## Re: Gravitational Propulsion, A Rotation with Translation Movement is standalone natural phenomenon(version 2)

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• Hi I reply message from other group. It would be helpfull for you. Have a fun. Alex ... Hi Alex, I m not sure to understand all what you mean and may be I m
Message 1 of 4 , Aug 17, 2009
Hi
I reply message from other group.
It would be helpfull for you.
Have a fun.
Alex
...
Hi Alex,

I'm not sure to understand all what you mean and may be I'm wrong on some
points.
So here is my simulation to clarify the problem (a hard work :-):
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com.wm2d

If you have not Workingmodel software (it is free), here is the video (6Mb,
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com.avi

Here are 3 jpg snapshots. Nï¿½0 is the initial position. Nï¿½1 is the state just
after the pulse. Nï¿½2 is some seconds later.
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com0.jpg
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com1.jpg
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com2.jpg

The speed vector of each rod is displayed in the animation. The impulse force it
not displayed. It provides 100 N during 0.01s between the rods, at points of
same coordinates (= the position of the center of mass of the green rod).

In the red and green small windows, you can monitor the speeds, kinetic energies
and momenta of each rod. The default equations of the program are used.

In the small blue windows, I have put the equations to obtain the angular speed,
the moment of inertia and the angular momentum of each rod, calculated from the
common center of mass of the system.
This center of mass is displayed in the simulation (it is calculated by the
program, it's not a fixed point. It is at rest because there is no external
force acting onto the system. It is the mid-point of the line joining the
centers of mass of the two rods).
To simplify the calculi, the positions of the rods are chosen in order the
common center of mass to be at position x,y = 0,0.

With this simulation, I have discovered that from the common center of mass C,
each rod possesses a "hidden" angular momentum: each rod flies horizontally away
one another but one is above and the other under the horizontal x axis
containing their common center of mass. Thus the angle delimited by the
horizontal x axis and the line joining the centers of mass of the two rods
(crossing at C), decreases when the distance of the rods from the origin
increases. This variation of the angle is to be considered as an angular
velocity. Then from it, we can calculate the moment of inertia of each rod and
its angular momentum. The result is displayed in the small blue windows.

IMPORTANT : We see that the sum of the angular momenta of the two rods,
calculated in the referential frame of their common center of mass, is equal but
with opposite sign, to the angular momentum of the only rotating rod, calculated
in its proper frame.
Thus by adding these two momenta, we find zero. I guess the key of the problem
lies around this, but it is not yet completely clear for me.

Franï¿½ois

--- In gravitationalpropulsionstevenson@yahoogroups.com, Katriel Porth <batl4etrnity@...> wrote:

I have a few simple questions...
Very concisely, please briefly and simplisticly explain the following...

1. What is gravity? How does it work and what does it affect?

2. What is antigravity? How does it work and what does it affect?

Shalom and thank you.
-Katriel
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