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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 Le Repteux

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Re: Why do we have two high tides a day?
« Reply #280 on: 13/09/2018 15:41:31 »
Quote from: rmolnav on 13/09/2018 08:51:50
Are you sure you fully understand the term "inertial", and its different ways it can manifests itself ??
Yes, I do know what inertia means, and I'm even studying a mechanism that might very well be the cause of both mass and inertia.. The inertial inertial bulge I'm talking about is similar to the equatorial one, it is also due to rotation, that I call inertial to distinguish it from an orbital one. The problem is that it is the same gravitational pulling that produces the two kinds of rotation, inertial or gravitational. At first, David's explanation seemed logical, so I thought it was right, then since you looked so convinced, I decided to really study your proposition, and I chose to take a closer look at the transition between an equatorial and an orbital bulge.

Quote from: rmolnav on 13/09/2018 08:51:50
1) "Tidal bulge" is a kind of "redundancy" ... The "bulges" we are referring to are always one of the manifestation of tides, therefore they are always "tidal".
An equatorial bulge is not a tide, so maybe you don't really understand what inertial means after all! ( just joking!) Your analysis (which seems to be the mainstream one) is perfectly logical, but David's one is also perfectly logical, so to be able to find the truth, I think we have to do what I did, which is to look for a way to link the two explanations. That's what I'm doing in my last posts, but it doesn't seem to interest either of you.

I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon. By chance, that single explanation seems to contain your two explanations: if we stop the tangential speed of the earth/moon system, it collapses but the tides don't disappear as in David's explanation, and as in your explanation, tangential speed is also what produces the tides, except that if we reversed the daily rotation of the earth, the tangential orbital speed of the two halves of the earth would change, and it would necessarily affect the tides. If we could manage so that the two halves would be going at the right orbital speed all the time for instance, then there would be no tides, but if we stopped the orbital speed of the earth/moon system, it would not prevent them to rebuild.

« Last Edit: 13/09/2018 16:05:04 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 #281 on: 13/09/2018 20:02:59 »
Quote from: Le Repteux on 13/09/2018 15:41:31
I think we have to do what I did, which is to look for a way to link the two explanations. That's what I'm doing in my last posts, but it doesn't seem to interest either of you.

I couldn't follow what you were describing, but we already have an answer which makes the linkage between the two explanations fully clear. My explanation treats the gravitational force as the only force involved, but rmolnav's explanation introduces an artificial complication by dividing the gravitational force into two components of the same force, one of which is labelled as centripetal force while the other simply remains as gravitational force. The component that he calls centripetal force is the component that makes parts of a body move along the orbital path that they naturally "want" to follow (if no other force was applied to them), and that material is able to follow that path because the material further out (which wants to pull away outwards) and the material further in (which wants to pull inwards) cancel each other out, pulling equally on the material near the middle (which is therefore not pulled off the path it wants to follow). Having made this artificial distinction between gravity (non-centripetal) and gravity (centripetal), he then uses the former component to complete the fake explanation of events on the inside (the stronger gravity there pulling up a "bulge"), and more of the latter component to complete the fake explanation of events on the outside (with the extra centrifugal force throwing out a "bulge"). It is a wonderful example of an exercise in self-deception where people (including good scientists) lose sight of the basic physics and create imaginary mechanisms that are far more complex than the simple reality behind what's going on. In this case, straight-line differential gravity accounts for all the action.

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I said that while the earth's C.G. was going at the right orbital speed around the moon

It isn't - the part that is is near the centre of gravity, but it's slightly nearer to the other body due to differential gravity varying more over the near side (of the Earth to the moon) than the far side.

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if we stop the tangential speed of the earth/moon system, it collapses but the tides don't disappear as in David's explanation, and as in your explanation, tangential speed is also what produces the tides...

And with rmolnav's explanation, if you stop the Earth and moon, the component of gravity that is labelled as centripetal force loses that label and becomes reunited with the rest of the gravitational force as gravitational force. That's the game he's playing: there is only one force involved throughout, but he takes part of it and rebrands it as something else and then makes an artificial mixture of the two which, unsurprisingly adds up to the same amount of force as in my explanation because there is only one force involved.
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #282 on: 13/09/2018 20:37:10 »
Quote from: David Cooper on 13/09/2018 20:02:59
And with rmolnav's explanation, if you stop the Earth and moon, the component of gravity that is labelled as centripetal force loses that label and becomes reunited with the rest of the gravitational force as gravitational force.
It is so with rmolnav's explanation, but not with mine. After you began to answer, I added a few lines on the part you quoted, so please read them and tell me if it changes anything.
Quote from: Le Repteux
By chance, that single explanation seems to contain your two explanations: if we stop the tangential speed of the earth/moon system, it collapses but the tides don't disappear as in David's explanation, and as in your explanation, tangential speed is also what produces the tides, except that if we reversed the daily rotation of the earth, the tangential orbital speed of the two halves of the earth would change, and it would necessarily affect the tides. If we could manage so that the two halves would be going at the right orbital speed all the time for instance, then there would be no tides, but if we stopped the tangential speed of the earth/moon system, it would not prevent them to rebuild.

Quote from: David Cooper on 13/09/2018 20:02:59
It isn't - the part that is is near the centre of gravity, but it's slightly nearer to the other body due to differential gravity varying more over the near side (of the Earth to the moon) than the far side.
I don't get it, can you change your wording please?
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #283 on: 13/09/2018 21:03:18 »
Quote from: Le Repteux on 13/09/2018 20:37:10
It is so with rmolnav's explanation, but not with mine. After you began to answer, I added a few lines on the part you quoted, so please read them and tell me if it changes anything.

Your last edit to it was made hours before I read it, so it doesn't change anything, but I haven't managed to follow your explanation. I was hoping that my latest post would help you understand what's going on well enough to apply that to your own thought experiment and fix whatever issues you have there.

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Quote from: David Cooper on 13/09/2018 20:02:59
It isn't - the part that is is near the centre of gravity, but it's slightly nearer to the other body due to differential gravity varying more over the near side (of the Earth to the moon) than the far side.
I don't get it, can you change your wording please?

The gravitational pull of the moon on the near side of the Earth is strongest, while the pull on the far side is weakest, but the difference in its strength between the near side and centre is greater than the difference in strength between the centre and far side, so the material that follows its natural path is not at the centre of the Earth, but a small distance away from there towards the near side, and it's further away from the centre the closer together the Earth and moon are. This is a very minor point, but it's worth considering if you want to put the artificial divide (between gravitational force labelled as gravitational force and gravitational force labelled as centripetal force) in the right place, and indeed it's crucial if you want to get the explanation of the tides wrong with high precision.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #284 on: 14/09/2018 07:39:37 »
Quote from: Le Repteux on 13/09/2018 20:37:10
Quote from: David Cooper on Yesterday at 20:02:59
And with rmolnav's explanation, if you stop the Earth and moon, the component of gravity that is labelled as centripetal force loses that label and becomes reunited with the rest of the gravitational force as gravitational force.
It is so with rmolnav's explanation, but not with mine.
I really can´t understand D.C. "blind" stand ...
" ... the component of gravity that is labelled as centripetal force" (AS A FUNCTION, NOT AS ITS ESSENCE), logically " ...if you stop the Earth and moon ... "losses that label", because if no revolving, NO CENTRIPETAL FORCE FUNCTION IS NECESSARY any more ...
But it doesn´t "become reunited with the rest of the gravitational force as gravitational force".... because it NEVER LOST ITS GRAVITATIONAL FORCE ESSENCE ...
On a couple of previous replies to D.C. I referred to that kind of philosophical BASIC point, to no avail !!
Similarly to what I already said, when Donald Trump acts as U.S. President (FUNCTION), he doesn´t loose his ESSENCE of man !!
Am I exaggerating when I say D.C. has problems relative to not only his "sight" and basic Physics, but also to basic Logics and Philosophy (or even to his own Language) ... ??
The FUNCTION of centripetal force can be assumed by ESSENTIALLY different forces: gravity, the tension of a string, the inward push of a locomotive wheel rim, the friction between road and a car tires ... But none of them looses its essence when bending the trajectory of the moving object !!
No wonder first thing he said about pretty clear concept of "centripetal force" was that it is a "grey" area, and that many people out there say things not quite rational about it  ... !!
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #285 on: 14/09/2018 15:11:18 »
Quote from: rmolnav on 14/09/2018 07:39:37
Am I exaggerating when I say D.C. has problems relative to not only his "sight" and basic Physics, but also to basic Logics and Philosophy (or even to his own Language) ... ??
From my viewpoint, you are effectively exaggerating, but most people would think you're not, because most of us don't realize that we are all blind to other's arguments. What we usually do is chose our party and stick to it. It took a while before I decide to study correctly your proposition, and I could only do that while looking for a way to unite the two thinking. It's chance that permits us to change our mind, nothing else than chance, and it takes time for chance to do its duty, a lot more time than we think, so it is useless to think that it is others that don't understand, it doesn't lead to understanding, and saying it just aggravates the problem.

There might be something important to discover if we try to unite your two propositions, something we don't yet know about gravitation, so I'll go on trying to find the link. David is right and you are too, but there seems to be no link between the two and it is impossible, so let's find that link instead of pulling constantly on our own side like orbiting lonesome cowboys. :0) The pulling is normal, it's how things work, but let's not aggravate it.

I made a proposition that seemed to reconcile the two thinking, but neither of you commented it. Who will be the first to comment? You're the only ones that can tell if it seems to dovetail yours or not. Here is that proposition again:
Quote from: Le Repteux
I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon. By chance, that single explanation seems to contain your two explanations: if we stop the tangential speed of the earth/moon system, it collapses but the tides don't disappear as in David's explanation, and as in your explanation, tangential speed is also what produces the tides, except that if we reversed the daily rotation of the earth, the tangential orbital speed of the two halves of the earth would change, and it would necessarily affect the tides. If we could manage so that the two halves would be going at the right orbital speed all the time for instance, then there would be no tides, but if we stopped the orbital speed of the earth/moon system, it would not prevent them to rebuild.
« Last Edit: 14/09/2018 18:46:41 by Le Repteux »
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #286 on: 14/09/2018 17:18:30 »
Quote from: David Cooper on 13/09/2018 20:02:59
My explanation treats the gravitational force as the only force involved, but rmolnav's explanation introduces an artificial complication by dividing the gravitational force into two components of the same force, one of which is labelled as centripetal force while the other simply remains as gravitational force. The component that he calls centripetal force is the component that makes parts of a body move along the orbital path that they naturally "want" to follow (if no other force was applied to them)
The orbital path is due to two different kinds of motion, direct and tangential, so both of you need those two components to explain it. Using only the direct one like you do doesn't explain the path, and using a combination of the two motions the way rmolnav uses it doesn't explain why the tides don't need tangential motion to build up. The only way that seems to work is using orbital motion as I did, which is while considering that the two halves of the earth are not going at the right speed during their transit. It produces the same bulges during the transit, which is what rmolnav defends, and the bulges stay there if we stop the tangential motion, which is what you defend.
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #287 on: 14/09/2018 21:26:06 »
Quote from: David Cooper on 13/09/2018 21:03:18
The gravitational pull of the moon on the near side of the Earth is strongest, while the pull on the far side is weakest, but the difference in its strength between the near side and centre is greater than the difference in strength between the centre and far side, so the material that follows its natural path is not at the centre of the Earth, but a small distance away from there towards the near side, and it's further away from the centre the closer together the Earth and moon are. This is a very minor point, but it's worth considering if you want to put the artificial divide (between gravitational force labelled as gravitational force and gravitational force labelled as centripetal force) in the right place, and indeed it's crucial if you want to get the explanation of the tides wrong with high precision.

I've missed a factor there though, because I divided the globe into two equal halves around the centre of gravity instead of making the section between them curved such that every point on the join is the same distance away from the moon's centre of gravity, and that should cancel out some of the previously mentioned effect, and I can't rule out the possibility that it cancels out 100% of it - there's some fun maths to do on that to find out, though I can't justify putting the time into it myself at the moment.
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #288 on: 14/09/2018 22:14:24 »
Quote from: rmolnav on 14/09/2018 07:39:37
But it doesn´t "become reunited with the rest of the gravitational force as gravitational force".... because it NEVER LOST ITS GRAVITATIONAL FORCE ESSENCE ...

There's only one gravitational pull being applied by the moon to the Earth (or by the Earth to the moon). It pulls most strongly on the near side and least strongly on the far side. That's differential gravity - the force diminishing over distance. What you're doing is rebranding the amount of that force applied at the centre of the Earth as centripetal force, and then at the near side where the gravitational force is stronger, you're attributing the excess to differential gravity, while at the far side where the gravitational force is weaker, you're attributing the bulge there to centrifugal force flinging the material outwards.

In a case where there's no orbit (because the two bodies are falling directly towards each other, you have no centripetal force (or nothing that can currently be described as such by current definitions of the word), so all the action has to be attributed to differential gravity.

This means that when the Earth and moon are at a certain distance apart, such as 250,000 miles, there is an amount of gravity being applied by the moon to the centre of the Earth which we can call n. If we were to stop the system to remove the orbit, the amount of gravity applying there would still be n (for a moment). It's identical in both cases,  but in the former case, you're artificially reducing that value from n to 0 and you're then rebranding the gravitational force as centripetal force. On the near side where the gravitational force is n+p, you continue to treat the n as 0 (because you say the centripetal force is n instead), and you count p as gravitational force. On the far side where the gravitational force is n-q, you continue to treat the n as zero (because you say the centripetal force is n instead), and the q becomes the centrifugal force that tries to throw material outwards (or it becomes a centripetal force of -q). But all you're actually doing is chopping up gravitational force into artificial chunks with silly labels attached to them which pretend there's something more complicated going on than the actual reality of it. You say it never lost its gravitational force essence, but in reality it never stopped being gravitational force in any way at all.

Quote
On a couple of previous replies to D.C. I referred to that kind of philosophical BASIC point, to no avail !!
Similarly to what I already said, when Donald Trump acts as U.S. President (FUNCTION), he doesn´t loose his ESSENCE of man !!

Why are you telling me that when you're the one making the mistake of treating him as if he's no longer a man? With this tide business, we're looking for an explanation to the tidal forces, and what you're doing is splitting off a massive chunk of gravitational force and presenting it as centripetal force while you keep some of it back and call it gravitational force, though only applying that at the near side, and then you introduce a centrifugal force at the far side which is also a deficit of centripetal force (not enough of it to hold that material in as strongly as it holds material at the centre) and which is also a deficit of gravitational force due to gravity decreasing over distance.

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Am I exaggerating when I say D.C. has problems relative to not only his "sight" and basic Physics, but also to basic Logics and Philosophy (or even to his own Language) ... ??

The problems are all at your end, and all the problems you mention certainly do apply there.

Quote
The FUNCTION of centripetal force can be assumed by ESSENTIALLY different forces: gravity, the tension of a string, the inward push of a locomotive wheel rim, the friction between road and a car tires ... But none of them looses its essence when bending the trajectory of the moving object !!
No wonder first thing he said about pretty clear concept of "centripetal force" was that it is a "grey" area, and that many people out there say things not quite rational about it  ... !!

The reason I don't like gravity being described as centripetal force is precisely that people like you are so misled by that usage that you end up treating the centripetal force as a different mechanism from gravitational force. The only mechanism in play here is straight-line differential gravity - the other mechanism is a ludicrous fabrication which comes out of people like you confusing themselves with words and dodgy labels.
« Last Edit: 14/09/2018 22:17:27 by David Cooper »
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #289 on: 14/09/2018 22:55:49 »
Quote from: Le Repteux on 14/09/2018 15:11:18
...but there seems to be no link between the two...

I've shown you the link between the two - they produce the same end result because they are doing the same thing, my approach getting there directly by recognising a single thing as itself (so that there is only one mechanism needed), and rmolnav's approach where he splits that same single thing into two things which he then treats as separate mechanisms, bringing in a whole lot of unnecessary complexity which is only needed because of his error in splitting something up that should be considered as the single thing that it actually is.

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I made a proposition that seemed to reconcile the two thinking, but neither of you commented it. Who will be the first to comment?

Quote from: Le Repteux
If we accelerate a satellite which is already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon.

A satellite is moving relative to the Earth, so slowing it down relative to the Earth makes it fall. Let's suppose we have two satellites crossing over the near point (nearest to the moon) in opposite directions. If we slow them down, they both fall, even though one is going faster than the Earth along the Earth's orbit and the other satellite is going slower than the Earth along that path. You are merely adding confusion to things by providing an illustration that doesn't shed light on the issue under discussion here.

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By chance, that single explanation seems to contain your two explanations:

It doesn't relate to either of them. You need imagine your satellites as stationary relative to a non-rotating Earth and then think about how they'll fall. The one on the near side will fall down more slowly than it would if it was 90 degrees further round the planet at the side because the moon's gravitational pull on it will be slightly higher than it is on the Earth as a whole, whereas the one on the far side will fall down more slowly too because the moon's gravity is pulling the Earth away from it slightly more strongly than it is pulling the satellite.

Quote
...except that if we reversed the daily rotation of the earth, the tangential orbital speed of the two halves of the earth would change, and it would necessarily affect the tides. If we could manage so that the two halves would be going at the right orbital speed all the time for instance, then there would be no tides,

That is the bit I can't make sense of. What is this right orbital speed that will eliminate the tides?

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but if we stopped the orbital speed of the earth/moon system, it would not prevent them to rebuild.

And I don't know how you're getting there either, which is why I didn't comment before.

Quote from: Le Repteux on 14/09/2018 17:18:30
The orbital path is due to two different kinds of motion, direct and tangential, so both of you need those two components to explain it.

I don't need it at all, and relativity tells you that - that's why I was so sure that I was right when I began questioning rmolnav about the involvement of centripetal force - the perpendicular movement is very clearly a complete irrelevance to the mechanism. Whatever perpendicular movement there is, it simply applies to all the components equally, just like when you throw a ball to and fro across a train carriage (while sitting inside it at opposite sides from a friend) and you can explain the entire mechanism behind each throw without considering the movement of the train even though the train may be doing 200mph. If it's a round carriage (which, while impractical, could certainly be made), we don't need to factor in the tangential movement when the ball reaches either side because that tangential movement is in the same direction as the one the train is moving in. There is no tangential force outwards away from the direction the train is moving in, so why would you want to ram it into the calculations?
« Last Edit: 14/09/2018 22:57:58 by David Cooper »
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #290 on: 15/09/2018 08:18:07 »
Quote from: Le Repteux on 12/09/2018 14:01:57
What about the NOAA explanation then? Can we show that their calculation or their drawings are wrong for instance? Unfortunately, there is no calculation to support their drawings, and the drawings themselves are only qualitative.
I have to further "work" on last posts, before replying them. But one thing now: you and D.C. talk about rmolnav theory ... It is not just "mine":
It is true that since my very first post here (#20, 29/5/2015 !!), I´ve had quite clear the importance of earth circular movement "dancing" with the moon, and inherent centripetal (function exerted by moon´s gravity and internal stresses) and centrifugal forces. Then I said:
"1) “Centrifugal” force is NOT a forbidden word whatsoever: it is just a poorly understood and poorly explained force. I´m not going to deliver now any further explanation about why I say so, perhaps in another post if found convenient" ...
But a couple of aspects of my stand did change, especially in relation with my discussion with the NOAA scientist I already mentioned, and I even included part of our discussion.
Initially I didn´t agree 100% with them , but from our discussion, and further ruminating of facts, I got to fully understand and agree with THEIR stand  ...
Regarding with what quoted, I imagine you refer to NOAA work:
 https://tidesandcurrents.noaa.gov/restles3.html
But we (Colin2B and I) have also mentioned here:
https://tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf
a more than 300 page thorough work on tides, as referred to on #268.
Neither on what quoted "there is (any) calculation to support their (drawings) theory", as you say in relation to the shorter work.
But the long one goes through every imaginable detail affecting tides, using all available tools such as astronomical calculations (based on those forces they consider are actually acting on earth material stuff), real measurements in situ and from satellites, etc.
E.g., just on page 11 of the "Introduction" Dr. Parker says:
"The details of how these overtides (higher harmonics of the astronomical constituents) and compound tides (new tidal constituents resulting from the interaction of two or more astronomical constituents) are actually generated have often been ignored. In some cases the mechanisms that produced them were not even fully understood, and the shallow-water constituent frequencies were merely determined by adding or subtracting astronomical frequencies. Not understanding the nonlinear hydrodynamics that generate these constituents can cause problems in tidal prediction accuracy …"
Despite all that thorough work, and all the prizes he got (see #268), could Mr. Parker be wrong on the very core of tidal generating forces ... ??
Well, theoretically he could ... But the odds would be, I guess, less than 1:1000 !! 

 
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #291 on: 15/09/2018 12:11:54 »

Quote from: Le Repteux on 14/09/2018 15:11:18
Quote from: rmolnav on Yesterday at 07:39:37
Am I exaggerating when I say D.C. has problems relative to not only his "sight" and basic Physics, but also to basic Logics and Philosophy (or even to his own Language) ... ??
From my viewpoint, you are effectively exaggerating, but most people would think you're not, because most of us don't realize that we are all blind to other's arguments
O.K. Naturally you are quite free to being so understanding D.C. ...
As I´ve posted here many times, I´ve discussed the issue on many sites, and even directly by email with many scientists, always giving logically MY arguments, but I´ve changed some details as said on my last post a few hours ago.
But D.C. time and again has loftly despised:
A) What said by eminent scientists, such what NOAA´s one I interchanged emails with ... That scientist had told me:
""The publication you are referring to is "Our Restless Tides", a 10-page pamphlet developed in the 1950's to provide a basic description of the forces which create the tides.  It's intended audience were the grade school children and adults of that time.  It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.  Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force”",
I quoted that here (#155), and D.C. replied:
"Here, we see that the scientist may not have a firm understanding of the tides himself though - all he's doing here is trying to get rid of the idea of centrifugal force by referring to the real mechanism that is behind centrifugal force, but he hasn't realized that this isn't the mechanism behind the tides”  ...
and, as I´ve also said many times, he is the author (or, at least one of them) of
https://tidesandcurrents.noaa.gov/restles3.html ...
...where they say:
"1. The Effect of Centrifugal Force. It is this little known aspect of the moon's orbital motion which is responsible for one of the two force components creating the tides. As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force. And, since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon. This fact is indicated by the common direction of the arrows (representing the centrifugal force Fc) at points A, C, and B in Fig. 1, and the thin arrows at these same points in Fig. 2.
It is important to note that the centrifugal force produced by the daily rotation of the earth on it axis must be completely disregarded in tidal theory. This element plays no part in the establishment of the differential tide-producing forces"
B) He also despise first line dictionary definitions of "centripetal force" ... (Again: NOT as a "new" force, independent from others, but as a FUNCTION, a kind of "job", essentially real forces (such as gravity) can exert when acting "transversely" (not with null component perpendicular to the trajectory) to an object moving with a "not null" speed ...)
C) When I asked D.C.:
"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 ??"
he replied:
"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. If a force is generated by rotation (??)..."
and he continued with more absurdities ...
We both may be kind of stubborn, but he is also impervious to properly learning basic Physics he utterly misunderstand !!
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #292 on: 15/09/2018 14:34:33 »
Quote from: David Cooper on 14/09/2018 22:14:24
There's only one gravitational pull being applied by the moon to the Earth (or by the Earth to the moon). It pulls most strongly on the near side and least strongly on the far side. That's differential gravity - the force diminishing over distance. What you're doing is rebranding the amount of that force applied at the centre of the Earth as centripetal force, and then at the near side where the gravitational force is stronger, you're attributing the excess to differential gravity, while at the far side where the gravitational force is weaker, you're attributing the bulge there to centrifugal force flinging the material outwards.
A) "What you're doing is rebranding the amount of that force applied at the centre of the Earth as centripetal force" ...
NO REBRANDING WHATSOEVER ! That gravitational force there matches exactly with the centripetal force necessary for the revolving, and does that "job" without any force excess or deficit.
Or, as on NOAA scientist stand, centrifugal force there is equal but opposite to moon´s pull, and adding up both vectors there is no spare force to do anything else than the very revolving.
B) "... and then at the near side where the gravitational force is stronger, you're attributing the excess to differential gravity..."
 NOT AL ALL !! Don´t you also say:
"There's only one gravitational pull being applied by the moon to the Earth" ...?
At the near side ONLY part of that REAL pull ("differential" is just a mathematical tool of our minds) is necessary for the revolving of stuff there, and the rest remains kind of "free", and "tries" to accelerate ("straight line" acceleration) > > > near hemisphere tides.
Or, as on NOAA scientist stand, adding up centrifugal force and moon´s pull there, last one prevails > > > tides on hemisphere closer to moon.

C) " ... while at the far side where the gravitational force is weaker, you're attributing the bulge there to centrifugal force flinging the material outwards" ...
Well, just "trying" to fling the material outwards !!
It couldn´t be otherwise: moon´s pull there is insufficient for the FUNCTION of centripetal force required to produce on material stuff the centripetal acceleration they are being "forced" to get by the rest of the planet ... And INERTIA reveals itself that way > > > further hemisphere tides !!


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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #293 on: 15/09/2018 15:39:54 »
Quote from: rmolnav on 15/09/2018 12:11:54
We both may be kind of stubborn
To me, it's not a voluntary behavior, so there is no name for it in our actual language. First, I think we are blind to other's ideas because we automatically resist to a change, but more specifically, because our mind doesn't necessarily have the background ideas needed to understand some of them. Not because some ideas are more difficult to learn than others, but because we simply don't all learn the same things. To change our mind about something we believe right is then a chance issue: It may happen that one of our ideas suffer what I call a mutation, or it may happen that an argument is so close to what we had in mind that we suddenly understand it. It's as if the two ideas would then suddenly interfere. That's what happened to me when I understood the beaming phenomenon while I was studying David's simulation of the Michelson/Morley experiment: all of a sudden, relativity was clear. Thanks to David's phlegm repeating the same thing over until I understood it. That's what he does here too, he doesn't seem able to get involved with emotions, and that's precisely the right thing to do even if we think we are right otherwise we wouldn't ever be able to change our mind on anything and there would be no intellectual evolution. You seem quite phlegmatic too otherwise the discussion wouldn't have lasted that long. I still can't chose between you two though so I'll go on trying to find the link.
« Last Edit: 15/09/2018 18:04:25 by Le Repteux »
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #294 on: 15/09/2018 16:54:38 »
Quote from: David Cooper on 14/09/2018 22:55:49
I don't need it at all, and relativity tells you that - that's why I was so sure that I was right when I began questioning rmolnav about the involvement of centripetal force - the perpendicular movement is very clearly a complete irrelevance to the mechanism. Whatever perpendicular movement there is, it simply applies to all the components equally, just like when you throw a ball to and fro across a train carriage (while sitting inside it at opposite sides from a friend) and you can explain the entire mechanism behind each throw without considering the movement of the train even though the train may be doing 200mph. If it's a round carriage (which, while impractical, could certainly be made), we don't need to factor in the tangential movement when the ball reaches either side because that tangential movement is in the same direction as the one the train is moving in. There is no tangential force outwards away from the direction the train is moving in, so why would you want to ram it into the calculations?
Orbital motions are not considered as rest frames, so I think that, in your example, the 200mph speed should be considered as an orbital speed. This way, the rotation speed of your round carriage would add or subtract to the orbital speed of the train when they coincide, what should affect the trajectory of the ball the same way accelerating or slowing down a satellite affects its trajectory. The ball is free to move radially, but the carriage is not, so it won't, but it could move tangentially if the ball does, and I think it should because a satellite on a circular path does when we change its speed: if we slow it down, it accelerates tangentially until it gets at its perigee, and if we accelerate it, it slows down tangentially until it gets at its apogee. Notice that this motion would slow down the earth's daily rotation because it opposes it, which gives us a true physical mechanism to explain that observation. Friction due to tides cannot slow down the earth's daily rotation unless it develops a force that opposes to the direction of that rotation, and there is no way for such a friction to produce such a force, which is probably why there is nothing about force in that standard explanation.
« Last Edit: 15/09/2018 18:54:13 by Le Repteux »
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Offline PmbPhy

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Re: Why do we have two high tides a day?
« Reply #295 on: 15/09/2018 18:41:01 »
Quote from: Colin2B on 12/09/2018 21:09:47
French Tides
Quote from: Colin2B on 12/09/2018 21:09:47
Quote from: PmbPhy on 11/09/2018 23:01:26
You're most welcome. See:
............
Thanks Pete, I’m not in good WiFi area at moment, when I am I will read through.
If I have time I’ll read through this thread and see what these guys are arguing about.
You're welcome. Notice the emphasis on inertial forces, in particular the centrifugal force of the earth orbiting the center of mass of the earth-moon system.
Quote
PS French Tides sounds a bit dodgy, didn’t think you were into that  ;)
Call me naive, but I don't get it this joke. ????
« Last Edit: 16/09/2018 02:12:44 by PmbPhy »
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #296 on: 16/09/2018 00:26:34 »
Quote from: Le Repteux on 15/09/2018 16:54:38
Orbital motions are not considered as rest frames,

Think about my train experiment again, but let's put it in space and turn it into a spaceship designed to resemble a train. We can put a very strong magnet at one side of the carriage and then play with an iron ball at the other side. We have a zero-G environment (technically microgravity - we can effectively ignore all gravity, and we can also remove all the air to avoid air currents messing things up). If I push the iron ball very gently towards the window at my side of the carriage, the magnet at the far side of the carriage will slow it down, then begin to pull it away from the window again. The carriage may be stationary in space, or it may be moving at close to the speed of light, but the behaviour of the iron ball that I see is the same in both cases. It may be that the iron ball is simply moving along a straight line and back, or it may be moving through space along a curved path (which is nearly perpendicular to the straight line that it would follow in a stationary system), but the perpendicular component of the movement is completely irrelevant to the cause of the ball being slowed by the magnet and then accelerated back towards it.

Now repeat the above with a magnet that gradually moves along its side of the carriage to vary the direction that it pulls on the ball. What difference does this make? Does it break relativity? No - the behaviour that I see as I move with the carriage is still the same regardless of the speed of the carriage, and if the magnet is directly opposite the ball at the point when the ball stops moving towards the window and starts to move away from it again, the force being applied to it is identical to the force being applied in the previous case without the magnet moving, as is the differential nature of the force, becoming less strong at greater distance. Relativity shows those of us who understand the relevant physics that there is no causal role for any perpendicular component of the movement - it all comes down to a straight-line pull.

Here's another idea that might help you understand this. Imagine that we have a moon that can ping in and out of existence at the press of a button. What happens to the tidal forces then? We make the moon appear, the Earth starts to accelerate a little towards it, the nearside sea is pulled more strongly towards it and the farside water is pulled less strongly, so we have tidal forces. Then we make the moon disappear again, and the forces disappear because the acceleration has been removed. If we start with a stationary Earth, we can make the moon appear several times in the same place (some distance away from the Earth), and each time we do so, tidal forces appear on the Earth due to differential gravity. Each time we remove the moon, the tidal forces go away. The Earth's now moving towards the place where the moon kept appearing, so let's now make the moon appear in a new location to the side instead. Tidal forces are again generated - they are caused by differential gravity as before, and the movement of the Earth doesn't do anything weird to this at all because we simply get "bulges" on opposite sides aligned with where the moon is. If we keep changing the places where we make the moon appear, we can repeatedly change the direction the Earth's moving in, and we can make it go round in something that approximates to a circle: a polygon with thousands of straight sides. The straight sides are made when the Earth isn't being accelerated, and the corners are made at moments when the moon exists to apply a force. At any of those moments when the moon exists, that force is straight-line gravity, diminishing in strength over distance. We can make these moments shorter and shorter, and we can also use more and more of them to make the Earth follow a path closer and closer to a circle or ellipse, but all we ever have at any moment is straight-line acceleration, and as soon as the moon isn't there, that acceleration is gone and there are no tidal forces. If the moon is there all the time, and constantly moving rather than appearing repeatedly in slightly different places, nothing has changed - it's still applying a series of straight-line accelerations towards its current position.

If we use a black hole instead of a moon, we can make a planet follow a square "orbit" by making the black hole appear for a moment four times per orbit in different places. Each time there will be tidal forces generated for a moment, but again we just have momentary straight-line accelerations. Every time the planet reaches a corner, a momentary acceleration force is applied which changes its course, while the perpendicular component of its movement is completely unchanged. That perpendicular component is shared with the water in the planet's ocean, so it retains that component of its motion too. Here we have a planet making a 90 degree turn, but the perpendicular (to the force) component of its motion is not changed at all. The water on the near side is pulled by the acceleration with greater force, so it tries to lift slightly, and the water at the far side is pulled with less force, so it tries to hold back slightly, but all this does is lead to pressure differences and slight leads and lags in reaction as those pressure differences are established, and then the whole caboodle (Earth plus water) accelerates and makes the 90 degree turn together. Switch frame of reference and you can view this as a 180 degree change of direction, completely ruling out any role for the perpendicular component of the planet's movement.

Now, I have no doubt that some minds are incapable of processing such ideas, and even if I was to go to the trouble of writing a program to illustrate how this square orbit works and how the tidal forces play out within it, such minds would not understand it and would simply regard the whole thing as nonsense (due to its exotic nature) and assert that I don't understand basic physics, but those with more flexible minds should be able to see that what I have just set out is entirely correct and that it shows more than adequately why the perpendicular movement has no role.
« Last Edit: 16/09/2018 00:30:44 by David Cooper »
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #297 on: 16/09/2018 01:06:28 »
Quote from: rmolnav on 15/09/2018 14:34:33
NO REBRANDING WHATSOEVER ! That gravitational force there matches exactly with the centripetal force necessary for the revolving, and does that "job" without any force excess or deficit.
Or, as on NOAA scientist stand, centrifugal force there is equal but opposite to moon´s pull, and adding up both vectors there is no spare force to do anything else than the very revolving.

And in a case where two bodies are moving directly towards each other? What then? You stop calling it centripetal force and use a more fundamental description of it instead, at which point you should recognise that straight-line differential gravity is the better explanation.

Quote
B) "... and then at the near side where the gravitational force is stronger, you're attributing the excess to differential gravity..."
 NOT AL ALL !! Don´t you also say:
"There's only one gravitational pull being applied by the moon to the Earth" ...?
At the near side ONLY part of that REAL pull ("differential" is just a mathematical tool of our minds) is necessary for the revolving of stuff there, and the rest remains kind of "free", and "tries" to accelerate ("straight line" acceleration) > > > near hemisphere tides.

If you're disagreeing with me, then that can only mean that the excess there isn't just down to differential gravity because you're also labelling a smaller amount of the gravity applying there as centripetal force than you do at the centre.

Quote
C) " ... while at the far side where the gravitational force is weaker, you're attributing the bulge there to centrifugal force flinging the material outwards" ...
Well, just "trying" to fling the material outwards !!
It couldn´t be otherwise: moon´s pull there is insufficient for the FUNCTION of centripetal force required to produce on material stuff the centripetal acceleration they are being "forced" to get by the rest of the planet ... And INERTIA reveals itself that way > > > further hemisphere tides !!

¿And you still can't see that your approach is just an unnecessarily complex way of explaining something that is more directly accounted for by straight-line differential gravity (which works in every case, including the one where your explanation breaks) where there's absolutely no need to make an artificial distinction between components of gravity that are labelled as centripetal force and those that aren't?

It would be interesting to see if your NOAA scientist is capable of understanding this.
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #298 on: 16/09/2018 17:12:12 »
Quote from: David Cooper on 16/09/2018 00:26:34
Think about my train experiment again, but let's put it in space and turn it into a spaceship designed to resemble a train. We can put a very strong magnet at one side of the carriage and then play with an iron ball at the other side.
You didn't comment my example and I'm not satisfied with your new explanation, so let me go on explaining my point while using your own example in case it would help. If we add to your orbiting train a circular carriage rotating in the same direction than its orbital trajectory, the train will be orbiting at the right speed all the time, but not the two halves of the carriage. Let's now replace the two rotating halves by two balls rotating at the two ends of an elastic. In principle, while the balls will transit on the line between the earth and the train, the one that is going too fast on its orbital trajectory around the earth will decelerate and get away from the earth a bit, and the one that is going too slow will accelerate and get closer to the earth a bit, what will slow down the rotation speed of the balls with time since those two accelerations are both opposed to the rotational direction of the balls. Of course, the elastic will prevent the balls from moving away from one another as much as if they were orbiting around one another because the more they pull on the elastic, the more its force increases, which is the inverse of how gravitation works, but the elastic should nevertheless stretch a bit each time the balls would transit, and shrink a bit each time they would cross over the train. If you think I make a mistake, tell me where so that I can change my mind, and if you think I'm right, then tell me why you think I cannot apply that principle to the tides.
« Last Edit: 16/09/2018 18:25:55 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 #299 on: 16/09/2018 20:35:44 »
Quote from: Le Repteux on 16/09/2018 17:12:12
You didn't comment my example...

I can't follow it well enough to be able to comment on it in any useful way, and the same applies to the new stuff. You seem to be introducing complications which obscure things rather than simplifying things to isolate the different factors.

Quote
...and I'm not satisfied with your new explanation,

All it is is a simple proof that relativity does apply and that it shows the perpendicular component to be irrelevant. Think about how my program handles the orbits. It applies forces from moon to planet and planet to moon many times a second and uses those forces to adjust the directions and speeds of the moon and planet. Both bodies move in a series of short, straight lines, and their orbital paths are polygons with many thousands of sides. The shorter you make those straight lines (which you can do by applying smaller forces at higher frequency), the closer the orbits get to being smooth, curved paths rather than polygons, and the simulation becomes more accurate if you do this. Ideally you would apply a near-infinite number of tiny forces, but whatever size they are, they always behave just like the bigger, less frequent applications of forces.

If you go the other way and make the straight lines longer by applying forces less often but at higher strength, you reduce the number of sides of the polygons, and they can go right down to 5, 4, 3 or even just 2 sides - you lose more and more accuracy compared to real planets and moons as you do this, but the principle as to how the forces are applied remains identical. Each momentary force applied leads to a change in direction of travel of the planet/moon and sets a new speed for it. Every time this happens, that change in direction which may be anything from a tiny fraction of a degree to a substantial turn can be converted into a precise 180 degree turn simply by viewing it with the right reference frame. When you do this, you see that relativity absolutely does apply to this issue and that there can be no role for the perpendicular component of movement in affecting the tidal forces. Once you understand that, you know that any thought experiment that attempts to show a role for the perpendicular component of motion to affect the forces along the line which the gravitational force acts along is doomed to fail unless it can break relativity (including both SR and LET) and thereby earn a Nobel prize.
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