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We have just found how much of the Moon’s orbit the Moon eats up by accelerating into it. That is what precession is, by the postulate of curvature. Now let us find a number for 19 years. 4850.1 arc sec/yr = 92,152 arcsec/19yr But we have a remainder of about 2 hours, remember, so we have to make a small correction. That 2-hour gap is found at the end of a period of precession, and is the distance from precise conjunction. Therefore the gap is not defined by precession, but by normal orbital motion. The exact gap is 2.082 hours. At 1023m/s the Moon travels (in a curve) 7,667,600m in 2.082 hours. During that same time, the Earth has moved as well, pulling the Moon with it. At 29,780m/s the Earth travels 223,200,000m in 2.082hr. To solve, we must know where in its orbit the Moon is at Metonic conjunction, since we must know whether its relative motion is with the Earth’s motion or against it. It could be either one, and we must know in order to know whether to subtract or add at this point. It turns out that at Metonic conjunction, the Moon is nearer the Sun, so that we must subtract. [But this is only an estimate for use in this paper. A precise calculation requires a bit more rigor in this part than I feel required to do.] So, in two hours, relative to the Sun, the Moon moves about 223,200,000 – 7,667,600 = 215,530,000m = 297.2 arcsec We take the difference of those two corrections and subtract them from the 19-year number to get 91,854 arcsec/19yr Now if we divide one full rotation (1,296,000 arcsec/orbit) by that number, we obtain 14.109. In 19 years, the Moon eats up about 1/14th of its orbit. But as we have seen, in this case "its orbit" means its orbit around the Sun. The Moon is precessing relative to the Sun, since the angle we found initially is an angle to the Sun. This means that we must divide the Tropical year by 14, to continue our calculations. 365.242/14.109 = 25.887 days The Moon precesses an amount of curvature equivalent to almost 26 days every 19 years. That amount of curvature is relative to the Sun, therefore the curvature is (approximately) on the main orbital path of the Earth. But the Moon does not travel on that path. It travels a path about the Earth which curves much more quickly than the Earth’s path. We could do a lot of math to compare the two curves, but a quick analysis shows that 26 days of curvature on one path must equal 26 days of curvature on the other, since Moon takes the two paths simultaneously. Therefore, we apply the 26 days of curvature to the Moon's real path, and so appear to already have a full month by some reckoning. But what reckoning is it? That is not one month by either the sidereal or lunar reckoning, so we need one more step. What we need here is the Draconic calendar, which is a direct comparison of the Moon and Sun. It is also called the eclipse calendar, since it tells us when the Moon and Sun conjoin. Obviously this is the calendar we need, and the Draconic year is 346.62 days, as I said above. If we divide that into the Tropical year, we get 1.054, and if we divide 14.109 by 1.054, we get 13.386. Now, there are 13.368 Sidereal months in a Tropical year, so we have completed the math. In 19 years, the Moon precesses an amount of curvature equivalent to one full month, returning to its original position relative to both Sun and Earth. [25.9 days x 13.368 = 1 Draconic year, therefore 25.9 days as we found it above is a sort of Draconic month whose background is the Tropical year. My final step allows us to assign the 19-year curvature to the Draconic year directly, so that we obtain the correct number for the Moon relative to the Sun (instead of the Moon relative to the Earth or the Earth relative to the Sun)]. The small margin of error is caused by two things. 1) Using mean values. 2) In the correction of 2 hours, using the speed of the Moon relative to the Earth in one curve to estimate its distance travelled on a different curve (its orbit around the Sun). The two curves aren’t equivalent, therefore the distances aren’t exactly equal. We used the Draconic year as a correction, since we needed numbers that compared the Moon and Sun directly. I could have used Draconic numbers from the beginning, but the problem had been defined by Sidereal and Tropical numbers, and those numbers are what most people know. Therefore I have preferred to make the correction at the end, with a single manipulation. Of course I could have also used lunar or synodic numbers, and made other transforms, but the method I used above is the fastest. In conclusion, what we found is that the Moon precesses the equivalent of one full month every 19 years, minus 2 hours, and that this precession can be shown to be caused by the curvature of General Relativity and the time separation supplied us by the absolute speed of light. Since this precession causes one full orbit, relative to both Earth and Sun, the Moon must return to the same spot it occupied 19 years earlier. Although I have used a variant math to simplify the calculations, my math is strictly equivalent numerically and mechanically and axiomatically to that of Einstein. However, I have shown that it is much easier to apply, as well as being much more transparent as a matter of kinematics and dynamics. I might add here at the end that this is the first time in history that anyone has shown the mechanical cause of the Metonic Cycle. Since the time of Einstein, the world has been in possession of the tools to do so, but the difficulties of the tensor calculus made the solution of such a complex problem too daunting. Even with my simplified math, I needed several pages of equations and commentary. The tensor calculus would have turned this paper into 50-page treatise, like other GR solutions, and made it completely impervious to mechanical explanation or comprehension. I know that some will not understand how the Moon’s precession due to curvature can be so much larger than Mercury’s, but these readers simply haven’t yet understood Einstein’s field. GR, due to its axioms, must take into account two related but basically separate functions. Einstein represents both by curvature. The first is Newton’s acceleration, which Einstein expresses by curvature. The second is the time differential, which comes out of SR, and Einstein also expresses this relationship with curvature, since it is not linear once it is imported into the curved field. But only the first curvature disminishes with distance. Only in calculating this first curvature would you expect the Moon to experience less curvature relative to the main field. But using the time differential, you must expect the reverse. In this differential field, curvature increases with distance, for the simple reason that it has more time to curve. Or, the angle of curvature is greater since light bends all along the path. This is why the Moon’s curvature is greater than Mercury’s, regarding precession. This becomes very clear with my math above.
This sort of precession is not a phenomenon of the Newtonian field, which is why Newton couldn’t calculate it. It is a phenomenon of the time differential field, which increases with distance. That is why it is relativistic, and why it is added as a vector to the curvature of the Newtonian field. This second curvature is a differential field compared to the first (a rate of change of the first), is much smaller than the first, and increases with distance rather than decreases. If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |