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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 #440 on: 23/10/2018 01:28:49 »
Quote from: rmolnav on 22/10/2018 07:39:33
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 ??

I read it, but something doesn't add up for me which has been blocking my ability to take it in properly, but perhaps you can help by explaining what I'm missing. Your first partial equation suggests that the centripetal force is the same for two cases where the Earth-moon separation is the same but with them in one case moving closer together and in the other case moving further apart. Your second partial equation suggests that the centrifugal force is the same in both cases too (if the angular speed means the component of speed perpendicular to the centripetal force), in which case it doesn't matter whether they're getting closer or moving further apart because the outward acceleration is the same for both cases - in one case it leads to an inward deceleration and in the other to an outward acceleration. That would indeed put your division location in the centre of the moon (or Earth) if these forces always match. But I'm puzzled by this, because if the centrifugal force and centripetal force are always equal there, I can't see any mechanism to accelerate the two bodies further apart and then have them accelerate closer together again. For two things to move further apart, then stop moving apart, then move closer together, then stop moving closer together and then move further apart again, you need a changing acceleration with the centrifugal force sometimes being stronger than the centripetal force and sometimes being weaker than it, but you can't have any such variation if I'm understanding your equations correctly - if one of the forces changes, the other changes to match, so if the moon is moving further away, it should spiral further and further out on every orbit, or if it's moving closer to the Earth, it should spiral inwards. I'd like to understand how you prevent that.

Quote
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" ...

The way I see it, you need to have times when the centripetal and centrifugal force don't balance for the centre of the body, because that's necessary if it's to follow an ellipse rather than a spiral path - this can move the point of balanced forces a long way away from that centre of mass. If we take a nearly-straight-line case with the moon falling almost directly towards the Earth (though still following an elliptical orbit and avoiding a collision), the centripetal force is clearly stronger than the centrifugal force, so they are nowhere close to being in balance at the centre of the moon, or indeed anywhere inside or near it.

[Note for those reading this who are interested in reality: bear in mind that centrifugal force only provides the illusion of existing in rotating frames while it doesn't exist in the real universe, so if something is moving round the centre at a fixed distance, the two apparent forces must match exactly for the rotating frame to maintain the same separation of the objects (or to maintain the rate at which they're moving together or apart). In reality, there is no centrifugal force and the only force acting is gravity - if there were two actual forces in balance, the moon would move in a straight line and not orbit the Earth at all. But let's return to the rotating frame: for centrifugal force to be used in a calculation, the centripetal force that's calculated to go with it will presumably need to be stronger than the actual amount of gravity involved in order to compensate for this imagined force outwards. If I've got this right (and I may not have - I'm no expert in the rules of rotating frames), that means that the way this centripetal force is divided into non-centripetal force and centripetal force is not the way I have been picturing up until now - the non-centripetal component of gravity must be negative rather than positive, and part of the centripetal force is therefore not gravity, but another imaginary force invented to balance the imaginary centrifugal force.]
« Last Edit: 23/10/2018 01:34:28 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 #441 on: 23/10/2018 01:46:25 »
Quote from: rmolnav on 22/10/2018 18:51:19
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.

All of it was correct. If you think otherwise, you clearly don't understand very much physics at all.

Quote
As I´ve told you several times, you not only have serious problems with Physics: neither Logics is your forte ...

Logic is my speciality - your problem is that you don't understand any logic, and that affects your ability to tell correct physics from voodoo.

Quote
One thing is to be one of the "necessary conditions" for something to happen, and another to be the "cause".

Having an ability to move is not the same as actually moving. If the ball has an ability to move but isn't moving, it isn't going to collide with the other ball and no forces will be generated by the collision that doesn't happen. I can assure you that the first ball actually moves quite in addition to being capable of moving, and that specific movement causes the collision in a way that it's mere ability to move does not cause the collision.

Quote
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 !!

Being able to breathe does not cause you to run, and being able to move does not cause the ball to move. If you run, your running causes you to breathe more, and if the ball moves, it is carrying momentum which can be transferred to something else via a force. The movement causes the force. You really ought to consider studying logical reasoning so that you have some kind of clue as to how to apply it.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #442 on: 24/10/2018 08:10:09 »
I am utterly baffled ...
Somebody who tells me:
Quote from: David Cooper on 23/10/2018 01:46:25
Logic is my speciality - your problem is that you don't understand any logic, and that affects your ability to tell correct physics from voodoo.
... and who had also previously told me (in relation to the snooker case):
 
Quote from: David Cooper on 19/10/2018 23:46:27
... 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.
[/b]
(by the way, Isaac Newton did "agree with me" regarding that; did he also become "stuck in place"?)
... yesterday, among many other bizarre things, he also said:
Quote from: David Cooper on 23/10/2018 01:28:49
Your second partial equation (?) suggests that the centrifugal force is the same in both cases too (if the angular speed means the component of speed perpendicular to the centripetal force), in which case it doesn't matter whether they're getting closer or moving further apart because the outward acceleration is the same for both cases
What I had said was:
Quote from: rmolnav on 21/10/2018 12:31:20
It is actually at mentioned "earth-related center of gravity" where 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.
Should I laugh? ... Should I cry? ... Should I "throw the towel ??
Where is the global moderator? Should he, at least, suggest him to start some new threads such as "What is angular speed?", or "What is movement?", or "What is a force?" ...? (though, I´m afraid, to no avail ...).
But, watch out! Just that measure wouldn´t avoid that, eventually,  could get next level of Naked Scientists Forum GOD !!
Would that be good for our forum ? Or, even more important, is it good for our forum to have in this thread such a lot of ideas contrary to basic Physics principles, and repeated time and again ? 
I also wonder, where are other members who used to send comments?? Do they agree with all that rubbish, or are they just fed up and don´t want to waste any more time ??
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #443 on: 24/10/2018 19:11:54 »
Quote from: rmolnav on 24/10/2018 08:10:09
Should I laugh? ... Should I cry? ... Should I "throw the towel ??
I think you should call for a pause and say that time will tell. I'm pretty sure David would accept that kind of compromise.
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #444 on: 24/10/2018 20:02:44 »
Quote from: rmolnav on 24/10/2018 08:10:09
Should I laugh? ... Should I cry? ... Should I "throw the towel ??

If you think I'm not understanding the symbols you're throwing at me, why not try illustrating their usage with actual numbers and clear descriptions.

Quote
Should he, at least, suggest him to start some new threads such as "What is angular speed?", or "What is movement?", or "What is a force?" ...? (though, I´m afraid, to no avail ...).

All you have to do is tell my what I'm misunderstanding. Is that really too difficult for you? (I looked up angular speed to check it's meaning and it appeared to mean what I said.) Or are you just running away from a problem you can't fix and engaging in diversion tactics to hide it?

Here's the point: if the centripetal force is always equal to the centrifugal force for the moon as a whole (or Earth, or anything else), you have no mechanism for keeping it in an elliptical orbit because it will either spiral out and never come back or spiral in collide with the other body.

Why don't you respond to that issue and explain how the position of equality of centripetal and centrifugal force remains at the middle of the body instead of varying in the manner it would need to to support an elliptical orbit.

Quote
Or, even more important, is it good for our forum to have in this thread such a lot of ideas contrary to basic Physics principles, and repeated time and again ?

This forum is a place that helps educate people. You are pushing fake physics and that needs to be challenged. I may be getting some things wrong here and there, and I expect people to point that out so that I can improve my understanding of things. I actually learn from that and gain, correcting any mistakes that I've been making. You don't though - you just keep on pumping out the same old bilge over and over again. There is no such thing as centrifugal force - it's an illusion. (There is a reactive centrifugal force which does exist, but it is completely absent in cases involving gravity.)

Quote
I also wonder, where are other members who used to send comments?? Do they agree with all that rubbish, or are they just fed up and don´t want to waste any more time ??

You may have forgotten now, but you spent a long time talking to yourself early on (after the thread had reached its natural conclusion and found the right answer) - no one wanted to engage with your bloated drivel so they kept out of it and simply left you to it after that. You annoyed people though (myself included) because everyone who's ever posted in this thread got a notification every time you added another chapter to your dismal book of warped physics. It eventually occurred to me though that you might be an interesting subject of study because it's important to understand how people think, and to see how flexible/inflexible they are in that regard. Can you recognise your mistakes? How much work does it take to get you to recognise any of them? I decided to turn an irritation into a learning opportunity by studying your mind. It's also been an opportunity to learn more physics though at the same time, and I've gained a lot by thinking through different aspects of the tides along the way. What have you gained though? You don't learn - you just go on and on being wrong long after being proved wrong, and you keep rejecting lots of correct physics that I provide, writing it all off as rubbish. You have absolutely no idea what's good physics and what's rubbish because your judgement is nowhere near up to the task.

One of my opponents from another forum found this thread and told you that he agreed that the correct answer is differential gravity. You then started treating him the same way as you treat me, throwing bold print and exclamation marks at him as if he was a moron. Why do you think he stopped engaging with you? He's one of the sharpest minds posting on this forum, and he knows a lost cause when he encounters one. You are incapable of learning.
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #445 on: 24/10/2018 21:05:27 »
Quote from: David Cooper on 24/10/2018 20:02:44
You are incapable of learning.
He looks intelligent, so he's probably able to learn, but as I always say in this case, he can no more change his mind instantly than a ball can accelerate instantly. The problem is not intelligence, thus conscious thinking, but unconscious one. We unconsciously resist to change ideas because our memory is about keeping our ideas the same. Our ideas change by chance, not by will, and not because they are exposed to others ideas if those ideas are too far from ours. Our ideas change with time the same way it takes time for chance to produce any durable thing.
« Last Edit: 24/10/2018 23:00:21 by Le Repteux »
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #446 on: 24/10/2018 23:05:04 »
Quote from: Le Repteux on 24/10/2018 21:05:27
The problem is not intelligence, thus conscious thinking, but unconscious one.

It's very rarely an intelligence issue. It's usually more about running the wrong algorithm. If you put too much trust in a rule with a fault in it, that sabotages any thinking you do that is dependent on that rule, regardless of how faultless your thinking is when applying that rule. A lot of the rules are deeply hidden though, and that makes them hard to test and to fix if they're faulty.

Quote
Our ideas change with time, because it takes time for chance to change anything durably.

There are some people who change position rapidly though - the slightest hint of a possible fault and they check everything they're doing, mend the broken part and then jump straight to an improved position. What is it that allows some people to fly while others can only drag themselves through the dust on their stomach?
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #447 on: 25/10/2018 14:30:52 »
Quote from: David Cooper on 24/10/2018 23:05:04
There are some people who change position rapidly though
I've been about ten years on the forums questioning Relativity. We were probably thousands out there doing that, but the only guy I saw changing his mind is me, and it still took ten years. :0) People that change their mind fast enough for the phenomenon to be readily observable are probably the exception. When I saw the photon traveling sideways in the laser of your simulation, I considered myself as a very lucky exception even if knew you helped a lot with your phlegmatic attitude. You had to repeat a few times before I saw the light, exactly like what you do here with Rmolnav, but considering that chance isn't part of the game as far as understanding ideas is concerned is no good for another kind of understanding: the one about feelings. We can easily get upset repeating things that we know right if we consider that anything else than chance is involved. On the other hand, if we know it's also a chance issue, we know it's a time one, and we can more easily let it work all by itself for a while. After all, ideas are not about immediate issues like being threatened by a firearm for instance.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #448 on: 26/10/2018 09:02:57 »
Quote from: David Cooper on 24/10/2018 20:02:44
Here's the point: if the centripetal force is always equal to the centrifugal force for the moon as a whole (or Earth, or anything else), you have no mechanism for keeping it in an elliptical orbit because it will either spiral out and never come back or spiral in collide with the other body.
Why don't you respond to that issue and explain how the position of equality of centripetal and centrifugal force remains at the middle of the body instead of varying in the manner it would need to to support an elliptical orbit.
However well anybody could explain that to you, you wouldn´t get it at all !!.
I´ve already said the question is far more complex than what you think:
- In most of earth´s locations, the center of curvature of its elliptical orbit is not exactly in the direction of the barycenter.
- At locations where the ellipse is more/less curved than if it were a circle, the radius of curvature is
smaller/bigger respectively.
- The required centripetal force (per unit of mass) is ω²r, so is associated inertial centrifugal force.
It is absolutely impossible for you to understand all that stuff, due to utter ignorance on basic Physics.
You yourself said a few days ago:
"(if the angular speed means the component of speed perpendicular to the centripetal force)"
and following day you said:
Quote from: David Cooper on 24/10/2018 20:02:44
(I looked up angular speed to check it's meaning and it appeared to mean what I said.)
[/b]
You utterly ignore what a quite basic concept, angular speed, is. Educated adults learnt that as teenagers.
But you, even after looking up its meaning (something I can´t remember if I ever had to do at all), you keep being wrong !! Do you have an own flawed dictionary, matching those unbelievably erroneous  ideas of you ??
Because,
"Angular velocity, (is the) time rate at which an object rotates, or revolves, about an axis, or at which the angular displacement between two bodies changes.
In ... physics, angles are usually expressed in radians and angular velocities in radians per second. These measures are related through the following conversion factors: 1 degree equals π/180 (about 0.0175) radian; 1 rpm equals π/30 (about 0.105) radian per second.
In many situations, an angular velocity—usually symbolized by the Greek letter omega (ω) …”
(Encyclopaedia Britannica).
And you dare say:
Quote from: David Cooper on 24/10/2018 20:02:44
You are pushing fake physics and that needs to be challenged. I may be getting some things wrong here and there, and I expect people to point that out so that I can improve my understanding of things. I actually learn from that and gain, correcting any mistakes that I've been making. You don't though ...
Your quite wrong basic ideas, deeply rooted in your mind, actually prevent you any correct learning.
Its not my problem if you have to say:
Quote from: David Cooper on 24/10/2018 20:02:44
If you think I'm not understanding the symbols you're throwing at me, why not try illustrating their usage with actual numbers and clear descriptions.
The problem is completely yours ... Why would you understand me better than definitions on dictionaries ??
By the way, you also said you had recently looked up the definition of "centripetal force", your "grey area" (what I even had previously included for you on one of my posts ...). Surely you also got it wrong, as you ignore what angular speed is !!
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Offline Le Repteux

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Re: Why do we have two high tides a day?
« Reply #449 on: 26/10/2018 13:44:15 »
Shake hands and say goodby, the game is over and nobody scored, I mean nobody succeeded to convince anybody that wasn't already convinced. I even tried to change viewpoints once in case I would understand better what the other camp was understanding, and it didn't work. As far as understanding ideas is concerned, good will doesn't even help. It does as far as feelings are concerned though, so smile, you're on candid camera!  :0)
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Offline jimbobghost

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Re: Why do we have two high tides a day?
« Reply #450 on: 26/10/2018 17:07:13 »
I was interested in this topic, as an old sailor; altho I have little knowledge in the science as discussed.
i would like to thank all those who participated in a robust discussion, in which i learned very much (altho the depth of the science remains over my head.

i was impressed by the vigor of the discussion, without anyone coming to blows :)...and the tolerance of the moderator for not arbitrarily deleting comments or in any way moving them. (i have been in other forums in which the moderators enjoyed displaying their power)

i look forward to following other topics with as much to offer.
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #451 on: 26/10/2018 22:57:24 »
Quote from: rmolnav on 26/10/2018 09:02:57
Quote from: David Cooper on 24/10/2018 20:02:44
Here's the point: if the centripetal force is always equal to the centrifugal force for the moon as a whole (or Earth, or anything else), you have no mechanism for keeping it in an elliptical orbit because it will either spiral out and never come back or spiral in collide with the other body.
Why don't you respond to that issue and explain how the position of equality of centripetal and centrifugal force remains at the middle of the body instead of varying in the manner it would need to to support an elliptical orbit.
However well anybody could explain that to you, you wouldn´t get it at all !!.

I'm not convinced that you are able to do the maths. I produced a computer program which generated the right numbers for an orbiting planet and moon and the tidal forces generated. I asked you to produce your maths routine for me to fit into the program in place of mine. You still haven't taken up that invitation, which is a pity, because producing a program would force you to produce your maths, and that would help everyone understand your position properly. We could write code to show diagrams of how the point of balance between centripetal and centrifugal force moves relative to the centre of the moon and planet throughout each orbit.

Quote
I´ve already said the question is far more complex than what you think:

...which is presumably why you can't do the maths. And it never occurs to you that the simplest explanation which doesn't produce all that complexity (and which fits all the situations where your explanation breaks down) might be superior (i.e. correct rather than wrong).

Quote
- In most of earth´s locations, the center of curvature of its elliptical orbit is not exactly in the direction of the barycenter.

It sounds as if there's some interesting maths involved there which would be worth exploring, so why hide it? And if it's hard, I'm sure there are some experts here who can help. It sounds as if that bit's just a matter of calculating the gradient of a line perpendicular to the tangent to the ellipse at a specific point.

Quote
- At locations where the ellipse is more/less curved than if it were a circle, the radius of curvature is smaller/bigger respectively.

More maths help needed for that then.

Quote
- The required centripetal force (per unit of mass) is ω²r, so is associated inertial centrifugal force.

So, are you applying this from the centre of curvature then instead of from where the actual centripetal force is coming from?

Quote
It is absolutely impossible for you to understand all that stuff, due to utter ignorance on basic Physics.

Not at all - I can understand things fine once you put numbers to them, or spell out your methods. I'd never have guessed that you were cheating on the direction in which the centripetal and centrifugal force are being applied, but it's beginning to look as if you might be. If so, then your physics is even more fake than I thought. That's why you need to lay all your cards on the table so that everyone can see exactly what kind of game you're playing here.

Quote
You yourself said a few days ago:
"(if the angular speed means the component of speed perpendicular to the centripetal force)"
and following day you said:
Quote from: David Cooper on 24/10/2018 20:02:44
(I looked up angular speed to check it's meaning and it appeared to mean what I said.)
[/b]

And at any single moment in time, what direction is that speed measured in? Can it be anything other than perpendicular to the line in which the centripetal force is acting? Clearly if you're measuring it over 5 degrees, it's more complex, but we're dealing here with a moment in time in which two forces (centripetal and centrifugal) are to be calculated for a moon or planet - we don't need to wait for it to move even a millionth of a degree to measure that because all we need's a value for that instant, and the resulting speed is equal to the perpendicular component of movement.

Quote
You utterly ignore what a quite basic concept, angular speed, is. Educated adults learnt that as teenagers.
But you, even after looking up its meaning (something I can´t remember if I ever had to do at all), you keep being wrong !! Do you have an own flawed dictionary, matching those unbelievably erroneous  ideas of you ??

It was Wikipedia, but it doesn't discuss the issue of measuring the angular speed for an instant as opposed to over a distance. There must be an angular speed for zero degrees though, and so far as I can see, that should be the perpendicular component of the movement. Is that wrong? If so, tell me what the correct answer is so that I can learn something.

Quote
Because,
"Angular velocity, (is the) time rate at which an object rotates, or revolves, about an axis, or at which the angular displacement between two bodies changes.
In ... physics, angles are usually expressed in radians and angular velocities in radians per second. These measures are related through the following conversion factors: 1 degree equals π/180 (about 0.0175) radian; 1 rpm equals π/30 (about 0.105) radian per second.
In many situations, an angular velocity—usually symbolized by the Greek letter omega (ω) …”
(Encyclopaedia Britannica).

Ah, I get it now - I had misunderstood that in a big way. So it would be the same angular speed for two objects with one twice as far away and moving twice as fast. A simple misunderstanding which can be corrected with ease. Thanks for your help. But what does this do to the balance between centripetal and centrifugal force? Does it put them out of balance at some points of the orbit, and if so, how far out of balance can it go? This is what I want to see, and it's why it would really help if you'd provide some actual numbers to use as illustrations.

Quote
And you dare say:
Quote from: David Cooper on 24/10/2018 20:02:44
You are pushing fake physics and that needs to be challenged. I may be getting some things wrong here and there, and I expect people to point that out so that I can improve my understanding of things. I actually learn from that and gain, correcting any mistakes that I've been making. You don't though ...
Your quite wrong basic ideas, deeply rooted in your mind, actually prevent you any correct learning.

I make a few mistakes here and there, and like I said, I correct them as soon as they become clear to me. This rotating frame stuff isn't something I've explored before as it's warped physics, so I expect to get things wrong here and there, and there may be places where I'm mixing rotating and non-rotating frames too, leading to further errors. I'll keep fixing those errors whenever they show up. None of that will alter the fact that this is warped physics though - centrifugal force is not real, and using it as part of a mechanism in an explanation of tidal forces is a very bad idea indeed. I'm still interested in exploring it though just because it's an interesting abstraction, and that's why I'd like to put it into the computer program. I can't do that though until I understand the algorithm you're running.

Quote
Its not my problem if you have to say:
Quote from: David Cooper on 24/10/2018 20:02:44
If you think I'm not understanding the symbols you're throwing at me, why not try illustrating their usage with actual numbers and clear descriptions.
The problem is completely yours ... Why would you understand me better than definitions on dictionaries ??

You're the one who has the algorithm that I'm trying to extract. I assume that you understand it, and I want to see how the centrifugal and centripetal forces do battle for the moon and planet to make them follow their elliptical paths. Up until now, I took from your words that they are always equal at the centre of the body in question, but that would lead to a spiral orbit. I'd have thought you'd be keen to show how that is avoided. Is there some reason why you don't want to provide any illustrations through numbers, or are you just unable to calculate them due to complications? If it's the latter, how do you know that they aren't so unequal that the point of balance is sometimes outside of the moon or planet entirely (as I suggested earlier)?

Quote
By the way, you also said you had recently looked up the definition of "centripetal force", your "grey area" (what I even had previously included for you on one of my posts ...). Surely you also got it wrong, as you ignore what angular speed is !!

What has understanding angular speed got to do with the fundamental difference between a centripetal force that's balanced by reactive centrifugal force and a centripetal force that isn't balanced by anything?
« Last Edit: 26/10/2018 23:01:05 by David Cooper »
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #452 on: 27/10/2018 12:24:33 »
I need more time to go slowly through your last post, and reply properly ...
But, in relation to the definition of "angular speed", I wonder how you can say:
Quote from: David Cooper on 26/10/2018 22:57:24
It was Wikipedia, but it doesn't discuss the issue of measuring the angular speed for an instant as opposed to over a distance. There must be an angular speed for zero degrees though, and so far as I can see, that should be the perpendicular component of the movement

Their article is long. I had a look over it minutes ago, and haven´t seen what you say ...
Its beginning is:
"In physics, the angular velocity of a particle is the rate at which it rotates around a chosen center point: that is, the time rate of change of its angular displacement relative to the origin (i.e. in layman's terms: how quickly an object goes around something over a period of time - e.g. how fast the earth orbits the sun). It is measured in angle per unit time, radians per second in SI units, and is usually represented by the symbol omega (ω, sometimes Ω).
By the way, "ω", one of the symbols you said I´m "throwing at you" ... when what I had said was:
"- ω²r (angular speed ω and distance to barycenter r)".
Do you need a kind of safety helmet for such "dangerous" cases?
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #453 on: 27/10/2018 18:45:05 »
Quote from: jimbobghost on 26/10/2018 17:07:13
I was interested in this topic, as an old sailor; altho I have little knowledge in the science as discussed.
I would like to thank all those who participated in a robust discussion, in which i learned very much (altho the depth of the science remains over my head.
I was impressed by the vigor of the discussion, without anyone coming to blows :)...and the tolerance of the moderator for not arbitrarily deleting comments or in any way moving them. (i have been in other forums in which the moderators enjoyed displaying their power)
I look forward to following other topics with as much to offer.
You are welcome!
I´m usually not that "vigorous" ... The bottom of the question is rather tricky, and I do understand other side stand. But I can´t understand why people with serious misconceptions in basic Physics can boldly despise, not my stand (that obviously could be erroneous) but basic Physics principles and works of eminent scientists (see #421).
I´m not going to "invite" you to join the discussion, because if you have little knowledge in what discussed, most probably you would say erroneous things.
But had you any question about basic details, don´t hesitate and ask. I´ll try and do my best.
When I started here (more than three years ago !!) I sent posts trying to put clear rather simple things, a necessary kind of foundation for any further discussion (you can see, e.g., #35 and 38).
And I had to repeat things ... Nobody could now read all that, but I can help telling people on which posts some particular question was dealt with.
By the way, we have not been continuously discussing all that time. There were a couple of "stand by" periods of several months. And most posts were sent in last few months (discussing with people who had not intervened previously, apart from a few cases) ... 
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #454 on: 27/10/2018 21:14:45 »
@ rmolnav

Quote
I´ve already said the question is far more complex than what you think:
- In most of earth´s locations, the center of curvature of its elliptical orbit is not exactly in the direction of the barycenter.
- At locations where the ellipse is more/less curved than if it were a circle, the radius of curvature is smaller/bigger respectively.

Because of the first line of that, I misunderstood where you were going with those comments last time. In reality, you were actually just stating the really obvious: these "complications" aren't complications at all and should cause us no difficulty whatsoever. The barycentre is exactly where the barycentre is expected to be and the moon and planet are exactly where they should be. If we're going to put your method into a simulation, we'll always know where these three things are, how they're moving and their direction of travel - we don't need to make any calculations about the shape of the orbit as it will come out automatically if the rules are right. What we need to do is run the simulation using your rules, but I need clarification from you as to how they apply.

Quote
- ω²r (angular speed ω and distance to barycenter r)
are in balance with each other.

This can be applied by working out the moon/Earth's component of speed perpendicular to the direction of the centripetal force, then working out the angle the object would move in a given length of time as viewed from the barycentre. Do you agree with that? (We always have coordinates for these three things and vectors for their movements, so we simply work from those.)

What I most want to know though is how you would calculate the centripetal force. Is it done by applying it from the barycentre too? If you apply the Earth's pull on the moon from the Earth in a conventional way, it will be too weak once you've subtracted the centrifugal force from it, but centrifugal force is an imaginary phenomenon which is only relevant to rotating frames, so do you move this (the source of the centripetal force) to the barycentre to compensate? If so, then that means you'd also have to move the origin of the moon's pull on the Earth to the barycentre as well.

What I want to do is add this on top of my existing simulation to see if your method makes the planet and moon there follow the same orbit as is generated by the existing code. Once that works, we can then continue to apply your method to calculate the tidal forces and see if they too match up to the ones calculated by differential gravity, and we can also look to see how the centripetal and centrifugal forces compare from one side of each body to the other. I'm sure that the numbers will match up when we've got it right, but this is something that needs to be done as it will clarify exactly what you're doing.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #455 on: 28/10/2018 10:38:32 »
Quote from: David Cooper on 26/10/2018 22:57:24
We could write code to show diagrams of how the point of balance between centripetal and centrifugal force moves relative to the centre of the moon and planet throughout each orbit.
You haven´t got it right yet … No wonder, as centripetal force is your “grey area” … (one of them, actually).
For any considered centripetal force, causing the turning of an object linear speed, there is an opposite centrifugal force equal in size, as a manifestation of inertia. There is never any imbalance between them.
You must be calling centripetal force what is not exactly (=mathematically) that.
Quote from: David Cooper on 26/10/2018 22:57:24
Quote
- In most of earth´s locations, the center of curvature of its elliptical orbit is not exactly in the direction of the barycenter.
It sounds as if there's some interesting maths involved there which would be worth exploring
Let us suppose, e.g., the orbiting object is at one of the extremes of the minor axis of its elliptical orbit. It is obvious that the perpendicular to the orbit is mentioned axis, and the center of curvature has to be a point of that axis.
But the other object which causes the gravitational field is on one of the focusses of the ellipse, separated from minor axis (unless it were the singular case of a circle, where the two focusses join to be its “center”).
Quote from: David Cooper on 26/10/2018 22:57:24
Quote
- At locations where the ellipse is more/less curved than if it were a circle, the radius of curvature is smaller/bigger respectively.
More maths help needed for that then.
I already referred to those mathematical concepts (#432):
"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).
Quote from: David Cooper on 26/10/2018 22:57:24
Quote
- The required centripetal force (per unit of mass) is ω²r, so is associated inertial centrifugal force.
So, are you applying this from the centre of curvature then instead of from where the actual centripetal force is coming from?
Where do you think the "actual centripetal force" is pointing at?
Quote from: David Cooper on 26/10/2018 22:57:24
But what does this do to the balance between centripetal and centrifugal force? Does it put them out of balance at some points of the orbit, and if so, how far out of balance can it go?
See what above ...
Quote from: David Cooper on 26/10/2018 22:57:24
I want to see how the centrifugal and centripetal forces do battle for the moon and planet to make them follow their elliptical paths.
Once more: you got it wrong. Those forces don´t “battle for the moon and planet to make them follow their elliptical paths" whatsoever ...
Quote from: David Cooper on 26/10/2018 22:57:24
I took from your words that they (centripetal and centrifugal f.) are always equal at the centre of the body in question, but that would lead to a spiral orbit
Please kindly tell us where “my words” you took that from are … I´ve never said they are equal ONLY at the center of the body: They are ALWAYS equal (each considered pair) !!
And, is that “spiral orbit” the result of one of your so good computer programs??
Quote from: David Cooper on 26/10/2018 22:57:24
how do you know that they (centripetal and centrifugal f.) aren't so unequal that the point of balance is sometimes outside of the moon or planet entirely (as I suggested earlier)?
Come on ! How deeply rooted are your misconceptions !!
Quote from: David Cooper on 26/10/2018 22:57:24
Quote
By the way, you also said you had recently looked up the definition of "centripetal force", your "grey area" (what I even had previously included for you on one of my posts ...). Surely you also got it wrong, as you ignore what angular speed is !!
What has understanding angular speed got to do with the fundamental difference between a centripetal force that's balanced by reactive centrifugal force and a centripetal force that isn't balanced by anything?
My last sentence applies here too ...
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #456 on: 28/10/2018 12:27:11 »
Quote from: jimbobghost on 26/10/2018 17:07:13
I was interested in this topic, as an old sailor; altho I have little knowledge in the science as discussed.
Youtube video of a conference by eminent astronomer mentioned by me on #421 titled:
What if the Moon Didn't Exist? — Neil F. Comins
could interest you.
You can see he is really an authority just googling "Neil F. Comins".
It´s rather long, because audience, mainly young students (by the way, not "too" knowledgable), is asked questions by NFC, and he answers them ...
In case you don´t want to be watching it for so much time, he refers to the issue we are discussing here (causes of tides) since app. time 15:40.
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Offline David Cooper

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Re: Why do we have two high tides a day?
« Reply #457 on: 28/10/2018 16:14:28 »
Quote from: rmolnav on 28/10/2018 10:38:32
Quote from: David Cooper on 26/10/2018 22:57:24
We could write code to show diagrams of how the point of balance between centripetal and centrifugal force moves relative to the centre of the moon and planet throughout each orbit.
You haven´t got it right yet … No wonder, as centripetal force is your “grey area” … (one of them, actually).
For any considered centripetal force, causing the turning of an object linear speed, there is an opposite centrifugal force equal in size, as a manifestation of inertia. There is never any imbalance between them.
You must be calling centripetal force what is not exactly (=mathematically) that.

If there is never any imbalance between them, how do you prevent spiral orbits? A circular orbit might work, but as soon as you have an elliptical one, your balanced forces are unable to accelerate or decelerate an object that's moving outwards or inwards, so it will keep moving outwards or inwards and will never switch from one to the other.

Quote
Once more: you got it wrong. Those forces don´t “battle for the moon and planet to make them follow their elliptical paths" whatsoever ...

In your model then, do they follow elliptical paths by magic? You're pushing a particular explanation which is based on the physics of an abstraction rather than on real physics, but if it deserves to be called physics at all it should still work mathematically, so it should be possible to simulate it. Do you not want to see it being simulated and functioning as claimed? Is there something about the way it works that you want to hide? You have a programmer here who is willing to build this for you for free - an opportunity not to be missed. All I need from you is a clear description of your mechanisms.

Quote
Quote from: David Cooper on 26/10/2018 22:57:24
I took from your words that they (centripetal and centrifugal f.) are always equal at the centre of the body in question, but that would lead to a spiral orbit
Please kindly tell us where “my words” you took that from are … I´ve never said they are equal ONLY at the center of the body: They are ALWAYS equal (each considered pair) !!

It's hard to work out what you mean when so many of the things you say seem to contradict other things elsewhere. If you have centrifugal force winning out on the far side and forming a tidal bulge, how can it be equal to the centripetal force there? If centrifugal force is losing out on the near side and extra gravitational pull is forming a tidal bulge there too, again how can the centrifugal force be equal to the centripetal force there? That's why I want to see actual numbers so that I can get a clear picture of what you're talking about, but better still would be clear rules of how they'd be calculated so that I can write a program to do all the hard work for you for a multitude of cases.

Quote
And, is that “spiral orbit” the result of one of your so good computer programs??

I'm predicting that your program will produce spiral orbits if you have the two forces in balance all the time. It will only fail to produce a spiral orbit if it's a circular orbit. Let's put that to the test by writing the actual program. All I need is your rules. You've given me enough to calculate the centrifugal force, but I need to know how you calculate the centripetal force and where it is applied from (meaning from the barycentre or directly from the moon/planet as in real physics).

Quote
Quote from: David Cooper on 26/10/2018 22:57:24
how do you know that they (centripetal and centrifugal f.) aren't so unequal that the point of balance is sometimes outside of the moon or planet entirely (as I suggested earlier)?
Come on ! How deeply rooted are your misconceptions !!

Are you unable to explain how to put my misconceptions right? If the centripetal and centrifugal forces are in balance, there is no outward or inward acceleration, but an elliptical orbit requires times when there is an inward or outward acceleration. Can you not see the problem there? Does your mechanism only work for circular orbits?

Quote
Quote
What has understanding angular speed got to do with the fundamental difference between a centripetal force that's balanced by reactive centrifugal force and a centripetal force that isn't balanced by anything?
My last sentence applies here too ...

No it doesn't - you appear to be blinded by the fake physics of rotating frames, imagining that they provide a view of reality. In real physics, gravity is not balanced by any opposing force.
« Last Edit: 28/10/2018 16:16:39 by David Cooper »
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #458 on: 29/10/2018 07:54:21 »
Sorry, but I´m running out of patience ...
After hundreds of posts discussing the issue, you keep altering my words and mixing up concepts time and again. This way we might reach post #1000 !
E.g.: I´ve directly referred to the cause of antipodal bulge tens and tens of times ... Please kindly tell us where "on earth" I said what in italics: in its supposed "battle" with centripetal force we...
Quote from: David Cooper on 28/10/2018 16:14:28
... have centrifugal force winning out on the far side and forming a tidal bulge, how can it be equal to the centripetal force there?

I regret to have to say it again, but I´m afraid you´ve got a problem either with your vision or with your reason ... or both.
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Offline rmolnav

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Re: Why do we have two high tides a day?
« Reply #459 on: 29/10/2018 18:31:44 »
Quote from: jimbobghost on 26/10/2018 17:07:13
I was interested in this topic, as an old sailor; altho I have little knowledge in the science as discussed.
As a continuation of my reply of yesterday, just to tell you that before than the timing I gave, NFC starts to say interesting things about tides, app. at 06:00. I had the other figure on a file with the reference since months ago, but this morning I watched the video again, and learnt that.
In fact, he starts talking about another "show" or conference by defenders of the idea that differential gravity is what actually causes tides, and includes a clip of a video ... He clearly says that is wrong, and makes other comments worthy to hear.
E.g., regarding the so common idea that the moon orbits the earth, he also clearly says it is wrong ... And he explains the joint movement of moon and earth as something quite different than a kind of "mutual orbiting" (last wording is mine), not following their tangents thanks to mutual pull, but not falling directly onto each other thanks to inertial "outward" forces (as he prefers to call them instead of the controversial term "centrifugal force") ...
By the way, that´s an idea I brought up here several times, though not qualifying so strongly the other model as wrong ... Last time perhaps on #364, 2nd part of my series "MY ULTIMATE GO ?"
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