[Jaaanosik (OP)]

Do we have a net forward momentum in the background frame or not?

I'll answer your question once you've answered mine:

Might you know the answer to this question?

So, what must the overall linear momentum be at the end of the stroke in order for the total momentum at the end of the stroke to match the total momentum before the stroke?

0 angular momentum + 0 linear momentum = 0 angular momentum + ? linear momentum. What is the value of the question mark?

Kryptid,
Right,
0am + 0lm = 0am + 0lm
... however...
7am + 3lm = 3am + ?lm

Back to you,
Jano

[Kryptid]

0am + 0lm = 0am + 0lm

So then you agree that a complete lack of net momentum at the beginning of a stroke means that there must also be a complete lack of net momentum at the end of a stroke as well.

7am + 3lm = 3am + ?lm

That would be 7. But, of course, your ship starts out with zero total momentum before the engine turns on. So those numbers should all still be zero.

Now about your newest design: I'm not sure whether the round things are balls or whether they are reels. I'm going to assume they are reels given your description. In that case, the ship will gain forward momentum as the CMGs are reeled in and the CMGs will gain momentum equal in magnitude and opposite in direction so that the total momentum remains zero.

[Jaaanosik (OP)]

0am + 0lm = 0am + 0lm

So then you agree that a complete lack of net momentum at the beginning of a stroke means that there must also be a complete lack of net momentum at the end of a stroke as well.

7am + 3lm = 3am + ?lm

That would be 7. But, of course, your ship starts out with zero total momentum before the engine turns on. So those numbers should all still be zero.

Now about your newest design: I'm not sure whether the round things are balls or whether they are reels. I'm going to assume they are reels given your description. In that case, the ship will gain forward momentum as the CMGs are reeled in and the CMGs will gain momentum equal in magnitude and opposite in direction so that the total momentum remains zero.

What reference frame?
Jano

[Kryptid]

What reference frame?

The barycenter's.

[Jaaanosik (OP)]

What reference frame?

The barycenter's.

The barycenter moves in the background frame.
This not the barycenter of plain linear momentum when barycenter is stationary in the background frame,
Jano

[Bobolink]

Sorry Jano, but there some laws you just can't break.  Your attempt is doomed for failure. 

[Jaaanosik (OP)]

Sorry Jano, but there some laws you just can't break.  Your attempt is doomed for failure. 

Bobolink,
here is the problem that Kryptid and it appears you too do not understand.

The system is a system of particles.
Writing 0am + 0lm = 0am + 0lm is wrong.
This is not a description of the system.

The systems has many particles so writing
0am_1 + 0lm_1 + 0am_2 + 0lm_2 + ... + 0am_n + 0lm_n = 0am_1 + 0lm_1 + 0am_2 + 0lm_2 + ... + 0am_n + 0lm_n
is wrong as well.

1. What is the left side of the equation?
- it is the system, let us take all the particles in the universe

2. What is the right side of the equation?
- it is the same system but after a delta time.

If we put zeros for all the particles on the left side then we do not have motion, we do not have particles.
We have nothing.
Ooops, ... we do not have nothing, we do not exist!!!!!
This is nothingness!!!!!!!!!

I hope you see the big problem!
We have to write some values on the left side ... and we have to have some values on the right side!!!

7am_1 + 3lm_1 + 5am_2 + 5lm_2 + ... + 2am_n + 8lm_n = 3am_1 + 7lm_1 + 5am_2 + 5lm_2 + ... + 8am_n + 2lm_n


I hope this helps,
Jano

[Kryptid]

The barycenter moves in the background frame.
This not the barycenter of plain linear momentum when barycenter is stationary in the background frame,
Jano

So the background frame is that of the CMGs? In that case, what you are doing is abandoning an inertial frame and entering an accelerating frame (for a moment), then entering a different inertial frame once the acceleration ceases. In the new inertial frame, the ship will have net forward momentum from the point of view of the CMGs. However, there can't be a contradiction between different frames: a device cannot look like reactionless propulsion in one frame but not another. So analyzing the system from the point of view of the barycenter is equally valid as analyzing it from the point of view of the CMGs.

I hope you see the big problem!
We have to write some values on the left side ... and we have to have some values on the right side!!!

7am_1 + 3lm_1 + 5am_2 + 5lm_2 + ... + 2am_n + 8lm_n = 3am_1 + 7lm_1 + 5am_2 + 5lm_2 + ... + 8am_n + 2lm_n

You're ignoring the fact that the sum of the momentum of all of the constituent particles is zero for an object that isn't moving. So although it's perfectly true that individual molecules will have momentum, we know that the total momentum of all the particles must add up to zero because the stationary object has no net momentum.

Before the engine turns on, the net linear and angular momentum of all of the particles in the system is zero because every piece of machinery in the ship is motionless relative to every other piece of machinery. And because of conservation of momentum, the net momentum must remain zero for every step of the engine's action. So it isn't possible to gain net linear momentum without violating conservation of momentum.

[alancalverd]

At the risk of being labelled a pedant, may I remind Jano that the dimensions of momentum are MLT^-1 and angular momentum ML²T^-1. They are not the same thing, and not being dimensionally equivalent, cannot be added.

But there's very little difference between elementary physics and pendantry. 

[Kryptid]

At the risk of being labelled a pedant, may I remind Jano that the dimensions of momentum are MLT-1 and angular momentum ML2T-1. They are not the same thing, and not being dimensionally equivalent, cannot be added.

I'm sure he'll find some excuse to find this "wrong"...

[Jaaanosik (OP)]

Kryptid,
the CMG is group of rotating wheels, gyros, there is a center of axis of rotations.
The intersect of axes of rotation is like a potential well because any rotational disturbance to the CMG will bring the intersect back to the center.
It is like ECI (Earth Center Inertial) reference frame.
This point is pretty stable in space, just an oscillation. It can be extrapolated one point that is 'fixed' to the background.
This point does not change during the startup or shutdown of the rotation.
Having said that, when the CMG does not rotate then the potential well does not exist.
The collection of wheels will behave as a body of parts.

If you look here:


You are saying that v_L - v_R = 0
Right.

So go ahead, grab the device with the trapped balls inside, point to a bottle of beer and open one end.
There is no linear momentum, so nothing happens, correct?

I am not going to do that, I'd rather enjoy my beer.
The momentum can 'disappears' for days and then it can appear out of nothing?
That is some serious 'magic'.
Not only that but when one ball is shot out it will have higher velocity than it came in.
True miracle! Increase of speed out of nothing!

If v_L - v_R = 0   then v_L = v_R is true as well. Where did 0 go?
Jano

[Kryptid]

There is no linear momentum, so nothing happens, correct?

Just because the net momentum is zero doesn't mean that nothing happens. The ball will be shot out by the spring and the rest of the device will recoil in the opposite direction. The total momentum, however, is still zero.

Not only that but when one ball is shot out it will have higher velocity than it came in.
True miracle! Increase of speed out of nothing!

There is no miracle. All that has happened is that potential energy was converted into kinetic energy. If you calculate the momentum, you will see that it is conserved. The momentum of the recoil is equal to the momentum of the ball.

If v_L - v_R = 0   then v_L = v_R is true as well. Where did 0 go?

It's still there.

[Jaaanosik (OP)]

There is no linear momentum, so nothing happens, correct?

Just because the net momentum is zero doesn't mean that nothing happens. The ball will be shot out by the spring and the rest of the device will recoil in the opposite direction. The total momentum, however, is still zero.

Not only that but when one ball is shot out it will have higher velocity than it came in.
True miracle! Increase of speed out of nothing!

There is no miracle. All that has happened is that potential energy was converted into kinetic energy. If you calculate the momentum, you will see that it is conserved. The momentum of the recoil is equal to the momentum of the ball.

If v_L - v_R = 0   then v_L = v_R is true as well. Where did 0 go?

It's still there.

Is the increased tension in spring, atoms moving faster caused by something?
Where did atoms momentum come from?
Jano

[Jaaanosik (OP)]

The problem is delta time. We can talk all day about energy/momentum, where they are.
Force, m*dv/dt and impulse are essential for momentum, energy transfer between systems.

If you say the momentum is 0 then can we have dp/dt?
p - momentum

Jano

[Kryptid]

Is the increased tension in spring, atoms moving faster

If the atoms move any faster, it would be due to the conversion of the ball's kinetic energy into heat energy. But that's not where the spring's tension is coming from, because you can let the spring cool back down to room temperature and the potential energy will still be there.

caused by something?

It's caused by the conversion of kinetic energy into potential energy.

Where did atoms momentum come from?

They don't receive any net increase in momentum.

The problem is delta time. We can talk all day about energy/momentum, where they are.
Force, m*dv/dt and impulse are essential for momentum, energy transfer between systems.

If you say the momentum is 0 then can we have dp/dt?
p - momentum

Jano

Conservation of momentum will not be broken no matter how long you wait.

[Jaaanosik (OP)]

...
It's caused by the conversion of kinetic energy into potential energy.
...

Kryptid,
I have no idea why it is so difficult to admit that linear momentum is like kinetic energy and angular momentum is like potential energy.
If we convert kinetic energy to potential then we have to convert linear momentum to angular momentum.

If a magnet flies in space around a rotating magnetic field generator.
When the generator is off then nothing happens the magnet flies straight ahead.
If we turn on the magnetic field then we trap the magnet into a 'circular orbit', potential energy, the magnet has angular momentum.
The magnet can orbit the generator multiple times around their barycenter.
When the generator is turned off the magnet flies away, the potential energy is transformed into kinetic energy.
The angular momentum into the linear momentum.
Is flying out magnet going to have the same linear momentum?
Not necessarily, there are flyby accelerations and decelerations.

How come you admit kinetic energy to potential energy conversion but you deny linear momentum to angular momentum conversion?
You keep talking that the system HAS TO BE DEFINED in a way that linear momentum is 0.
The system definition can be changed. We can change the boundaries - space and time.
Initial and end conditions...

So let say:
generator mass m=10
magnet mass m=1

IN
lm - linear momentum
magnet_lm = 10
generator_lm = 0

OUT
magnet_lm = 20
generator_lm = 2

Can we get the numbers above?
Sure we can, it all depends on the generator design, how it will increase the angular momentum.
Jano

[Kryptid]

I have no idea why it is so difficult to admit that linear momentum is like kinetic energy and angular momentum is like potential energy.

That is incorrect. An object with either linear momentum or angular momentum will have kinetic energy. Potential energy does not need anything to be moving. None of the components of a spring need to be moving for it to have potential energy. You can suppress its temperature down to absolute zero and it will still have potential energy.

f we convert kinetic energy to potential then we have to convert linear momentum to angular momentum.

No.

If a magnet flies in space around a rotating magnetic field generator.
When the generator is off then nothing happens the magnet flies straight ahead.
If we turn on the magnetic field then we trap the magnet into a 'circular orbit', potential energy, the magnet has angular momentum.
The magnet can orbit the generator multiple times around their barycenter.
When the generator is turned off the magnet flies away, the potential energy is transformed into kinetic energy.

Orbital kinetic energy is a form of kinetic energy, not potential energy.

The angular momentum into the linear momentum.

No. The angular momentum was there before the magnetic field was ever turned on.

Is flying out magnet going to have the same linear momentum?

The system as a whole will have the same linear momentum, but, as you say, the individual components of the system can have different momenta. It just all has to add up to the original momentum.

How come you admit kinetic energy to potential energy conversion but you deny linear momentum to angular momentum conversion?

Maybe if you actually went back and read that link I posted where it was stated that linear momentum and angular momentum had separate conservation laws, you'd know. But as I said before, converting angular momentum into linear momentum won't help you because both linear and angular momentum start at zero before the engine is turned on. Then there is the fact that your machine is symmetrical: any angular momentum possessed by the left gyroscope will be cancelled out by that of the right gyroscope. So starting out by assuming that the gyroscopes are rotating won't help either.

You keep talking that the system HAS TO BE DEFINED in a way that linear momentum is 0.
The system definition can be changed. We can change the boundaries - space and time.
Initial and end conditions...

Before the engine is turned on, all of the internal components are stationary. You realize that means zero linear and zero angular momentum, don't you? Both linear and angular momentum require motion.

So let say:
generator mass m=10
magnet mass m=1

IN
lm - linear momentum
magnet_lm = 10
generator_lm = 0

OUT
magnet_lm = 20
generator_lm = 2

Can we get the numbers above?

Not unless you violate conservation of momentum. 10 + 0 does not equal 20 + 2. If the total momentum of the system starts out at 10, then it can never be anything other than 10 (unless you add momentum in from an outside source).

[Bobolink]

Bobolink,
here is the problem that Kryptid and it appears you too do not understand.

I think we both understand that you want your conjecture to be correct so you are ignoring all attempts to help you understand the reality of the situation.  It is clear that you won't listen, because that would destroy your fantasy that you have figured out something that scientists say is not possible.  I don't think you are a troll, you are just someone who has a conjecture based on a limited and flawed understanding and are excited that you think you have made a great discovery.  So, I guess all I can say is have fun living this fantasy. [Shrug]

[Jaaanosik (OP)]

...
Not unless you violate conservation of momentum. 10 + 0 does not equal 20 + 2. If the total momentum of the system starts out at 10, then it can never be anything other than 10 (unless you add momentum in from an outside source).

Kryptid,
exactly my point!
We are adding only angular momentum from the generator!!!
We are not adding linear momentum!!!
Jano

[Jaaanosik (OP)]

Bobolink,
here is the problem that Kryptid and it appears you too do not understand.

I think we both understand that you want your conjecture to be correct so you are ignoring all attempts to help you understand the reality of the situation.  It is clear that you won't listen, because that would destroy your fantasy that you have figured out something that scientists say is not possible.  I don't think you are a troll, you are just someone who has a conjecture based on a limited and flawed understanding and are excited that you think you have made a great discovery.  So, I guess all I can say is have fun living this fantasy. [Shrug]

Bobolink,
Sailboats can sail 3x faster than wind.
Please, explain, how is that possible,
Jano