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

The Time-Delay:
The Time-Delay is here of course as tall as in the other Cruise of Einstein. Be-
cause the 4-thread in the fourth room-direction is here as long as in the other
Einstein-Cruise. That is why time will past here as well √ 1 – ( V3 / C) 2 times slower
than the “real time” (time of an not through the ether moving object).
The Room-Contraction:
In this room-direction you will find no Room-Contraction. Because in this
direction the ether has in all room-directions normal thickness.
The Velocity of light:
In this Einstein-Cruise now, the light in all room-directions has the same
velocity. To the left side for example, it has the velocity:
room /time = √(c 2 – v 3 2 ) / √(c 2 – v3 2 = 1 /1 = 1 or C
And in the opposite direction the light has the velocity:
room /time = √(c 2 – v 3 2 ) / √(c 2 – v3 2 = 1 /1 = 1 or C
So in these room-directions you as well will measure the velocity of light as C between two mirrors.
Further Room-directions (Ergänzung von mir):
And in all further room-directions you will get as well the same result for the velocity of light between two mirrors. All this
you can see at the following equations and pictures:
_______ c
√ c 2 – v3 2 P2
va v3 α
P1
z • x
vb ___ ___
y with : vb = P1P3 and va = P1P2
and with: α є (0, 90°)
P3
______________________
Then is : cos(α) = X /(v3/C) x = cos(α) • (v3/C) va = y + z = √1–cos(α) 2 •(v3/C) 2 + sin(α)•(v3/C)
sin(α) = Z /(v3/C) z = sin(α) • (v3/C)_____ √1 –(v3/C) 2 .
x 2 + y 2 = c 2 y = √1 – cos(α) 2 •(v3/C) 2 vb = y - z = √1 – cos(α) 2 •(v3/C) 2 - sin(α)•(v3/C)
√1 –(v3/C) 2
________
Addition: The factor √1 –(v3/C) 2 comes from the constant time-contraction in the with the velocity v3 moving
object. The thickness p of the ether in the watched moving-direction is then:
p = va + vb = 2 √1–cos(α) 2 •(v3/C) 2
2 2 √1 –(v3/C) 2
And the velocity of light in the direction of moving is then between two mirrors:
ttog. = S /va + S /vb = s (va + vb) = 2 s √1–cos(α) 2 •(v3/C) 2 (cause va • vb ≡ 1)
2 2 va vb 2 √1 –(v3/C) 2
That means that the light has between two mirrors for different angles α (different moving directions) always the
velocity v(α) ≈ c • √1 –(v3/C) 2 ____ . Because of the the field-thickness p = √1–cos(α) 2 •(v3/C) 2 in these “α”-
√1–cos(α) 2 •(v3/C) 2 √1 –(v3/C) 2
directions light then always only can move with the light-velocity C from one mirror to an other mirror and back
in these room-directions. Cause v(α) • p ≡ 1 for all angles α! So as you see, everywhere in the room one only can
measure the velocity of light between two mirrors as C and as nothing else. For this it does not matter, if you
have a velocity V * in a 5. or 6. room-direction or not. That is why c is always equal to the velocity K in your own
object-time.
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