*From*: "Martin G. Selbrede" <mselbrede@xxxxxxxxxxxxx>*To*: geocentrism@xxxxxxxxxxxxx*Date*: Fri, 9 May 2008 16:50:14 -0500

Hi Allen,

Martin On May 9, 2008, at 4:39 PM, Allen Daves wrote:

I would like to point out however, that the tidal bulges are onboth sides of the earth & earth's oceans at the same time, theearth is flatened around it's whole cercumfrence supposedly due tothe centrifical force that is cused by inertia wrt the stars notjust the tidal actions of the sun/moon.....Even that explinationwould not reslove the the fact that the acceleration is detectablebut it does not address the overall inertial paradox that exist forboth the tides as well as any given orbiting body in a elipticalorbit where the accelerations advance and slow down wrt the verysame causes of inertial reaction when not in orbit. (the wholepoint of the diagram)Therfore that (sun only) explination is notonly too simplistic but kinda irrelevant to the issue as a wholethat I am trying to point out.----- Original Message ---- From: Martin G. Selbrede <mselbrede@xxxxxxxxxxxxx> To: geocentrism@xxxxxxxxxxxxx Sent: Friday, May 9, 2008 2:25:19 PM Subject: [geocentrism] Re: TidesI understood Regner to be mounting an argument based on relativedifferential force (the entire premise of tidal forces in the firstplace). It is this differential force that supposedly gave rise tothe rings of Saturn, when a sufficiently large chunk of materialwas disrupted as it passed within the Roche limit (meaning theforce between the near-side and far-side parts of the body exceededthe tensile strength of the material comprising that body). Thistidal disruption causes the alleged shattering of the body into amyriad of small chunks, each finding a unique orbit around Saturndue to inter-body collisions over time.I believe Regner mounted a Sun-only explanation for the sake ofdidactic simplicity, and he even went so far as to point this out.I saw nothing faulty in his analysis given the limits he himselfprovided to frame his discussion.Martin On May 9, 2008, at 4:14 PM, Neville Jones wrote:Regner,Your analysis of the cause of the tides is invalid, becausegravitational attraction between parts of extended objects beingexactly equivalent to all mass of each object being concentratedinto mathematical points is valid for these objects being rigid,not flexible. You cannot assume the World to be rigid on the onehand and then subtract the theoretical force on a rigid World fromboth sides of a flexible World in order to 'explain' the tides.Where would these forces on the oceans be acting? At the centre ofmass of the ocean, or on individual water molecules, or at thecentre of mass of the World?The pull would vary as the inverse-square of the distance, whichwould result in very serious tidal waves arising as the Moon-sideocean was pulled more than the other was 'left behind'.Rather than being<- | -> as depicted, the force would be proportional tox*x^(-1) (x*x+2xR+R*R)^(-1) (x*x+4xR+4R*R)^(-1)where x is the distance from the centre of mass of the Moon to theclosest tip of the ocean nearest to the Moon and R is the radiusof the World, and x >> R in the heliocentric model.Ocean swell would also vary with the Moon's orbit.The analysis is thus far too simplistic. As with other aspects ofthe so-called 'solar system', such forces would lead very quicklyto disorder and destruction.Neville. -----Original Message----- From: art@xxxxxxxxxxxxxx Sent: Wed, 07 May 2008 15:18:40 +1000There has been a couple of references to tides, lately, and muchconfusionon the subject. I'll do what I can to clear it up a bit. Please take the time to read the whole post before commenting. The force of gravity moves objects The Moon and the Earth orbit each other due to their gravitational interaction. It can be shown that you can simplify this problem:Every single bit of the Moon interacting gravitationally withevery singlebit of the Earth. is exactly equivalent toAll the mass of the Moon being located at the centre-of-mass ofthe Moon,and all the mass of the Earth being located at the centre-of-mass of theEarth.From such an analysis we find that the two objects will orbittheir commoncentre-of-mass in elliptical orbits. Tidal forces deform objectsBut the Earth and the Moon are extended objects so what effectdoes thathave? Well, the Moon is still close enough to the Earth, that thegravityfrom the Moon will change over the diameter of the Earth: Moon Near Centre Far side of Earth side O <-------- <------- <------Okay, if we now subtract the part that actually moves the Earth,that iswe subtract <------- which is the same as adding -------> toall threevectors (arrows) above, and we get the tidal forces: Moon Near Centre Far side of Earth side O <- | -> So looking at the Moon-Earth system from the outside we see that theMoon-side of Earth experience a larger gravitational pull from theMoonthan the centre of the Earth, and the opposite side of Earthexperienceexperience even less gravitational pull from the Moon. Seeing itfrom theEarth point of view, we have a Moon-ward force on the Moon-side ofEarthand an opposite force on the opposite side of Earth. This is whatcausesthe tidal bulge, and this is why there are (approximately) twohigh- and twolow-tides a day. The oceans, being a liquid, obviously respondsmuch strongerthan the bedrock, but the Earth's crust is flexing too, as hasbeen measured.The Earth rotates under this tidal bulge, which makes it adynamical phenomenon.The gravitational force is proportional to 1/r² where r is thedistancebetween the Earth and the Moon. The tidal force is the difference ingravitational force over a certain distance, which means it can beexpressedas the r-derivative of the gravitational force. The tidal force isthereforeproportional to 1/r³. That means the tidal force falls off morerapidly withdistance than the gravitational force. This is the reason that thegravitationalforce of the Sun dominates the orbit of the Earth, but the Moon'stidal forceson Earth are larger than the Sun's.The Sun also contributes a significant tide (about half that ofthe Moon)and when the two are aligned at new Moon or full Moon, we get morepowerful tides called spring tides. At either half Moon we getneap tides.The ellipticity (eccentricity) of the Moon-Earth and the Sun-Earthorbitsalso causes variations in the strength of tides on Earth. The Moon experience (about 22 times) stronger tides from the Earth because of the Earth/Moon mass ratio of 81 and the size ratio of 3.7.The Lunar tide from the Sun is 3.7 times weaker than here onEarth, dueto the 3.7 times smaller diameter of the Moon.The tides would of course be exactly the same in a geocentricUniverse.Allen's concerns in the "Acceleration thread"To account for tidal influences, the centre-of-mass of the test-mass ofthe accelerometer, should coincide with the centre-of-mass of thefree-falling laboratory. Tides would have NO effect on such a set-up.But then the tidal effects on such a lab are very small in thefirst place:Less than a billionth per meter of the gravitational force keepingthe labAND the test-mass in orbit.That's all for now (sorry for such a long post), and I hope itclarified afew things. - Regner Receive Notifications of Incoming MessagesEasily monitor multiple email accounts & access them with a click.Visit www.inbox.com/notifier and check it out!-------- Martin G. Selbrede Chief Scientist Uni-Pixel Displays, Inc. 8708 Technology Forest Place, Suite 100 The Woodlands, TX 77381281-825-4500 main line (281) 825-4507 direct line (281) 825-4599fax (512) 422-4919 cellmselbrede@xxxxxxxxxxxxx / martin.selbrede@xxxxxxxxxxxx

-------- Martin G. Selbrede Chief Scientist Uni-Pixel Displays, Inc. 8708 Technology Forest Place, Suite 100 The Woodlands, TX 77381

mselbrede@xxxxxxxxxxxxx / martin.selbrede@xxxxxxxxxxxx

**References**:**[geocentrism] Re: Tides***From:*Allen Daves

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