Skip to main content

TR Memescape

  • TalkRational - guys do you really want to be a part of the socratic method with a retard acting as professor?

Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

Messages - cold one

1
Science / Re: The Tides ... Take 5
And one more question. I am assuming when he talks about the moon on a tower there is no rotation involved.  But he doesn't specify that.  If there is a rotating earth with the moon on a tower, isn't that a different issue?

You mean, the moon is on a tower planted on the earth, which rotates (so the moon stays over a fixed location on the earth's surface)?  Or the moon is on a tower planted on something else, while the earth rotates nearby (so the moon's position above the earth's surface changes)?
I somehow missed your post before.

I mean if the moon is attached by a tower to the earth.

I think we covered this, but one more time.

Several different situations.  If the moon is on a tower, so it is exactly still in regards to the earth.  One bulge?

But if on a tower, and the earth is rotating?  24 hour a day rotation that is? One bulge or two?

If the moon is on a tower standing on the earth, there would just be one bulge centered on the base of the tower.  If the earth is rotating, the effect of the rotation would be to squash the ocean's surface along the axis of rotation.  Those are two separate effects, and they would just sum.

Quote
If the moon is on a tower, no 24 hour spinning, but the two are rotating around their shared center of gravity monthly?  What then? (is that even possible?)

If you mean orbiting around their center of gravity, the tower isn't doing anything and you're back to the usual two bulges.
2
Science / Re: The Tides ... Take 5
OK look, here's a way to describe this gravity vs inertia story.  Consider a point mass (zero radius) orbiting the sun.  You can go to a rotating coordinate system in which the mass is at rest.  In those coordinates, the force of the sun's gravity precisely cancels the "fictitious" centrifugal force (which we can loosely term "inertia") induced by the rotating coordinates (this cancellation is why the mass is at rest in these coordinates).

Now consider a finite-radius planet rather than a point mass.  Relative to the center, the part of the planet closest to the sun has a stronger gravity force and a weaker centrifugal force/inertia, so the net force is towards the sun, creating a bulge pointing towards the sun.  The opposite is true on the part farthest from the sun, so there's a bulge there pointing away from the sun. 

Described this way, both bulges are affected by both gravity and "inertia", but gravity is stronger on the point closest to the sun, and inertia/centrifugal force is stronger at the point farthest from the sun.
3
Science / Re: The Tides ... Take 5
That is exactly what I want to know about. "Sometimes people talk about gravity producing the near side bulge and "inertia" producing the far side bulge" So is it inertia or not?

I'm not sure how to cleanly separate gravity from inertia here.  Einstein's great insight was that gravity and acceleration are equivalent, in a sense.  But that equivalence doesn't apply in any simple way to tidal forces...  I'll think about it, but I'm not sure there's a simple answer to that one.

Quote
  Is the slow monthly orbit around the earth/moon center of gravity causing the tides?  That's the only orbit that is involved in the lunar tide.  Right?

That and the rotation of the earth, yes.  Again, if the earth rotated with the same period as the moon's orbit, so that the same side of the earth was always facing the moon ("tidally locked"), there'd be two bulges but no tides per se. 

You can say that gravity+orbits+inertial causes two bulges (that's the "equilibrium" part), and the earth's rotation (plus continents, finite/variable ocean depth, etc) turns those would-be bulges into actual tides (that's the "dynamic" part).
4
Science / Re: The Tides ... Take 5
To me the solar orbit tide bulge is easy to visualize.  But it seems to involve inertia as well.  The part of the planet farther away orbits more slowly than the near side, causing tidal forces, which stretch the planet.  If the planet wasn't solid it would turn the planet into a ring. Right?

But the lunar orbit tidal force is harder to mentally grasp.  Even when it is obviously much stronger.

Well, they are very similar.  Maybe it helps to remember that in all cases, both bodies orbit the center of mass of the system.  The point in both cases is that the orbital speed of the center of mass isn't the same as the would-be orbital speed of (most parts of) the surface, so there is an internal stress that causes a deformation. 

Is that deformation due to inertia or gravity?  Well, it's gravity that leads to the different would-be orbital speeds, so ultimately it's all gravity... 
5
Science / Re: The Tides ... Take 5
The reason that doesn't create two bulges when the moon is on a tower is that it ignores the main component of the force, the part acting on the center of mass, which is directed towards the moon at all points on the earth's surface.  You can (and should) ignore that part when the two bodies are in free fall, but not when they are connected by a tower or held in place in some other way.
That's the part I want you to explain.

OK, two explanations.

(1) The surface of the ocean should be an equipotential surface.  With the moon and earth fixed in place, you can easily compute the shape of that surface.  Relative to the case with no moon (where the equipotential surface would be spherical, at least assuming the earth itself is) the surface is deformed towards the moon everywhere.  So there is a bulge under the moon, and a depression opposite the moon.

(2) The moon's gravity is pulling the earth's ocean towards it.  The ocean is not solid so it can deform, while the earth's surface and the moon's are held fixed by the tower.  Obviously, the earth's ocean will be pulled towards the moon by the moon's gravity, so it will stretch towards the moon on the near side and get pushed towards it on the far side.

Clear?
6
Science / Re: The Tides ... Take 5
To cut to the chase, if the tidal effect is from motion (which it seems has to be the case), the the explanation for the second bulge using the gravity gradient as the reason does not make sense.  If it was just due to gravity there would be no second bulge.  Right?

No, not really right.  Remember, the (orbital) motion is itself due to gravity.

With just gravity (which implies orbital motion), there are two bulges because of the gradient in the gravitational force.  With gravity plus another force - the tension in the tower, for instance - you have to also take that other force into account, and it swamps the gradient and produces only one bulge.

Sometimes people talk about gravity producing the near side bulge and "inertia" producing the far side bulge.  That's not a good way to explain it, but I wouldn't say it's totally wrong either - after all, there is an equivalence between gravity and acceleration.
7
Science / Re: The Tides ... Take 5
This also is what is being said about the sun.  If the earth was not orbiting the sun there would be just one bulge from the sun as well.

Correct?

If they were both held in place, yes.  It's just like the moon.
That is exactly what I thought.

Now, the following is a new issue (as far as I can recall) in this discussion.  And it's difficult to explain with just words.  The simplest start would be those vector lines for the gravity gradient.  When they are shown (to explain the bulges), it isn't just the gravity being shown,  It includes the rotation as well.  Correct?

If I'm thinking of the same picture you are (like Figure 2 here https://en.wikipedia.org/wiki/Tidal_force), then no, it's just the gravity gradient (actually, it's the vector difference between the force at the center and the force at some point on the surface).  The reason that doesn't create two bulges when the moon is on a tower is that it ignores the main component of the force, the part acting on the center of mass, which is directed towards the moon at all points on the earth's surface.  You can (and should) ignore that part when the two bodies are in free fall, but not when they are connected by a tower or held in place in some other way.
8
Science / Re: The Tides ... Take 5
This of course applies to the solid earth tide as well.  The deformation of the solid earth (which we do know is twin bulges, so to speak), is from the orbiting motion.  Both the moon earth orbit, and the earth orbiting the sun.

Yes, and since the earth's interior is fairly symmetric those are simpler than ocean tides (no continents in the way or variable ocean depths).  I suppose there are also atmospheric tides, which again should be fairly simple.
9
Science / Re: The Tides ... Take 5
This also is what is being said about the sun.  If the earth was not orbiting the sun there would be just one bulge from the sun as well.

Correct?

If they were both held in place, yes.  It's just like the moon.
10
Science / Re: The Tides ... Take 5
So the motion creates the tides, not just the gravity gradient? 

Of course - if the tidal forces were constant in time (as they would be without motion), the response would also be constant in time.

Quote
Not the earth spinning, but the orbit of the moon and earth? 

It's a combination of both spin and orbit.  The closest "realistic" situation to the moon on a tower is the moon in a geosynchronous orbit, where the earth's spin period equals the moon's orbital period (and they are in the same plane, let's say).  In that case there would be no lunar tides - there would just be two static (with respect to the earth's surface) bulges.  But if the moon's orbit is not geosynchronous, there will be dynamical lunar tides.
11
Science / Re: The Tides ... Take 5
And one more question. I am assuming when he talks about the moon on a tower there is no rotation involved.  But he doesn't specify that.  If there is a rotating earth with the moon on a tower, isn't that a different issue?

You mean, the moon is on a tower planted on the earth, which rotates (so the moon stays over a fixed location on the earth's surface)?  Or the moon is on a tower planted on something else, while the earth rotates nearby (so the moon's position above the earth's surface changes)?
I'm not saying anything.  I am asking what you think he is saying.



The moon on a tower (so it is not orbiting) seems to be a different issue than non rotating, but in either case, what do you think?

If the moon is on a tower, so no orbit, what would happen?  He clearly says one bulge.

Same for non orbiting of the sun, one bulge.  It seems motion is needed for tides.

If the moon is on a tower standing on the earth, there would just be one bulge centered on the base of the tower.  If the earth is rotating, the effect of the rotation would be to squash the ocean's surface along the axis of rotation.  Those are two separate effects, and they would just sum.
12
Science / Re: The Tides ... Take 5
He is saying with no rotation there is one bulge,  If you asked me if I agree with that, I would have said yes.  Because I simply believe what an expert wrote about it.  But I want your view on it.

I agree with this - if the earth and moon were held fixed in place by some external force, there would only be a bulge on the near side, and no bulge on the far side of the earth (actually there would be an "anti-bulge" there).

However if the moon and earth were in a geosynchronous orbit, there would be two bulges, stationary with respect to the earth's surface, one under the moon and one on the opposite side.
13
Science / Re: The Tides ... Take 5
And one more question. I am assuming when he talks about the moon on a tower there is no rotation involved.  But he doesn't specify that.  If there is a rotating earth with the moon on a tower, isn't that a different issue?

You mean, the moon is on a tower planted on the earth, which rotates (so the moon stays over a fixed location on the earth's surface)?  Or the moon is on a tower planted on something else, while the earth rotates nearby (so the moon's position above the earth's surface changes)?
14
Science / Re: The Tides ... Take 5
He is saying with no rotation there is one bulge,  If you asked me if I agree with that, I would have said yes.  Because I simply believe what an expert wrote about it.  But I want your view on it.

That's a good question - it's actually trickier than the real case where the earth and moon are in free fall.  I'm thinking about it.
15
Science / Re: The Tides ... Take 5
The following is a different matter.
It's exactly why I asked you if the tide would be the same in the canal, if it was 21,000 m deep, instead of 4000 m deep.  And then I asked about a canal 10 meters deep.
And I answered that, years ago.  The tide will not be the same, because the depth determines the free wave speed, and the free wave speed determines the phase for a given driving frequency (and affects the amplitude).  However, for all depths the tidal wave speed will be the same, as always for a driven oscillation in steady state.
I understand exactly what you are claiming, that the depth only changes the timing of the tide bulges in a canal. That shallow means the bulges will be "out of phase", but they will still be moving at 1000 mph.

Correct.  Do you or do you not agree with that?

Quote
I mean you wrote " for all depths the tidal wave speed will be the same", so that much is clear.  What you avoid answering is how and why you think this would be the case.

I never avoided that either.  I've explained it multiple times over the years.  In the end, it's the same phenomenon as a driven damped harmonic oscillator, a system studied by every physics student.  If you've never studied physics and/or can't understand math I can still provide an intuitive explanation for it.  But we never got even close to that, because you constantly bukkake the thread with "because physics" trolling.
16
Science / Re: The Tides ... Take 5
If there is no energy lost from bottom friction, what would happen?  Either in a real ocean or a magic canal? 
As always, you avoid simply answering this straight forward question.  The answer was mentioned in several books I linked to,  It's directly part of the argument you are putting forward, especially in regards to the hypothetical canal circling the earth.
And this new thing may or may not be connected to the canal problem.  Which you still have avoided answering.

I've never avoided answering anything - but keep right on trollin'.
17
Science / Re: The Tides ... Take 5
Keep on trollin'!
18
Science / Re: The Tides ... Take 5
Your own source says forced waves can go with a speed corresponding to that of the force.
You haven't read all of the text, that much is obvious.  It's a shame it's not all online, but if you read Chapter 14 you might realize there is more to the issue than a" one sentence hypothetical statement" that is the part of a very long discussion.

Dunno if that was directed at me, but I have read (most of) Chapter 14.  It agrees completely with me, and Butikov. 
19
Science / Re: The Tides ... Take 5
Now, F X, are you finally going to concede?  Or are you just going to hide from the burn for a few days and come back with more trolling?  The choice is yours....

Looks like option B.  What a shock.
20
Science / Re: The Tides ... Take 5
Quote from: Biology and the Mechanics of the Wave-Swept Environment
[so far] we have dealt with free waves...[that] travel under their own steam.  In contrast, tidal waves are forced...this requires the crest of a tidal wave to travel...every 12 hours and 25 minutes....a free wave would travel with celerity of 196 m/s and require 19 hours...Thus, the tractive forces must coerce the tidal wave to move faster than it otherwise would.
(my bold)

Now, F X, are you finally going to concede?  Or are you just going to hide from the burn for a few days and come back with more trolling?   The choice is yours....

21
Science / Re: The Tides ... Take 5
The right way to understand things is to start simple and gradually add complexity.
Stating something that is well known is a meaningless statement. 

Here's a simple start, which I used years ago, to explain why a tide can't move at 1000 mph.  (I usually say "because physics")



See?  That you think I came up with this is hilarious.

You're such a fool.  There's really no other word for it. 

That book is an excellent source, and I agree completely with everything I read in it - which directly contradicts you.

(The page you reproduced is 111.  Did you read p112, by any chance?)
22
Science / Re: The Tides ... Take 5
So what causes the other bulge?

If you are actually asking, I'd be happy to explain it.

Quote
And since there isn't actually two bulges, why are there so many different sources explaining a non-existent two bulge situation? 

The right way to understand things is to start simple and gradually add complexity.

Quote
And along with  all that, why are there different claims about the bulges?  Some claim they are ahead of the moon, others claim they are lagging behind the moon.  It's all so fucked up.  It really is.

There are certainly a lot of wrong things on the internet.  The irony here is, by posting them because they support you, you're helping me correct them so they refute you...
23
Science / Re: The Tides ... Take 5
Quote
On the equator where the planet's circumference is about 38,700 km (24,000 mi), the tide crest would travel at about 1600 km (1000 mi) per hr. On the real Earth, continents break up the ocean into separate basins and the ocean has a finite depth. Tides are shallow water waves so that wave celerity depends on water depth. For an average ocean depth of 4000 m (13,000 ft), the tidal celerity is about 200 in per sec (444 mi per hr).
http://oceanmotion.org/html/background/tides.htm

That's a bit vague, but probably wrong.

Quote
Quote
From this Earth-centric reference frame, in order
for the sea surface to "keep up" with
the forcing, the sea level bulges need
to move laterally through the ocean.
The signal propagates as a surface
gravity wave (influenced by rotation)
and the speed of that propagation is
limited by the shallow water wave
speed, C = gH , which at the
equator is only about half the speed
at which the forcing moves.
https://faculty.washington.edu/luanne/pages/ocean420/notes/tidedynamics.pdf

That one is definitely wrong.  Well done!  I'll see if I can get that one corrected too, like the oceanography textbook.
24
Science / Re: The Tides ... Take 5
You know F X, if you were correct about the speed of tidal waves, it would mean that in the canal earth the number of high tides per day would depend entirely on the depth.  If the canal was pretty shallow, the wave speed would be small and the high tide line would move slowly.

As a numerical example, suppose the canal is 100 meters deep (uniformly).  Then the period of a free wave would be about 23 days - so someone living by the canal would observe one high tide every 11.5 days.  Change the depth, and you can get any period you want.  Getting two (or one) high tides a day would require the depth to be exactly tuned to the right number. 

This would also mean that the standard harmonic analysis of tides used on the real earth - which breaks tides down into harmonic constituents at discrete frequencies determined by the period of motion of the moon and sun - would fail completely, since the periods would be determined by ocean depth, not by astronomy.

Doesn't that bother you a little?  Feel some doubt creeping in?

Of course no such strange periods are observed anywhere on earth, including in shallow water near coasts, because tidal wave frequencies are not determined by the depth, but by the driving force(s).
25
Science / Re: The Tides ... Take 5
Bottom friction.  The thing Butikav leaves out of the equations. 

OK, that's at least a coherent statement.  Unfortunately for you, Butikov does not leave friction out.  Instead, he writes

Quote from: Butikov
If we ignore friction (dissipation of mechanical energy in the excited oscillation),
the forced motion occurs exactly in phase with the driving force, provided the driving
period T is longer than the natural period (T > T0). Otherwise (if T < T0) the forced
motion occurs in the opposite phase with respect to the driving force.

In equations (18) of http://butikov.faculty.ifmo.ru/TidesOD.pdf, you can see there is a parameter gamma - which would be zero if Butikov ignored friction.

Care to try again?