Why conical boulders

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Why conical boulders

Postby Dave Wood on Fri Sep 06, 2013 8:00 am

One question that has come to mind that doesn't seem to have been asked,having searched through the archives and not found anything is, why were conical boulders used for marking out Nolan's Cross and not spherical boulders which would have been far easier to manoeuvre into position.
Over the past 5 years I have been delving into the geometric mysteries of the Cross as well as the Triangle and their encryptions placed within both by a very clever 17th century mathematician who I would dearly love to know the name of.
Having found two that involved Cone A(Kether) but both involving different measurements an idea came to mind as to why those boulders were conical,for example place the Cone A at the required distance of 147ft from the centre of the Cross then tilt the boulder slightly away from the centre to give the distance from the Cross centre to the apex of the Cone to give 147ft 2inch and 7/10inch.The one boulder then gives two measurements,one at the base and another at the apex.
Plausible?

Dave
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Re: Why conical boulders

Postby Dave Wood on Fri Sep 06, 2013 8:32 am

The distance of 147ft gives the encryption and the second measurement of 147ft 2inchand 7/10inch enables the construction of a regular pentogram using the North-South line through the centre of the Cross.
Using the centre of the Cross and drawing a circle of 360ft radius(the centre to Cone C) and extending the stem of the Cross through Cone A to that circle gives the radius of 360ft.360ft - 147ft = 213ft and the other radius of 360ft along the stem towards Cone D giving the figure along that diameter of 213147360.
It probably doesn't mean anything to you,but when I realised what it was I sat in awe looking at the calculator saying to myself "Beautiful" in its simplicity.
213147360 as 2131473-60 = 2131413. 2=B 1=A 3=C 14=O 13=N Bacon.

Dave
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Re: Why conical boulders

Postby Dave Wood on Fri Sep 06, 2013 9:14 am

The construction technique for the regular pentagram(5 sides) using Cone A at the quoted distance,in that 360ft radius circle draw the True North-South line through the centre of the cross and where it intersects the circle at North draw a line from that point through the apex of Cone A to the other side of the circle and from that point draw a line to the centre.
The angle between that line and the stem of the Cross is precisely 72deg. and the resulting pentagram has an area of 308142sq.ft.which is the value of the name Christian RosyCross using the 16th century alphabet,note he hasn't used the name Rosenkreutz as spelt in the Fama Fraternitatis.
3-C 8H Rosy 72 + Cross 70=142 308142.
The name of RosyCross is also encrypted into the Welling Triangle which for personal gratification I have renamed Bacon's Headstone.

Dave
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Re: Why conical boulders

Postby Dave Wood on Sun Sep 22, 2013 2:08 am

Using the 16th century alphabet and the Simple Code the value of Francis is 6171133918 and Bacon 2131413 who was born in 1561 and died in 1626.I had a hunch that somehow and somewhere the designer of Nolan's Cross had managed to encrypt those figures as a whole and it took me 5 years to find it using two figures that had me stumped until recently, and the distances between the boulders of the Cross as revised by Petter Amundsen,they being from Malkuth to Cone E,282' then 294' to Cone D,429' to the centre of the Cross and 147' to Cone A.
294 x 429 = 126126 and 429 x 147 = 63063 are the two figures and the 63063 reading as 2x63's ie.263 added to the 126126 =126389.
My location is 132' from the stem of the Cross and 174' from the Cross arm formed by Cone B placing it below Cone A at 135'.
Replacing the 147' to Cone A with 174' then multiplying those figures along the stem 282x294x429x174= 6188750568 then adding the 126389 = 6188876957.Francis 6171133918+2131413+15611626=6188876957.
The other reason for those two figures and the distance to Malkuth of 282' is for the very ingenious construction of an extremely accurate Heptagon(7 sides).

Dave
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Re: Why conical boulders

Postby n4n224ccw on Mon Sep 23, 2013 1:32 am

Dave Wood wrote:The distance of 147ft gives the encryption and the second measurement of 147ft 2inchand 7/10inch enables the construction of a regular pentogram using the North-South line through the centre of the Cross.
Using the centre of the Cross and drawing a circle of 360ft radius(the centre to Cone C) and extending the stem of the Cross through Cone A to that circle gives the radius of 360ft.360ft - 147ft = 213ft and the other radius of 360ft along the stem towards Cone D giving the figure along that diameter of 213147360.
It probably doesn't mean anything to you,but when I realised what it was I sat in awe looking at the calculator saying to myself "Beautiful" in its simplicity.
213147360 as 2131473-60 = 2131413. 2=B 1=A 3=C 14=O 13=N Bacon.

Dave


I'm trying to follow along here.... why did you take 213,147,360 and remove the last two digits which resulted with 2,131,473 , then subtract the 60?

From all appearances it seems arbitrary, but I've followed your calculations long enough to know you are careful with number crunching, that is why I am at a loss to understand that manipulation or step.

Cheers
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Re: Why conical boulders

Postby Dave Wood on Mon Sep 23, 2013 2:22 am

Hey Paul,
If youhave followed me for that long you will know I don't explain myself very well,for want of uploading drawings.As you will see in further encryptions he has made use of the distances as a whole number as shown and used it as an equation by the addition of a +,-,/ or x,and in this case a -.213147360 as 2131473-60 = 2131413.In other cases he has mostly used x,for example which I will explain later he has taken the distances of 3071,218 and 287,combined them producing 3071218287 and and changing to 30712x18287=561630344 whichyou may recognise as a compass heading.
Hope that helps.
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Re: Why conical boulders

Postby Dave Wood on Thu Sep 26, 2013 12:32 am

Before I move on and explain why that particular spot was chosen for the location I have given which I have dubbed "F"(Francis) I would like to explain another example of encryption used.He has used distances again and by the inclusion of O's either added or subtracted both figures to attain the figure required.
The location F ,from the centre of the total geometry is 1486' at 23d26m West of True North and Cone C at 1967' at 30d W of True North.1486 as 1040806 and 1967as 1090607 added = 2131413(Bacon).
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Re: Why conical boulders

Postby Dave Wood on Thu Sep 26, 2013 1:39 am

The location "F" at 174' from the arm of the cross formed by Cone B and 132' from the stem of the cross has been positioned there for a number of reasons,one being a reason that I did not expect.Through interest I calculated the distances to each boulder forming the cross,they being from F to Cone A-135',B-287',C-522',D-617'and E-907'.
Cones A+B+C+D+E = 1561' which is the birth year of Francis Bacon had me wondering what else I would find by calculating the distances to the other boulders found buried by Petter Amundsen and the total of those distances worked out at 4047'.
The total of the distances to the cross come to 2468' and to it's centre 218' resulting in 6733' in total.
6733 is the numerical value of Francis Bacon-Francis 67 and Bacon 33.
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Re: Why conical boulders

Postby Dave Wood on Thu Sep 26, 2013 1:43 am

I hoped this would'nt happen but it eventually did,an amendment to the above.Cones A,B,Cand D total 1561'
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Re: Why conical boulders

Postby Dave Wood on Thu Sep 26, 2013 2:37 am

The distance to Cone D interested me the most as 617 is the beginning of Francis-6171133918 and from that I knew that the remainder of his name was somewhere in the geometry of the cross but over quite a period of time was unable to find it no matter what I tried.
When I discovered his technique for encryption as shown in the first example I found it within a matter of hours and it involved the distance to Cone C and the 174' to the cross arm from the location at F.As shown the total of the distances is 6733' + 174'=6907.Along with the 522' to C and the 617 to D,the figure of 5226907617 is produced and using that technique changing it to 5x226907-617=1133918.Along with the 617 gives 6171133918(Francis).
Incidentally the two distances from the cross centre and Cone B at 218' and 287' with 1 added to each,2181 x 2871 =6261651(1561626 in reverse.1561-1626)
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