free book - Impulse Drive.

[eccentrically1]

The vectors forces appear to net sum zero even in the 3 ball asymmetry, if the compression and retraction springs located behind the cue ball and in between the two target balls are supposed to be returning things to the original state. The asymmetry is the electric force accelerating the cue ball. That's the useful product.

[Kirk]

I provided a step by step analysis re the momentum and changing vectors. If you are telling me the 3 ball collision is equivalent to just letting the cue ball proceed to the opposite end you are not even logical.

[eccentrically1]

It would be equivalent if the balls were perfectly elastic. But since they aren't, the 3 ball situation will have less impulse than just letting the cue ball proceed to the opposite end.

[Kirk]

and the vector means nothing?

[eccentrically1]

The 2 target balls won't have the same momentum as the cue ball. Some is lost to deformation and sound. So the vector will change direction, and the sum of the target balls' arrows' length (for magnitude) will be less than the cue ball arrow length.

[Kirk]

you are on track. Did not the book say the pair share the cue ball's momentum? Whats more their vector is 45 degrees from the old vector thus only one half is the original vector. This leaves half the cue balls original momentum to subtract from the original momentum leaving half as useful thrust

[eccentrically1]

"Only one half is the original vector"? No, none of the original vector is left.

Half the momentum is useful thrust? What happened to elastic collision?
You're not making sense Kirk.

[Kirk]

Assume the transducer imparts an mv of 8 to the cue ball. An equal but opposite vector force is imparted to the transducer which is fastened to our vehicle. The cue ball strikes 2 equivalent mass stationary balls. The 8 units of momentum are now shared and we have two balls with an mv of 4. Their vector, however is at 45 degrees to the original vector, That is because they move 1 unit of distance at 90 degrees to the original vector and 1 unit of distance along the original vector. That at 90 degrees cancels when the 2 balls complete their travel and the momentum of half the original velocity along the cue ball vector is transferred as deceleration to the vehicle. Since 2 balls with 2 units of momentum is 4 and the original input via the cue ball was 8 we see 4 units are still in the vehicle. You transferred 50% of the original momentum into output. Effects such as spring losses and elastic deformation of the balls is second order.

[honza]

would be nice if it worked with water but divergence is not supplied by a collision, rather it is supplied by a knife or wedge which transmits the force to the enclose. As a result it won't work. you need 3 elastic objects.

You are right Kirk.
Have found the formulas for calculating reaction force at a plate inclined to the water jet and the reaction force would always equal the nozzle reaction force (independently of the plate inclination angle).

It seems that the only way to create the required asymmetry is by your impulse vector splitting idea.

[greendoor]

Brilliant work - thank you for this Kirk. When I first found this site and tried to crack the Bessler problem I have believed that antigravity or a "reactionless space drive" is exactly the same problem, just in reverse.

I'm not sure all viewers would appreciate what this is all about, so here is my crude attempt at explaining why I think this is so important:

Newtons Third Law of physics is that every action has an equal and opposite reaction. So far, nobody has ever been able to prove otherwise.

This imposes limitations on space travel, because rockets have to eject mass in order to travel - they run out of fuel. There is abundant energy in space, e.g. solar power - so it would be fantastic if there was someway of turning energy into a reaction force just like burning rocket fuel, but not requiring mass.

But if we throw mass around in space, the equal and opposite reaction force cancels all movement out. In theory, if a spaceman was lost in space he could not "swim" his way back, because every motion of his limbs would be canceled by the reaction force. If he was ejected from his spacecraft with the smallest velocity - he is doomed to continue out into space forever at that same velocity. Without a rocket pack with fuel in it, he is doomed. Ejecting mass seems to be the only solution - but is it really??

NASA has obviously done a lot of zero gravity research, and I would love to know the truth about whether a spaceman can "swim" in space ... but I don't expect to hear any truth from NASA anytime soon. War is war, and NASA is about war, not truth.

Force = Mass x Acceleration.

IF it is possible to turn energy into a force with no reaction then we can freely accelerate mass on demand, without having to eject mass.

It doesn't matter how efficient this is - we have unlimited free solar energy, so any device at all that can produce force without an equal an opposite reaction will do the trick. This would effectively be an anti-gravity lifter in our earth environment, or a spacecraft propulsion unit in space.

The reverse should also be true ... just as we can turn an electric motor into a generator by running it backwards, if we could place such a device in a constant force field (e.g. earth gravity is a constant downwards force) then we would be able to turn some of that force into energy. The efficiency doesn't matter - it can afford to be VERY inefficient, because the force is freely available so any energy output is free. I believe this is what Bessler achieved.

[greendoor]

Imagine a high pressure fire hose mounted to a trolley on an ice lake. The ejecting water propels it like a rocket, but we quickly run out of hose.

Now imagine we have a high pressure diesel powered pump and tank of water so we don't need the hose. This should function like a rocket - we could propel the trolley until we ran out of water.

Now imagine we try to recycle the water so we don't run out ... that's where it fails. Or does it?

I would wonder if we could pump two high pressure jets of water into two counter-rotating turbines. The torque of each turbine would cancel out (otherwise the trolley would spin in circles). These turbines could drive generators connected to a load of some sort, so energy would be removed from the streams of water.

Would this result in unresolved force in one direction? I'm guessing probably not ... the backwards thrust of the water leaving the nozzels would be counteracted by the forward thrust of the water hitting the turbine blades ...

Or is it possible to configure this in any way so the forces are asymmetric?

I'm not convinced that elastic balls are much better than water molecules in this regard ... still pondering.

Why is elastic desirable anyway? If the goal is to create unbalanced force, wouldn't it be better to have very inelastic material that would turn the kinetic energy into heat before the mass it got a chance to hit the opposing surface?

[Art]

.

Lets suppose the black box is shaped like a sphere with axis A .

The Drive Cue Ball is shot out from A towards the perimeter and impacts the other Cue Ball Pair at about ½ radius from the axis A , say at point B.

A total cycle of the cue ball process involves : -

1. Drive Cue Ball shot from A to impact Cue Ball Pair at B .

2. Reaction force at A drives the “black box Sphere� to new position A + d .
The 3 cue balls are doing their thing at B but are not connected in any way yet with the black box sphere. They are acting like expelled mass.

3. The mass of the 3 cue balls have to be now returned to their original positions in relation to the black box centre at axis A .

4. But A is now at A+d and B is now at a corresponding B+d , so the combined masses of the three cue balls have to be accelerated to their new positions at A+d and B+d

>>I suspect that the force required to do this will just exactly cancel the free force obtained in the cue ball collision .


Does this make any sense ? I don’t think I am up to demonstrating this mathematically because I just plain don’t trust me with numbers .

Would somebody else like to have a go ?

.

[Kirk]

Imagine a high pressure fire hose mounted to a trolley on an ice lake. The ejecting water propels it like a rocket, but we quickly run out of hose.

Now imagine we have a high pressure diesel powered pump and tank of water so we don't need the hose. This should function like a rocket - we could propel the trolley until we ran out of water.

Now imagine we try to recycle the water so we don't run out ... that's where it fails. Or does it?

I would wonder if we could pump two high pressure jets of water into two counter-rotating turbines. The torque of each turbine would cancel out (otherwise the trolley would spin in circles). These turbines could drive generators connected to a load of some sort, so energy would be removed from the streams of water.

Would this result in unresolved force in one direction? I'm guessing probably not ... the backwards thrust of the water leaving the nozzels would be counteracted by the forward thrust of the water hitting the turbine blades ...

Or is it possible to configure this in any way so the forces are asymmetric?

I'm not convinced that elastic balls are much better than water molecules in this regard ... still pondering.

Why is elastic desirable anyway? If the goal is to create unbalanced force, wouldn't it be better to have very inelastic material that would turn the kinetic energy into heat before the mass it got a chance to hit the opposing surface?

that would work too I think but thermal management would be a bear.

[Kirk]

.

Lets suppose the black box is shaped like a sphere with axis A .

The Drive Cue Ball is shot out from A towards the perimeter and impacts the other Cue Ball Pair at about ½ radius from the axis A , say at point B.

A total cycle of the cue ball process involves : -

1. Drive Cue Ball shot from A to impact Cue Ball Pair at B .

2. Reaction force at A drives the “black box Sphere� to new position A + d .
The 3 cue balls are doing their thing at B but are not connected in any way yet with the black box sphere. They are acting like expelled mass.

3. The mass of the 3 cue balls have to be now returned to their original positions in relation to the black box centre at axis A .

4. But A is now at A+d and B is now at a corresponding B+d , so the combined masses of the three cue balls have to be accelerated to their new positions at A+d and B+d

>>I suspect that the force required to do this will just exactly cancel the free force obtained in the cue ball collision .


Does this make any sense ? I don’t think I am up to demonstrating this mathematically because I just plain don’t trust me with numbers .

Would somebody else like to have a go ?

.

Did you get a copy of the book, Art?

[eccentrically1]

Assume the transducer imparts an mv of 8 to the cue ball. An equal but opposite vector force is imparted to the transducer which is fastened to our vehicle. The cue ball strikes 2 equivalent mass stationary balls. The 8 units of momentum are now shared and we have two balls with an mv of 4. Their vector, however is at 45 degrees to the original vector, That is because they move 1 unit of distance at 90 degrees to the original vector and 1 unit of distance along the original vector. That at 90 degrees cancels when the 2 balls complete their travel and the momentum of half the original velocity along the cue ball vector is transferred as deceleration to the vehicle. Since 2 balls with 2 units of momentum is 4 and the original input via the cue ball was 8 we see 4 units are still in the vehicle. You transferred 50% of the original momentum into output. Effects such as spring losses and elastic deformation of the balls is second order.

Doesn't some of the 8 units stretch the retraction spring? In order to return the cue ball back to the transducer, the spring has to have enough energy to do that, and it can only get it from the cue ball.