jaanuar 22, 2013

About the speed of light, encoded in the Great Pyramid

In Estonian:
Pidasin vajalikuks oma tähelepanekud valguse kiiruse kohta avaldada ka ingliskeelsetena.

The figure above concerns speed of light and how this is encoded in the Great Pyramid.

The square  there is base square of the Great Pyramid.

There are 2 circles, L1 is circumscribed circle of base square and L2 is inscribed circle of base square.

As pointed out by Jonathan Langdale, in


subtracting circumferences L1 and L2 gives us 1/1000000 from the speed of light in meters per second.  
But he did not notice, that this means, this distance is
1/1000000 from light second, it is independent of measurement system of lenghts, but requires knowledge about seconds.
Additionally in this Figure there are 2 other circles, L3 and L4, constructed by myself. L3 is circle, inscribed into triangle ABC. L4 is constructed by dividing diameter of L3 with square root of 2, using
right angled isosceles triangle PTR.

The idea for constructing  this circles is to get circle with circumference, what is exactly

1/1000000 light seconds. This construction is logical, not artificial and the circumference of L4 is exactly 1/10**6 light seconds.

Notice, circumferences of circles L1, L2, L3 and L4 are (sqrt(x) is square root of x in google calculator):

C(L1) = (sqrt(2)+2)/10**6 light seconds
C(L2) = (sqrt(2)+1)/10**6 light seconds
C(L3) = (sqrt(2)/10**6 light seconds
C(L4) = 1/10**6 light seconds.
This results are geometrical conclusions from the fact, that difference between circumferences L1 and L2 is 1/10**6 light seconds.
Using the fact, that side length of the base square of the Great Pyramid equals 440 royal cubits and 1 royal cubit equals pi/6 meters (remarkable  fact!)
I will  write the formulas for calculation and you can use Google calculator for checking them. Results are already expressed in micro light seconds (for convienency)
C(L1) = 440*pi/6*sqrt(2)*pi/299.792458 = 3.41425117
                                                      Sqrt(2) + 2 = 3.41421356
C(L2) = 440*pi/6*pi/299.792458 = 2.41424015
                                       Sqrt(2) + 1 = 2.41421356
C(L3) = (440**2)/(440+440+sqrt(2)*440)*pi/6*2*pi/299.791458 = 1.41423386
                                                                                                       Sqrt(2) = 1.41421356
C(4) is obviously  1.000   (C(L3) divided by sqrt(2)

Now I bring elementary, but more technical proof of the fact, that when difference between lenghts of circles L1 and L2 is 1/1000000 light seconds, then circumferences of L1,L2, L3, L4 - C(L1), C(L2), C(L3) and C(L4) are the expressions:
C(L1) = sqrt(2) + 2
C(L2) = sqrt(2) + 1
C(L3) = sqrt(2)
C(L4) = 1 micro light seconds.
Sqrt() denotes square root, it is used in Google calculator for calculating this function, as well in other mathematics oriented computer programs (R for example). ** denotes powering (2**4 = 16)
Let D1, D2, D3 and D4 be the diameters of the circles L1…. L4, measured in light seconds.
Let D1 – D2 = s  (scale factor, in this case 1/pi * 1/10**6 )
As D2 = D1*sqrt(2)/2,
D1*(1-sqrt(2)/2) = s
D1 = s / (1—sqrt(2)/2) =s*(1+ sqrt(2)/2)/(1/2) = s*(2 + sqrt(2)
D2 = 1/sqrt(2) * D1 = s * (1 + Sqrt(2))
From formula D = 4*S/(a + b + c), where S is area of the triangle, a,b,c lenghts of sides of the triangle,
and D is diameter of the circle, inscribed into the triangle      
we get
D3 = 2*D2**2/D2*(2+sqrt(2)) = s*2*(1+sqrt(2)/(2+sqrt(2)=s*sqrt(2)
D4 is D3/sqrt(2), D4 = s
But, as is well known:
 “the lightspeed latitude” with degree 29.9792458  goes through centre of „Great Gallery“ inside the Great Pyramid.
The informations is available in

But this is also the centre of L4, T !
This coincidence is really amazing and I suppose, this is not casual, but causal coincidence.