The 'New Berlin' base: Nazis in the Antarctic - Project Camelot

In direction of moving the Cruise of Einstein then looks like that:
v4, object-time
v4, r.t.
v3
The two-leveled ether
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The Cruise of Einstein
When now the velocity v 5 (the length of the second green thread in the 5.th room-direction) is not changing by the time (is as
big as v4, r.t ) one can identify v 4, r.t. as well as v 5 with the velocity of light C. Cause then all photons will always move with
the same velocity C = K through the ether in their own object-time (v 4, r.t = V 3, maximal in the room of interactions) . But
nevertheless velocities much higher than C or K are possible when you are able to influence your very individual objecttime….
(Compare next chapter). The Cruise of Einstein then looks somehow like on this picture:
c – v3 c + v3
c The two-leveled ether
√ c 2 – v3 2 v4, object-time c c
v3
The Time-Delay: _________
The Object-time of the moving object is then √ 1 – ( V3 / C) 2 shorter than the time of an not moving object (= “real-time”).
The time of the moving object you can measure with the length of his 4-thread in the forth room-direction.
The Room-Contraction: ______ __________
In direction of moving the object sees the ether-threads in a (c -v3)/√c 2 – v 3 2 = √ C -V3 /C + V3 lower distance than normal.
And in the opposite direction it sees the ether-threads in a (c +v3)/√c 2 – v 3 2 = √ C +V3 /C - V3 bigger distance than normal.
So summing up a light-beam which moves forewards and backwards between two mirrors in a with the velocity v 3 moving
object sees the ether-threads in the room in the middle in the distance ½ [ √ C -V3 /C + V 3 + √ C +V3 /C - V 3 ]= 1 /√ 1 – ( V3 / C) 2 .
As well one could say, that the room must be contracted in the direction of moving cause the light only canmove with the
velocity c * √ 1 – ( V3 / C) 2 backwards and forewards between the mirrors.
The Velocity of light:
Light moves through the ether in all directions always as fast as possible. So through his own two-leveled ether it will move
with the velocity C = V4, r.t.. Light from an moving object has then in the direction of moving the velocity:
va = room / time = (c -v3)/√c 2 2 C -V3
– v3 = √ /C +V3
vb = room / time = (c +v3)/√c 2 2 C + V3
– v3 = √ /C -V3 (in the opposite direction)
In the direction of moving it will need then for an distance s the time ta = S /va . Because s = va • ta . And in the oppsite
direction it will need for the same distance s the time tb = S /vb . All together it will need then for the way from one mirror and
back the time ttog = ta + tb = S /va + S /vb = 2 S C /√c 2 2
– v3 . That is why it only can have on this way the velocity vtog. = C *
√1 – ( V3 / C) 2 . When now the whole room in this direction is 1 /√1 – ( V3 / C) 2 times thicker than normal, the light will put back
distances in this direction 1 /√1 – ( V3 / C) 2 times “faster” than without room-contraction. That is why you always will measure
the velocity of light as C between two mirrors in this direction. Now watch the Cruise of Einstein in the direction in which
the object is not moving: _______ _______
√ c 2 – v3 2 √ c 2 – v3 2
_______
√ c 2 – v3 2
v4, object time
v3 = 0
The two-leveled ether of the object
The Cruise of Einstein