Oceans in RET

What? Weight has nothing to do with it, the water can't sink down if there's solid earth below it. The fact remains that the water at the Equator has more gravitational potential energy per unit mass than that at the poles, according to RET.

But there's not solid earth below the surface of the oceans. There's a fluid, water.

Why wouldn't the water near the poles be in the same frame of reference as the water at the equator?  If you're talking about one body of  water, then it only makes sense that it should be in one FoR, doesn't it?

In RET, the acceleration vectors of the two frames of reference meet at a right angle. If you want to argue that these are the same vector, I'd love to see your working.

But there's not solid earth below the surface of the oceans. There's a fluid, water.

And what is under that water?

let's imagine a flat earth, just for fun.

all g vectors are parallel, but at the center g is bigger than at the edges. how would water behave there? same level everywhere, more at the center or more at the edges?

The Earth is flat, and these oceans are not impressed to move because the acceleration keeping them in place is the same everywhere.

If that is the case then why do we have tides? Because in your FET the earth rocking back and forth do not explain why the tides always move in the same direction

RE explains perfectly

If that is the case then why do we have tides? Because in your FET the earth rocking back and forth do not explain why the tides always move in the same direction

The tides in FET are caused by the gravitation of the Moon and the Anti-Moon.

And what is under that water?

Under a RE model, it would be a thin crust followed by slow flowing molten rock.

Iznih does bring up a good point. If we just lay the Earth flat and put a stronger downwards force in the middle, what would you expect to happen?

The tides in FET are caused by the gravitation of the Moon and the Anti-Moon.

Ah, the Anti-Moon. If the FE model is allowed to have things that have no explanation for their existence, why do you so strongly need an explanation as to why the oceans are where they are in a RE model?

let's imagine a flat earth, just for fun.

all g vectors are parallel, but at the center g is bigger than at the edges. how would water behave there? same level everywhere, more at the center or more at the edges?

Given the information you have provided, it would collect near the centre. However, all g vectors are not parallel in FET.

Under a RE model, it would be a thin crust followed by slow flowing molten rock.

So how is the water supposed to go down?

Ah, the Anti-Moon. If the FE model is allowed to have things that have no explanation for their existence, why do you so strongly need an explanation as to why the oceans are where they are in a RE model?

The oceans in RET violate the laws of physics; the anti-moon in FET does not.

I smell a nobel prize for you Robosteve.

In what way does the oceans in RET violate the laws of physics?

I smell a nobel prize for you Robosteve.

In what way does the oceans in RET violate the laws of physics?

Have you been following the thread at all? That's kind of the whole premise of this discussion.

Quote from: iznih on April 22, 2009, 07:39:28 AM

let's imagine a flat earth, just for fun.

all g vectors are parallel, but at the center g is bigger than at the edges. how would water behave there? same level everywhere, more at the center or more at the edges?

Given the information you have provided, it would collect near the centre. However, all g vectors are not parallel in FET.

that was  not aimed at fet, only to make it more simple.

we already agreed that the surface of water forms an equipotential surface. so with more water at the centre the equipotential surface would be higher above ground level than at the edges, correct?

Given the information you have provided, it would collect near the centre. However, all g vectors are not parallel in FET.

Let's consider it another way. You have a U shaped tube, fixed upright, half filled with water. You connect one end of the tube to air at a slightly higher pressure than normal, and leave the other end open. The high pressure air will exert a larger downwards force on the water than the open end, much like gravity at the poles exerting a large downwards force than at the equator. What's going to happen to the water?

So how is the water supposed to go down?

This thing seem to be trailing off, but the original point I was planing on making was that it can go to the side.

The oceans in RET violate the laws of physics; the anti-moon in FET does not.

The anti-moon does not violate the laws of physics simply because nobody knows anything about it yet. If we knew a bit more about it than 'it's something that explains the tides' then I'm sure there would be laws of physics that it violates.

Quote from: DD2014 on April 22, 2009, 01:11:32 PM

If that is the case then why do we have tides? Because in your FET the earth rocking back and forth do not explain why the tides always move in the same direction

The tides in FET are caused by the gravitation of the Moon and the Anti-Moon.

Your posts are hilarious! Please tell us more about the anti-moon!

Is there an anti sun too?! Maybe there's an anti-fail too. You should try and find it.

Your posts are hilarious! Please tell us more about the anti-moon!

Is there an anti sun too?! Maybe there's an anti-fail too. You should try and find it.

If all you're going to do is poke fun, don't expect a serious response. Also, stay the fuck out of Debate & Discussion. There's a place for people like you, it's called Angry Ranting.

we already agreed that the surface of water forms an equipotential surface. so with more water at the centre the equipotential surface would be higher above ground level than at the edges, correct?

Yes, in your simplified model. Now, how does this relate to reality?

Let's consider it another way. You have a U shaped tube, fixed upright, half filled with water. You connect one end of the tube to air at a slightly higher pressure than normal, and leave the other end open. The high pressure air will exert a larger downwards force on the water than the open end, much like gravity at the poles exerting a large downwards force than at the equator. What's going to happen to the water?

It's going to move towards the low pressure end until the force applied to it by the pressure perfectly counteracts the fictitious force felt due to gravitation. On the Earth, there are not two forces that try to balance each other, but rather one (fictitious) downwards force. Since it is an attractive force which is stronger at the poles, the water should collect there. Your analogy does not work.

This thing seem to be trailing off, but the original point I was planing on making was that it can go to the side.

It can. But why would it move away from where it is being attracted most strongly?

The anti-moon does not violate the laws of physics simply because nobody knows anything about it yet. If we knew a bit more about it than 'it's something that explains the tides' then I'm sure there would be laws of physics that it violates.

It is a large mass orbiting the North Countercelestial Pole. I don't see how this is in violation of any known law of physics.

On the Earth, there are not two forces that try to balance each other, but rather one (fictitious) downwards force.

There's not just a single force acting down on the water in a RE model. You've got the water that it's displaced pushing it back up, as well.

It can. But why would it move away from where it is being attracted most strongly?

Here's the problem. Water all over the world is not being attracted towards the poles.

It is a large mass orbiting the North Countercelestial Pole. I don't see how this is in violation of any known law of physics.

Where is this 'North Countercelestial Pole'? Never heard of that one.

I suppose I could say that it's violating Newton's law of universal gravitation, as it's pulling water towards it yet is failing to be pulled towards the water.

Quote from: roadhumper on April 25, 2009, 07:39:50 AM

Your posts are hilarious! Please tell us more about the anti-moon!

Is there an anti sun too?! Maybe there's an anti-fail too. You should try and find it.

If all you're going to do is poke fun, don't expect a serious response. Also, stay the fuck out of Debate & Discussion. There's a place for people like you, it's called Angry Ranting.

There is no anti moon, and if there was it still wouldn't account for how tides are observed. You're trying to invoke attraction between masses, which would immediately cause the moon to fall out of the sky (and likewise the anti moon to "whump" the backside of the earth. Lastly there is no clear explanation of how the moon and anti moon are able to maintain synchronisation.

Hence:

If you can't give a serious answer, don't expect a serious response.

Can't you back that statement up at all?

Quote from: iznih on April 23, 2009, 07:45:06 AM

we already agreed that the surface of water forms an equipotential surface. so with more water at the centre the equipotential surface would be higher above ground level than at the edges, correct?

Yes, in your simplified model. Now, how does this relate to reality?

just out of curiosity, do you think it's the same for any other sort of scalar potential?

Robo steve, the ocean has already occupied the spaces of where the gravity is stronger thus it can't all move to that one spot LOL.

There's not just a single force acting down on the water in a RE model. You've got the water that it's displaced pushing it back up, as well.

Granted, my mistake. But these forces are balanced in any configuration where the water density is left free to settle to its natural state. I don't see how this helps your argument.

Here's the problem. Water all over the world is not being attracted towards the poles.

It is being more strongly attracted at the poles than at the Equator. Therefore, the gravitational field vectors should always be slightly angled towards the poles, except at the Equator and the poles themselves. Over a long period of time, this would cause the water to settle at the poles.

Where is this 'North Countercelestial Pole'? Never heard of that one.

Directly beneath the North Celestial Pole.

I suppose I could say that it's violating Newton's law of universal gravitation, as it's pulling water towards it yet is failing to be pulled towards the water.

Incorrect. Its surface area to mass ratio is low enough that Dark Energy accelerates it slightly less than the Earth. This compensates for the upward attraction of the water.

There is no anti moon, and if there was it still wouldn't account for how tides are observed. You're trying to invoke attraction between masses, which would immediately cause the moon to fall out of the sky (and likewise the anti moon to "whump" the backside of the earth. Lastly there is no clear explanation of how the moon and anti moon are able to maintain synchronisation.

There's a perfectly clear explanation. But if you're so infernally convinced that there isn't an anti-moon to the point that you reject any arguments which involve it, get the fuck out of the Flat Earth debate forums.

just out of curiosity, do you think it's the same for any other sort of scalar potential?

It should be, yes. Fluids will always settle to their state of lowest potential.

Robo steve, the ocean has already occupied the spaces of where the gravity is stronger thus it can't all move to that one spot LOL.

Water can exist on top of other water, you know.

Granted, my mistake. But these forces are balanced in any configuration where the water density is left free to settle to its natural state. I don't see how this helps your argument.

It's the idea that water where it's heavier due to the downwards force will displace water more than where it's light. Thus the heavy water will push the light water up. Much like if you put something heavy on one end of a water bed, the other side goes up.

It is being more strongly attracted at the poles than at the Equator. Therefore, the gravitational field vectors should always be slightly angled towards the poles, except at the Equator and the poles themselves. Over a long period of time, this would cause the water to settle at the poles.

But the gravitational field vectors are not angled towards the poles.

Directly beneath the North Celestial Pole.

Incorrect. Its surface area to mass ratio is low enough that Dark Energy accelerates it slightly less than the Earth. This compensates for the upward attraction of the water.

Wait, what does this anti moon do? Does it attract or repel water? Where is it in comparison to the Moon we see? I'm also impressed that you know the surface area to mass ratio of an object nobody has ever seen or detected beyond it's tidal effect.

The fact that I've never seen any laws that relate to the surface area/mass of an object with regards to it's acceleration due to dark energy also bring up my prior point, that we don't know enough about how it should operate to tell if it's violating any laws.

Robo steve, the ocean has already occupied the spaces of where the gravity is stronger thus it can't all move to that one spot LOL.

He already gave examples of places with higher gravity that were not occupied by the ocean according to round earth theory. Please pay attention.

and another question: let's have a look at the distance between two aquipotential surfaces. is the distance bigger at the point of highest attraction or at the point of lowest attraction?

Quote from: Mammon on April 25, 2009, 03:27:11 PM

Robo steve, the ocean has already occupied the spaces of where the gravity is stronger thus it can't all move to that one spot LOL.

He already gave examples of places with higher gravity that were not occupied by the ocean according to round earth theory. Please pay attention.

I don't understand.. So if one part of the ocean has a higher gravity than another then you'll find more water in that spot? That's sounds ridiculous, and if there were places that the ocean didn't occupy, us RET call it LAND.

Quote from: Raist on April 26, 2009, 06:23:38 AM

Quote from: Mammon on April 25, 2009, 03:27:11 PM

Robo steve, the ocean has already occupied the spaces of where the gravity is stronger thus it can't all move to that one spot LOL.

He already gave examples of places with higher gravity that were not occupied by the ocean according to round earth theory. Please pay attention.

I don't understand.. So if one part of the ocean has a higher gravity than another then you'll find more water in that spot? That's sounds ridiculous, and if there were places that the ocean didn't occupy, us RET call it LAND.

...... You are a miracle of evolution. And no, I don't mean that in a good way.

I don't understand.

Sig'dFT.

imo robo is either trolling or he has a flawed understanding of potential. but before you ask, i'm way to lazy to pull out some maths on that.  if he's interested in it he can always look it up in a physics book, if he doesn't care there's no point in arguing about it either

It's the idea that water where it's heavier due to the downwards force will displace water more than where it's light. Thus the heavy water will push the light water up. Much like if you put something heavy on one end of a water bed, the other side goes up.

So you're arguing that the water should be pushed to where it is lightest? Why doesn't it all just fall upward into space?

But the gravitational field vectors are not angled towards the poles.

They are, and this can easily be shown with a diagram. I will draw one up later today.

Wait, what does this anti moon do? Does it attract or repel water? Where is it in comparison to the Moon we see? I'm also impressed that you know the surface area to mass ratio of an object nobody has ever seen or detected beyond it's tidal effect.

It attracts matter, and therefore water accumulates above it. It orbits exactly opposite the Moon, but underneath the Earth instead of above it.

The fact that I've never seen any laws that relate to the surface area/mass of an object with regards to it's acceleration due to dark energy also bring up my prior point, that we don't know enough about how it should operate to tell if it's violating any laws.

We could just make up something to explain it if it turns out that it is in violation of a law. Kind of like REers did when they found out that galaxies didn't rotate the way they were expected to.

and another question: let's have a look at the distance between two aquipotential surfaces. is the distance bigger at the point of highest attraction or at the point of lowest attraction?

I'm afraid I don't understand what you're trying to ask. Could you please rephrase the question?

So you're arguing that the water should be pushed to where it is lightest? Why doesn't it all just fall upward into space?

Because once enough water has moved to a certain place, that place is no longer going to be the 'lightest' place, even if a kilogram of water weighs less there. A larger pile of lighter water will weigh much the same as a smaller pile of heavier water.

They are, and this can easily be shown with a diagram. I will draw one up later today.

That should be interesting.

It attracts matter, and therefore water accumulates above it. It orbits exactly opposite the Moon, but underneath the Earth instead of above it.

So does that mean that the normal Moon repels matter, given it's on the other side of the Earth than the anti-Moon, and also creates a bulge?

We could just make up something to explain it if it turns out that it is in violation of a law. Kind of like REers did when they found out that galaxies didn't rotate the way they were expected to.

But the anti-moon is already something that's been made up to explain the tides. Nothing else suggests it should exist. It's all well and good making a few things up, and any model is going to have a few things in it that aren't confirmed to exist, as no model is an exact replica of reality. But when the model starts to have vast numbers of these made up things, and does exactly the same thing as one with a fraction of the number, it begins to get a bit silly.

Because once enough water has moved to a certain place, that place is no longer going to be the 'lightest' place, even if a kilogram of water weighs less there. A larger pile of lighter water will weigh much the same as a smaller pile of heavier water.

But there is a (relatively) constant quantity of liquid water on the surface of the Earth, so why should it not all move to where it is lightest? A large pile of lighter water still weighs less than a slightly smaller pile of lighter water plus a small pile of heavier water, totalling the same mass.

That should be interesting.

The green ellipse represents the Earth with its eccentricity exaggerated, with the black circle being a set of points equidistant from the centre of mass. The arrow is the gravitational field vector at the surface of the Earth. It is perpendicular to the black circle, but notice that to an observer on the surface of the green circle it is angled towards the North Pole (and the same would be true for the South Pole in the southern hemisphere). Of course, in reality the angle would be much smaller as I have exaggerated the equatorial bulge, but you get the point I'm sure.

So does that mean that the normal Moon repels matter, given it's on the other side of the Earth than the anti-Moon, and also creates a bulge?

No, the Moon also attracts matter. Gravitation is always attractive.

But the anti-moon is already something that's been made up to explain the tides. Nothing else suggests it should exist. It's all well and good making a few things up, and any model is going to have a few things in it that aren't confirmed to exist, as no model is an exact replica of reality. But when the model starts to have vast numbers of these made up things, and does exactly the same thing as one with a fraction of the number, it begins to get a bit silly.

Yes, RET is very silly, what with its men walking on the moon and dark matter and nuclear fusion in stars.