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HECHOS 12:12 -JUAN MARCOS ES EL GRIAL: 911 (IMPRESIONANTE NEXO CON EL NUMERO PI=3.14 Y EL DIA DE SAN MARCOS 25/4)
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Reply  Message 1 of 110 on the subject 
From: BARILOCHENSE6999  (Original message) Sent: 20/02/2013 02:48
 
milky way in Simple Gematria Equals: 119 ( m 13 i9 l 12 k 11 y 25 0 w 23 a1 y 25 )
queen mary in Simple Gematria Equals: 119 ( q 17 u 21 e5 e5 n 14 0 m 13 a1 r 18 y 25  
hebrew calendar in Simple Gematria Equals: 119 ( h8 e5 b2 r 18 e5 w 23 0 c3 a1 l 12 e5 n 14 d4 a1 r 18
mary magdalene in Simple Gematria Equals: 119 ( m 13 a1 r 18 y 25 0 m 13 a1 g7 d4 a1 l 12 e5 n 14 e5  
new york in Simple Gematria Equals: 111 ( n
14
e
5
w
23

0
y
25
o
15
r
18
k
11
)
york in Simple Gematria Equals: 69 ( y
25
o
15
r
18
k
11
)
O+R=15+18=33
 
En Hechos 12:12 (1+2,1+2:3,3 o 33) es PEDRO el que aparece con Maria la madre de Juan Marcos.
 Hechos 12:12: Y habiendo considerado esto, llegó a casa de María la madre de Juan, el que tenía por sobrenombre MARCOS, donde muchos estaban reunidos orando.
Y/K=2+5/11=7/11=SEPTIEMBRE/11
PERIODO 1/1-11/9=254 DIAS
PERIODO 11/9-31/12=111 DIAS
254 TIENE NEXO CON EL 25/4 (DIA DE JUAN MARCOS)
PERIODO 1/1-25/4=115 DIAS (116 SI ES BISIESTO)
PERIODO 11/9-25/4=226 DIAS (227 SI ES BISIESTO)
227=22/7 ES UNA BUENA APROXIMACION DE PI=3.14
NUMERO PI
 


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Reply  Message 2 of 110 on the subject 
From: BARILOCHENSE6999 Sent: 26/02/2013 15:58
 
Squaring the Circle - The Great Pyramid

The square represents the physical. The circle represents the spiritual. All sacred geometers have attempted the impossible: to square the circle (create a square who's perimeter is equal to the circumference of a circle.)

 

Here is the first of two valiant attempts:
This squaring of the circle works with a right triangle that represents the apothem (ZY) - (a line drawn from the base of the center of one of the sides to top of the pyramid), down to the center of the base (ZE), and out to the point where the apothem touches the Earth (EY).

The Great Pyramid of Egypt
(Sphinx in foreground)

Now let's look at this in 2D,
from directly above.


For the purpose of this exercise,
the side (AB) of the base equals 2.

(ABCD) is the base of the Great Pyramid.


This is lettered similarly to the wire frame version (above).


For the purpose of this exercise,
the side (AB) of the base equals 2.

Construct square (i JKD),
thus creating double square (JKE f).

Create diagonal (EK)
which intersects (i D) at (l).


iD = 1,
therefore the diameter
of the circle is also 1.


(EK) = (5) = .618 + 1 + .618

Put the point of your compass at (E) and extend it along the diagonal (EK) to point (m) where the circle intersects (EK), and draw the arc downward to intersect (KD f C) at (n).


If (EK) = (5), and (l m/l D) and l i = .5, the diameter of this circle is 1.

 

This makes (E m) = .618 + 1, or 1.618.

(E m) is the apothem.

Draw (E n) which intersects (A i l D ) at (o).


Put compass point at (f) and extend it to (n). Again put your point at (E) and draw the circle which happens to have the radius (E o).

(f n) is the height of the Great Pyramid.

This circle comes remarkably close to having the same circumference as the perimeter of the base (ABCD).

 

Let's go back to the original right triangle (EYZ)
(EY) = .5

(YZ) = phi

(EZ) = (phi)


EY = .5, The apothem is phi/1.618. This makes the 51 degree + degree angle.

Using a² + b² = c², this makes the height the square root of phi.


Squaring the Circle - The Earth & the Moon

Create a square (ABCD) with (AB) = 11


Create diagonals (AC) and (BD) crossing at center point (E)


Construct a circle which is tangent to square (ABCD) at f

Construct two 3 . 4 . 5 right triangles, with the 4 . 5 angles at (A) and (D).


Connect the 5 . 3 angles creating square (abcd) with side (ab) = 3


{4 + 3 + 4 = 11, or side (AD) of square (ABCD)}


Create diagonals (ac) and (bd) centering at (e)


Create a circle that is tangent to square (abcd) at four places.

Draw line (Ee) which intersects side (AD) at (F)


(EF) = the radius of the larger circle and (eF) = the radius of the smaller circle


The larger circle thus created is to the smaller circle as the moon is to the Earth!

With your compass point at (E), create a circle with radius (Ee)


This creates a circle whose circumference is equal to the perimeter of square (ABCD)!

 

The Math:
 

 1

(AB) = 11
(EF) = 1/2 of (AB) = 5.5

(ab) = 3
(eF) = 1.5

Therefore   5.5 + 1.5 = 7

The circumference of a circle is equal to two times the radius (the diameter) times pi (3.1416).

C= 14 x 3.1416
C= 43.9824

 2

In Square (ABCD), (AB) = 11
The perimeter of a square is four times one side. 11 x 4 = 44

According to the Cambridge Encyclopedia, the equator radius of the Earth is 3963 miles. The equator radius of the Moon is 1080.

The claim is that the smaller circle (in square abcd) is to the larger circle (in square ABCD) as the Moon is to the Earth.

 3

(EF) = 5.5
(F e) = 1.5
5.5 : 1.5 :: 3963 : 1080
5.5 / 1.5 = 3.66666
3963 / 1080 = 3.6694 - (if it had been 3960, it would have been exact!)


Reply  Message 3 of 110 on the subject 
From: BARILOCHENSE6999 Sent: 26/02/2013 18:03
 

 
john in Simple Gematria Equals: 47 ( j
10
o
15
h
8
n
14
)
grail in Simple Gematria Equals: 47 ( g
7
r
18
a
1
i
9
l
12
)

  

gize in Simple Gematria Equals: 47 ( g
7
i
9
z
26
e
5
)
john in Simple Gematria Equals: 47 ( j
10
o
15
h
8
n
14
)
grail in Simple Gematria Equals: 47 ( g
7
r
18
a
1
i
9
l
12
)
 
EN JOHN HAY ALGO CURIOSO. LAS PRIMERAS LETRAS SUMAN 33 y luego con la N se suma 47,
 
 

 

  

Put the point of your compass at (E) and extend it along the diagonal (EK) to point (m) where the circle intersects (EK), and draw the arc downward to intersect (KD f C) at (n).


If (EK) = (5), and (l m/l D) and l i = .5, the diameter of this circle is 1.

 

This makes (E m) = .618 + 1, or 1.618.

(E m) is the apothem.

Draw (E n) which intersects (A i l D ) at (o).


Put compass point at (f) and extend it to (n). Again put your point at (E) and draw the circle which happens to have the radius (E o).

(f n) is the height of the Great Pyramid.

This circle comes remarkably close to having the same circumference as the perimeter of the base (ABCD).

 

Let's go back to the original right triangle (EYZ)
(EY) = .5

(YZ) = phi

(EZ) = (phi)


EY = .5, The apothem is phi/1.618. This makes the 51 degree + degree angle.

Using a² + b² = c², this makes the height the square root of phi.

NUMERO 47

Reply  Message 4 of 110 on the subject 
From: BARILOCHENSE6999 Sent: 01/03/2013 16:41
 
ENTRE EL 25/4 (25 DE ABRIL) Y EL 8 DE DICIEMBRE (INMACULADA CONCEPCION) TENEMOS 227 DIAS, Y DESDE ESTE ULTIMO HASTA EL 22/7 (DIA DE MARIA MAGDALENA), SI ES BISIESTO O CON COMPUTO INCLUSIVO TAMBIEN TENEMOS 227 DIAS. ¿MUCHA CASUALIDAD NO? CURIOSO ES QUE EL DOGMA DE "INMACULADA CONCEPCION", AUNQUE ES FALSO, ES ESOTERICO DE UN MENSAJE MAYOR. YHWH NOS DA UN MAYOR MENSAJE DETRAS DEL CULTO TRADICIONAL Y ESTA FUERTEMENTE INTERRELACIONADO CON EL SANTO GRIAL.
 
Desde el 11 de Febrero (Virgen de Lourdes) hasta el 21/22 de diciembre (CONSTELACION DEL AGUILA Y DEL CISNE/SOLSTICIO DE INVIERNO EN EL HEMISFERIO NORTE) tenemos 314 dias (NUMERO PI). LA PIRAMIDE DE LOUVRE TIENE NEXO ESPIRITUAL CON LA CONCEPCION DEL NUEVO MESIAS. ESTO EXPLICA TEOLOGICAMENTE EL NEXO DE LA PIRAMIDE TRUNCADA DEL BILLETE DOLAR E INCLUSO LA PIRAMIDE MAYA EN EL CONTEXTO AL 21/22 DE DICIEMBRE DEL 2012.
NUMERO PI

Reply  Message 5 of 110 on the subject 
From: BARILOCHENSE6999 Sent: 01/03/2013 18:27

25 de abril

De Wikipedia, la enciclopedia libre
 

El 25 de abril es el 115.º (centésimo decimoquinto) día del año en el calendario gregoriano y el 116.º en los años bisiestos. Quedan 250 días para finalizar el año.

[editar] Acontecimientos

 
http://es.wikipedia.org/wiki/25_de_abril


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