Propellant less space engine

[John Lowe]

I have a drawing now which should help clarify. Shafts and stands are omitted for clarity.

All the wheels spin in direction 'A'. All the wheels are attached to a circular shaft (not shown) and all the wheels rotate together in direction 'B' (they are spinning in direction 'A' at the same time).

The outside edge of each wheel will have a linear velocity 'V1' which will be larger than the inside edge velocity 'V2'. Thus the outside edge of each wheel will gain slightly more relativistic mass than the inside edge. This will cause an imbalance/force in the upward direction as the outside edge is going upwards.

Wheel_Layout.pdf

[Bufofrog]

31 minutes ago, John Lowe said:

This will cause an imbalance/force in the upward direction as the outside edge is going upwards.

I still do not see why this would result in a net force in the upwards direction.  Could you sketch out a force diagram showing how this setup would result in an net upward force?

Edited March 20, 2019 by Bufofrog

[swansont]

39 minutes ago, John Lowe said:

I have a drawing now which should help clarify. Shafts and stands are omitted for clarity.

All the wheels spin in direction 'A'. All the wheels are attached to a circular shaft (not shown) and all the wheels rotate together in direction 'B' (they are spinning in direction 'A' at the same time).

The outside edge of each wheel will have a linear velocity 'V1' which will be larger than the inside edge velocity 'V2'.

This will be the case regardless of the rotation of an individual wheel.

39 minutes ago, John Lowe said:

Thus the outside edge of each wheel will gain slightly more relativistic mass than the inside edge. This will cause an imbalance/force in the upward direction as the outside edge is going upwards.

Wheel_Layout.pdf

"Inside edge" and "outside edge" are in-plane. IOW, the mass distribution moves radially outward. There is no reason that the center of mass would move up.

The wheel edge is moving upwards, but the mass differential is not.  The wheel edge direction doesn't matter.

 

[John Lowe]

3 minutes ago, Bufofrog said:

I still do not see why this would result in a net force in the upwards direction.  Could you sketch out a force diagram showing how this setup would result in an net upward force?

Not sure about a force diagram but my thinking is this:-

If we don't spin the whole unit in direction 'B' and before we spin all the wheels we attach a weight to the outside edge of each wheel. We then spin all the wheels together and at the same speed so all the weights are going up on the outside and down on the inside out the same time. This whole unit would then be imbalanced and if we put it on some scales, I think it would read less when the weights are moving up and less when the weight are moving down.

There would have to be forces because that's what we need to correct when we balance wheels.

What my aim is to remove the weights and cause a permanent imbalance in the upward direction caused by the extra relativistic mass which would stay on the outside pointing up.

Forces would be small, but scaleable and useful in space.

[Q-reeus]

1 hour ago, John Lowe said:

I have a drawing now which should help clarify. Shafts and stands are omitted for clarity.

All the wheels spin in direction 'A'. All the wheels are attached to a circular shaft (not shown) and all the wheels rotate together in direction 'B' (they are spinning in direction 'A' at the same time).

The outside edge of each wheel will have a linear velocity 'V1' which will be larger than the inside edge velocity 'V2'. Thus the outside edge of each wheel will gain slightly more relativistic mass than the inside edge. This will cause an imbalance/force in the upward direction as the outside edge is going upwards.

Wheel_Layout.pdf

OK with the configuration as shown in Wheel_Layout.pdf, there is no variation in angular speed about 'directions A' - unlike my impression of the earlier case with single concentrated masses on each wheel rim. You are relying on the combined, orthogonal velocities, of inner wheel rims about 'A directions', plus convective carousel motion of the wheels in 'B' direction(s).
That's a fairly messy situation to evaluate as it requires integrating over the appropriate components of relativistic momentum acting along carousel rotation axis. Not so straightforward as you may think.

One immediate fact to consider is that even if there is a net momentum along the carousel rotation axis by that integration, it could only give rise to an initial impulse during the build up phase to a final steady state motion. Since F = dp/dt, and there is no change in net p in a steady state. Hence no steady propulsive force is ever possible - at best a tiny drift speed conceivably has been induced.
But even that assumes there has been no back-reaction going on during speed build up phase that exactly cancels any such assumed net linear momentum residing within the wheels assembly per se.
Newton's 3rd law has to be respected in all this! You will have to do the arduous relativistic formulae sums and determine analytically the actual case one way or the other.
I wouldn't attempt it as there are just too many factors to consider in total.

The case of rotating pendulum I proposed earlier is much more tractable though still not easy. Attempting that scenario may give you some valuable insights.

[PS: there is something else of interest going on akin to your latest setup that is a real head scratcher. Sorry won't say what but investigated it many years ago and never found a conventional resolution. I add this only to have it on the record, in case there is any chance of a claim down the track that I 'stole or at least got initial inspiration from someone else's idea'. Can say this much - it relates to the concepts of hidden momentum (real enough), and so-called 'stored field momentum' (doubtful imo).
Look them both up - it will doubtlessly broaden your understanding. I believe your latest incarnation likely does involve 'hidden momentum' residing in the wheels assemblage.]

Edited March 20, 2019 by Q-reeus
added PS notes

[Bufofrog]

13 minutes ago, John Lowe said:

Not sure about a force diagram but my thinking is this:-

If we don't spin the whole unit in direction 'B' and before we spin all the wheels we attach a weight to the outside edge of each wheel. We then spin all the wheels together and at the same speed so all the weights are going up on the outside and down on the inside out the same time. This whole unit would then be imbalanced and if we put it on some scales, I think it would read less when the weights are moving up and less when the weight are moving down.

There would have to be forces because that's what we need to correct when we balance wheels.

What my aim is to remove the weights and cause a permanent imbalance in the upward direction caused by the extra relativistic mass which would stay on the outside pointing up.

Forces would be small, but scaleable and useful in space.

I see what you are saying now, I believe.  So the bottom line to what you are saying is that the rotating wheels have a higher mass on the side away from the center of the apparatus than the side of the wheel closer to the center of the apparatus.  So there would be more momentum on the outward side of the wheel in the up direction than the momentum in the inward side of the wheel in the down direction.  Is that what you are saying? 

[swansont]

30 minutes ago, John Lowe said:

Not sure about a force diagram but my thinking is this:-

If we don't spin the whole unit in direction 'B' and before we spin all the wheels we attach a weight to the outside edge of each wheel. We then spin all the wheels together and at the same speed so all the weights are going up on the outside and down on the inside out the same time. This whole unit would then be imbalanced and if we put it on some scales, I think it would read less when the weights are moving up and less when the weight are moving down.

There would have to be forces because that's what we need to correct when we balance wheels.

What my aim is to remove the weights and cause a permanent imbalance in the upward direction caused by the extra relativistic mass which would stay on the outside pointing up.

Forces would be small, but scaleable and useful in space.

In the case of the imbalanced wheel, the extra up and down forces are equal. In the relativistic case, there is no vertical imbalance to begin with. It's not the motion of the wheels you need to look at, it's the motion of the additional mass, and there is no motion. It's static. Mass doesn't "point"

[John Lowe]

14 minutes ago, Bufofrog said:

I see what you are saying now, I believe.  So the bottom line to what you are saying is that the rotating wheels have a higher mass on the side away from the center of the apparatus than the side of the wheel closer to the center of the apparatus.  So there would be more momentum on the outward side of the wheel in the up direction than the momentum in the inward side of the wheel in the down direction.  Is that what you are saying? 

Yes.

[John Lowe]

If you spin a balanced wheel is space it will stay spinning around it's centre. If it is imbalanced, there is more mass on one side  than the other, therefore, more momentum in one direction than the other. This will cause the wheel to oscillate, i.e. the axis will scribe a circle in space as the extra mass goes round with the wheel.

The extra momentum in my system is always going up. So why won't my system have a force upwards? 

[Q-reeus]

6 minutes ago, John Lowe said:

If you spin a balanced wheel is space it will stay spinning around it's centre. If it is imbalanced, there is more mass on one side  than the other, therefore, more momentum in one direction than the other. This will cause the wheel to oscillate, i.e. the axis will scribe a circle in space as the extra mass goes round with the wheel.

The extra momentum in my system is always going up. So why won't my system have a force upwards? 

Did you read and try and absorb the content of my last post?

[John Lowe]

7 minutes ago, Q-reeus said:

Did you read and try and absorb the content of my last post?

It is very in depth for my level of expertise. That is why I am trying to break it down into simple points for people to answer, so I can understand easier.

[Q-reeus]

16 minutes ago, John Lowe said:

It is very in depth for my level of expertise. That is why I am trying to break it down into simple points for people to answer, so I can understand easier.

Uh huh. Well regarding your unbalanced wheel spinning/wiggling in space, is it not obvious that all one has is rotation about a common center of mass?
Going back to your more interesting 'doughnut Mk II', a basic fact to have to consider is the universal validity of Newton's 3rd law. Action & reaction are always exactly paired - sometimes statically, sometimes dynamically. Every now and then even 'experts' claim violations of that 3rd law, but that's only an indication they have an artificially restricted definition/application. All considered, there can logically never be a violation of that 3rd law - properly and comprehensibly applied. Must go.

[John Lowe]

1 hour ago, Q-reeus said:

 All considered, there can logically never be a violation of that 3rd law - properly and comprehensibly applied. Must go.

That's the whole point, trying to get around that and the conversation of momentum law. Trying to understand why it won't work instead of "you just can't coz it's the law"

[StringJunky]

57 minutes ago, John Lowe said:

That's the whole point, trying to get around that and the conversation of momentum law. Trying to understand why it won't work instead of "you just can't coz it's the law"

If some parameter is said to be conserved then that means any total must add up to that conserved figure. Some things about nature are axiomatic and just need to be accepted as being the way nature behaves. 

[Janus]

2 hours ago, John Lowe said:

If you spin a balanced wheel is space it will stay spinning around it's centre. If it is imbalanced, there is more mass on one side  than the other, therefore, more momentum in one direction than the other. This will cause the wheel to oscillate, i.e. the axis will scribe a circle in space as the extra mass goes round with the wheel.

The extra momentum in my system is always going up. So why won't my system have a force upwards? 

With the "imbalanced" wheel, all you've done is shift the center of mass away from the geometrical center of the wheel.  The wheel isn't oscillating in terms of any shift in its center of mass. Even though it might "look" like it's oscillating.

If I had two weights of different masse on the ends of a rod like this:

o-------------------------O

and spun it end over end, it is obvious that it will spin around a point closer to the larger mass as marked here:

o---------------^--------O

But you wouldn't say that the rod is oscillating back and forth.  Is is rotating around a fixed center of  mass.

You could visually hide the fact that one end is more massive than the other, so that it would look like the rod is moving back and forth, but you you aren't actually changing the fact that it is still rotating around a fixed center of mass.

[swansont]

3 hours ago, John Lowe said:

 The extra momentum in my system is always going up. So why won't my system have a force upwards? 

Momentum isn't mass.

And I'm not sure you can say there is extra momentum going up, since you haven't analyzed length contraction, any other possible relativistic effects, with all of these motions in play.

 

 

1 hour ago, John Lowe said:

That's the whole point, trying to get around that and the conversation of momentum law. Trying to understand why it won't work instead of "you just can't coz it's the law"

Violation of a conservation law tells us there is a mistake in the analysis. We're looking for the mistake, though there's almost no actual analysis present.

[John Lowe]

16 hours ago, Janus said:

With the "imbalanced" wheel, all you've done is shift the center of mass away from the geometrical center of the wheel.  The wheel isn't oscillating in terms of any shift in its center of mass. Even though it might "look" like it's oscillating.

If I had two weights of different masse on the ends of a rod like this:

o-------------------------O

and spun it end over end, it is obvious that it will spin around a point closer to the larger mass as marked here:

o---------------^--------O

But you wouldn't say that the rod is oscillating back and forth.  Is is rotating around a fixed center of  mass.

You could visually hide the fact that one end is more massive than the other, so that it would look like the rod is moving back and forth, but you you aren't actually changing the fact that it is still rotating around a fixed center of mass.

The imbalanced wheel wants to spin around it's center of mass. When it can't because the axle is fixed, it causes a force on the axle, which will move around with the extra mass. In my system the force on the axle would always be up. 

[Q-reeus]

1 hour ago, John Lowe said:

The imbalanced wheel wants to spin around it's center of mass. When it can't because the axle is fixed, it causes a force on the axle, which will move around with the extra mass. In my system the force on the axle would always be up. 

From earlier post https://www.scienceforums.net/topic/118458-propellant-less-space-engine/?do=findComment&comment=1098095
"...Since F = dp/dt, and there is no change in net p in a steady state. Hence no steady propulsive force is ever possible..."
Evidently you reject that for some unstated reason. Why? At very best, your arrangement may carry a tiny amount of 'hidden momentum' implying a very tiny drift velocity has been induced. Generated - just maybe - only during build-up phase when angular velocities are changing. Nothing more beyond that is logically conceivable, and most would insist even that much is impossible. Time to start listening and give imagination a vacation.

[swansont]

2 hours ago, John Lowe said:

The imbalanced wheel wants to spin around it's center of mass. When it can't because the axle is fixed, it causes a force on the axle, which will move around with the extra mass. In my system the force on the axle would always be up. 

You keep claiming this but you have not shown that it is true.

[John Lowe]

55 minutes ago, Q-reeus said:

From earlier post https://www.scienceforums.net/topic/118458-propellant-less-space-engine/?do=findComment&comment=1098095
"...Since F = dp/dt, and there is no change in net p in a steady state. Hence no steady propulsive force is ever possible..."
Evidently you reject that for some unstated reason. Why? At very best, your arrangement may carry a tiny amount of 'hidden momentum' implying a very tiny drift velocity has been induced. Generated - just maybe - only during build-up phase when angular velocities are changing. Nothing more beyond that is logically conceivable, and most would insist even that much is impossible. Time to start listening and give imagination a vacation.

This sounds a bit circular and chicken and the egg. I understand F=dp/dt or F=ma, if there is no acceleration then no force. But you are saying there is no change in net p, so no force. I am saying that the fact that one side of the wheel has more momentum than the other side, this will cause a force. (this same force is felt on the axle of spinning, unbalanced wheel). This force will cause an acceleration.

[Q-reeus]

30 minutes ago, John Lowe said:

This sounds a bit circular and chicken and the egg. I understand F=dp/dt or F=ma, if there is no acceleration then no force. But you are saying there is no change in net p, so no force. I am saying that the fact that one side of the wheel has more momentum than the other side, this will cause a force. (this same force is felt on the axle of spinning, unbalanced wheel). This force will cause an acceleration.

See, from the start your diagrams vs commentary create confusion since in both diagrams the carousel axis is shown horizontal, whereas your commentary referencing 'up' only makes sense if 'up' is really 'sideways/horizontal' as per illustrations. Clarify that please. Also, your 'directions A' and 'directions B' are totally non-standard notation.

Regardless of which way your 'up' is relative to carousel axis, maximum 'relativistic mass' in each convected wheel rotating about the carousel axis, lies in the plane normal to that axis and which bisects each wheel. Thus any net momentum (for now ignoring my earlier input re role of stress!) in each convected wheel is directed along the carousel axis. There is zero net rate-of-change of momentum along that axis by reason of the symmetries of wheel circular motions. Hence no force along that axis. The symmetries of rim motions guarantees no net axial force can exist, given we are talking about a steady-state situation. Your whole thinking is that the carousel assembly will accelerate parallel to the carousel rotation axis - yes?
Which cannot happen. If that is not blindingly obvious by now, you have a real problem.

[John Lowe]

1 hour ago, Q-reeus said:

See, from the start your diagrams vs commentary create confusion since in both diagrams the carousel axis is shown horizontal, whereas your commentary referencing 'up' only makes sense if 'up' is really 'sideways/horizontal' as per illustrations. Clarify that please. Also, your 'directions A' and 'directions B' are totally non-standard notation.

On the "front view" all the wheels rotate in the direction of arrow 'A'. This means that the outside edge of each wheel goes up and the inside edge goes down, towards the center.

Then on the "plan view", all the wheel are spun (they have a common cirular shaft through them all) in direction 'B'. The act of spinning all the wheels in direction 'B' is what gives the outside edge of each wheel more mass, as it has a faster velocity V1, than the inside edge, V2.

This extra mass is always going up, because it is always on the part of the wheel that is going fastest (in direction 'B' not 'A')

1 hour ago, Q-reeus said:

Regardless of which way your 'up' is relative to carousel axis, maximum 'relativistic mass' in each convected wheel rotating about the carousel axis, lies in the plane normal to that axis and which bisects each wheel. Thus any net momentum (for now ignoring my earlier input re role of stress!) in each convected wheel is directed along the carousel axis. There is zero net rate-of-change of momentum along that axis by reason of the symmetries of wheel circular motions. Hence no force along that axis. The symmetries of rim motions guarantees no net axial force can exist, given we are talking about a steady-state situation. Your whole thinking is that the carousel assembly will accelerate parallel to the carousel rotation axis - yes?
Which cannot happen. If that is not blindingly obvious by now, you have a real problem.

If it was a carousel and we remove the horses and replace them with wheels that are pointed and spinning towards the center then as the wheel turn and the carousel turn the lift will be upwards.

[Q-reeus]

19 minutes ago, John Lowe said:

On the "front view" all the wheels rotate in the direction of arrow 'A'. This means that the outside edge of each wheel goes up and the inside edge goes down, towards the center.

Then on the "plan view", all the wheel are spun (they have a common cirular shaft through them all) in direction 'B'. The act of spinning all the wheels in direction 'B' is what gives the outside edge of each wheel more mass, as it has a faster velocity V1, than the inside edge, V2.

This extra mass is always going up, because it is always on the part of the wheel that is going fastest (in direction 'B' not 'A')

If it was a carousel and we remove the horses and replace them with wheels that are pointed and spinning towards the center then as the wheel turn and the carousel turn the lift will be upwards.

Still didn't answer my question. But I can hazard a guess and assume your 'up' is indeed directed along the carousel spin axis.
What you keep failing to see is that in order to gain that extra speed thus naively computed 'relativistic mass' gain at outer rim locations, any given element of mass in a wheel rim had to be 'pushed' somehow to achieve the extra speed there. Which implies a back reaction force on whatever did the pushing. Newton's 3rd law. I believe that has been mentioned before - numbers of times in fact.

Every wheel rim mass element undergoes an overall helical motion, and depending on the exact ratio between wheel and carousel angular speeds, will sooner or later form a closed path. Especially if the carousel base is firmly anchored to terra firma, you must explain how closed helical paths can possibly generate a time-averaged net axial thrust.
Think about that some. Must go.

[John Lowe]

15 minutes ago, Q-reeus said:

What you keep failing to see is that in order to gain that extra speed thus naively computed 'relativistic mass' gain at outer rim locations, any given element of mass in a wheel rim had to be 'pushed' somehow to achieve the extra speed there. Which implies a back reaction force on whatever did the pushing. Newton's 3rd law. I believe that has been mentioned before - numbers of times in fact.

If you stand near the center of a revolving carousel you are moving slower (same RPM less m/s) than if you stand on the outside edge. Therefore the outside edge of each wheel is travelling faster than the inside edge.

I don't understand what you mean by "..... 'pushed' somehow to achieve the extra speed there."

[Q-reeus]

1 minute ago, John Lowe said:

If you stand near the center of a revolving carousel you are moving slower (same RPM less m/s) than if you stand on the outside edge. Therefore the outside edge of each wheel is travelling faster than the inside edge.

I don't understand what you mean by "..... 'pushed' somehow to achieve the extra speed there."

Look up Coriolis force - and think about that some. Oh, and that mass element is then 'pulled' back down to a lower speed - and so on cyclically. Nothing changes over a complete cycle.

If you think otherwise, try and prove using maths based on relativistic dynamics. Good luck!