Friday, 10 October 2008
St. Mary's Chapel, Mount Lothian grid contd.
So, having established That there is a grid in Lothian, based on the the exact same dimensions as that found at Bornholm, with St. Mary's Chapel and St. Baldred's Chapel on The Bass Rock forming the diagonal, which links with the side of a square through Dunsappie hill-fort and extends to Seafield Tower between Kinghorn and Kirkcaldy, I shall now show the other diagonal, north-west of St. Mary's Chapel. This diagonal can be fixed by Blackness Castle, and a natural sub-division is found, namely the side of the square fixed by Seafield Tower, becomes the diagonal of a smaller nested square, which shall then be used in the next quite astonishing development. For this exercise though, I shall use the North Berwick Law orientation, confirming that in the landscape, both are relevant. See below!
3275.00 6570.19 St. Mary's Chapel
3055.66 6792.48 Blackness Castle
Using Pythagoras' Theorem: 312.29 O.S grid units(100 metre), which converts to:
19.4046 miles(E), which is some 337 yards short of the exact figure of 19.596 miles(E)! Now as Blackness juts out into the Firth of Forth, the exact point is in fact off-shore, just like at Seafield Tower.
Through calculation the exact grid reference is found to be :
3275.00 6570.19 St. Mary's Chapel
3055.71 6796.88 Blackness(calculated point)
again, using Pythagoras' theorem: 315.40 grid units of 100 metres, which converts to: 19.598 miles(E), and allowing for the small rounding off in the calculations is good to 0.002 miles! (I have restricted the figures to two decimal places for convenience here, and to 3 decimal places in the final miles calculation! A discrepancy of some 10 feet. I trust this is acceptable! I normally work to 10 figures on the calculator!)
And the angle to grid north is 44.05 degrees west of grid north, which corresponds to ninety degrees difference to the North Berwick line, 45.95 degrees east of grid north.
The grid squares having diagonal St.Mary's Chapel to Blackness are shown below, with sides equal to 19.598 miles(E)/square root 2 = 13.858 miles(E):
The next section uses this smaller square as the grid unit for the next stage, which extends this grid in all directions.
Now, a final point for now, as I have just realized:
This grid square side length of 13.858 miles(E) is equal to 12.36 miles(S)(33/37 is the onversion factor, see explanation in previous posts, and why I here, always distinguish between the two systems by the (E) and (S).
Now, half of 12.36 miles(S) is 6.18, a harmonic of phi, or little phi, or 1/Phi!!!
This is a new finding, although I may have it in my notes, but I don't recall having found this previously in relation to this system, and has to be of significance, to my mind! Quite astonishing, but then again, that's nothing new in this whole research!
I shall work on this, and see what else is to be found!
Good grief, the time on my computer at this exact moment is 6.19am, BST!!!
I had just done some calculations, one of the dogs barked, and I checked the clock!!!!
This is what I found:
The Bornholm grid axis is 16*square root three miles(E). This equates to 24.71683315miles(S). This divided by 40 gives 0.617920828. The reciprocal is 1.6183303, which squared is 2.61899296. Now that is an approximation of Phisquared, or Phi^2. This multiplied by 6/5; or 1.2 is 3.142791552, which multiplied by 7 = 21.99954086, which is 99.998% of 22. 22/7 is a rough, and often used form of Pi.
So, this grid, and hence the Bornholm grid, in Scottish measure is based on a common form of Pi, and Phi! I had found some correlation with Scottish measure when I was working on the Bornholm grid, but nothing so convincing!
So the full factors involved must all resolve in some way:
[(16*sq.rt3*33/37*40)^2]*5/6 = 7/22; so; (16^2,*3,*33^2,*5,*22)/(37^2,*40^2,*7) = 1;
which resolves to (2^2,*3^2,*11^3)/(5,*37^2) = 47916/47915; which equals 1.00002087, the reciprocal of which is 0.99997913, equivalent to 99.9979% of 1.
Interesting exercise! Or perhaps I should get a life!?
Then again, astonishing find!