Fluid Space Drive (V3) explained.

 

 

 

 

              

                    

                                             Fig 1 Principal elements of a Fluid Space Drive.

 

We have:

A pressurized structure/spacecraft (1) in a micro-gravity environment, inside the spacecraft is a 100k mass (2) we call Ram Mass Assembly or RMA.

The RMA (2) has freedom to move across the length (not breadth) of the spacecraft.

Inside the pressurized structure we have a forward ramming piston (3a) and a rear ramming piston (3b)

We also have a probe (5) we wish to accelerate.

And a radioisotope thermoelectric generator (RTG, RITEG) (4) to provide power to the system.

 

For this presentation we shall assume that the assembly composed of pressurized structure (1), probe (5), rear and forward ramming pistons (3a and 3b) and secondary control systems (not shown) have a mass of 800k (mass of RMA (2) is not included in this total).

 

                                                                                                                                    Fig 2

 

Fig 2 illustrates the RMA (M2), it has a series of flaps attached to servo motors, when the flaps are closed they form the shape of an open cone that functions as an air brake increasing the RMA’s drag coefficient (note Dd) therefore slowing it’s relative velocity inside the pressurized structure (M1).

 

Note: The illustration does not show the ideal aerodynamically shape for low drag, see note 3 for a better design

 

So how does it work?

 

              

  Cycle 0

We start the cycle with the RMA (2) positioned against the rear ramming piston (3a).

 

 

Cycle 1

 

The rear ramming piston (3a) expands with sufficient force to accelerate the RMA (2) to a velocity (V2) of 1m/s in the +X direction.

P2 is the RMA’s momentum, (momentum = Mass x Velocity =100k x 1m/s =100P) in the +X direction so P2 =100p.

The spacecraft’s momentum (P2) in the –X direction is equal but in the opposite (-X) direction or -100p.

Therefore the velocity of the spacecraft in the –X direction is -0.125m/s.

V1 = Velocity of pressurized structure/spacecraft (1) in m/s.

V2 = Velocity of RMA(2) in m/s

P1 = Momentum of pressurized structure/spacecraft (1).

P2 = Momentum of RMA (2) mass x V2

 

Cycle 2

As the spacecraft’s travels in the –X direction with a constant velocity of 0.125m/s, the RMA is traveling inside the spacecraft with a constant velocity of 1m/s in the +X direction.

 

Although the spacecraft is pressurized (air or other suitable gas at normal atmospheric pressure), as the RMA’s flaps are open, the air drag is minimal and does not exert sufficient force to noticeably slow it’s +X velocity.

 

 

Cycle 3

 

The spacecraft (1) and the RMA (2) travel in opposite directions until they collide at the inner forward end of the spaceship.

As both the spaceship (1) and the RMA (2) have the equal momentum (P1 = -100p   P2 = 100p P1 + P2 = 0) the system comes to a full stop.

 

Cycle 4

 

The forward ramming piston (3b) expands with sufficient force to accelerate the RMA (2) to a velocity (V3) of 1m/s in the -X direction.

P3 is the RMA’s momentum, (momentum = Mass x Velocity =100k x 1m/s =100p) in the +X direction so P3 =-100p.

The spacecraft’s momentum (P4) in the –X direction is also 100p therefore the velocity of the spacecraft in the +X direction is 0.125m/s.

 

  

V3 = Velocity of RMA (2) in m/s

V4 = Velocity of pressurized structure/spacecraft (1) in m/s.

P4 = Momentum of pressurized structure/spacecraft (1).

P3 = Momentum of RMA (2) mass x V2

 

 

Cycle 5

 

The RNA’s flaps close (see fig 2), greatly incrementing the drag force slowing the RMA’s –X velocity without greatly affecting the +X velocity of the spacecraft (M1).

 

This effect (the velocity of M2 is affected in greater magnitude that the velocity of M1) is very counter intuitive to many (maybe everybody), why it works is explain here, but more important the effect can be observed with an experiment set up using the same elements available in a physics classroom to demonstrate the conservation of momentum. (See here)

 

 

 

 

Cycle 6

 

The spacecraft (1) and the RMA (2) travel in opposite directions until they collide at the inner forward end of the spaceship.

When the RMA collides with the spacecraft’s inner wall it’s –X final velocity (V3f) is LESS than it’s –X initial velocity (before the flaps were closed) it does not have the same momentum (it lost velocity) therefore the collision is not sufficient to cancel M1’s +X acceleration (gained in cycle 4) and the cycle finishes with a net gain of velocity in the +X direction for the spacecraft.

The gain in +X velocity will increase every time the cycle (1 to 6) is repeated

 

 

 

 

INSERTION from simple description (Version 3) using two astronauts and batman (Adam West version).

 

At this point I generally get the following comment:

 

This will not work because all the impulse of the mass (steel ball), that you "get rid of" by using airbrakes, is actually transferred to the air inside the cylinder which pushes against the wall of the cylinder and accelerates it. It is all a closed system, so there will be no change in impulse and no acceleration at all.

 

This idea that the airbrake pushes against the AIR and the AIR pushes against the rear wall of the cylinder is illustrated in fig 6a.

 

Description: http://www.wjetech.cl/u/index_files/image017.jpg

Fig 6a

 

But that is not really how drag works, the airbrake does not push against an “AIR”, it pushes (collides) against individual fast moving gas molecules (fig 6b 1), for every collision of molecule against air brake there are two equal and opposite forces (no problem with Newton’s Laws), and the molecules that have been hurled in the general direction of the rear wall of the cylinder encounter billions and billions of fast moving molecules (fig 6b 2), as air has a tendency to expand in every direction to fill the container (equalizing local differences in air pressure) some of the force is apparently diverted towards the other walls of the container(1).

 

 Description: http://www.wjetech.cl/u/index_files/image019.jpg

Fig 6b

 

There are also some that say when the airbrake collides with the air, the air in front of the airbrake is compressed (fig26c 1) and that it will travel to the rear wall of the cylinder like a Kamehameha (2). However as one of the basic characteristic of a gas is its tendency to equalize its pressure when contained (we do not see a tendency for higher pressures inside a container at rest) part of the momentum will be diverted to the other walls of the cylinder.

Description: http://www.wjetech.cl/u/index_files/image021.jpg

Fig 26c

 

(1) Actually I do not really know with absolute certainty how the air molecules behave for I cannot directly observe them (I do not have a Schlieren set up available), I can expect air to behave in a certain manner because of the kinetic theory of gases. But what I do know is what I can observe, and the experiment described in http://www.wjetech.cl/e/ , and the experiment described in a DIY test (figs 13 to 21) work as described.

 

(2) Totally unnecessary dragon ball reference.