The mystery of where the opposing centrifugal force is from has more or less been resolved. If I do not quite understand it, no doubt I am dense. But first I have to admit that on the page "Sir George Darwin corercted Galileo on tides" I gave an explanation which was in fact Lalande's, and was the one I had been given in my physics classes in high school. No doubt our physics teacher thought that Sir George Darwin's explanation was too much above our heads. As I am reading it (pp.63 ff. in J. Rouch's book) I wonder if he was not pretty right about that.
Moon and Earth revolve around a common centre inside Earth (and this explanation is apparently valid for Sun and Earth too, since in both cases Earth has a centrifugal force equal to the attraction of the star at the centre of Earth!) and that revolution of Earth around its centre with the Moon gives rise to a centrifugal force. This is the centrifugal force which is on each side roughly but not exactly equal to the centripetal force exercised on particles (such as water) on either side of Earth, Moonwards or opposite Moon (and never mind that on Moonwards side the attraction of the Moon would be centrifugal as to Earth quite as much as the centrifugal power of Earth itself!) - and for that matter Sunwards and opposite Sun. Which may coincide, may be opposite, may be at straight angles and anything in between. But so far Sir George Darwin as referred to by J. Rouch is only giving the explanation for the Moon.
J. Rouch mentioned that English scientists, as opposed to French ones, do not feel they are degrading science by divulging it to the public. I wonder if he had degraded the renommé of science if he had tried to divulge equally the explanation valid for the Sun.
I wonder if my science teacher did not give me Lalande's explanation because it is back in favour?
So, I could not derive from M/r2 - M/(r+a)2 he other formula 2Ma/3.
I am not bright enough in algebra for that. But I can test their equivalence. As I did for my surprising results which were however correct, though they gave me not the formula I was looking for.
Instead of inserting just any values, I will insert the right ones - that is the right ones according to astronomy. The distances r and a and therefore r+a and r-a are no problem. That has been measured by trigonometry that I admit is correct or, at least, not far off. It is a question of accuracy, not of possible wrong approach. And neither do I offer to make the values more accurate, nor do I see any point in doing so.
The kg values are however problematic for me who do not believe the heliocentric system, therefore not believe that Kepler's laws simply calque cosmographic reality and are simply explainable by Newtonian / Laplacean physics based on the celebrated formula F=GMm/r2. But problematic or not, I go along with it, so I have a value with which to test the accuracy of the algebra in reducing the formulas.
73,470,000,000,000,000,000,000 kg (M)
_____________________________________________
147,761,591,201 km2 (square of r)
minus
73,470,000,000,000,000,000,000 kg (M)
___________________________________________________
152,706,741,884.41 km2 (square of r+a)
Should give the same value as:
937,198,014,000,000,000,000,000,000 kgkm (2Ma)
______________________________________________________
56,799,407,896,273,199 km3 (r3)
I will confide to the computer whether they do or not.
The first formula gives 1610.16 g/m2 or 1.61016 kg/m2.
The second formula gives 1.65001 g/m2! Ouch, what is this? Did the calculator cut off a few zeros from the first number arbitrarily?
Better check it out by hand! Or wait, I might just shorten the zeros by dividing with a million on either side:
937,198,014,000,000,000,000 kgkm (2Ma)
______________________________________________________
56,799,407,896.273,199 km3 (r3)
OK, this time I get the result far too large instead: 165.001 kg/m2. Whatever!
Probably you can get the maths right even if I cannot. Or if my calculator cannot when I copy above and below values into it, deleting commas for the thousands and replacing the decimal point with a comma. Try it yourself, you are more bright than I. Let's go on.
Actually, I think I will test the equivalence of the formulas on simpler values. M=1,2,3; r=1,2,3; a=1,2,3 but r>a
1, 1, 1; M/r2 - M/(r+a)2 = 1 - 1/4 = 3/4
1, 1, 1; 2Ma/r3 = 2/1 ! Stuck at even first try !
Of course, the author added it works - only? - while a/r is very small, is that the clue, it being an approximation?
Let's make a=1 but M and r each=1000.
1000, 1000, 1; M/r2 - M(r+a)2 = 1000/1000000 - 1000/1002001 =?
1000, 1000, 1; 2Ma/r3 = 2000/1000000000 = 2/1000000
And to resolve first:
1000/1000000 - 1000/1002001
= 1002001000/1002001000000 - 1000000000/1002001000000
= 1002001/1002001000 - 1000000/1002001000 = 2001/1002001000
.... which is close to 2/1000000. But not exactly equal.
I will now stick to the first formula. Tidal force for Moon on object far side of Earth is then 1.6061 kg/m2. Times mass of earth times gravitational constant. Presumably the force on object same side of Earth is pretty equal. And tidal force of Sun is to it as 26/59, that is smaller.
And measuring tidal forces is of course a child's play. Seeing that any port has a tide, all you have to do is measure the tide and the amount of water to measure the tidal force. Right?
Oh, why measure the tidal force at all, we have after all calculated it from the known masses as well as known distances of Sun and Moon to Earth. Why bother to look up reality and see if it corresponds, since the theory is so neat? I mean a child could understand such reasonings as we have been dealing with, couldn't it?
Well, because it is a matter of science. And science means to look it up in reality, remember.
So, you measure the generating forces of the tides by looking at the tides in the ports. A smaller variation occurs every truly local noon hour and midnight hour. A larger variation - sometimes coinciding with the smaller - occurs every time the moon is straight above or straight below the waters. And when that is, obviously is a matter of when in the lunar month we are dealing with. Newmoon and fullmoon there is only one large high tide at noon and one large tide at midnight. That is called spring tide. At half moons - for astronomers known as the quarters - there is a smaller high tide at morning and at evening while the high tides of noon and midnight are somewhat smaller still. That is what is called neap tides.
And in between the high tide will be at nine o'clock morning and evening or at three o'clock after noon and in the morning. With a somewhat lesser peak at noon and at midnight.
This is what we see in every port, right?
Of course, J. Rouch does not actualy say that, but this is what the theory tends to predict right?
What J. Rouch actually says about the portal tides is so funny that I suppose (if we wish to defend this theory of tides that is, either Lalande's or Sir George Darwin's) I must have misunderstood it.
There is the Semidiurnal tide, like in Brest or on either side of the English channel. Fig. 5 on page 38 (for Brest) puts the peaks at 6 and 18 hours and the lowest levels at noon and at midnight. Of course that could be that particular day in the month, but J. Rouch does not mention that fact.
There is the Diurnal tide anywhere in the Mexican Gulf, or for that matter in Copenhagen. The diagram given for Pensacola puts peaks at six o'clock one day and starting six o'clock next day but lasting further on to noon. Lowest points somewhere between 18 and midnight.
There is the Mixed type which gives really funny patterns for San Diego, San Francisco and Honolulu, all of them in the Pacific Ocean. I mean, if the usual explanations about what happen at tides are given in San Francisco, to young people who can then observe the tides, I do not wonder at all if New Age abounds in San Francisco. Seeing that conventional Christianity is not there and now (or for that matter here and now) in the habit of giving spiritual explanations to material phenomena except at a very far distance when God set creation started. Because, what the diagram shows for San Francisco is at least as compatible with Mermaids sensing the passages of Sun and Moon and dancing to it, with some freedom of movement, as it is with gravitation causing it all.
You see, the tides that really correspond to the aboriginal tidal forces to which we have now given the formula as far as I was able to make it through the algebra and the explanations as far as I was able to follow them are the tides on the Ocean. There is where we measure how high or low water is directly due to Sun and Moon. What happens in ports is a resonance phenomenon, something related to the tides over deep Oceans but not identical to them.
And if the resonance for a Moon tide moving around the clock around the lunar month is funny insofar as it is not moving around the clock around the lunar month (unless I got that wrong, that is), it is at least possible.
Think of the formants due to which we have the vowel sounds. They are very clearly defined, whichever be the tone we are singing at. If F1 and F2 are far apart we get an Uh or an Ah as in Cup or Father, for instance. If both are low we get an OO, if both are high we get an EE. And so on. And we can sing any vowel on any tone of the scale, remember.
Something like that may be the case in portal tides. Unless of course I misinterpreted the information in the book of J. Rouch.
But if measuring the portal tides does not confirm heliocentric astronomy with its gravitational explanations very directly, at least measuring the tides in the Oceans will do it, right?
And I suppose that has been done by satellites, right? Or maybe deduced from currents?
Because tide meters are no use at all in the middle of the Ocean. A floater on top of the water needs a thing attached to the bottom to use as a measuring rod. You get that in ports but not in the Mid Atlantic. And a pressure chamber at the bottom needs to send its measurings to someone above water. That is also more feasible in ports than in Mid Pacific outside the Island Belts. Plus the fact that it is easier to measure a difference in water pressure between 6 and 8 metres of depth than between 10.000 metres and 10.010 metres of depth. At least I think it is.
So, the distances Earth-Sun and Earth-Moon are known. If we assume the masses are correct for Earth, Moon and Sun as well, not forgetting gravitational constant, we can explain by F=GMm/r2 the attraction of Sun an Earth and of earth on Moon keeping them in orbit and the primary tidal forces by GMm/(r-a)2 - GMm/r2 on close side and GMm/r2 - GMm/(r+a)2 on far side, but do not know really whether the difference of GMm/r2 stands directly for the heavenly body's attraction on Earth as a whole or whether it stands for a centrifugal force in Earth. Neat enough, but neither side of this neatness can be very accurately independently confirmed as far as I have so far seen.
And that is somewhat less neat for people pretending that it is child's play to prove from tides and from planetary and satellitary orbits that the bodies have such and such masses and from masses that the gravitational explanation is both true and requiring Earth to move around the Sun.
Actually, it is a pretty hard science insofar as it is science at all. In between the theoretical primary tidal forces and actually observed tides, you get at least five variables:
- Principal semidiurnal lunar wave or M2, 28.98° per hour, 12.42 h. between them
- Principal semidiurnal solar wave or S2, 30.00° per hour, 12.00 h. between them
- Elliptic mean lunar semidiurnal wave or N2, 28.44° per hour, 12.66 h. between them
- Diurnal lunisolar wave or declinational or K1, 15.04° per hour, 23.93 h. between them
- Diurnal lunar wave or O1, 13.94° per hour, 25.82 h. between them.
And when you have the relative strengths of those for any port, you can calculate the tides with a certain accuracy.
I would have understood this much better if I had been offered a real course on tides when I was asking. All I got was some ten minutes to half an hour about Lalande's explanation. I asked how it applied to neap tides ... I was not satisfied and the teacher had no time to go on. That is what comes of scheduling all pupils to same curriculum, whatever their interests, school becomes unsatisfactory since never making a profound transfer of interesting knowledge.
Now, with issues as complex as these, there is a certain chance or risk or whatever that we are being bamboozled. Just as we very probably must trust the scientists not only on facts they start out with but also on calculations, unless we are good at mathematics, we are supposed to accept that we must similarily trust the scientist on the side of logic.
No, we are quite free to ask whether these tides, hard science as they are in themselves, are not very flimsy as proof of gravitation being the true explanation of both Heliocentrism (which is therewith supposed to be truly explained and truly proven) and of Tides. Because tides as we see them are a good deal removed from the astronomical factors.
But if you do wish to indulge in gravitational explanations, I will give you another video about a theory of Geocentrism where that is taken into account: Robert A. Sungenis builds on a theory of a hollow universe actually working and adds that by divine intervention Earth is in that void in the middle.
The alternative seems to be a cosmology quite unacceptable with the three dimensions of space "wrapped around a fourth one".
Hans-Georg Lundahl
Georges Pompidou Library
St Lateran's Inauguration day
which is Mother and Head of
All Churches in Rome and on Earth
9-XI-2012
Video with Sungenis:
http://www.youtube.com/watch?v=EMr8lb2tYvo
Video with Phil Plait:
http://www.youtube.com/watch?v=nw5W3CszeAI
To me Sungenis makes at least as much sense, except that he is presupposing gravitational explanations are really and truly proven, where I am doubtful, and he does not mention spiritual explanations although to us as Christians believing in God Creator of the Visible and Invisible things that explanation is fully available and as for stars at least favoured by the book of Baruch.
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