[Jaaanosik (OP)]

...
All youv'e done here is fail to account for all the momentum transfers between balls and tubes.  You will get no net gain of momentum with this set up. 

I always amazes me that people believe that they are the first person to think up such an idea.  I can't tell you how many times I've seen the same type of idea tossed around.  The reason we don't already have devices that use this type of arrangement is that they just don't work. 

Janus,
Please, have a look at the second stroke.
That's the case where the net momentum is 0 because it is linear.

The first stroke has the original momentum changed to angular momentum of the balls and the momentum that hits sideways.
Therefore not 100% of the original momentum is applied in the linear way and that's the reason forward net momentum is generated,
Jano

[Kryptid]

That's the case where the net momentum is 0 because it is linear.

You're wrong. Newton's third law states that for every action there is an equal and opposite reaction. If the balls are being redirected by the curved paths towards the front of the device, then that means a force is being applied to the balls to change their path. In turn, that means the balls are producing an equal and opposite force against the curved paths (and therefore the device as a whole). This second stroke is equal and opposite to the first stroke and thus produces an equal and opposite force. There is no net movement.

[Jaaanosik (OP)]

That's the case where the net momentum is 0 because it is linear.

You're wrong. Newton's third law states that for every action there is an equal and opposite reaction. If the balls are being redirected by the curved paths towards the front of the device, then that means a force is being applied to the balls to change their path. In turn, that means the balls are producing an equal and opposite force against the curved paths (and therefore the device as a whole). This second stroke is equal and opposite to the first stroke and thus produces an equal and opposite force. There is no net movement.

Kryptid,
your first stroke analysis is wrong. The forces change direction at the back of the ship.
It is better to analyze momentum or energy.
If you just think about the angular momentum of the balls. Where does it come from?
The amount of the angular momentum is going to be missing in equalization of the momentum from the front where the first stroke started. Don't you think?
Jano

[Jaaanosik (OP)]

All,
if the balls fly out with 10m/s at the beginning of the first stroke.
What would be the velocity when they hit the side of the spaceship?  It is not going to be 0m/s for sure.
If it is 5m/s then how much momentum went to slowing down the ship?
Still, remember that some linear momentum went to the angular momentum of the balls.
10m/s = v for slowing down momentum + v for angular momentum + 5m/s
10m/s > v for slowing down momentum
... we have a net momentum forward,
Jano

[Kryptid]

Kryptid,
your first stroke analysis is wrong. The forces change direction at the back of the ship.

And?

If you just think about the angular momentum of the balls. Where does it come from?
The amount of the angular momentum is going to be missing in equalization of the momentum from the front where the first stroke started. Don't you think?

There is no net angular momentum. The balls end up spinning in opposite directions, so they cancel each other out.

What would be the velocity when they hit the side of the spaceship?

Depends on the reference frame. Is this the frame of an outside observer or the frame of the machine?

The very act of firing the balls backwards will push the rest of the assembly forward with momentum equal and opposite to that of the balls. As the balls begin to push against the curved track, their backwards momentum slowly starts to cancel out the forward momentum of the machine as a whole. Once they reach the wall, their backwards momentum is gone and so is the forward momentum of the machine. The sideways momentum of the two balls cancel each other out because they are moving in opposite directions.

There is no net movement.

[Jaaanosik (OP)]

...
All youv'e done here is fail to account for all the momentum transfers between balls and tubes.  You will get no net gain of momentum with this set up. 

I always amazes me that people believe that they are the first person to think up such an idea.  I can't tell you how many times I've seen the same type of idea tossed around.  The reason we don't already have devices that use this type of arrangement is that they just don't work. 

Hi Janus,
If we would shoot just the balls on the left side then the conservation of the momentum would start to rotate the spaceship.
We shoot the balls at the same time so the horizontal components cancel out.
They are still there though.
This is the reason why the backward component is not equal to 100% of the original momentum.
There is going to be a delta front net momentum, plus some momentum goes to the rotation of the balls,
Jano

[Jaaanosik (OP)]

This solution for a propulsion is not practical at the moment... but in the future.
There could be a spaceship with this type of propulsion.
The spaceship could spread its sails, collect the interstellar hydrogen, burn it for the propulsion,
Jano

[Janus]

...
All youv'e done here is fail to account for all the momentum transfers between balls and tubes.  You will get no net gain of momentum with this set up. 

I always amazes me that people believe that they are the first person to think up such an idea.  I can't tell you how many times I've seen the same type of idea tossed around.  The reason we don't already have devices that use this type of arrangement is that they just don't work. 

Hi Janus,
If we would shoot just the balls on the left side then the conservation of the momentum would start to rotate the spaceship.
We shoot the balls at the same time so the horizontal components cancel out.
They are still there though.
This is the reason why the backward component is not equal to 100% of the original momentum.
There is going to be a delta front net momentum, plus some momentum goes to the rotation of the balls,
Jano

Wrong.  Momentum is still full conserved.  Just because you don't see it, doesn't mean that it doesn't occur.  Your system will not, and can not, produce net momentum, no matter how hard you want to believe otherwise.  The fact that by your analysis, it does, just means that your analysis is incomplete.

[Jaaanosik (OP)]

...
Wrong.  Momentum is still full conserved.  Just because you don't see it, doesn't mean that it doesn't occur.  Your system will not, and can not, produce net momentum, no matter how hard you want to believe otherwise.  The fact that by your analysis, it does, just means that your analysis is incomplete.

Janus,
Well, let's talk about conservation of energy.
The 2 balls have the weight of the spaceship, so we can ignore the mass.
The beginning of the first stroke, balls fly out at v and the spaceship -v, agreed?
At the end of the first stroke, the balls stopped the ship to zero for no net gain and they have additional energy to hit the side walls and the balls have additional rotational energy.
Where is the conservation of energy in this?
What do I miss?
Jano

[Kryptid]

Janus,
Well, let's talk about conservation of energy.
The 2 balls have the weight of the spaceship, so we can ignore the mass.
The beginning of the first stroke, balls fly out at v and the spaceship -v, agreed?
At the end of the first stroke, the balls stopped the ship to zero for no net gain and they have additional energy to hit the side walls and the balls have additional rotational energy.
Where is the conservation of energy in this?
What do I miss?
Jano

We can use the principles of conservation of momentum and conservation of energy to figure this out.

We start off with the balls being equal in mass to the rest of the ship. So when they are sent backwards, an equal amount of kinetic energy is received by both the balls and the ship. Thus, the balls are sent backwards at -v and the ship is sent forward at +v. The momentum of the balls is equal and opposite to that of the ship, thus giving the ship/ball system a total momentum of zero (equal to when it was at rest).

Now, the balls encounter the curved paths. Since the curved paths are attached to the ship, they are moving forward at a velocity +v. This extra velocity component accelerates the balls as they change direction. In turn, the ship is decelerated as kinetic energy is transferred from the ship to the balls via the curved paths. Once the balls have completed the curve and hit the wall, they have gained all of the kinetic energy that was once a part of the ship, thus bringing the ship's forward velocity down to zero. The rotational energy of the balls comes from part of the existing kinetic energy, so that energy is not new.

The balls themselves are, just before hitting the wall, traveling at a right angle to the ship's initial heading. So the balls no longer have a backwards momentum component either. Since the balls no longer have rearward momentum and the ship no longer has forward momentum, then the total momentum at the end of the cycle is still zero. The total energy of the system has also remained constant because all of the kinetic energy that was once in the ship has been transferred to the balls. If the collision of the balls with the walls is inelastic, then all of the kinetic energy in the balls is converted into heat energy.

Both momentum and energy are conserved.

[Jaaanosik (OP)]

Kryptid,
Thanks, good response.
Question:

If a ball is sliding on ice and it hits the red carpet.
The friction at the bottom will cause the ball to roll.
The roll decreases the velocity v and this decreases the forward kinetic energy that is available to slow down the spaceship.

The rotational energy of the balls comes from part of the existing kinetic energy, so that energy is not new.

Isn't this energy going to be missed?
Jano

[Kryptid]

Isn't this energy going to be missed?

No, because the same force that is slowing the ball's forward kinetic energy will act in an equal and opposite manner on the carpet. Remember, the red carpet is attached to the ship, so any forces applied to the carpet will also be applied to the ship. If a force of X slows down the ball, the same force of X will slow down the ship by the same amount (since you specified that their masses are equal).

[Jaaanosik (OP)]

All,
here is the spaceship, the ball mass is the same as the mass of the spaceship.
If the ball is pushed down on the left side the ball and the spaceship will rotate around the barycenter.
The spaceship does not move anywhere when v_L = v_B and the v is constant all around the rotation.
If v_B < v_L then we have a forward (up in the figure) net momentum gain, considering the mechanism is coupled with symmetrical wheel/ball in opposite direction.
Here is a little secret, it is better to use hollow cylinders instead of balls for the original design where cylinder v_L kinetic energy is transformed to cylinder rotational energy at the bottom of the spaceship because more kinetic energy will be absorbed by the hollow cylinder and v_L, v_B delta will be bigger.
Jano

[Kryptid]

Unfortunately, I don't really understand your diagram or set-up.

[Jaaanosik (OP)]

Unfortunately, I don't really understand your diagram or set-up.

Kryptid,
that's very cool you said that.
Does this help?

The barycenter is important for the inertial propulsion system,
Jano

[Kryptid]

Kryptid,
that's very cool you said that.
Does this help?

The barycenter is important for the inertial propulsion system,
Jano

I know what a barycenter is. What is causing the balls to move in the diagram? Where is the spaceship at?

I can say this much: if you think that this will cause a net change in momentum, then you are trying to break conservation of momentum. The laws of physics do not allow that. For this reason, I'm considering this a "new theory" and thus moving it to that board.

[Jaaanosik (OP)]

...
I know what a barycenter is. What is causing the balls to move in the diagram? Where is the spaceship at?

I can say this much: if you think that this will cause a net change in momentum, then you are trying to break conservation of momentum. The laws of physics do not allow that. For this reason, I'm considering this a "new theory" and thus moving it to that board.

Kryptid,
The point A is an axle through the spaceship, balls rotate. The L ball mass plus R ball mass is equal to the spaceship mass.
The position A, also the geometrical center of the spaceship, was at B when the balls were horizontal.
When the L/R balls get pushed down at the same time the spaceship moves up/forward.
The point B is the barycenter.
The balls get stopped at the bottom, the dotted position, two balls - one behind the other.

Now, nothing moves, everything is steady. The new dotted position is the 'displaced' spaceship that got moved ahead.
Agreed?

Jano

[Kryptid]

Kryptid,
The point A is an axle through the spaceship, balls rotate. The L ball mass plus R ball mass is equal to the spaceship mass.
The position A, also the geometrical center of the spaceship, was at B when the balls were horizontal.
When the L/R balls get pushed down at the same time the spaceship moves up/forward.
The point B is the barycenter.
The balls get stopped at the bottom, the dotted position, two balls - one behind the other.

Now, nothing moves, everything is steady. The new dotted position is the 'displaced' spaceship that got moved ahead.
Agreed?

Jano

As long as you agree that the barycenter didn't move, sure.

[Jaaanosik (OP)]

...
As long as you agree that the barycenter didn't move, sure.

How is my post above different from the first stroke here?
Do you agree that the spaceship is going to be displaced after the first stroke when everything is static?
Jano

[Kryptid]

How is my post above different from the first stroke here?
Do you agree that the spaceship is going to be displaced after the first stroke when everything is static?
Jano

The ship is displaced, but it stops moving as soon as the balls stop moving. The barycenter hasn't moved. No net momentum is generated. As soon as the second stroke is completed, the ship has moved back into its original position.