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So the real question is: "When/where on circumference circle does the percentage decrease in AC = % decrease in AB, because past that point the angle ABC must steepen?" That point is not D. It is very close to C. It is the rate of curvature of circle that is of interest here = pi

BUT!, e and Phi are here. 'e' being easier as I am still sure that at D, the equal change in length point defines 'e' either with respect to the ratio created by perpendicular DE with horizontal Diameter, OR the ratio of circumference BD/DC.

So where is my Phi ratio?

(2-16-08))Redid my 'eyeball' drawing of how we find 'e' and Phi/phi = 1 by shortening ABC triangle in equal steps. ABC is based on the left quadrant of any circle. Got same result = We have 'e' defined geometrically by drawing perpendicular up from horizontal diameter to point on circumference where unit decrease in AC(vertical of triangle ABC) = decrease in AB(base of triangle ABC). Phi is another matter.

I did not go for this result, I backed into it when I saw that D, the equal change in length of base and vertical sides of our triangle ABC point; could define 'e' using perpendicular up from initial horizontal diameter that formed base of initial triangle. How serendipitous and how massively important to mathematics and modeling our universe. PLEASE VERIFY NOW!!!!!! (2-17-08) I think that after more 'eyeball' checking with pencil/paper/compass that it is the ratio of the circumference lengths AD/DC = 2.718. When true, we have navigation and location in a moebius strip shaped universe = disconnection from A-M debt.

(2-15-08) It is when the base AB of triangle ABC formed by left hand quadrant of any circle shortens faster as height AC decreases that is useful. Where on Circumference of circle this 'change in AB' = 'Change in AC' occurs defines the necessary logical/causal order of events on and between two sides of Moebius surface. They arrive at zero = surface of M simultaneously but AB must shorten(move) faster in last units of time to do so. This rate of change must be reflected in straight line of sight 'repulsive force' between matter lumps on opposite sides of surface.

New 1st sentence P 1 says better:"Help prove Geometric property of Moebius navigation equation. Proven true, we have FTL navigation to the stars. See eq. log article inside. (2-15-08)"

(2-14-08))"Moebius navigation discovery to be tested. Proven true, we have FTL navigation to the stars. See eq. log article inside. (2-14-08)". This Page 1 first sentence says what needs to be done: Outside testing and verification.

My original equation for predicting the maximum surface area of a Moebius(MSA) is 2DDPhi = MSA where Phi = 1.618 and D is the diameter of the circle the Moebius is based upon. When tested, it was off by 1/10,000. I quickly realized the test had assumed a zero thickness Moebius. That could not be true for a Moebius. It must have two sides that could not touch one another = not see through = not occupy same point in space at the same time. Since a M has a double circumference this implied a width of 1/20,000 of Circumference of M for the width between the two sides.

The real question is: How does a Moebius know what Phi is geometrically? Phi must be built into its base geometry? I have figured out the answer and we may also have a direct geometric link to Euler's Number 'e' built in.

I visualized the problem this way. the mid line of a 180 degree twist Moebius is a circle. Assuming a double surface at mid line, then looking along the mid line Circumference of Moebius, i.e. between the two sides along the mid line, one side would appear to be convex to the 'upper surface'.

1. Next draw a tangent to this convex mid line that goes equal amounts in both directions until it hits upper surface at both ends of tangent line.

2. From tangent point extend line ar right angles to tangent up to upper surface.

3. Draw the hypotenuse and we have created a right triangle with two unknown other angles.

QUESTION 1: Is the 'height' of triangle up to upper surface = to the 1/20,000 width????

QUESTION 2: Is the 'hypotenuse' of triangle = 1.618 of the 'Height" line? or is the "base line' of this right triangle = 1.618 of the 'Height' line?

HOW WE CAN TEST THESE ?s: 1. Draw a circle with two lines intersecting as a perfect square cross at 90 degrees, and use the intersection of 'cross' as center of circle

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2. Look at left hand quadrant and draw a hypotenuse that intersects ends of cross where they touch the 'circle'. This creates a classic 45 degree right triangle.

3. Lets denote the center of circle where the 90 degree corner of our triangle is 'A'. Next, denote horizontal base end touching circle as 'B'. And last denote upper end of triangle where it touches circle as 'C. Ergo we have created triangle ABC. With AB as your base of triangle and AC as verical height of triangle, and BC is hypotenuse.

What we are going to do is shorten triangle height AC in equal increments and see what happens to the two angles that touch the circle; and also see how the length of the 'base' AB changes as 'height' AC shortens in equal jumps. Finally we can of course, using the Pathgorean Theorem, deduce the hypotenuse BC of our triangle ABC.

WHAT DO WE DISCOVER AS the height AC shortens in equal jumps until it goes to zero at C on the circles(Moebius) circumference. Very interesting indeed.

First as AC, the vertical height of triangle shortens, AB our base also shortens but to a much lesser degree. In fact, not until around 5/6ths to 7/8ths of our quarter Circumference of our circle is traversed by the B end of our baseline AB of triangle ABC, does an equal unit shortening of AC, the height, result in an equal shortening of baseline AB. Lets call this 'equal shortening point' D. Only after D is reached does continuing shortening of AC result in an proportionately greater shortening of AB.

Strangely enough, as AC goes to zero, AB shortens with ever more rapidity, necessarily rapidly increasing the angle ABC to 90 degrees at AC = zero. Counter-intuitive at first.

REPEAT QUESTIONS: What actual percentage of initial 'radius' of circle = initial start length of AC, is AC when hypotenuse reaches 1.618 of shortened AC? Same question for when AB = 1.618 of a shortened AC?

About 'e' Euler's Number 2.718. Run a perpendicular off of AB baseline of triangle up to 'D' our equal change in length point. Denote bottom of this line as E. QUESTION: What is proportion of AD to initial AB? What is proportion of BE to diameter of circle? Looks very close to 'e' to me.

I will draw this up and put in coppermine shortly. I do confess not taking trigonometry, and my calculus is 30 plus years unused; but I visualize just fine. A 1.618 times AC hypotenuse is quite a steep angle implying quite thin distance between two surfaces.

A suggestion on method, as the length of AC approaches C, you also see the B end of AB approach A in bigger and bigger jumps, and both go to zero simultaneously. Reminds of Leibniz and Newton tiny little jumps to approach a minima and maxima. Here it is a double, but one stays the same while other increases in jump size. What is the constant ratio of change between these two line length changes? e?

Who will answer the questions first? Who will verify?

(2-12-08) My original equation for Maximum Surface Area(MSA) of a Moebius implied that a 'Moebius' somehow knew directly and geometrically the Golden Ratio Phi/phi = 1, it looks like I found it, how it works, and the implication for a straight line repulsive force between particles on opposite sides of surface. This force is implied by the No! No! requirement that +/- particles can not be perpendicular to each other, but the force is abruptly and cleanly stopped by any interposing of the surface between a straight line of sight between the particles.

Test; Dark matter thus would rotate around plus matter and vice versa as gravity is not affected, but the repulsive force would keep them separated from perpendicularity. I will have mathematicians test this first before i publish here. It is a common geometric property of a moebius and a circle as MSA of a Moebius is based upon a circle. We will see.

(2-08-08) The fixed point of reference the equation rests upon is the crucial element in tying together the equations 'elements', e.g. c, Euler's Number, zero, phi/Phi, unmoved mover, division by zero as a function (f) of x, take turns, in/out, odd/even defining inside and boundary, what the basic geometric +/- Not Do is, 'geometric origin of Moebius'(bowtie point), ratio of 'side edge to edge width' TO '2Circumferences'(is it two and why, see description of MSA Moebius below), and above all, THE FLOAT unit of time where we balance the equation using faster a than c speed between the two sides of the surface(this speed may vary with distance apart of +/- particles, the Circumference limit apart of any pair particle = there is always a C path between any +/- pair and nothing longer is possible across the surface of a moebius. I will add others as I remember them, but these are crucial.

Damn, that looks complicated, so lets cut to the chase and answer the Question, what is the fixed point of refeerence common to all of these 'real abstractions' (symbolic descriptions made out of 'reality' and used to describe ourselves. Answer is obvious, our physical spirit is, our consciousness, where we sit.

An equation is an equality, so what is it we are equating as spirits so as to keep our Life Forms = the body you are presently trotting down the road Alive? In order for us to be making bodies move, we split particles and send the minus half to our A-M debt, where we are anchored so we have real leverage to Cut the plus particle anchored in our body by our physical Spirit. ERGO, The two sides must be kept in balance, in equality, through time using a fixed float where ins and outs can balance regardless of when they arrived during float. In short a short period of time where we can rob peter to pay paul as long as we get the money back before the end of the "FLOAT". Thank God for free lunches. Enough for today. I did not realize I had all those pieces floating around in my head. Get us a ship people, the equation is coming together.

(2-06-08)

Today's Blog comment on how & why FR acts the way it does is crucial to finishing eq. QUESTION: What existing common plus particle can we split into Matter & Anti-Matter and then replace from the Sun?

(2-02-08) The eq can only be written in one way, left to right, top to bottom, its entry point is the 'float' = ins and outs must balance in less than one unit of time. So we start with defining zero as gatekeeper and how we use f(0) in that position. Remember we must take turns in going through zero as only one direction along a vector passing thru zero is open at a time.

(1-26-08) Proof of Constant. It is literally a physical movement geometric proof. It shows that all Perpendicular pair particles(P+/-) opposite each other on a Moebius surface are exactly the Circumference of the circle apart across the surface. Further no P+/- particles can ever be more than C apart in any direction. Required for a Present to exist in a conserved, yet expanding universe. (?) In fact, a present requires that they always be C apart along some vector across surface. (is this necessary?) Ergo, since C of a circle is known; then any difference form C apart along some vectors = when +/- particles are not perpendicular; then there actual locations are a function of C.

(1-25-08) All fundamental eq. are based upon a constant. e = mcc where c is the constant. I have discovered my constant and now can see all the implications for navigation etc. It is of course an objective characteristic of a +/- pair particle on a Moebius. It is a geometric relationship shared by all particle pairs and shows how the 'present' works and implies the speed between the two sides of surface. We denote our constant C versus c for speed light. With this constant we can tie together the equation using Euler's Identity and some judicious substitutions of Phi/phi and e - mcc and an upside down i for other side of complex plane and now C whioch is tied to Phi/phi. Plus I have discovered some more properties of a Moebius. i.e take a perpendicular paid particle on a Moebius. put on mid line have them separate in opposite directions and the pair will meet 180 degrees away on mid line. If they couyld continue they would meet again as a perpendicular pair only each would be on the opposite side from whence they started. Now send them in same direction and they do not meet but again end up as a perpendicular pair openly each is on the other side from whence they started. Ergo, an on off switch or a Nogo/Go switch.

(1-20-08)This article gave me the clue for the form of a Moebius eq. When I first realized in early 1991 we lived in a Moebius universe I went down to math dept. at UW to find the eq. for a Moebius in some math book. They did not have such an eq. I was right, it was in a math book somewhere, just not recognized as such.