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George,

I was under the impression that you were arguing there would not be 200
meter shifts in sea level even in the presence of such fluctuations in
potential, due to currents, wind, etc.  Upon rereading your original
post, I see you were not.  Combined with this one, I can at least claim
understanding of what you're saying, which implies agreement, for what
that's worth coming from me.

I had thought that g was constant on the equipotentials, but that is
another discussion.

OK, now, are the shifts 200 meters or 20?  200 sounds way to much.

Fred

On Monday, Aug 18, 2003, at 16:56 US/Eastern, George Huxtable wrote:

> Fred Hebard said-
>
>> I might note,
>> however, that work would be done moving across a gradient in the
>> gravitational field even if there were no change in sea-surface
>> elevation.
>
> We have to be careful here. The Earth is immersed in its own
> gravitational
> field. This consists of a series of shells called equipotentials, a
> long
> word which means nothing more than that there is NO work done at all
> against gravity in moving from any point to any other on the same
> equipotential. These shells fit inside each other like Russian dolls,
> each
> one having a different potential, so work is done (or regained) if you
> move
> from one shell to another, as you do when climbing (or descending) a
> mountain.
>
> Each of these shells is roughly ellipsoidal, the combination of the
> effect
> of gravitational attraction ang the centifugal effects of a rotating
> Earth;
> but because of the uneven mass distribution within the Earth, there are
> some superimposed dips and valleys, the objects we have been
> discussing. At
> any point on an equipotential shell, its surface is exactly at right
> angles
> to the local direction of gravitational attraction,"g" (including those
> centrifugal effects into the gravity). This simply has to be so: if it
> wasn't, then as you moved along that surface, there would be some
> component
> of g causing a force along your direction of travel, and you would be
> giving-up or gaining energy, which would defeat the definition of an
> equipotential.
>
> The most important of these equipotentials for our purpose, is that at
> sea-level. In the absence of disturbing factors, such as driving winds
> or
> barometric differences or local waves or tides, sea-level adjusts
> itself to
> conform to the sea-level equipotential. If not, one patch would find
> itself
> with more gravitational energy than another, which would cause a
> water-flow
> to adjust the levels until there was no longer a difference. When they
> say
> "water finds its own level", that's the level it finds. Because that
> equipotential  surface has dips and bumps, so does the sea-level.
>
> However, those dips and bumps are just differences in the distance to
> the
> earth's centre. They are not like the hills and valleys on the Earth's
> solid surface. Because they are exactly on an equipotential, there is
> no
> work involved in travelling from any one point on the Earth's ocean
> surface
> to any other. They are NOT dips and bumps in the gravitional sense:
> there's
> no uphill and no downhill. And that's just as true for the ship
> floating in
> the sea as it is for the sea surface itself. Except for those minor
> disturbing factors listed above, there are NO changes of gravitational
> energy AT ALL when moving around ANYWHERE on the ocean surface.
>
> So how does all this relate to "g", which describes the STRENGTH of the
> Earth's gravitational pull? Well, g is related to the spacing between
> the
> equipotentials, rather in the same way as on a land-map, in which the
> ground-slope corresponds to the closeness of the contours. The closer
> the
> equipotentials are (which happens near the poles) the greater is the
> value
> of g. Near the equator, the potential contours are more widely spaced,
> and
> so g is less. You could if you wished draw out a set of shells of equal
> values of g but it would not be very helpful to do so, because such a
> shell
> of equal g is NOT itself an equipotential, and the sea surface does NOT
> follow it.
>
> It's possible to go from one value of g to another, always following an
> equipotential path, and there will be changes in elevation above the
> Earth's centre along that path, but being an equipotential path, there
> will
> be no work done against gravity. None at all. In fact, this is exactly
> what
> happens when a ship travels from one latitude to another and from one
> value
> of g to another, over the oceans.
>
> So when Fred Heberd states-
>
>> I might note,
>> however, that work would be done moving across a gradient in the
>> gravitational field even if there were no change in sea-surface
>> elevation.
>
> It's not entirely clear to me what he means here, but whatever it is, I
> think it must be wrong.
>
> Then he adds-
>
>> It would seem that accounting for a 400-meter-equivalent change in
>> gravity might be a worthwhile consideration in ship routing.
>
> I ask Fred what exactly he means by "a 400-meter-equivalent change in
> gravity".
>
> I would like to point out that the matters I have been describing are
> not
> those of any scientific controversy. They have been accepted and
> understood
> rather well for over 100 years, and are not a matter of current
> bickering
> or dispute. However, like all scientific theory, it's always open to
> challenge by anyone who can show measurements or logical arguments that
> contradict it.
>
> George Huxtable.
>
> ================================================================
> contact George Huxtable by email at george@huxtable.u-net.com, by
> phone at
> 01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
> Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
> ================================================================
>
>
------------------------------------------------------------------------
Frederick V. Hebard, PhD                      Email: mailto:Fred@acf.org
Staff Pathologist, Meadowview Research Farms  Web: http://www.acf.org
American Chestnut Foundation                  Phone: (276) 944-4631
14005 Glenbrook Ave.                          Fax: (276) 944-0934
Meadowview, VA 24361

   

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