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
--

"Donut" wrote

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

I agree!

Well, you're wrong!

"Scott Vernon" wrote in message
...
"Donut" wrote

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

I agree!

Well, you're wrong!

Perhaps you're right.

Can you justify your position?



Regards


Donal
--

"Donal" wrote in message
...

"Scott Vernon" wrote in message
...
"Donut" wrote

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

I agree!

Well, you're wrong!

Perhaps you're right.

Can you justify your position?

Sitting down, facing NNW.

Scotty

"Scott Vernon" wrote in message
...

"Donal" wrote in message
...

Can you justify your position?

Sitting down, facing NNW.

Dammit! I thought that you were facing EWE!!


Regards


Donal
--

Jeff,

If CF has an effect on the Tides, And this is a question) why isn't the
bulge on the far side of the earth, opposite the Moon in a single bulge?
Again a question, what makes the Liquid (Tide) stop at edges of the
globe and not continue to the off side of the globe?

Why isn't it a simple matter of Liquid Displacement due to the greater
amount of Liquid gathered for the High Tide under the Moon?

These are simple Layman's questions, that to me seem logical.

Ole Thom

At the risk of starting this all over again ...

The two bulges (near side & far side) are not caused by centrifugal force. The bulges
are caused by the fact that the Moon's pull is stronger on the near side and weaker on
the far side. This is call differential gravity and it works out to be inversely
proportional to distance cubed, which is why the Moon's effect is somewhat stronger
than the Sun's.

Centrifugal Force is not needed to work this out. However, many people are perplexed
at how the far side bulge is created - to some it appears to be negative gravity. (In
reality, it is just less positive gravity.) For these people, it is easy to say that
Centrifugal Force causes the far side bulge.

One important point about CF is that it is a "fictional force," that is, it is only to
needed to make things work out for an observer in a "non-inertial reference frame."
For example, for someone on a merry-go-round, there is an outward force. However,
there is no force pushing outward; that is just an artifact of being on the rotating
frame. To an outside observer, the kid's inertia is pulling in a straight line, and
the ride is trying to pull the kid around in a curve.

The formulas that I worked out a few posts ago show the actual numerical values. One
interesting thing in the math is that the centrifugal force is exactly equal and
opposite to the gravitational force. Whether or not you use CF depends on whether
you're working in an "Earth-centric" reference frame. Either way, the part of the
math that varies with distance is the gravitational force itself.

Hope this helps.

-jeff



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

If CF has an effect on the Tides, And this is a question) why isn't the
bulge on the far side of the earth, opposite the Moon in a single bulge?
Again a question, what makes the Liquid (Tide) stop at edges of the
globe and not continue to the off side of the globe?

Why isn't it a simple matter of Liquid Displacement due to the greater
amount of Liquid gathered for the High Tide under the Moon?

These are simple Layman's questions, that to me seem logical.

Ole Thom

Thank Jeff,

You're the man!!

OT