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.

I wanted to run the numbers again and realized I had garbled the equations somewhat.
I didn't realize the Nav had slipped in the equation for Earth's gravity into the mix.
Since that's 10,000 times stronger than any of the other forces involved, it isn't
really relevant, unless you're trying to calculate the actual tide heights.

Here's the numbers I'll use. These don't really have enough precision to do this
properly:

Gravitational Constant G 6.67x10^-11
Mass of Moon M 7.35x10^22 kg
Distance to Moon R 3.8x10^8 m
Radius of Earth r 6.3x10^6 m
Earth center to E-M barycenter s 4.641x10^6


The proper formulas for the Moon's effect a

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

Near side:
GmM/(R-r)^2 - m s omega^2

This does not count the Earth direct force, which is huge, and use a constant
Centrifugal Force.


I'll work this in terms of the acceleration felt by a body, which is numerically the
same as the force on a one kg body. All numbers are m/sec^2.

The CF comes out to 3.29x10^-5 . This is using an orbital period of 27.3 days (not
the 28 nav mentions)

The gravitation acceleration using GM/r^2 is 3.39x10^5. These numbers are actually
the same, differing only due to the rough approximations used. They must be the same,
because the orbital velocity is determined by the gravitational pull. Thus, we can
use the number computed with GM/r^2.

So, the nearside acceleration becomes:
GM/(R-r)^2 - GM/r^2 = 3.51x10^-5 - 3.39x10^5 = 1.2x10^-6

for the farside:
GM/(R+r)^2 - GM/r^2 = 3.28x10^-5 - 3.39x10^5 = -1.1x10^-6

Thus, the net force is almost equal and opposite, and in agreement with the
traditional values. (The difference between near and farside may even be less, given
roundoff issues.)

If, however, we used Nav's equations, the number are radically different.

The Nearside CF for Nav is (r-s)omega^2, or 1.17x10^-5
The Farside CF is (r+s)omega^2 or 7.74x10^5

Nav said this must be subtracted from the Grav force, which would result in:
Nearside 3.51x10^-5 - 1.17x10^-5 or 2.34x10^5
Farside 3.39x10^-5 - 7.74x10^-5 or -4.46x10^-5

Note that this is a serious imbalance between near and farside, which is certainly not
supported by observation. My hunch is that Nav would prefer to ADD the nearside grav
and CF, which would result in 4.68x10^-5, which is close enough to the farside to
actually make some sense. However, this is roughly 40 times the traditional value
that has been used for many years. Many calculations have been presented that show
the traditional values generate the observed tides; Nav has not presented any credible
explanation for how the experts could be off by a factor of 40.

But, lets take this one more step. What is the contribution from the Sun? First,
running the net gravitational and Centrifugal forces at the Earth's center yields
5.93x10^-3 for grav, and 5.94x10^-3 for CF. Again, these numbers must actually be
equal.

The next step is to calculate the differential gravity. Frankly, since that is so
small compared to the direct gravity that using these approximations would be futile.
I'm content to accept the traditional value: the Sun's differential gravity is a bit
under half of the Moon's, which would be about 5x10^-7.

Finally, to calculate the Centrifugal Force from the Earth-Sun system according to
Nav's formula, we can take advantage of the fact that it is linear with distance. The
delta from the CF at the Earth's center, according to Nav, will be small, 2.5x10-7.
On the near side the tides would be reduced by this amount, on the far side they would
be increased. When added to the Differential Gravity, the net result is the farside
is increased to about 7.5x10^-7, and the nearside is reduced to 2.5x10^-7. This
presents the problem that the night tides have triple the contribution from the Sun as
the day tides.

Using Nav's formulas for the tides pass one test: the near and far side contributions
from the Moon are roughly equal. However, in all other regards they fail miserably.
The are 40 times the accepted and well studied values for tidal forces. But worse,
the Moon's contribution is 100 times the contribution from the Sun. I don't there is
any way to reconcile these discrepancies, and Nav doesn't seem willing to explain it.

Ah ****;

I wonder just how many Tides will come and go before we even agree on a
formular to figure God only know what. When, and if, a formular is
arrived at, I really wonder if and what it will address?
Will it have anything at all to do with the original question; "How can
I sail a Tide Ride?"

I DON'T THINK SO!?!?

Was it worth a "Donal 1/4 Point?"

I DON'T THINK SO?????

Did we learn anything? I DON'T KNOW??
Maybe. Is It correct;

I DON'T THINK SO!?!?

Signing off,

Ole Thom

Sorry Thom, you really don't have read through the math if you don't want to. I have
to wear my sea boots to wade through the muck of political discussions and worse here,
so you can humor my occasional technical obsession.

As I've mentioned before, this is the kind of work I did before retiring, and although
I don't miss the nine to five, I do miss the intellectual challenge of working through
the problems.

As for agreeing on anything, its a basic fact of human nature that there will always
be disagreement, even on the most obvious truths. I view this a good thing, but it
does mean that there will always be people who refuse to accept what others take as a
given.

This however, doesn't explain why some people are always wrong!


"Thom Stewart" wrote in message
...
Ah ****;

I wonder just how many Tides will come and go before we even agree on a
formular to figure God only know what. When, and if, a formular is
arrived at, I really wonder if and what it will address?
Will it have anything at all to do with the original question; "How can
I sail a Tide Ride?"

I DON'T THINK SO!?!?

Was it worth a "Donal 1/4 Point?"

I DON'T THINK SO?????

Did we learn anything? I DON'T KNOW??
Maybe. Is It correct;

I DON'T THINK SO!?!?

Signing off,

Ole Thom

Jeff,

Hey, my friend, you're welcome to all the "net space you want. Just hope
you don't mind being reminded about the reason the question was asked. I
know if I had to calculate all those figures (Right or wrong) the Tide
would have at least turn once if not more.

Have at it my friend, I'll try to keep up but remember, I have a Web TV.
No computer

Ole Thom

"Jeff Morris" wrote in message
...

As for agreeing on anything, its a basic fact of human nature that there
will always
be disagreement, even on the most obvious truths. I view this a good
thing, but it
does mean that there will always be people who refuse to accept what
others take as a
given.

This however, doesn't explain why some people are always wrong!

I agree!




Regards


Donal
--