From the 2nd page of the Foreword to WD Gann's "The Tunnel Thru the Air" - I call this "Robert Gordon's 7 days challenge." Invert the above equivalence; If one solves the mystery of RG's 7 days, then one will understand why Jesus rose on the third day and rested on the seventh [didn't God rest the seventh day, a meaningful nuance perhaps] and man's last great enemy, Death, will be overcome. Quite a challenge for Robert Gordon.
Background; the inscrutable and intentionally misnumbered Chapter 39. The hero of WD Gann's TTTTA, Robert Gordon, circumnavigates the earth in less than a week (6 days and 5 hours according to a person sitting on a park bench in NYC watching him depart and return, but 5 days and 5 hours according to the number of sunrises RG would have seen) after the final battle of the Great War and destroys buildings of the great cities of the enemies of the US. Including both end points (NYC departure and NYC return), there were 22 stops along the circumnavigation route as RG stopped, destroyed buildings of enemies or conveyed appreciation among allies. When one analyzes the "crow flies" distance of his trip and then parse it into East/West miles and North/South miles, you end up constructing 1) an isosceles triangle if measured at the latitude of NYC and 2) a right triangle if measured relative to the equator. RG's 7 days is, on the surface, a study in Euclidian geometry measured according to Cartesian coordinates (a "system of DesCartean coordinates). The math is so perfect as to defy one who might deny Mr. Gann's conscious contrivance. Following is the math I developed in posts beginning several months ago:
Robert Gordon's 7 day trip covered 33,543 (columns 'c', red cell) miles as the 'crow flies' computed using both the Haversine formula for spherical distance and according to the many distance calculators available on the Internet. Parsed by absolute direction and absolute miles, he traveled 18,213 miles East/West and 18,495 North/South (see the two green cells), again computed both by Haversine and Internet calculators. With the E/W distance so closely corresponding the N/S distance, we have an approximate 18/18/33 conceptual isosceles triangle….or a coincidence so saith the denier. [One might note the distance around the earth at the latitude of NYC is 18,842 according to the Haversine formula.]
Let's eliminate coincidence. RG's trip began and ended at NYC (NYSE latitude 40.7069) so all of the N/S calcs above were from latitude 40.7069. Thinking as an Einsteinian relativist would (I think), time and space are different at the equator relative to the 40th parallel. As the earth turns, a person standing on the floor of the NYSE is traveling at a speed of 785 MPH (18,842 miles / 24 hours) relative to a person standing exactly on the equator who is traveling at a speed of 1,037 MPH (24901 miles / 24 hours). Obviously, the person at the equator travels farther in 24 hours than the person at the 40th parallel. There's a nuance there, I'm sure. We know what happens when distances are computed at the both parallel, so what happens if distance is measured at the equator. So let's measure RG's N/S mileage relative to the equator and we find it to be 47,117 miles (bottom of column "Miles to Equator" in the above table). Not, in itself, meaningful.
Contrivance. On pages 400 and 401 found in the inscrutable chapter, Mr. Gann gives us the numbers 100 and 60 which produce a ratio of 1.666. Nice number in its own right. But just for fun, use 1.666 to reduce the Equatorial distance of 47,117 to 28,271 miles….no reason, just trying to find a use for those extraneous numbers Mr. Gann gives us. Now we have E/W mileage of 18,213 and N/S mileage of 28,271. Since E/W and N/S are perpendicular relations, we have two sides of a right triangle and the Pythagorean hypotenuse is 33,630. Recall the very first figure in RG's 7 days was his 'crow flies' miles of 33,543. Mr. Gann contrived RG's 7 days to provide a Euclidian (3 dimension) and Pythagorean right triangle.
Mr. Gann spent a LOT of time picking the cities he picked for RG's travels to create an isosceles triangle AND a right triangle that culminated at the end of RG's trip. I cannot imagine running the numbers to do this without an Internet web site or a calculator with advanced functions to calculate the Haversine formula. That's what he faced in writing TTTTA. There must be an important lesson in RG's 7 days as promised by the Foreword.
You are now up to speed on what I had discovered many weeks/months/days ago. But then I discovered something more about two days ago and realized this morning that it infers the cubic structure I've been looking for. The isosceles and right triangles are in planar geometry….X, Y and Z. But, if we dig deeper, the sheer numbers may infer more than planar geometry. Perhaps X, Y, Z and T?
We are dealing with a sphere and time according to the rotation of that sphere. And we are dealing with two triangles in planar geometry…angles as opposed to curves. Of course, spherical trig distances by the Great Circle method are determined by bisecting the globe with a plane. But that's still the sphere and the plane. I'm looking for the sphere and cube. I know the sphere is the earth but where did Mr. Gann bury the cube in RG's 7 days? I expect he did, somewhere.
Let's do some math on the isosceles triangle. Make it a perfect isosceles by averaging the 18,213 and 18495 to gives you and average side length of 18,354. Like any isosceles triangle, when you bisect the angle between the equal sides, that bisector intersects the opposite line at 90 degrees and you have the right triangle…perfectly one half of the isosceles triangle. On either side of our new right triangle we have 18,354 and 9177 (half of 18,354). And we use Pythagorus to calculate the hypotenuse at 20,520:
To recap above, the computed hypotenuse of 20,520 divided by the smallest side of 9,127 is 2.23607 which happens to be the the diagonal of a 2X1 rectangle. If you square 2.23607 you get 5.0000 to as many decimals as you want to expand your Excel spreadsheet. [If you use the pre averagee numbers of 9106 and 18213, you get 5.0004398.] This isn't a coincidence, IMO, any more than was the isosceles and right triangles. It is contrived.
I realized the implication of the 2.236 this morning, April 30. I am not sure where this is going because a rectangle is twice the cube I was expecting to find. Relative to a sphere that has a circumference of 18,213 or even the circumference at the Earth's center of 24901, this rectangle would seem to be far larger.
On the other hand, RG's Great Conceptual Triangle infers an angular geometric solid. Four equal sized rectangles stacked contiguously and aligned, after all, gives you the 4X4 cube and far bigger than a sphere with an 18,354 diameter. Still, its an intriguing thread. There's an answer, and when I find it, I'll call it Robert Gordon's "Earth Cube."
With, then, the East/West sphere of the earth calculating time and the square of the North/South cube inferring space (matter, price, etc.), perhaps we have the four dimensional calculator we need to determine time and price.
Hmmm, and does not all this involve the square, the circle and the triangle, the logo of Lambert-Gann? The sphere and the cube? Where's the Einstein Essay Contest Editor when you need him? But that's another essay thread.