Matrix Message 355

A "Matrix-Valid" Meter? ..

Michael Lawrence Morton Chronos Finally Jogged me Enough .. Sat Nov 23 15:04:37 2002 205.188.209.11 Well, Chronos .. you've nudged me to the obvious !! Please notice what Chronos wrote, from 2 postings immediately-preceding this one, on "GridPoint" BBS .. << Now being a little confused, I could give this some leeway, and we could be looking at anything from, say, the Grid Point of Mount Rushmore to 1.6666666666... and in that range, we find, for example, 1.641403174... 656.5612701 / 400 1.663089126... (1.641403174 / (Pi^2)) x 10 1.653668090... 89298.07632 / 54000 1.643745154... 887.6223994 / 54 = 2.22222222 / 1.351926225 Those are some of them... the last one is pretty tempting, knowing that 54 is fairly popular around the ancient UK landscape, and knowing that this is probably an imperfect from where square root of Volume of a Sphere 887.6223994 should apply just as it does at Gop Cairn's "older cousin," Silbury Hill... If I "graded" candidates on A: recognizability (the less I have to look something up, the better), B: ability to talk to Giza, and C: versatility, 1.644934066 may be the rightful winner... I also threw on extra points for doing a "Stonehengey" thing and being the square root of something, and because that is part of the relationship to Giza, really... And of course we can probably half of the candidates out of parts lying around... We may be crazy (because the cairn is not this deformed on the drawn map, and may never have been), but we have these 100m x 3.281 = 328.1 ft = intended R^2 / 10? = 328.280635 68m x 3.281 =223.108 ft = intended dh 72 Pi^3? = 223.2451921 12m x 3.281 = 39.372 ft = intended (2 Pi)^2 = 39.4784176 And I say this because the main thing I'm looking for help from, is the leader in this genera, Silbury Hill, Grid Point = dh Reciprocal Squared Megalithic Yard = 1.351926225 (WGS 84)... 328.280635 x 1.351926225 = 4438.111996 = dh Grid Longitude Stonehenge (WGS 84) / 2 223.2451921 x 1.351926225 = 3018.110298 / 10 = Perimeter Great Pyramid / 10 39.4784176 / 1.351926225 = 29.20160647 = Grid Point Stonehenge (WGS 84) x 10 What a nice bunch that is! And it would be so easy, to go 328.280635 / 2 = minor radius = 164.1403175, and say, aha! 100 times our Grid Point! Just like it would be so easy to say, Grid Longitude Silbury Hill 51566.20159 / Pi = 16414.03715... but that's exactly how I got into trouble over at Hatfield Barrow, isn't it? :-) You know what is worse, it that 51566.20159 / (Pi^3) = dh 1.663089126, which is also the ratio between the Grid Latitude of Silbury Hill 69713.70025 and 4191.819859, which more and more looks like the "keeper" on Hatfield Barrow, courtesy Michael L. Morton... And the Squared Radian / 10 is even more tempting when you consider what a series you can beat out of using the 1.351926625 Grid Point of Silbury 328.280635 x 1.351926225 = 4438.111996 328.280635 x (1.351926225^2) = 6000 328.280635 x (1.351926225^3) = 8111.557345 328.280635 x (1.351926225^4) = 10966.2271 Okay, maybe it *is* 328.280635 ?!?!?! >> Why haven't I realized .. I wonder .. (??) .. until "NOW" .. (??) .. that our "metre" ("meter" in American) .. that our "metre" .. was/is probably "intended" as .. simply the Square of the Generic Radian(deg) .. divided-by .. (10^3) ??!!! in Regular Feet ??!! I don't know. I_have_wondered about this .. or .. I've been "beating around this bush" for years. Somehow .. I've been "allowing" a mere 2 "thousandths-of-a-Foot" .. to "discourage" me from proposing this !!?? I find this to be incredible !! After all .. the "usual" figure .. to 3 decimal places .. in terms of conversion from metres to regular feet .. is 3.281 ... right ? Well .. if it "turns-out" to be 3.283 to the nearest digit in terms of 3 decimal places .. that's only a difference of 2 "thousandths" of a foot .. correct ? So; right now, as I sit here wondering about this .. I'm really perplexed as to "why" a mere 2 "thousandths-of-a-foot" .. has "stopped" me .. until NOW .. ??? .. from realizing that 3.28280635 Regular Feet .. is APPARENTLY the "true-and-intended" Metre !! I think Chronos .. in his above-referenced-posting .. has jogged my mind enough to recognize this !! Chronos .. I can't thank you enough !! The credit should not go to me alone !! We've apparently found the "true-and-intended" conversion between Metres and Regular Feet !! Chronos .. what do you think ? What does anyone think ? -- Michael Lawrence Morton(c) 2002 by mailto:Milamo@aol.com Michael Lawrence Morton ~ Archeocryptographer.

Randy Palmer 13 inches/ 33 centimeters Wed Dec 11 15:49:56 2002 137.28.62.35 Hi Michael, I thought that I'd compare your ideal metric conversion to the my ideal "Masonic" metric conversion which is based on 13 inches equalling 33 centimeters (2.5384615 cm/in = .393939393in/cm)(.39393939 x 33 = 12.999987 inches)and the ideal meter would be 3.2828282 feet long. I compared your idealized measurement: 3.28280635 feet per meter which yields .39393675 in/cm. Multiply that by 33 and you get 12.999911 inches - a difference of .000076 inches. 13/33 would make a handy conversion between the two units of measure either way - but would it be perhaps - too obvious? Randy Palmer

Michael Lawrence Morton Makes GREAT Sense !! .. Wed Dec 11 18:19:45 2002 205.188.209.168 Randy .. This is excellent !! Congratulations !! Wonderful .. yes .. 13 and 33 !! .. how very appropriate ! And the difference is "0.000076" of an Inch !!! How can anyone quibble with THAT ??!! Both of them work !! (IMO). -- MLM

Michael Lawrence Morton Look at the Royal Cubit , now .. Sun Nov 24 00:08:57 2002 152.163.188.2 Take the Royal Cubit {"Egyptian"} .. in Regular Inches ("British" Inches) .. and divide-by .. The Meter, in terms of Regular {"British"} FEET .. (20.62648063 / 3.28280635) = 6.283185309 You get a "high-precision" (2Pi) figure. Is this strong evidence, or what ?! NOW .. take the "Ideal" Fine Structure Constant .. its reciprocal ("inverse"), that is .. and multiply by The Meter .. (137.0778389 * 3.28280635) = 450 .. which is the_exact_ratio between .. the Grid LAT of The Great Pyramid of Giza {Munck, 1993} and the Grid LONG of the Greenwich Observatory in England !!! (89298.07684 / 198.4401707) = 450 OK .. now here's South Galactic Node .. (6.981317008 * 3.28280635) = (72 / 3.141592654) .. = W.Cydonia Longitude of "Cydonia City Square Center" on Mars !!! This is .. 22.91831181(min) W.Cydonia. Think about this .. you have The Meter, in Regular Feet, equivalent to .. the_Square_of the Radian-arc numerical value .. divided-by (10^3) .. in the_360_system !!!! So; you have Base Ten and (360 / 2Pi)^2 involved .. (360 / 2Pi)^2 / (10^3).. in Regular Feet !! This is very, very "suggestive" of 'standard' methodology involved in the calculation of Areas and Volumes, of course. How do you get the Area of a Circle, for example ? You_Square_the "radius" .. and then multiply by Pi. If I divide The Meter (in regular feet) by VEGA .. (3.28280635 / 3.183098862) = 1.031324031 .. a decimal- harmonic OF the Generic Area of a Circle .. assuming the Generic Radian(deg) AS "radius" numerical length. And .. another one, involving Rosslyn Chapel .. (2.908882087 * Pi * 3.28280635) = 30. I think we got it, Chronos !!! -- MLM P.S. .. OK .. I'm sure it won't take long for the word to get out to Koppel. I'll have to go get a haircut and a new shirt. Maybe a pair of pants. And underwear, while I'm at it. -- MLM

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