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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 #420 on: 17/10/2018 00:30:44 »
Quote from: Le Repteux on 15/10/2018 22:44:17
So, if we consider that there is no centrifugal force in the case of the tides, then I think we should also consider that the equatorial bulge suffers no such force.

Well, officially there's no such force as centrifugal force - it's a pseudo force (unless it's reactive centrifugal force operating the opposite way to centripetal force in cases like the one with a ball on a string), so it's technically wrong already. We already have two radically different cases though, so let's see if we now have a third one:-

(1) A ball on a string when moving round in circles generates the centripetal force in the string. The force acting the opposite way in the string is called reactive centrifugal force, and that's a real force. Stop the ball and the centripetal force and reactive centrifugal force both disappear. Let's also use a transparent ball half full of water though so that we can think about what happens to the water - we would see it being flung into the outer half of the ball opposite the string by an apparent centrifugal force, but the apparent force flinging it there is really just the water trying to go in a straight line.

(2) A moon going round a planet doesn't generate centripetal force, but it goes round the planet because of gravity, and while that gravity can be described as centripetal force, it is really just gravity. Stop the moon and the force which had been described as centripetal continues to act in full, completely untransformed by losing the centripetal label. Note that there is also no reactive centrifugal force involved - that is a clue to the reality that we're dealing with a radically different category of "centrifugal" when dealing with gravity. Now replace the moon with the little transparent ball half full of water and the water will gather all round the sides to leave an air-filled globe in the middle - it doesn't get flung out to the side furthest from the Earth because the Earth's gravity is pulling on it with the same strength as it's pulling on the ball, so there isn't any apparent centrifugal force.

(3) The material on the surface of a planet going round the planet due to the planet's rotation doesn't generate centripetal force either, so it's similar to (2). That material is sitting there because it's moving too slowly to take of into an orbit, so it's like a special case of (2) in which there happens to be contact due to insufficient speed to take off. So is it centrifugal force that makes it rise as part of an equatorial bulge? Obviously it can't be as the only real centrifugal force is reactive centrifugal force, and we don't have any of that acting here.

But let's look at a case without any gravity pulling to the centre. Place balls on the edge of a roundabout (not a road junction, but the rotating disc thing that children play on in playgrounds) and spin it - the balls will roll off. That's like case (3), but without gravity preventing them from leaving the disc. If that's centrifugal force, then it's centrifugal force in case (3) too. Again there's no centripetal force generated here though, so there can be no reactive centrifugal force. The balls just fall off the edge. They are flung away from the centre of rotation, but not directly away from it - the paths they're moving along lead back to the edge of the disc at a tangent rather than a line passing through the centre, and the acceleration force that was applied to them only ever acted at 90 degrees to a line passing through the centre of the disc, so that force isn't centrifugal. Where is the centrifugal force? It only exists from the warped perspective of rotating frames with the balls appearing to accelerate away from the centre, but the balls are not accelerating at all as they leave the disc. In explaining what's happening there, centrifugal force is a false mechanism, but we can still describe this as centrifugal force as a convenient shorthand for what's happening with things moving out away from the centre - we just need to understand that that's a misleading description and translate from that word back to the real mechanism. Crucially, the only kind of centrifugal force that's ever part of a real mechanism is reactive centrifugal force.

If a child's actively hanging onto the edge of the roundabout, then there's centripetal and reactive centrifugal force acting there. With (3) we have the illusion of that same situation created by gravity holding the material on the planet, but there is no reactive centrifugal force, and the material lifts more than it would if there was no rotation, but rather than the material feeling as if it's being flung outwards, it merely feels that it's being pulled down less, and that's a different thing. Should we call it centrifugal? It's a convenient term to use, but it is a different category of centrifugal from the other cases. The complexities here are all in the labels and not in the physics, so the proper cure is simply to avoid using the awkward labels. Centrifugal force only looks fully appropriate in case (1) where centripetal force is generated by something following a curved path and reactive centrifugal force is generated in the opposite direction to match. Centripetal force is also only fully appropriate in case (1). In the other cases, the use of both words becomes more of a metaphor.

[The words are a mess though even in case (1) because they don't include the straight-line cases like when you slam the brakes on in a car, or when a chained dog is jerked to a halt while running at the postman, even though the fundamental mechanism is the same. (If the dog runs at an angle, starting some way to the side of the attachment point of the chain, it may be flung sideways when the chain tightens, that's centripetal force, but the straight-line case is just a special case of this more general case, so rotation or curving should not be a requirement of the definition.) They are both bad labels, and we can see from this thread that they can disrupt people's understanding of the physics involved in some cases.]

Think about a moon in an elliptical orbit. When its speed is too high for it to stay the same distance from the planet that it's going round, we could argue that centrifugal force makes it move further out. (I don't know what it would be called when the moon's moving closer in - there seems to be a vocabulary vacuum there.) That conflicts with rmolnav's usage where he only calls it centrifugal for parts of the moon further out than the centre of mass and he ignores the issue of whether the moon's increasing or decreasing its distance away from the Earth. In the first case we're using a rotating frame in which we measure an acceleration outwards by the moon relative to the Earth, so when we see the moon moving further away, we can put that down to centrifugal force, although we understand that this is not a real mechanism - this would be consistent with our use of the word centrifugal when describing the equatorial bulge forming, and with the balls rolling off the edge of the rotating roundabout. In the other case (rmolnav's), he's using different rules in which he ignores the outward or inward acceleration by the moon and focuses instead on the part further out than the centre of gravity of the moon, calling it centrifugal force on the basis that that material at FE (the point on the moon furthest from the Earth) would follow a different orbit if the rest of the moon wasn't there due to the speed of that material being too high to maintain the same size of orbit as the moon as a whole. This is a very different usage of the word centrifugal - the rules aren't consistent for the different cases. This illustrates again that the complexities are in the words and not in the actual physics - the physics itself is really simple, but we use misleading labels which are riddled with ambiguities, running different rules for them in different cases.
« Last Edit: 17/10/2018 00:42:59 by David Cooper »
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #421 on: 17/10/2018 08:06:08 »
D.C.´s posts #415, 416 and 420 are "the last straw ...", but in a positive sense!
I have finally realized he knows better than many scientists from institutions such as:
-NOAA (visit "Tides and Water Levels" - National Ocean Service).
-NASA (visit https://tidesandcurrents.noaa.gov/restles3.html,  and https://tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf
- Univ. of Maine ( "What if the Moon Didn't Exist? — Neil F. Comins", youtube video), who even has publications on "Sources of Misconceptions in Astronomy"...
- Univ. of Ohio ( "phisics.ohio-state.edu”, "Dynamics of Uniform Circular Motion" (Chapter 5, 5.3 Centripetal Force).
- Univ. of Louisville (Centripetal Force - Physics 298 - Department of Physics and Astronomy, www.physics.louisville.edu/cldavis/phys298/notes/centripetal.html 
- Several Physics academies, mentioned on my post # 413.
- Merriam Webster dictionary (“movement”, “to move”, “force”…)
- Oxford dictionary (“centripetal force”) 
- http://www.newenglandphysics.org/other/French_Tides.pdf,
among others.
D.C. has discovered they consider (and even much worse, they teach) quite erroneous things, such as:
- Centrifugal forces (or “outward” forces) intervene on tides (on a revolving celestial body such as our planet)
- Gravity is what generates the centripetal force required for the elliptical orbits of celestial objects.
- Other strange things about “centripetal force” in general (such as it “Is the force that makes a moving object change direction. It is not a particular force, but the name given to whatever force or combination of forces is responsible for a centripetal acceleration”.
- A fraction of moon´s gravitational pull on some earth´s material stuff can be "responsible for its centripetal acceleration”, and the rest for other effects such as deformation and tides.
If you read carefully what above, you can clearly see there are a lot of grave errors … Ah! I forgot most important:
D.C. discovered (#184, and he also refers to this on his last post) that:
"If a force is generated by rotation, that is clearly centripetal force - a force that comes into play because of the rotation ...”.
Not even the great Isaac Newton was able to discover that. According to him (2nd of his universal Motion Laws) things are the other way around: forces generate movements, or change their speed vectors (at a pace given by the acceleration caused by the force, a=F/m). Unbelievable!

And, even more remarkable , according to D.C. (#184):
"I am always more interested in actual science than error-ridden authorities. I haven't seen anyone in science support my position (primarily because I haven't looked for that) - what I'm saying is based 100% on what I see when I look directly at the physics involved in this”.
Definitely, all our group should be proud of having such a genius as a "Naked Science Forum King!”...
I´m going to suggest somebody in charge of The Naked Scientists forum to apply for an Amateur Nobel Prize on Physics for D.C.  Perhaps Global Moderator Colin2B ?
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #422 on: 17/10/2018 19:13:27 »
Quote from: rmolnav on 17/10/2018 08:06:08
D.C.´s posts #415, 416 and 420 are "the last straw ...", but in a positive sense!
I have finally realized he knows better than many scientists from institutions such as:...

It's good to know that you've finally understood that scientists can often be wrong. You've cherry-picked quotes from scientists who back your position, but the mainstream backs my position on this issue.

Quote
D.C. has discovered they consider (and even much worse, they teach) quite erroneous things, such as:
- Centrifugal forces (or “outward” forces) intervene on tides (on a revolving celestial body such as our planet)

Pseudo force.

Quote
- Gravity is what generates the centripetal force required for the elliptical orbits of celestial objects.

Gravity is simply gravity - it isn't generating a centripetal force, but is merely having the "centripetal" label attached to it, with the result that some people are misled into thinking that this makes some kind of mechanistic difference distinct from it simply being gravity.

Quote
- Other strange things about “centripetal force” in general (such as it “Is the force that makes a moving object change direction. It is not a particular force, but the name given to whatever force or combination of forces is responsible for a centripetal acceleration”.

Lovely, but you're failing to understand that it has no magic properties when it's associated with gravity. Gravity is simply gravity and the division that you're making which leads to all the complexity that you have such difficulty working with is artificial and unnecessary - your contrived mechanism is fake.

Quote
- A fraction of moon´s gravitational pull on some earth´s material stuff can be "responsible for its centripetal acceleration”, and the rest for other effects such as deformation and tides.

An upside-down understanding. It's just total gravitational pull on each particle (taking into account directions of pull) and you have everything you need.

Quote
If you read carefully what above, you can clearly see there are a lot of grave errors … Ah! I forgot most important:
D.C. discovered (#184, and he also refers to this on his last post) that:
"If a force is generated by rotation, that is clearly centripetal force - a force that comes into play because of the rotation ...”.

It's hardly a discovery - it's simply another attempt to get you to understand that there are different cases with different physics involved. Some centripetal force disappears when the movement stops because it's generated by that movement, but not when it's gravity labelled as centripetal force. You still have a magical view of gravity being divided into two components, centripetal and non-centripetal, with them carrying out different mechanistic roles, and that's why you're stuck where you are getting things wrong again and again and again.

Quote
Not even the great Isaac Newton was able to discover that. According to him (2nd of his universal Motion Laws) things are the other way around: forces generate movements, or change their speed vectors (at a pace given by the acceleration caused by the force, a=F/m). Unbelievable![/b]

You're misapplying his laws. Set up a heavy object connected by a string to a ball in space (something like a barbell with the string attached to a loop that goes round the handle) and just watch the centripetal force set the ball moving round and round! You believe in magic, but I don't - I say you have to start the ball moving first, and then the centripetal force is generated as a consequence of that movement. Every mainstream scientist will tell you the same thing, and Newton not have disagreed with that either.

Quote
And, even more remarkable , according to D.C. (#184):
"I am always more interested in actual science than error-ridden authorities. I haven't seen anyone in science support my position (primarily because I haven't looked for that) - what I'm saying is based 100% on what I see when I look directly at the physics involved in this”.

Apart from definitions of "centripetal", I only looked one thing up, and that was yesterday when I wanted to know what the force in the opposite direction in a string was (reactive centrifugal force). That's how I work - I avoid research as much as possible because research leads you into taking on other people's errors without thinking them through properly for yourself. I prefer to work out how things work first without any outside help, and then look to see what other people think afterwards.

Quote
Definitely, all our group should be proud of having such a genius as a "Naked Science Forum King!”...
I´m going to suggest somebody in charge of The Naked Scientists forum to apply for an Amateur Nobel Prize on Physics for D.C.  Perhaps Global Moderator Colin2B ?

Awards of that kind are part of the problem, setting up imperfect authorities and leading people to be lazy and fail to question the science properly. You've already illustrated how easy it is to fall into a wrong position and fix yourself there - you spend your time selectively gathering claims from authorities to back it up and you appear to imagine that that's how points are proved. But it isn't. Science should be about ideas doing battle with each other without any bias from any authority tied to either side. Reason should dictate the winner. The trouble with that though is that most people can't reason correctly, so until we have AGI acting as a perfect impartial judge, we're stuck with troops of monkeys making collective guesses and making some major mistakes. On this issue though, they haven't made a mistake - it's only a minority of scientists who are carelessly putting out incorrect information because they haven't thought things through properly, and you're allowing yourself to be misled by them.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #423 on: 18/10/2018 12:48:08 »
Quote from: David Cooper on 17/10/2018 19:13:27
Quote
Not even the great Isaac Newton was able to discover that. According to him (2nd of his universal Motion Laws) things are the other way around: forces generate movements, or change their speed vectors (at a pace given by the acceleration caused by the force, a=F/m). Unbelievable!
[/b]
You're misapplying his laws. Set up a heavy object connected by a string to a ball in space (something like a barbell with the string attached to a loop that goes round the handle) and just watch the centripetal force set the ball moving round and round! You believe in magic, but I don't - I say you have to start the ball moving first, and then the centripetal force is generated as a consequence of that movement. Every mainstream scientist will tell you the same thing, and Newton not have disagreed with that either.
What follows is for other possible readers rather than for D.C., incapable of "seeing" correctly physical phenomena.
Newton´s Motion Laws are universal: it is a force what ALWAYS causes movement (or changes its already existing speed)
And I don´t "believe in magic" whatsoever ...
To properly grasp physical phenomena our "vision" is not sufficient. We would need a magnifying lens, both for space and somehow "for time" (a modern camera capable of taking thousands of pictures would do). Or, at least, a reasonable degree of imagination.
An initial linear speed is logically necessary. But it can´t produce directly any force ... At the very first instant a string or wire hanging from a pole is connected to a hook on an object with a linear speed (e.g,. by "superman"), the string or cable has an initial tension (given by superman).
It is absolutely necessary (2nd Newton´s Motion Law) that, during an infinitesimal lapse of time, that tension initiates the change of the speed of the object, what causes the inertial reaction of the object pulling outwards the cable end (3rd Newton´s Motion Law).
Since that very instant we have a centripetal force (that initial cable tension), and the outwards reaction, centrifugal force. And the rotating movement starts.
It is THAT centrifugal force what, in its "turn", actually makes the wire tension increase (the centripetal force), and during following infinitesimal lapses of time, a chain of equal phenomena happens, with progressive velocity vector curving, with corresponding increases of centripetal and centrifugal force (for a period more or less long depending on the momentum mv of the object) ...
Eventually, if no energy added (either directly to the object, or from the inner end of the cable), the object will fall (that fall actually starts at the very first instant, because the tension of the wire always has a vertical downwards component). 
Therefore, it is not the speed (the movement as D.C. says) what causes the centripetal force. Without an initial FORCE, the mentioned initial tension, and those successive and increasing inertial FORCES (centrifugal), we couldn´t have any centripetal force, that is, NO CURVED MOVEMENT could be generated.
So, initial speed don´t generate the rotational movement ... Those mentioned forces change the direction of the initial speed, transforming a rectilinear movement into a curved one.
 So, I don´t need to ...
 
Quote from: David Cooper on 17/10/2018 19:13:27
  ... understand that there are different cases with different physics involved.
and it is not "magical" my ...
Quote from: David Cooper on 17/10/2018 19:13:27
... view of gravity being divided into two components, centripetal and non-centripetal, with them carrying out different mechanistic roles,
If I push very strongly on somebody´s breast, part of the force will move him backwards, but due to his inertia (tendency to remain still) some of his ribs may break ("different mechanism roles) ...
Who is ...
Quote from: David Cooper on 17/10/2018 19:13:27
... getting things wrong again and again and again ??
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #424 on: 18/10/2018 21:37:50 »
Quote from: rmolnav on 18/10/2018 12:48:08
What follows is for other possible readers rather than for D.C., incapable of "seeing" correctly physical phenomena.
Newton´s Motion Laws are universal: it is a force what ALWAYS causes movement (or changes its already existing speed)

When a tennis ball hits the net, does the force of the collision cause movement of the ball that leads to the collision? No. Even you, I hope, are sufficiently intelligent to realise that it was a different force that set the ball moving. Why can't you see that with the case of a ball on the end of a string where its movement energy generates the centripetal force rather than the other way round?

Quote
And I don´t "believe in magic" whatsoever ...

Good. Stop using magical in your mechanisms then. Put the causation in the correct order instead of having some of it working backwards.

Quote
An initial linear speed is logically necessary. But it can´t produce directly any force ... At the very first instant a string or wire hanging from a pole is connected to a hook on an object with a linear speed (e.g,. by "superman"), the string or cable has an initial tension (given by superman).
It is absolutely necessary (2nd Newton´s Motion Law) that, during an infinitesimal lapse of time, that tension initiates the change of the speed of the object, what causes the inertial reaction of the object pulling outwards the cable end (3rd Newton´s Motion Law).
Since that very instant we have a centripetal force (that initial cable tension), and the outwards reaction, centrifugal force. And the rotating movement starts.

The rotation doesn't add any movement energy to the ball - it just changes its direction of travel. What you keep failing to understand is that forces don't just appear without being caused by something else. A force can cause a movement, and that movement can later cause another force - it works in both directions. While the tennis ball is heading for the net, there is no force. When it hits the net, a force is generated, and it isn't generated by magic.

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It is THAT centrifugal force what, in its "turn", actually makes the wire tension increase (the centripetal force)...

No - it's the movement of the ball and the ball's mass (the movement energy) that pushes the tension up.

Quote
Eventually, if no energy added (either directly to the object, or from the inner end of the cable), the object will fall (that fall actually starts at the very first instant, because the tension of the wire always has a vertical downwards component).

Don't confuse yourself with gravity - put the experiment in deep space to eliminate extraneous input forces. If the ball is stationary, no centripetal force is generated and the ball remains motionless. We apply a force to the ball to set it moving (not centripetal force), and the ball's movement then generates the centripetal force in the string. As I said weeks ago, we can separate out these events by starting with a ball that isn't attached to the string, so the force applied to get it going is separated in time from the generation of centripetal force - the ball can hook onto the end of the string when it reaches it, and then it's continued movement initiates the centripetal force which modifies the ball's direction of travel.

Quote
Therefore, it is not the speed (the movement as D.C. says) what causes the centripetal force.

Of course it is - no movement of the ball means no generation of the centripetal force. The movement of the ball (the energy of it moving) causes the centripetal force - not the other way round. You've learned some laws, but only half understand them and don't know how to apply them. That is one of the hazards of bad education where people are given rules to memorise without being tested on their understanding and correct application and where they are allowed to pass exams with scores of less than 100%.

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So, initial speed don´t generate the rotational movement ...

The movement energy (initial speed) generates the centripetal force that causes the change in direction of the movement. This is really basic physics, but you don't understand it because you've been taught to misapply rules and "understand" things on the basis of conformity to misapplied rules. If there are any physics teachers reading this, be careful not to program your students to make the same mistakes because some of them will be stuck with them for life.

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Those mentioned forces change the direction of the initial speed, transforming a rectilinear movement into a curved one.[/b]

Maybe it's gradually getting through to you though. That bit's right.

Quote
If I push very strongly on somebody´s breast, part of the force will move him backwards, but due to his inertia (tendency to remain still) some of his ribs may break ("different mechanism roles)

Forget the breakage part. If you push him and part of that input of energy would lead to him moving at the same speed as someone who's walking past while the rest of the input makes him go faster than that, that would be an artificial division of the force. That is the kind of artificial division that you're making when you call part of a gravitational pull "centripetal force" and the rest as non-centripetal gravity. I said earlier that you aren't making an arbitrary division and that it's merely an artificial one, but one of my recent posts reveals that your division point is actually arbitrary. I can argue that the centripetal component is the part that would make the planet follow a circle rather than an ellipse, whereas you argue that it's the part that makes it follow the ellipse that it's following. These are two sensible division points, but they are different. Which one of these division points does the actual mechanism tell us is the correct one? The answer is, neither of them - the division is artificial and has no mechanistic role.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #425 on: 19/10/2018 08:32:37 »
Sorry, but RUBBISH, AND MORE RUBBISH ...
I´m not going to waste too much time refuting all your errors again and again! Just a couple of things ...
Did you read point 5 of my four days ago post?:
Quote from: rmolnav on 15/10/2018 11:29:28
A centripetal force (Fc) is the force that makes a moving object change direction.
It is not a particular force, but the name given to whatever force or combination of forces is responsible for a centripetal acceleration
( Centripetal Force - Summary – The Physics Hypertextbookhttps://physics.info/centripetal/summary.shtml )
Similar definitions can be find "all over the world".
I would add, as I previously did, that "name (is) given" because, for the period that centripetal acceleration does exist, the "JOB" of the that force (whatever its ESSENCE") is to cause it. And that "combination of forces" can include not "complete" forces (only one of their components).
And you say:
 
Quote from: David Cooper on 18/10/2018 21:37:50
... one of my recent posts reveals that your division point is actually arbitrary. I can argue that the centripetal component is the part that would make the planet follow a circle rather than an ellipse, whereas you argue that it's the part that makes it follow the ellipse that it's following.
NONE  of your posts reveals any "arbitrary" division point of me !
Please kindly tell us where do I "argue" what quoted. And don´t reply you have no time to look for it ...
It´s you who don´t understand, or don´r read carefully!
I´ve always had quite clear that an initial speed of the orbiting object, and gravity attraction from other nearby celestial object, are needed for the rotation ...
But, at any point of the orbit, gravitational pull vector can be perpendicular to the object´s speed vector, or not (apart from the case of circular orbit). And that pull vector can be divided into two components: one perpendicular, and the other tangential.
The "JOB" of the perpendicular component is to change the direction of the speed vector, towards the "center" of the orbit (more exactly towards the center of curvature of the orbit at that point) ... what ALL physicists (and most laymen) call CENTRIPETAL FORCE (obviously, without loosing its "gravitational" ESSENCE !!).
And the tangential component makes tangential speed size increase or decrease.
That is something an average teenager knows ... Is it too complicated for you to understand ??
I said what follows months ago. I´m afraid your so frequent errors are due not only to rather poor knowledge on physics Science, but also on Philosophy and Logics, or simply on the use of your own language: in your "arguments", you have been mixing the concepts of "essence" and "function" for months (at least) ! 
 

Apart from that ...
Quote from: David Cooper on 18/10/2018 21:37:50
When a tennis ball hits the net, does the force of the collision cause movement of the ball that leads to the collision?
Please kindly, when discussing physics (as a Science), use clearly identifiable terms. Otherwise confusion is almost unavoidable ... What do you exactly mean with "the force of the collision" ?? 
Because several "scientifically" called "forces" are present in that scenario !
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #426 on: 19/10/2018 12:02:23 »
Quote from: David Cooper on 18/10/2018 21:37:50
I can argue that the centripetal component is the part that would make the planet follow a circle rather than an ellipse.
In the singular case of a circular orbit same principles necessarily apply ... In this case gravitational pull keeps constantly perpendicular to the orbit, and therefore its "tangential" component is always null. Velocity vector doesn´t change in size, only in direction, due (as when an elliptical orbit) to the gravitational pull, whose "function" keeps now being 100% as centripetal force (what, together with the somehow initially given velocity, originates that "uniform circular movement", as it is called in this case) !
There are no different Physics whatsoever, as you have argued several times (not finding other "arguments" ...). Same basic Physics laws applied to different circumstances ... That simple !
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Offline RobC

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Re: Why do we have two high tides a day?
« Reply #427 on: 19/10/2018 15:11:03 »
This thread is almost as long-winded as the MH370 disappearance on the pilot's forum.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #428 on: 19/10/2018 18:08:20 »
Quote from: rmolnav on 15/10/2018 11:29:28
5)
A centripetal force (Fc) is the force that makes a moving object change direction
is not a particular force, but the name given to whatever force or combination of forces is responsible for a centripetal acceleration ( Centripetal Force - Summary – The Physics Hypertextbookhttps://physics.info/centripetal/summary.shtml )
When I chose that definition to include on my post #413, I didn´t realize that the inclusion of "or combination of forces" alludes to cases such as farther from the moon high tide.
At earth CM, moon´s pull vector and centripetal force required for earth´s revolving are equal: 100% of that gravitational pull exerts the function of centripetal force.
But at farthest hemisphere moon gravitational is smaller, there is a deficit (compared to required centripetal force) and an additional pull in the same sense is necessary to "combine" with moon´s pull to counter the deficit. That pull, on any considered portion of earth stuff, only can be exerted by the rest of our planet.
That additional pull is exerted both through contiguous material (especially on solid earth), and by own earth gravity.
Where solid earth, opposite pulls appear (exerted by the considered portion on the rest of the planet, through contiguous material), according to 3rd Newton´s Motion Law. All those internal stresses stretch solid earth.
On the oceans, as water follows its curved path (its revolving around the barycenter) partially due to its own weight, its inertial tendency to follow the tangent kind of diminishes (though very slightly) its weight, and water moves towards where that weight is smaller, and the farthest bulge builds.
In that last detail is where the phenomenon is very similar (though much, much smaller) to equatorial bulge formation, as far their causes (mentioned inertial effect) are concerned.
In both cases that "diminishing" of water weight can also be called a centrifugal force, because it is equivalent to a force vector, opposed to own earth´s pull on the water ...   
 
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #429 on: 19/10/2018 23:20:21 »
Quote from: RobC on 19/10/2018 15:11:03
This thread is almost as long-winded as the MH370 disappearance on the pilot's forum.

I hope you didn't wade through the whole thing. Maybe a warning needs to be added at the top to guide people to the right answer and away from centripetal farce explanation.
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #430 on: 19/10/2018 23:46:27 »
Quote from: rmolnav on 19/10/2018 08:32:37
NONE  of your posts reveals any "arbitrary" division point of me !
Please kindly tell us where do I "argue" what quoted. And don´t reply you have no time to look for it ...
It´s you who don´t understand, or don´r read carefully!

If you want to claim that I misrepresented your position, that's fine - it's hard to work out exactly what your position is when your wording of everything is so weird, all built into bloated masses of messy text decorated with bold print and exclamation marks, but your situation is hopeless. There is an arbitrary choice to be made as to how much of the force should be counted as centripetal depending on whether you consider it to be the amount that would result in a circular orbit or if you consider it to be the amount that maintains the elliptical orbit at any point in time, and whichever one you choose, you'd be allocating different amounts of the lifting of material further out to centrifugal force. Your artificial division point is arbitrary. That provides two rival proposed mechanisms (both of which are equally wrong), and it doesn't matter which of them you decide is yours.

Quote
I said what follows months ago. I´m afraid your so frequent errors are due not only to rather poor knowledge on physics Science, but also on Philosophy and Logics, or simply on the use of your own language: in your "arguments", you have been mixing the concepts of "essence" and "function" for months (at least) !

The errors are yours, darling. You're the one mauling the physics over and over again while believing that you understand it on the basis of your faulty application of half-learned rules.

Quote
Quote from: David Cooper on 18/10/2018 21:37:50
When a tennis ball hits the net, does the force of the collision cause movement of the ball that leads to the collision?
Please kindly, when discussing physics (as a Science), use clearly identifiable terms. Otherwise confusion is almost unavoidable ... What do you exactly mean with "the force of the collision" ??

I'm asking you if the force from the collision between ball and net causes the ball to hit the net (which would be reversed causality) or if the ball hitting the net causes the collision and generates the forces. The point here is that you somehow imagine that forces cause movement and that movement can't cause forces, but the consequence of that would be that in a case where a tennis ball is heading for a net, no forces can be generated by any collision between the two because you demand that the force happens first.

Take another case. In snooker, one ball is moving and another is stationary. The first ball hits the second one and stops, setting the other ball moving. I say that the movement of the first ball leads to a force being generated when the two balls collide, and that the force transfers the energy to the other ball, causing it to move. Movement causes a force and the force causes further movement. If you never have the original movement, the force is never generated. Quite why you need this explained to you is a mystery, but you've somehow become stuck in a place where you don't understand that movement can generate forces, thinking instead that it only works the other way round. It's a symmetric process.
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #431 on: 20/10/2018 00:34:40 »
To clarify the arbitrary divide further, in the case the centrifugal force making the material at the equator rise, we call it centrifugal force because there is movement away from the centre. With a moon moving away from the barycentre (because it's on an elliptical orbit rather than a circular one), when we use this same definition of centrifugal, we must regard the movement of the centre of mass of the moon away from the centre of the system as being driven by centrifugal force rather than having that description apply solely to the material further out than the centre of the moon, so the amount of declared centrifugal force depends on whether you start it at zero for the centre of mass or if you give it a positive value there, in which case there may sometimes be centrifugal force applying to every particle of the moon. At another point of the orbit where the moon's getting nearer to the barycentre rather than further away, the zero centrifugal point moves the other way from the centre of mass, and in some cases there may be no centrifugal force applying to any particle of the moon at all. This means that in some cases the tidal forces across the entire moon are presumably "explained" by differential gravity dictating how much centrifugal force applies at any point, while in other cases they are "explained" by differential gravity dictating how much negative differential centrifugal force applies at any point.

If we only start counting it as centrifugal force from the centre of mass, again it is differential gravity that dictates how much centrifugal and negative centrifugal force is in play at any given point, and it should be obvious to anyone that the whole business of counting up the centrifugal force is entirely superfluous, having absolutely no role in the mechanism.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #432 on: 20/10/2018 10:54:23 »
As  yesterday, I won´t waste too much time trying to make you understand many things you say are utterly wrong. I know it´d be to no avail ...
Just one thing, relative to one of the main reasons of your quite erroneous stand, that could help others learn that (just in case somebody doesn´t have it clear yet).
Quote from: David Cooper on 19/10/2018 23:46:27
... your situation is hopeless. There is an arbitrary choice to be made as to how much of the force should be counted as centripetal depending on whether you consider it to be the amount that would result in a circular orbit or if you consider it to be the amount that maintains the elliptical orbit at any point in time
"An arbitrary choice" when deciding "how much of the force should be counted as centripetal ...?" :
NOT AT ALL.
What quoted shows, once again, your really low education in Physics and Maths.
A much simpler case:
If we throw upwards and vertically a stone, all Physics variables will always have same vertical direction (speed, gravitational force, acceleration ...). Speed vector changes in size, but not in direction. Not necessary, I hope, to elaborate any further ...
But if we throw the stone with some inclination, earth´s gravitational pull vector (logically always vertical as before) produces two quite different effects: it changes speed size (as in above case), and also its direction.
It can easily be deduced that the trajectory will be parabolic, thanks to basic mathematical analysis, considering (at any point) separately the effect of the component of the weight in the direction of speed vector (which changes the size of the later), and the effect of the component perpendicular to the speed (which bends the trajectory).
That last component exerts what is called a centripetal force on the stone, because in an infinitesimal lapse of time any curved line is an infinitesimal arc of a circle, with same curvature as the line at that point ...
As somebody could not clearly understand that concept of “curvature” (most probably, you one of them):
"Curvature, in mathematics, (is) the rate of change of direction of a curve with respect to distance along the curve. At every point on a circle, the curvature is the reciprocal of the radius; for other curves (and straight lines, which can be regarded as circles of infinite radius), the curvature is the reciprocal of the radius of the circle that most closely conforms to the curve at the given point (see figure)”.
(Encyclopedia Britannica).
So, there is no arbitrary choice whatsoever, because it´s not me who choses "how much of the force should be counted as centripetal depending on whether you (I) consider ..."
It is just a rational way to analyze a physical phenomenon, really simple but not just "gravity" that only could be considered as a whole.
Educated adults learnt that as early as when teenagers !!
Whose situation is hopeless ??




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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #433 on: 21/10/2018 00:11:16 »
Quote from: rmolnav on 20/10/2018 10:54:23
As  yesterday, I won´t waste too much time trying to make you understand many things you say are utterly wrong. I know it´d be to no avail ...

You are the one who needs to understand them, but you're not open to learning.

Quote
"An arbitrary choice" when deciding "how much of the force should be counted as centripetal ...?" :
NOT AT ALL.
What quoted shows, once again, your really low education in Physics and Maths.

You may be right, but if so, that just makes it all the worse that you keep pushing the wrong answer on this issue while I was able to identify the right one for myself (differential gravity) without even researching it. They didn't teach much about centrifugal force though because they didn't regard it as a real force, so it should be no surprise if I make some mistakes in analysing your fake physics and your terrible wording used to describe it. I now realise that even when the moon is following part of its elliptical orbit where it's moving closer to the Earth, you probably still imagine centrifugal force to be acting on it, whereas I was thinking you counted it as zero at the centre of mass of the moon, but I suspect now that you don't. If you place the moon at the same distance from the Earth in part of its orbit where it's moving further out rather than further in, you presumably assert that there's the same amount of centripetal force in both cases, but this time you'll have a greater amount of centrifugal force, and that's where I see an arbitrary division point being selected. I may not have found the right way to explain how there is a difference here that leads to you making an arbitrary divide, but I can see very clearly that you are making an arbitrary divide there, so I'll have another go:-

In cases where the moon is maintaining distance from the Earth rather than moving further in or out, the two forces (centripetal and centrifugal) will be in balance (from the rotating frame's perspective). What's happening then when the moon's moving out and the two forces aren't in balance at the moon's centre of mass? Why would you chose the centre of mass as your division point for your artificial divide between the two mechanisms when the forces are no longer in balance there? If the centrifugal force is stronger than the centripetal force at this centre of mass, it is also stronger than the centripetal force nearer to the Earth than the moon's centre of mass, and this must apply to the entire moon, so it seems odd to make a distinction between the two halves of the moon. In this particular situation, you could assert that centrifugal force wins out over centripetal force all the way through rather than just from the centre out to the far side. In the opposite case where the moon is moving closer to the Earth, you would likewise have to say that centrifugal force is losing out to centripetal force the whole way through the moon and not just from the centre to the near side. In one case, the tidal forces would then be explained in their entirety by centrifugal force out-gunning centripetal force and doing so more strongly further out, so you don't need the near side to be lifted at all as the centre of the moon is being lifted away from it in the opposite direction. In the other case, the tidal forces would need to be explained in their entirety by the Earth's stronger pull closer to it, which is differential gravity. What you're doing though is locking the division point to the centre of mass, but what justification do you have for putting it there rather than cutting the cake the way I've just outlined where the whole thing can be accounted for by centrifugal force out-powering the centripetal force in some parts of the orbit and the whole thing being accounted for by differential gravity at other parts of the orbit? Both ways of cutting the cake are equally valid (and they're both equally wrong as mechanistic explanations of events).

Quote
So, there is no arbitrary choice whatsoever, because it´s not me who choses "how much of the force should be counted as centripetal depending on whether you (I) consider ..."
It is just a rational way to analyze a physical phenomenon, really simple but not just "gravity" that only could be considered as a whole.
Educated adults learnt that as early as when teenagers !!
Whose situation is hopeless ??

I was imagining before that you were giving different values to the size of the centripetal force in two cases where the moon is at the same distance form the Earth but in one case moving away and the other case moving nearer, and the reason I thought you were doing that is that you always make your divide at the centre of mass of the moon. Now, I assume that you aren't asserting different amounts of centrifugal force for those two cases, but that makes your division point questionable. There remains room for a rival place to cut the cake which renders your chosen place for doing it arbitrary.

Imagine three objects replacing the moon such that they aren't gravitationally tied together, one of them at the centre of mass, another at NE (nearest the Earth) and the third at FE (furthest from the Earth). A different amount of centrifugal and centripetal force should be applied to each rather than applying the CoG's value of centripetal force to all three - to apply the CoG's value to NE and FE is clearly wrong in this new situation. If you make this sudden replacement (replacing the moon with these three objects), you might find that the centripetal force is stronger than the centrifugal force for NE while the centrifugal force is stronger than the centripetal force for FE, but if you do it at other points of the moon's orbit, there will be times when the centrifugal force is stronger than the centripetal force for both FE and NE, or the opposite.

Now imagine more than just three objects on our straight line - we could have an infinite series of them all the way from just above the Earth to far out beyond the moon. In every case, there will be one object which has a match between the centripetal and centrifugal forces that apply to it, and that  object is in the correct location for making your artificial divide. It is not the same as the centre of gravity of the moon, so we have two rival places for making an artificial divide, and our decision to choose one over the other is arbitrary.
« Last Edit: 21/10/2018 00:14:58 by David Cooper »
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #434 on: 21/10/2018 08:17:26 »
What a confusing and misleading exposition !
And you even told me:
Quote from: David Cooper on 19/10/2018 23:46:27
... it's hard to work out exactly what your position is when your wording of everything is so weird, all built into bloated masses of messy text ...??
Surely my expositions could be better. In your case it is much worse... After all, English is your mother tongue, but not mine !
Later on another post I´ll try and explain, especially for others who may be interested, the singular case of the rotation of the moon around earth-moon barycenter  Compared to earth´s revolving, there are some differences, and I want to elaborate them on with more time. I´ll make an effort to put it as concise and clear as possible ...
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #435 on: 21/10/2018 12:31:20 »
Quote from: rmolnav on 21/10/2018 08:17:26
In cases where the moon is maintaining distance from the Earth rather than moving further in or out, the two forces (centripetal and centrifugal) will be in balance (from the rotating frame's perspective). What's happening then when the moon's moving out and the two forces aren't in balance at the moon's centre of mass? Why would you chose the centre of mass as your division point for your artificial divide between the two mechanisms when the forces are no longer in balance there? If the centrifugal force is stronger than the centripetal force at this centre of mass, it is also stronger than the centripetal force nearer to the Earth than the moon's centre of mass, and this must apply to the entire moon, so it seems odd to make a distinction between the two halves of the moon. In this particular situation, you could assert that centrifugal force wins out over centripetal force all the way through rather than just from the centre out to the far side
Inertia (and its manifestation as centrifugal force), especially when elliptical orbits, is rather tricky to grasp. And you often mix up forces with "movements", what doesn´t make things any easier ...
In the first place, some "singularities" of our moon are:
1) It´s center of mass it is not located where its geometric center (if it were perfectly spherical, the center of the sphere). Visible part of its crust is slightly denser than the rest, and the center of mass is some 2km closer to earth than the geometric center.
2) That center of mass would also be the center of "gravity", if in relation to a uniform gravitational field. That would "almost" be the case if the celestial object causing the gravitational field were a far distant star ... But, leaving aside sun´s gravity, earth is sufficiently close to the moon to cause a varying pull, with a negative gradient in relation to distance.
Therefore, the earth-related center of gravity is even a little closer than above mentioned 2 km.
3) And that doesn´t change in time, whatever the movement of the moon, precisely because it is tidal locked to earth.
When considering moon´s dynamics and saying closer/farther hemispheres, to be 100% precise we should say e.g. "moon´s parts closer/farther than that earth-related CG). But the differences are relatively very small, it would make difficult the exposure ... and nobody does.
THAT SAID, for the sake of simplicity, let us now suppose moon´s orbit were a perfect circle.
It is actually at mentioned "earth-related center of gravity" were gravitational pull per unit of mass:
- GM/d² (Gravitational universal constant, earth´s Mass and distance to earth´s CM)
and centrifugal force per unit of mass:
- ω²r (angular speed ω and distance to barycenter r)
are in balance with each other.
On moon´s farther parts not only gravity is smaller: centrifugal force, on the contrary, is bigger.
And on moon´s closer parts gravity is bigger, and centrifugal force is smaller.
On farther parts centrifugal forces prevail, and on closer parts gravitational pulls prevail.
That is why moon got stretched long ago ...
As the post is already rather long, I´ll leave the actual case of an elliptical orbit for another post.
 
 
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #436 on: 21/10/2018 17:33:14 »
Quote from: David Cooper on 21/10/2018 00:11:16
In cases where the moon is maintaining distance from the Earth rather than moving further in or out, the two forces (centripetal and centrifugal) will be in balance (from the rotating frame's perspective). What's happening then when the moon's moving out and the two forces aren't in balance at the moon's centre of mass? Why would you chose the centre of mass as your division point for your artificial divide between the two mechanisms when the forces are no longer in balance there? If the centrifugal force is stronger than the centripetal force at this centre of mass, it is also stronger than the centripetal force nearer to the Earth than the moon's centre of mass, and this must apply to the entire moon, so it seems odd to make a distinction between the two halves of the moon. In this particular situation, you could assert that centrifugal force wins out over centripetal force all the way through rather than just from the centre out to the far side.
Quote from: rmolnav on 21/10/2018 12:31:20
Therefore, the earth-related center of gravity is even a little closer than above mentioned 2 km.
3) And that doesn´t change in time, whatever the movement of the moon, precisely because it is tidal locked to earth.
Minutes ago I happened to be ruminating on that, and I realized that it is not fully correct: in the actual case of elliptical orbit, things change (though very, very slightly).
The center of mass (as referred to on point 1 of my last post) is not always exactly at its closest to earth´s location.
Moon´s tidal locking was asymptotical: its previous spinning angular speed got its actual value (2π radians/some 28 days) at an ever smaller pace, partially due to the high moon´s angular momentum.
And, as far as I can understand, same reason doesn´t let earth-related tidal effects to change significantly that angular speed in only weeks ...
That must be one of the reasons of moon´s relative wobbling (the visible part of the moon is not always exactly the same) ...
I´ll think it over a little longer. In any case, I´m afraid I´m trying to be too precise, just to refute a kind of bizarre argument of D.C. If, in a certain moment, the exact area where earth´s pull and centrifugal forces balance each other is a little closer or farther, that has no importance for the validity of "my" model ... Closer and farther "hemispheres" (where either gravitational pulls or centrifugal forces prevail) would be slightly bigger or smaller. But that doesn´t invalidate the model whatsoever !
There wouldn´t be any "arbitrary" election whatsoever, just a lack of information and mathematical tools to be "almost 100%" precise !!
You seem to be running out of "arguments".
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #437 on: 21/10/2018 23:36:59 »
Running out of arguments? Hardly. Your model's already been disproved in multiple ways (such as in cases where the actual acceleration's acting in the opposite direction to the one where you supposedly have centrifugal force working, and in straight-line cases where there is no centripetal force). What I'm doing now is looking at your broken argument and finding an arbitrary aspect to it (quite in addition to its well-established brokenness).

Quote
On moon´s farther parts not only gravity is smaller: centrifugal force, on the contrary, is bigger.
And on moon´s closer parts gravity is bigger, and centrifugal force is smaller.
On farther parts centrifugal forces prevail, and on closer parts gravitational pulls prevail.

You assert that for one side, centrifugal force wins out over gravitational force, while at the other side it loses out to it, but if you cut the cake in a different way, you could say that centrifugal force wins out the whole way through on some occasions, while on other occasions it loses out the whole way through. The chosen location of the point beyond which winning turns to losing is arbitrary - you always seem to put it in the middle, but it could be further in or further out than the whole moon (or planet, if you'd rather work the other way round - they both follow the same shape of orbit, so whatever applies to one of them in relation to this point also applies to the other).

For a correct analysis of what's going on, you should be looking at the amount of gravitational pull (from the other body) applying at any given point and going by that alone - considering the centrifugal force is then superfluous as it doesn't change the result, but you're determined to include it. The only way you can make the centrifugal force appear relevant is by fiddling the amount of gravitational force that you're counting as centripetal force at different distances from the source, and then by using that artificial value instead of using the actual gravitational force (so that the net effect is the same as you get just by considering the gravitational force alone). We already know that you're making an artificial division in this way and that your explanation is broken, but I'm now pointing out that you're doing it in an arbitrary way too (in terms of where you cut the cake).
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #438 on: 22/10/2018 07:39:33 »
Quote from: David Cooper on 21/10/2018 23:36:59
You assert that for one side, centrifugal force wins out over gravitational force, while at the other side it loses out to it, but if ... The chosen location of the point beyond which winning turns to losing is arbitrary - you always seem to put it in the middle, but it could be further in or further out than the whole moon (or planet, if you'd rather work the other way round - they both follow the same shape of orbit, so whatever applies to one of them in relation to this point also applies to the other).
The scientifically "chosen location", obviously matches with basic Physics laws: it is where
 
Quote from: rmolnav on 21/10/2018 12:31:20
...gravitational pull per unit of mass:
- GM/d² (Gravitational universal constant, earth´s Mass and distance to earth´s CM)
and centrifugal force per unit of mass:
- ω²r (angular speed ω and distance to barycenter r)
are in balance with each other.
... as I already said yesterday. You didn´t even read it, did you ??
Once again, on farther and closer parts that balance can´t exist (basic Maths):
- On farther parts centrifugal force prevails.
- On closer parts gravitational pull prevails.
NOW IT IS YOUR TURN:
Please kindly give us an example of location (and "occasions") where ...
 
Quote from: David Cooper on 21/10/2018 23:36:59
... you (could) cut the cake in a different way,  (and) you could say that centrifugal force wins out the whole way through on some occasions, while on other occasions it loses out the whole way through.
... and explain your "reasons" !!
Otherwise, anybody could tell you just "invent" false scenarios due to your "hopeless situation" ...
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #439 on: 22/10/2018 18:51:19 »
A couple of days ago I started a post saying::
Quote from: rmolnav on 20/10/2018 10:54:23
As  yesterday, I won´t waste too much time trying to make you understand many things you say are utterly wrong. I know it´d be to no avail ...
Just one thing, relative to one of the main reasons of your quite erroneous stand, that could help others learn that (just in case somebody doesn´t have it clear yet).
Another of the main reasons of your quite erroneous stand is the "bizarre" and erroneous relation you suppose between force and movement, as when you say, e.g.:
Quote from: David Cooper on 19/10/2018 23:46:27
Take another case. In snooker, one ball is moving and another is stationary. The first ball hits the second one and stops, setting the other ball moving. I say that the movement of the first ball leads to a force being generated when the two balls collide, and that the force transfers the energy to the other ball, causing it to move. Movement causes a force and the force causes further movement. If you never have the original movement, the force is never generated. Quite why you need this explained to you is a mystery, but you've somehow become stuck in a place where you don't understand that movement can generate forces, thinking instead that it only works the other way round. It's a symmetric process.
What in bold is correct, but not the rest. As I´ve told you several times, you not only have serious problems with Physics: neither Logics is your forte ...
One thing is to be one of the "necessary conditions" for something to happen, and another to be the "cause".
One of the conditions required for me to run is, e.g., to be able to breath. If I couldn´t breath it´d be impossible for me to run ...
But we couldn´t say that breathing is what causes me to run !!
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