Exploring Asymmetry Momentum Transfer for Space Propulsion

[terrance_teoh]

Momentum Drive
===============
Just playing with some ideas on rocket propulsion in space based on asymmetry momentum transfer.
Any reason why this is not workable?

1.
Let
m1 = 100 kg robot arm at one end of a cylinder;
m2 = 100 kg robot arm at opposite one end of the cylinder;
m3 = 10 kg canon ball;

2.
Case 1:
=======
Let m1 and m2 throw the canon ball between them.
The gain in movement is zero, although there should be some oscillation:
Mass Velocity Momentum Duration Distance Total Distance
==== ======== ======== ======== ======== ==============
i
m1 = 100 kg; v1 = 0 m/s; p1 = 0 kg m/s; t = 0 s; d = 0 m; d_sum = 0 m
m2 = 100 kg; v2 = 0 m/s; p2 = 0 kg m/s;
m3 = 10 kg; v3 = 0 m/s; p3 = 0 kg m/s;

ii
m1 = 100 kg; v1 = -0.5 m/s; p1 = -50 kg m/s; t = 10 s; d = -5 m; d_sum = -5 m
m2 = 100 kg; v2 = -0.5 m/s; p2 = -50 kg m/s;
m3 = 10 kg; v3 = 10 m/s; p3 = 100 kg m/s;

iii
m1 = 100 kg; v1 = 0 m/s; p1 = 0 kg m/s; t = 0 s; d = 0 m; d_sum = -5 m
m2 = 100 kg; v2 = 0 m/s; p2 = 0 kg m/s;
m3 = 10 kg; v3 = 0 m/s; p3 = 0 kg m/s;

iv
m1 = 100 kg; v1 = 0.5 m/s; p1 = 50 kg m/s; t = 10 s; d = 5 m; d_sum = 0 m
m2 = 100 kg; v2 = 0.5 m/s; p2 = 50 kg m/s;
m3 = 10 kg; v3 = -10 m/s; p3 = -100 kg m/s;

v
m1 = 100 kg; v1 = 0 m/s; p1 = 0 kg m/s; t = 0 s; d = 0 m; d_sum = 0 m
m2 = 100 kg; v2 = 0 m/s; p2 = 0 kg m/s;
m3 = 10 kg; v3 = 0 m/s; p3 = 0 kg m/s;


3.
Case 2:
=======
Let m1 and m2 throw the canon ball between them.
But this time place a revolving door like a turbine; when the cannon ball goes only from m1 to m2.
The idea is to drain the kinetic energy (1/2 mv^2) from the canon ball.
Let us assume we can take out about 51% of that energy.
(http://en.wikipedia.org/wiki/Wind_turbine#Efficiency)
This should slow down the cannon ball, hence reducing its momentum when it hits m2.
From the calculations below we can see that it is possible to convert energy to velocity and still recycle the mass being used for propulsion.

Mass Velocity Momentum Duration Distance Total Distance Energy
==== ======== ======== ======== ======== ============== ======
i
m1 = 100 kg; v1 = 0 m/s; p1 = 0 kg m/s; t = 0 s; d = 0 m; d_sum = 0 m
m2 = 100 kg; v2 = 0 m/s; p2 = 0 kg m/s;
m3 = 10 kg; v3 = 0 m/s; p3 = 0 kg m/s;

ii
m1 = 100 kg; v1 = -0.5 m/s; p1 = -50 kg m/s; t = 10 s; d = -5 m; d_sum = -5 m
m2 = 100 kg; v2 = -0.5 m/s; p2 = -50 kg m/s;
m3 = 10 kg; v3 = 10 m/s; p3 = 100 kg m/s;

e3 = 500 kg m2/s2

iii
m1 = 100 kg; v1 = -0.5 m/s; p1 = -50 kg m/s; t = 0 s; d = 0 m; d_sum = -5 m
m2 = 100 kg; v2 = -0.5 m/s; p2 = -50 kg m/s;
m3 = 10 kg; v3 = 7 m/s; p3 = 70 kg m/s;

e3 = 245 kg /s2;
Energy Drain = -255 kg m2/s2 (about 51%)

iv
m1 = 100 kg; v1 = -0.15 m/s; p1 = -15 kg m/s; t = 0 s; d = 0 m; d_sum = -5 m
m2 = 100 kg; v2 = -0.15 m/s; p2 = -15 kg m/s;
m3 = 10 kg; v3 = 0 m/s; p3 = 0 kg m/s;

v
m1 = 100 kg; v1 = 0.35 m/s; p1 = 35 kg m/s; t = 10 s; d = 3.5 m; d_sum = -1.5 m
m2 = 100 kg; v2 = 0.35 m/s; p2 = 35 kg m/s;
m3 = 10 kg; v3 = -10 m/s; p3 = -100 kg m/s;

vi
m1 = 100 kg; v1 = -0.15 m/s; p1 = -15 kg m/s; t = 0 s; d = 0 m; d_sum = -1.5 m
m2 = 100 kg; v2 = -0.15 m/s; p2 = -15 kg m/s;
m3 = 10 kg; v3 = 0 m/s; p3 = 0 kg m/s;

Note:
=====
p = mv;
e = 0.5mv^2;

 

[jbriggs444]

Momentum is always conserved. It seems likely that you have ignored the momentum going into the revolving door.

 

[Drakkith]

Just playing with some ideas on rocket propulsion in space based on asymmetry momentum transfer.
Any reason why this is not workable?

Conservation of momentum. Reactionless drives are similar to perpetual motion/free energy devices in that it requires breaking known scientific laws. That means it violates PF posting policy, and because of this I am locking the thread.

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