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Propellant less space engine


John Lowe
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I am trying to break the law of conservation of momentum, and design a mode of transport through space without using a propellant. I know most people will stop reading now and say it can't be done, but if somebody could at least explain what is wrong with my logic I would appreciate it.

Propellant less space engines using relativistic mass.

 

Method 1 – Shotgun

The engine is basicaIIy accelerating two “cannon balls” down twin barrels and then stopping them before they reach the end of the “spaceship”. Before we fire them we spin them to a high speed so they gain a portion of relativistic mass, then we stop the spinning during the journey down the barrel, then we decelerate to zero the linear motion.

The act of accelerating a heavier mass than we are decelerating will give is a net velocity gain. The method of "spinning" and the design of the "cannon ball" can be sorted after the theory has been proved.

 

Method 2 – Doughnut

The engine is a number of wheels arranged in a circle to look like a doughnut. Spin all the wheels the same direction, then spin the whole doughnut to a high speed. As the doughnut is spinning the outside edge of each wheel will have a faster speed than the inside edge that is at the centre of the doughnut. Thus the outside edge of each wheel will have slightly more relativistic mass gain than the inside edge and as the wheel are spinning this will cause an imbalance in each wheel causing a thrust in the direction of the outside edge tangential motion.

An imbalanced wheel causes forces which follow the excess mass around the wheel. On this system the imbalance is always at the same place pointing in the same direction.

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16 minutes ago, John Lowe said:

I am trying to break the law of conservation of momentum, and design a mode of transport through space without using a propellant. I know most people will stop reading now and say it can't be done, but if somebody could at least explain what is wrong with my logic I would appreciate it.

Propellant less space engines using relativistic mass.

 

"Relativistic mass" is just a proxy for energy, and is not some magic elixir that will let you violate the laws of motion. It's rare that a serious work actually discusses relativistic mass.

16 minutes ago, John Lowe said:

 

Method 1 – Shotgun

The engine is basicaIIy accelerating two “cannon balls” down twin barrels and then stopping them before they reach the end of the “spaceship”. Before we fire them we spin them to a high speed so they gain a portion of relativistic mass, then we stop the spinning during the journey down the barrel, then we decelerate to zero the linear motion.

The act of accelerating a heavier mass than we are decelerating will give is a net velocity gain. The method of "spinning" and the design of the "cannon ball" can be sorted after the theory has been proved.

Why do you think that stopping the spinning won't require a force along the axis? 

16 minutes ago, John Lowe said:

 

Method 2 – Doughnut

The engine is a number of wheels arranged in a circle to look like a doughnut. Spin all the wheels the same direction, then spin the whole doughnut to a high speed. As the doughnut is spinning the outside edge of each wheel will have a faster speed than the inside edge that is at the centre of the doughnut. Thus the outside edge of each wheel will have slightly more relativistic mass gain than the inside edge and as the wheel are spinning this will cause an imbalance in each wheel causing a thrust in the direction of the outside edge tangential motion.

A diagram would be helpful, and the caveat about relativistic mass still applies.

16 minutes ago, John Lowe said:

An imbalanced wheel causes forces which follow the excess mass around the wheel. On this system the imbalance is always at the same place pointing in the same direction.

Your analysis is incomplete, and very hand-wavy. You only look at one section of the device. There will be speed difference for all of the wheels. Also, as this is rotating, I don't see how you can claim that the alleged imbalance doesn't move.

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55 minutes ago, John Lowe said:

I am trying to break the law of conservation of momentum, and design a mode of transport through space without using a propellant. I know most people will stop reading now and say it can't be done,

I stopped reading at this point because it can't be done.;)

Actually, that was a nice try but your analysis is based on a faulty assumption as swansont pointed out. 

For a rotating wheel of 10 m in diameter at a speed of 2/3 c, the wheel would need to spin at about 400,000,000 rpm.  I think to keep the wheel from flying apart it would need be made from Doesntexistium.

Edited by Bufofrog
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23 minutes ago, swansont said:

"Relativistic mass" is just a proxy for energy, and is not some magic elixir that will let you violate the laws of motion. It's rare that a serious work actually discusses relativistic mass.

Why do you think that stopping the spinning won't require a force along the axis? 

A diagram would be helpful, and the caveat about relativistic mass still applies.

Your analysis is incomplete, and very hand-wavy. You only look at one section of the device. There will be speed difference for all of the wheels. Also, as this is rotating, I don't see how you can claim that the alleged imbalance doesn't move.

Is "Relativistic mass" real? Does a spinning top weigh more than a stationary one.

The force stopping the spinning is at 90 deg to the axis so cant be resolved in the direction motion.

Working on a diagram.

The imbalance is only caused by the extra relativistic mass which is always always on the outside of each wheel.

It is not the practicality that I'm worried about, just the theory. Even a small amount of thrust would be useful. Look at all the fuss that went into the EM drive.

3 minutes ago, Bufofrog said:

I stopped reading at this point because it can't be done.;)

Actually, that was a nice try but your analysis is based on a faulty assumption as swansont pointed out. 

For a rotating wheel of 10 m in diameter at a speed of 2/3 c, the wheel would need to spin at about 400,000,000 rpm.  I think to keep the wheel from flying apart it would need be made from Doesntexistium.

It is not the practicality that I'm worried about, just the theory. Even a small amount of thrust would be useful. Look at all the fuss that went into the EM drive.

Here is a simplified version with only 2 wheels. 

Place a weight on each wheel at position A and spin them both in synchronisation. The scales will read less when the weights are at position A than when they are at position B. This is because the wheels are imbalanced. Now remove the weights and spin the wheels. Scales will read constant. Now if we the spin the whole assembly on the roundabout the wheel will always be heavier at point A because of a slight gain in mass than at point B and cause the scales to read less.

Diagram A.pdf

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52 minutes ago, John Lowe said:

Is "Relativistic mass" real? Does a spinning top weigh more than a stationary one.

Yes, it has more mass. There are very few situations where this would be the proper analysis.

52 minutes ago, John Lowe said:

The force stopping the spinning is at 90 deg to the axis so cant be resolved in the direction motion.

Is it? The object is moving. You can't analyze it in the rest frame of the object, since that's not the frame where you are analyzing conservation of momentum. 

52 minutes ago, John Lowe said:

Working on a diagram.

The imbalance is only caused by the extra relativistic mass which is always always on the outside of each wheel.

You have asserted this. You have not shown this via analysis.

52 minutes ago, John Lowe said:

Here is a simplified version with only 2 wheels. 

That's not consistent with your setup, though, where everything is rotating n the same direction, so it's not really a simplified version.

52 minutes ago, John Lowe said:

Place a weight on each wheel at position A and spin them both in synchronisation. The scales will read less when the weights are at position A than when they are at position B. 

This is because the wheels are imbalanced. Now remove the weights and spin the wheels. Scales will read constant. Now if we the spin the whole assembly on the roundabout the wheel will always be heavier at point A because of a slight gain in mass than at point B and cause the scales to read less.

Diagram A.pdf

Not the same situation, though. The mass reads a greater amount in the first case because the extra mass is at one point on the wheel. The system oscillates. In the second, the extra mass is at the same point on the wheel. There is no oscillation. It's at steady state.

This is functionally the same as the spinning top you referred to earlier. The mass is larger. That's it. There's no force it exerts on its own, resulting from that.

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Both wheels (or all in the doughnut version) are rotating and also spinning around a central point. As they are spinning around a central point the outside edge of each wheel will have slightly more relativistic mass, and this extra mass will cause an imbalance. But, unlike when we put an extra weight on the wheel and it oscillated, this mass is always on the outside edge of each wheel with them all pointing in the same direction, causing a force in that direction.

 

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3 hours ago, John Lowe said:

I am trying to break the law of conservation of momentum, and design a mode of transport through space without using a propellant. I know most people will stop reading now and say it can't be done, but if somebody could at least explain what is wrong with my logic I would appreciate it.

Propellant less space engines using relativistic mass.

 

Method 1 – Shotgun

The engine is basicaIIy accelerating two “cannon balls” down twin barrels and then stopping them before they reach the end of the “spaceship”. Before we fire them we spin them to a high speed so they gain a portion of relativistic mass, then we stop the spinning during the journey down the barrel, then we decelerate to zero the linear motion.

The act of accelerating a heavier mass than we are decelerating will give is a net velocity gain. The method of "spinning" and the design of the "cannon ball" can be sorted after the theory has been proved.

 

Method 2 – Doughnut

The engine is a number of wheels arranged in a circle to look like a doughnut. Spin all the wheels the same direction, then spin the whole doughnut to a high speed. As the doughnut is spinning the outside edge of each wheel will have a faster speed than the inside edge that is at the centre of the doughnut. Thus the outside edge of each wheel will have slightly more relativistic mass gain than the inside edge and as the wheel are spinning this will cause an imbalance in each wheel causing a thrust in the direction of the outside edge tangential motion.

An imbalanced wheel causes forces which follow the excess mass around the wheel. On this system the imbalance is always at the same place pointing in the same direction.

You deserve more than just skepticism void of some level of detailed physical reasoning behind such. Regarding your 1st idea, it's enough to know that absence of any axial force gaurantees axial i.e. linear momentum is constant. But it can be established another way. Applying a mid-flight spin down torque means applying a purely transverse force (seen from the 'stationary' gun barrel frame) on each mass element in the ball not lying eaxactly on the spin axis. How that can be done in practice we don't worry about here!
That transverse force is acting on a mass element at some finite radius from the spin axis, thus having both transverse and axial motion.

It's a fact of relativistic transformation laws that acceleration and force are not collinear if force and velocity are not collinear. You guessed it - transverse deceleration i.e. spin down is also accompanied by an axial acceleration that exactly maintains linear momentum constant.
I'm not going to work through the tedious specifics for a worked example confirming that claim, but will point you to an article, which YOU can then use to verify that momentum conservation holds: https://arxiv.org/abs/1806.08680

Regarding your 2nd example, you need to be aware that stress, both as cause of elastic energy and stress 'all by itself', is a source of inertial mass in relativity. Unlike the generally much smaller elastic energy contribution, which is a quadratic function of stress, inertial mass owing to stress 'all by itself' changes sign with a change of sign of stress. At the locations where the masses have maximum KE thus 'relativistic mass', there is also maximum tensile (negative!) inertial mass owing to tension in the spokes holding each mass. Again, I won't bother computing the competing effects, but suggest you ponder the situation more fully.
What looks cut and dried initially is actually fairly complex.

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1 hour ago, John Lowe said:

Both wheels (or all in the doughnut version) are rotating and also spinning around a central point. As they are spinning around a central point the outside edge of each wheel will have slightly more relativistic mass, and this extra mass will cause an imbalance. But, unlike when we put an extra weight on the wheel and it oscillated, this mass is always on the outside edge of each wheel with them all pointing in the same direction, causing a force in that direction.

I don't see an imbalance in your diagram. It's symmetrical. The outside edge is furthest from the center, and there is one on each side.

 

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There is an imbalance in each wheel because it has extra weight on the outside edge.

If you held one wheel in each out stretched hand and spun them with a weight on each of them, then you would feel the imbalance. Likewise, I suggest that if you held balanced wheels, spun them, them yourself rotated, the imbalance would be at the outside edge and you would feel an upward force.

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1 hour ago, John Lowe said:

There is an imbalance in each wheel because it has extra weight on the outside edge.

If you held one wheel in each out stretched hand and spun them with a weight on each of them, then you would feel the imbalance. Likewise, I suggest that if you held balanced wheels, spun them, them yourself rotated, the imbalance would be at the outside edge and you would feel an upward force.

But you have an imbalance in the wheel opposite it, so the system is symmetric.

If there is an asymmetry, then a more detailed analysis is warranted. But, as Q-reeus has observed, things are not simple in these problems. Exerting a torque in such a system may involve a linear force as well, counterbalancing the possible mass shift. You have to actually do the physics if you want the answer.  

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System is symmetric, but both wheels give an upward thrust. 

When you have a wheel with an imbalance the force follows the weight around, causing an oscillation. In my system the imbalance is always on the outside edge of both (or all) wheels causing an upward thrust. 

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3 hours ago, John Lowe said:

System is symmetric, but both wheels give an upward thrust. 

When you have a wheel with an imbalance the force follows the weight around, causing an oscillation. In my system the imbalance is always on the outside edge of both (or all) wheels causing an upward thrust. 

Perhaps I'm not understanding your diagram, but if the wheels gain mass, it happens all the way around the rim. Now you rotate the whole thing. I assumed, from your diagram, that it's the central axis that rotates. That adds another velocity vector, perpendicular to the existing ones. No change in the symmetry that I see to shift the mass vertically. And, as has been stated, you really need to do a careful analysis to ensure there's nothing weird about trying to exert the required torque, since the wheels are in a relativistic state, and you want to put the superstructure into one. 

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From John Lowe's first post: "...but if somebody could at least explain what is wrong with my logic I would appreciate it..."
Actually, you seem quite unappreciative. Why have you totally ignored my attempt to point out basic flaws in your reasoning? Obsessed? Totally sure you are right? Unwilling or unable to apply the formulae in article linked to? Not up to doing your own search for role of stress in relativistic physics? Or just ignorant maybe?

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14 hours ago, swansont said:

Perhaps I'm not understanding your diagram, but if the wheels gain mass, it happens all the way around the rim. Now you rotate the whole thing. I assumed, from your diagram, that it's the central axis that rotates. That adds another velocity vector, perpendicular to the existing ones. No change in the symmetry that I see to shift the mass vertically. And, as has been stated, you really need to do a careful analysis to ensure there's nothing weird about trying to exert the required torque, since the wheels are in a relativistic state, and you want to put the superstructure into one. 

I think that the wheels gain more mass on the outside edge than on the inside edge, because the outside edge has a faster linear speed than the inside edge (larger diameter, same revs)

6 hours ago, Q-reeus said:

From John Lowe's first post: "...but if somebody could at least explain what is wrong with my logic I would appreciate it..."
Actually, you seem quite unappreciative. Why have you totally ignored my attempt to point out basic flaws in your reasoning? Obsessed? Totally sure you are right? Unwilling or unable to apply the formulae in article linked to? Not up to doing your own search for role of stress in relativistic physics? Or just ignorant maybe?

Many thanks for the link and info. I am reading and trying to understand the post in the link you sent but it is quite detailed, and will take time to sink in. I am not "unappreciative, obsessed or ignorant", and I think it is quite rude to say so on the basis of me not replying quick enough. Especially from a senior member.

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1 hour ago, John Lowe said:

I think that the wheels gain more mass on the outside edge than on the inside edge, because the outside edge has a faster linear speed than the inside edge (larger diameter, same revs)

Which is true of both wheels, and is in-line with the rest of the disk. That doesn't move the center-of-mass upward.

Plus, you would have to apply the equations, and not just hand-wave this, to determine the correct answer.

 

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1 hour ago, John Lowe said:

..and I think it is quite rude to say so on the basis of me not replying quick enough. Especially from a senior member.

'Senior' as designation doesn't take long to accrue here. Point is, you could have said something. Even a simple acknowledgement.
Anyway, re your doughnut arrangement: as has been pointed out, your illustration is symmetric wrt dynamics. I took it you really meant an arrangement where each inner wheel had the same rotational speed about it's axis as that of wheel(s) about carousel axis, but where phasing of weights led to maximal speed at a single angular location wrt carousel axis. Only then could one even hope for a net force or rather impulse. Your diagram needs amending accordingly. But it won't eventuate even then, if all factors are considered:
https://arxiv.org/abs/gr-qc/0510041v1
https://arxiv.org/abs/physics/0609144

There are consistency issues with stress as source of gravity/inertial mass, but it probably works out 'as normally expected' for your scenario.

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2 hours ago, Q-reeus said:

'Senior' as designation doesn't take long to accrue here. Point is, you could have said something. Even a simple acknowledgement.
Anyway, re your doughnut arrangement: as has been pointed out, your illustration is symmetric wrt dynamics. I took it you really meant an arrangement where each inner wheel had the same rotational speed about it's axis as that of wheel(s) about carousel axis, but where phasing of weights led to maximal speed at a single angular location wrt carousel axis. Only then could one even hope for a net force or rather impulse. Your diagram needs amending accordingly. But it won't eventuate even then, if all factors are considered:
https://arxiv.org/abs/gr-qc/0510041v1
https://arxiv.org/abs/physics/0609144

There are consistency issues with stress as source of gravity/inertial mass, but it probably works out 'as normally expected' for your scenario.

Apologies for my shortness, it was my first day and limited to 5 posts. Thanks for your help and I do appreciate all comments and help.

I am struggling explaining and drawing exactly what I mean, let me try again.

The drawing was the most simplified version of my idea where there are only 2 wheels 180 deg apart on a "T" stand. This easily could be expanded to 4 wheel at 90 deg apart. The more wheels you add, the more it starts to look like the doughnut I described. So then we spin each wheel so they are all going up on the outside and down on the inside.

If we put a weight at the same place on each wheel so that each weight was going up on the outside at the same time and going down on the inside at the same time then the whole assembly would wobble up (when weights going up) and Down(when weights going down).

Now if we remove the weights, spin the wheels, then spin the whole "doughnut". Now the outside edge of the doughnut is going faster than the inside edge and as such will gain (slightly) more relativistic mass than the inside edge and as such there will be a slight upward force (it will constantly "wobble" up).

When the weights were on, the wobble followed the weights around, but now the extra weight is always on the outside, due to the extra relativistic mass.

If stress is like a spring or potential energy, would this not cause more mass on the outside, not cancel it out?

 

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1 minute ago, John Lowe said:

 Now if we remove the weights, spin the wheels, then spin the whole "doughnut". Now the outside edge of the doughnut is going faster than the inside edge and as such will gain (slightly) more relativistic mass than the inside edge and as such there will be a slight upward force (it will constantly "wobble" up).

Outside edge is in-line with the wheel. There is no asymmetry of speed with respect to that plane.

 

 

 

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5 minutes ago, swansont said:

Outside edge is in-line with the wheel. There is no asymmetry of speed with respect to that plane.

 

 

 

I agree, the speed is symmetrical, but the the weight on each wheel is imbalanced due to the extra mass on the outside edge, and this extra mass will cause the force. On wheels with a weight on them the doughnut will wobble up and down. When the wheels have no weight on them, and we spin the whole doughnut of wheels, the imbalance is due to the extra relativistic mass that appears on the outside edge, then there will be a net force upwards. 

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5 minutes ago, John Lowe said:

I agree, the speed is symmetrical, but the the weight on each wheel is imbalanced due to the extra mass on the outside edge, and this extra mass will cause the force. On wheels with a weight on them the doughnut will wobble up and down. When the wheels have no weight on them, and we spin the whole doughnut of wheels, the imbalance is due to the extra relativistic mass that appears on the outside edge, then there will be a net force upwards. 

Key here is outside edge. IOW, aligned with the vertical center-of-mass. This does not change. The center-of-mass does not move.

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2 minutes ago, swansont said:

Key here is outside edge. IOW, aligned with the vertical center-of-mass. This does not change. The center-of-mass does not move.

If the doughnut wobbles up and down when the wheels have an imbalanced weight on them, why does it not move upwards when the extra weight is always on the outside.

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27 minutes ago, John Lowe said:

If the doughnut wobbles up and down when the wheels have an imbalanced weight on them, why does it not move upwards when the extra weight is always on the outside.

The mass distribution is symmetric with respect to the center of mass. It's not moving around.

 

Why would extra mass on the outside make it move up? Those directions are orthogonal.

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4 minutes ago, swansont said:

The mass distribution is symmetric with respect to the center of mass. It's not moving around.

If the whole system was on scales then, when the wheels are imbalanced (with weights on) the scales would read less when the weights were on the outside moving up, and more when the weights were going down on the inside. Surely, then, when the whole unit spinning, (without the weights) it would read slightly less as the slight extra weight (relativistic) is always on the outside going up. 

Is it not the imbalanced weight that is moving the centre of mass. In my case, this would be constantly up.

 

 

18 minutes ago, swansont said:

Why would extra mass on the outside make it move up? Those directions are orthogonal.

There is a doughnut shape made out of lots of wheels. The doughnut is spinning, but also each wheel is spinning so that the outside edge of each wheel is going up and the inside edge is going down. The extra mass is on the outside edge of each wheel which is going up. The spinning of the whole doughnut shape is to give the extra mass.

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1 hour ago, John Lowe said:

If the whole system was on scales then, when the wheels are imbalanced (with weights on) the scales would read less when the weights were on the outside moving up, and more when the weights were going down on the inside. Surely, then, when the whole unit spinning, (without the weights) it would read slightly less as the slight extra weight (relativistic) is always on the outside going up. 

Is it not the imbalanced weight that is moving the centre of mass. In my case, this would be constantly up.

The mass depends on the speed. The speed at any point on any given wheel, at the same distance from the center of its rotation, is the same. It is symmetric. There is no change in the center of mass of any wheel, even though the mass increases.

A section on the edge has the same mass, regardless of location. It doesn't matter if it's moving up or down. Only that it's moving with some speed, v. v dictates the size of the effect.

1 hour ago, John Lowe said:

There is a doughnut shape made out of lots of wheels. The doughnut is spinning, but also each wheel is spinning so that the outside edge of each wheel is going up and the inside edge is going down. The extra mass is on the outside edge of each wheel which is going up. The spinning of the whole doughnut shape is to give the extra mass.

Without the doughnut spinning, the analysis above applies.

Now, spin the doughnut. The motion is in the plane of the CoM. The effect on any part of the system, at the same distance from the center, is the same. The CoM does not move, even though the mass increases.

 

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8 hours ago, John Lowe said:

 Apologies for my shortness, it was my first day and limited to 5 posts. Thanks for your help and I do appreciate all comments and help.

Ah ok then it was a misunderstanding as I had no idea about a 5 post initial limit. All get on that score.

Quote

I am struggling explaining and drawing exactly what I mean, let me try again.

The drawing was the most simplified version of my idea where there are only 2 wheels 180 deg apart on a "T" stand. This easily could be expanded to 4 wheel at 90 deg apart. The more wheels you add, the more it starts to look like the doughnut I described. So then we spin each wheel so they are all going up on the outside and down on the inside.

If we put a weight at the same place on each wheel so that each weight was going up on the outside at the same time and going down on the inside at the same time then the whole assembly would wobble up (when weights going up) and Down(when weights going down).

Now if we remove the weights, spin the wheels, then spin the whole "doughnut". Now the outside edge of the doughnut is going faster than the inside edge and as such will gain (slightly) more relativistic mass than the inside edge and as such there will be a slight upward force (it will constantly "wobble" up).

When the weights were on, the wobble followed the weights around, but now the extra weight is always on the outside, due to the extra relativistic mass.

If stress is like a spring or potential energy, would this not cause more mass on the outside, not cancel it out?

Last part first. No - as I explained in initial response post. The 'spring energy' contribution from stress is of an entirely different character and magnitude to that of stress as component(s) of the stress-energy-momentum tensor in GR. See the articles I linked to last post.

To avoid confusion as to what is meant by 'up on outside edge' vs 'down on inside edge' etc., use the carousel axis (lets label it z) as a spatial reference. Hence 'up on outside edge means '+z velocity'? etc. At any rate I agree with other poster here that your arrangement is symmetric and no wobble can occur for that reason alone.

Try a simplest case. Just a single 'inner wheel' best thought of as a pendulum - mass on a light but rigid arm free to pivot about a shaft whose axis is oriented normal to Earth's gravity. Give the mass a sufficient initial boost that it rotates continually about it's axis, but with an obvious periodic variation in angular speed owing to gravity.
Assume negligible 'wobble' of the shaft which is housed in a very rigid support frame itself firmly attached to terra firma.
We know that based solely on Newtonian physics, the net vertical impulse through one complete cycle must equal just the product of pendulum mass m and period T.
Your task is to apply relativistic corrections to the mass and speed, and determine precisely how much the impulse per cycle is different to Newtonian case. It will be greater for the obvious reason that 'relativistic mass' has increased over the rest mass, and further because the period T will be greater (less average speed than for Newtonian case). What is relevant is whether there is an additional 'dynamical component' that would correspond to your notion of propellantless propulsion. In other words, if the properly computed net impulse per cycle differs from m'.T, where m' is the averaged 'relativistic mass' gamma.m.  If it is, then you will realize that stress in the pendulum arm is probably a crucial missing factor.

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Edited by Q-reeus
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