Oceans in RET

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.

Where would there be a smaller pile of lighter water plus a small pile of heavier water? If you move light water to a place where water's heavy, it's going to become heavy, and vice versa.

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.

What it seems you've done is drawn the shape of the Earth when there are two apparent forces acting on the surface, and then only represented one of those forces at the surface with your arrow. It's normal to the black circle, as that the shape you get when there's just that force. It's not normal to the green ellipse, as that's not the shape you get when there's just the one force. It's the shape you get when there's two apparent forces, a gravitational one and a centrifugal one.



The green arrow represents the apparent centrifugal 'force', and the blue arrow is the combination of the two forces, which is normal to the green ellipse, on which the two forces are acting.

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

Ah, wonderful, I see how it all works now.

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

Likewise the FE model, with it's elaborate conspiracy, unexplainable galaxies and spontaneous matter-to-energy in the Sun. The question is which has the least of these little odds and ends.

Where would there be a smaller pile of lighter water plus a small pile of heavier water? If you move light water to a place where water's heavy, it's going to become heavy, and vice versa.

Sigh. You're claiming that some water is moved to where it is lighter. I'm illustrating why even a small amount of water left where it's heavier would cause the total weight to be more than if it all moved to where it was lighter.

What it seems you've done is drawn the shape of the Earth when there are two apparent forces acting on the surface, and then only represented one of those forces at the surface with your arrow. It's normal to the black circle, as that the shape you get when there's just that force. It's not normal to the green ellipse, as that's not the shape you get when there's just the one force. It's the shape you get when there's two apparent forces, a gravitational one and a centrifugal one.



The green arrow represents the apparent centrifugal 'force', and the blue arrow is the combination of the two forces, which is normal to the green ellipse, on which the two forces are acting.

Can you show that the fictitious centrifugal force always has a magnitude appropriate to create such an effect?

Sigh. You're claiming that some water is moved to where it is lighter. I'm illustrating why even a small amount of water left where it's heavier would cause the total weight to be more than if it all moved to where it was lighter.

What does the total weight matter? The two places just need to weigh the same.

Can you show that the fictitious centrifugal force always has a magnitude appropriate to create such an effect?

I'm not sure why you would think it wouldn't. It's what you would expect to happen in a spinning body of water.

What does the total weight matter? The two places just need to weigh the same.

Why?

I'm not sure why you would think it wouldn't. It's what you would expect to happen in a spinning body of water.

But the Earth is not a spinning body of water. It is a spinning rock (according to RET) with water on its surface. This water should gravitate to the poles.

Quote from: NTheGreat on April 29, 2009, 06:39:47 AM

The green arrow represents the apparent centrifugal 'force', and the blue arrow is the combination of the two forces, which is normal to the green ellipse, on which the two forces are acting.

Can you show that the fictitious centrifugal force always has a magnitude appropriate to create such an effect?

I would say that for the oceans to behave the way they do, the centrifugal 'force' would have to have a magnitude appropriate to create the observed effect.  If it were any greater, or less, a different effect would be observed.
I realise this is incredibly circular reasoning, but I don't know how else to word it.

Why?

Well, if one weighed more overall, the weight of water in that part would push water into where it weighed less overall.

Think of a upright U-tube. If you pour a lot of water into one side, the weight of all the water on that side is going to push down until both sides are level.

But the Earth is not a spinning body of water. It is a spinning rock (according to RET) with water on its surface. This water should gravitate to the poles.

And? Most of that rock is a fluid, and that's how you would expect a fluid to behave.

Sorry to intervene, but I find this to be a very interesting topic, and I think I can add something to it. I must say I never asked myself your question, Robosteve, so I had to wrap my head around this in order to be able to explain what's going on.

You say that the RET is faulty, because it says that gravity varies from place to place, which would lead to the ocean water flooding certain places and leave others dry. Well, in fact, the mean sea levels have quite a respectable difference in different places on the earth because of local differences in gravity. Thus, the sea level at places with higher gravity due to more mass within the earth, such as the indian ocean, actually is 85 meters higher than the worldwide average. In places with a less dense core, such as northeast of Australia (thinner magma => less mass => less gravity), the sea level is 100 meters lower than the worldwide average sea level.

Now, why doesn't the water flow from the indian ocean to above Australia then? Well, why does water level out anyway? It does because of gravity. Gravity states that a water plane has to be level, because it pulls the water towards the earth's surface equally. But if you have local differences in gravity, you will have local differences in the water plane as well. Therefore, the mean sea level is not entirely constant over the earth.

Haven't followed this thread in a while. Has RET accounted for its "crawling oceans" yet?

Haven't followed this thread in a while. Has RET accounted for its "crawling oceans" yet?

Please read the post above yours.

Quote from: Proleg on April 30, 2009, 08:46:28 PM

Haven't followed this thread in a while. Has RET accounted for its "crawling oceans" yet?

Please read the post above yours.

No thanks. I can think of much more pleasurable ways of killing brain cells.

Well, if you choose to ignore arguments, how do you expect to be convinced of anything?

Well, if you choose to ignore arguments, how do you expect to be convinced of anything?

Whoa...I might just place this in my signature.

With regard to the problem, it's probably in that 'both sides think they're right' status. I think we've sorted that differing strengths of apparent downward force at sea level in different places is fine, and we're now at the idea that the surface of the ocean is perpendicular to the direction of the apparent downwards force. The FE side holds that the oceans are not perpendicular to the force, and the RE side holds they are.

One piece of evidence that you Round Earthers often like to flaunt as supporting your theory is the way that the measured value of g varies from place to place around the surface of the Earth.

Actually, this is the first time I've heard this, however I can see how that could change as much as maybe a thousanth of a g?

I'd rather simply look at the countless photos we have of a round earth, then challenge a FE'er to come up with one single photo of a flat earth.  Just one!

The clock is ticking.

 

Quote from: Robosteve on April 17, 2009, 10:05:27 PM

One piece of evidence that you Round Earthers often like to flaunt as supporting your theory is the way that the measured value of g varies from place to place around the surface of the Earth.

Actually, this is the first time I've heard this, however I can see how that could change as much as maybe a thousanth of a g?

have a look at the wiki article in the first post, the variation is quite big.

and nthegreat is right, we won't come to an agreement in that case (like in any other case )

Quote from: chREes on May 01, 2009, 01:55:50 AM

Well, if you choose to ignore arguments, how do you expect to be convinced of anything?

Whoa...I might just place this in my signature.

Feel free to do so.

With regard to the problem, it's probably in that 'both sides think they're right' status. I think we've sorted that differing strengths of apparent downward force at sea level in different places is fine, and we're now at the idea that the surface of the ocean is perpendicular to the direction of the apparent downwards force. The FE side holds that the oceans are not perpendicular to the force, and the RE side holds they are.

So if I get this right, FE says that the mean sea level is exactly the same all over the earth's surface, because the only force acting on it is the earth accelerating upwards?

So if I get this right, FE says that the mean sea level is exactly the same all over the earth's surface, because the only force acting on it is the earth accelerating upwards?

Well, you have the Moon and anti-Moon acting on it as well, giving the two tidal bulges, and I'm not entirely sure on whether the FE model currently allows a little local variation in apparent downwards force. There may well be a myriad of other little things acting on a FE ocean as well, given the number of forces flying round in a FE model.

But I suppose in essence, yes. The sea would be almost perfectly level across the whole of the FE disc, if you just had the Earth accelerating up.

i think robo's point was that the variations in g should cause the earth's radius at the equator being smaller than the radius towards any of the poles what would be contrary to the assumed shape of earth by ret.

i think robo's point was that the variations in g should cause the earth's radius at the equator being smaller than the radius towards any of the poles what would be contrary to the assumed shape of earth by ret.

For what reason would the earth have a smaller radius at the equator?

There isn't one. But that isn't what Robo said so it's kind of irrelevant.

Let's come back to Robosteve's actual argument then. He says that the oceans can not be following gravity, since that would lead to the water flowing away from certain places and flood others. The thing he ignores is that the oceans have always been following gravity and have adapted to it as soon as they evolved. Cities like Helsinki and Singapore don't drown, because they were founded on landmasses that had already been defined by the water level of the surrounding ocean, which in itself had been defined by local gravity.

Robosteve's argument is that water once was perfectly level, then such cities were built, then gravity was introduced, and now those cities should drown or dry out. Actually, gravity has always been there and has been affecting sea levels from the beginning, so the cities close to the seas were founded on solid conditions of a stable local mean sea level.

let me rephrase it: robo thinks that the sea level should be higher at places with higher attraction than at places with lower g and not vice versa as ret states.

But the water level is higher at places with higher attraction (= gravity), and lower where there's lower attraction. The mean sea level actually isn't level all over the world due to differences in local gravity. Where does RET state the opposite?

i quote myself:

imo it's quite easy:

the tendency is that we have a lower g when approaching the equator (fictitious forces at work etc)

g_equ<g_pol (1)

the potential energy is the same at every point of the water suface:

E_equ=E_pol

E=mgh (good enough for this problem)

g_equh_equ= g_polh_pol

(1)=> h_equ>h_pol

matches shape of the earth as predicted by re

if you find an error let me know, a change in the FOR should cause no problems here cause it's only classical mechanics. a change in g with growing height has been neglected. look up the shape of earth. water distributes the same way like the earth is bulged at the equator.

Okay, I think I got it now. I must say this is an excellent question and shall be a riddle to be solved for me. I will get back to you guys.

I'm not ignoring anything, except that I am now going to ignore you on this subject, because you keep saying the same thing over and over regardless of how many times I explain myself.

You don't explain anything, you only state that it isn't proven and ignore totally the current technological state.

Quote from: zork on April 21, 2009, 05:54:51 AM

How difficult is to understand that g is not dependent from the sea level? It is dependent from the distance of center of the earth or I guess we can say radius of the earth. So, what is the radius of the earth at both places?

I'm aware of what factors into the value of g at a particular location. What I'm not aware of is how you think this is in any way relevant to my point.

Because your g factors doesn't include anything about distance from earth center. They include acceleration and pull from stars.

Quote from: zork on April 21, 2009, 05:54:51 AM

No, because now it seems to me that you say that the celestial gears have gravitation, not stars. And all this celestial gear stuff is not in any way proven, measured or observed. So, no not agreed.

I did not use the phrase "celestial gears" in my post. Please don't put words into my mouth.

You used phrase "celestial plane". I don't see much difference between gear or plane in FE context, so, you just play with words again to dodge actual explaining.

No. You are so infernally convinced that my equation is wrong, yet can't even provide one which you think is better, or suggest how it may be improved. If anybody has anything to prove here, it is you.

Sorry, I suggested how you can make it better or improve it - Go to the physic professors(or whoever you have there) who you have by your side at school every day and ask from them.  But you refuse to accept the advise and bother me instead. I wonder why.

One piece of evidence that you Round Earthers often like to flaunt as supporting your theory is the way that the measured value of g varies from place to place around the surface of the Earth. I will now show how this is, in fact, in direct opposition to the ridiculous hypothesis that the Earth is round, and can only be properly explained by Flat Earth Theory.

It is a natural state of affairs that the Universe will always tend towards its lowest energy state. In order for the world's oceans to achieve their lowest possible energy state, the measured value of g should be the same at every point corresponding to the surface of the ocean, if indeed they are being held in place by gravitation and not acceleration. If it is higher in some places than in others, the water should flow from the places with low values of g to those with higher values until g is equal everywhere at the surface. This is basic physics; the water is more weakly attracted where g is lower and will move to where gravitation is stronger.

Now, this table indicates that the measured value of g in Singapore is just 9.781 m s^-2, while in Helsinki it is measured to be 9.819 m s^-2. This is a difference of nearly 0.4%. Were the oceans truly bound to the surface of the Earth gravitationally, those nearer the Equator should be seeking out the lower energy states that exist in the polar regions and accumulating there. Coastal cities such as Helsinki and Copenhagen would be completely flooded, while there would be much more dry land in Oceania and Central America.

Why is this not the case? The answer, of course, is simple. The Earth is flat, and these oceans are not impressed to move because the acceleration keeping them in place is the same everywhere. The simple fact that the oceans are positioned where they are, ladies and gentlemen, is thus damning evidence that there exists a Conspiracy, and that the Conspiracy has succeeded in keeping the layman so ignorant to the true shape of the Earth that he believes the variance in g across the Earth's surface to be supportive of his beliefs, rather than in total contradiction to them.

Just quickly and I know it's petty (but so are most of these arguments) but 9.819-9.781 = 0.038, not 0.4. This is a big difference and if we cant trust him to make a simple mathematical subtraction, why bother reading his arguments on more advanced fluid mechanics and physics?

Quote from: Robosteve on April 17, 2009, 10:05:27 PM

One piece of evidence that you Round Earthers often like to flaunt as supporting your theory is the way that the measured value of g varies from place to place around the surface of the Earth. I will now show how this is, in fact, in direct opposition to the ridiculous hypothesis that the Earth is round, and can only be properly explained by Flat Earth Theory.

It is a natural state of affairs that the Universe will always tend towards its lowest energy state. In order for the world's oceans to achieve their lowest possible energy state, the measured value of g should be the same at every point corresponding to the surface of the ocean, if indeed they are being held in place by gravitation and not acceleration. If it is higher in some places than in others, the water should flow from the places with low values of g to those with higher values until g is equal everywhere at the surface. This is basic physics; the water is more weakly attracted where g is lower and will move to where gravitation is stronger.

Now, this table indicates that the measured value of g in Singapore is just 9.781 m s^-2, while in Helsinki it is measured to be 9.819 m s^-2. This is a difference of nearly 0.4%. Were the oceans truly bound to the surface of the Earth gravitationally, those nearer the Equator should be seeking out the lower energy states that exist in the polar regions and accumulating there. Coastal cities such as Helsinki and Copenhagen would be completely flooded, while there would be much more dry land in Oceania and Central America.

Why is this not the case? The answer, of course, is simple. The Earth is flat, and these oceans are not impressed to move because the acceleration keeping them in place is the same everywhere. The simple fact that the oceans are positioned where they are, ladies and gentlemen, is thus damning evidence that there exists a Conspiracy, and that the Conspiracy has succeeded in keeping the layman so ignorant to the true shape of the Earth that he believes the variance in g across the Earth's surface to be supportive of his beliefs, rather than in total contradiction to them.

Just quickly and I know it's petty (but so are most of these arguments) but 9.819-9.781 = 0.038, not 0.4. This is a big difference and if we cant trust him to make a simple mathematical subtraction, why bother reading his arguments on more advanced fluid mechanics and physics?

If we can't trust you to notice a percent sign, then why should we trust any of your criticism of his argument?

Quote from: Infinity on May 08, 2009, 01:39:07 AM

Quote from: Robosteve on April 17, 2009, 10:05:27 PM

One piece of evidence that you Round Earthers often like to flaunt as supporting your theory is the way that the measured value of g varies from place to place around the surface of the Earth. I will now show how this is, in fact, in direct opposition to the ridiculous hypothesis that the Earth is round, and can only be properly explained by Flat Earth Theory.

It is a natural state of affairs that the Universe will always tend towards its lowest energy state. In order for the world's oceans to achieve their lowest possible energy state, the measured value of g should be the same at every point corresponding to the surface of the ocean, if indeed they are being held in place by gravitation and not acceleration. If it is higher in some places than in others, the water should flow from the places with low values of g to those with higher values until g is equal everywhere at the surface. This is basic physics; the water is more weakly attracted where g is lower and will move to where gravitation is stronger.

Now, this table indicates that the measured value of g in Singapore is just 9.781 m s^-2, while in Helsinki it is measured to be 9.819 m s^-2. This is a difference of nearly 0.4%. Were the oceans truly bound to the surface of the Earth gravitationally, those nearer the Equator should be seeking out the lower energy states that exist in the polar regions and accumulating there. Coastal cities such as Helsinki and Copenhagen would be completely flooded, while there would be much more dry land in Oceania and Central America.

Why is this not the case? The answer, of course, is simple. The Earth is flat, and these oceans are not impressed to move because the acceleration keeping them in place is the same everywhere. The simple fact that the oceans are positioned where they are, ladies and gentlemen, is thus damning evidence that there exists a Conspiracy, and that the Conspiracy has succeeded in keeping the layman so ignorant to the true shape of the Earth that he believes the variance in g across the Earth's surface to be supportive of his beliefs, rather than in total contradiction to them.

Just quickly and I know it's petty (but so are most of these arguments) but 9.819-9.781 = 0.038, not 0.4. This is a big difference and if we cant trust him to make a simple mathematical subtraction, why bother reading his arguments on more advanced fluid mechanics and physics?

If we can't trust you to notice a percent sign, then why should we trust any of your criticism of his argument?

I did notice and the percentage difference is even smaller (0.00387%) so I was being kind to him assuming that he meant otherwise. You probably should have done the simple calculation yourself before bothering to respond.

lol it's not 0,00387%, you get % after multiplying with 100 , so 0,387% and that is about 0,4%