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Planck's constant(h) is key to discover M coor system.

Author Message
Dan
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Joined: Dec 31, 1969
Posts: 315
Location: USA

PostPost subject: Planck's constant(h) is key to discover M coor system.
Posted: Wed Nov 07, 2012 8:50 pm
Reply with quote

edit 2-26-13 including title.

Dan: 07-Nov-2012 11:44:24 from MT
Planck's constant h is the key to discovering its own implied coordinate system . The first Moebius has to have it as its base measure. That must be its radius. Since a M is a circle based geometry, then 2pih = 2piradius. Hence all subsequent measurements and ratios of all sub divisions of our initial m will reflect 2pih. Ergo, we can deduce why c is 300,000 kms and what the width between the two sides of the surface of our present M universe by determining what it's generation number is. I suspect over 100. Determining why c = 300,000 kms will determine generation number.

11-25-12 I have been comparing Edge Width(EW) = h, my first hypothesis to letting the radius = h. Using EW = h would make the radius quite long as the angle from the center of a circle to the two opposite edges of the sides of the M surface, where the EW is 1/20,000 of the Maximum surface area(= 2DDPhi D= diameter) for a fixed circumference(conserved) Moebius whose opposite surface edges separate in a straight line until the fixed circumference stops them from going further; then let this 1/20,000 of MSA = h. However now the radius would be some very high multiple of h. This seems most unlikely, but we can use trigonometry to figure out the actual number of EW units in the radius since we are positing EW = h.

Necessarily the perpendicular bisector of this long thin triangle should be a rational number of h units implying the two long sides are not rational.

Now we can let the radius = h which simplifies things. With the above trigonometry calculation we can deduce the actual % of h that = EW. In turn I know that the actual EWs fraction of h must be 1/20,000 of the Maximum Surface Area(MSA) of a 2pih circumference Moebius. i.e. Does 1/20,000th MSA = result of our trigonometric exercise?

After I figured out the Max Surface Area Moebius equation, it was tested by embedding it in a cylinder(a coordinate system assumed from outside any Moebius). When so tested it was short by a 10,000th of the surface area. Since we have a double circumference the EW must equal 1/20,000th and the EW surface area is 1/20,000 times 2C or 1/10,000th of MSA.

Conclusions: A M is a circle based geometry that expands by self-division beginning with a radius of Planck's constant denoted h.

Its surface represents a simultaneous NOW which keeps track of matter by holding it 'close' with Gravity where G force is transmitted at < h bar.

Questions: There must be a base energy input that drives the expansion which keeps the two sides of the surface embodied in Matter/Anti-Matter separated

Isn't anybody going to check my reasoning?

Old title: Planck's constant( h) is the key. Addon: 11-25-12

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