 »
Quote from: rmolnav on 04/08/2018 14:15:49
But in the real case, M and A neither are free whatsoever, nor “know” how much E is pulled by the moon, let alone that material stuff “knows” how to subtract pull vectors, in order to directly react to gravitational differences !!

Picture three massive balls, M, E and A, which attract each other using gravity. They are aligned in a straight line with where we're going to place the moon. M and A are both sitting on springs, so they don't contact E directly, but they are indirectly sitting on it, and each is sitting at height h over the surface of E. Now we introduce the moon into the system. The three balls begin to accelerate towards the moon, and each tries to go at its own speed, but the local pull on M and A from E feels stronger than the pull from the moon, so they will remain close to E. What happens though is that M settles to a different altitude slightly higher than h above the surface of E because part of the Earth's pull on it is cancelled out by the extra pull on M that E isn't getting (that extra being the result of E not being pulled with quite as much force as M). A also settles to an altitude slightly higher than h above the surface of E because E is being pulled towards the moon more strongly than A is. Those are the tidal bulges (and in this case they are actual bulges rather than just pressure reductions).

As the three balls fall towards the moon, the difference in force between the moon and each successive ball grows stronger, so the altitudes of M and A above the surface of E will increase. None of the balls need to know anything - they simply react directly to the forces being applied to them. This is the mechanism behind the tides and it does not depend on rotation from orbits.
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #173 on: 04/08/2018 19:30:38 »
Quote from: rmolnav on 04/08/2018 14:52:44
I regret to have to say: more, and more, and more rubbish ...

Then you need to learn to stop posting it.

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One of the many differences between our stands is that you say ONLY differential pull causes the bulges, but I don´t say ONLY inertial effects related to earth revolving can cause the bulges ...

Practically all the bulge involved in tides is produced by the mechanism I've described. Any potential additions to it will be trivial, but I haven't seen you propose anything that could serve as an addition. You magically have water wanting to follow the tangent while the Earth follows the curve in a case with an even force being applied, and that's plain bonkers (loco). In reality, for water to want to follow the tangent, it would have to have zero force applied to it by gravity from the other body, and that certainly isn't going to happen. What we actually have is a lesser force being applied by the distant body's gravity at the far side and a greater one at the near side, and those directly pull up the bulge on one side and pull the Earth away from what becomes the bulge at the other side. The orbital path is irrelevant and has no role in the causation, and you can tell that this is the case because the bulges are the same size at that separation distance whether there is an orbit or not. It is a massive error to attribute the bulges to a rotation when they are identical to the ones that are there without the rotation - the rotation clearly cannot have a role if it makes no difference to them. The biggest question that remains here for this thread though is how many more pages of posts will it take before you recognise that simple fact.

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And with your absurd logic, if you find another scenario where deformations similar to the bulges occur without any curved movement, you wrongly deduce your stand is the right one!!

It is an extension of the same scenario. The moon's orbit is growing in size, and some day it will become more linear, eventually reaching a point where it collides with the Earth (if the sun hasn't swallowed both of them first). At no point during this transition from near-circular orbit to more linear orbit and ultimately to a nearly straight line path to impact does the mechanism for the bulges change - it remains the exact same mechanism throughout.

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E.g.:

Quote from: David Cooper on 04/08/2018 00:25:45
The bulges are not caused by the path that either body is moving in. If you want to create a bulge, you could actually do it by grabbing the Earth and forcing it to move in a straight line while the water (free to move) tries to continue to follow the orbit.

Whatever the imagined way you did that, it would be a different case, where same reason (inertia) would have similar effects because you change the speed vector of a massive body ...
But that doesn´t mean those inertial effects wouldn´t happen changing only speed vector direction, what is continuously happening to the revolving earth !!

Like I said, it's the same case and the mechanism doesn't change. The orbit aspect of it is irrelevant other than as a means to delay the collision. Going back to the three balls with springs holding M and A off the surface of E, the height of the balls above the surface of E is dependent on one single factor, and that is the distance to the moon (because it's that distance that determines how much extra force is applied to M over that applied to E, and how much more is applied to E over that applied to A.

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As I thought yesterday, perhaps it would be good for you (and for many others, I´m afraid ...) to go back to basics.

It would be a good idea if you took your own advice. Picture a centrifuge used to separate out a mixture of heavy and light material. It goes round and round to throw the material outwards, and the heaviest material ends up collecting together at the outside. Now picture a scientist with a poor grasp of physics who, in the interests of reducing maintenance and energy costs, tries to make a centrifuge using a black hole to make the container move round in the same size of circle at the same speed as in the earlier centrifuge. The container whirls round and round this circuit just like the other one, but the material inside it is weightless, so it simply doesn't separate out - it does not serve as a centrifuge, and the reason for that is that there is no force on the content generated by the rotation: no centrifugal/centripetal force is acting on it. It's the same with the water sitting on the Earth - there is no centrifugal/centripetal force acting on it either.

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If e.g you google "phisics.ohio-state.edu" and go to "Dynamics of Uniform Circular Motion" (Chapter 5, 5.3 Centripetal Force), you can read (apart from many other things and formulas):
"...Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration.
The centripetal force is the name given to the net force required to keep an object moving on a circular path
".

It doesn't matter how much justification you can find for calling it centripetal force. This is a grey area in which some usages of the word aren't fully rational. What is absolutely certain though is that centrifugal/centripetal force is not the cause of the bulges because the bulges remain when the rotation is removed. It is plain wrong to attribute them to a rotation which is not causing them.
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #174 on: 04/08/2018 22:29:20 »
I have a question about the tides that has not been answered yet, so I might as well post it here.

While the earth and the moon execute their common orbital trajectory with the sun, their common barycenter goes at the right speed all the time, but not their respective gravity centers: those are too fast when they transit on the far side, and too slow when they transit on the near side. They should thus behave as if they were at an aphelia on the far side, and at a perihelia on the near one, slowing down on their orbital trajectory with the sun and getting away from it progressively during the first half of their far side transit (and then doing the inverse during the second half), and accelerating and getting closer to the sun during the first half of their close side transit (and doing the inverse also during the second half), what should progressively increase and decrease the distance between them during their transit. I never heard of such an observation, but I can't find a flaw in the logic. Would it be too small to be observable?

Now if we apply this mechanism to the tides, we find that it would coincide with the height they get in the middle of the oceans, and that it would also coincide with the horizontal speed they get near the shores. This time, it may be observable, but how to distinguish it from the effect of the other mechanism if it only adds to it?
« Last Edit: 04/08/2018 22:31:21 by Le Repteux »
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #175 on: 04/08/2018 23:45:14 »
Quote from: Le Repteux on 04/08/2018 22:29:20
While the earth and the moon execute their common orbital trajectory with the sun, their common barycenter goes at the right speed all the time, but not their respective gravity centers: those are too fast when they transit on the far side, and too slow when they transit on the near side.

They would be too fast or slow if they weren't also pulling against each other. If the moon is furthest away from the earth at the same time as it's directly between the sun and Earth, it will be moving relatively slowly such that if the Earth suddenly disappeared, the moon would then follow a new orbit that would take it closer to the sun. The Earth can't just disappear though, so the moon is held up by it and accelerated back up to speed as it continues to orbit around it. While the moon is between the Earth and sun, the Earth is further out and moving faster than the barycentre, so if the moon suddenly disappeared, the Earth would follow a new orbit that would take it further away from the sun. Again though, the presence of the moon slows it back down and all the speeding up and slowing down is balanced out.

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They should thus behave as if they were at an aphelia on the far side, and at a perihelia on the near one,

A body moves fastest at perihelion and slowest at aphelion, but here we have them slower when closer to the sun and faster when further from it.

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... slowing down on their orbital trajectory with the sun and getting away from it progressively during the first half of their far side transit (and then doing the inverse during the second half), and accelerating and getting closer to the sun during the first half of their close side transit (and doing the inverse also during the second half), what should progressively increase and decrease the distance between them during their transit.

If you imagine a case where the moon's orbit round the Earth is circular, it won't accelerate relative to the Earth at all, but it will be accelerating relative to the sun throughout the outer part of its orbit (when the Earth is nearer to the sun than the moon is), and decelerating relative to the sun throughout the inner part (when it is nearer to the sun than the Earth is). If you're imagining that the moon's greater movement towards the sun during part of its orbit round the sun will lead to the sun accelerating it, that won't happen because the sun's pull on the moon has already been applied to set a course for the Earth and moon collectively to travel around it, so it can't then add some more to accelerate the moon even more towards it - the movement of the moon inwards is driven by the Earth, and the Earth's gravity serves to decelerate that inward movement as soon as the moon is nearer the sun than the Earth is.

The place where a complication comes in is that the sun's pull on the moon is stronger when the moon is between it and the Earth, and weaker when the moon's furthest out, so that must have some impact on the shape of the orbit, but I certainly can't model that in my head - it needs to be simulated in a computer to see what the effect of that is, although exaggerated proximity to the sun in such a simulation would likely be necessary to make that effect visible to the eye.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #176 on: 05/08/2018 10:59:08 »
Quote from: Le Repteux on 04/08/2018 15:30:00
Let my three balls be accelerated by the earth's gravitation while falling directly to it, and add mass to them so that they are also accelerated towards one another. We can give them a large mass so that they stay together while falling, then reduce it so that they get away from one another just a bit, or reduce it so much that they get away from one another a lot. The just a bit case represents the tides' case, with no input from forces due to rotation.
Main lack of freedom in the tides´real case is due to the fact that M and A are pulled by own earth gravity with forces some billion times bigger than the differential moon´s pull, NOT to rigidity when they are part of a solid, compact object ...
Your "redesigned" case leaves them free, though initially with a pseudo-equilibrium between pulls between each other, and earth´s gravity. Quite a different scenario !!
 
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #177 on: 05/08/2018 13:57:42 »
I need more time to analyze all said here in last 24 hours!
But going again to basics, just some short comments ...
When I started here more than three years ago, I said "Centrifugal force is not a forbidden word whatsoever" ... I couldn´t imagine that, if not forbidden, it is so a controversial term.
Mainly because in some moment physicists decided to define it as the "fictitious" force applied to the object following a curved path if we analyze the movement using a so called "non inertial frame of reference" ... Any other "outward" force, in the direction of "flying" (sense of "fugal") from a "center", shouldn´t be called centrifugal force, according to them ...
O.K. It´s just a question of terminology. If the force does exist, and produces their effects according to laws of Physics, it does´t matter much the term we use.
But in my wildest dreams I could have thought I would have to discuss about the term "centripetal force".
Yesterday I posted what said on a "ohio.state.edu" page:
"The centripetal force is the name given to the net force REQUIRED to keep an object moving on a circular path".
Similar things are said on a lot of other sites ...
But one can read now:
 
Quote from: David Cooper on 04/08/2018 19:30:38
It doesn't matter how much justification you can find for calling it centripetal force. This is a grey area in which some usages of the word aren't fully rational
Could you please send some link, from a University or similarly reliable source, where one could see that is not another absurd idea of yours ??
There could be some not fully rational usages of the term ... Absurd things are said by many people. But do you really think that term is a "grey area"??
I sincerely imagine you would find it crystal clear area, if it reinforced your fixed idea that ONLY differential pull from moon (apart from sun´s similar effects) counts for tides !!
By the way, the so called "non inertial frame of reference" is something I have in my back log (?), I mean some day I have to try and convey some own ideas about it, surely controversial ...
But I´m sure it won´t be easy at all ... seeing that we can´t agree on even really basic terms !! 
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #178 on: 05/08/2018 14:43:21 »
By definition, a force is felt when we try to accelerate a massive object, and it is so when we throw a ball, swing it in circles, or hold it against gravity. If there is no force felt, then there is no force by definition. In free fall, no force is felt, and whether we are rotating or not is unimportant since we can't feel any anyway. That's why I prefer using the idea that, when no force is felt and we still have a motion, then that motion must depend on an internal mechanism that is not yet elucidated. The curved space/time mechanism is not physical so it can't be studied, but the idea that this kind of motion may be driven by an exchange of information can: we just have to study the way light could be exchanged between two bonded particles during their acceleration, and move them so that they stay on sync all the time.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #179 on: 05/08/2018 17:25:22 »
Quote from: David Cooper on 04/08/2018 18:40:02
Picture three massive balls, M, E and A, which attract each other using gravity. They are aligned in a straight line with where we're going to place the moon. M and A are both sitting on springs, so they don't contact E directly, but they are indirectly sitting on it, and each is sitting at height h over the surface of E. Now we introduce the moon into the system. The three balls begin to accelerate towards the moon, and each tries to go at its own speed, but the local pull on M and A from E feels stronger than the pull from the moon, so they will remain close to E.
That´s right …
But those springs would act as a kind of “information” transmission chain, what in real case is done as internal stresses, transmitting action/reaction forces between contiguous material, a real “mechanism” you´ve rejected several times … You said many times just differential gravity is the mechanism !
But if we count with internal stresses, we have to include not only those, but the ones originated by revolving related inertia ...
The springs would also slightly change their lengths due to the revolving of the earth, even if moon attraction were the same on M and on A …
M and A, at their maximum distances from the moon (within their own revolving trajectories), with tangential speeds perpendicular to line between moon´s and earth´s centers, would require a centripetal force in both cases towards the moon. Otherwise they could not follow their circular paths.
Same happens on E, where the centripetal acceleration is provided by moon´s pull there: they exactly match with each other.
Where M moon´s pull is stronger than required centripetal force: part of it does accelerate M towards the moon (the required centripetal acceleration for revolving of M), and the rest is what actually causes the bulge there.
But where A moon´s pull is smaller, insufficient for the centripetal force necessary to make it follow its circular path. Own earth´s pull has to "supply" the difference. Water kind of become slightly lighter, and a bulge builds also there, same way as previously explained for the formation of the equatorial bulge:
"The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non rotating earth, man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation”.
(https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm)
 
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