A riddle for Tony Pagano: Tides

[J. J. Lodder]

A riddle for Tony Pagano: Tides
Let us assume that the earth is really stationary
at the centre of the universe,
and that the moon orbits around it.
The moon will cause tides, by its gravitational attraction,
and the tides will follow the moon.

Riddle: Why are there two tides a day, and not one?
Why is there also a high ride on the side of the earth
that is farthest away from the moon?
(where the moon's gravity is weaker)

Hint: Newtonians explain the high tide on the far side of th earth
by noting that the earth and the moon both orbit
around their common centre of mass, (and next do the sums)
but that explanation is impossible if the earth is truly stationary.

So please explain how there can be two tides a day
in a purely geostationary model,

Jan

[John Harshman]

I don't understand why this is an objection to geocentrism, or what the
center of gravity has to do with tides.

[J. J. Lodder]

Then you don't understand the tides.
The problem has a long history, going back to Galileo.
G. thought that he could prove the motion of the earth
by means of his theory of the tides.
The idea was sound in principle,
but G. lacked an adequate understanding of mechanics,
and his theory of the tides was wrong.

The crucial point is that there are -two- high tides a day.
There is a high tide directly 'under' the moon,
but there is also a high tide on the opposite side of the earth,
as far away from the moon's gravity as you can get.

The challenge is to explain that.
Newton (and of course also Einstein) could,
using a moving earth.
(see any textbook for how it is done)

I don't think it can be done on a strictly stationary earth,
where you expect only one tide per day.
(high tide 'under' the moon,
low tide on the opposite side of the earth,

where the moon's gravity is weaker)

Jan

[John Harshman]

On 3/16/18 7:30 AM, J. J. Lodder wrote:
> John Harshman <jhar...@pacbell.net> wrote:
>
>> On 3/16/18 3:12 AM, J. J. Lodder wrote:
>>> A riddle for Tony Pagano: Tides
>>> Let us assume that the earth is really stationary
>>> at the centre of the universe,
>>> and that the moon orbits around it.
>>> The moon will cause tides, by its gravitational attraction,
>>> and the tides will follow the moon.
>>>
>>> Riddle: Why are there two tides a day, and not one?
>>> Why is there also a high ride on the side of the earth
>>> that is farthest away from the moon?
>>> (where the moon's gravity is weaker)
>>>
>>> Hint: Newtonians explain the high tide on the far side of th earth
>>> by noting that the earth and the moon both orbit
>>> around their common centre of mass, (and next do the sums)
>>> but that explanation is impossible if the earth is truly stationary.
>>>
>>> So please explain how there can be two tides a day
>>> in a purely geostationary model,
>>
>> I don't understand why this is an objection to geocentrism, or what the
>> center of gravity has to do with tides.
>
> Then you don't understand the tides.

Perhaps, but an alternative hypothesis is that you don't understand the

tides.

> The problem has a long history, going back to Galileo.
> G. thought that he could prove the motion of the earth
> by means of his theory of the tides.
> The idea was sound in principle,
> but G. lacked an adequate understanding of mechanics,
> and his theory of the tides was wrong.

How is that relevant?

> The crucial point is that there are -two- high tides a day.
> There is a high tide directly 'under' the moon,
> but there is also a high tide on the opposite side of the earth,
> as far away from the moon's gravity as you can get.
>
> The challenge is to explain that.

The explanation is simple enough, but I don't see what it has to do with
a moving earth. A stationary earth, given a moving moon, would have the
same tides. On the near side, the water, being closer to the moon than
the rest of the earth, is pulled more toward the moon. On the far side,
the rest of the earth, being closer to the moon than the water on that
side, is pulled more toward the moon, in both cases causing a tidal bulge.

> Newton (and of course also Einstein) could,
> using a moving earth.
> (see any textbook for how it is done)

Which text book did you get the need for a moving earth from? I'm dubious.

> I don't think it can be done on a strictly stationary earth,
> where you expect only one tide per day.
> (high tide 'under' the moon,
> low tide on the opposite side of the earth,
> where the moon's gravity is weaker)

I think you're the one who misunderstands the tides. Low tide would in
all cases be half-way around, not the opposite side. Tides are the same
given the same relative motions, regardless of whether the moon's
apparently motion is real or caused by earth's rotation.

[erik simpson]

Tides are a manifestation of the derivative of the "gravitational force"
(apologies to GR, it's actually the derivative of space-time curvature), and
is present whether or not the earth is moving or rotating. The rotation has
the effect of "delaying" the tidal bulge, mainly through hydrodynamic effects,
so the bulges aren't seen directly under the moon (and sun). It's interesting
to note that the magnitude of the tidal effects depend only on the solid angle
subtended by the attracting object and its density (proportional to M/R^3). The
sun's mean density is 1.41 g/cm^3, while the moon's is 3.34 g/cm^3, accounting
for their relative strength of tidal interactions.

[John Harshman]

Say what? I'd say that the dependence of tidal strength on m/d^3
accounts almost entirely for the relative strength of moon's and sun's
tidal interactions with earth. And wouldn't tides be quite accurately
determined if the moon and sun were treated as point masses?

It also seems to me that the observed magnitude of ocean tides depends
strongly on the relative deformability (or whatever other word is
commonly used) of water and rock. The water is free to respond fully to
tidal forces, while most of the planet is not.

[erik simpson]

I expressed that clumsily. Point sources for sun and moon would do nicely, but
since angle*distance ~ radius, m/radius^3 can be relevant. Venus and Jupiter
sometimes have very comparable angular sizes, but the much higher density of
Venus means its tide (on earth) are much stronger.

[jillery]

Two tides per day are not a consequence of Earth's rotation. From:

<https://en.wikipedia.org/wiki/Tidal_force>

*****************************
Tidal acceleration does not require rotation or orbiting bodies; for
example, the body may be freefalling in a straight line under the
influence of a gravitational field while still being influenced by
(changing) tidal acceleration.
*****************************

The Moon causes two water bulges on opposite sides of the Earth
because the Moon attracts the Earth as well as the water on the Earth.
Look at figures 2 and 4 for illustration.

Nevertheless, some places on Earth experience only one Lunar tide per
day. From:

<https://en.wikipedia.org/wiki/Tide>

*******************************
Semi-diurnal tides dominated coastline, but some areas such as the
South China Sea and the Gulf of Mexico are primarily diurnal.
*******************************

Also, the Lunar tidal bulge isn't always directly underneath the Moon:

******************************
Although it may seem that tides could be predicted via a sufficiently
detailed knowledge of instantaneous astronomical forcings, the actual
tide at a given location is determined by astronomical forces
accumulated over many days. In addition, precise results require
detailed knowledge of the shape of all the ocean basins—their
bathymetry, and coastline shape.
******************************

And of course, the Sun also causes tides, but that's not relevant to
this discussion.

--
I disapprove of what you say, but I will defend to the death your right to say it.

Evelyn Beatrice Hall
Attributed to Voltaire

[John Harshman]

Are you perhaps using angular size * density as a proxy for mass? I
don't understand you here. The relative strength of tidal forces is
equal to the relative values of m/d^3. Venus exerts a stronger tidal
force because it's much closer than Jupiter even though it's much less
massive. My calculations using only mass and distance show Venus at
closest approach exerting 8 times the tidal force on earth as Jupiter at
closest approach. Is that wrong?

[John Harshman]

On 3/16/18 3:12 AM, J. J. Lodder wrote:

While I don't think that works at all, here's an interesting question
that comes to mind (though Tony will never respond): Why has the
universe's (or the ether's, if you prefer) rotation been slowing down
over the past several hundred million years? The usual explanation is
tidal friction exerted by the moon and the sun, but of course that can't
apply to the universe, can it?

I forget whether Tony rejects an old earth; that might be his out.

[erik simpson]

Exactly.

> don't understand you here. The relative strength of tidal forces is
> equal to the relative values of m/d^3. Venus exerts a stronger tidal
> force because it's much closer than Jupiter even though it's much less
> massive. My calculations using only mass and distance show Venus at
> closest approach exerting 8 times the tidal force on earth as Jupiter at
> closest approach. Is that wrong?

No. But if you knew only the angular size and guessed the density you'd be
close.

[T Pagano]

***********************TIDES******************************************
There are plenty of good explanations for tides; what's your point? If
you think the GeoCentric Model can't explain tides then provide the
details of the failure. Otherwise I have nothing to explain.

The geometries and relative motions of the heliocentricism and
geocentricism are IDENTICAL. So why would you assume without providing
specific details of a failure that geoCentricism can't explain tides?

The claims are easy, the details always seem to be elsewhere.





*********************CLOCK STABILITY************************************
In another thread lodder previously claimed:
[BEGIN LODDER QUOTE]
.. . .the idea of a non-rotating earth is unambigeously falsified by all
observations in which the clock stability is better than 10^-8. (it is
better than 10^-14 nowadays).

We know empirically, by direct bservation, that the rotation rate of the
earth is quite variable on all time scales. So the universe (if it would
be doing that) must rotate differentially at different distances. (for
example much faster in the past) This has the absurd consequences already
outlined in my previous posting.
[END LODDER QUOTE]



1. The sidereal rotation rate has virtually no variation with respect to
an immobile earth.

2. Lodder hasn't shown exactly how variations in rotation rate on the
order of magnitude of 10^-8 are significant or relevant to the GeoCentric
model.

3. Furthermore the GeoCentric model can be idealized as a rotating
Newton's Sphere, with a dense, fluid like universal ether that can be
idealized as a rigid body. That is, every part of the sphere is rotating
at 15 degrees per hour around an immobile Earth.

4. If there are "timing" problems Lodder has failed to explain them.


5. Finally atheists in general and Lodder in particular have yet to prove
that the Earth moves. That challenge hasn't been met in 500 years.
Perhaps JJ lodder can pull that rabbit out of his hat.

[J. J. Lodder]

T Pagano <notmya...@dot.com> wrote:

> On Fri, 16 Mar 2018 11:12:10 +0100, J. J. Lodder wrote:
>
> > A riddle for Tony Pagano: Tides Let us assume that the earth is really
> > stationary at the centre of the universe,
> > and that the moon orbits around it.
> > The moon will cause tides, by its gravitational attraction,
> > and the tides will follow the moon.
> >
> > Riddle: Why are there two tides a day, and not one?
> > Why is there also a high ride on the side of the earth that is farthest
> > away from the moon?
> > (where the moon's gravity is weaker)
> > Hint: Newtonians explain the high tide on the far side of th earth by
> > noting that the earth and the moon both orbit around their common centre
> > of mass, (and next do the sums)
> > but that explanation is impossible if the earth is truly stationary.
> >
> > So please explain how there can be two tides a day in a purely
> > geostationary model,
> >
> > Jan
>
>
> ***********************TIDES******************************************
> There are plenty of good explanations for tides; what's your point?

Just one will do, and Newtonians supplied it.
(Laplace had the final word, iirc)

> If
> you think the GeoCentric Model can't explain tides then provide the
> details of the failure. Otherwise I have nothing to explain.

You must explain why there are -two- tidal bulges,
one where the moon is high in the sky, aka at zenith,
and the other one where the moon is op the opposite side of the earth,
aka at nadir.
(perhaps offset by some hours by local basin geometry)

> The geometries and relative motions of the heliocentricism and
> geocentricism are IDENTICAL. So why would you assume without providing
> specific details of a failure that geoCentricism can't explain tides?

Newtonians explain the second tidal bulge from the fact
that the moon is not only orbiting the earth,
the earth is also orbiting the moon.
In fact they both orbit around their common centre of mass.
In other words, the second bulge is an inertial effect,
caused by the motion of the earth.

If you insist that the earth is not moving at all
you must come up with some other explanation
for there being two tidal bulges,

Jan

[J. J. Lodder]

I have a posting on that from way back that I never posted.
I'll reviveit sometime.

> I forget whether Tony rejects an old earth; that might be his out.

Doesn't matter.
The earth slowed down quite noticably already in historical times.
Records of historical eclipses from all over the world prove this.
See for example
<http://www.sciencemag.org/news/2016/12/ancient-eclipses-show-earth-s-ro
tation-slowing>
The earth has lost about six hours since biblical times.
(or the universe, for Tony)

And for Tony, No not even slowing down the universe
will save you here.
The orbital motions of the moon and the sun just go on.
They are not affected by earth or universe slowdown,
so these historical eclipses really are observed in the wrong place,

Jan

[J. J. Lodder]

Indeed, real tides are a very complicated subject.
The idealised model with two tidal bulges
applies to an idealised all-ocean earth only.
(and to the solid tides of the crust)

In reality, continents get in the way of the tidal wave,
and basisns will have resonances.

Nevertheless, all those local tidal waves
can be explained only from semi-diurnal forcing,
(that is, twice per lunar day)

Jan

[Ernest Major]

It's not clear to me that the tides per se are a problem for Tony. He
claims that the net gravitational force of the whole universe on the
earth is zero (ignoring the missing mass problem, while decrying dark
matter and dark energy as explanations of observational data).
Qualitatively, on one side of the earth the relativity proximity of the
moon adds to the force giving one tidal bulge. On the other side of the
earth the relative distance of the moon means that the net force from
the rest of the universe isn't quite cancelled out, giving a second
tidal bulge. That is, the ocean surface seeks to adopt a gravitational
equipotential surface.

Tony's problem is explaining how all his dark matter rearranges itself
on a monthly basis as the sun and moon move from conjuction to
opposition and back again, without disrupting the orderly motions of the
solar system.

Tony's general dynamical problem, assuming that the earth is
instaneously simultaneously at the centre of mass and centre of gravity
of the universe, is reconciling the observed distribution of matter in
the universe with the two points remaining coincident and the earth
remaining at either of these points.

He could arbitrarily select a reference frame in which the earth is
non-moving and non-rotating, but he has to introduce all sorts of
fictional forces (a general geocentric force, and all sorts of
epicircular forces), in additional to centrifugal and coriolis forces,
preserving the relative motions. I hate to think what it does to the
rest of physics, such as electrodynamics.

Or he could just admit to throwing out Newtonian physics.

--
alias Ernest Major

[J. J. Lodder]

T Pagano <notmya...@dot.com> wrote:
[reply part 2]

> *********************CLOCK STABILITY************************************
> In another thread lodder previously claimed:
> [BEGIN LODDER QUOTE]
> .. . .the idea of a non-rotating earth is unambigeously falsified by all
> observations in which the clock stability is better than 10^-8. (it is
> better than 10^-14 nowadays).
>
> We know empirically, by direct bservation, that the rotation rate of the
> earth is quite variable on all time scales. So the universe (if it would
> be doing that) must rotate differentially at different distances. (for
> example much faster in the past) This has the absurd consequences already
> outlined in my previous posting.
> [END LODDER QUOTE]
>
>
>
> 1. The sidereal rotation rate has virtually no variation with respect to
> an immobile earth.

That is precisely where you are wrong.
The irregularities in the rotation of the earth
are established by observing Mach's 'distant stars',
(nowadays quasars, and by VLBI)
Your point was valid, before WWII, but it is untenable
with the millionfold increase in accuracy since.

> 2. Lodder hasn't shown exactly how variations in rotation rate on the
> order of magnitude of 10^-8 are significant or relevant to the GeoCentric
> model.

Once you know that the rotation rate of the earth is quite irregular,
for well understood physical reasons,
the idea of a stationary earth is untenable.

> 3. Furthermore the GeoCentric model can be idealized as a rotating
> Newton's Sphere, with a dense, fluid like universal ether that can be
> idealized as a rigid body. That is, every part of the sphere is rotating
> at 15 degrees per hour around an immobile Earth.

As long as you cannot observe to better than 10^-8.
(which is the stability of the rotation of the earth)

> 4. If there are "timing" problems Lodder has failed to explain them.

No, you have failed to understand them.

> 5. Finally atheists in general and Lodder in particular have yet to prove
> that the Earth moves. That challenge hasn't been met in 500 years.
> Perhaps JJ lodder can pull that rabbit out of his hat.

Science doesn't supply proofs.
It merely shows that some explanations of the phenomena are reasonable,
and that others (yours in particular) are utterly ridiculous,

Jan

[J. J. Lodder]

John Harshman <jhar...@pacbell.net> wrote:

> Say what? I'd say that the dependence of tidal strength on m/d^3
> accounts almost entirely for the relative strength of moon's and sun's
> tidal interactions with earth. And wouldn't tides be quite accurately
> determined if the moon and sun were treated as point masses?

Correct. The moon's gravitational field is rather irregular,
but I don't think that the higher multipole moments
can be observed in the tides.
As a matter of historical fact they were unknown (except by theory)
before the Apollo missions.

> It also seems to me that the observed magnitude of ocean tides depends
> strongly on the relative deformability (or whatever other word is
> commonly used) of water and rock. The water is free to respond fully to
> tidal forces, while most of the planet is not.

There are solid earth tides as well.
(with a typical amplitude of decimeters)
Your idea of deformability is wrong,
material strength is irrelevant on a whole earth scale,

Jan

[John Harshman]

You just contradicted yourself. Solid earth tides are much smaller than
oceanic tides. Why? Because rock is less deformable than water.

[John Harshman]

I don't think you need the rest of the universe at all. A moon that goes
around the earth in a little bit less than 24 hours is all you need to
explain both tidal bulges. Of course you do need the rest of the
universe to explain the moon's extra motion, but the two tidal bulges
aren't a problem for Tony at all.

[Jonathan]

On 3/16/2018 6:12 AM, J. J. Lodder wrote:

> A riddle for Tony Pagano: Tides
> Let us assume that the earth is really stationary
> at the centre of the universe,
> and that the moon orbits around it.
> The moon will cause tides, by its gravitational attraction,
> and the tides will follow the moon.
>

In a geocentric model there wouldn't be any change
in the tides, as it's just a frame of reference change
from a heliocentric model.

But if you mean the Earth didn't rotate about it's
axis also, then we'd be in a real pickle as
the ocean bulge at the equator of some 8km
due to centrifugal forces would be lost and
cause the oceans to flow to the poles, and the
land would be at the equator.

Not to mention a day would be 8760 hours long.
That would give new meaning to the five day
work week.

Makes me tired just thinking about it.

> Riddle: Why are there two tides a day, and not one?
> Why is there also a high ride on the side of the earth
> that is farthest away from the moon?
> (where the moon's gravity is weaker)
>
> Hint: Newtonians explain the high tide on the far side of th earth
> by noting that the earth and the moon both orbit
> around their common centre of mass, (and next do the sums)
> but that explanation is impossible if the earth is truly stationary.
>
> So please explain how there can be two tides a day
> in a purely geostationary model,
>
> Jan
>

--

"To paraphrase the Buddha — Three things cannot be long hidden:
the sun; the moon; and the truth. ‬

~ Former FBI Director James Comey (12-1-17)


s

[Bob Casanova]

On Sat, 17 Mar 2018 10:21:38 +0100, the following appeared
in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
Lodder):

<snip to the point>

>The earth slowed down quite noticably already in historical times.
>Records of historical eclipses from all over the world prove this.
>See for example
><http://www.sciencemag.org/news/2016/12/ancient-eclipses-show-earth-s-ro
>tation-slowing>
>The earth has lost about six hours since biblical times.
>(or the universe, for Tony)

Wait; what? Are you saying that the length of a day was 30
hours in Biblical times? If not, what *are* you saying?
--

Bob C.

"The most exciting phrase to hear in science,
the one that heralds new discoveries, is not
'Eureka!' but 'That's funny...'"

- Isaac Asimov

[John Harshman]

On 3/17/18 11:22 AM, Bob Casanova wrote:
> On Sat, 17 Mar 2018 10:21:38 +0100, the following appeared
> in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
> Lodder):
>
> <snip to the point>
>
>
>> The earth slowed down quite noticably already in historical times.
>> Records of historical eclipses from all over the world prove this.
>> See for example
>> <http://www.sciencemag.org/news/2016/12/ancient-eclipses-show-earth-s-ro
>> tation-slowing>
>> The earth has lost about six hours since biblical times.
>> (or the universe, for Tony)
>
> Wait; what? Are you saying that the length of a day was 30
> hours in Biblical times? If not, what *are* you saying?
>

Don't know what he means, but the article says that the total time since
that biblical date is 6 hours more than it would have been if the earth
weren't slowing down.

[Ernest Major]

On 17/03/2018 18:22, Bob Casanova wrote:
> On Sat, 17 Mar 2018 10:21:38 +0100, the following appeared
> in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
> Lodder):
>
> <snip to the point>
>
>
>> The earth slowed down quite noticably already in historical times.
>> Records of historical eclipses from all over the world prove this.
>> See for example
>> <http://www.sciencemag.org/news/2016/12/ancient-eclipses-show-earth-s-ro
>> tation-slowing>
>> The earth has lost about six hours since biblical times.
>> (or the universe, for Tony)
>
> Wait; what? Are you saying that the length of a day was 30
> hours in Biblical times? If not, what *are* you saying?
>

He's saying that solar eclipes occurred 6 hours (either earlier or
later) and hence a quarter of the world away from where we would
calculate them to have occurred if we assumed the rotational period of
the earth was unchanging.

--
alias Ernest Major

[riskys...@gmail.com]

I don't know what you mean by "virtually", but the Earth does in fact vary in rotational speed. Slightly? Sure.

>
> 2. Lodder hasn't shown exactly how variations in rotation rate on the
> order of magnitude of 10^-8 are significant or relevant to the GeoCentric
> model.

Many people have, including Carlip ... and even me.

>
> 3. Furthermore the GeoCentric model can be idealized as a rotating
> Newton's Sphere, with a dense, fluid like universal ether that can be
> idealized as a rigid body. That is, every part of the sphere is rotating
> at 15 degrees per hour around an immobile Earth.

And that model is directly contradicted by the observed variations in rotational speed. Why, you might ask?

>
> 4. If there are "timing" problems Lodder has failed to explain them.

Here it is in a nutshell:

Assume the Earth does not rotate. We then observe the rest of the Cosmos to rotate around us, with all objects except those in our own Solar System appearing to retain a fixed relationship to each other. If the rotational speed is perfectly unchanging, that model might kind-of work.

But what if there are variations, which there are?

Let's take John's question first: We observe the rotation to be slowing down. That kills your "rigid body" model. Because the rotational rate we measure for Alpha Centauri today reflects its motion a scant 4 years ago, while for even the nearest quasars we are measuring their motion from a billion years ago.

Yet they match each other. Both are slowing down, but the quasar must have slowed to today's rate a billion years ago, while Alpha Centauri got to that rate only in 2013. So they cannot both be part of the same "dense, fluid like universal ether that can be idealized as a rigid body", because at any given time, objects at different distances from Earth are rotating around us at different speeds.

But it's worse than that. The cosmos isn't merely slowing down; our more accurate clocks show that the cosmos varies up and down in speed on even short time scales. And every object in the cosmos appears to vary in perfect synchronicity, despite the fact that in order to do so they all need to follow exactly the same pattern, but separated in time by years, or thousands of years, or millions of years, or billions of years.

> 5. Finally atheists in general and Lodder in particular have yet to prove
> that the Earth moves. That challenge hasn't been met in 500 years.
> Perhaps JJ lodder can pull that rabbit out of his hat.

Perhaps rather than uncountable objects in the universe each doing its own decelerating and slightly jerky little dance around the Earth in order to create the illusion of synchronicity, it is simply one object, the Earth, that rotates; slowed by well-understood physical factors such as tha moon and subject to slight variations up and down as physical objects are wont to do.

[William Hyde]

On Saturday, March 17, 2018 at 2:25:03 PM UTC-4, Bob Casanova wrote:
> On Sat, 17 Mar 2018 10:21:38 +0100, the following appeared
> in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
> Lodder):
>
> <snip to the point>
>
>
> >The earth slowed down quite noticably already in historical times.
> >Records of historical eclipses from all over the world prove this.
> >See for example
> ><http://www.sciencemag.org/news/2016/12/ancient-eclipses-show-earth-s-ro
> >tation-slowing>
> >The earth has lost about six hours since biblical times.
> >(or the universe, for Tony)
>
> Wait; what? Are you saying that the length of a day was 30
> hours in Biblical times? If not, what *are* you saying?

Babylonian astronomy was extremely accurate and therefore it's implausible to believe that they could get the timing of eclipses so very wrong, as they do. Calling, indeed, for solar eclipses after sunset.

It didn't take long for people to speculate that the Babylonians were, in fact, accurate, (in fact, I bet someone said "That's funny") and that the assumption of a uniform rotation rate for the earth was in error so that our estimated times for these eclipses were in error, not in an absolute sense, but relative to local solar time.

About thirty years ago the pattern of isostatic rebound after the last ice age was worked out. Rebound changes the inertial properties of the earth ever so slightly, and hence the rotation rate. I cannot cite the exact paper (by Peltier and Wu, I think), but IIRC when the earth's rotation was adjusted for its isostatic rebound history, the Babylonians proved to be correct within fifteen minutes.

William Hyde

[jillery]

On Sat, 17 Mar 2018 10:21:38 +0100, nos...@de-ster.demon.nl (J. J.

If you agree with the Wiki article, then you refute your claim that
the double bulge is a consequence of Earth's rotation or some kind of
inertial effect from Earth's motion.

[Bob Casanova]

On Sat, 17 Mar 2018 20:28:27 -0700 (PDT), the following
appeared in talk.origins, posted by William Hyde
<wthyd...@gmail.com>:

Thanks.

I guess I should have just read the article...

[Bob Casanova]

On Sat, 17 Mar 2018 13:44:09 -0700, the following appeared
in talk.origins, posted by John Harshman
<jhar...@pacbell.net>:

Aha! OK; thanks. So the "6 hours" assumes an absolute
duration for an "hour", not "1/24 of a solar day".

[Bob Casanova]

On Sat, 17 Mar 2018 21:09:16 +0000, the following appeared
in talk.origins, posted by Ernest Major
<{$to$}@meden.demon.co.uk>:

OK; got it. Thanks.

[John Harshman]

On 3/18/18 11:10 AM, Bob Casanova wrote:
> On Sat, 17 Mar 2018 13:44:09 -0700, the following appeared
> in talk.origins, posted by John Harshman
> <jhar...@pacbell.net>:
>
>> On 3/17/18 11:22 AM, Bob Casanova wrote:
>>> On Sat, 17 Mar 2018 10:21:38 +0100, the following appeared
>>> in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
>>> Lodder):
>>>
>>> <snip to the point>
>>>
>>>
>>>> The earth slowed down quite noticably already in historical times.
>>>> Records of historical eclipses from all over the world prove this.
>>>> See for example
>>>> <http://www.sciencemag.org/news/2016/12/ancient-eclipses-show-earth-s-ro
>>>> tation-slowing>
>>>> The earth has lost about six hours since biblical times.
>>>> (or the universe, for Tony)
>>>
>>> Wait; what? Are you saying that the length of a day was 30
>>> hours in Biblical times? If not, what *are* you saying?
>
>> Don't know what he means, but the article says that the total time since
>> that biblical date is 6 hours more than it would have been if the earth
>> weren't slowing down.
>
> Aha! OK; thanks. So the "6 hours" assumes an absolute
> duration for an "hour", not "1/24 of a solar day".
>

Now you're just being contrary for the fun of it. If an hour were
defined by that, it would be impossible for a day to have other than 24
hours. Find a definition and stick with it. To be clear: an hour is 3600
seconds. A second has a quite precise definition, which you could look up.

[T Pagano]

On Sat, 17 Mar 2018 10:21:38 +0100, J. J. Lodder wrote:

> T Pagano <notmya...@dot.com> wrote:
>
>> On Fri, 16 Mar 2018 11:12:10 +0100, J. J. Lodder wrote:
>>
>> > A riddle for Tony Pagano: Tides Let us assume that the earth is
>> > really stationary at the centre of the universe,
>> > and that the moon orbits around it.
>> > The moon will cause tides, by its gravitational attraction,
>> > and the tides will follow the moon.
>> >
>> > Riddle: Why are there two tides a day, and not one?
>> > Why is there also a high ride on the side of the earth that is
>> > farthest away from the moon?
>> > (where the moon's gravity is weaker)
>> > Hint: Newtonians explain the high tide on the far side of th earth by
>> > noting that the earth and the moon both orbit around their common
>> > centre of mass, (and next do the sums)
>> > but that explanation is impossible if the earth is truly stationary.
>> >
>> > So please explain how there can be two tides a day in a purely
>> > geostationary model,
>> >
>> > Jan
>>
>>
>> ***********************TIDES******************************************
>> There are plenty of good explanations for tides; what's your point?
>
> Just one will do, and Newtonians supplied it.
> (Laplace had the final word, iirc)

While Laplace took into account several different relevant factors to get
a more precise understanding of tides the theory still depends
principally upon Newtonian gravitational Forces.

>
>> If you think the GeoCentric Model can't explain tides then provide the
>> details of the failure. Otherwise I have nothing to explain.
>
> You must explain why there are -two- tidal bulges,
> one where the moon is high in the sky, aka at zenith,
> and the other one where the moon is op the opposite side of the earth,
> aka at nadir.
> (perhaps offset by some hours by local basin geometry)

Again you haven't explained why the GeoCentric Model fails to explain the
two high tides on opposite sides of the Earth.

>
>> The geometries and relative motions of the heliocentricism and
>> geocentricism are IDENTICAL. So why would you assume without providing
>> specific details of a failure that geoCentricism can't explain tides?
>
> Newtonians explain the second tidal bulge from the fact that the moon is
> not only orbiting the earth,
> the earth is also orbiting the moon.

1. Not only is this false it has no relevance to the GeoCentric model.
The problem with Newton's tidal explanation is NOT with Newtonion
Mechanics, per se, but a failure to recognize ALL the net forces at the
surface of the Earth.

2. Newton (along with some today) incorrectly explained the two high
tides (on opposite sides of the Earth) using the differential Tidal Force
vectors along the Earth-Moon line----one vector on the side closest to
the moon and the other vector on the opposite side of the Earth.

3. Taking the Earth's center as an inertial frame and examining the
gravitational forces only along the Earth-Moon line, Newton (and others)
incorrectly explained that:
(a) there is a relatively larger Tidal Force vector on the
Earth's surface closer to the moon (pointing towards the moon) and a
smaller Tidal Force vector on the opposite side of the Earth pointing
away from the moon. Newton (and still a few today) argue that these
Tidal Force vectors explain the high tides on opposite sides of the Earth.
(b) Unfortunately while no one disputes the Tidal Force vectors
due to the differential gravity along the Earth-Moon line those forces
are too small to explain the high tides.
(c) While the relative motions of Earth and Moon have relevance
to how the high tides move over the surface over the Earth over time (at
the Earth-Moon line) that relative motion (between Earth and Moon) had no
relevance to Newton's explanation for the high tides themselves.

> In fact they both orbit around their common centre of mass.
> In other words, the second bulge is an inertial effect,
> caused by the motion of the earth.

The claim that the tides where caused by the rotation of the Earth and
its revolution around the Sun was the nonsense foisted by Galileo as
proof that the Catholic Church was wrong. Unfortunately Galileo's theory
was known to be false almost before the ink dried on the claim.

>
> If you insist that the earth is not moving at all you must come up with
> some other explanation for there being two tidal bulges,
>
> Jan

1. You FAIL miserably.

2. I warned you REPEATEDLY that the relative motions between the Earth
and Moon in the GeoCentric Model are IDENTICAL to those in the
heliocentric system. The fact that the Earth is immobile does NOT
preclude the relative motions between the two systems from being
identical. YOU FAIL.

3. As a result any successful explanatory power attributed to the Earth-
Moon relative motion in the *heliocentric model* is automatically passed
on to the *geocentric model*. Likewise, any failure to explain
attributed to their relative motion in the geocentric model would count
as a failure in the heliocentric one.

4. While there are a great number of complications that can be added to
the theory of tides (Laplace included several: friction, Sun
gravitational additions/subtractions and others) they can be ignored to
come up with an idealized model in order to produce a basic understanding
for the high tides on opposite sides of the Earth.

5. The currently accepted model has almost nothing to do with the
relative motions between the Earth and Moon.

[scienceci...@gmail.com]

J. J. Lodder wrote 3/17,

>Indeed, real tides are a very >complicated subject. The
>idealised model with two
>tidal bulges applies to an >idealised all-ocean earth
>only. (and to the solid tides
>of the crust)

>In reality, continents get in
>the way of the tidal wave,
>and basisns will have
>resonances.

>Nevertheless, all those local >tidal waves can be explained
>only from semi-diurnal
>forcing, (that is, twice per
>lunar day)
>Jan

Like this. See NASA video below. Bulging when the syringe pulls and releases. But the moon's gravity is not like this. It does not turn on & off, not even by being directly over continents. Or else tides would recede at night and rise at morning. Like you say, Jan, " continents get in the way of the tidal wave"...

https://m.youtube.com/watch?v=HMXpoFVQ_Pw

SC RED

[Mark Isaak]

On 3/19/18 8:38 AM, T Pagano wrote:
[paraphrasing]
> Everyone is wrong about the cause of tides except me, and I have no
> idea what the cause is.

--
Mark Isaak eciton (at) curioustaxonomy (dot) net
"Ignorance, allied with power, is the most ferocious enemy justice can
have." - James Baldwin

[Bob Casanova]

On Sun, 18 Mar 2018 11:18:21 -0700, the following appeared

No intent to be contrary at all; I was perfectly serious, if
not pedantically correct.

> If an hour were
>defined by that, it would be impossible for a day to have other than 24
>hours. Find a definition and stick with it. To be clear: an hour is 3600
>seconds. A second has a quite precise definition, which you could look up.

No need; I'm aware of the definition, which has nothing to
do (now) with the length of a day. It was the idea of the
"loss of 6 hours" that had me a bit confused; I could only
interpret that as the day being 6 hours shorter now. And the
idea that the total time interval could be 6 hours less
directly contradicts the definitions of the second and the
hour, which have *not* changed (by the current definition)
since the era mentioned. So it looks to me as if I wasn't
the one using multiple definitions.

[John Harshman]

But it does have to do withe the length of an hour, which was the
immediate topic.

> It was the idea of the
> "loss of 6 hours" that had me a bit confused; I could only
> interpret that as the day being 6 hours shorter now. And the
> idea that the total time interval could be 6 hours less
> directly contradicts the definitions of the second and the
> hour, which have *not* changed (by the current definition)
> since the era mentioned. So it looks to me as if I wasn't
> the one using multiple definitions.

I have no idea what that means. Try this: if the year is one second
longer than you think it is, then any event that happened a thousand
years ago happened 17 minutes earlier than you might suppose. That's the
sort of thing the article is talking about, except that slight changes
have accumulated rather than being a single difference.

[John Harshman]

On 3/19/18 11:06 AM, Mark Isaak wrote:
> On 3/19/18 8:38 AM, T Pagano wrote:
> [paraphrasing]
> > Everyone is wrong about the cause of tides except me, and I have no
> > idea what the cause is.
>

Now, now, he didn't say he has no idea. He just failed to mention the
correct explanation. I for one would like to know what it is. Of course
Tony has lost the habit of replying to people.

[Ernest Major]

On 19/03/2018 15:38, T Pagano wrote:
> 2. I warned you REPEATEDLY that the relative motions between the Earth
> and Moon in the GeoCentric Model are IDENTICAL to those in the
> heliocentric system. The fact that the Earth is immobile does NOT
> preclude the relative motions between the two systems from being
> identical. YOU FAIL.
>
> 3. As a result any successful explanatory power attributed to the Earth-

> Moon relative motion in the*heliocentric model* is automatically passed
> on to the*geocentric model*. Likewise, any failure to explain

> attributed to their relative motion in the geocentric model would count
> as a failure in the heliocentric one.

That's not true.

You can describe the universe in any arbitrary reference frame. But
that's not the same as a physical model with a particular frame, and in
particular your preferred frame, having the same explanatory power.

When passing from a consensus physics model (in the earth orbits a point
near the sun) to a geocentric model you have to introduce multiple
fictional forces, especially if you insist that the earth is not
rotating. In a Newtonian physics model the gross motions of the solar
system are explained by the Newton's laws of motion and gravity. (There
are a few minor corrections involving radiation pressure and mass loss.)
In a more modern version these are replaced by the relativistic
equations. You have to introduce extremely strong geocentric forces to
stop the rest of the universe disapppearing due to retreating at speeds
greater than light. You have to introduce a very carefully tuned
distribution of dark matter to keep the net gravitational force on the
earth at zero - I'm not sure that it's even possible, as the forces due
to the sun and moon are much the largest gravitational forces acting on
the earth, and their net force varies greatly over the duration of a
month, as the move between conjunction and opposition and back again.

In a consensus physics model changes to the earth's rotation rate are
explained by changes to its moment of inertia (and conservation of
angular momentum), resulting from changes in its distribution of mass,
from earthquakes, and volcanic eruptions, and motion of air masses, and
growth and retreat of glaciers and ice caps, and on a longer timescale
tidal friction. Your geocentric model has to explain the past changes to
the rotation rates of various bits of the universe at various times in
the past to achieve that simultaneous appearance of changes to rotate
rate to an earth-based observer. I've never seen you even attempt to
address that challenge.

Your model rejects special relativity. That means you've lost the power
to explain observed feature of the universe such as time dilation,
relativistic mass, and mass-energy equivalence. Losing that means that
you've also lost an explanation for energy generation in stars. And in
general if you follow all the threads all sorts of other bits of physics
unravel, further reducing your explanatory power.

--
alias Ernest Major

[scienceci...@gmail.com]

To No One in Particular.

To prove the earth is moving, I think I would start with what causes sloshing hydrodynamics in man-made satellites and conclude with the same answer that makes the sloshing Seas on Earth. The hydrodynamics are the same.

From what we know about sloshing hydrodynamics in man-made satellites there must be fluid in a container, there must be centripetal center mass and centrifugal force caused by rotation; and the Earth must be acting like a satellite in orbit around the sun just as our man-made satellites containing fluids orbit around the Earth.

The Earth is a container by it's basins and its escape velocity--the forces of a centripetal center mass and a centrifugal rotational force. And finally it is in orbit because the equivalency principal is also a factor in satellite hydrodynamics.

Sloshing hydrodynamics caused instability in the positioning accuracy in satellites. NASA engineers had missed predicting it, so research grants went out. It was solved by applied mathematics. Needless to say there was a tension between the engineers and the mathematicians hired, but the problem got solved.

RS RED

[T Pagano]

On Sat, 17 Mar 2018 13:41:37 +0100, J. J. Lodder wrote:

> T Pagano <notmya...@dot.com> wrote:
> [reply part 2]
>
>> *********************CLOCK
>> STABILITY************************************ In another thread lodder
>> previously claimed:
>> [BEGIN LODDER QUOTE]
>> .. . .the idea of a non-rotating earth is unambigeously falsified by
>> all observations in which the clock stability is better than 10^-8. (it
>> is better than 10^-14 nowadays).
>>
>> We know empirically, by direct bservation, that the rotation rate of
>> the earth is quite variable on all time scales. So the universe (if it
>> would be doing that) must rotate differentially at different distances.
>> (for example much faster in the past) This has the absurd consequences
>> already outlined in my previous posting.
>> [END LODDER QUOTE]
>>
>>
>>
>> 1. The sidereal rotation rate has virtually no variation with respect
>> to an immobile earth.
>
> That is precisely where you are wrong.
> The irregularities in the rotation of the earth are established by
> observing Mach's 'distant stars',
> (nowadays quasars, and by VLBI)
> Your point was valid, before WWII, but it is untenable with the
> millionfold increase in accuracy since.

1. I should point out that Einstein himself stated unequivocally that
there was no difference between placing a reference frame with the sun as
the center or the Earth. BOTH WERE PERFECTLY VALID CHOICES. This is
fatal to your position.

2. Sagnac's interferometer experiments proved absolute rotation. That
is, his experiment proved that either the Earth was rotating on its axis
OR the star field was rotating around a fixed Earth. NEITHER ARE
PRECLUDED AS A Physical POSSIBILITY. This is FATAL to your claim.

3. In the GeoCentric Model the Earth is located at the Center of Gravity
(COG) / Center of Mass (COM) for a rotating universe. There are no net
forces or torques applied to the Earth at the COG/COM. The rotating
GeoCentric model is inherently stable regardless of minor variations in
system rotation or precession. Such variations have NO EFFECT on the
Earth at the COG/COM and hence such variations are irrelevant. This is
FATAL to your unexplained/unsupported claim.

4. After at least three postings you have yet to explain (with some
detail) why minor variations in rotation rate of the magnitude 10^-8 are
significant to the GeoCentric model. Since you cannot explain, this is
FATAL to your position.

5. You definitely have failed, refused or are impotent to explain why
this rotational variation would prohibit a central Earth. Since you
cannot explain, this is FATAL to your position.

6. All future postings which fail to explain *in detail* how rotational
variations are relevant to the GeoCentric Model will be ignored as a
waste of time.

[Mark Isaak]

On 3/19/18 8:00 PM, T Pagano wrote:
> [...]

> 3. In the GeoCentric Model the Earth is located at the Center of Gravity
> (COG) / Center of Mass (COM) for a rotating universe. There are no net
> forces or torques applied to the Earth at the COG/COM. The rotating
> GeoCentric model is inherently stable regardless of minor variations in
> system rotation or precession. Such variations have NO EFFECT on the
> Earth at the COG/COM and hence such variations are irrelevant. This is
> FATAL to your unexplained/unsupported claim.

It's time to remind people that Pagano has demonstrated, by his
cowardice, that he is unable to calculate for a simple two-body problem
what is the center of mass and where net gravitational forces are zero.

Go ahead, Pagano. Tell us what you calculate for masses of 1 and 100kg
separated by 100m. (I don't need to tell you to show your work, because
you already have, and it is nonexistent.)

[scienceci...@gmail.com]

To T Pagano,

Ancient art depictions of the seas all over the world were waves. Waves are still seen all over the world sloshing on shore lines. Basins are cups for the seas. The abstract, in physics, is sloshing in a container is caused by motion. The Earth was and still is in motion known by our observation of sloshing waves since ancient times.

What is the abstract for sloshing without motion, that would challenge the current physics of sloshing hydrodynamics?

SC RED

[Bob Casanova]

On Mon, 19 Mar 2018 11:41:07 -0700, the following appeared

Sure. But the definition of the hour, which is derived from
the length of the second, is identical now to what it was in
Biblical times *using the current definition*. And if we use
the old definition neither the second nor the hour was a
constant, and it was indeed impossible for the day to have
other that 24 hours.

What we wind up with, using the current definition, is a
requirement to adjust the length of the second periodically
if we want the day to continue to comprise 24 hours of 3600
seconds each, since the length of the day, on an absolute
basis, *is* changing constantly (if not necessarily
smoothly). Not a problem, of course.

>> It was the idea of the
>> "loss of 6 hours" that had me a bit confused; I could only
>> interpret that as the day being 6 hours shorter now. And the
>> idea that the total time interval could be 6 hours less
>> directly contradicts the definitions of the second and the
>> hour, which have *not* changed (by the current definition)
>> since the era mentioned. So it looks to me as if I wasn't
>> the one using multiple definitions.
>
>I have no idea what that means. Try this: if the year is one second
>longer than you think it is, then any event that happened a thousand
>years ago happened 17 minutes earlier than you might suppose. That's the
>sort of thing the article is talking about, except that slight changes
>have accumulated rather than being a single difference.

IOW, it's a "perception thing", and needs to be addressed on
a regular basis as noted above. OK.

[Bob Casanova]

On Mon, 19 Mar 2018 11:44:10 -0700, the following appeared

in talk.origins, posted by John Harshman
<jhar...@pacbell.net>:

He had that habit? When?

[John Harshman]

On 3/20/18 11:29 AM, Bob Casanova wrote:
> On Mon, 19 Mar 2018 11:44:10 -0700, the following appeared
> in talk.origins, posted by John Harshman
> <jhar...@pacbell.net>:
>
>> On 3/19/18 11:06 AM, Mark Isaak wrote:
>>> On 3/19/18 8:38 AM, T Pagano wrote:
>>> [paraphrasing]
>>> > Everyone is wrong about the cause of tides except me, and I have no
>>> > idea what the cause is.
>>>
>> Now, now, he didn't say he has no idea. He just failed to mention the
>> correct explanation. I for one would like to know what it is. Of course
>> Tony has lost the habit of replying to people.
>
> He had that habit? When?
>

I mean "replying" in the sense of attaching a post to a previous post,
not in any substantive way. He used to make "replies"; now he just
starts a new thread.

[John Harshman]

I don't understand what your point is, if any. Now of course what we do,
in practice, is to occasionally add a leap second to a year. Surely
you've heard of this. But all this has nothing to do with the 6-hour
difference in eclipse timing.

>>> It was the idea of the
>>> "loss of 6 hours" that had me a bit confused; I could only
>>> interpret that as the day being 6 hours shorter now. And the
>>> idea that the total time interval could be 6 hours less
>>> directly contradicts the definitions of the second and the
>>> hour, which have *not* changed (by the current definition)
>>> since the era mentioned. So it looks to me as if I wasn't
>>> the one using multiple definitions.
>>
>> I have no idea what that means. Try this: if the year is one second
>> longer than you think it is, then any event that happened a thousand
>> years ago happened 17 minutes earlier than you might suppose. That's the
>> sort of thing the article is talking about, except that slight changes
>> have accumulated rather than being a single difference.
>
> IOW, it's a "perception thing", and needs to be addressed on
> a regular basis as noted above. OK.

I have no idea what you mean by "perception thing". I was using "longer
than you think it is" in order to construct an example that's simpler
than "longer than it used to be, and getting very slightly longer all
the time".

[J. J. Lodder]

John Harshman <jhar...@pacbell.net> wrote:

> On 3/17/18 5:41 AM, J. J. Lodder wrote:
> > John Harshman <jhar...@pacbell.net> wrote:
> >
> >> Say what? I'd say that the dependence of tidal strength on m/d^3
> >> accounts almost entirely for the relative strength of moon's and sun's
> >> tidal interactions with earth. And wouldn't tides be quite accurately
> >> determined if the moon and sun were treated as point masses?
> >
> > Correct. The moon's gravitational field is rather irregular,
> > but I don't think that the higher multipole moments
> > can be observed in the tides.
> > As a matter of historical fact they were unknown (except by theory)
> > before the Apollo missions.
> >
> >> It also seems to me that the observed magnitude of ocean tides depends
> >> strongly on the relative deformability (or whatever other word is
> >> commonly used) of water and rock. The water is free to respond fully to
> >> tidal forces, while most of the planet is not.
> >
> > There are solid earth tides as well.
> > (with a typical amplitude of decimeters)
> > Your idea of deformability is wrong,
> > material strength is irrelevant on a whole earth scale,
>
> You just contradicted yourself. Solid earth tides are much smaller than
> oceanic tides. Why? Because rock is less deformable than water.

Really Harshman, you may understand some things in biology,
(but you make me begin to doubt it)
but celestial mechanics is not your thing.

Solid tides (aka earth tides) have been understood since Darwin,
(that is Sir George, son of guess who)
and the subject has been developed in great detail by now.
See
<https://en.wikipedia.org/wiki/Earth_tide>
for more than you want to know.

The dominant component is of course semi-diurnal,
just like the ocean tides.
And again, the solid tides cannot be understood
in a purely geocentric theory.

Jan

[J. J. Lodder]

That's a nice one, thanks.
Indeed, in principle one can demonstrate tidal forces also in free fall.
(but the micro-gravity on the space station will spoil it)

Read Larry Niven's 'Neutron Star' for the serious stuff,

Jan

[J. J. Lodder]

Ernest Major <{$to$}@meden.demon.co.uk> wrote:

> On 17/03/2018 09:21, J. J. Lodder wrote:
> > T Pagano <notmya...@dot.com> wrote:
> >

> >> On Fri, 16 Mar 2018 11:12:10 +0100, J. J. Lodder wrote:
> >>
> >>> A riddle for Tony Pagano: Tides Let us assume that the earth is really
> >>> stationary at the centre of the universe,
> >>> and that the moon orbits around it.
> >>> The moon will cause tides, by its gravitational attraction,
> >>> and the tides will follow the moon.
> >>>
> >>> Riddle: Why are there two tides a day, and not one?
> >>> Why is there also a high ride on the side of the earth that is farthest
> >>> away from the moon?
> >>> (where the moon's gravity is weaker)
> >>> Hint: Newtonians explain the high tide on the far side of th earth by
> >>> noting that the earth and the moon both orbit around their common centre
> >>> of mass, (and next do the sums)
> >>> but that explanation is impossible if the earth is truly stationary.
> >>>
> >>> So please explain how there can be two tides a day in a purely
> >>> geostationary model,
> >>>
> >>> Jan
> >>
> >>
> >> ***********************TIDES******************************************
> >> There are plenty of good explanations for tides; what's your point?
> >
> > Just one will do, and Newtonians supplied it.
> > (Laplace had the final word, iirc)
> >

> >> If
> >> you think the GeoCentric Model can't explain tides then provide the
> >> details of the failure. Otherwise I have nothing to explain.
> >
> > You must explain why there are -two- tidal bulges,
> > one where the moon is high in the sky, aka at zenith,
> > and the other one where the moon is op the opposite side of the earth,
> > aka at nadir.
> > (perhaps offset by some hours by local basin geometry)
> >

> >> The geometries and relative motions of the heliocentricism and
> >> geocentricism are IDENTICAL. So why would you assume without providing
> >> specific details of a failure that geoCentricism can't explain tides?
> >
> > Newtonians explain the second tidal bulge from the fact
> > that the moon is not only orbiting the earth,
> > the earth is also orbiting the moon.

> > In fact they both orbit around their common centre of mass.
> > In other words, the second bulge is an inertial effect,
> > caused by the motion of the earth.
> >

> > If you insist that the earth is not moving at all
> > you must come up with some other explanation
> > for there being two tidal bulges,
> >
> > Jan
> >

> It's not clear to me that the tides per se are a problem for Tony. He
> claims that the net gravitational force of the whole universe on the
> earth is zero (ignoring the missing mass problem, while decrying dark
> matter and dark energy as explanations of observational data).

Tony might be able to get the correct predictions for the tides
by letting the whole universe rotate, in phase with the moon,
around the centre of mass of the earth-moon system,
instead of around the centre of the earth.
(and on a smaller epicycle centerd on that to account for the sun's
effect, and...)
His handwaving would need to become even more frantic,
(shouting Lense-Thirring won't do)
and his whole 'theory' would become even more ludicrous,

Jan

[J. J. Lodder]

Bob Casanova <nos...@buzz.off> wrote:

> No need; I'm aware of the definition, which has nothing to
> do (now) with the length of a day. It was the idea of the
> "loss of 6 hours" that had me a bit confused; I could only
> interpret that as the day being 6 hours shorter now. And the
> idea that the total time interval could be 6 hours less
> directly contradicts the definitions of the second and the
> hour, which have *not* changed (by the current definition)
> since the era mentioned. So it looks to me as if I wasn't
> the one using multiple definitions.

A matter of approximating in stages.
Three thousand years ago the day was slightly shorter
than 86400 seconds,
but still indistinguishable from 86400 seconds
for all practical purposes.
Your wrist watch couldn't have told you the difference,

Jan

[J. J. Lodder]

Bob Casanova <nos...@buzz.off> wrote:

> On Sat, 17 Mar 2018 10:21:38 +0100, the following appeared
> in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
> Lodder):
>
> <snip to the point>
>
>
> >The earth slowed down quite noticably already in historical times.
> >Records of historical eclipses from all over the world prove this.
> >See for example
> ><http://www.sciencemag.org/news/2016/12/ancient-eclipses-show-earth-s-ro
> >tation-slowing>
> >The earth has lost about six hours since biblical times.
> >(or the universe, for Tony)
>
> Wait; what? Are you saying that the length of a day was 30
> hours in Biblical times? If not, what *are* you saying?

You haven't understood eclipse predictions.
It is quite possible to predict with exquisite precision
where the earth, the moon, and the sun will be thousands of years hence,
so to predict that the moon's shadow cone will intersect
with the surface of the earth.

It is not possible to predict where the surface of the earth
will be at that moment.
(because it cannot be predicted accurately enough
how much the earth will have rotated in the meantime)

The -cumulative- effect of the speed-up of the earth in the past
amounts to about 6 hours error, back in biblical times.
So the longitude of the path of totality of eclipses
(as predicted on basis of constant rotation of the earth)
is off by about ninety degrees.

Even today, with all the information about irregularities
of the earth's rotation that we have
it is not possible to predict for example
the moment of first contact of an eclipse occuring ten years hence
to the nearest second.
(see the irrugularity with which leap seconds must be declared)

Jan

[J. J. Lodder]

John Harshman <jhar...@pacbell.net> wrote:\

> I have no idea what that means. Try this: if the year is one second
> longer than you think it is, then any event that happened a thousand
> years ago happened 17 minutes earlier than you might suppose. That's the
> sort of thing the article is talking about, except that slight changes
> have accumulated rather than being a single difference.

If a clock is slowing down linearly (like the earth does)
its error accumulates quadratically,

Jan

[J. J. Lodder]

John Harshman <jhar...@pacbell.net> wrote:

> On 3/16/18 7:30 AM, J. J. Lodder wrote:
> > John Harshman <jhar...@pacbell.net> wrote:
> >

> >> On 3/16/18 3:12 AM, J. J. Lodder wrote:
> >>> A riddle for Tony Pagano: Tides
> >>> Let us assume that the earth is really stationary
> >>> at the centre of the universe,
> >>> and that the moon orbits around it.
> >>> The moon will cause tides, by its gravitational attraction,
> >>> and the tides will follow the moon.
> >>>
> >>> Riddle: Why are there two tides a day, and not one?
> >>> Why is there also a high ride on the side of the earth
> >>> that is farthest away from the moon?
> >>> (where the moon's gravity is weaker)
> >>>
> >>> Hint: Newtonians explain the high tide on the far side of th earth
> >>> by noting that the earth and the moon both orbit
> >>> around their common centre of mass, (and next do the sums)
> >>> but that explanation is impossible if the earth is truly stationary.
> >>>
> >>> So please explain how there can be two tides a day
> >>> in a purely geostationary model,
> >>

> >> I don't understand why this is an objection to geocentrism, or what the
> >> center of gravity has to do with tides.
> >
> > Then you don't understand the tides.
>
> Perhaps, but an alternative hypothesis is that you don't understand the
> tides.

Really Harshman, you may be competent on some aspects of biology,
but Newtonian mechanics isn't your thing.

> > The problem has a long history, going back to Galileo.
> > G. thought that he could prove the motion of the earth
> > by means of his theory of the tides.
> > The idea was sound in principle,
> > but G. lacked an adequate understanding of mechanics,
> > and his theory of the tides was wrong.
>
> How is that relevant?

Galileo' intuition that the tides show that the earth is 'really' moving
is fundamentally correct.
All of his attempts to work this out
into a theory of the tides were wrong,
because G. lacked adequate understanding of the mechanics involved.
He didn't know about gravitation and his theory of inertia was wrong.

> > The crucial point is that there are -two- high tides a day.
> > There is a high tide directly 'under' the moon,
> > but there is also a high tide on the opposite side of the earth,
> > as far away from the moon's gravity as you can get.
> >
> > The challenge is to explain that.
>
> The explanation is simple enough, but I don't see what it has to do with
> a moving earth. A stationary earth, given a moving moon, would have the
> same tides. On the near side, the water, being closer to the moon than
> the rest of the earth, is pulled more toward the moon. On the far side,
> the rest of the earth, being closer to the moon than the water on that
> side, is pulled more toward the moon, in both cases causing a tidal bulge.

You are as good as Pagano about it.

> > Newton (and of course also Einstein) could,
> > using a moving earth.
> > (see any textbook for how it is done)
>
> Which text book did you get the need for a moving earth from? I'm dubious.

So you admit to not understanding the first things about it.
Hint: it can only be understood dynamically, not statically,
because the earth is also orbiting the moon.

> > I don't think it can be done on a strictly stationary earth,
> > where you expect only one tide per day.
> > (high tide 'under' the moon,
> > low tide on the opposite side of the earth,

> > where the moon's gravity is weaker)
>

> I think you're the one who misunderstands the tides. Low tide would in
> all cases be half-way around, not the opposite side. Tides are the same
> given the same relative motions, regardless of whether the moon's
> apparently motion is real or caused by earth's rotation.

Papagano again, and wrong,

Jan

[John Harshman]

I don't understand what you think I'm not understanding here. How does
anything in that Wikipedia article contradict what I said or support
what you said?

[John Harshman]

Then show me where I'm wrong.

>>> The problem has a long history, going back to Galileo.
>>> G. thought that he could prove the motion of the earth
>>> by means of his theory of the tides.
>>> The idea was sound in principle,
>>> but G. lacked an adequate understanding of mechanics,
>>> and his theory of the tides was wrong.
>>
>> How is that relevant?
>
> Galileo' intuition that the tides show that the earth is 'really' moving
> is fundamentally correct.
> All of his attempts to work this out
> into a theory of the tides were wrong,
> because G. lacked adequate understanding of the mechanics involved.
> He didn't know about gravitation and his theory of inertia was wrong.

Again, how is this relevant?

>>> The crucial point is that there are -two- high tides a day.
>>> There is a high tide directly 'under' the moon,
>>> but there is also a high tide on the opposite side of the earth,
>>> as far away from the moon's gravity as you can get.
>>>
>>> The challenge is to explain that.
>>
>> The explanation is simple enough, but I don't see what it has to do with
>> a moving earth. A stationary earth, given a moving moon, would have the
>> same tides. On the near side, the water, being closer to the moon than
>> the rest of the earth, is pulled more toward the moon. On the far side,
>> the rest of the earth, being closer to the moon than the water on that
>> side, is pulled more toward the moon, in both cases causing a tidal bulge.
>
> You are as good as Pagano about it.

Please explain where I'm wrong.

>>> Newton (and of course also Einstein) could,
>>> using a moving earth.
>>> (see any textbook for how it is done)
>>
>> Which text book did you get the need for a moving earth from? I'm dubious.
>
> So you admit to not understanding the first things about it.
> Hint: it can only be understood dynamically, not statically,
> because the earth is also orbiting the moon.

Not an answer.

>>> I don't think it can be done on a strictly stationary earth,
>>> where you expect only one tide per day.
>>> (high tide 'under' the moon,
>>> low tide on the opposite side of the earth,
>>> where the moon's gravity is weaker)
>>
>> I think you're the one who misunderstands the tides. Low tide would in
>> all cases be half-way around, not the opposite side. Tides are the same
>> given the same relative motions, regardless of whether the moon's
>> apparently motion is real or caused by earth's rotation.
>
> Papagano again, and wrong,

Again, no answer.

[jillery]

Ironic that you cite a Wiki article which refutes your original claim,
that the existence of ocean tides depends on Earth's rotation.

--
I disapprove of what you say, but I will defend to the death your right to say it.

Evelyn Beatrice Hall
Attributed to Voltaire

[Don Cates]

Jan, check out
<https://en.wikipedia.org/wiki/Tidal_force>
The tidal bulges are due strictly to the gravitational field. Rotation
(moon around the earth ~once a day east to west or earth on its axis
west to east) is only necessary to ~twice a day tides at a specific
location on the earth.

--
--
Don Cates ("he's a cunning rascal" PN)

[jillery]

Great minds cite alike.

[J. J. Lodder]

Right. I was suggesting you think for yourself,

Jan

[J. J. Lodder]

JH: Solid earth tides are much smaller than oceanic tides. Why? Because

rock is less deformable than water.

Wikip: The Earth tide encompasses the entire body of the Earth and is
unhindered by the thin crust and land masses of the surface, on scales
that make the rigidity of rock irrelevant.

JJL: Harshman doesn't understand tides,

Jan

[J. J. Lodder]

Yes, of course.
The tides are caused by the earth and moon orbiting each other.
The axial rotation of the earth has nothing to do with it.
(beyond making the tides unmistakable)
In the far future, if the axial rotation of the earth
gets tidally locked to the moon the tidal bulges will still be there.
(but you will need accurate satillite altimetry to detect them)

> Rotation
> (moon around the earth ~once a day east to west or earth on its axis
> west to east) is only necessary to ~twice a day tides at a specific
> location on the earth.

Twice a day is crucial here.
Newtonian mechanics predicts two tidal bulges, hence two tides a day.
Pagano's geostationarity (if combined with Newtonian gravity)
predicts one tidal bulge, hence one tide a day.

Pagano is wrong,

Jan

[Bob Casanova]

On Tue, 20 Mar 2018 23:02:23 +0100, the following appeared

in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
Lodder):

OK; thanks.

[Bob Casanova]

On Tue, 20 Mar 2018 23:02:21 +0100, the following appeared

in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
Lodder):

>Bob Casanova <nos...@buzz.off> wrote:
>
>> On Sat, 17 Mar 2018 10:21:38 +0100, the following appeared
>> in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
>> Lodder):
>>
>> <snip to the point>
>>
>>
>> >The earth slowed down quite noticably already in historical times.
>> >Records of historical eclipses from all over the world prove this.
>> >See for example
>> ><http://www.sciencemag.org/news/2016/12/ancient-eclipses-show-earth-s-ro
>> >tation-slowing>
>> >The earth has lost about six hours since biblical times.
>> >(or the universe, for Tony)
>>
>> Wait; what? Are you saying that the length of a day was 30
>> hours in Biblical times? If not, what *are* you saying?
>
>You haven't understood eclipse predictions.

Nope, what I haven't done is go back to the reference. Sorry
'bout that...

>It is quite possible to predict with exquisite precision
>where the earth, the moon, and the sun will be thousands of years hence,
>so to predict that the moon's shadow cone will intersect
>with the surface of the earth.
>
>It is not possible to predict where the surface of the earth
>will be at that moment.
>(because it cannot be predicted accurately enough
> how much the earth will have rotated in the meantime)
>
>The -cumulative- effect of the speed-up of the earth in the past
>amounts to about 6 hours error, back in biblical times.
>So the longitude of the path of totality of eclipses
>(as predicted on basis of constant rotation of the earth)
>is off by about ninety degrees.
>
>Even today, with all the information about irregularities
>of the earth's rotation that we have
>it is not possible to predict for example
>the moment of first contact of an eclipse occuring ten years hence
>to the nearest second.
>(see the irrugularity with which leap seconds must be declared)

OK; got it. As I noted in the past, I *definitely* should
have at least skimmed the article instead of zeroing in on
the somewhat ambiguous phraseology "The earth has lost about

six hours since biblical times".

Hell, even the title should have clued me that the issue was
variance in eclipse locations from "predicted" (postdicted?)
due to the slowing of the Earth's rotation, not an actual
"loss" of anything.

Thanks, and sorry for the misunderstanding (mine).

[Bob Casanova]

On Wed, 21 Mar 2018 15:48:09 +0100, the following appeared

in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
Lodder):

>Don Cates <cate...@hotmail.com.invalid> wrote:

<snip>

>> Jan, check out
>> <https://en.wikipedia.org/wiki/Tidal_force>
>> The tidal bulges are due strictly to the gravitational field.

>Yes, of course.
>The tides are caused by the earth and moon orbiting each other.
>The axial rotation of the earth has nothing to do with it.
>(beyond making the tides unmistakable)
>In the far future, if the axial rotation of the earth
>gets tidally locked to the moon the tidal bulges will still be there.
>(but you will need accurate satillite altimetry to detect them)
>
>> Rotation
>> (moon around the earth ~once a day east to west or earth on its axis
>> west to east) is only necessary to ~twice a day tides at a specific
>> location on the earth.
>
>Twice a day is crucial here.
>Newtonian mechanics predicts two tidal bulges, hence two tides a day.
>Pagano's geostationarity (if combined with Newtonian gravity)
>predicts one tidal bulge, hence one tide a day.
>
>Pagano is wrong,

Of course, which is why he will continue to ignore this
thread.

[Bob Casanova]

On Tue, 20 Mar 2018 11:41:36 -0700, the following appeared

If I'd actually looked at the referenced article, or even
the title, I'd have realized that "lost 6 hours" was not
intended literally (nothing was actually "lost"), but was a
reference to the "postdiction" of eclipse locations in the
past being not as expected due to the slowing of the Earth's
rotation.

My fault entirely; thanks for bearing with me.

>>>> It was the idea of the
>>>> "loss of 6 hours" that had me a bit confused; I could only
>>>> interpret that as the day being 6 hours shorter now. And the
>>>> idea that the total time interval could be 6 hours less
>>>> directly contradicts the definitions of the second and the
>>>> hour, which have *not* changed (by the current definition)
>>>> since the era mentioned. So it looks to me as if I wasn't
>>>> the one using multiple definitions.
>>>
>>> I have no idea what that means. Try this: if the year is one second
>>> longer than you think it is, then any event that happened a thousand
>>> years ago happened 17 minutes earlier than you might suppose. That's the
>>> sort of thing the article is talking about, except that slight changes
>>> have accumulated rather than being a single difference.
>>
>> IOW, it's a "perception thing", and needs to be addressed on
>> a regular basis as noted above. OK.
>
>I have no idea what you mean by "perception thing". I was using "longer
>than you think it is" in order to construct an example that's simpler
>than "longer than it used to be, and getting very slightly longer all
>the time".

[Bob Casanova]

On Tue, 20 Mar 2018 11:42:55 -0700, the following appeared

....or ignores it entirely if he lacks the bafflegab required
to make it look like he might be able to refute it.

And, of course, repeatedly declares victory while wondering
why he's holding his head in his hand.

But point taken.

[Don Cates]

On 2018-03-21 11:18 AM, Bob Casanova wrote:
> On Wed, 21 Mar 2018 15:48:09 +0100, the following appeared
> in talk.origins, posted by nos...@de-ster.demon.nl (J. J.
> Lodder):
>
>> Don Cates <cate...@hotmail.com.invalid> wrote:
>
> <snip>
>
>>> Jan, check out
>>> <https://en.wikipedia.org/wiki/Tidal_force>
>>> The tidal bulges are due strictly to the gravitational field.
>
>> Yes, of course.
>> The tides are caused by the earth and moon orbiting each other.
>> The axial rotation of the earth has nothing to do with it.
>> (beyond making the tides unmistakable)
>> In the far future, if the axial rotation of the earth
>> gets tidally locked to the moon the tidal bulges will still be there.
>> (but you will need accurate satillite altimetry to detect them)
>>
>>> Rotation
>>> (moon around the earth ~once a day east to west or earth on its axis
>>> west to east) is only necessary to ~twice a day tides at a specific
>>> location on the earth.
>>
>> Twice a day is crucial here.
>> Newtonian mechanics predicts two tidal bulges, hence two tides a day.

>> Pagano's geostationarity (if combined with Newtonian gravity)
>> predicts one tidal bulge, hence one tide a day.

I believe you are wrong about this. In the absence of orbital motion you
would still get two tidal bulges. (check out that link above, carefully)

>>
>> Pagano is wrong,
>
Well, yes. But not because of the nature of tides.

> Of course, which is why he will continue to ignore this
> thread.
>


--

[John Harshman]

How so? You have already agreed that a tidally locked earth would have
two tidal bulges. The reason there are two tides per day is that the two
tidal bulge pass through the same spot 12 hours apart. Whether it's the
rotation of the earth, the motion of the moon, or a combination of the
two is irrelevant to the phenomenon. What does the center of gravity of
the earth-moon system have to do with any of it?

> Pagano is wrong,

But not for the reason you suggest here.

[John Harshman]

Thanks for that. Actually quoting your source instead of just citing a
long article most of which is irrelevant to your point is a better way
to go. But I think you misunderstand the point there. I think they're
talking about the rigidity of crust as being irrelevant, because the
scale is such that nothing has to bend much.

What I don't see there is an explanation of why the earth tide is so
much smaller than the ocean tide. Do you know?

> JJL: Harshman doesn't understand tides,

You seem unusually hostile. Why do you think there are two high tides
per day? Why do you think the earth tide is smaller than the ocean tide?

[John Harshman]

Thinking for myself has convinced me that I'm not wrong, so that's no
good. You will have to help me.

[John Bode]

On Friday, March 16, 2018 at 5:15:03 AM UTC-5, J. J. Lodder wrote:
> A riddle for Tony Pagano: Tides
> Let us assume that the earth is really stationary
> at the centre of the universe,
> and that the moon orbits around it.
> The moon will cause tides, by its gravitational attraction,
> and the tides will follow the moon.
>
> Riddle: Why are there two tides a day, and not one?
> Why is there also a high ride on the side of the earth
> that is farthest away from the moon?
> (where the moon's gravity is weaker)
>
> Hint: Newtonians explain the high tide on the far side of th earth
> by noting that the earth and the moon both orbit
> around their common centre of mass, (and next do the sums)
> but that explanation is impossible if the earth is truly stationary.
>
> So please explain how there can be two tides a day
> in a purely geostationary model,
>

> Jan

Exactly the same way there are two tides a day in a rotating Earth model.

I don't understand why you think this refutes a geostationary model. The secondary
tidal bulge on the far side is strictly due to tidal forces and has nothing to do with the
Earth's rotation (or lack of same).

What would be a *relevant* argument against a geostationary model is the relationship
between tidal forces and the gradual lengthening of the day. Days *are* getting longer
over time. In the rotating Earth model, tidal forces from the Moon are slowing the Earth's
rotation; to conserve angular momentum, the Moon is receding from the Earth.

I'm curious how a geostationary model would explain these phenomena (lengthening days
plus lunar recession).

[J. J. Lodder]

John Bode <jfbod...@gmail.com> wrote:

> On Friday, March 16, 2018 at 5:15:03 AM UTC-5, J. J. Lodder wrote:
> > A riddle for Tony Pagano: Tides
> > Let us assume that the earth is really stationary
> > at the centre of the universe,
> > and that the moon orbits around it.
> > The moon will cause tides, by its gravitational attraction,
> > and the tides will follow the moon.
> >
> > Riddle: Why are there two tides a day, and not one?
> > Why is there also a high ride on the side of the earth
> > that is farthest away from the moon?
> > (where the moon's gravity is weaker)
> >
> > Hint: Newtonians explain the high tide on the far side of th earth
> > by noting that the earth and the moon both orbit
> > around their common centre of mass, (and next do the sums)
> > but that explanation is impossible if the earth is truly stationary.
> >
> > So please explain how there can be two tides a day
> > in a purely geostationary model,
> >
> > Jan
>
> Exactly the same way there are two tides a day in a rotating Earth model.

Without a severe case of handnwaving?

> I don't understand why you think this refutes a geostationary model. The
> secondary tidal bulge on the far side is strictly due to tidal forces and
> has nothing to do with the Earth's rotation (or lack of same).

So you think that you can explain why there are two tidal bulges
---in Pagano's geostationary model---
Please explain howto.

> What would be a *relevant* argument against a geostationary model is the
> relationship between tidal forces and the gradual lengthening of the day.
> Days *are* getting longer over time. In the rotating Earth model, tidal
> forces from the Moon are slowing the Earth's rotation; to conserve angular
> momentum, the Moon is receding from the Earth.
>
> I'm curious how a geostationary model would explain these phenomena
> (lengthening days plus lunar recession).

Yes, we covered that, but it is a different argument.

Jan

[J. J. Lodder]

If there had been an atomic clock up and running for thee thousand years
the difference with the earth rotation (considered as a clock)
and atomic would have been about six hours.
There wasn't, but there is no real need.
Orbital motions do keep real (that is atomic or universe) time.
Hence the discrepamcy in eclipse observations.

This is a practical point, and a spurce of a persistent quarrel.
Astronomers want to keep atomic time (aka TAI)
in step with UTC (almost the same is Greenwich time)
To this end leap seconds are inserted when needed,
to keep UTC within a second of TAI.

The astronomers are right, IMHO, for the universe
(from planets to distant quasars) keeps real time,
that is TAI, not the slowing down earth rotation time.

GPS time is equal to TAI, up to a fixed offset of 19 seconds.
GPS sats mpove in real time too, and the short term errors
in the rotation of the earth are determined by measuring
where the surface of the earth is wrt the GPS constellation.

Computer geeks (having been more than a little bit silly
in defining their UNIX network time) want to abolish leap seconds.
So if they have their way sunset will be at noon, (by TAI and GPS)
sometime in the future.

The solution is obvious of course,
computer networks should keep TAI, internally,
and take the leap second offset in software,
(just like they do with daylight saving time changes)
instead of demanding that the rest of the world
should adapt to their historical errors of thirty years ago.

Jan

[Ernest Major]

You haven't explained why think that the tides are a problem for a
geocentric model.

Tony hasn't offered an explanation of the tides, and he may well not
have one - he doesn't seem to understand much physics; if he did he'd
have a better understanding of how much observationally supported
physics he's throwing out.

He also hasn't offered a fleshed out physical model for his geocentric
universe - he seems to confuse the ability to mathematically describe
the motion in any reference frame with the existence of a physical model.

If his model was that the earth is magically fixed in place by God's
will, then semidiurnal tides would indeed seem to be a problem. But Tony
has gone so far as to claim that unspecified masses elsewhere in the
universe cancel out the effects of the sun and moon's gravity on the
earth. (His problem is that those masses are invisible and not even
detectable by their gravitational effects on other bodies - and he then
has the chutzpah to cast shade on dark matter.) If was pass over the
problem of the missing masses, we have the prediction that while the
forces cancel out at the center of the earth, they fail to cancel out,
in opposite directions, at the sublunar and antilunar points, giving two
tidal bulges. So that geocentric model does seem to agree qualitatively
with observation on this point. I don't see any obvious reason why it
wouldn't agree quantitatively as well.

--
alias Ernest Major

[scienceci...@gmail.com]

J. J. Lodder wrote on 3/20/18,

>>> <scienceci...@gmail.com>
>>> wrote:
>>> J. J. Lodder wrote 3/17,
>>>
>>> Indeed, real tides are a
>>> very complicated subject.
>>> The idealised model with
>>> two tidal bulges applies to
>>> an idealised all-ocean
>>> earth only. (and to the
>>> solid tides of the crust)
>>>
>>> In reality, continents get in
>>> the way of the tidal wave,
>>> and basisns will have
>>> resonances.
>>
>>> Nevertheless, all those
>>> local tidal waves can be
>>> explained only from
>>> semi-diurnal forcing, (that is, >>> twice per
>>> lunar day)
>>>> Jan
>>
>> Like this. See NASA video
>> below. Bulging when the
>> syringe pulls and releases.
>> But the moon's gravity is not
>> like this. It does not turn on
>> & off, not even by being
>> directly over continents. Or
>> else tides would recede at
>> night and rise at morning.
>> Like you say, Jan,
>> "continents get in the way of
>> the tidal wave"...
>>
https://m.youtube.com/watchv=HMXpoFVQ_Pw

> That's a nice one, thanks.
> Indeed, in principle one can
> demonstrate tidal forces
> also in free fall.
> (but the micro-gravity on the
> space station will spoil it)

> Read Larry Niven's 'Neutron
> Star' for the serious stuff,
> Jan

I read the wiki plot notes. All the 'keywords' like 'tidal force' for the abstract principle are there, and the cover of the book has bulging letters. Lol.

Comment about star and blob of water:
Unlike the release of the syringe in the weightless blob of water on the space station, for a neutron star with no exothermic fusion reactions to generate either radiation pressure or hydrodynamic pressure with which to withstand the gravitational forces, the star will collapse until it is supported by the electron Fermi pressure. If the syringe is like unwithstandable gravitational force and continue to pull on one end of the water, differential force would make the water elongate. The middle of the blob of water would be narrower than the ends.

SC RED

[jillery]

Your quote from Wikipedia refers to the rigidity of the crust, which
is an insignficant fraction of the diameter of the Earth. Most of the
interior of the Earth is plastic, and so responds to the uneven
gravitational pull of the Moon and Sun, although not nearly to the
same degree and frequency as the far more fluid oceans. Even so, the
viscosity of the "solid" Earth makes it virtually impossible for its
tidal bulges to follow celestial motions. Instead, any appreciable
bulge would have to develop over time through resonances.

[jillery]

On Wed, 21 Mar 2018 23:24:07 +0100, nos...@de-ster.demon.nl (J. J.
Lodder) wrote:

>John Bode <jfbod...@gmail.com> wrote:
>
>> On Friday, March 16, 2018 at 5:15:03 AM UTC-5, J. J. Lodder wrote:
>> > A riddle for Tony Pagano: Tides
>> > Let us assume that the earth is really stationary
>> > at the centre of the universe,
>> > and that the moon orbits around it.
>> > The moon will cause tides, by its gravitational attraction,
>> > and the tides will follow the moon.
>> >
>> > Riddle: Why are there two tides a day, and not one?
>> > Why is there also a high ride on the side of the earth
>> > that is farthest away from the moon?
>> > (where the moon's gravity is weaker)
>> >
>> > Hint: Newtonians explain the high tide on the far side of th earth
>> > by noting that the earth and the moon both orbit
>> > around their common centre of mass, (and next do the sums)
>> > but that explanation is impossible if the earth is truly stationary.
>> >
>> > So please explain how there can be two tides a day
>> > in a purely geostationary model,
>> >
>> > Jan
>>
>> Exactly the same way there are two tides a day in a rotating Earth model.
>
>Without a severe case of handnwaving?

With no handwaving, severe or otherwise.

>> I don't understand why you think this refutes a geostationary model. The
>> secondary tidal bulge on the far side is strictly due to tidal forces and
>> has nothing to do with the Earth's rotation (or lack of same).
>
>So you think that you can explain why there are two tidal bulges
>---in Pagano's geostationary model---
>Please explain howto.

Since you said you accept Wikipedia's explanation of two tidal bulges,
as a result of differential gravity, which has nothing to do with
Earth's motion, the burden is on you to explain how Pagano's
geostationary model would *not* also predict two tidal bulges.

>> What would be a *relevant* argument against a geostationary model is the
>> relationship between tidal forces and the gradual lengthening of the day.
>> Days *are* getting longer over time. In the rotating Earth model, tidal
>> forces from the Moon are slowing the Earth's rotation; to conserve angular
>> momentum, the Moon is receding from the Earth.
>>
>> I'm curious how a geostationary model would explain these phenomena
>> (lengthening days plus lunar recession).
>
>Yes, we covered that, but it is a different argument.
>
>Jan

[J. J. Lodder]

But I have. Again, in somewhat more words:
In Tony's geostatic model the only forces acting on the oceans
are the earth's gravity, which is sperically symmetric,
and the moon's gravity, which is strongest under the moon,
and weakest away from it.
So the water moves towards the side where the moon is,
and away from the far side.
So the geostatic model predicts one tidal bulge,
instead of the two that are observed.
Therefor the geostatic model is wrong.

Jan

PS on your snipped text: Tony's centrifugal force
has no effect on the tides, for it has cylinder symmetry
around the polar axis.

[J. J. Lodder]

Minor detail: electron Fermi pressure cannot support a neutron star.
That's precisely why it is a neutron star,

Jan

[J. J. Lodder]

There is a cross-over point at which material strength
can no longer hold against gravity.
It is at about the size of the largest asteroids.
Any larger body will be gravity dominated, and hence spherical.
(elipsoidal with rotation)

> What I don't see there is an explanation of why the earth tide is so
> much smaller than the ocean tide. Do you know?
>
> > JJL: Harshman doesn't understand tides,
>
> You seem unusually hostile. Why do you think there are two high tides
> per day? Why do you think the earth tide is smaller than the ocean tide?

Don't blame me, you started it.
As for your question: the oceans are at the surface,
so they experience the strongest tidal pull.
The solid tide is a body effect,
and most of the body experiences much smaller tidal forces,
zero even, at the right point.

Jan

[jillery]

That's called the Roche Limit, a consequence of the differential
gravitational attraction across a body. Not sure what you think that
has to do with this discussion.

>> What I don't see there is an explanation of why the earth tide is so
>> much smaller than the ocean tide. Do you know?
>>
>> > JJL: Harshman doesn't understand tides,
>>
>> You seem unusually hostile. Why do you think there are two high tides
>> per day? Why do you think the earth tide is smaller than the ocean tide?
>
>Don't blame me, you started it.
>As for your question: the oceans are at the surface,
>so they experience the strongest tidal pull.
>The solid tide is a body effect,
>and most of the body experiences much smaller tidal forces,
>zero even, at the right point.
>
>Jan

[jillery]

On Thu, 22 Mar 2018 08:54:10 +0100, nos...@de-ster.demon.nl (J. J.
Lodder) wrote:

Once again, the only force required to create two tidal bulges on a
celestial body is the gravitational force from another celestial body.
That gravitational force is the same regardless of celestial model.
You can ignore this fact all you want, but it proves you wrong on this
point. Tony's "geostatic model", whatever that is, is probably wrong,
but not because of any prediction about tidal bulges.

>Jan
>
>PS on your snipped text: Tony's centrifugal force
>has no effect on the tides, for it has cylinder symmetry
>around the polar axis.

What have you been smoking?

[Bill Rogers]

I think you are incorrect. Consider some mass of water nearest the moon. That water is acted on by a gravitational force greater than the one acting on an equal mass of iron at the center of the earth. That water therefore accelerates towards the moon with a greater acceleration than does the center of the earth. Therefore, there's a tidal bulge on the side closest to the moon.

Now consider an equal mas of water on the side of the earth farthest from the moon. That water is acted on by a gravitational force less than that acted on an equal mass at the center of the earth. Therefore, the center of the earth accelerates towards the moon with a greater acceleration than that of an equal mass of water on the side farthest from the moon. So the water on the farther side is left behind, a bit, causing a tidal bulge on the side farthest from the moon.

Two tidal bulges, twelve hours apart, and the earth needn't rotate to cause them.

[Ernest Major]

Tony has not offered a worked out geocentric model, but I'm pretty sure
that his model has more forces operating that the earth's gravity, the
moon's gravity and a centrifugal force. (I agree that the centrifugal
force is not relevant.)

My text that you snipped had no references to centrifugal forces. It
contained an agreement that tides are a problem for a geostatic model in
which the only forces are those you mention, and an observation that
Tony's model includes more forces.

--
alias Ernest Major

[Ernest Major]

Jan is referring to a model in which the earth is not accelerated
towards the moon at all. (This is physically unrealistic. While Tony's
model is also physically unrealistic, I don't think Tony's model is the
same.)

>
> Two tidal bulges, twelve hours apart, and the earth needn't rotate to cause them.
>
>
>>
>> Jan
>>
>> PS on your snipped text: Tony's centrifugal force
>> has no effect on the tides, for it has cylinder symmetry
>> around the polar axis.
>
>

--
alias Ernest Major

[jillery]

On Thu, 22 Mar 2018 14:03:50 +0000, Ernest Major

If Lodder's model isn't the same as Pagano's, or at least similar
enough in the essentials, then that would moot Lodder's "riddle" as a
strawman.

Alternately, AIUI Pagano's model assumes equivalent accelerating
forces from a rotating universe, to those of an accelerating Earth.
Based on that, I would expect Pagano to assume that whatever
accelerating forces applied in a heliocentric model, which caused the
dual tidal bulge, to equivalently exist in his geocentric model, with
equivalent effects.

>> Two tidal bulges, twelve hours apart, and the earth needn't rotate to cause them.
>>
>>
>>>
>>> Jan
>>>
>>> PS on your snipped text: Tony's centrifugal force
>>> has no effect on the tides, for it has cylinder symmetry
>>> around the polar axis.
>>
>>

--

[John Bode]

Looking at the acceleration the Moon exerts on the Earth (arrows represent vectors, longer
vectors mean stronger acceleration):

-> Earth ---> Moon

The side of the Earth nearest the Moon feels greater acceleration than the opposite side.
The difference in acceleration basically amounts to this:

<- Earth ->

Hence, two tidal bulges on opposite sides of the Earth.

In a geostationary model, the Moon is orbiting the earth not quite once a day, and the
tidal bulges follow the Moon's movement. Two bulges, two tides per day.

Now, whether this complies with *Pagano's* model of a stationary Earth - I don't know.
I'll be honest that I've never felt that compelled to truly understand Pagano's model.
If he maintains that the Moon doesn't exert an acceleration on the Earth, then yeah,
tides of any sort (once or twice a day) are a problem.

[J. J. Lodder]

At last, someone here who posts a qualitatively correct understanding
of tides.
And 'with a greater acceleration than does the center of the earth'
is just it, for it implies that the earth is moving.
(just what it can't do in Pagano's geocentrism)

> Now consider an equal mas of water on the side of the earth farthest from
> the moon. That water is acted on by a gravitational force less than that
> acted on an equal mass at the center of the earth. Therefore, the center
> of the earth accelerates towards the moon with a greater acceleration than
> that of an equal mass of water on the side farthest from the moon. So the
> water on the farther side is left behind, a bit, causing a tidal bulge on
> the side farthest from the moon.

Entirely correct again. Or to say the same thing in other words:
the earth orbits the moon, just as much as the moon orbits the earth.
Actually they both orbit on opposite sides
of their common centre of mass.
(which is inside the earth, but of course not at the centre)
Or is still other words: orbiting is a free fall that lasts forever.
The moon forever falls towards the earth,
the earth falls towards the moon.
In still other words: the tides cannot be understood
without allowing the earth to accelerate, hence to move.

> Two tidal bulges, twelve hours apart, and the earth needn't rotate to
> cause them.

Also as said many times already:
the axial rotation of the earth has nothing to do with the tides,
beyond making them unmistakable.
A spin-orbit locked, aka tidally locked, planet or moon
still has tidal bulges.

Jan

[J. J. Lodder]

Nothing. Just another of your misunderstandings,

Jan

[J. J. Lodder]

There is still Coriolis force, also irrelevant for the tides.

> My text that you snipped had no references to centrifugal forces. It
> contained an agreement that tides are a problem for a geostatic model in
> which the only forces are those you mention, and an observation that
> Tony's model includes more forces.

It's up to Tony to explain what extra tide-causing forces
there may be in his model.
If he can't, his geocentrism stands falsified by the tides,

Jan

[Ernest Major]

Tony pays lip-service to Newtonian physics. (Or at least he appeals to
it, and doesn't cancel out his appeals with explicit denial as in the
case of his appeals to relativity.)

--
alias Ernest Major

[J. J. Lodder]

> The opposite side. difference in acceleration basically amounts to this:

>
> <- Earth ->
>
> Hence, two tidal bulges on opposite sides of the Earth.

Correct, just like Bill Rogers.
The tides can only be understood if the earth accelerates, hence moves.
(as it must do in Newtonian mechanics, and can't according to Pagano)

> In a geostationary model, the Moon is orbiting the earth not quite once a
> day, and the tidal bulges follow the Moon's movement. Two bulges, two
> tides per day.

There are no two tidal bulges in a geostationary model,
there is only one.

Jan

[John Harshman]

This too seems like an irrelevant interjection intended to display your
knowledge on some subject.

>> What I don't see there is an explanation of why the earth tide is so
>> much smaller than the ocean tide. Do you know?
>>
>>> JJL: Harshman doesn't understand tides,
>>
>> You seem unusually hostile. Why do you think there are two high tides
>> per day? Why do you think the earth tide is smaller than the ocean tide?
>
> Don't blame me, you started it.

No, you did, by giving an erroneous objection to Tony's geocentrism.

> As for your question: the oceans are at the surface,
> so they experience the strongest tidal pull.
> The solid tide is a body effect,
> and most of the body experiences much smaller tidal forces,
> zero even, at the right point.

But the oceans aren't at the surface to any greater extent than the
continental crust. Anything above sea level, by your logic, should be
experiencing a greater tidal effect than the oceans.

[J. J. Lodder]

Ernest Major <{$to$}@meden.demon.co.uk> wrote:

> > The the opposite side. difference in acceleration basically amounts to
> > The this:

> >
> > <- Earth ->
> >
> > Hence, two tidal bulges on opposite sides of the Earth.
> >
> > In a geostationary model, the Moon is orbiting the earth not quite once
> > a day, and the tidal bulges follow the Moon's movement. Two bulges, two
> > tides per day.
> >
> > Now, whether this complies with *Pagano's* model of a stationary Earth -
> > I don't know. I'll be honest that I've never felt that compelled to
> > truly understand Pagano's model. If he maintains that the Moon doesn't
> > exert an acceleration on the Earth, then yeah, tides of any sort (once
> > or twice a day) are a problem.
> >
>
> Tony pays lip-service to Newtonian physics.

Hoist by one's own petard, as the bard said.
Geostationarity combined with Newtonian gravity,
yields an obviously false prediction of one tide per day,

Jan

[Ernest Major]

On 22/03/2018 14:55, J. J. Lodder wrote:
> Correct, just like Bill Rogers.
> The tides can only be understood if the earth accelerates, hence moves.
> (as it must do in Newtonian mechanics, and can't according to Pagano)

Consider an asymmetric linear 3 body system with one body at the centre
of gravity. The net force on this body is zero. It seems to me that
there are still net tidal forces at the surface of this body as the
gradient of the gravitational forces from the other two bodies is different.

--
alias Ernest Major

[John Harshman]

I don't understand why. Doesn't Newtonian relativity allow us to choose
a coordinate system in which the earth is stationary in the earth-moon
system? If we consider the center of the earth to be stationary, aren't
the sides near and far from the moon still affected?

[John Harshman]

Are you perhaps, like Tony, confusing the center of gravity with the
position of zero net force? The net force at the center of gravity of
the earth-moon system accelerates any body there toward the earth.

[erik simpson]

Of course it does, although some Dutchmen may disagree. Any coordinate system
chosen doesn't affect reality, but a particularly badly chosen one makes
calculations very difficult. Einstein's famous falling elevator coordinate
system was chosen for two reasons: it's an inertial frame, making caluclations
simple, and it illustrated the fact that "gravitation" as a force is
problematic, since it can be eliminated by a choice of coordinates. Tidally
induced strain is real, can't be transformed away. Water doesn't support strain
well, it moves instead. Rock supports strain quite well (up to limits), but
gets hot from being flexed. This keeps the interiors of Jupiter's moons hot.

[Ernest Major]

Yes. (Reading Tony Pagano is not good for remaining unconfused about
terminology.) I was inferring a distinction between the centre of mass
and the centre of gravity - there is one - but I misinferred (and didn't
check) the actual definition of centre of gravity.

I may have known the correct definition 40 years ago, but that's plenty
of time for my physics to get rusty.

--
alias Ernest Major