NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Navigating Around Hills and Dips in the Ocean
From: George Huxtable
Date: 2003 Aug 18, 21:56 +0100
From: George Huxtable
Date: 2003 Aug 18, 21:56 +0100
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. ================================================================