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Sacred Geometries powered by Google Maps
Locate weird geometrical shapes on the places you love
Tool created by Mariano Tomatis
The tool here provided can be used to find weird (maybe sacred?) geometries on maps provided by Google®, starting from two points defined by the user.
1) Specify in decimal format latitudes and longitudes of the two key points of your geometry (use a DMS-to-decimal converter if you need it).
You can verify the correctness of your coords by clicking "View" buttons.
2) Select one of the four shapes from the icons on the left and click on it, to draw it on the map starting from the two points.
How to get latitude and longitude from Google Maps
Center the map on the point you want and paste in the URL bar of your browser this command:
javascript:void(prompt('',gApplication.getMap().getCenter()));
Default settings
Default settings refer to the Magdala Tower in Rennes-le-Château and the Blanchefort Castle, both in the French region of Aude. According to Henry Lincoln, the mountain top of Rennes-le-Château and of four other adjacent mountain tops precisely mark out a pentagram, the tips of a five-pointed star that stretches across the landscape.
Coordinates
Starting from the two points specified by the user, the calculation of vertices is performed with trigonometrical formulas, with a proper correction because of the spherical surface of the Earth.
At first, the longitude of each point is divided by cosine of latitude. Then, the coordinates of other points are calculated. At the end, all points are multiplied by cosine of latitude.
Triangles
Two points on a map define two distinct equilateral triangles.
Given the coordinates of the two points (ll1,lt1) and (ll2,lt2), points #3 are defined this way:
ll3 = cos(120°)×(ll2-ll1) - sin(120°)×(lt2-lt1) + ll2
lt3 = sin(120°)×(ll2-ll1) + cos(120°)×(lt2-lt1) + lt2
ll3 = cos(120°)×(ll1-ll2) - sin(120°)×(lt1-lt2) + ll1
lt3 = sin(120°)×(ll1-ll2) + cos(120°)×(lt1-lt2) + lt1
Squares
Two points on a map define three distinct squares (they may define a side in two different ways A,B or its diagonal in one way C).
For triangle C, it is useful to define the median point (llM,ltM) as ((ll1+ll2)/2,(lt1+lt2)/2).
ll3 = cos(90°)×(ll2-ll1) - sin(90°)×(lt2-lt1) + ll2
lt3 = sin(90°)×(ll2-ll1) + cos(90°)×(lt2-lt1) + lt2
ll4 = cos(90°)×(ll3-ll2) - sin(90°)×(lt3-lt2) + ll3
lt4 = sin(90°)×(ll3-ll2) + cos(90°)×(lt3-lt2) + lt3
ll3 = cos(90°)×(ll1-ll2) - sin(90°)×(lt1-lt2) + ll1
lt3 = sin(90°)×(ll1-ll2) + cos(90°)×(lt1-lt2) + lt1
ll4 = cos(90°)×(ll3-ll1) - sin(90°)×(lt3-lt1) + ll3
lt4 = sin(90°)×(ll3-ll1) + cos(90°)×(lt3-lt1) + lt3
ll3 = cos(90°)×(llM-ll1) - sin(90°)×(ltM-lt1) + llM
lt3 = sin(90°)×(llM-ll1) + cos(90°)×(ltM-lt1) + ltM
ll4 = cos(90°)×(llM-ll2) - sin(90°)×(ltM-lt2) + llM
lt4 = sin(90°)×(llM-ll2) + cos(90°)×(ltM-lt2) + ltM
Pentacles
Two points on a map define four distinct pentacles.
ll3 = cos(72°)×(ll2-ll1) - sin(72°)×(lt2-lt1) + ll2
lt3 = sin(72°)×(ll2-ll1) + cos(72°)×(lt2-lt1) + lt2
ll4 = cos(72°)×(ll3-ll2) - sin(72°)×(lt3-lt2) + ll3
lt4 = sin(72°)×(ll3-ll2) + cos(72°)×(lt3-lt2) + lt3
ll5 = cos(72°)×(ll4-ll3) - sin(72°)×(lt4-lt3) + ll4
lt5 = sin(72°)×(ll4-ll3) + cos(72°)×(lt4-lt3) + lt4
ll3 = cos(72°)×(ll1-ll2) - sin(72°)×(lt1-lt2) + ll1
lt3 = sin(72°)×(ll1-ll2) + cos(72°)×(lt1-lt2) + lt1
ll4 = cos(72°)×(ll3-ll1) - sin(72°)×(lt3-lt1) + ll3
lt4 = sin(72°)×(ll3-ll1) + cos(72°)×(lt3-lt1) + lt3
ll3 = cos(144°)×(ll2-ll1) - sin(144°)×(lt2-lt1) + ll2
lt3 = sin(144°)×(ll2-ll1) + cos(144°)×(lt2-lt1) + lt2
ll4 = cos(144°)×(ll3-ll2) - sin(144°)×(lt3-lt2) + ll3
lt4 = sin(144°)×(ll3-ll2) + cos(144°)×(lt3-lt2) + lt3
ll5 = cos(144°)×(ll4-ll3) - sin(144°)×(lt4-lt3) + ll4
lt5 = sin(144°)×(ll4-ll3) + cos(144°)×(lt4-lt3) + lt4
ll3 = cos(144°)×(ll1-ll2) - sin(144°)×(lt1-lt2) + ll1
lt3 = sin(144°)×(ll1-ll2) + cos(144°)×(lt1-lt2) + lt1
ll4 = cos(144°)×(ll3-ll1) - sin(144°)×(lt3-lt1) + ll3
lt4 = sin(144°)×(ll3-ll1) + cos(144°)×(lt3-lt1) + lt3
© 2024 Mariano Tomatis Antoniono • Dharma Initiative • Mathematical Forecasting Initiative • Italian Division (Turin)