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

the room. So when you use an correct room-mechanics-Theory you maybe get the correct results for the moving of the Universe
through the room. This I have tried to do here.
By doing this I was able to get an easy sinus-function for the moving of the room through the ether. That was years-long hard
work! I had to learn many things about physics before I was able to do that. But I really wanted to know what’s all about with the
things they have teached me at home. That is why I have engaged me with these problems in the past very often. Now I think
that I have found the same equations the others in our home had.
How I was able to get this Sinus-equation for the moving of the Universe through the ether you can see at these equations
(Many of the things you can read here I have out of usual physic-books. Not everything seemed to be wrong what you can read
there!):
s(t) = ½ a t 2
v(t) = a t
With s = Length of the way you have moved with the velocity v, v = Your velocity, a = your acceleration. At these equations you
can see now that the Length of the wave also can be:
v(t) = ∫ a(t)
s(t) = ∫ ∫ a(t)
and also :
s(t) ’ = v(t)
s(t) ’’ = a(t)
With the help of these functions s(t) and v(t) you can calculate now way- or velocity-functions who are not ½ a t 2 or a• t, but
something else. For example they also can be ½ a t 4 or sin(t) or cos(t).
Now look at the force who plays the most important part in the whole Universe. Of course this force is the gravitational
interaction. Because she holds all masses in the Universe together. The gravitation force is:
FGravitation = G M m , with G = Gravitational Constant, M = Whole mass in the Universe, m = Mass of an
R 2 object in the room, R = Distance between M and m, F = Force between M and m
FGravitation = G M m = a m , because F = m a (mass multiplicated with the acceleration of the mass)
R 2
FGravitation ≈ G • (M = in the interaction involved mass < range of the field) • m • (room-mechanic-effects)
(R = distance) 2
In this equation is:
{Range of the field}
G M m ≈ ∫ G • M(β) • m • dβ
R 2 R(β) 2
To find out how big M(β) and R(β) are in this equation, you must find out how big the Gravitation-field in the Universe really is.
To find out all that, of course you must consider that the 3-dimensional room in a Universe is the surface of an 4-dimensional
hollow-ball. If all that still is an physical fact or not doesn’t matter. For getting the correct gravitation-field for an sin-function for
the moving of the Universe you must guess that the whole room has the form of an 4-dimensional hollow-ball. Other cases can’t
be correct, because you get there wrong results for the field. So it must be an two-planed hollow-ball like they have told it me in
our base. How big the gravitation-field in such a room always is, you can see at this picture:
4
2
dM(β) 1
3
room s(t)= r β (range of the field)
sin(β) r
β ( “ )
357
mass m
dM(β)
In this curved field is now the field-Distance (dM(β); m) not β /360° • r (way along the curved surface), but ≈ sin(β)• r. The reason
for that is, that the field of dM(β) achieves in m as if it would come from a point who has only the distance sin(β) r from m. All
that you can measure at the diameter of the field-hollow balls who arrive at m. Their diameter is not β /360° • r, but only sin(β)• r.
All that is why the Gravitation-Potenzial between m and dM(β) is:
Gravitation-Potenzial between dM and m is: G dM m _1_
sin(β)
Gravitation-Force between dM and m is: G dM m cos(β) , F(r) = d M
sin(β) 2 dr r