As I said, I was leaving this as an exercise for the reader.

I have heard tides on the far side of the Earth described in terms of the
centrifugal force caused by the Earth's rotation around Earth-Moon system.
Although this is a consistent way of describing it, I've never liked using
"fictional" forces. Here's a site that uses that approach:

http://co-ops.nos.noaa.gov/restles3.html


"Nav" wrote in message
...
Jeff, you really cannot explain two tides a day unless you also include
the centripetal forces of the earth moon pair -this is the key that is
seems repeatedly lost.

Cheers


Jeff Morris wrote:

"DSK" wrote in message
...

Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:

On both sides the change in gravitational pull from the moon reduces
or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the moon is
culminating and one when it is on the other side.

I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the water
level. But it would be far less than when it's overhead, and the water
has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly harmonics.


Gravity is the force that drives it, harmonics determines the timing.

Here's a site that describes the Differential Gravity in a fairly simple
way:
http://burro.astr.cwru.edu/Academics...ity/tides.html

Now you're actually claiming you did it purpose to be sarcastic?

While its true that you mentioned once that you had lived in Baltimore, you
ignored that fact that I (and others) mentioned Boston as the home of the
Constitution a number of times. And we said a number of times that you seemed
to be describing a different ship from what we all knew. Any reasonable person
would have stopped to consider if they were making a mistake. But not you,
Navvie!

And it wasn't that you made a simple mistake, it was the insisted that you knew
more about the Constitution than everyone else! Classic!



"Nav" wrote in message
...
And your point is what, that I was describing a different ship with a
very similar name? Big deal. It's hilarious that you still don't see the
sarcasm in my refence to her "motor". But if it makes you feel good I'll
say it again, I _was_ talking about the Constellation in _BALTIMORE_ and
not the Constitution -although why you seemed to continually miss the
fact that I was talking about a ship in Baltimore is beyond me. I guess
you only see what you want. At least I can acknowlege when I'm wrong.

Now why don't you tell us again about how great an engineer you are?

Cheers

DSK the wiper wrote:


http://groups.google.com/groups?hl=e...com%26rnum%3D2


Or since long URLs can be troublesome

http://tinyurl.com/3uc5a

The discussion was originally about "Old Ironsides" and went on for a
dozen or more posts about the "USS Constitution" ver specifically.

Amazing what just a few seconds in the archives can reveal.



Nav wrote:

Constellation actually, little man.




Are you sure?


Nav wrote:

Yes, and with each snipe you get smaller.


And with each denial and/or backpedal, you get....

DSK

OzOne wrote:

On Fri, 01 Oct 2004 09:57:01 +1200, Nav
scribbled thusly:


The key to understanding resides in where the center of mass of the
earth-moon system resides.

Cheers


Nav, you need to go sailing....without the textbooks OK?


Hey, I learn't that long ago. The amizing thing is the number of
explanations out there that try to come up with the right answer just
using a gravity argument -even in school textbooks!

Cheers

Hey, we were talking at crossed prurposes. I mentioned Baltimore several
times -whether (or not) the ship might have gone to Boston would have
not helped let me see we were talking about different ships with very
similar names would it? I think it's much more revealing how certain
people crow on about an innocent mistaken identity don't you?

Cheers


Jeff Morris wrote:

Now you're actually claiming you did it purpose to be sarcastic?

While its true that you mentioned once that you had lived in Baltimore, you
ignored that fact that I (and others) mentioned Boston as the home of the
Constitution a number of times. And we said a number of times that you seemed
to be describing a different ship from what we all knew. Any reasonable person
would have stopped to consider if they were making a mistake. But not you,
Navvie!

And it wasn't that you made a simple mistake, it was the insisted that you knew
more about the Constitution than everyone else! Classic!



"Nav" wrote in message
...

And your point is what, that I was describing a different ship with a
very similar name? Big deal. It's hilarious that you still don't see the
sarcasm in my refence to her "motor". But if it makes you feel good I'll
say it again, I _was_ talking about the Constellation in _BALTIMORE_ and
not the Constitution -although why you seemed to continually miss the
fact that I was talking about a ship in Baltimore is beyond me. I guess
you only see what you want. At least I can acknowlege when I'm wrong.

Now why don't you tell us again about how great an engineer you are?

Cheers

DSK the wiper wrote:


http://groups.google.com/groups?hl=e...com%26rnum%3D2


Or since long URLs can be troublesome

http://tinyurl.com/3uc5a

The discussion was originally about "Old Ironsides" and went on for a
dozen or more posts about the "USS Constitution" ver specifically.

Amazing what just a few seconds in the archives can reveal.



Nav wrote:


Constellation actually, little man.




Are you sure?


Nav wrote:


Yes, and with each snipe you get smaller.


And with each denial and/or backpedal, you get....

DSK

Yep it's spot on. I like the pointed quote:

"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."

The question is how many sites that try to explain the two tide problem
ignore this? Answer -almost all!!!! Even the one from your astronomy
Professor! That the tidal problem can be repeated incorrectly so many
times really annoys to me. I'd say it is not beyond the ability of most
children to understand the correct answer is it?

Cheers

Jeff Morris wrote:

As I said, I was leaving this as an exercise for the reader.

I have heard tides on the far side of the Earth described in terms of the
centrifugal force caused by the Earth's rotation around Earth-Moon system.
Although this is a consistent way of describing it, I've never liked using
"fictional" forces. Here's a site that uses that approach:

http://co-ops.nos.noaa.gov/restles3.html


"Nav" wrote in message
...

Jeff, you really cannot explain two tides a day unless you also include
the centripetal forces of the earth moon pair -this is the key that is
seems repeatedly lost.

Cheers


Jeff Morris wrote:


"DSK" wrote in message
. net...


Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:


On both sides the change in gravitational pull from the moon reduces
or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the moon is
culminating and one when it is on the other side.

I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the water
level. But it would be far less than when it's overhead, and the water
has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly harmonics.


Gravity is the force that drives it, harmonics determines the timing.

Here's a site that describes the Differential Gravity in a fairly simple

way:

http://burro.astr.cwru.edu/Academics...ity/tides.html

Jeff,

I think the term centrifuigal is appropriate in this context. It is a
term in the system that can be appreciated without needing to consider
Newtonian forces. To understand centripetal forces is a lot harder than
just demonstrating the effect. similarly gravity can be demonstrated
without maths. Thus the explanation becomes really simple e.g.:

The earth-moon body rotates around a common point and water tries to
move away from the center (water in a bucket swung on a rope analogy).
The moon exerts gravity which is stronger on the side of the moon. Thus
water forms two bulges on opposite sides and makes two tides as the
earth rotates...


Cheers


Jeff Morris wrote:

As I said, I was leaving this as an exercise for the reader.

I have heard tides on the far side of the Earth described in terms of the
centrifugal force caused by the Earth's rotation around Earth-Moon system.
Although this is a consistent way of describing it, I've never liked using
"fictional" forces.

A "standard" tide prediction uses (at least) 32 factors. The most
important being the difference in graviton from the moon over the
earth surface. That leaves out at least 31 factors of practical
importance. One of these is the difference in centrifugal forces due
to the rotation around the mass-centre of the moon-earth system. There
are still 30 left. I was not trying to give the full explanation of
the tide generating forces.

I was answering a question as to why the influence of the sun on the
tides was smaller than the influence of the moon. I think that the
fact that while the gravitation of the sun is larger than that of the
moon, the difference of the gravitation over the earth surface is
smaller, answers that question fairly well.

Peter S/Y Anicula

"Nav" skrev i en meddelelse
...


DSK wrote:
Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:

On both sides the change in gravitational pull from the moon
reduces
or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the
moon is
culminating and one when it is on the other side.


I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the
water
level. But it would be far less than when it's overhead, and the
water
has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly
harmonics.


The key to understanding resides in where the center of mass of the
earth-moon system resides.

Cheers

OK. I think you mean that as the gravitational field is flatter for the
Earth Sun pair compared to the Earth Moon pair the differential effect
of sun position is smaller?

Cheers

Peter S/Y Anicula wrote:

A "standard" tide prediction uses (at least) 32 factors. The most
important being the difference in graviton from the moon over the
earth surface. That leaves out at least 31 factors of practical
importance. One of these is the difference in centrifugal forces due
to the rotation around the mass-centre of the moon-earth system. There
are still 30 left. I was not trying to give the full explanation of
the tide generating forces.

I was answering a question as to why the influence of the sun on the
tides was smaller than the influence of the moon. I think that the
fact that while the gravitation of the sun is larger than that of the
moon, the difference of the gravitation over the earth surface is
smaller, answers that question fairly well.

Peter S/Y Anicula

"Nav" skrev i en meddelelse
...


DSK wrote:

Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:


On both sides the change in gravitational pull from the moon

reduces

or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the

moon is

culminating and one when it is on the other side.


I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the

water

level. But it would be far less than when it's overhead, and the

water

has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly

harmonics.

The key to understanding resides in where the center of mass of the
earth-moon system resides.

Cheers

You may be right, certainly proponents of this explanation use "centrifugal
force."

However, differential gravity can be explained a number of ways. For example,
the Moon's pull causes the Earth to accelerate towards the Moon. That portion
of the Earth closer feels more force, and thus falls faster; that portion on the
far side feels less force, and thus falls slower. These differences cause the
bulges on the near and far sides.

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.



"Nav" wrote in message
...
Jeff,

I think the term centrifuigal is appropriate in this context. It is a
term in the system that can be appreciated without needing to consider
Newtonian forces. To understand centripetal forces is a lot harder than
just demonstrating the effect. similarly gravity can be demonstrated
without maths. Thus the explanation becomes really simple e.g.:

The earth-moon body rotates around a common point and water tries to
move away from the center (water in a bucket swung on a rope analogy).
The moon exerts gravity which is stronger on the side of the moon. Thus
water forms two bulges on opposite sides and makes two tides as the
earth rotates...


Cheers


Jeff Morris wrote:

As I said, I was leaving this as an exercise for the reader.

I have heard tides on the far side of the Earth described in terms of the
centrifugal force caused by the Earth's rotation around Earth-Moon system.
Although this is a consistent way of describing it, I've never liked using
"fictional" forces.

Jeff Morris wrote:

You may be right, certainly proponents of this explanation use "centrifugal
force."

However, differential gravity can be explained a number of ways. For example,
the Moon's pull causes the Earth to accelerate towards the Moon. That portion
of the Earth closer feels more force, and thus falls faster; that portion on the
far side feels less force, and thus falls slower. These differences cause the
bulges on the near and far sides.


I like it! That's a way of putting it I've not heard before and it is
quite elegant (provided the listener can accept that the Earth is
falling toward the Moon!)


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.


I could be devious and say we are all in an accelerating frame! But you
are quite right about the artifice of a virtual force. Nevertheless,
children want to know about tides and for them centripetal force can be
experienced more easily than the idea they are on an earth that is
falling...

Cheers

Nav wrote:
The earth-moon body rotates around a common point and water tries to
move away from the centre (water in a bucket swung on a rope
analogy).
The moon exerts gravity which is stronger on the side of the moon.
Thus
water forms two bulges on opposite sides and makes two tides as the
earth rotates...


You make it sound as if the gravitational forces explains the bulge
under the moon and the centrifugal forces explains the bulge on the
side of the earth that turns away from the moon.
That is not right.

The gravitational difference alone can explain that there are bulges
on both side of the earth. That's why it is sometimes the only factor
mentioned when trying to keep the explanation simple.
The centrifugal element can only explain that there is a bulge on the
part of the earth that turns away from the moon.
That is why it is one of the elements (and there are others), that is
sometimes left out of the explanation.

While I think that in some cases it is a good idea to include the
centrifugal element in the explanation, I don't know exactly how many
elements one should include to make it a good explanation - but I
haven't yet seen a complete explanation in a popular publication.

Peter S/Y Anicula





"Nav" skrev i en meddelelse
...
Jeff,

I think the term centrifuigal is appropriate in this context. It is
a
term in the system that can be appreciated without needing to
consider
Newtonian forces. To understand centripetal forces is a lot harder
than
just demonstrating the effect. similarly gravity can be demonstrated
without maths. Thus the explanation becomes really simple e.g.:

The earth-moon body rotates around a common point and water tries to
move away from the center (water in a bucket swung on a rope
analogy).
The moon exerts gravity which is stronger on the side of the moon.
Thus
water forms two bulges on opposite sides and makes two tides as the
earth rotates...


Cheers

Nav wrote:
And your point is what, that I was describing a different ship with a
very similar name?

No, my point is that you have no clue.

Again.

DSK

Nav wrote:
Hey, we were talking at crossed prurposes.

No, you just plain didn't know WTF you were talking about.

... I mentioned Baltimore several
times

Actually, you didn't. Jeff and I both mentioned Boston several times and
you apparently either missed it (poor reading skills) or were trying to
fake your way out of it.

... I think it's much more revealing how certain
people crow on about an innocent mistaken identity don't you?

You mean, certain people (or one person in particular) crowing about
their (his) tremendous knowledge & expertise when they've made...
repeatedly... a very obvious mistake in identity, and clearly don't know
diddley-squat?

DSK

DSK wrote:

Nav wrote:

Hey, we were talking at crossed prurposes.


No, you just plain didn't know WTF you were talking about.

... I mentioned Baltimore several times


Actually, you didn't. Jeff and I both mentioned Boston several times and
you apparently either missed it (poor reading skills) or were trying to
fake your way out of it.

... I think it's much more revealing how certain people crow on about
an innocent mistaken identity don't you?


You mean, certain people (or one person in particular) crowing about
their (his) tremendous knowledge & expertise when they've made...
repeatedly... a very obvious mistake in identity, and clearly don't know
diddley-squat?


Now if you think my mistaking the Constitution for the Constellation is
a big deal then so be it. I think to anyone else it would be an obvious
minor error -after all their names are very similar and they were both
rotting on the East coast at the same time. But to you it seems like
such a big thing to try to make me look ignorant. Now why would that be?

Look Doug, get used to it, I have a wonderful rich life and enjoy a good
argument. In fact I even deliberately provoke it. But your childish
scampering around this playground does you no good. You seem to like to
BS almost all the time but it's so easy to catch you out. So just try to
stick to tips on sail trim and boat cleaning and you'll do much better.

This advice really is in your best interest -you don't want your blood
pressure to get worse do you? People would miss you if you had a cardio.

Cheers

Yes, how about that? Care to comment Donal?

Cheers

Thom Stewart wrote:

Donal,


Any ideas


How about the Increase of Tidal Flow and height during Hurricanes?

Ole Thom

You mean, certain people (or one person in particular) crowing about
their (his) tremendous knowledge & expertise when they've made...
repeatedly... a very obvious mistake in identity, and clearly don't
know diddley-squat?


Nav wrote:
Now if you think my mistaking the Constitution for the Constellation is
a big deal then so be it.

No, I think your mistaking the Constitution... and Boston... when both
are spelled out at least a half dozen times... for the Constellation &
Baltimore, is typical for you. To be charitable, it could be an example
of poor reading skills.

To then go on & crow about your superior knowledge, and freely insult
everyone who disagrees with you, and then boldly deny you ever made a
mistake at all, is the kind of behavior that ill-mannered children
should be corrected for. It's a shame that you suffer under the handicap
of never learning better.

DSK

Nav,

Don't you think you are overstating the obvious? The Tide is a dynamic
action of water caused by the moon. The height of the Tide and the Tidal
flow are dependent on one another. The location of the rise in the Tide
is determined by the speed of the earths rotation and the revolving of
the moon. The height of the Tide determined by the Phase of the Moon.
The higher the Tide rise the greater the Tidal Flow

I hope this meets with your approval. I'm very sure we all have
overstated what Scott wanted to know about riding the Tide.

Ole Thom
P/S Donal give him the full point. He gave the answer to the moving
tide, making a tide ride possible

Nav,

Why wouldn't you use a RMS relationship if it close to a sine wave?

Ole Thom

"Nav" wrote in message
...
Yes, how about that? Care to comment Donal?


Hey, I ain't an expert on these things.

Why don't you tell us what might have caused it.

I've already expressed the sum total of my knowledge on this subject. Ie,
it might have been caused by pressure systems.

If you think that my opinion is stupid, then you should have the courage to
say so.

If you actually know what caused it, then I'd be grateful if you would share
your knowledge with us. After all, this is a "group". Are you are member,
or not?


Regards


Donal
--

SDK wrote:



To then go on & crow about your superior knowledge, and freely insult
everyone who disagrees with you, and then boldly deny you ever made a
mistake at all, is the kind of behavior that ill-mannered children
should be corrected for. It's a shame that you suffer under the handicap
of never learning better.


You think I crow or is it could it be that I question in a way designed
to disclose ignorance? If you mistake my comments for crowing so be it,
but I think that just reflects your jealousy of the wonderful life I
lead. Now, I freely admitted my simple mistake about the Constellation
when Jeff identified it. It was you that then "crowed" about it like a
child isn't it? In fact I've always acknowledged when I'm wrong -but the
discussant has to show why I am and not simply naysay (like you). I
don't resort to ad hominems or insults until I'm insulted (I treat like
with like and if people are polite then so am I)-but the record shows
that you frequently start insulting people who disagree with you. So
what is this post except more bile from you? Like I said, _you_ are the
one who makes yourself look bad.

Cheers

I'm sorry Thom I don't follow your thought about RMS. How would
geometric mean be related to the time and size of current flow? If the
tide took twice as long to rise and fall, the RMS would be the same but
the actual current would be smaller would it not?

Cheers


Thom Stewart wrote:

Nav,

Why wouldn't you use a RMS relationship if it close to a sine wave?

Ole Thom

Donal wrote:

"Nav" wrote in message
...

Yes, how about that? Care to comment Donal?


Hey, I ain't an expert on these things.

Why don't you tell us what might have caused it.

I have no idea if there was hurricane there or what the tide difference
was? Only you can tell us that.

I've already expressed the sum total of my knowledge on this subject. Ie,
it might have been caused by pressure systems.

If you think that my opinion is stupid, then you should have the courage to
say so.

Why are you being defensive -you were there weren't you?

If you actually know what caused it, then I'd be grateful if you would share
your knowledge with us. After all, this is a "group". Are you are member,
or not?

I can express an opion based on my Nav but I would need to know more
about what the condions were -how different the tides were etc.

Cheers

Nav,

RMS would be .707 of the Sine wave (Or there abouts) The Tide cycle
being 12 hrs. If we round off .707 to 70% times 12 hrs we would get 8.4
hrs. this would represent the straight line, or Max flow time. The
curved section of the sine wave RMS wise would be 30%. Sense we are only
concerned with a 1/2 wave (Tide Run in one direction) it would Be 15% on
the top (Start of the run) and 15% on the end.
1.8 from slack, 8.4 max flow 1.8 to slack
12hrs total.

Over the years, I've found that this describes the tide flow damn close.
As I explained in a previous post, The slack is the problem. It can last
sometimes for over an hour and at time it doesn't even seem to exist but
the hour and 40minutes still holds up pretty well to max low.

So, as the tide runs start to slow down, you have about 1hr3/4 to make
your inlet on the slack.

As I said before; BE EARLY

Ole Thom

i'm sure you are right but it's interesting to explore the possibilities
is it not? I find it more interesting than guns and GWB (for example).

Cheers

Thom Stewart wrote:

Nav,

Don't you think you are overstating the obvious? The Tide is a dynamic
action of water caused by the moon. The height of the Tide and the Tidal
flow are dependent on one another. The location of the rise in the Tide
is determined by the speed of the earths rotation and the revolving of
the moon. The height of the Tide determined by the Phase of the Moon.
The higher the Tide rise the greater the Tidal Flow

I hope this meets with your approval. I'm very sure we all have
overstated what Scott wanted to know about riding the Tide.

Ole Thom
P/S Donal give him the full point. He gave the answer to the moving
tide, making a tide ride possible

"Nav" wrote in message
...


Donal wrote:

"Nav" wrote in message
...

Yes, how about that? Care to comment Donal?


Hey, I ain't an expert on these things.

Why don't you tell us what might have caused it.

I have no idea if there was hurricane there or what the tide difference
was? Only you can tell us that.

I've already expressed the sum total of my knowledge on this subject.
Ie,
it might have been caused by pressure systems.

If you think that my opinion is stupid, then you should have the courage
to
say so.

Why are you being defensive -you were there weren't you?

Perhaps I misunderstood your question - my apologies!

The fact that I was there doesn't help much. I encountered tides that
appeared to be an hour early. However, if the current was also *stronger*
than forecast, then it is possible that the tide may only have been 1/2 an
hour early. Everything about the trip suggested that the tide turned an
hour early. I don't think that there was a major difference in current.



If you actually know what caused it, then I'd be grateful if you would
share
your knowledge with us. After all, this is a "group". Are you are
member,
or not?

I can express an opion based on my Nav but I would need to know more
about what the condions were -how different the tides were etc.

As stated, I'm fairly sure that the only noticeable difference was that the
tide was early.

After the Cap De la Hague we aimed for about five miles offshore - where we
expected to find about 3 kts behind us. When we discovered that we had 2
kts against us we headed back inshore(where there would be less contrary
tide). As we are familiar with the area, we didn't cough keep an accurate
log.


Regards


Donal
--

Nav,

Guns and "W" Spare me.
You're absolutely right!

Ole Thom

Thom Stewart wrote:

Nav,

Guns and "W" Spare me.
You're absolutely right!


LOL

Don't tell that to Doug!

Cheers

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


In the middle (when the moon is in the horizon) the moons
pulls with the same as on the earth as a whole, and there is no
vertical component, so here the water-molecule is "unaffected" by the
moon.

Wrong! This is when LW occurs. By definition, LW is lower than the
average sea level. Therefore, there must be a force that is dragging it
down.


Regards


Donal
--

"Martin Baxter" wrote in message
...
Donal wrote:
"Thom Stewart" wrote in message
...

Marty,

Right you are! Tell Donal to mark you up for One Point.

The Tides use the Moon Calender:^)
That is also the reason that Tide rides are possible.


I'm afraid that I can only award Marty 0.75 of a point.

I'm about to dissappear for a couple of days, so I'll explain when I get
back.

[hint] The sun has a much greater gravitational effect on the Earth
than
the moon. So why does the moon seem to have a greater impact on the
tides?

Well duh! Remember F=G*(m'*m")/(d^2),

Emmm... Huh?

What the hell does that mean in English?

d is very much smaller in the earth moon case than the sun earth case.
IOW, it's because the moon is closer.

Don't tell me! "d" stands for "distance"???


Give my point to Joe, he needs all the help he can get. ;-o

You haven't earned a point - yet! Joe will just have to wait.


Regards


Donal
--

Donal wrote:
So why does the moon seem to have a greater impact on the

tides?

Well duh! Remember F=G*(m'*m")/(d^2),


Emmm... Huh?

What the hell does that mean in English?

Did you not take physics in school? Look up Newton's Law of Gravitation.


d is very much smaller in the earth moon case than the sun earth case.
IOW, it's because the moon is closer.


Don't tell me! "d" stands for "distance"???

Ok, I won't.

Cheers
Marty

Donal wrote:

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

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?

Cheers

"Nav" wrote in message
...


Donal wrote:

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

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.

This force can be viewed as having two components, one is on the Earth-Moon (or
Earth-Sun) axis, and the other is off axis. The on axis component varies
inversely with R cubed, outwards from the center of the Earth, and is usually
cited as cause of the two bulges. The other component is pulling towards the
Moon, but since its off-axis, from the an observer on Earth sees this as
somewhat downward. When the total gravitation force is subtracted from this,
the result is a downward pull. This is greatest at "low tide points," half way
around from the bulges, but also in a ring around Earth that includes the poles.
The net resulting idealized surface is a "prolate ellipsoid," squeezed around
the middle and out at the ends.

As a simple "reductio ad absurdum" consider that the tides at the poles are
always "Low" because the Moon is always pulling a bit downward on the poles.
If you only consider the centrifugal force, there is no component of that which
pulls the poles inward.

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.

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

Peter S/Y Anicula wrote:

Nav wrote:



You make it sound as if the gravitational forces explains the bulge
under the moon and the centrifugal forces explains the bulge on the
side of the earth that turns away from the moon.
That is not right.



The gravitational difference alone can explain that there are bulges
on both side of the earth. That's why it is sometimes the only factor
mentioned when trying to keep the explanation simple.

The trouble is that the gravity difference does not "explain" the two
tides -it may seem to but that is not the case. The correct explanation
resides in the difference between the inertial force and gravity.
Interestingly, it predicts larger tides than are observed (and predicted
by the differerntial model). However, that is because the land masses
and friction reduce the tide height (to what is actaully observed). That
differential gravity appears to produce the "right" answer shows how
shallow (pardon the pun) that "explanation" really is!

Does this make sense?

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" wrote in message
...
"Martin Baxter" wrote in message
Donal wrote:
So why does the moon seem to have a greater impact on the

tides?

Well duh! Remember F=G*(m'*m")/(d^2),


Emmm... Huh?

What the hell does that mean in English?

Did you not take physics in school?

Sadly, I didn't!

With hindsight, I suspect that I had a poor teacher who managed to make the
subject appear much duller than it really is.

My high school physics teacher was possibly the worst teacher I ever had - a
true nut case who shouldn't have been left alone with children. Fortunately I
found much better teachers in college.



I've still got a suspicion that if we expand your equation, we will find
that the sun has a greater gravitational influence on the earth than the
moon does.

Yes, its does. The direct gravitational pull of the Sun is enormous, much
larger than the Moon's. However, the tides are caused by the difference in pull
between the near side and the far side. Since the Moon is a lot closer, that
difference is more significant. If you remember any calculus, you'd know that
differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of
the Sun's distance becomes a tiny number compared to the Moon's. The net
result is that the Moon's effect on the tides is 2.2 stronger than the Sun's.

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

"Jeff Morris"

Yes, its does. The direct gravitational pull of the Sun is enormous, much
larger than the Moon's. However, the tides are caused by the difference in pull
between the near side and the far side. Since the Moon is a lot closer, that
difference is more significant. If you remember any calculus, you'd know that
differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of
the Sun's distance becomes a tiny number compared to the Moon's. The net
result is that the Moon's effect on the tides is 2.2 stronger than the Sun's.

Yeah Donal!

Have ya ever seen any body on earth affected by a full sun?

Sheeze... even a 13 year old girl knows it the moon that shakes things up.


Joe

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