You mean you don't have one of those on your boat? Shame on you, mind
you rough water plays havoc when the weights start swinging.

Cheers

Jeff Morris wrote:
If you're not prepared to derive the tides from first principles, you're not a
real sailor!

In the old days, this is what we had to pack in our ditch bag:
http://ftp.arl.army.mil/ftp/historic...gif/eniac1.gif

but it was a step up from this puppy:
http://co-ops.nos.noaa.gov/predmach.html

OzOne wrote in message ...

Jeez, I'm glad I can just look at a tide chart :-)
Or use J tides
http://vps.arachnoid.com/JTides/index.html
unless of course I lived in the UK where GovCo has decided that they
OWN tide data ....


On Tue, 05 Oct 2004 18:09:58 +1300, Nav
scribbled thusly:


Ok, then we are almost there ! The final answer on that good site
shows that it is the _imbalance_ between the radial component of gravity
and the centripetal term. You seem to have missed the importance (and
cause) of the term that raises the radius of the earth (r) to the fourth
power.

h~Mr^4 cos^2 theta/ER^3.

Cheers



Jeff Morris wrote:


I see your point, but I keep looking at the final answer. When all the

terms

are balanced, and the minor effects ignored, what is the left is 2GmMr
cos^2/R^3, which comes from the radial component of the moon's gravity on a
piece if the Earth m. All of the other forces, including all of the
centrifugal forces have been balanced out. The cos^2 term is the only

thing

left that varies with latitude, which means that explains why the bulges

are at

the equator, and the pole's tides are depressed.


"Nav" wrote in message
...


Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not interested
in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy minima due
to the system rotation. Note the first two terms of the force balance
equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include the
centripetal force term (mr omega^2) across the diameter of the earth is
simply not a correct analysis. Furtherore, if that "explanation" is used
to show two tides it's bogus (it is clear the earth's radius does not
cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:




For anyone who wants to follow this through, here's a pretty complete

version.


http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on individual

parts

of


the Earth, then to add in the centrifugal force and the Earth's pull and

the

Earth's daily rotation component. However, the centrifugal force is

derived

from the total gravitation pull, so that component is certainly not

ignored.

As is often the case, it gets messy before terms start to cancel, but the

end


result is that the "differential gravity" is exactly symmetrical on the

near

and


far side from the Moon. It would not be fair to say the the near side

component


is caused by one force, and the far side by another. In fact, all of the
"latitude dependent" forces are caused by the differential pull from the

Moon.


It is when you subtract out the net pull (or add the centrifugal) that

this

becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered except
that for simplicity I did not consider the earth's rotation. It also
supports my quick and dirty proof or there being two energy minima due
to the system rotation.






Oz1...of the 3 twins.

I welcome you to crackerbox palace,We've been expecting you.