Bob Ware (5 Feb 2007)
"Lord Jesus Christ and the 400th Anniversary of Jamestown Cross"


 
When the circumference of a circle equals the sum of the Jewish day numbers for the two simultaneous dates: 4/5/07 and 4/6/07 (Good Friday) then the diameter equals 2211 > 2 Kings 2:11 > Elijah taken up.
 
7 circles will fit perfectly within this circle.
Each of these 7 circles has a diameter of 737.099 and a circumference of 2315.667.
 
Make 2315.667 the perimeter of a Golden Rectangle and draw a circle around this Golden Rectangle.
The area of each one-degree segment of this circle is 1544.
1544 is the sum of the ASCII codes for 'Lord Jesus Christ'.
 
Place a triangle and a square within this circle.
Here are the lengths of each arc opposite these sides:
 
        Triangle side > 880.939 > 881 > 153rd prime number and the exact center of the 'Prime List'.
        Square side > 660.704 > 661 > the 10th Star of David and last number in column two of the 'Prime List'.
        Short side of Golden Rectangle > 465.685.
 
                Sum of these three arcs: 880.939 + 660.704 + 465.685 = 2007.328 > 4/29/2007.
       
The arc opposite the long side of the Golden Rectangle is 856 which equals 8 x 107 > the rapture number.
856 + 661 = 1517 > the year the Ottoman Turks took control of Jerusalem and held it for 400 years.
On 4/29/2007 (Julian date?) it will be 400 years since Rev. John Hunt erected a cross at Jamestown and dedicated the settlement to the Lord. I believe that the corrected Gregorian date of this Jamestown cross anniversary is 5/14/2007. This will be the 59th anniversary of Israel's rebirth.
 
*******
 
http://bobware1998.ameranet.com