Jeff Morris wrote:

"Nav" wrote in message
...


Donal wrote:


"Peter S/Y Anicula" wrote in message
.dk...


The moon are a lot closer than the sun. Therefore the gravitational
force of the moon varies more over the earth's surface. It is the
variation in the gravitational force and not the force in itself that
creates the tides.

The moons pull on a water-molecule directly under the moon is larger
than
on a molecule on the far side of the earth, actually it is larger than
"the average pull on the whole earth", and here the moon pulls away

from the earth.

On the far side of the earth (seen from the moon) the gravitation from
the moon is less than average and at this point the moon pulls toward
the earth.


On the far side the tide is "high", ... just the same as at the near side.
If the moon's gravity was pulling the water, then you would expect LW to be
opposite the moon.


Quite so. The tidal gravity force is in the direction of the moon. This
is the potential energy in the system. So, there must be another force
present. The moon has kinetic energy in it's orbital velocity. From
Newton's first law:

F=m r omega^2

It is the difference in the two forces (and the resulting energy minima)
that causes two tides. Simple no? Why invoke something new like
"differential gravity"? Could it be to avoid saying inertial force?


Before I thought you were just arguing philosophically how much we should credit
centrifugal force, but now it appears you haven't really looked at the math at
all. The reason why "differential gravity" is invoked is because it represents
the differing pull of the Moon on differing parts of the Earth. Although this
force is all obviously towards the Moon, when you subtract off the centrifugal
force this is what is left. It is this differing pull that causes the two
tides.


Again you say that gravity causes the two tides -but I say that is not
correct. It is the DIFFERENCE between the centripetal term and the
gravity that causes the tide (How many times do I have to say this?).
You may try to malign me by saying I've not looked at the math but I
have -much closer than you have I think. Here's something for you to try
during coffee:

1) How big is Differential gravity? (1e-6 m/s/s ? It's on the web)
2) How big is the difference in centrifigal acceleration on each side of
the barycenter?

Centrifugal acceleration = r omega^2
The moon orbits the earth every 28 days so omega = 2 pi/ (28 * 24 * 60 *
60).

The barycenter is about at 3/4 r (1/4 r under the earth surface) so the
difference in r from one side to the other makes the imabalance.

The difference in centrifugal acceleration is therefo

1.5 * r * (2 pi/ 28 * 24 * 60 * 60)^2

r is 6.4e^6 m

Get out your calculator and work it out for yourself. It's easy and I
think you'll be surprised at the answer.

Or is my equation for centrifugal acceleration wrong?



You can handwave the centrifugal force causes the outward bulge, but
mathematically, the idealized shape of the Earth is caused specially by the
differential forces. Trying to explain it all by "inertia" is just making it
simple for young children, it doesn't really explain what's going on.


But I never tried to explain it all by intertial forces, Jeff. I always
said it was the _difference_ between inertial forces and gravity. I'd
say it's you who is trying to explain it all by gravity instead!

What this shows is that without land masses to block tidal flow the
tides would be much bigger than they are. That differential gravity
produces the "correct" answer really just shows how it is not the
correct sole explanation (if it had predicted higher tides that would be
expected). Perhaps you would like to think about that approximation used
in the gravity term that "allows" it to cancel the centripetal term. Is
it correct to use this when we are dealing with very very small
fractions of the total acceleration??

Cheers

"Nav" wrote in message
...
Jeff Morris wrote:
....

Before I thought you were just arguing philosophically how much we should
credit
centrifugal force, but now it appears you haven't really looked at the math
at
all. The reason why "differential gravity" is invoked is because it
represents
the differing pull of the Moon on differing parts of the Earth. Although
this
force is all obviously towards the Moon, when you subtract off the
centrifugal
force this is what is left. It is this differing pull that causes the two
tides.


Again you say that gravity causes the two tides -but I say that is not
correct. It is the DIFFERENCE between the centripetal term and the
gravity that causes the tide (How many times do I have to say this?).
You may try to malign me by saying I've not looked at the math but I
have -much closer than you have I think. Here's something for you to try
during coffee:

1) How big is Differential gravity? (1e-6 m/s/s ? It's on the web)

roughly correct

2) How big is the difference in centrifigal acceleration on each side of
the barycenter?

Centrifugal acceleration = r omega^2
The moon orbits the earth every 28 days so omega = 2 pi/ (28 * 24 * 60 *
60).

The barycenter is about at 3/4 r (1/4 r under the earth surface) so the
difference in r from one side to the other makes the imabalance.

The difference in centrifugal acceleration is therefo

1.5 * r * (2 pi/ 28 * 24 * 60 * 60)^2

r is 6.4e^6 m

Get out your calculator and work it out for yourself. It's easy and I
think you'll be surprised at the answer.


So, you you're claiming the lunar tidal forces are 65 times the accepted values.

Now, get out your calculator and run the same numbers for the Sun. The distance
to the Sun (and the E-S barycenter) is 1.5e^11 meters. The result is about 100
times less than your result for the Moon. So you're claiming that the Sun has
negligible effect on the night time tides?



Or is my equation for centrifugal acceleration wrong?


Actually, applying it in this context is your problem. Centrifugal acceleration
is constant, it doesn't vary across the surface of the Earth as you claim.
Remember, it doesn't even exist, its actually a reference frame shift.




You can handwave the centrifugal force causes the outward bulge, but
mathematically, the idealized shape of the Earth is caused specially by the
differential forces. Trying to explain it all by "inertia" is just making
it
simple for young children, it doesn't really explain what's going on.


But I never tried to explain it all by intertial forces, Jeff. I always
said it was the _difference_ between inertial forces and gravity. I'd
say it's you who is trying to explain it all by gravity instead!

Gravity is actually the only force at work. Any explanation must be consistent
with that. Centrifugal forces is "ficticious," it doesn't really exist. The
reality is that the Earth is in free fall towards the E-M barycenter.



What this shows is that without land masses to block tidal flow the
tides would be much bigger than they are.

Absoulute nonsense! The land masses build up the tides, they don't reduce
them. And there is no major landmass on the equator for almost half of the
Earth' circumfrence - there is plenty of room for the tides to fully devolope.
Your theory predicts island in the Pacific would be hit by 100 foot tides every
day.

That differential gravity
produces the "correct" answer really just shows how it is not the
correct sole explanation (if it had predicted higher tides that would be
expected).

Double talk - you made up a silly explanation for why we don't have 100 foot
tides and then fault the accepted explantion for not predicting the same thing.

Perhaps you would like to think about that approximation used
in the gravity term that "allows" it to cancel the centripetal term. Is
it correct to use this when we are dealing with very very small
fractions of the total acceleration??

Perhaps you'd like to explain why your approach shows that the Sun has
negligible contribution to the tides. Sorry Nav, this is looking like the
Constellation all over again.

-jeff

Donal,

I hope you're satisfied! For a damned 1/4 of a point!? All this
differential and centrifugally, how does this Knowledge help a sailor to
ride the tides? That was the original question. Remember?
I hope you're satisfied (g)

Now, I wonder if Scot did any Tide Riding while he has been on Vacation
Cruise?

Ole Thom

"Thom Stewart" wrote in message
...
Donal,

I hope you're satisfied! For a damned 1/4 of a point!? All this
differential and centrifugally, how does this Knowledge help a sailor to
ride the tides? That was the original question. Remember?

IIRC, I gave him my honest opinion.

I hope you're satisfied (g)

I'll admit that I'm pleased.


Now, I wonder if Scot did any Tide Riding while he has been on Vacation
Cruise?


It will be interesting to find out how his trip went. Does anybody know
when we can expect to hear from him?


Regards


Donal
--

Jeff Morris wrote:




So, you you're claiming the lunar tidal forces are 65 times the accepted values.


No I'm not saying that. The tidal forces are what they are ("accepted
values?"). I can see you are very stubborn. The point is that the
outward component due to rotation is much larger than the apparent
outward force due to the change in distance and gravity. It's a fact
-you've calculated it for yourself!


Or is my equation for centrifugal acceleration wrong?



Actually, applying it in this context is your problem. Centrifugal acceleration
is constant, it doesn't vary across the surface of the Earth as you claim.
Remember, it doesn't even exist, its actually a reference frame shift.



It is a much larger force than differential gravity but you want to
ignore it? You are wrong Jeff, it does vary across the surface of the
earth Jeff -the Barycenter is at ~3/4 r! On the moon side it's ~1/7 as
large. Finally, (repeating yet again) it is the ____DIFFERENCE______
between inertial and gravity forces that make the tides. To say it's
only "differential gravity" (I shudder at that term) is clearly wrong -
this was a simple proof.

Cheers

OK, Nav, its clear you're not going to get this without some help. You keep
claiming the centrifugal force varies across the Earth. However, that is not
the case. Your assumption is that the Earth is rotating around the E-M
barycenter, and that because that is offset from the Earth center, the
centrifugal force is unbalanced. (Or more precisely, you claim the "r" in the
centrifugal force equation is different on the near and far sides of the Earth.)

However, if we remove the daily rotation, the Earth does not move around the
barycenter quite like you think. Only the center of the Earth describes a
circle around the barycenter. A point on the surface rotates around a point
offset by an Earth radius towards that point. Thus, the "r" in the centrifugal
force equation is the same for all points on the Earth.

I know this is hard concept to grasp at first, but its really quite simple once
you see it. To help visualize, rub your hand around your tummy, holding it
horizontal. The center of your hand rotates around the center of your stomach,
perhaps with a two inch radius. Your fingertips will also describe a two inch
circle, offset to the side. All points on your hand will describe the same
circle, and feel the same centrifugal acceleration.

Given that, your argument falls apart. The centrifugal force is exactly the
same on all points of the Earth, and (not by coincidence) is exactly opposite
the net gravitational force. What is left over is the differential gravity.


Now that's a simple proof.

So tell us Nav, why did you chose to ignore the Sun's contribution? You deleted
my comments that following your arguments, the Sun's contribution is 1% of the
Moon's; this is clearly at variance with reality.



"Nav" wrote in message
...


Jeff Morris wrote:




So, you you're claiming the lunar tidal forces are 65 times the accepted
values.


No I'm not saying that. The tidal forces are what they are ("accepted
values?"). I can see you are very stubborn. The point is that the
outward component due to rotation is much larger than the apparent
outward force due to the change in distance and gravity. It's a fact
-you've calculated it for yourself!


Or is my equation for centrifugal acceleration wrong?



Actually, applying it in this context is your problem. Centrifugal
acceleration
is constant, it doesn't vary across the surface of the Earth as you claim.
Remember, it doesn't even exist, its actually a reference frame shift.



It is a much larger force than differential gravity but you want to
ignore it? You are wrong Jeff, it does vary across the surface of the
earth Jeff -the Barycenter is at ~3/4 r! On the moon side it's ~1/7 as
large. Finally, (repeating yet again) it is the ____DIFFERENCE______
between inertial and gravity forces that make the tides. To say it's
only "differential gravity" (I shudder at that term) is clearly wrong -
this was a simple proof.

Cheers

Jeff Morris wrote:

OK, Nav, its clear you're not going to get this without some help. You keep
claiming the centrifugal force varies across the Earth. However, that is not
the case. Your assumption is that the Earth is rotating around the E-M
barycenter, and that because that is offset from the Earth center, the
centrifugal force is unbalanced. (Or more precisely, you claim the "r" in the
centrifugal force equation is different on the near and far sides of the Earth.)


I think it's you that does not understand that the rotation force is
based on the lunar cycle -28 days!

However, if we remove the daily rotation,

Great idea -not based in reality of course.

the Earth does not move around the
barycenter quite like you think.

Like I think?

Only the center of the Earth describes a
circle around the barycenter.

So the Earth does wobble (now you are getting really close to the whole
story -where harmonics of all the orbital periods give a complete
answer). Now, don't all points on the surface move similarly around the
barycenter and if they do, what is the difference in their orbital path
to that of a circle? Now as I see it, from the math, the differential
gravity model takes no account of this, exploiting the idea that the
orbital motion of every point on the surface is perfectly circular
around the moon (thereby allowing a cancellation of the centrigal
component), a point that you seem to be having a bit of trouble grasping.

By the way, the differential gravity idea first came from Newton. It's
correct as far as it goes but the orbital mechanics of the Earth-Moon
pair are more complicated (as far as I've been able to read, Newton only
saw the free falling body aspect in his tidal proposal). The Devil _is_
in the details and you can't ignore the system rotation. This really is
my last post on this. If you still haven't got the idea then I really
can't make it any clearer and you'll just have to ponder why University
Departments of Oceanographics (and NOAA etc.) all say that THE TIDES ON
EARTH are due to the difference between centrifugal and gravity forces.

Cheers

"Nav" wrote in message
...


Jeff Morris wrote:

OK, Nav, its clear you're not going to get this without some help. You keep
claiming the centrifugal force varies across the Earth. However, that is
not
the case. Your assumption is that the Earth is rotating around the E-M
barycenter, and that because that is offset from the Earth center, the
centrifugal force is unbalanced. (Or more precisely, you claim the "r" in
the
centrifugal force equation is different on the near and far sides of the
Earth.)


I think it's you that does not understand that the rotation force is
based on the lunar cycle -28 days!

However, if we remove the daily rotation,

Great idea -not based in reality of course.

We are looking for the dominant effect - the daily rotation is not a contributor
to that. There are a variety of effects we're ignoring.


the Earth does not move around the
barycenter quite like you think.

Like I think?

OK, as you claim.


Only the center of the Earth describes a
circle around the barycenter.

So the Earth does wobble (now you are getting really close to the whole
story -where harmonics of all the orbital periods give a complete
answer). Now, don't all points on the surface move similarly around the
barycenter

No. Only the center of the Earth revolves around the barycenter. Other points
on the Earth revolve around other points nearby. Stop arguing and just work it
out with a model. If you "wobble" a disk, all points wobble the same way.
You're claiming that some points describe small circles, and some point describe
large circles, but that clearly can't happen unless you rotate the disk.

and if they do, what is the difference in their orbital path
to that of a circle?

All paths are the same. That's my point - stop talking and wobble a plate on
the table without rotating it. All points on it trace the same circle. Hold a
pencil on each side and look at the circles they trace. They all have the same
radius, and therefore the same centrifugal force.

Now as I see it, from the math, the differential
gravity model takes no account of this, exploiting the idea that the
orbital motion of every point on the surface is perfectly circular
around the moon (thereby allowing a cancellation of the centrigal
component), a point that you seem to be having a bit of trouble grasping.


There is no need to grasp it. You entire argument is different points feel
different rotation. But they don't - all points on the Earth feel the same
centrifugal force.

But your argument fails another test: The centrifugal force is "fictional" -
it is just a convenience to simplify some problems. The only real force at play
here is gravity, so any alternate approach must yield the same answer as a
"gravity only" solution.

By the way, the differential gravity idea first came from Newton. It's
correct as far as it goes but the orbital mechanics of the Earth-Moon
pair are more complicated (as far as I've been able to read, Newton only
saw the free falling body aspect in his tidal proposal). The Devil _is_
in the details and you can't ignore the system rotation. This really is
my last post on this. If you still haven't got the idea then I really
can't make it any clearer and you'll just have to ponder why University
Departments of Oceanographics (and NOAA etc.) all say that THE TIDES ON
EARTH are due to the difference between centrifugal and gravity forces.

As I've said, it's very easy to find numerous sites that scoff at the NOAA site
you've mentioned, its regularly cited as "bad science." And I don't deny that
there are a handful of sites that "handwave" that centrifugal force is the cause
of the second bulge, but there are dozens that refute that in great detail.
And virtually every published text supports my view.

And you still haven't responded to the obvious flaws in the formula you
proposed. If the centrifugal force from the Moon is as you claim, why does your
math show that the Sun's contribution is only 1% of the Moon's? You when to
great pains to show the math, but when I showed it was bogus you got very quiet
on that front.

"Nav" wrote in message
...
....
If you still haven't got the idea then I really
can't make it any clearer and you'll just have to ponder why University
Departments of Oceanographics (and NOAA etc.) all say that THE TIDES ON
EARTH are due to the difference between centrifugal and gravity forces.

I have never claimed that you can't derive the tides by computing the difference
between gravitational pull and the centrifugal force. My point has been that
centrifugal force is constant across the Earth, and thus does not explain why
the bulges are in opposite directions. The centrifugal force is exactly
canceled by the average gravitational pull, and what is left over is the
differential force. Your claim has been that the centrifugal force varies
across the Earth, and that's what I've taken exception to.

You keep citing the NOAA page,
http://www.co-ops.nos.noaa.gov/restles3.html
So I went back and read that in detail. In it is the "disclaimer":

"While space does not permit here, it may be graphically demonstrated that, for
such a case of revolution without rotation as above enumerated, any point on the
earth will describe a circle which will have the same radius as the radius of
revolution of the center-of-mass of the earth around the barycenter. Thus, in
Fig. 1, the magnitude of the centrifugal force produced by the revolution of the
earth and moon around their common center of mass (G) is the same at point A or
B or any other point on or beneath the earth's surface. Any of these values is
also equal to the centrifugal force produced at the center-of-mass (C) by its
revolution around the barycenter."

it goes on to develop differential gravity:

"While the effect of this centrifugal force is constant for all positions on the
earth, the effect of the external gravitational force produced by another
astronomical body may be different at different positions on the earth because
the magnitude of the gravitational force exerted varies with the distance of the
attracting body."

In other words, this site actually agrees with what I have been saying.
Frankly I owe an apology to the site's author, since I maligned it without
reading carefully. In fact, although it leads with a provocative line about a
"little known aspect of the moon's orbital motion," and has a rather confusing
diagram, its basic approach is correct and in full agreement with my claim.
Apparently the site was actually changed at some point about two years ago
because of complaints on another board.

Jeff Morris wrote:



In other words, this site actually agrees with what I have been saying.
Frankly I owe an apology to the site's author, since I maligned it without
reading carefully. In fact, although it leads with a provocative line about a
"little known aspect of the moon's orbital motion," and has a rather confusing
diagram, its basic approach is correct and in full agreement with my claim.
Apparently the site was actually changed at some point about two years ago
because of complaints on another board.


The site certainly agrees with me. Don't confuse rotation of the earth
with rotation about the system center. Here is the exact quote and it's
concurrance with my view is as clear as day:

"1. The Effect of Centrifugal Force. It is this little known aspect of
the moon's orbital motion which is responsible for one of the two force
components creating the tides. As the earth and moon whirl around this
common center-of-mass, the centrifugal force produced is always directed
away from the center of revolution. All points in or on the surface of
the earth acting as a coherent body acquire this component of
centrifugal force. And, since the center-of-mass of the earth is always
on the opposite side of this common center of revolution from the
position of the moon, the centrifugal force produced at any point in or
on the earth will always be directed away from the moon. This fact is
indicated by the common direction of the arrows (representing the
centrifugal force Fc) at points A, C, and B in Fig. 1, and the thin
arrows at these same points in Fig. 2."

Note the "one of the two forces".

Cheers