Further clarification with tides

Perhaps how I understand physics is wrong.  It wouldn't be the first time, and I hope it's not the last.

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.

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.

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

Quote from: 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 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.

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?

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.

Quote from: 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 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.

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.

Quote from: 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?

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.

Quote from: MDCharlatan on July 27, 2007, 10:24:15 PM

Quote from: 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 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.

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'.

Quote from: MDCharlatan on July 27, 2007, 10:36:40 PM

Quote from: 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?

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)

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.

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.

Quote from: CommonCents on July 27, 2007, 10:36:15 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 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...

True, my pool has no tides.

Quote from: Gulliver on July 27, 2007, 10:55:41 PM

Quote from: CommonCents on July 27, 2007, 10:36:15 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.

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...

The concepts I can work with....

the math makes me dizzy though. :p

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.

Quote from: 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...

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.

Quote from: Gulliver on July 28, 2007, 03:49:16 PM

Quote from: 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...

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.

Ah, I see what I did wrong (sorta).  Thanks for sticking through and helping me out.  ^_^

Ah, I see what I did wrong (sorta).  Thanks for sticking through and helping me out.  ^_^

You're welcome. I enjoyed the discussion. Thanks.

... I'm confused. Did we ever come to a conclusion one way or the other? :p

... 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.

Quote from: 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

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.

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.