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

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  • abelov0927
    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 8:59 PM
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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 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,
      probably long to download):
      http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com.avi

      Here are 3 jpg snapshots. N�0 is the initial position. N�1 is the state just
      after the pulse. N�2 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.

      Fran�ois

      --- 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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