"Nav" wrote in message ...


F= mr omega^2. The distance from the barycenter to all points on earth
is NOT the same.

As the site expalins in the next paragraph, only the center of the Earth rotates
around the barycenter. Other points rotate around neighboring points.

Anyway, the site clearly shows in Fig. 2 that it is the
_DIFFERENCE_ between gravity and centrifugal force that makes the tides,
not gravity alone.

We never disagreed on this point. My issue has always been that since Centrifugal
Force is constant, it doesn't directly explain the two bulges. It is useful for
people who would have trouble with the seeming "negative gravity" of the far side
bulge, but it is also possible to fully describe the tides quite simply without
Centrifugal Force at all.

There really is no point going on any more is there?
For myself, I'm sick of repeating the same point over and over.

But your point seems to change on each post. First you claim CF varies, then you
claim to be in complete agreement with a site that says its constant. Now you're
saying it varies again. You provide your formula and challange me to work it out,
then never explain why it give bogus answers.

Constitution Constitution CONSTITUTION!!!!! opps .... constellation


EOT

Cheers


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:

"Nav" wrote in message ...


F= mr omega^2. The distance from the barycenter to all points on earth
is NOT the same.


As the site expalins in the next paragraph, only the center of the Earth rotates
around the barycenter. Other points rotate around neighboring points.


Anyway, the site clearly shows in Fig. 2 that it is the
_DIFFERENCE_ between gravity and centrifugal force that makes the tides,
not gravity alone.


We never disagreed on this point.


HOLY BACKPEDAL!!!!!!

Cheers

"Nav" wrote in message ...
Jeff Morris wrote:
"Nav" wrote in message
...
F= mr omega^2. The distance from the barycenter to all points on earth
is NOT the same.

As the site expalins in the next paragraph, only the center of the Earth rotates
around the barycenter. Other points rotate around neighboring points.

Anyway, the site clearly shows in Fig. 2 that it is the
_DIFFERENCE_ between gravity and centrifugal force that makes the tides,
not gravity alone.

We never disagreed on this point.

HOLY BACKPEDAL!!!!!!

I'm not sure you really want to go back over this thread - your record is rather
shaky. Mine, however, has been quite consistent. Remember, I started by posting
sites with differing approaches to show that this problem can be looked at in
different ways. I then made my first comment about Centrifugal force with:

"Remember that Centrifugal Force may be a handy explanation, but it is a
"fictional force" that only appears real to an observer in an accelerating frame
of reference. Therefore, whenever it is used to explain something, there must
be another explanation that works in a non-accelerating frame."

but then you started claiming that differential gravity wasn't needed, I responded
with:

"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."

or, in other words, exactly what I just said above.

a few posts later:
"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."

The bottom line here is that the tides are properly described by the differential
gravity equation. Centrifugal force can be used to explain how an outward force can
be generated, but it is not needed, and it does not yield the equation that describes
the tides.

Frankly, your the one who started this by claiming that the traditional explanation of
tides is fundamentally flawed, and that the differential force normally cited is not
what causes the tides. You really haven't produced any coherent evidence to support
this claim.

Now why try to distort the truth Jeff? I never ever said differential
gravity was not needed. I always said that it's the difference between
gravity and centrifugal forces. You do understand the connotations of
the DIFFERENCE between forces don't you? It does not mean that either
component is zero and actually implies that both are important. Shesh!
Still it's nice to see that you now agree that centrifugal forces should
not be ignored (as they are in the gravity only model). As I've said so
many times, the key to understanding is that the system rotates about
the barycenter and it is not just a gravity field problem. The rotation
actually provides the energy needed to power the daily tides -think
about it OK?

Cheers

Jeff Morris wrote:

"Nav" wrote in message ...

Jeff Morris wrote:

"Nav" wrote in message

...

F= mr omega^2. The distance from the barycenter to all points on earth
is NOT the same.

As the site expalins in the next paragraph, only the center of the Earth rotates
around the barycenter. Other points rotate around neighboring points.


Anyway, the site clearly shows in Fig. 2 that it is the
_DIFFERENCE_ between gravity and centrifugal force that makes the tides,
not gravity alone.

We never disagreed on this point.

HOLY BACKPEDAL!!!!!!


I'm not sure you really want to go back over this thread - your record is rather
shaky. Mine, however, has been quite consistent. Remember, I started by posting
sites with differing approaches to show that this problem can be looked at in
different ways. I then made my first comment about Centrifugal force with:

"Remember that Centrifugal Force may be a handy explanation, but it is a
"fictional force" that only appears real to an observer in an accelerating frame
of reference. Therefore, whenever it is used to explain something, there must
be another explanation that works in a non-accelerating frame."

but then you started claiming that differential gravity wasn't needed, I responded
with:

"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."



a few posts later:
"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."

The bottom line here is that the tides are properly described by the differential
gravity equation. Centrifugal force can be used to explain how an outward force can
be generated, but it is not needed, and it does not yield the equation that describes
the tides.

Frankly, your the one who started this by claiming that the traditional explanation of
tides is fundamentally flawed, and that the differential force normally cited is not
what causes the tides. You really haven't produced any coherent evidence to support
this claim.

No, you never said it wasn't needed, but you did minimize its importance, claiming
that it can't be used alone to predict the tides. You said:

"Even worse, the site then goes on to use the _differential_ expression
to calculate the ratio of forces between the moon and sun!"

You even produced a bogus formula to support your claim that centrifugal force varies
across the Earth. This would predict a tidal force 65 times stronger than
differential gravity. If that isn't minimalizing differential gravity, I don't know
what is.

You demonstrated a further lack of understanding with:

If the moon stopped its rotation around the earth and the earth and
the moon was "falling" toward each other, there would still be two
bulges.

"YES but how big would they be (Hint: Smaller than the tides?)"

Actually, the tides would be the same.

One fundamental difference that we have is that you insist that taking centrifugal
force into account is the *only* way to look at the problem. As I've said a number of
times, centrifugal force is a "fictional force" that is only needed if you wish to
work in the accelerating Earth-centric reference frame. In fact, it is required that
there must be an alternate approach that does not use "fictional" forces.

It is quite easy to explain the tides without Centrifugal force - I mentioned the
approach in one of my first posts. There is a gravitational force pulling the Earth
and Moon together. This creates an acceleration such that the Earth is in free fall,
and no net force is felt at the Earth's center. However, there is a larger force on
the Moon side, and a smaller force on the far side. Since the average force has been
accounted for it must be subtracted from these two forces, and the smaller force ends
up being the same magnitude as the larger, but in the opposite direction. Hence, two
tidal bulges, both caused solely by gravity.

Another fundamental difference we have is that I agree with the traditional value for
the tidal force. Ignoring minor effects, the result predicted by Differential Gravity
(whether or not you use centrifugal force as part of the explanation) that is about 2
feet for both the near and far side bulges. (The Sun's contribution is about half of
the Moon's.) The land masses and shallow water tends to "pile up" the water to create
tides that are somewhat higher.

You, however, have claimed that the variation of the centrifugal force from the near
side to the far creates a force that dominates the tides. Your formula predicts a
force 65 times greater than the differential formula, which would seem to create tides
100 feet or more. Your explanation that land masses dampen these tides runs counter
to common experience. And, your formula predicts the Sun's contribution is only 1% of
the Moon's, which is clearly not the case.

As to your claim that the rotation around the barycenter "powers" the tides, well,
that's not true. In fact, even if the Earth and Moon were not rotating around the
barycenter, the tides would be the same, at least, until the Earth and Moon collided.
Frankly, it clear that you still do not accept the fact that CF is constant, exactly
equals the average gravitational force, and thus has no interesting contribution to
the tides.






"Nav" wrote in message ...
Now why try to distort the truth Jeff? I never ever said differential
gravity was not needed. I always said that it's the difference between
gravity and centrifugal forces. You do understand the connotations of
the DIFFERENCE between forces don't you? It does not mean that either
component is zero and actually implies that both are important. Shesh!
Still it's nice to see that you now agree that centrifugal forces should
not be ignored (as they are in the gravity only model). As I've said so
many times, the key to understanding is that the system rotates about
the barycenter and it is not just a gravity field problem. The rotation
actually provides the energy needed to power the daily tides -think
about it OK?

Cheers

Jeff Morris wrote:

"Nav" wrote in message
...

Jeff Morris wrote:

"Nav" wrote in message

...

F= mr omega^2. The distance from the barycenter to all points on earth
is NOT the same.

As the site expalins in the next paragraph, only the center of the Earth rotates
around the barycenter. Other points rotate around neighboring points.


Anyway, the site clearly shows in Fig. 2 that it is the
_DIFFERENCE_ between gravity and centrifugal force that makes the tides,
not gravity alone.

We never disagreed on this point.

HOLY BACKPEDAL!!!!!!


I'm not sure you really want to go back over this thread - your record is rather
shaky. Mine, however, has been quite consistent. Remember, I started by posting
sites with differing approaches to show that this problem can be looked at in
different ways. I then made my first comment about Centrifugal force with:

"Remember that Centrifugal Force may be a handy explanation, but it is a
"fictional force" that only appears real to an observer in an accelerating frame
of reference. Therefore, whenever it is used to explain something, there must
be another explanation that works in a non-accelerating frame."

but then you started claiming that differential gravity wasn't needed, I responded
with:

"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."



a few posts later:
"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."

The bottom line here is that the tides are properly described by the differential
gravity equation. Centrifugal force can be used to explain how an outward force
can
be generated, but it is not needed, and it does not yield the equation that
describes
the tides.

Frankly, your the one who started this by claiming that the traditional
explanation of
tides is fundamentally flawed, and that the differential force normally cited is
not
what causes the tides. You really haven't produced any coherent evidence to
support
this claim.

Jeff Morris wrote:
No, you never said it wasn't needed, but you did minimize its importance, claiming
that it can't be used alone to predict the tides. You said:

"Even worse, the site then goes on to use the _differential_ expression
to calculate the ratio of forces between the moon and sun!"

Yes that's right for that site. The differential of the gavity equation
does not give a force it gives the rate of chnage of force with
distance. That's basic calculus.

You even produced a bogus formula to support your claim that centrifugal force varies
across the Earth. This would predict a tidal force 65 times stronger than
differential gravity. If that isn't minimalizing differential gravity, I don't know
what is.


Nothing bogus abot F=m r omaga^2. It's high school physics.


It is quite easy to explain the tides without Centrifugal force - I mentioned the
approach in one of my first posts. There is a gravitational force pulling the Earth
and Moon together. This creates an acceleration such that the Earth is in free fall,
and no net force is felt at the Earth's center. However, there is a larger force on
the Moon side, and a smaller force on the far side. Since the average force has been
accounted for it must be subtracted from these two forces, and the smaller force ends
up being the same magnitude as the larger, but in the opposite direction. Hence, two
tidal bulges, both caused solely by gravity.


Jeff, stop a moment and put what you just said into an equation.
According to you subtraction of a constant from both sides of an
inequality makes an equality?




You, however, have claimed that the variation of the centrifugal force from the near
side to the far creates a force that dominates the tides. Your formula predicts a
force 65 times greater than the differential formula, which would seem to create tides
100 feet or more.

That's because you've forgetten that the tiode is due to the difference
between forces -how many times do I have to repeat this????


Your explanation that land masses dampen these tides runs counter
to common experience. And, your formula predicts the Sun's contribution is only 1% of
the Moon's, which is clearly not the case.

No it doen't. It's your miscomprehension about the force balance.

As to your claim that the rotation around the barycenter "powers" the tides, well,
that's not true.

Jeff, don't be hysterical. Of course it's true. Furthermore, the
barycenter is not the same distance from all points on earth so the
centrifugal force varies across the earth.

Here is the equation for the force on a mass m of water on the far side
for a non-rotating earth (to keep it clear):

GmM/(R+r)^2 + GmE/r^2 - m(r+s)omega^2


On the near side:

GmM/(R-r)^2 + GmE/r^2 - m(r-s) omega^2

M is the moon mass, E the earth mass, s the distance from the center of
the earth to the barycenter, omega the angular velocity of the
earth-moon pair G the gravitational constant.


Now look very carefully at the three terms in each equation. The first
two are gravity, the third centrifugal. Two terms are different in both
cases. The gravity term is smaller on the far side and the centrifugal
term is bigger. On the near side the gavity is bigger and the
cenbtrifugal is smaller. That is the proof of my argument.

Cheers

"Nav" wrote in message ...


Jeff Morris wrote:
No, you never said it wasn't needed, but you did minimize its importance, claiming
that it can't be used alone to predict the tides. You said:

"Even worse, the site then goes on to use the _differential_ expression
to calculate the ratio of forces between the moon and sun!"

Yes that's right for that site. The differential of the gavity equation
does not give a force it gives the rate of chnage of force with
distance. That's basic calculus.

Simply differentiating the gravitational force is not enough. But the equation
normally given for differential gravity is dimensionally correct.


You even produced a bogus formula to support your claim that centrifugal force
varies
across the Earth. This would predict a tidal force 65 times stronger than
differential gravity. If that isn't minimalizing differential gravity, I don't
know
what is.


Nothing bogus abot F=m r omaga^2. It's high school physics.

That formula is correct. Applying it with a varying "r" is bogus. That's the issue
here, but you haven't noticed it.





It is quite easy to explain the tides without Centrifugal force - I mentioned the
approach in one of my first posts. There is a gravitational force pulling the
Earth
and Moon together. This creates an acceleration such that the Earth is in free
fall,
and no net force is felt at the Earth's center. However, there is a larger force
on
the Moon side, and a smaller force on the far side. Since the average force has
been
accounted for it must be subtracted from these two forces, and the smaller force
ends
up being the same magnitude as the larger, but in the opposite direction. Hence,
two
tidal bulges, both caused solely by gravity.


Jeff, stop a moment and put what you just said into an equation.
According to you subtraction of a constant from both sides of an
inequality makes an equality?

I have no idea what you're saying.

If the force at the center is X, and on the near side is X+D, and on the far side is
X-D, then if you subtract the central force, the near side is D and the far side
is -D.






You, however, have claimed that the variation of the centrifugal force from the
near
side to the far creates a force that dominates the tides. Your formula predicts a
force 65 times greater than the differential formula, which would seem to create
tides
100 feet or more.

That's because you've forgetten that the tiode is due to the difference
between forces -how many times do I have to repeat this????


I just tried to follow all your instructions. You challenged me to work the math, I
did, it produced bogus answers. The difference in the centrifugal force that your
equation predicts is 65 times more that the difference in the gravitational force, so
subtracting still won't help much. Why don't you work out the math? And don't forget
to do it for the Sun also.



Your explanation that land masses dampen these tides runs counter
to common experience. And, your formula predicts the Sun's contribution is only
1% of
the Moon's, which is clearly not the case.

No it doen't. It's your miscomprehension about the force balance.

I'm just reporting the numbers your equation predicts.



As to your claim that the rotation around the barycenter "powers" the tides, well,
that's not true.

Jeff, don't be hysterical. Of course it's true. Furthermore, the
barycenter is not the same distance from all points on earth so the
centrifugal force varies across the earth.

This is your basic mistake. Even the web site you pointed to for support explicitly
says this is not true.
Let me repeat it again
http://www.co-ops.nos.noaa.gov/restles3.html

"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. This fact
is indicated in Fig. 2 by the equal lengths of the thin arrows (representing the
centrifugal force Fc) at points A, C, and B, respectively."

it continues with:
"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."

Go on. Read it, think about it. You were quick to cite this page when you thought it
supported you.



Here is the equation for the force on a mass m of water on the far side
for a non-rotating earth (to keep it clear):

GmM/(R+r)^2 + GmE/r^2 - m(r+s)omega^2


On the near side:

GmM/(R-r)^2 + GmE/r^2 - m(r-s) omega^2

M is the moon mass, E the earth mass, s the distance from the center of
the earth to the barycenter, omega the angular velocity of the
earth-moon pair G the gravitational constant.

Nope. This is incorrect. The proper equations a

Farside:
GmM/(R+r)^2 + GmE/r^2 - m s omega^2

On the near side:
GmM/(R-r)^2 + GmE/r^2 - m s omega^2

Further, since GmE/r^2 = m s omega^2, we are left simply with:
Farside: GmM/(R+r)^2
On the near side: GmM/(R-r)^2

This leads to the traditional differential equation 2GmMr/R^3 as shown in
http://mb-soft.com/public/tides.html


Now look very carefully at the three terms in each equation. The first
two are gravity, the third centrifugal. Two terms are different in both
cases.

Nope. Only one term is different.

The gravity term is smaller on the far side and the centrifugal
term is bigger. On the near side the gavity is bigger and the
cenbtrifugal is smaller. That is the proof of my argument.

The proof fails because of a faulty assumption. Sorry Nav, Centrifugal Force is
constant. You can use it if you chose, but it doesn't really change the math and
isn't particularly interesting. The part of the equation that actually produces the
two tides is the differential gravity.