Matrix Message 379

4R and "Full NIBIRU Cross" Find NIBIRU's Grid LONG ..

Michael Lawrence Morton 4R and "Full NIBIRU Cross" Find NIBIRU's Grid LONG .. Tue Feb 18 14:28:01 2003 64.12.97.10 If you take the "Full-Octave NIBIRU Cross" .. which is found in the_line-of-sight_'Y2K' data, encoded_in_the ASM .. you can "find" NIBIRU's 'Y2K' Ecliptic-Grid LONG. By "Full-Octave NIBIRU Cross" .. I mean .. the 5 'nodes' of_line-of-sight_sky-location .. at 'Y2K' .. using the Ecliptic-Sidereal Zodiac data in the "Astrolog 5.30f" fixed-stars software .. http://users.cwnet.com/~sidereal/mag/astfixst.htm {{ Please see my explanations of how to treat that data, in order to translate it to 'ASM' values .. https://matrix-messenger.tripod.com/index }} The 5 nodes of NIBIRU's orbit-path (line-of-sight) are .. 1) Aphelion-SIRIUS .. 2.368705056 2) Incoming-ecliptic-crossing .. @ W.Galactic Cross-Node .. 7.5 3) Perihelion .. intersection of ecliptic-latitude of ALTAIR (in AQUILA constellation) and ecliptic- longitude of VEGA .. (180000 / 17225.70927) = 10.44949716 4) Outgoing-ecliptic-crossing .. @ E.Galactic Cross-Node .. 281.4477322 5) Back to SIRIUS, again .. completing the orbit-cycle .. 2.368705056 TOTAL Multiplied-Composite for this full orbit-cycle, (line-of-sight) .. encoded as of 'Y2K' in the ASM .. (2.368705056 * 7.5 * 10.44949716 * 281.4477322 * 2.368705056) = 123758.8838 That's the "Full-Octave NIBIRU Cross" figure. ---------------------------- Now .. think of this full NIBIRU orbit-cycle as encompassing 4 "mean" AUs .. because this "mean" AU in my 'prediction' is_57.29577951_"Mean" AU-distance for NIBIRU. From SIRIUS .. (line-of-sight) .. the "round-trip" encompasses 4 AUs .. if you envision this as "segments" of AUs .. elliptical, yes .. but the end-result or *destination* still "traverses" 4 AUs, in terms of segments. (4 * 57.29577951) = 229.183118 = (720 / 3.141592654). And_720_is the total "corner-angles", in the 360 system, on the_surface_of a Tetrahedron. This is 4 equal faces .. assuming a Regular Tetrahedron .. made-up of 4 equilateral-triangles of 180 deg each .. on the surface. So; (4 * 180) = 720. At Cydonia (MARS) .. I've proposed a W.Cydonia longitude of .. (72 / 3.141592654) at the_center_of the "Cydonia City Square". This is in terms of_arc-minutes_to the West of the_center_of "The D&M Pyramid" at Cydonia. [Munck found the "D&M" to be the MARS prime meridian 'marker' in the ASM/Pyramid Matrix (1993)]. Now .. notice .. (123758.8838 / 229.1831181) = 540. And 540 is the Ecliptic-Grid LONG at 'Y2K' for .. the sky-location of NIBIRU .. according to my 'prediction'. AND .. 540 is encoded at Cydonia .. as the Grid Point Value at the_center_of "The Fortress" .. {Morton, 2000; Internet}. {{ Ya don't suppose the 'Anunnaki' had anything to do with what's at Cydonia, do 'ya ??!! }} -- Michael Lawrence Morton ------------ Michael Lawrence Morton Bit of clarification on "4 AU", here .. Wed Feb 19 23:52:45 2003 205.188.209.16 Just thought I'd add a bit, here .. hopefully as clarification on the "4 AU" in the immediately- preceding post. Think of an Ellipse .. representing the orbit-path of NIBIRU. At one "end" of this Ellipse .. think of a "dot" .. and let's say that's SIRIUS .. which is also the "line-of-sight" aphelion-marker-star .. as seen on line-of-sight_from_'EARTH'. Now .. visualize NIBIRU going 'clockwise' .. from the "dot" .. along this Ellipse .. and your "viewing-perspective" is 'from-north', here .. you are way-up-north of the Solar-ecliptic, now .. not on EARTH. You are "looking-down". OK .. now .. visualize NIBIRU moving around to the 'opposite-end' of this Ellipse. How else_could_you have gotten there, via "drawing a straight line" ? Yes .. by simply drawing a straight-line_diameter_from your starting-place to the opposite-end of this Ellipse. That distance .. that_diameter_you 'drew' .. is 2R .. '2 Radians' in AUs .. (57.29577951 * 2) = 114.591559 AUs. (That is; according to my recent 'prediction'). Sol .. the Sun .. would be right in the middle of this Ellipse .. right at the_mid-point_of that_diameter_there. I've predicted that the "mean AU" distance of NIBIRU .. from SOL .. is 57.29577951 AUs. Now .. back to the 'opposite-end' .. where I left you. You now complete the "orbit" .. going clockwise .. back to your starting-place. You_could_have turned around, at the 'opposite-end' .. and done a "180" .. drawing a straight-line_diameter_back along the 'same' straight- line path. You "drew" two straight-line segments .. out, and back. How many_AUs_did you traverse ? Out .. was 2, and back was 2. That's 4 AUs .. (4 * 57.29577951) = 229.183118 I hope that helps clarify what I was getting-at, in the immediately-preceding post. -- Michael Lawrence Morton ------------ Michael Lawrence Morton *CORRECTION* .. sorry !! .. please make note .. .. Thu Feb 20 00:05:45 2003 205.188.209.171 Sorry about that !! Sol is_NOT_in the middle of that Ellipse !! Sol is_THE OPPOSITE-END POINT_there !!! (I wish I could literally erase that mistake). The_MID-POINT_on that "diameter" straight-line .. is 57.29577951 AUs from Sol .. and is also 57.29577951 AUs from the "dot" (starting-place) which, as viewed on_line-of-sight_from EARTH .. aligns with SIRIUS .. i.e. only on line-of-sight .. (it does NOT go out anywhere near SIRIUS). SOL is 114.591559 AUs from the "dot". I hope that is clear. -- Michael Lawrence Morton ------------ Michael Lawrence Morton MORE Correction needed ! .. Please bear with me .. Thu Feb 20 01:04:32 2003 205.188.209.51 Well; I'll get it eventually, I think .. I hope. I need to correct this 'scene', again .. sorry. (Sometimes this happens .. on the road to truth). OK. Back to the Ellipse .. as visualized in the preceding parts of this thread. Re-draw your Ellipse, first of all. Now; put the "dot" in the same place you had it, before .. at one end of the Ellipse. (Either end). We'll call that the *aphelion* node of NIBIRU's orbit. Next; visualize NIBIRU moving clockwise .. on the Ellipse, around to the 'opposite' end of the Ellipse. Now .. *that* point is the *perihelion* node of NIBIRU's orbit path. SOL's location .. is_INSIDE_this point .. INSIDE that point of the orbit-path. How_far_inside ? I've predicted .. by 2.354491542 AUs inside that perihelion-node of NIBIRU's orbit. Then; understand .. that the_straight-line_distance from_SOL_to the "dot" (the aphelion node) .. is .. 2R .. (2 * 57.29577951 AUs) or .. 114.591559 AUs. So; the TOTAL straight-line distance, in AUs .. from the "dot" to the *perihelion node* .. is .. (114.591559 + 2.354491542) AUs. I hope this is (finally) clear .. and, my apologies. Now, then .. please notice that the_2.354491542_is my predicted "Perihelion AUs" for NIBIRU's orbit-path. Because this portion of NIBIRU's orbit (according to my prediction) is "on the other side of Sol" .. albeit north (above) the Solar-ecliptic .. whereas the "dot" is south (below) the Solar-ecliptic .. I will "symbolically-mathematically" follow the "suggestion", here .. by now_DIVIDING_the "4R" portion of this orbit .. which is "from the 'dot' to Sol, and *back* again" in 2 straight-line segments .. (shortest distance between_Sol_itself and aphelion-node) .. by .. the distance of the "perihelion AUs". (I'm not "going" to-and-fro the aphelion part .. I'm only 'gauging' its distance as I do a little on-the-node "180" at Sol .. (-; .. heh-heh .. you like the way I dance around THAT ? .. (-; ..) Now .. "Are we getting there ?" .. (229.183118 / 2.354491542) = 97.33868822 Talk about hitting a biggie !!! That's the "Inner" {R.Carl, 2003; Internet}_DIAMETER_of the *Sarsen Circle* of STONEHENGE in Regular Feet .. {Munck, 1993}. AND .. 97.33868822 is_also_the "Giza Pyramids Composite" Ratio {Morton, 2001; Internet} .. which is .. (Cheops * Mycerinus) / Chephren .. = (248.0502134 * 2261.946711) / 5764.166073 .. = (561076.3643 / 5764.166073) .. = 97.33868822 {Munck, 1993 .. individual Grid Point Values for these 3 main Giza pyramids .. http://www.pyramidmatrix.com} -- Michael Lawrence Morton ------------(c) 2003 by mailto:Milamo@aol.com Michael Lawrence Morton ~ Archeocryptographer.

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