The Flat Earth Society

Flat Earth Discussion Boards => Flat Earth Q&A => Topic started by: Ferruccio on June 28, 2007, 07:48:14 PM

Title: Further clarification with tides
Post by: Ferruccio on June 28, 2007, 07:48:14 PM
Q: "What about tides?"

A: The tides exist due to a slight see-saw effect on the earth. As it goes back and forth, the water rushes to the side that is lower. Note, this is a very slight wobble. Remember, these wobbles are created by very minor earthquakes. They keep the tides in check. Notice that large earthquakes result in large tides or "tsunami".


How, then, do the tides correspond with where the moon is in the sky?

Additionally, if gravity of the FE moon is the real culprit of the tides, what makes the sun different, especially if both are approximately the same distance in the sky?  "Because the sun isn't the moon" is going to be a guess as to what's going to be said.
Title: Re: Further clarification with tides
Post by: Gulliver on June 28, 2007, 08:13:10 PM
Q: "What about tides?"

A: The tides exist due to a slight see-saw effect on the earth. As it goes back and forth, the water rushes to the side that is lower. Note, this is a very slight wobble. Remember, these wobbles are created by very minor earthquakes. They keep the tides in check. Notice that large earthquakes result in large tides or "tsunami".


How, then, do the tides correspond with where the moon is in the sky?

Additionally, if gravity of the FE moon is the real culprit of the tides, what makes the sun different, especially if both are approximately the same distance in the sky?  "Because the sun isn't the moon" is going to be a guess as to what's going to be said.
Also, how does FE moon cause two high tides daily, one under the Moon and one when the Moon is not in the sky about 12 hours later?
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 26, 2007, 07:41:01 PM
Q: "What about tides?"

A: The tides exist due to a slight see-saw effect on the earth. As it goes back and forth, the water rushes to the side that is lower. Note, this is a very slight wobble. Remember, these wobbles are created by very minor earthquakes. They keep the tides in check. Notice that large earthquakes result in large tides or "tsunami".


How, then, do the tides correspond with where the moon is in the sky?

Additionally, if gravity of the FE moon is the real culprit of the tides, what makes the sun different, especially if both are approximately the same distance in the sky?  "Because the sun isn't the moon" is going to be a guess as to what's going to be said.

According to FE, the UA pushes the earth 'upward'... if the whole thing wobbles... wouldn't stuff generally roll in the direction of the side of the disc that is lower? if it's acceleration that holds us to the disc, then logically things should slide along the tilt, right?
Title: Re: Further clarification with tides
Post by: CommonCents on July 26, 2007, 08:19:14 PM
Q: "What about tides?"

A: The tides exist due to a slight see-saw effect on the earth. As it goes back and forth, the water rushes to the side that is lower. Note, this is a very slight wobble. Remember, these wobbles are created by very minor earthquakes. They keep the tides in check. Notice that large earthquakes result in large tides or "tsunami".


How, then, do the tides correspond with where the moon is in the sky?

Additionally, if gravity of the FE moon is the real culprit of the tides, what makes the sun different, especially if both are approximately the same distance in the sky?  "Because the sun isn't the moon" is going to be a guess as to what's going to be said.

According to FE, the UA pushes the earth 'upward'... if the whole thing wobbles... wouldn't stuff generally roll in the direction of the side of the disc that is lower? if it's acceleration that holds us to the disc, then logically things should slide along the tilt, right?

Yes, if by 'things should slide along the tilt' you mean 'tides should occur'.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 26, 2007, 08:24:46 PM
Q: "What about tides?"

A: The tides exist due to a slight see-saw effect on the earth. As it goes back and forth, the water rushes to the side that is lower. Note, this is a very slight wobble. Remember, these wobbles are created by very minor earthquakes. They keep the tides in check. Notice that large earthquakes result in large tides or "tsunami".


How, then, do the tides correspond with where the moon is in the sky?

Additionally, if gravity of the FE moon is the real culprit of the tides, what makes the sun different, especially if both are approximately the same distance in the sky?  "Because the sun isn't the moon" is going to be a guess as to what's going to be said.

According to FE, the UA pushes the earth 'upward'... if the whole thing wobbles... wouldn't stuff generally roll in the direction of the side of the disc that is lower? if it's acceleration that holds us to the disc, then logically things should slide along the tilt, right?

Yes, if by 'things should slide along the tilt' you mean 'tides should occur'.

That is the thread we are talking in... yes. But why is it that water is the only thing that the tilt applies to? Shouldn't there be plenty of other things like... I don't know, well say a wind chime. I've never heard of anything hanging from a rope being on a tilt that corresponds with the tides in reference to the force applied by the UA.
Title: Re: Further clarification with tides
Post by: CommonCents on July 26, 2007, 08:34:51 PM
Q: "What about tides?"

A: The tides exist due to a slight see-saw effect on the earth. As it goes back and forth, the water rushes to the side that is lower. Note, this is a very slight wobble. Remember, these wobbles are created by very minor earthquakes. They keep the tides in check. Notice that large earthquakes result in large tides or "tsunami".


How, then, do the tides correspond with where the moon is in the sky?

Additionally, if gravity of the FE moon is the real culprit of the tides, what makes the sun different, especially if both are approximately the same distance in the sky?  "Because the sun isn't the moon" is going to be a guess as to what's going to be said.

According to FE, the UA pushes the earth 'upward'... if the whole thing wobbles... wouldn't stuff generally roll in the direction of the side of the disc that is lower? if it's acceleration that holds us to the disc, then logically things should slide along the tilt, right?

Yes, if by 'things should slide along the tilt' you mean 'tides should occur'.

That is the thread we are talking in... yes. But why is it that water is the only thing that the tilt applies to? Shouldn't there be plenty of other things like... I don't know, well say a wind chime. I've never heard of anything hanging from a rope being on a tilt that corresponds with the tides in reference to the force applied by the UA.

Funny, the Moon doesn't pull these things in the RE theory, so why would an equal 'force' pull them in FE theory?
Title: Re: Further clarification with tides
Post by: Gulliver on July 26, 2007, 08:48:01 PM
Q: "What about tides?"

A: The tides exist due to a slight see-saw effect on the earth. As it goes back and forth, the water rushes to the side that is lower. Note, this is a very slight wobble. Remember, these wobbles are created by very minor earthquakes. They keep the tides in check. Notice that large earthquakes result in large tides or "tsunami".


How, then, do the tides correspond with where the moon is in the sky?

Additionally, if gravity of the FE moon is the real culprit of the tides, what makes the sun different, especially if both are approximately the same distance in the sky?  "Because the sun isn't the moon" is going to be a guess as to what's going to be said.

According to FE, the UA pushes the earth 'upward'... if the whole thing wobbles... wouldn't stuff generally roll in the direction of the side of the disc that is lower? if it's acceleration that holds us to the disc, then logically things should slide along the tilt, right?

Yes, if by 'things should slide along the tilt' you mean 'tides should occur'.

That is the thread we are talking in... yes. But why is it that water is the only thing that the tilt applies to? Shouldn't there be plenty of other things like... I don't know, well say a wind chime. I've never heard of anything hanging from a rope being on a tilt that corresponds with the tides in reference to the force applied by the UA.

Funny, the Moon doesn't pull these things in the RE theory, so why would an equal 'force' pull them in FE theory?
Do tell us how you reached the amazing conclusion that there would be an equal 'force' (whatever that means). Do tell us how your reached the conclusion that the Moon doesn't pull these thing in RE theory. Finally do tell us how you make the connection that a force parallel to the FE surface would be akin to an upward force from the Moon.

Or better yet: if you're going to play games with the newcomers, at least get your argument straight.
Title: Re: Further clarification with tides
Post by: CommonCents on July 26, 2007, 08:51:49 PM
He said there these things aren't noticeably pulled.  Considering that the tides in FE and RE have to match, the same 'force' has to be made by the tilting FE as the Moon pulling the water in the RE.  Since in the RE we don't notice these things being pulled (as he stated), there is absolutely no reason to expect them to be pulled in the FE.

<sarcasm>That was so hard I think my brain is bleeding.</sarcasm>
Title: Re: Further clarification with tides
Post by: Gulliver on July 26, 2007, 09:01:07 PM
He said there these things aren't noticeably pulled.  Considering that the tides in FE and RE have to match, the same 'force' has to be made by the tilting FE as the Moon pulling the water in the RE.  Since in the RE we don't notice these things being pulled (as he stated), there is absolutely no reason to expect them to be pulled in the FE.

<sarcasm>That was so hard I think my brain is bleeding.</sarcasm>
Wow! That would be amazing, if you made any sense. Of course, you don't.

The same effect by different mechanism does not imply equal forces.

He said that the items weren't tilted. You said the item weren't pulled. He's right; You're wrong.

His reasoning far surpasses yours. For a newcomer, he is sure making you look bad, fast.
Title: Re: Further clarification with tides
Post by: CommonCents on July 26, 2007, 09:04:19 PM
He said there these things aren't noticeably pulled.  Considering that the tides in FE and RE have to match, the same 'force' has to be made by the tilting FE as the Moon pulling the water in the RE.  Since in the RE we don't notice these things being pulled (as he stated), there is absolutely no reason to expect them to be pulled in the FE.

<sarcasm>That was so hard I think my brain is bleeding.</sarcasm>
Wow! That would be amazing, if you made any sense. Of course, you don't.

The same effect by different mechanism does not imply equal forces.

He said that the items weren't tilted. You said the item weren't pulled. He's right; You're wrong.

His reasoning far surpasses yours. For a newcomer, he is sure making you look bad, fast.

Why would moving the same mass in the same direction by the same amount not imply equal force?

Tilted items would show that the FE is PULLING them at a different angle.  Keep up there, TomG.

He's not making me look bad.  You're trying, and failing, to make me look bad.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 26, 2007, 09:06:07 PM
Thank you Gulliver.

How about we ignore that point for a moment then...

There is a difference between the tides in a lake, and the tides in an ocean. Can you explain to me (please don't just say why I'm stupid, indulge me) why there is a difference?

I can explain it in an RE model, it's because the mass of the lake is so small compared to the mass of an ocean. Which makes perfect sense in the model considering that the smaller masses of water aren't affected by the gravitational pull of the moon as much.

Shouldn't they be about the same according to the tilting explanation?
Title: Re: Further clarification with tides
Post by: CommonCents on July 26, 2007, 09:08:12 PM
I'm not sure how the tilting FE answers this.  Perhaps TomB can enlighten us.
Title: Re: Further clarification with tides
Post by: Gulliver on July 26, 2007, 09:12:02 PM
He said there these things aren't noticeably pulled.  Considering that the tides in FE and RE have to match, the same 'force' has to be made by the tilting FE as the Moon pulling the water in the RE.  Since in the RE we don't notice these things being pulled (as he stated), there is absolutely no reason to expect them to be pulled in the FE.

<sarcasm>That was so hard I think my brain is bleeding.</sarcasm>
Wow! That would be amazing, if you made any sense. Of course, you don't.

The same effect by different mechanism does not imply equal forces.

He said that the items weren't tilted. You said the item weren't pulled. He's right; You're wrong.

His reasoning far surpasses yours. For a newcomer, he is sure making you look bad, fast.

Why would moving the same mass in the same direction by the same amount not imply equal force?

Tilted items would show that the FE is PULLING them at a different angle.  Keep up there, TomG.

He's not making me look bad.  You're trying, and failing, to make me look bad.
It's your assertion. The burden of proof is yours.

Here's an example to demonstrate your fallacy. Consider a sled on a snow covered lake on a winter's day. For a constant mass on the sled, say your little brother, is equal force required to pull as to push the sled to obtain the same acceleration? No. Force and energy will be diverted from the work based on the mechanism. That's high school physics. Not understanding high school physics does make you look bad indeed.

You've stated his claim very well. Are you now agreeing with him? You said the Moon isn't pulling on these item "in RE". I'm quite sure that you're wrong. But do tell us, and this is my second request, how you determined that the Moon is not pulling on the wind chime or the rope.

A newcomer seems to have tripped you up on his (or her) first question!
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 26, 2007, 09:14:54 PM
Second one actually, I got shot down pretty quick in The Conclusive Categorical Conspiracy Compendium. It hurt. :p
Title: Re: Further clarification with tides
Post by: Gulliver on July 26, 2007, 09:17:18 PM
Thank you Gulliver.

How about we ignore that point for a moment then...

There is a difference between the tides in a lake, and the tides in an ocean. Can you explain to me (please don't just say why I'm stupid, indulge me) why there is a difference?

I can explain it in an RE model, it's because the mass of the lake is so small compared to the mass of an ocean. Which makes perfect sense in the model considering that the smaller masses of water aren't affected by the gravitational pull of the moon as much.

Shouldn't they be about the same according to the tilting explanation?
The tidal effects must be developed over time, about six hours. The water must be free to move along "under" the Moon for hours for the effect to become substantial.

I believe that you have pointed out another shortcoming of the "tilting-causing-tides" theory. You should also demand that the theory predict the tides, not just explain them as well as RE does.

All that said, I believe that the FEers have retreating from both the earthquake and tilting theories. They currently have no substantial answer to explain or predict twice daily high tides.

Oh, and you're most welcome. I really am happy that you're arguing so well so quickly. Good job.
Title: Re: Further clarification with tides
Post by: CommonCents on July 26, 2007, 09:20:03 PM
Yea, I'm all tripped up here...what with me not knowing what to say...oh wait...


Shouldn't there be plenty of other things like... I don't know, well say a wind chime. I've never heard of anything hanging from a rope being on a tilt that corresponds with the tides in reference to the force applied by the UA.

That's why I said the Moon is not pulling them in the RE model.  He used it for his own argument.


Where, exactly, is the energy going in the FE tilting model?  Sounds to me that you're saying it is just 'lost' because the mechanism is different.



Yes, I am agreeing with his latest point because it is valid.  His previous one, however, was not.  You've seen me post long enough to know I don't care which side someone's on.  If (s)he makes an invalid claim, or a claim I think is invalid, I will call him/her out on it.  If (s)he can prove themselves, I will admit that I was wrong.  This is something I have yet to see you do when you're wrong.
Title: Re: Further clarification with tides
Post by: divito the truthist on July 26, 2007, 09:26:32 PM
Not sure how you guys didn't understand this.

The Moon affects the tides in the RE. This has no noticeable effect on the objects that were mentioned, like the wind chime.

The tilting of the FE affects the tides (even though it's a lame explanation). This also has no noticeable effect on the objects that were mentioned, obviously. People seem to enjoy proposing things that are obviously contrary to observation as if that makes their point more valid.

Shouldn't they be about the same according to the tilting explanation?

Obviously not.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 26, 2007, 09:31:37 PM
Yea, I'm all tripped up here...what with me not knowing what to say...oh wait...


Shouldn't there be plenty of other things like... I don't know, well say a wind chime. I've never heard of anything hanging from a rope being on a tilt that corresponds with the tides in reference to the force applied by the UA.

That's why I said the Moon is not pulling them in the RE model.  He used it for his own argument.


Where, exactly, is the energy going in the FE tilting model?  Sounds to me that you're saying it is just 'lost' because the mechanism is different.



Yes, I am agreeing with his latest point because it is valid.  His previous one, however, was not.  You've seen me post long enough to know I don't care which side someone's on.  If (s)he makes an invalid claim, or a claim I think is invalid, I will call him/her out on it.  If (s)he can prove themselves, I will admit that I was wrong.  This is something I have yet to see you do when you're wrong.

The moon can't pull on the object because there isn't enough mass.. The point of the first argument is this:

The FE is tilted to one side, but the force applied on it and the direction in which is moves remains constant. This is the solution offered to account for the tides...
If this is true for tides, then a free-hanging object like a rope in a gymnasium should also be tilted, even if only by a couple degrees.

The RE model shows that the gravitational force of the moon is what pulls on the masses of water to make a tide. But the mass of a rope or a windchime is /far/ too small for the same pull to apply.

I think I explained it better that time. :/

It's something that should be easy to demonstrate with and experiment, even on a small scale.
Title: Re: Further clarification with tides
Post by: Gulliver on July 26, 2007, 09:34:19 PM
Yea, I'm all tripped up here...what with me not knowing what to say...oh wait...


Shouldn't there be plenty of other things like... I don't know, well say a wind chime. I've never heard of anything hanging from a rope being on a tilt that corresponds with the tides in reference to the force applied by the UA.

That's why I said the Moon is not pulling them in the RE model.  He used it for his own argument.


Where, exactly, is the energy going in the FE tilting model?  Sounds to me that you're saying it is just 'lost' because the mechanism is different.



Yes, I am agreeing with his latest point because it is valid.  His previous one, however, was not.  You've seen me post long enough to know I don't care which side someone's on.  If (s)he makes an invalid claim, or a claim I think is invalid, I will call him/her out on it.  If (s)he can prove themselves, I will admit that I was wrong.  This is something I have yet to see you do when you're wrong.
(S)he is quite clear: tilting vice pulling up. You loose.

Look it's your assertion that the forces would be equal. That's ridiculous of you, and you should be thoroughly embarrassed. You know that the FE and RE are different in size, shape, and mass. You know that tilting the FE would require tilting the entire surface of the FE, while RE accelerates the entire RE toward the Moon, though the near ocean more and the far ocean less. You have different masses, different directions, and probably even different magnitude of accelerations. You are not just moving the same water to the same place.

I have not seen you do anything respectable, so don't try to convince me with "I am humbler than thou" argument. You seem to enjoy attacking arguments, rather than educating. You seem to want to reduce the help we have for newcomers, arguing against the RE Primer. You attack people almost as much as you attack arguments. Nah, I'm not buying what you're selling.
Title: Re: Further clarification with tides
Post by: Gulliver on July 26, 2007, 09:37:12 PM
...This also has no noticeable effect on the objects that were mentioned, obviously. ...
Do you have proof of your assertion? What angle is required to raise the tide two feet twice a day and lower it twice a day? You'd have to know in order to make your claim. Personally, I think you have no idea.
Title: Re: Further clarification with tides
Post by: CommonCents on July 26, 2007, 09:39:07 PM
The moon can't pull on the object because there isn't enough mass.. The point of the first argument is this:

The FE is tilted to one side, but the force applied on it and the direction in which is moves remains constant. This is the solution offered to account for the tides...
If this is true for tides, then a free-hanging object like a rope in a gymnasium should also be tilted, even if only by a couple degrees.

The RE model shows that the gravitational force of the moon is what pulls on the masses of water to make a tide. But the mass of a rope or a windchime is /far/ too small for the same pull to apply.

I think I explained it better that time. :/

It's something that should be easy to demonstrate with and experiment, even on a small scale.


You did do a much better job explaining that time, but I still don't quite agree with you.  I don't see why the greater mass would be pulled more.  We know that all objects regardless of mass will 'fall' at the same rate in a vacuum.  Why would the rope move any less than the tides if it's accelerated by the same amount?  The force changes, yes, but the acceleration would remain constant, wouldn't it?
Title: Re: Further clarification with tides
Post by: divito the truthist on July 26, 2007, 09:42:07 PM
Do you have proof of your assertion?

Proof that there is no noticeable effect on objects? Why would I need to provide proof for something we already know? Oh, you misunderstood what I typed.

I don't think the tilting is an adequate explanation for the tides. I still haven't thought of anything just yet to be good enough as an explanation to explain the tides.
Title: Re: Further clarification with tides
Post by: Gulliver on July 26, 2007, 09:45:16 PM
Do you have proof of your assertion?

Proof that there is no noticeable effect on objects? Why would I need to provide proof for something we already know? Oh, you misunderstood what I typed.

I don't think the tilting is an adequate explanation for the tides. I still haven't thought of anything just yet to be good enough as an explanation to explain the tides.
Are you clear that MD argues that the FE tilting for tides would cause these objects, wind chime, rope, to tilt?
Title: Re: Further clarification with tides
Post by: divito the truthist on July 26, 2007, 09:51:25 PM
Are you clear that MD argues that the FE tilting for tides would cause these objects, wind chime, rope, to tilt?

I know what he's saying.

How high are the highest tides and where are they?
Title: Re: Further clarification with tides
Post by: Gulliver on July 26, 2007, 09:56:14 PM
Are you clear that MD argues that the FE tilting for tides would cause these objects, wind chime, rope, to tilt?

I know what he's saying.

How high are the highest tides and where are they?
Your country, Bay of Fundy. Wiki (http://en.wikipedia.org/wiki/Bay_of_Fundy)
Title: Re: Further clarification with tides
Post by: divito the truthist on July 26, 2007, 10:02:36 PM
"Folklore in the Mi'kmaq First Nation claims that the tides in the Bay of Fundy are caused by a giant whale splashing in the water. "

Works for me!  ;D
Title: Re: Further clarification with tides
Post by: Gulliver on July 26, 2007, 10:05:20 PM
"Folklore in the Mi'kmaq First Nation claims that the tides in the Bay of Fundy are caused by a giant whale splashing in the water. "

Works for me!  ;D
Another FE problem solved! Tides are caused by the giant whales, not related to the four elephants or turtle.
Title: Re: Further clarification with tides
Post by: divito the truthist on July 26, 2007, 10:23:18 PM
It's an animal conspiracy!
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 26, 2007, 10:25:49 PM
The moon can't pull on the object because there isn't enough mass.. The point of the first argument is this:

The FE is tilted to one side, but the force applied on it and the direction in which is moves remains constant. This is the solution offered to account for the tides...
If this is true for tides, then a free-hanging object like a rope in a gymnasium should also be tilted, even if only by a couple degrees.

The RE model shows that the gravitational force of the moon is what pulls on the masses of water to make a tide. But the mass of a rope or a windchime is /far/ too small for the same pull to apply.

I think I explained it better that time. :/

It's something that should be easy to demonstrate with and experiment, even on a small scale.


You did do a much better job explaining that time, but I still don't quite agree with you.  I don't see why the greater mass would be pulled more.  We know that all objects regardless of mass will 'fall' at the same rate in a vacuum.  Why would the rope move any less than the tides if it's accelerated by the same amount?  The force changes, yes, but the acceleration would remain constant, wouldn't it?

That lsecond to last thing is exactly the question I am posing... According to the FE model.. the rope /should/ move in conjunction to the tides. The acceleration remains constant... but it is being applied on what is now a /tilted/ disc. The definition of down (which is opposite of the force applied by UA) shifts slightly in accordance with the tilt according to the FE model. The rope doesn't move on a RE model, because the mass is too small to be affected by the gravity of the moon.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 26, 2007, 11:28:00 PM
Hello? I know it's late... but I can see topics getting updated so /someone/ has to be online.. :/
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 06:51:20 AM
I don't see how its mass can be too small to be affected by the Moon.  It'll be pulled with less 'force', yes, but should still accelerate by the same amount and in the same direction in the RE.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 10:30:26 AM
You're missing the point again. You are looking at this under the assumption that the force applied by each model to produce tides is equal, and therefor must be in essence the same...

But it's not.

The RE Model states that the gravity of the moon is what is pulling on the large bodies of water, which causes a large tide in oceans and a very reduced tide in smaller bodies of water like a lake. This is due to the principles of gravity, which are based heavily around *mass* as I understand them. (Again, I know... I am no scientist.)

Now... with the FE model, mass should not, in any way, matter. The entire planet tilts to one side and the other, but the direction in which the planet travels and the UA maintaining that movement remain constant. That gives us the tides in your model. HOWEVER, were it to be true, there would be an equal shift in ocean levels AND lake levels. And it would expand to a free hanging body like a rope, which is hanging downward. NOT because of the rope's mass, it's not considered at all in this model because gravity doesn't exist as it does in the RE model. The concept of /down/ changes.
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 10:38:38 AM
Now... with the FE model, mass should not, in any way, matter. The entire planet tilts to one side and the other, but the direction in which the planet travels and the UA maintaining that movement remain constant. That gives us the tides in your model. HOWEVER, were it to be true, there would be an equal shift in ocean levels AND lake levels. And it would expand to a free hanging body like a rope, which is hanging downward. NOT because of the rope's mass, it's not considered at all in this model because gravity doesn't exist as it does in the RE model. The concept of /down/ changes.

I disagree on this point.  Even if mass were not a factor, if this "tilting" was indeed what caused the tides, one would expect more water to be displaced from a body of water as large as an ocean than from one as relatively small as a large lake.  In fact, this particular aspect of the nature of tides seems to support the FE model.

To see this in action, fill an 8 ounce cup halfway with water, and a 32 ounce cup.  Mark the water line.  Tilt each cup exactly 30 degrees. Measure how much higher the water level is as a result of the tilting.  You should see that the "tide" is higher in the 32 ounce cup than in the 8 ounce cup.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 10:50:07 AM
Now... with the FE model, mass should not, in any way, matter. The entire planet tilts to one side and the other, but the direction in which the planet travels and the UA maintaining that movement remain constant. That gives us the tides in your model. HOWEVER, were it to be true, there would be an equal shift in ocean levels AND lake levels. And it would expand to a free hanging body like a rope, which is hanging downward. NOT because of the rope's mass, it's not considered at all in this model because gravity doesn't exist as it does in the RE model. The concept of /down/ changes.

I disagree on this point.  Even if mass were not a factor, if this "tilting" was indeed what caused the tides, one would expect more water to be displaced from a body of water as large as an ocean than from one as relatively small as a large lake.  In fact, this particular aspect of the nature of tides seems to support the FE model.

To see this in action, fill an 8 ounce cup halfway with water, and a 32 ounce cup.  Mark the water line.  Tilt each cup exactly 30 degrees. Measure how much higher the water level is as a result of the tilting.  You should see that the "tide" is higher in the 32 ounce cup than in the 8 ounce cup.

Now.... Note that tilt in the experiment. If that tilt is what produces the tides, wouldn't I be able to measure it with a creative experiment? I could, for example, tie a string to a nail in my ceiling, and a pen to the other end of the string. Then all I would need to to do is either A: Take photographs (using a tripod) at hour intervals throughout the day and then compare then, or just use photoshop to make them opaque and overlap. There would be a discrepancy over an interval because of the same force that causes the tides. OR .. B: Place a piece of paper under the same pen, so it just touches the paper without resting on it. After a day of hanging there will be a dot on an RE model, or a small line on a FE model.

There are a few more things I could think up with a pendulum that could also demonstrate that tilt. I think it's a simple experiment that could easily prove the FE model's tilting right, or wrong, if you have an appropriate place to do it that wouldn't allow for outside sources on the string.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:50:34 AM
You're missing the point again. You are looking at this under the assumption that the force applied by each model to produce tides is equal, and therefor must be in essence the same...

But it's not.

The RE Model states that the gravity of the moon is what is pulling on the large bodies of water, which causes a large tide in oceans and a very reduced tide in smaller bodies of water like a lake. This is due to the principles of gravity, which are based heavily around *mass* as I understand them. (Again, I know... I am no scientist.)

Now... with the FE model, mass should not, in any way, matter. The entire planet tilts to one side and the other, but the direction in which the planet travels and the UA maintaining that movement remain constant. That gives us the tides in your model. HOWEVER, were it to be true, there would be an equal shift in ocean levels AND lake levels. And it would expand to a free hanging body like a rope, which is hanging downward. NOT because of the rope's mass, it's not considered at all in this model because gravity doesn't exist as it does in the RE model. The concept of /down/ changes.

Gravitation does have a lot to do with mass, yes.  However the water's mass affects the Moon, not the water.  More mass in the water pulls the moon towards it (not by much, but it does).  The acceleration of the water still should be the same as the acceleration of the rope, the rope just won't pull the Moon as much as the water will.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 10:54:41 AM
You're missing the point again. You are looking at this under the assumption that the force applied by each model to produce tides is equal, and therefor must be in essence the same...

But it's not.

The RE Model states that the gravity of the moon is what is pulling on the large bodies of water, which causes a large tide in oceans and a very reduced tide in smaller bodies of water like a lake. This is due to the principles of gravity, which are based heavily around *mass* as I understand them. (Again, I know... I am no scientist.)

Now... with the FE model, mass should not, in any way, matter. The entire planet tilts to one side and the other, but the direction in which the planet travels and the UA maintaining that movement remain constant. That gives us the tides in your model. HOWEVER, were it to be true, there would be an equal shift in ocean levels AND lake levels. And it would expand to a free hanging body like a rope, which is hanging downward. NOT because of the rope's mass, it's not considered at all in this model because gravity doesn't exist as it does in the RE model. The concept of /down/ changes.

Gravitation does have a lot to do with mass, yes.  However the water's mass affects the Moon, not the water.  More mass in the water pulls the moon towards it (not by much, but it does).  The acceleration of the water still should be the same as the acceleration of the rope, the rope just won't pull the Moon as much as the water will.

... That's exactly why it support the RE model.. but I am trying to get you to apply it to the FE model for me. :/
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:56:38 AM
You're missing the point again. You are looking at this under the assumption that the force applied by each model to produce tides is equal, and therefor must be in essence the same...

But it's not.

The RE Model states that the gravity of the moon is what is pulling on the large bodies of water, which causes a large tide in oceans and a very reduced tide in smaller bodies of water like a lake. This is due to the principles of gravity, which are based heavily around *mass* as I understand them. (Again, I know... I am no scientist.)

Now... with the FE model, mass should not, in any way, matter. The entire planet tilts to one side and the other, but the direction in which the planet travels and the UA maintaining that movement remain constant. That gives us the tides in your model. HOWEVER, were it to be true, there would be an equal shift in ocean levels AND lake levels. And it would expand to a free hanging body like a rope, which is hanging downward. NOT because of the rope's mass, it's not considered at all in this model because gravity doesn't exist as it does in the RE model. The concept of /down/ changes.

Gravitation does have a lot to do with mass, yes.  However the water's mass affects the Moon, not the water.  More mass in the water pulls the moon towards it (not by much, but it does).  The acceleration of the water still should be the same as the acceleration of the rope, the rope just won't pull the Moon as much as the water will.

... That's exactly why it support the RE model.. but I am trying to get you to apply it to the FE model for me. :/

No, that's exactly why it supports neither model.  The ACCELERATION on the rope is the same as the ACCELERATION on the water.  The force is less, yes, but that is because it takes less force to accelerate a smaller mass the same amount.  Reread it and see if it makes more sense.
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 10:57:33 AM
There are a few more things I could think up with a pendulum that could also demonstrate that tilt. I think it's a simple experiment that could easily prove the FE model's tilting right, or wrong, if you have an appropriate place to do it that wouldn't allow for outside sources on the string.

Unfortunately, I don't think such a place exists.  :(
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 11:11:48 AM
There are a few more things I could think up with a pendulum that could also demonstrate that tilt. I think it's a simple experiment that could easily prove the FE model's tilting right, or wrong, if you have an appropriate place to do it that wouldn't allow for outside sources on the string.

Unfortunately, I don't think such a place exists.  :(
I believe that there is such a place. The device the Eotvos used with its mechanical resonance works in most any laboratory free of drafts. It would detect such motion, easily
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 11:13:34 AM

No, that's exactly why it supports neither model.  The ACCELERATION on the rope is the same as the ACCELERATION on the water.  The force is less, yes, but that is because it takes less force to accelerate a smaller mass the same amount.  Reread it and see if it makes more sense.

Take your two glasses of water from earlier and set them on a platter of some sort, now tape a piece of string to the bottom. Now tilt the entire platter 30 degrees to one side (might want to tape them to the platter too...) and note that the same results you saw earlier. HOWEVER, observe the affect on the string. Do you see what I'm trying to get at?
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 11:16:28 AM

No, that's exactly why it supports neither model.  The ACCELERATION on the rope is the same as the ACCELERATION on the water.  The force is less, yes, but that is because it takes less force to accelerate a smaller mass the same amount.  Reread it and see if it makes more sense.

Take your two glasses of water from earlier and set them on a platter of some sort, now tape a piece of string to the bottom. Now tilt the entire platter 30 degrees to one side (might want to tape them to the platter too...) and note that the same results you saw earlier. HOWEVER, observe the affect on the string. Do you see what I'm trying to get at?

I know what you're trying to say.  I also see that you don't know what I'm trying to say.  I'm saying that in RE the same thing should be observed because the acceleration on the string in the same relative direction is THE SAME.  I'm not supporting FE's tide theory, I'm merely pointing out that your example of why it's flawed is not valid.
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 11:18:41 AM
There are a few more things I could think up with a pendulum that could also demonstrate that tilt. I think it's a simple experiment that could easily prove the FE model's tilting right, or wrong, if you have an appropriate place to do it that wouldn't allow for outside sources on the string.

Unfortunately, I don't think such a place exists.  :(
I believe that there is such a place. The device the Eotvos used with its mechanical resonance works in most any laboratory free of drafts. It would detect such motion, easily

Interesting.  Explain exactly how we would use Eotvos' device for this experiment.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 11:21:39 AM
The same thing wouldn't be observed in RE, because the mass of a rope or a windchime is simply far too small for it to work with the moon in such a way. Given, if we had a rope with the mass of a lake and it was somehow freehanging... we would probably be able to see the same results that I propose in an FE model. You're comparing a mouse to a mountain, so to speak, and telling me that forces would work on each towards the same results :/
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 11:24:26 AM
The same thing wouldn't be observed in RE, because the mass of a rope or a windchime is simply far too small for it to work with the moon in such a way. Given, if we had a rope with the mass of a lake and it was somehow freehanging... we would probably be able to see the same results that I propose in an FE model. You're comparing a mouse to a mountain, so to speak, and telling me that forces would work on each towards the same results :/

I really don't think you've been reading what I've posted so far.


Acceleration towards a body caused by gravitation does not care about the mass of the body being accelerated.  A feather and a bowling ball in a vacuum fall at the same speed, correct?

Keeping the above in mind, why would the rope not accelerate towards the Moon at the same rate as the water?
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 11:28:51 AM
The same thing wouldn't be observed in RE, because the mass of a rope or a windchime is simply far too small for it to work with the moon in such a way. Given, if we had a rope with the mass of a lake and it was somehow freehanging... we would probably be able to see the same results that I propose in an FE model. You're comparing a mouse to a mountain, so to speak, and telling me that forces would work on each towards the same results :/

I really don't think you've been reading what I've posted so far.


Acceleration towards a body caused by gravitation does not care about the mass of the body being accelerated.  A feather and a bowling ball in a vacuum fall at the same speed, correct?

Keeping the above in mind, why would the rope not accelerate towards the Moon at the same rate as the water?

Yes... but the earth is /not/ a vacuum. The RE model explains that the earth has it's own gravitational pull, which is what keeps the smaller mass of the rope from shifting with the tides. The mass of the water is great enough to interact with the moon in such a way as to produce tides, but the mass of the rope isn't enough to defy the gravitational force of the earth itself.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 11:32:48 AM
The same thing wouldn't be observed in RE, because the mass of a rope or a windchime is simply far too small for it to work with the moon in such a way. Given, if we had a rope with the mass of a lake and it was somehow freehanging... we would probably be able to see the same results that I propose in an FE model. You're comparing a mouse to a mountain, so to speak, and telling me that forces would work on each towards the same results :/

I really don't think you've been reading what I've posted so far.


Acceleration towards a body caused by gravitation does not care about the mass of the body being accelerated.  A feather and a bowling ball in a vacuum fall at the same speed, correct?

Keeping the above in mind, why would the rope not accelerate towards the Moon at the same rate as the water?

Yes... but the earth is /not/ a vacuum. The RE model explains that the earth has it's own gravitational pull, which is what keeps the smaller mass of the rope from shifting with the tides. The mass of the water is great enough to interact with the moon in such a way as to produce tides, but the mass of the rope isn't enough to defy the gravitational force of the earth itself.

I believe you missed the point where the mass of the object being pulled doesn't matter.  That's the point of the 'dropping objects in a vacuum' experiment.  The rope still should be attracted to the Moon, just not as much as it is towards the Earth.  I say that the Moon DOES pull the rope, just you don't notice it because it's so slight.  I also say that the FE tilting is so slight you don't notice the rope's slant.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 11:47:16 AM
consider the difference in mass between the rope in the gym and the pacific ocean, or one of the great lakes if you'd like to use a smaller body of water.

The difference between the models is that gravity applies in RE, and is a product of the upward acceleration provided by the UA.. We can agree on this?

Now, in the RE model the mass of the Moon, the Earth, and <insert body of water here> are what causes the tides.

In the FE model it's the tilt of the planet in relation to the direction that the UA is pushing the earth.

Just a short review, partially for my own purposes. :p

NOW... you contest that the gravitational attraction cannot affect the rope in the same way that it does the larger mass of something like a large lake. This is due to the power of the Earth's gravity, the rope cannot be pulled enough for a noticeable difference.

I contest the same, BUT... in the FE model there is a tilt that makes the bodies of water move and cause tides, and larger bodies have more water to move and therefor a greater shift in ocean level at the shores. That is derived with thought to your experiment with the cups. But where there should be no noticeable (or perhaps even measurable) shift in the rope on the RE model, I am saying that there should be some measurable level of difference in the rope on an FE model. That is due to the fact that 'down' is changing over time. The mass of the rope has nothing to do with it in the FE model, where gravity doesn't exist, and it should be something that could be measured in a laboratory (or even a less structured) environment.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 11:51:46 AM

Now, in the RE model the mass of the Moon, the Earth, and <insert body of water here> are what causes the tides.

This proves that you don't know what I'm saying.  The mass of the body of water PULLS THE MOON, it does not PULL ITSELF.  The mass of the Moon and Earth part is correct.


That is, of course, unless you have a new theory of gravitation that's better than the current one?
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 12:00:24 PM

Now, in the RE model the mass of the Moon, the Earth, and <insert body of water here> are what causes the tides.

This proves that you don't know what I'm saying.  The mass of the body of water PULLS THE MOON, it does not PULL ITSELF.  The mass of the Moon and Earth part is correct.


That is, of course, unless you have a new theory of gravitation that's better than the current one?

Gravitation is a natural phenomenon by which all objects attract each other.*

The moon pulls on the body of water, even as the body of water pulls on the moon... I fail to see how that makes my argument faulty.. :/

* - http://en.wikipedia.org/wiki/Gravitation if you want to contest my source (I know people love to bash wiki here...)

Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 12:03:46 PM

Now, in the RE model the mass of the Moon, the Earth, and <insert body of water here> are what causes the tides.

This proves that you don't know what I'm saying.  The mass of the body of water PULLS THE MOON, it does not PULL ITSELF.  The mass of the Moon and Earth part is correct.


That is, of course, unless you have a new theory of gravitation that's better than the current one?

Gravitation is a natural phenomenon by which all objects attract each other.*

The moon pulls on the body of water, even as the body of water pulls on the moon... I fail to see how that makes my argument faulty.. :/

* - http://en.wikipedia.org/wiki/Gravitation if you want to contest my source (I know people love to bash wiki here...)



I'll try it one more time.  The mass of the water PULLS THE MOON TOWARDS THE WATER.  It does not PULL THE WATER TOWARDS THE MOON.  This is a major part of gravitation.  If you continue to fail to understand this, I will cease responding.  This is the only point I'm trying to get across, and if you don't understand it, I don't know what else I can do.

Oh, I've read the wiki gravitation articles, they agree with me on this point.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 12:06:30 PM
I've said a few times.. I'm no scientists.. :/

Could you explain it to me in some layman's terms? I'd appreciate the added material as it would help me in the debate.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 12:11:25 PM
I've said a few times.. I'm no scientists.. :/

Could you explain it to me in some layman's terms? I'd appreciate the added material as it would help me in the debate.

I've been trying to put this in layman's terms, and I apologize if you still can't understand.  You are aware that things falling in a vacuum towards a larger mass (feather and bowling ball on the moon) will accelerate at the same right, correct?


Knowing that, there is no reason why the rope should not be pulled towards the Moon.  In fact, the rope is pulled towards the Moon in RE, you just don't notice it.  Now why would the force required to move water noticeably but not move the rope noticeably in the RE require a tilt in the FE that would make the rope's movement noticeable?
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 12:16:29 PM
But in the RE model the gravity of the earth comes into play, providing a barrier that the moon/rope interaction have to overcome for there to be a notable movement.

in the FE model, no such gravity exists, and I still believe that a tilt, even a small one, that could provide for the tides would be measurable through experiment.

I do apologize if you find me frustrating, I know that I am without all the science... but I'm pretty sure my logic is solid. (pretty sure..)
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 12:18:44 PM
And why in the RE does the Earth not come into play when moving the water?  It does.


It's alright, man.  I'm just glad you're not a "ERTH IS ROUND, UR ALL STOOPID" noob ^_^
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 12:24:19 PM
but the principles in play are different by their very nature when you compare the RE and the FE reasoning behind tides. We can clearly see in the RE model that the rope /doesn't/ move, but it's  due to the properties and rules of gravity.

I need to head to work, but I'll think it over... probably all day. :/

I'll throw something up when I get home.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 12:28:51 PM
but the principles in play are different by their very nature when you compare the RE and the FE reasoning behind tides. We can clearly see in the RE model that the rope /doesn't/ move, but it's  due to the properties and rules of gravity.

I need to head to work, but I'll think it over... probably all day. :/

I'll throw something up when I get home.

Gravity exists in both models, but it does not exist as a force in either.  Gravity is merely an object's tendency to follow the curvature of spacetime, whether from acceleration (FE) or mass bending it (RE).
Title: Re: Further clarification with tides
Post by: narcberry on July 27, 2007, 12:34:52 PM
Gravitation, not gravity.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 12:36:04 PM
Gravitation, not gravity.

We've already proven that gravity is a term to relate to this effect.  It exists, just not as a force, in both models.
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 01:26:19 PM
There are a few more things I could think up with a pendulum that could also demonstrate that tilt. I think it's a simple experiment that could easily prove the FE model's tilting right, or wrong, if you have an appropriate place to do it that wouldn't allow for outside sources on the string.

Unfortunately, I don't think such a place exists.  :(
I believe that there is such a place. The device the Eotvos used with its mechanical resonance works in most any laboratory free of drafts. It would detect such motion, easily

Interesting.  Explain exactly how we would use Eotvos' device for this experiment.
Measure g over the course of a day at approximately hourly intervals. If the FE is tilting then g should vary with the tilting, increasing and then decreasing four times a day. As the FE acceleration is oblique when you're tilted (tide high or low), g should be lower than when you're flat (base tide).
 
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 01:34:27 PM
There are a few more things I could think up with a pendulum that could also demonstrate that tilt. I think it's a simple experiment that could easily prove the FE model's tilting right, or wrong, if you have an appropriate place to do it that wouldn't allow for outside sources on the string.

Unfortunately, I don't think such a place exists.  :(
I believe that there is such a place. The device the Eotvos used with its mechanical resonance works in most any laboratory free of drafts. It would detect such motion, easily

Interesting.  Explain exactly how we would use Eotvos' device for this experiment.
Measure g over the course of a day at approximately hourly intervals. If the FE is tilting then g should vary with the tilting, increasing and then decreasing four times a day. As the FE acceleration is oblique when you're tilted (tide high or low), g should be lower than when you're flat (base tide).
 

g = gram
g = acceleration due to gravitation
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 01:40:01 PM
Now... with the FE model, mass should not, in any way, matter. The entire planet tilts to one side and the other, but the direction in which the planet travels and the UA maintaining that movement remain constant. That gives us the tides in your model. HOWEVER, were it to be true, there would be an equal shift in ocean levels AND lake levels. And it would expand to a free hanging body like a rope, which is hanging downward. NOT because of the rope's mass, it's not considered at all in this model because gravity doesn't exist as it does in the RE model. The concept of /down/ changes.

I disagree on this point.  Even if mass were not a factor, if this "tilting" was indeed what caused the tides, one would expect more water to be displaced from a body of water as large as an ocean than from one as relatively small as a large lake.  In fact, this particular aspect of the nature of tides seems to support the FE model.

To see this in action, fill an 8 ounce cup halfway with water, and a 32 ounce cup.  Mark the water line.  Tilt each cup exactly 30 degrees. Measure how much higher the water level is as a result of the tilting.  You should see that the "tide" is higher in the 32 ounce cup than in the 8 ounce cup.
Roundy does make an excellent point, perhaps accidentally. You cannot create the required two high tides and two low tides by tilting. Furthermore, the demonstrated tides occur at the edge of the FE, not along the ocean shores of the middle and low latitudes.

It looks to me like Roundy has killed FE's explanation with one post. Good job!
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 01:44:48 PM
If the FE is tilting then g should vary with the tilting, increasing and then decreasing four times a day.
 

Why is that, exactly?
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 02:15:08 PM
If the FE is tilting then g should vary with the tilting, increasing and then decreasing four times a day.
 

Why is that, exactly?
FE's acceleration is up. Your 'up' doesn't align with the acceleration when you're tilted, reducing your gravity or causing you to lean. In order for there to be two high tides and two low tides almost daily, you must be tilted fours times a day and (producing less gravity) not tilted four times a day (producing normal gravity).
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 02:16:21 PM
I don't see why the acceleration would be any different whether tilted or not.
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 02:21:06 PM
I don't see why the acceleration would be any different whether tilted or not.
Then I can only suggest a good high school physics textbook. Concentrate on the vector nature of forces.
Title: Re: Further clarification with tides
Post by: narcberry on July 27, 2007, 02:22:22 PM
I don't see why the acceleration would be any different whether tilted or not.
Then I can only suggest a good high school physics textbook. Concentrate on the vector nature of forces.

I think you are underestimating him, like we underestimated the VC. You will send you best napalm and still you'll lose.
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 02:38:39 PM
I don't see why the acceleration would be any different whether tilted or not.
Then I can only suggest a good high school physics textbook. Concentrate on the vector nature of forces.

If you can't explain it how can you assert that it exists?
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 02:45:30 PM
I don't see why the acceleration would be any different whether tilted or not.
Then I can only suggest a good high school physics textbook. Concentrate on the vector nature of forces.

If you can't explain it how can you assert that it exists?
I have explained it, to everyone with a high school physics background. The rest will have to read up on the subject.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 03:15:42 PM
I don't have a high school physics background.. :p

Skipped it, probably shouldn't have.....

(on break... back to work!)
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 03:20:31 PM
I don't see why the acceleration would be any different whether tilted or not.
Then I can only suggest a good high school physics textbook. Concentrate on the vector nature of forces.

If you can't explain it how can you assert that it exists?
I have explained it, to everyone with a high school physics background. The rest will have to read up on the subject.

Where?  Can you at least provide a link?
Title: Re: Further clarification with tides
Post by: narcberry on July 27, 2007, 03:21:30 PM
If you talk with gulliver long enough, he'll argue your points for you.
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 03:21:58 PM
We're talking about a very slight shift in relation to the size of the earth.  Perhaps this is why g is measured differently in different places at different times?
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 03:28:56 PM
We're talking about a very slight shift in relation to the size of the earth.  Perhaps this is why g is measured differently in different places at different times?
Really? Please tell us how you calculated the angle of the tilt required to produce two high tides at the same time. Please tell us where you heard that g is measured differently at different times?
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 03:31:12 PM
We're talking about a very slight shift in relation to the size of the earth.  Perhaps this is why g is measured differently in different places at different times?
Really? Please tell us how you calculated the angle of the tilt required to produce two high tides at the same time. Please tell us where you heard that g is measured differently at different times?

Are you sure it's not?  ;D
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 03:32:35 PM
We're talking about a very slight shift in relation to the size of the earth.  Perhaps this is why g is measured differently in different places at different times?
Really? Please tell us how you calculated the angle of the tilt required to produce two high tides at the same time. Please tell us where you heard that g is measured differently at different times?

Are you sure it's not?  ;D
Yep.
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 03:34:35 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Title: Re: Further clarification with tides
Post by: narcberry on July 27, 2007, 03:37:25 PM
Incomming wiki...
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 03:37:56 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Yep. Your 32 oz. cup experiment does a fine job demonstrating FE's failure. The classic boy on a sled experiment also provides a great proof. You'll find it in almost all high school physics textbooks, usually in the chapter on friction.
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 03:39:19 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Yep. Your 32 oz. cup experiment does a fine job demonstrating FE's failure. The classic boy on a sled experiment also provides a great proof. You'll find it in almost all high school physics textbooks, usually in the chapter on friction.

How do any of these experiments prove that g never changes in any particular place?  ???

I think you're attacking a straw man, TomG!  Careful!  ;D
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 03:50:15 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Yep. Your 32 oz. cup experiment does a fine job demonstrating FE's failure. The classic boy on a sled experiment also provides a great proof. You'll find it in almost all high school physics textbooks, usually in the chapter on friction.

How do any of these experiments prove that g never changes in any particular place?  ???

I think you're attacking a straw man, TomG!  Careful!  ;D
I've never argued that g never changes in any particular place. Eotvos data show that it hasn't changed in the required four times daily cycle that FE's tilting argument would require. Therefore FE is wrong.
Title: Re: Further clarification with tides
Post by: Roundy the Truthinessist on July 27, 2007, 03:58:00 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Yep. Your 32 oz. cup experiment does a fine job demonstrating FE's failure. The classic boy on a sled experiment also provides a great proof. You'll find it in almost all high school physics textbooks, usually in the chapter on friction.

How do any of these experiments prove that g never changes in any particular place?  ???

I think you're attacking a straw man, TomG!  Careful!  ;D
I've never argued that g never changes in any particular place. Eotvos data show that it hasn't changed in the required four times daily cycle that FE's tilting argument would require. Therefore FE is wrong.

And this is (more or less) what I was asking for.  Thank you.  Wasn't so difficult, right?
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 04:01:44 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Yep. Your 32 oz. cup experiment does a fine job demonstrating FE's failure. The classic boy on a sled experiment also provides a great proof. You'll find it in almost all high school physics textbooks, usually in the chapter on friction.

How do any of these experiments prove that g never changes in any particular place?  ???

I think you're attacking a straw man, TomG!  Careful!  ;D
I've never argued that g never changes in any particular place. Eotvos data show that it hasn't changed in the required four times daily cycle that FE's tilting argument would require. Therefore FE is wrong.

And this is (more or less) what I was asking for.  Thank you.  Wasn't so difficult, right?
You're welcome.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 05:56:24 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Yep. Your 32 oz. cup experiment does a fine job demonstrating FE's failure. The classic boy on a sled experiment also provides a great proof. You'll find it in almost all high school physics textbooks, usually in the chapter on friction.

Your boy on a sled experiment proved nothing.  When I asked you where the 'missing energy' in the FE model goes, you did not respond.  In the boy on a sled experiment, the energy does not just disappear, it goes to a different location.
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 06:36:22 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Yep. Your 32 oz. cup experiment does a fine job demonstrating FE's failure. The classic boy on a sled experiment also provides a great proof. You'll find it in almost all high school physics textbooks, usually in the chapter on friction.

Your boy on a sled experiment proved nothing.  When I asked you where the 'missing energy' in the FE model goes, you did not respond.  In the boy on a sled experiment, the energy does not just disappear, it goes to a different location.
Well, equal forces was your asserting, so the burden to show that that is yours. If you don't understand that, then you'd don't debating. If you believe that all energy goes to work, then you don't understand physics. Oh, the little brother on the sled does prove that two different mechanisms can achieve the same work with different amounts of force.

Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 09:09:57 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Yep. Your 32 oz. cup experiment does a fine job demonstrating FE's failure. The classic boy on a sled experiment also provides a great proof. You'll find it in almost all high school physics textbooks, usually in the chapter on friction.

Your boy on a sled experiment proved nothing.  When I asked you where the 'missing energy' in the FE model goes, you did not respond.  In the boy on a sled experiment, the energy does not just disappear, it goes to a different location.
Well, equal forces was your asserting, so the burden to show that that is yours. If you don't understand that, then you'd don't debating. If you believe that all energy goes to work, then you don't understand physics. Oh, the little brother on the sled does prove that two different mechanisms can achieve the same work with different amounts of force.


Believe what you want TomG, but matter and energy cannot be created or destroyed last time I checked.  Even if it's a different mechanism, the energy still has to go somewhere.  You continue to not tell me where the energy goes.  I would like to know, because if I'm wrong then I can learn.  Could you explain where it went?
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 09:24:36 PM
Well... this thread just went above my head I think.. :/

That's what going to work gets me I guess.
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 09:39:44 PM
Can you provide a source that gives an scientifically sound experiment where this was proven?
Yep. Your 32 oz. cup experiment does a fine job demonstrating FE's failure. The classic boy on a sled experiment also provides a great proof. You'll find it in almost all high school physics textbooks, usually in the chapter on friction.

Your boy on a sled experiment proved nothing.  When I asked you where the 'missing energy' in the FE model goes, you did not respond.  In the boy on a sled experiment, the energy does not just disappear, it goes to a different location.
Well, equal forces was your asserting, so the burden to show that that is yours. If you don't understand that, then you'd don't debating. If you believe that all energy goes to work, then you don't understand physics. Oh, the little brother on the sled does prove that two different mechanisms can achieve the same work with different amounts of force.


Believe what you want TomG, but matter and energy cannot be created or destroyed last time I checked.  Even if it's a different mechanism, the energy still has to go somewhere.  You continue to not tell me where the energy goes.  I would like to know, because if I'm wrong then I can learn.  Could you explain where it went?
Well, let's start with your understanding of matter and energy being created and destroyed. Neither energy or matter can be created or destroyed in a closed system except from one to another as given by Einstein's famous equation: E=mc2.

Next, you're arguing that the force must be the same. You have the burden of proof.

Finally, let's go over work and energy. Work is the productive expenditure of energy. All mechanical systems are inefficient. Some energy is wasted, converted to heat through friction or to noise or to moving the item in the wrong direction. In the case of your little brother on the sled, you waste energy melting the ice under the runners. In the RE case of tides, the Moon "wastes energy" in creating tides by the friction of the ocean over the ocean floor. In the FE case of tides, the moving of the FE wastes energy tilting the whole Earth, not just the ocean. To argue that the force in both cases must be equal is supportable--until you calculate the energy required to tilt the FE (The energy wasted by the Moon has long been calculated.). I don't believe that you'll ever be able to determine the energy required to tilt the FE until FE states the mass of the FE.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 09:51:40 PM
OK, you're putting a lot of words in my mouth, and this is where the confusion is arising.  My only point is that the Moon pulls the rope as well as the ocean, and thinking that because you don't notice it in the RE means that you should notice it in the FE is wrong.  That's been my stance this whole time.

I never said the energy required to pull the oceans = the amount of energy to tilt the world.  I said the energy required to move the ocean in RE would be the same as is required to move the ocean in the same manner in the FE.  You told me this was wrong, and I have not seen a reason why yet.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 09:55:20 PM
Perhaps I stated it wrong.  Let me try a different way.

The force required to move the Ocean in the RE should be the same in the FE.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 09:59:23 PM
Perhaps I stated it wrong.  Let me try a different way.

The force required to move the Ocean in the RE should be the same in the FE.

But aren't the forces different by the laws of physics, with different sets of rules? The amount of force could be the exact same and it would /still/ interact differently with earth in an FE model and the earth in an RE model.

That or my understanding of the two sets of principles is flawed.. Which, I admit, is possible. :/
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:03:19 PM
Perhaps how I understand physics is wrong.  It wouldn't be the first time, and I hope it's not the last.
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 10:10:41 PM
Perhaps I stated it wrong.  Let me try a different way.

The force required to move the Ocean in the RE should be the same in the FE.
I would have to argue that FE tilting mechanism is not only about moving just the oceans, but also about lowering and raising coastlines. To get its effect, it would actually spend less force moving the oceans than RE spends moving the oceans. You simply can't divorce the mechanism from tilting the entire FE.

If we, just for a theoretical exercise, consider your new claim, I'd still have to ask to substantiate the claim. I believe that the force required to pull the Earth and the nearby ocean more than the far ocean the need distance toward the Moon is not the same as the force required to push the waters of the FE around laterally. Heck, the even the oceans are larger on FE than they are on RE, so there is more to more around.

I maintain that you can't calculate the force that the FE tidal mechanism employs, so you can't make any comparison to the force the RE mechanism employs.

BTW, I'm still rather sure than no FEer still supports the tilting mechanism of tides. It should be detectable, but isn't. It fails to position the tides in the low and middle latitudes. It fails to produce two high tides and two low tides daily.

Please never worry about asking me about these topics. I appreciate your questions and willingness to learn. Cheers.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:16:40 PM
What about the main part of my claim, that you shouldn't expect to observe the rope tilting in the FE?  I still think that stands, even if those explanations were flawed.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 10:24:15 PM
What about the main part of my claim, that you shouldn't expect to observe the rope tilting in the FE?  I still think that stands, even if my explanations why were flawed.

You shouldn't expect the rope to move a measurable distance in the RE model, no... that is due to the properties of gravitation...

but on the FE model, the entire disc is being tilted to one side, and the concept of gravitation is a myth... No matter where the rope is on the FE, it should tilt a measurable degree, even if it's a  small one.

Just flame me if I am repeating something that's been debunked already, :p
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:26:28 PM
What about the main part of my claim, that you shouldn't expect to observe the rope tilting in the FE?  I still think that stands, even if my explanations why were flawed.

You shouldn't expect the rope to move a measurable distance in the RE model, no... that is due to the properties of gravitation...

but on the FE model, the entire disc is being tilted to one side, and the concept of gravitation is a myth... No matter where the rope is on the FE, it should tilt a measurable degree, even if it's a  small one.

Just flame me if I am repeating something that's been debunked already, :p

No one on these forums says gravitation is a myth.

I don't think you understand RE gravitation...I've explained it as many ways as I can think...Perhaps TheEngineer can do a better job.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:29:03 PM
Let me try again, one point on one post.


Do you, MDCharlatan, understand that 2 different masses will be attracted to a larger mass at the same acceleration in a vacuum?
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 10:33:51 PM
What about the main part of my claim, that you shouldn't expect to observe the rope tilting in the FE?  I still think that stands, even if my explanations why were flawed.
I'm not sure about that. I suspect that you're right. It would probably take a device with mechanical resonance, like the Eotvos device, to detect the tilting. You probably could get that resonance by choosing a pendulum with a period a harmonic of the tides.

Let's try the math.
Assume the height of the tide of 2ft up and 2ft down (I'm using the Chesapeake City, MD data from the RE Primer as a base.)
Let's assume that we're at the Equator. Let's use the FE distance for the Equator of 39,113 miles from the RE Primer. Let's assume that the tilt deals with half of that distance. (There's a real problem here that I can't overcome. We did the tilting to give us two high tides on the FE on opposite sides of the Earth. I don't know how to tilt a disc to do that, so I'm just going to tilt the FE to create one high tide of 2ft.

The angle change for the bob (wind chime) would be (2ft + 2ft)/(39,113 miles/2) or  40 parts in a billion. Certainly nothing the naked eye could see in a wind chime. Certainly something a laboratory could measure readily. So it looks to me like the OP's concern about wind chimes is extremely overstated (as you said early on), but true.

Just for grins, note that if we have such a laboratory we could determine whether it's the Moon or the FE-tilting by measuring when the tilt occurred. RE would tilt toward the Moon at base tides. FE would tilt toward the ocean at high tide and away at low tide.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:36:15 PM
What about the main part of my claim, that you shouldn't expect to observe the rope tilting in the FE?  I still think that stands, even if my explanations why were flawed.
I'm not sure about that. I suspect that you're right. It would probably take a device with mechanical resonance, like the Eotvos device, to detect the tilting. You probably could get that resonance by choosing a pendulum with a period a harmonic of the tides.

Let's try the math.
Assume the height of the tide of 2ft up and 2ft down (I'm using the Chesapeake City, MD data from the RE Primer as a base.)
Let's assume that we're at the Equator. Let's use the FE distance for the Equator of 39,113 miles from the RE Primer. Let's assume that the tilt deals with half of that distance. (There's a real problem here that I can't overcome. We did the tilting to give us two high tides on the FE on opposite sides of the Earth. I don't know how to tilt a disc to do that, so I'm just going to tilt the FE to create one high tide of 2ft.

The angle change for the bob (wind chime) would be (2ft + 2ft)/(39,113 miles/2) or  40 parts in a billion. Certainly nothing the naked eye could see in a wind chime. Certainly something a laboratory could measure readily. So it looks to me like the OP's concern about wind chimes is extremely overstated (as you said early on), but true.

Just for grins, note that if we have such a laboratory we could determine whether it's the Moon or the FE-tilting by measuring when the tilt occurred. RE would tilt toward the Moon at base tides. FE would tilt toward the ocean at high tide and away at low tide.

Woot!  Would the rope also tilt in the RE?  I think it would.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 10:36:40 PM
Let me try again, one point on one post.


Do you, MDCharlatan, understand that 2 different masses will be attracted to a larger mass at the same acceleration in a vacuum?

Yep.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:39:35 PM
Let me try again, one point on one post.


Do you, MDCharlatan, understand that 2 different masses will be attracted to a larger mass at the same acceleration in a vacuum?

Yep.

OK, so then that proves that the amount the smaller object moves towards the larger object (by smaller I mean less massive, and larger I mean more massive) is not affected by the mass of the smaller object.  The 'small' ocean (small meaning less massive than the Moon) and the 'smaller' rope (smaller meaning less massive than the ocean) would both move towards the Moon, with different 'forces' but the same acceleration.


EDIT:  This is why I say observing a tilting or not tilting rope does not disprove either model.
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 10:41:01 PM
What about the main part of my claim, that you shouldn't expect to observe the rope tilting in the FE?  I still think that stands, even if my explanations why were flawed.

You shouldn't expect the rope to move a measurable distance in the RE model, no... that is due to the properties of gravitation...

but on the FE model, the entire disc is being tilted to one side, and the concept of gravitation is a myth... No matter where the rope is on the FE, it should tilt a measurable degree, even if it's a  small one.

Just flame me if I am repeating something that's been debunked already, :p

No one on these forums says gravitation is a myth.

I don't think you understand RE gravitation...I've explained it as many ways as I can think...Perhaps TheEngineer can do a better job.
I believe the point about the myth was that FE says the FE does not produce gravity, even though it's composed of the same matter that does produce gravity in the Cavendish experiment.

I also argue that talking about GR when dealing with FE is just silly. FE implies that GR is wrong; GR implies that FE is wrong.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:44:30 PM
FE implies a part of GR is wrong.  If Einstein never thought that part of Newtonian physics was wrong, he wouldn't have come up with the Equivalence Principle, GR, or SR.  Science involves questioning the 'foundations'.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 27, 2007, 10:44:54 PM
Let me try again, one point on one post.


Do you, MDCharlatan, understand that 2 different masses will be attracted to a larger mass at the same acceleration in a vacuum?

Yep.

OK, so then that proves that the amount the smaller object moves towards the larger object (by smaller I mean less massive, and larger I mean more massive) is not affected by the mass of the smaller object.  The 'small' ocean (small meaning less massive than the Moon) and the 'smaller' rope (smaller meaning less massive than the ocean) would both move towards the Moon, with different 'forces' but the same acceleration.


EDIT:  This is why I say observing a tilting or not tilting rope does not disprove either model.

I understand that much so far. (I know it sucks to educate me while it's being discussed)

But Gulliver is explaining a methodology by which the tilt, which would exist in both models, could be measured and used to infer which is the case.

(Note - That's me conceding to a crushing intellectual defeat, and passing the burden to a smarter man. :p)
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 10:52:55 PM
Heh, I don't mind helping teach someone.
Yes, Gulliver did come up with ways to tell the difference from tilting and mass attraction.  I don't think I said there was no difference, but it's been a long and strange day, so without actually going back and rereading it all I'm not going to say I didn't.
Title: Re: Further clarification with tides
Post by: Gulliver on July 27, 2007, 10:55:41 PM
Woot!  Would the rope also tilt in the RE?  I think it would.
Of course, it would. Now can we detect that outside of the lab is the question.

Let's do the math.
Mass of the Moon: 7.36 × 1022 kg
Distance to the Moon: 384,400 km
Formula for the force applied by the Moon on the surface of the Earth:

Gravitational force = (G * m1 * m2) / (d2)

Using the calculator in Google (thanks Google!) we get: (G * 7.36 × ((10^22) kg)) / ((384 400 km)^2) = 3.32377932 × 10-5 m / s2

That's compared to g, is about 4 parts in 100,0000, not something you'd measure on your back porch, but measurable in a lab.
Title: Re: Further clarification with tides
Post by: CommonCents on July 27, 2007, 11:23:14 PM
Woot!  Would the rope also tilt in the RE?  I think it would.
Of course, it would. Now can we detect that outside of the lab is the question.

Let's do the math.
Mass of the Moon: 7.36 × 1022 kg
Distance to the Moon: 384,400 km
Formula for the force applied by the Moon on the surface of the Earth:

Gravitational force = (G * m1 * m2) / (d2)

Using the calculator in Google (thanks Google!) we get: (G * 7.36 × ((10^22) kg)) / ((384 400 km)^2) = 3.32377932 × 10-5 m / s2

That's compared to g, is about 4 parts in 100,0000, not something you'd measure on your back porch, but measurable in a lab.

Ggrr, I got a different answer and used my only available paper...I assumed you used 1 kg to represent the rope's mass because you never mentioned the second mass again.  I have:


F = G * ((m1m2)/r2)


F = 6.67 X 10-11Nm2kg-2 * ((7.38 X 1022kg * 1kg)/(3.844 X 108)2m)

The answer I ended up with was 7.48 X 10-2N.  I'm not sure if the math's wrong though, it's rather late.  If it's still on my mind in the morning, I'll see if I can do it again.

EDIT:  Changed my end answer from 7.48 X 102 to what it currently says.  I left out that - sign.

EDIT AGAIN:  Added the unit, N, to my answer.

EDIT AGAIN AGAIN:  (Yea, lots of edits, but it's been a long day)  The acceleration, since the mass was assumed to be 1 kg (and the acceleration for any other mass moving towards the Moon will be the same anyway) is 7.48 X 10-2 m/s2...I think.
Title: Re: Further clarification with tides
Post by: TheEngineer on July 28, 2007, 01:52:19 AM
I can explain it in an RE model, it's because the mass of the lake is so small compared to the mass of an ocean. Which makes perfect sense in the model considering that the smaller masses of water aren't affected by the gravitational pull of the moon as much.
I noticed that TomG felt it was not necessary to point this out.  That's quite strange, as he is always first to point out an error most of the time.  It's probably that whole bias thing...

Title: Re: Further clarification with tides
Post by: divito the truthist on July 28, 2007, 04:23:37 AM
True, my pool has no tides.
Title: Re: Further clarification with tides
Post by: Gulliver on July 28, 2007, 01:31:44 PM
Woot!  Would the rope also tilt in the RE?  I think it would.
Of course, it would. Now can we detect that outside of the lab is the question.

Let's do the math.
Mass of the Moon: 7.36 × 1022 kg
Distance to the Moon: 384,400 km
Formula for the force applied by the Moon on the surface of the Earth:

Gravitational force = (G * m1 * m2) / (d2)

Using the calculator in Google (thanks Google!) we get: (G * 7.36 × ((10^22) kg)) / ((384 400 km)^2) = 3.32377932 × 10-5 m / s2

That's compared to g, is about 4 parts in 100,0000, not something you'd measure on your back porch, but measurable in a lab.

Ggrr, I got a different answer and used my only available paper...I assumed you used 1 kg to represent the rope's mass because you never mentioned the second mass again.  I have:


F = G * ((m1m2)/r2)


F = 6.67 X 10-11Nm2kg-2 * ((7.38 X 1022kg * 1kg)/(3.844 X 108)2m)

The answer I ended up with was 7.48 X 10-2N.  I'm not sure if the math's wrong though, it's rather late.  If it's still on my mind in the morning, I'll see if I can do it again.

EDIT:  Changed my end answer from 7.48 X 102 to what it currently says.  I left out that - sign.

EDIT AGAIN:  Added the unit, N, to my answer.

EDIT AGAIN AGAIN:  (Yea, lots of edits, but it's been a long day)  The acceleration, since the mass was assumed to be 1 kg (and the acceleration for any other mass moving towards the Moon will be the same anyway) is 7.48 X 10-2 m/s2...I think.
i appreciate the effort you've put into this.

I see one mistake in your math. The "m" in the denominator needs to move back into the previous expression.

Once you do so, your answer becomes:
6.67 X (10^(-11)) N (m^2) (kg^(-2)) * ((7.38 X ((10^22) kg) * (1 kg)) / ((3.84400 X ((10^8) m))^2)) = 3.33131299 × 10-5 newtons

Now to calculate the acceleration, we know the F=m1a or a =F/m1. So the mass drops out the equation, which the reason you didn't see me estimate it.

So we get a= 3.33*10-5 m/s2 and basic agreement.
Title: Re: Further clarification with tides
Post by: CommonCents on July 28, 2007, 02:58:22 PM
Isn't m2 already in the denominator to cancel out with the numerator's m2?  The equation's set up that way to begin with.  I did mess up the mass of the Moon though..stupid sleepiness.

I don't see why you would move an m2 from the denominator when it has to cancel out one in the numerator...
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 28, 2007, 03:15:09 PM
The concepts I can work with....

the math makes me dizzy though. :p
Title: Re: Further clarification with tides
Post by: Gulliver on July 28, 2007, 03:49:16 PM
Isn't m2 already in the denominator to cancel out with the numerator's m2?  The equation's set up that way to begin with.  I did mess up the mass of the Moon though..stupid sleepiness.

I don't see why you would move an m2 from the denominator when it has to cancel out one in the numerator...
It's just the equation is wrong. It's m2, not m, in the denominator at that point. Sure you can cancel m2 out from both numerator and denominator, either before or after that step. However, while you still have m2 in the numerator, you still need m2 in the denominator. As your equation stood, you only had m in the denominator.
Title: Re: Further clarification with tides
Post by: CommonCents on July 28, 2007, 03:55:24 PM
Isn't m2 already in the denominator to cancel out with the numerator's m2?  The equation's set up that way to begin with.  I did mess up the mass of the Moon though..stupid sleepiness.

I don't see why you would move an m2 from the denominator when it has to cancel out one in the numerator...
It's just the equation is wrong. It's m2, not m, in the denominator at that point. Sure you can cancel m2 out from both numerator and denominator, either before or after that step. However, while you still have m2 in the numerator, you still need m2 in the denominator. As your equation stood, you only had m in the denominator.

Ah, how do you get rid of m in the numerator then?

EDIT:  Sorry, I used to be good at this shit. >:(  I guess it's like anything else, lack of practice hurts.
Title: Re: Further clarification with tides
Post by: Gulliver on July 28, 2007, 04:28:49 PM
Isn't m2 already in the denominator to cancel out with the numerator's m2?  The equation's set up that way to begin with.  I did mess up the mass of the Moon though..stupid sleepiness.

I don't see why you would move an m2 from the denominator when it has to cancel out one in the numerator...
It's just the equation is wrong. It's m2, not m, in the denominator at that point. Sure you can cancel m2 out from both numerator and denominator, either before or after that step. However, while you still have m2 in the numerator, you still need m2 in the denominator. As your equation stood, you only had m in the denominator.

Ah, how do you get rid of m in the numerator then?

EDIT:  Sorry, I used to be good at this shit. >:(  I guess it's like anything else, lack of practice hurts.
Honestly I just type the equation into Google and let it figure it out. I believe that after you move m back into the squared term in the denominator you have (in terms of just units of measure) N * m2 Kg-2 Kg2/m2 which gets you to just N, which matches the need UoM.
Title: Re: Further clarification with tides
Post by: CommonCents on July 28, 2007, 04:31:43 PM
Ah, I see what I did wrong (sorta).  Thanks for sticking through and helping me out.  ^_^
Title: Re: Further clarification with tides
Post by: Gulliver on July 28, 2007, 04:42:36 PM
Ah, I see what I did wrong (sorta).  Thanks for sticking through and helping me out.  ^_^
You're welcome. I enjoyed the discussion. Thanks.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 28, 2007, 09:51:44 PM
... I'm confused. Did we ever come to a conclusion one way or the other? :p
Title: Re: Further clarification with tides
Post by: Gulliver on July 28, 2007, 10:03:10 PM
... I'm confused. Did we ever come to a conclusion one way or the other? :p
Yep.
Looking at a wind chime on the back porch... tilting (as needed to produce tides) would not be detectable.
In a good lab with some patience... detectable.
Title: Re: Further clarification with tides
Post by: CommonCents on July 28, 2007, 10:08:53 PM
... I'm confused. Did we ever come to a conclusion one way or the other? :p
Yep.
Looking at a wind chime on the back porch... tilting (as needed to produce tides) would not be detectable.
In a good lab with some patience... detectable.

But the tilt you see in a lab could just be the Moon pulling the windchime too.  Either way the experiment doesn't work, unless you work out exactly how much the windchime should tilt in RE, the amount is different significantly (enough not to be reduced to %error) from the FE one, put it in a vacuum, and measure carefully.
Title: Re: Further clarification with tides
Post by: MDCharlatan on July 29, 2007, 01:14:46 AM
Aww...

I did my best, a fun debate for the first decent argument I put up I guess. Thanks for the patience and fun Goodcents, and thanks for the backup and fun Gulliver.